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http://physics.stackexchange.com/tags/carrier-particles/new | # Tag Info
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Quite probably gravitons can cause similar effects as photons. The reason I believe so is that gravitons arise by quantizing linearized gravity (linear approximation to GR), the procedure is very similar to quantization of electromagnetism. Individual gravitons have the usual relativistic energy and momentum relation to frequency (energy-density is ...
0
One doesn't. We haven't even detected gravitational waves, much less single quanta of gravitational waves. As of now, gravitons are a theoretical idea derived by extending quantum mechanical ideas to general relativity.
1
From what I understand gravity is similarly quantized and transmitted via gravitons. Well, we don't know that. There is no accepted quantum theory of gravity, only approximations like semiclassical approaches. We cannot give you a "mental picture" at the moment because we don't have one. We can speculate all day, and extrapolate from all the other ...
0
Basically recoil, since photons can also carry momentum. The reason why this recoil can also lead to attraction instead of plain repulsion is that the exchanged virtual photon has a definite momentum, so it can not have a definite position at the same time (the position wave of a virtual photon is stretched out to infinity). The momentum can be transfered ...
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https://www.physicsforums.com/threads/can-anybody-help-with-group-velocity-simulations.762019/ | # Can anybody help with group velocity simulations?
1. ### mysearch
522
On first reading, the description of ‘group velocity [vg]’ appears to be quite straightforward. However, I also found a number of speculative explanations as to ‘how’ and ‘why’ the group velocity may exceed the ‘phase velocity [vp]’. Therefore, in order to get a better intuitive understanding of the issues, I am ‘attempting’ to simulate 4 different types of beat wave configurations:
• 1-way, non-dispersive phase waves
• 2-way, non-dispersive phase waves – see 2ND.gif attachment
• 1-way, dispersive phase waves
• 2-way, dispersive phase waves – see 2D.gif attachment
Each simulation uses the same basic approach in which the amplitude of two individual phase waves is added, and calculated, for all values of [x], which is then displayed as a single frame in the animation. The process is then repeated for incrementing values of time [t] to create the next frame within the animation. While the first simulation in the list above produced ‘sensible’ results, i.e. [vg=vp], all the subsequent permutations, based on the same algorithms, lead to ‘unexplained’ results. For example, the second simulation changes only the direction of one of the phase waves, but appears to suggest a group velocity that is 10x the phase velocity, although the actual value calculated was [vg=1]. My initial lines of thoughts:
• There is an mistake in the simulation. If so, it is not obvious to me, but see next bullets.
• The equation used to calculate [vg=dw/dk] might have to accommodate the direction of the phase wave velocity [vp]. For example, given that [vp=w/k], such that [k=w/vp], then treating [vp] as a vector having magnitude and direction might suggest that the sign of [k] must also be direction dependent. However, it is unclear whether this would explain the anomalous result associated with the 1-way simulations.
• The equation [vg=dw/dk] would be greater than [vp], when [dw>dk], which appears possible in dispersive media – see equation [1] in attached pdf for details. However, it is not clear this would explain the non-dispersive cases.
• The simulations are only showing an 'pattern shift'. For example, when the phase waves propagate with velocity [c] in opposite directions, the phase relationship between these waves constantly changes with time, i.e. on every frame. So while the ‘shift’ in the beat waveform might suggest a propagation velocity greater than [c], it has nothing to do with anything propagating through the media, rather it simply reflects the rate of change in phase shift between the phase waves.
I have attached ‘simulations.pdf’ in order to give some more details of the simulation results for anybody who might be interested in this topic. I have also attempted to attach two gif files to give ‘examples’ of the animations, although the upload size required 500 frames to be reduced to just 25 and so are they very jerky.
Anyway, I would appreciate any help on offer as I am sure somebody must have already resolve these issues. Thanks
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2. ### Greg Bernhardt
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
3. ### olivermsun
969
Since nobody seems to be picking up your questions, let me point out a couple things that I noticed about your simulation and derivation:
Yes, the phase vector ##k## is signed. Alternatively, you can view the back-propagating wave as having frequency ##-\omega_2##, which would explain why you are getting an envelope velocity which looks like ##\dfrac{\omega_1+\omega_2}{k_1-k_2}## instead of ##\dfrac{\omega_1-\omega_2}{k_1-k_2}.##
I believe so. Your wave phases are propagating in opposite directions, so there is no sense of any actual "group" propagating through the medium.
4. ### mysearch
522
I think we have to accept that the PF is not a guaranteed service, but sometimes it is worth a shot. Often I simply post some details for my own backup and cross-reference; plus it might help somebody else who is treading the same path at some later time.
While I agree with you, working on some updates to the simulation seems to suggest that the equations above require the ##\pm## sign to be assigned to ##\omega##, not k; otherwise the equations below do not seem to work for waves travelling in the same and opposite direction:
$ω_p = \dfrac{ω_1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{ω_p}{k_p}$
$ω_g = \dfrac{ω_1 - ω_2}{2}; k_g = \dfrac{k_1 - k_2}{2}; v_g = \dfrac{ω_g}{k_g}$
Why this is so is not clear to me, as I would argue that frequency in the form of ##\omega## is essentially direction independent, if you ignore negative time [t]. As such, it 'should' be the wavelength in the form of the wave number [k] that must change to match the requirements of the propagation media and the direction of the travelling waves in order to comply with the following equation; although the simulation suggests otherwise(!):
##v=f \lambda = \dfrac{\omega}{k} \Rightarrow \pm \omega = \pm v*k##
Anyway, I have attached some updated animations, which now show the relative phase and group velocity on the beat waves, see bottom two red and black waveforms, for the following configurations.
• 1-way, non-dispersive phase waves – see 1.gif
• 2-way, non-dispersive phase waves – see 2.gif
• 1-way, dispersive phase waves – see 3.gif
• 2-way, dispersive phase waves – see 4.gif
Given the upload limit I have restricted the animation to the first 50 frames of 500. Despite the reservations above, these animations seem to make sense, but maybe I am still missing some explanation of the issue above. Anyway, as indicated, the information posted might help somebody at a later time.
Yes, I still hold to this view. Appreciated your feedback
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• ###### 4.gif
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Last edited: Jul 22, 2014
5. ### olivermsun
969
I don't think it should matter where you put the sign. For positive wavenumber ##k##, then the wave should have the form
$$\cos(k(x-vt))$$ if it's forward propagating with phase velocity ##v##, and
$$\cos(k(x+vt))$$ if it's backward propagating with phase velocity ##-v##.
Then ##\omega = kv## in the forward case and ##\omega = -kv## in the backward case.
If you choose the convention that ##\omega## is positive, then the back-propagating wave can be written as
$$\cos(kx+kvt) = \cos(-kx-kvt),$$
which is the same as a forward-propagating wave with negative wavenumber ##-k## and positive frequency ##\omega = -kv##.
The signs for the sum and difference waves should all work out as long as you choose one convention and stick to it.
Last edited: Jul 22, 2014
6. ### mysearch
522
Again, I do not disagree with your reasoning, but my simulation results gave up 2 conflicting sets of results depending on whether using [w=v*k], which worked, as opposed to [k=w/v], which didn’t. The following equations are used in both cases below, but seem to give up different results based on the sign of ##[ \omega ] ## or [k] - see results below as an example of the 2-way non-dispersive case using both approaches.
$ω_p = \dfrac{ω_1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{ω_p}{k_p}$
$ω_g = \dfrac{ω_1 - ω_2}{2}; k_g = \dfrac{k_1 - k_2}{2}; v_g = \dfrac{ω_g}{k_g}$
‘ Common parameters
‘-------------------------
c = 1
w0 = 0.2
k0 = w0/c = 0.2
dw0 = w0*0.1 = 0.02
dk0 = dw0/c = 0.02
‘calculate [w] from [k]
‘this appears to work
------------------------
'wave-1:
k1 = k0+dk0 = 0.22
v1 = c*1 =1
w1 = v1*k1 = 0.22
'wave-2:
k2 = k0-dk0 = 0.18
v2 = -c*1 = -1
w2 = v2*k2 = -0.18
' deltas:
wp = (w1+w2)/2 = 0.02
kp = (k1+k2)/2 = 0.2
vp = wp/kp = 0.1
wg = (w1-w2)/2 = 0.2
kg = (k1-k2)/2 = 0.02
vg = wg/kg = 10
‘calculate [k] from [w]
‘this does not appear to work
-------------------------
'wave-1:
w1 = w0+dw0 = 0.22
v1 = c*1 =1
k1 = w1/v1 = 0.22
'wave-2:
w2 = w0-dw0 = 0.18
v2 = -c*1 = -1
k2 = w2/v2 = -0.18
' deltas:
wp = (w1+w2)/2 = 0.2
kp = (k1+k2)/2 = 0.02
vp = wp/kp = 10
wg = (w1-w2)/2 = 0.02
kg = (k1-k2)/2 = 0.2
vg = wg/kg = 0.1
Maybe there a mistake somebody can see. I argued for ##A=A_0cos( \omega t ± kx)## rather than ##A=A_0cos( kx ± \omega t)## because an oscillating charged particle might be said to define the frequency, independent of the media, which then propagates through a given media with a wavelength defined by the refractive index [n] in order to maintain ##[v= \omega/k] ##. In this context, the only way [ ## \omega ## t] is negative is when you are calculating the amplitude at some earlier point in time, which is not the issue in this simulation. Again, if somebody can spot an error it would be much appreciated as it is bugging me.
Hope this makes some sense as I am rushing at the end of my day. Thanks
7. ### olivermsun
969
In both cases the analysis shows that you have a product of two waves, a short wave with ##k = 0.2, w = 0.02, v = 0.1## and a long wave ##k= 0.02, w = 0.2, v = 10##. The only difference I can see is which wave you call ##g## and which wave you call ##p##. Maybe I'm missing something.
8. ### mysearch
522
Apologises if my rushed outline in post #6 was confusing. In an attempt to provide some greater clarity to the issue I am trying to resolve, I have attached a pdf file to this post that provides more detail. Some limited animations produced by the simulation are attached to post #4 with 2.gif showing the specific case detailed in the pdf file. While it is realised that nobody may be interested in working through all this detail, should anybody be able to resolve the apparent anomaly, it would be appreciated if they could let me know via the PF forum. Many thanks.
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9. ### olivermsun
969
In your pdf you mention several times that you apply your values to 2 sets of equations and get different values of ##v_p## and ##v_g##. The 2 sets of equations describe different situations, so they should get different values.
You have to choose either
a) a single equation and signed ##k## (or ##\omega##), to signify the direction of propagation, or
b) or two different equations depending on propagation direction and unsigned ##k, \omega##.
Also, you mention at the top of page 2 that both values appear to be wrong. Just looking at your simulation #2, it seems consistent with what I said in post #7, which is that you have a wave with ##k=0.2, v=\text{small}##, modulated by a wave with ##k=0.02, v=10##. What you've computed in the box is indeed choice a) above: using the single equation ##\cos(\omega_1 t - k_1 x) + \cos(\omega_2 t - k_2 x)## with a signed ##k_2##, and you seem to have the correct result for when the ##k=0.18## wave is propagating to the left.
Finally, the choice of the form ##\cos(\omega t\pm\ kx)## vs. the form ##\cos(kx \pm \omega t)## makes no difference at all because
$$\cos(\omega t- \ kx) = \cos(-(\omega t - \ kx)) = \cos(kx - \omega t).$$
10. ### mysearch
522
Again, appreciate the feedback as it is making me review all my assumptions. Some initial thoughts in a somewhat random order:
I do not disagree with the mathematically equivalence of the cosine functions shown. However, in isolation of the cosine, (wt-kx) is not equivalent to (kx-wt); especially when considered from the physical interpretation of a propagating wave. This, in part, was one of the points initially raised because the simulation only considers incrementing time [t]. If frequency [w] is always a positive, then [wt] must always be positive in the cases being considered. Therefore, [wt ## \pm ## kx] was assumed to be the appropriate starting point. On this basis, it was assumed that [ (## \pm ## k) = w/(## \pm ## v) ] not [ (## \pm ## w) = (## \pm ## v)*k ], although this appeared to lead to problems when calculating [wp,kp] and [wg,kg] as I tried to illustrate.
When I first did the derivation of the superposition beat equation, i.e.
[1] $A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)$
I started from the form [wt-kx] for the reasons explained above:
[2] $A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t – k_2 x) ]$
However, it seemed equally valid to start from
[3] $A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t + k_2 x) ]$
Although the derivations based on [2] or [3] both lead to [1], the definitions of [kp,kg] changes in sign depending on whether [2] or [3] is used. In-line with your advice in (a), the simulation has tried to use only the equations from one derivation, i.e. as derived from [2], where it was assumed that changing the sign of [k] would change the direction of the second plane wave [w2, -k2], which it does. The results I previously gave for the second option based on [3] were erroneous as they end up only reflecting the waves propagating in the same direction. However, based on option a) and [ (## \pm ## k) = w/(##\pm## v)], the following equations were assumed to reflect the values of [wp,kp] and [wg,kg] when [-k2] reflects the second wave propagating in the opposite direction:
$ω_p = \dfrac{ω_1 + ω_2}{2}; k_p = \dfrac{k_1 + (-k_2)}{2}; v_p = \dfrac{ω_p}{k_p}$
$ω_g = \dfrac{ω_1 - ω_2}{2}; k_g = \dfrac{k_1 – (-k_2)}{2}; v_g = \dfrac{ω_g}{k_g}$
As stated, this did not seem to provide consistent results. However, when I changed the sign dependency to [w], not [k], i.e. [(##\pm## w)=(##\pm##v)*k] everything appears to work and I believe I can physically explain the values of [vp] and [vg], see my quote below, but not necessarily [ ##\pm##w].
For the reasons I think we both agree, the simulation should be consistent, i.e. based on ‘choice a)’. However, I do not believe any of the simulation results on page 2 are right, when the direction of the second plane wave was defined by [-k]. Only the configuration based on [ (##\pm##w) = (##\pm##v)*k] at the bottom on page 2 and the results on page 3 appear consistent with my interpretation of the beat superposition, i.e.
“In this specific non-dispersive case, where [v1=+1, v2=-1], the rationale for [vp=0.1] can be explained as being analogous to a standing superposition waves. For as [Δω] approaches zero, the phase velocity of the superposition wave, i.e. the higher frequency component of the beat waveform must also approach zero. While a group velocity [vg>1] appears anomalous, it is argued that the group wave envelope is not actually propagating through space-time, as it more accurately reflects the phase shift between waves-1 and wave-2 occurring at all value of [x] simultaneously in time [t]. This phase shift effect, leading to the perception of [vg>1] is higher when wave-1 and wave-1 propagate in opposite directions. However, while now having to argue for the results above, it is still unclear why these equations require the directional [±] sign to be assigned to [ω] not [k].”
So while I have always had a simulation that works, based on option a), my main issue is that I cannot physically explain why the sign of second plane wave [w2, k2] has to be assigned to [w2] not [k2] to get the correct values of [vp] and [vg]. Maybe there is another [##\pm##] I am missing - I will keep checking but I have possibly become 'snow-blind' by looking at these equations for too long, which is why a second pair of 'fresh eyes' is helpful. Thanks
11. ### olivermsun
969
I guess I still don't understand why there's any difference in physical interpretation.
##\cos(kx-\omega t) = \cos\left(k\left(x-\dfrac{\omega}{k}t \right)\right)## is a (spatial) waveform with shape ##\cos(kx)## which is translating to the right at speed ##\omega/k## per unit time.
##\cos(\omega t - kx) = \cos \left(\omega\left(t - \dfrac{k}{\omega} x\right)\right)## is a (time) oscillation ##\cos(\omega t)## that is translating forward in time at ##\dfrac{k}{\omega}## per unit distance.
By mathematical equality,
##\cos(\omega t - kx) = \cos(-kx + \omega t) = \cos \left(-k\left(x - \dfrac{\omega}{k} t\right)\right) = \cos \left( k \left(x - \dfrac{\omega}{k} t \right)\right)##, which is exactly a wave ##\cos(kx)## that is translating to the right at speed ##\omega/k##.
You don't need to carry the ##\pm## signs everywhere. You just pick a convention, and then the signs work out.
Look at ##A(x, t) = \cos(\omega t - kx)##: when ##\omega## and ##k## are both positive, then ##v = \omega/k## is also positive, so you have rightward phase propagation. If ##\omega## and ##k## have different signs, then you will have leftward phase propagation. If you decide that ##\omega## is always positive, then the sign of ##k## is enough to know whether the wave is propagating to the right or the left.
As they should. In [3] the second wave is ##\cos( \omega_2 t + k_2 x) = \cos\left( k_2 x - \left(-\dfrac{\omega_2}{k_2}t \right)\right)##. This means that for ##\omega_2, k_2## both positive, the second wave is propagating to the left at ##\omega_2/k_2##. This approach is fine, but then you have to use different equations for different combinations of propagation directions.
However, when you use [3], you don't "double-count" the propagation direction by also using a negative ##k_2##. You use ##k_1, k_2## both positive and the leftward propagation is built into the equation.
So for example, on your page 2 in the box under
$$\cos(\omega_1 t - k_1 x) + cos(\omega_2 t + k_2 x)$$
you should have
\begin{align} k_p &= \dfrac{0.22 - 0.18}{2} = 0.02; \quad v_p = \dfrac{0.2}{0.02} = 10 \\ k_g &= \dfrac{0.22 + 0.18}{2} = 0.2; \quad v_g = \dfrac{0.02}{0.2} = 0.1 \end{align}
which would make the results consistent.
12. ### mysearch
522
Many thanks for post #11, it helped me focus on what I was trying to convey when referring to the mathematical and physical interpretation of these wave equations. Of course, you may not agree! However, before commenting on the last post, can I raise a direct question regarding the result of just one specific simulation previously cited as it might also help me understand where any discrepancy lies. The example is the non-dispersive case, where the two plane waves propagate in opposite directions.
[1] $Wave-1: \omega_1=0.22; k_1=0.22; v_1=+1$
[2] $Wave-2: \omega_2=0.18; k_2=0.18; v_2=-1$
I believe we agree that the following equation is able to represent the beat superposition:
[2] $A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)$
Where the value of [ ## \omega_p, k_p, v_p##] and [ ## \omega_g, k_g, v_g##] were defined by:
[3] $\omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}$
[4] $\omega_g = \dfrac{\omega _1 - ω_2}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}$
However, [3] and [4] were derived from [5] below, where the waves are travelling in the same direction:
[5] $A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t – k_2 x) ]$
As such, it would seem that if [3] and [4] are to work, the direction of wave-1 and wave-2 has to be accounted somewhere.
So what values of [##v_p, v_g##] do you think results in this specific case?
Can you also identify whether you assumed [##\omega_2 ##] or [##k_2##] to be negative or some other permutation?
Anyway, turning to the issue of physical interpretation.
As indicated, I found these equations to be really useful in reviewing the physical assumptions underpinning them for they explicitly help identify where the wave velocity [v] appears in these equations, e.g.
[6] ##\cos(kx-\omega t) = \cos\left(k\left(x-\dfrac{\omega}{k}t \right)\right) = cos\left(k\left(x-vt \right)\right) ##
[7] ##\cos(\omega t - kx) = \cos \left(\omega\left(t - \dfrac{k}{\omega} x\right)\right) =\cos \left(\omega\left(t - \dfrac{x}{v} \right)\right) ##
Now, in the context of [6] and [7], the vector quantity of velocity [v] that carries the direction [##\pm##] sign can be more clearly seen; although it does not necessarily explain how it should be inferred on [##\omega##] or [k], i.e.
[8] ## \pm v= \pm f λ = \pm \dfrac{\omega}{k} = \pm \dfrac{x}{t}##
Mathematically, based on [8], it would seem that either [f or ##\lambda##] or [ ##\omega## or k] or [x or t] may assume the [##\pm##] sign. However, I have argued that the idea of ‘negative’ frequency [f, ##\omega##] has no physical meaning, while the simulation has also constrained time [t] to only positive values. As a composite quantity with the restriction of [+t], negative velocity [v] seems to only makes sense in terms of [-x/+t], but which then implies that the negative sign has to extend to [ ##\lambda## and k], although by a somewhat circular argument, it might be said that the idea of negative wavelength is only inferred from the directional velocity.
Sorry that this is a bit long-winded, but I wanted to try to explain why I was interested in the physical interpretation of these equations as much as their mathematical consistency.
I entirely agree and I believe my initial simulation, based on [1] thru [5] was doing just this, except when applying the directional sign to [k] not [##ω##] I didn’t seem to get the right values of [##v_p, v_g##] from [3] and [4], hence the specific example/question at this begin of this post to cross-check what values you think result in this case. Thanks
Last edited: Jul 25, 2014
13. ### olivermsun
969
If you are going to use this equation and account for propagation direction using signed values of ##\omega## and/or ##k##, then your Wave-2 is inconsistent as written.
\begin{alignat}{3} \text{Wave 1:}\quad \lvert\omega_1\rvert &= 0.22; \quad \lvert k_1\rvert &= 0.22; \quad v_1 &= +1 \\ \text{Wave 2:}\quad \lvert\omega_2\rvert &= 0.18; \quad \lvert k_2\rvert &= 0.18; \quad v_2 &= -1 \end{alignat}
which implies that exactly one of ##\omega_2## or ##k_2## is negative.
If signed values of ##\omega, k## are allowed, then the waves in [5] do not necessarily travel in the same direction. Instead, ##v_p, v_g## in [3] and [4] have signs which are determined by the signs of ##\omega_p, k_p## and ##\omega_g, k_g##.
I assumed ##k_2## to be negative ##(k_2 = -\lvert k_2 \rvert)##. As I explained in my previous post, you get the same results either way, except that your ##p## and ##g## labels need to be swapped.
##\omega## isn't just a measure of "number of crests passing by per time." It's actually telling you the direction and rate of phase progression, in units of (radians)/time. Similarly, ##k## tells you the rate of phase progression in units of (radians)/length. Physically, the phase progression means the wave is rising or falling between crests and troughs, whether you measure the wave in time at a fixed point in space or whether you look down the x (or -x) direction at a snapshot in time.
I think the easiest way to understand this is to think of plane waves in more than 1 dimension, where ##\vec{k}## is clearly a vector pointing in the direction of positive phase propagation in time. The wavelength is ##{\lambda} = {2\pi}/{\lVert \vec{k} \rVert}##, which is always positive. That wave with positive phase propagation in the direction of ##\vec{k}## is exactly equivalent to the wave with the same wavelength but negative phase propagation in the direction of ##-\vec{k}##. This happens for basically the same reason why a vector of given length and direction is equal to the vector with negative length pointing in the opposite direction.
Last edited: Jul 25, 2014
14. ### mysearch
522
Sorry to belabour the issue, but your statement above seems to be the key point that I am missing. For it does seem that if I simply reverse the ##p## and ##g## labels, the 2-way simulations work, but I am not certain why just changing the sign of ##[ k_2]## changes the definition? Simply to recap, I am using following 1-way equation as the basis of the derivation:
[1] $A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t – k_2 x) ]$
This leads to the following equations used in the simulations:
[2] $A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)$
Where the value of [ ## \omega_p, k_p, v_p##] and [ ## \omega_g, k_g, v_g##] were defined by:
[3] $\omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}$
[4] $\omega_g = \dfrac{\omega _1 - ω_2}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}$
As they stand, these equations seem to give the correct results, when the 2 waves propagate in the same direction, i.e. the 1-way configurations.
Agreed. Therefore, in the case where the 2 waves propagate in different directions, I have argued that it should be ##(k_2 = -\lvert k_2 \rvert)## that is signed for reasons previously outlined. However, what is not clear to me is why just changing the sign of ##[k_2]## requires the definition of [3] and [4] reversed from ##p## to ##g##? For I assumed that changing the sign of ##[-k_2]## would simply lead to:
[3a] $\omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + [-k_2]}{2}; v_p = \dfrac{\omega _p}{k_p}$
[4a] $\omega_g = \dfrac{\omega _1 - ω_2}{2}; k_g = \dfrac{k_1 – [-k_2]}{2}; v_g = \dfrac{\omega _g}{k_g}$
While the suggestion appears to required the reversing of the ##p## and ##g## definition as follows:
[4b] $\omega _g = \dfrac{\omega _1 + ω_2}{2}; k_g = \dfrac{k_1 + [-k_2]}{2}; v_g = \dfrac{\omega _g}{k_g}$
[3b] $\omega_p = \dfrac{\omega _1 - ω_2}{2}; k_p = \dfrac{k_1 – [-k_2]}{2}; v_p = \dfrac{\omega _p}{k_p}$
The simulation of the 2-way cases, where ##[-k_2]## seems to support [3b, 4b]. As this is the central issue underpinning the simulations, I would really like to understand whether the equations, as stated, are correct and why. Just for the record, my original simulation appeared to work for all cases when I changed the sign of [##\omega_2##], i.e.
[3c] $\omega _p = \dfrac{\omega _1 + [-ω_2]}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}$
[4c] $\omega_g = \dfrac{\omega _1 – [-ω_2]}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}$
While I will try to work through the issues myself, it would be helpful if you indicate whether you support the (a), (b) or (c) form or some other variant. Thanks
Last edited: Jul 26, 2014
15. ### olivermsun
969
Let me change the way you label your definitions so you can see why the ##v## and ##p## labels get reversed:
Now which of ##\lvert k_+\rvert, \lvert k_-\rvert## is smaller if ##k_1, k_2 > 0##? That would normally be the envelope ##k_g##, since smaller ##\lvert k\rvert ## means longer wavelength.
What if ##k_1 > 0, k_2 < 0##?
16. ### mysearch
522
Well, it took me a bit longer than it should have, but I have finally worked through the permutations, which explains the reversal of the ##p## and ##g## and why this notation can be misleading in terms of the phase and group waves, when the waves propagate in opposite directions. For when deriving the following equations, it was not obvious, at least to me, that reversing the sign of [##k_2##] would cause the inference of ##p## and ##g## to also be reversed in terms of phase and group, as few references seem to cover this point:
[1] $A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)$
[2] $\omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}$
[3] $\omega_g = \dfrac{\omega _1 - ω_2}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}$
Purely, by way of reference should anybody else come down this road, the attached file shows the results when [##\omega=v*k##], which works for all values of [ ##\pm v##], and [##k=\omega/v##], which works for [##v_1=v_2##] but not [##v_1=-v_2##]. It is also highlighted that the normal definition of [##v_g=\Delta \omega / \Delta k##] looks suspect when [##v_2##] is negative and [##k=\omega /v##], but maybe this definition doesn't support the inclusion of direction? Anyway, splitting the definition of [1] into its two components:
[4] $A=2 A_0 cos( \omega_p t – k_p x)$
[5] $A=2 A_0 cos( \omega_g t – k_g x)$
Based on the argument that [##k=ω/v##], then [4] does describe the phase wave and [5] the group wave, but only when [##v_1=v_2##]. If [##v_1=-v_2##], then the descriptions are reversed, such that [4] now describes the group waves and [5] the phase wave. However, if you ignore the physical implications of negative time and allow [## \omega=v*k##], then [4] describes the phase wave and [5] the group wave, irrespective of the value of [##\pm v##]. In part, because this last approach worked and the notation the ##p## and ##g## suggested phase and group, I didn’t think that just changing the sign of [##k_2##] would change the definition as described, but it does when you sit down and work through the numbers.
While the initial response to this thread didn’t look too promising, I really appreciate the time and effort in helping me resolve the problems I was having with the simulations. While I should really leave matters here, there was one other issue that was touched on throughout this thread, which was summarised in post #13:
I think this is a really interesting summary. The normal assumption is that the relationship between the amplitude [A] and time [t] and space [x] can be anchored in the standard 1D solution of the wave equation:
[6] $\dfrac{ \partial A^2}{\partial t^2} = v^2 \dfrac{ \partial A^2}{\partial x^2}; \ where \ A=A_0sin(\omega t - kx)$
Again, I have deliberately put positive [##\omega t##] before [##\pm kx##] because I have argued that waves only physically propagate through time [t] in the positive direction. As such, I am arguing that negative time [-t] is only a mathematical concept for calculating the amplitude [A] of a wave at some earlier point in time, i.e. it is not a physical propagation mode. Therefore, while I agree with your description of ##\vec{k}## as a vector, especially when extended to 3 dimensions, the scope of frequency [f] or angular frequency [##\omega##] might be open to some debate, when considered in terms of a physical wave system. In this context, [f] or [##\omega##] might simply define the oscillations per unit time [t] without any specific inference on the subsequent propagation of a wave through a given ‘media’, which might be affected by both the refractive index of the media and Doppler effects due to relative motion of the observer, i.e. source and/or receiver. However, the point I am trying to make/clarify is that [f] or [##\omega##] is essentially a scalar quantity without direction, while some might argue that time [t] is a vector quantity, as it may appear to have a forwards and backwards direction, physical systems only appear to go in one direction, i.e. forwards. Therefore, the scope of time as a vector is fairly constrained. I realise this is possibly beyond the scope of this thread, but I would be interested in any alternative perspectives.
Again, really appreciated the help in sorting out my confusion over [## \pm k ##]. Many thanks.
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17. ### olivermsun
969
The 1-d wave equation is second order, so it has 2 solutions which have opposite sign relationship between space and time:
## \frac{1}{2}f_0(x - vt)## and ##\frac{1}{2} f_0(x + vt)##.
Once you realize that there are these two independent wave solutions, you can choose whatever convention you like for interpreting the solutions as sums of sine waves in ##k, \omega## with ##\pm kv = \omega > 0## if it's convenient for you.
Choosing to write ##\omega t - kx## instead of ##kx - \omega t## (or any other permutation) makes no difference at all to either the mathematics or the physics. The notation doesn't mean that time is progressing forward in the first case and backward in the second. You still put in the same positive ##t, \omega## and signed ##k## if you like, and get the same thing out.
In fact, the sinusoidal wave itself is just a mathematical concept. The physics are all right there in the wave equation and its solutions, and we just interpret them in various ways. It's similar to the way we interpret certain waveforms as "carriers" modulated by an "envelope," or we can interpret them as a superposition of (unrelated) sine waves; it doesn't matter one bit to the physics.
Okay.
Last edited: Jul 27, 2014
18. ### mysearch
522
Hi,
I originally raised some questions in this thread last year regarding phase and group velocity, but wanted to return to the subject in order to try to resolve any ‘physical’ interpretation associated with these velocities. In part, my issue relates to the maths and interpretation linked to the simulation attached. This simulation attempts to shows two waves (blue and green) propagating through a ‘non-dispersive’ medium, such that both waves have equal velocity magnitude [v=1], but different angular velocities [w] and wave numbers [k], but where [w/k=v=1]. Within the simulation, 2 different approaches were used to produce the superposition red and black traces, which takes the shape of a beat waveform. The red trace is produced by simply adding the 2 blue and green waveforms, while the black trace is calculated using the following formula:
$A=2 A_0 cos( w_p t – k_p x) cos( w_g t – k_g x)$
The values of $w_p, k_p, v_p$ and $w_g, k_g, v_g$ are determined from the parameters assigned to waves 1 & 2:
$w_p = \dfrac{w_1 + w_2}{2}; \quad k_p = \dfrac{k_1 + (\pm k_2)}{2}; \quad v_p = \dfrac{w_p}{k_p}$
$w_g = \dfrac{w_1 – w_2}{2}; \quad k_p = \dfrac{k_1 - (\pm k_2)}{2}; \quad v_p = \dfrac{w_p}{k_p}$
When blue and green waves propagate in the same direction, the sign of $k_1, k_2$ are the same, such that the $\pm$ can be ignored. However, when these waves propagate in opposite directions, as in this specific case, the wave number [k] effectively acts a vector quantity, which reflects the $\pm$ direction of propagation. Hence the following values for the simulation attached.
$w_p = \dfrac{0.22+0.18}{2}=0.2; \quad k_p = \dfrac{0.22 +(-0.18)}{2} = 0.02; \quad v_p = \dfrac{0.2}{0.02} = 10$
$w_g = \dfrac{0.22-0.18}{2}=0.02; \quad k_g = \dfrac{0.22 – (-0.18)}{2} = 0.2; \quad v_g = \dfrac{0.02}{0.2} = 0.1$
While the red and black outputs are identical, which appears to supports the maths used in both approaches, the values of the phase $v_p$ and group $v_g$ velocities appears difficult to reconcile within the simulation, which is reduced to the first 50 frames. The black dot on both the blue and green traces corresponds to the propagation velocity of these waves, e.g. [v=1]. Superimposed on the bottom black trace is a black and red dot that should correspond to the phase velocity $v_p=10$ and group velocity $v_g=0.1$ respectively. However, the faster black dot appears to tracking the wave packet envelope, which is normally associated with the group velocity, while slower red dot appears to be tracking successive peaks within the underlying phase wave.
Can anybody explain this anomaly?
My initial interpretation is that ‘information’ flowing between [A-B] and [B-A] is physically constrained by [v=1]. As such, any implied ‘superluminal’ inference of [v=10] simply reflects the change in the superposition amplitude at each value of [x] at any instance in time. Therefore, this is not a ‘real’ velocity, but rather a wave pattern that is changing at every position [x] on every tick of the clock. However, I assumed that the group velocity $v_g=0.1$ would have some physical interpretation, i.e. the velocity of the group wave packet, which does appear to be the case, at least, in this simulation. Would be grateful for any knowledgeable insights as to what might be going on. Thanks
P.S. I haven't used the forum in a while. Can you still reverse order the posts within a thread, i.e. latest first? Plus, its seems to take a while for the scrolling within a page to settle down - is this now normal?
Last edited: Jun 7, 2015
### Staff: Mentor
You might get a quicker response to these sorts of questions in the "Forum Feedback and Announcements" forum instead of buried in a highly technical thread on a completely different topic. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9056525230407715, "perplexity": 1296.573662594908}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398474527.16/warc/CC-MAIN-20151124205434-00280-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://conference.portonvictor.org/wiki/Algebraic_general_topology | Publish here. Donate
# Algebraic general topology
We will stick to the notation of the book, such as denoting binary join and meets as $\sqcap$ and $\sqcup$. However, we will denote arbitrary joins and meets as $\bigvee$ and $\bigwedge$ (due markup limitations of our wiki engine). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9747484922409058, "perplexity": 1090.1902378895272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987750110.78/warc/CC-MAIN-20191020233245-20191021020745-00034.warc.gz"} |
https://math.stackexchange.com/questions/1506034/show-that-the-petersen-graph-is-vertex-transitive/1509242 | # Show that the Petersen graph is vertex transitive
Show that the Petersen graph is vertex transitive.
This is my first opinion to show that the Petersen graph is vertex transitive :
The vertices of the petersen graph $G$ are labeled with 2-element subsets of {$1,2,3,4,5$} and two vertices are adjacent if and only if their intersection is empty. This is the Kneser graph labelling (1955). So that $S_5$ is contained in Automorphisms and so Graph $G$ is vertex transitive. In fact automorphism of $G$ = $S_5$.
High appreciated if someone can explain it clearly.
• You can also draw the Petersen graph and provide explicit maps, that map every vertex onto every other. This approach, too, can have an interesting effect: you see vertices move. Your approach above includes (for me at least) that the Petersen graph is isomorph to the construction process described. – Moritz Oct 31 '15 at 8:36
• @Moritz can you explain to me with another way, sorry i still can not show that Petersen graph is vertex transitive. I appreciate any answer, guidance or reference. Thank you – user273952 Nov 1 '15 at 16:36
Consider the Petersen graph in the image on the left. I have labeled the vertices via sets, i.e., $12$ is $\{1,2\} = \{2,1\}$ and $34$ is $\{3,4\} = \{4,3\}$ et cetera. If you apply the permutation $(1,4,5,2,3)$ to the vertices, you get a clockwise rotation, that is, $(1,4,5,2,3)$ takes $\{1,2\}$ and maps it to $\{4,3\}$ and so on. The cycle notation $(1,4,5,2,3)$ means $1 \mapsto 4$ and $4 \mapsto 5$ and $5 \mapsto 2$ and $2\mapsto 3$ and $3\mapsto 1$. All you have to do now is the following: If you find a permutation which maps the inner $5$ vertices onto the outer $5$ vertices, the you are finished because then you can mix the two permutations to map every vertex onto every other. Still unclear?
1. Show that, for any permutation $\pi$ of the set $\{1,2,3,4,5\},$ there is an isomorphism of the Petersen graph which maps each vertex $\{x,y\}$ to $\{\pi(x),\pi(y)\}.$
2. Show that, for any two vertices $\{x,y\}$ and $\{u,v\}$ of the Petersen graph, there is a permutation $\pi$ of $\{1,2,3,4,5\}$ such that $\pi(x)=u$ and $\pi(y)=v.$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9099576473236084, "perplexity": 147.75964531889068}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703517966.39/warc/CC-MAIN-20210119042046-20210119072046-00152.warc.gz"} |
https://math.stackexchange.com/questions/1727052/with-regards-to-theorem-10-35-of-rudins-real-and-complex-analysis | # With regards to Theorem 10.35 of Rudin's Real and Complex Analysis
Let $\Omega$ be the complex plane minus two paths. Do some closed paths $\Gamma$ in $\Omega$ satisfy assumption $(1)$ of Theorem 10.35 of Rudin's Real and Complex Analysis without being null-homotopic in $\Omega$?
Theorem 10.35. Suppose $f \in H(\Omega)$, where $\Omega$ is an arbitrary open set in the complex plane. If $\Gamma$ is a cycle in $\Omega$ that satisfies$$\text{Ind}_\Gamma(\alpha) = 0 \text{ for every }\alpha \text{ not in }\Omega,\tag*{(1)}$$then$$f(z) \cdot \text{Ind}_\Gamma(z) = {1\over{2\pi i}} \int_\Gamma {{f(w)}\over{w - z}}\,dw \text{ for }z \in \Omega - \Gamma^*\tag*{(2)}$$and$$\int_\Gamma f(z)\,dz = 0.\tag*{(3)}$$If $\Gamma_0$ and $\Gamma_1$ are cycles in $\Omega$ such that$$\text{Ind}_{\Gamma_0}(\alpha) = \text{Ind}_{\Gamma_1}(\alpha) \text{ for every }\alpha \text{ not in }\Omega,\tag*{(4)}$$then$$\int_{\Gamma_0} f(z)\,dz = \int_{\Gamma_1} f(z)\,dz.\tag*{(5)}$$
Let $\Omega = \mathbb C \setminus\{a,b\}$. Let $\Gamma$ be a closed curve in $\Omega$ so that $\Gamma$ is homotopic to $$\alpha* \beta *\alpha^{-1} *\beta^{-1},$$
where $\alpha$, $\beta$ are simple closed curves which wrap around $a$ and $b$ respectively. Then
$$\text{Ind}_\Gamma(a) = \text{Ind}_\Gamma (b) = 0$$
and $\Gamma$ is not null homotopic. Indeed, $\pi_1(\Omega)$ is freely generated by $\alpha$ and $\beta$ by the Van Kampen Theorem, so $\alpha* \beta *\alpha^{-1} *\beta^{-1}$ is not a trivial element in $\pi_1(\Omega)$.
• It would be helpful to add (for those not familiar with the fundamental group of $\mathbb C\setminus\{a,b\}$) that $\alpha$ and $\beta$ do not commmute. – Yiorgos S. Smyrlis Apr 4 '16 at 10:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9745272397994995, "perplexity": 78.282389347822}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00446.warc.gz"} |
https://economics.stackexchange.com/questions/10115/solow-model-steady-state-v-balanced-growth-path/10199 | # Solow Model: Steady State v Balanced Growth Path
Okay, so I'm having real problems distinguishing between the Steady State concept and the balanced growth path in this model:
$$Y = K^\beta (AL)^{1-\beta}$$
I have been asked to derive the steady state values for capital per effective worker:
$$k^*=\left(\frac{s}{n+g+ \delta }\right)^{\frac{1}{1-\beta }}$$
As well as the steady state ratio of capital to output (K/Y):
$$\frac{K^{SS}}{Y^{SS}} = \frac{s}{n+g+\delta }$$
I found both of these fine, but I have been also asked to find the "steady-state value of the marginal product of capital, dY/dK". Here is what I did:
$$Y = K^\beta (AL)^{1-\beta}$$ $$MPK = \frac{dY}{dK} = \beta K^{\beta -1}(AL)^{1-\beta }$$
Substituting in for K in the steady state (calculated when working out steady state for K/Y ratio above):
$$K^{SS} = AL\left(\frac{s}{n+g+\delta }\right)^{\frac{1}{1-\beta }}$$
$$MPK^{SS} = \beta (AL)^{1-\beta }\left[AL\left(\frac{s}{n+g+\delta }\right)^{\frac{1}{1-\beta }}\right]^{\beta -1}$$
$$MPK^{SS} = \beta \left(\frac{s}{n+g+\delta }\right)^{\frac{\beta -1}{1-\beta }}$$
Firstly I need to know whether this calculation for the steady state value of MPK is correct?
Secondly, I have been asked to Sketch the time paths of the capital-output ratio and the marginal product of capital, for an economy that converges to its balanced growth path "from below".
I am having problems understanding exactly what the balanced growth path is, as opposed to the steady state, and how to use my calculations to figure out what these graphs should look like.
Sorry for the mammoth post, any help is greatly appreciated! Thanks in advance.
This is when the attempt at accuracy creates confusion and misunderstanding.
Back in the day, growth models were not incorporating technological progress, and led to a long-run equilibrium characterized by constant per capita magnitudes. Verbally, the term "steady-state" seemed appropriate to describe such a situation.
Then Romer and endogenous growth models came along, which also pushed the older models to start including as a routine feature exogenous growth factors (apart from population). And "suddenly", per capita terms were not constant in the long-run equilibrium, but growing at a constant rate. Initially the literature described such a situation as "steady state in growth rates".
Then it appears the profession thought something like "it is inaccurate to use the word "steady" here because per capita magnitudes are growing. What happens is that all magnitudes grow at a balanced rate (i.e at the same rate, and so their ratios remain constant). And since they grow, they follow a path..." Eureka!: the term "balanced growth path" was born.
...To the frustration of students (at least), which have now to remember that for example, the "saddle path" is indeed a path in the Phase diagram, but the "balanced growth path" is only a point! (because in order to actually draw a Phase diagram and obtain a good old long-run equilibrium, we express magnitudes per effective worker, and these magnitudes do have a traditional steady-state. But we continue to call it "balanced growth path", because per capita magnitudes, which is what we are interested in, in our individualistic approach), continue to grow).
So "balanced growth path" = "steady state of magnitudes per efficiency unit of labor", and I guess you can figure out the rest for your phase diagram.
Following the conversation with user @denesp at the comments of my previous answer, I have to clarify the following: the usual graphical device we use related to the basic Solow growth model (see for example here, figure 2) is not a phase diagram, since reasonably we call "phase diagrams" those that contain zero-change loci, identify the crossing points of them as fixed points of a dynamical system, and examine their stability properties. And this is not what we do for the Solow model. So it was careless use of terminology from my part.
Nevertheless we can draw a "semi-Phase diagram" for the Solow growth model, in $(y,k)$ space. Understanding the symbols as "per efficiency unit of labor" we have the system of differential equations (while $y=f(k)$)
$$\dot k = sy - (n+\delta+g)k$$
$$\dot y = f'_k(k)\cdot \dot k$$ Writing the zero-change equation as a weak inequality to show also the dynamical tendencies, we have
$$\dot k \geq 0 \implies y \geq \frac {n+\delta+g}{s} k$$
$$\dot y \geq 0 \implies \dot k \geq 0$$
So this system gives a single zero change locus, a straight line. No crossing points to identify a fixed point What can we do? Draw also the production function in the diagram, since, in reality, the $(y,k)$ space is unidimensional, not an area, but a line. Then we get
The vertical/horizontal arrows indicating the dynamical tendencies come properly from the weak inequalities above (both $y$ and $k$ tend to grow when above the zero-change locus). Then, since $y$ and $k$ are constrained to move on the dotted line (which is the production function), it follows that they move towards their fixed point, no matter where we start. Here the production function graph represents essentially the path towards long-run equilibrium, since convergence is monotonic. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9399492740631104, "perplexity": 984.0044313152586}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738878.11/warc/CC-MAIN-20200812053726-20200812083726-00586.warc.gz"} |
https://pressbooks.online.ucf.edu/osuniversityphysics2/chapter/kirchhoffs-rules/ | Chapter 10. Direct-Current Circuits
# 10.3 Kirchhoff’s Rules
### Learning Objectives
By the end of the section, you will be able to:
• State Kirchhoff’s junction rule
• State Kirchhoff’s loop rule
• Analyze complex circuits using Kirchhoff’s rules
We have just seen that some circuits may be analyzed by reducing a circuit to a single voltage source and an equivalent resistance. Many complex circuits cannot be analyzed with the series-parallel techniques developed in the preceding sections. In this section, we elaborate on the use of Kirchhoff’s rules to analyze more complex circuits. For example, the circuit in Figure 10.19 is known as a multi-loop circuit, which consists of junctions. A junction, also known as a node, is a connection of three or more wires. In this circuit, the previous methods cannot be used, because not all the resistors are in clear series or parallel configurations that can be reduced. Give it a try. The resistors ${R}_{1}$ and ${R}_{2}$ are in series and can be reduced to an equivalent resistance. The same is true of resistors ${R}_{4}$ and ${R}_{5}$. But what do you do then?
Even though this circuit cannot be analyzed using the methods already learned, two circuit analysis rules can be used to analyze any circuit, simple or complex. The rules are known as Kirchhoff’s rules, after their inventor Gustav Kirchhoff (1824–1887).
### Kirchhoff’s Rules
• Kirchhoff’s first rule—the junction rule. The sum of all currents entering a junction must equal the sum of all currents leaving the junction:
$\sum {I}_{\text{in}}=\sum {I}_{\text{out}}.$
• Kirchhoff’s second rule—the loop rule. The algebraic sum of changes in potential around any closed circuit path (loop) must be zero:
$\sum V=0.$
We now provide explanations of these two rules, followed by problem-solving hints for applying them and a worked example that uses them.
### Kirchhoff’s First Rule
Kirchhoff’s first rule (the junction rule) applies to the charge entering and leaving a junction (Figure 10.20). As stated earlier, a junction, or node, is a connection of three or more wires. Current is the flow of charge, and charge is conserved; thus, whatever charge flows into the junction must flow out.
Although it is an over-simplification, an analogy can be made with water pipes connected in a plumbing junction. If the wires in Figure 10.20 were replaced by water pipes, and the water was assumed to be incompressible, the volume of water flowing into the junction must equal the volume of water flowing out of the junction.
### Kirchhoff’s Second Rule
Kirchhoff’s second rule (the loop rule) applies to potential differences. The loop rule is stated in terms of potential V rather than potential energy, but the two are related since $U=qV.$ In a closed loop, whatever energy is supplied by a voltage source, the energy must be transferred into other forms by the devices in the loop, since there are no other ways in which energy can be transferred into or out of the circuit. Kirchhoff’s loop rule states that the algebraic sum of potential differences, including voltage supplied by the voltage sources and resistive elements, in any loop must be equal to zero. For example, consider a simple loop with no junctions, as in Figure 10.21.
The circuit consists of a voltage source and three external load resistors. The labels a, b, c, and d serve as references, and have no other significance. The usefulness of these labels will become apparent soon. The loop is designated as Loop abcda, and the labels help keep track of the voltage differences as we travel around the circuit. Start at point a and travel to point b. The voltage of the voltage source is added to the equation and the potential drop of the resistor ${R}_{1}$ is subtracted. From point b to c, the potential drop across ${R}_{2}$ is subtracted. From c to d, the potential drop across ${R}_{3}$ is subtracted. From points d to a, nothing is done because there are no components.
Figure 10.22 shows a graph of the voltage as we travel around the loop. Voltage increases as we cross the battery, whereas voltage decreases as we travel across a resistor. The potential drop, or change in the electric potential, is equal to the current through the resistor times the resistance of the resistor. Since the wires have negligible resistance, the voltage remains constant as we cross the wires connecting the components.
Then Kirchhoff’s loop rule states
$V-I{R}_{1}-I{R}_{2}-I{R}_{3}=0.$
The loop equation can be used to find the current through the loop:
$I=\frac{V}{{R}_{1}+{R}_{2}+{R}_{2}}=\frac{12.00\phantom{\rule{0.2em}{0ex}}\text{V}}{1.00\phantom{\rule{0.2em}{0ex}}\text{Ω}+2.00\phantom{\rule{0.2em}{0ex}}\text{Ω}+3.00\phantom{\rule{0.2em}{0ex}}\text{Ω}}=2.00\phantom{\rule{0.2em}{0ex}}\text{A}.$
This loop could have been analyzed using the previous methods, but we will demonstrate the power of Kirchhoff’s method in the next section.
### Applying Kirchhoff’s Rules
By applying Kirchhoff’s rules, we generate a set of linear equations that allow us to find the unknown values in circuits. These may be currents, voltages, or resistances. Each time a rule is applied, it produces an equation. If there are as many independent equations as unknowns, then the problem can be solved.
Using Kirchhoff’s method of analysis requires several steps, as listed in the following procedure.
### Problem-Solving Strategy: Kirchhoff’s Rules
1. Label points in the circuit diagram using lowercase letters a, b, c, …. These labels simply help with orientation.
2. Locate the junctions in the circuit. The junctions are points where three or more wires connect. Label each junction with the currents and directions into and out of it. Make sure at least one current points into the junction and at least one current points out of the junction.
3. Choose the loops in the circuit. Every component must be contained in at least one loop, but a component may be contained in more than one loop.
4. Apply the junction rule. Again, some junctions should not be included in the analysis. You need only use enough nodes to include every current.
5. Apply the loop rule. Use the map in Figure 10.23.
Let’s examine some steps in this procedure more closely. When locating the junctions in the circuit, do not be concerned about the direction of the currents. If the direction of current flow is not obvious, choosing any direction is sufficient as long as at least one current points into the junction and at least one current points out of the junction. If the arrow is in the opposite direction of the conventional current flow, the result for the current in question will be negative but the answer will still be correct.
The number of nodes depends on the circuit. Each current should be included in a node and thus included in at least one junction equation. Do not include nodes that are not linearly independent, meaning nodes that contain the same information.
Consider Figure 10.24. There are two junctions in this circuit: Junction b and Junction e. Points a, c, d, and f are not junctions, because a junction must have three or more connections. The equation for Junction b is ${I}_{1}={I}_{2}+{I}_{3}$, and the equation for Junction e is ${I}_{2}+{I}_{3}={I}_{1}$. These are equivalent equations, so it is necessary to keep only one of them.
When choosing the loops in the circuit, you need enough loops so that each component is covered once, without repeating loops. Figure 10.25 shows four choices for loops to solve a sample circuit; choices (a), (b), and (c) have a sufficient amount of loops to solve the circuit completely. Option (d) reflects more loops than necessary to solve the circuit.
Consider the circuit in Figure 10.26(a). Let us analyze this circuit to find the current through each resistor. First, label the circuit as shown in part (b).
Next, determine the junctions. In this circuit, points b and e each have three wires connected, making them junctions. Start to apply Kirchhoff’s junction rule $\left(\sum {I}_{\text{in}}=\sum {I}_{\text{out}}\right)$ by drawing arrows representing the currents and labeling each arrow, as shown in Figure 10.27(b). Junction b shows that ${I}_{1}={I}_{2}+{I}_{3}$ and Junction e shows that ${I}_{2}+{I}_{3}={I}_{1}$. Since Junction e gives the same information of Junction b, it can be disregarded. This circuit has three unknowns, so we need three linearly independent equations to analyze it.
Next we need to choose the loops. In Figure 10.28, Loop abefa includes the voltage source ${V}_{1}$ and resistors ${R}_{1}$ and ${R}_{2}$. The loop starts at point a, then travels through points b, e, and f, and then back to point a. The second loop, Loop ebcde, starts at point e and includes resistors ${R}_{2}$ and ${R}_{3}$, and the voltage source ${V}_{2}$.
Now we can apply Kirchhoff’s loop rule, using the map in Figure 10.23. Starting at point a and moving to point b, the resistor ${R}_{1}$ is crossed in the same direction as the current flow ${I}_{1}$, so the potential drop ${I}_{1}{R}_{1}$ is subtracted. Moving from point b to point e, the resistor ${R}_{2}$ is crossed in the same direction as the current flow ${I}_{2}$ so the potential drop ${I}_{2}{R}_{2}$ is subtracted. Moving from point e to point f, the voltage source ${V}_{1}$ is crossed from the negative terminal to the positive terminal, so ${V}_{1}$ is added. There are no components between points f and a. The sum of the voltage differences must equal zero:
$\text{Loop}\phantom{\rule{0.2em}{0ex}}abefa:\phantom{\rule{0.5em}{0ex}}-{I}_{1}{R}_{1}-{I}_{2}{R}_{2}+{V}_{1}=0\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}{V}_{1}={I}_{1}{R}_{1}+{I}_{2}{R}_{2}.$
Finally, we check loop ebcde. We start at point e and move to point b, crossing ${R}_{2}$ in the opposite direction as the current flow ${I}_{2}$. The potential drop ${I}_{2}{R}_{2}$ is added. Next, we cross ${R}_{3}$ and ${R}_{4}$ in the same direction as the current flow ${I}_{3}$ and subtract the potential drops ${I}_{3}{R}_{3}$ and ${I}_{3}{R}_{4}.$ Note that the current is the same through resistors ${R}_{3}$ and ${R}_{4}$, because they are connected in series. Finally, the voltage source is crossed from the positive terminal to the negative terminal, and the voltage source ${V}_{2}$ is subtracted. The sum of these voltage differences equals zero and yields the loop equation
$\text{Loop}\phantom{\rule{0.2em}{0ex}}ebcde:{I}_{2}{R}_{2}-{I}_{3}\left({R}_{3}+{R}_{4}\right)-{V}_{2}=0.$
We now have three equations, which we can solve for the three unknowns.
$\begin{array}{}\\ \\ \\ \phantom{\rule{1em}{0ex}}\text{(1)}\phantom{\rule{0.2em}{0ex}}\text{Junction}\phantom{\rule{0.2em}{0ex}}b:{I}_{1}-{I}_{2}-{I}_{3}=0.\hfill \\ \phantom{\rule{1em}{0ex}}\text{(2)}\phantom{\rule{0.2em}{0ex}}\text{Loop}\phantom{\rule{0.2em}{0ex}}abefa:\phantom{\rule{0.5em}{0ex}}{I}_{1}{R}_{1}+{I}_{2}{R}_{2}={V}_{1}.\hfill \\ \phantom{\rule{1em}{0ex}}\text{(3)}\phantom{\rule{0.2em}{0ex}}\text{Loop}\phantom{\rule{0.2em}{0ex}}ebcde:{I}_{2}{R}_{2}-{I}_{3}\left({R}_{3}+{R}_{4}\right)={V}_{2}.\hfill \end{array}$
To solve the three equations for the three unknown currents, start by eliminating current ${I}_{2}$. First add Eq. (1) times ${R}_{2}$ to Eq. (2). The result is labeled as Eq. (4):
$\begin{array}{}\\ \left({R}_{1}+{R}_{2}\right){I}_{1}-{R}_{2}{I}_{3}={V}_{1}.\hfill \\ \\ \\ \phantom{\rule{1em}{0ex}}\text{(4)}\phantom{\rule{0.2em}{0ex}}6\phantom{\rule{0.2em}{0ex}}\text{Ω}{I}_{1}-3\phantom{\rule{0.2em}{0ex}}\text{Ω}{I}_{3}=24\phantom{\rule{0.2em}{0ex}}\text{V}.\hfill \end{array}$
Next, subtract Eq. (3) from Eq. (2). The result is labeled as Eq. (5):
$\begin{array}{}\\ {I}_{1}{R}_{1}+{I}_{3}\left({R}_{3}+{R}_{4}\right)={V}_{1}-{V}_{2}.\hfill \\ \\ \\ \phantom{\rule{1em}{0ex}}\text{(5)}\phantom{\rule{0.2em}{0ex}}3\phantom{\rule{0.2em}{0ex}}\text{Ω}{I}_{1}+7\phantom{\rule{0.2em}{0ex}}\text{Ω}{I}_{3}=-5\phantom{\rule{0.2em}{0ex}}\text{V}.\hfill \end{array}$
We can solve Eqs. (4) and (5) for current ${I}_{1}$. Adding seven times Eq. (4) and three times Eq. (5) results in $51\phantom{\rule{0.2em}{0ex}}\text{Ω}{I}_{1}=153\phantom{\rule{0.2em}{0ex}}\text{V},$ or ${I}_{1}=3.00\phantom{\rule{0.2em}{0ex}}\text{A}.$ Using Eq. (4) results in ${I}_{3}=-2.00\phantom{\rule{0.2em}{0ex}}\text{A}.$ Finally, Eq. (1) yields ${I}_{2}={I}_{1}-{I}_{3}=5.00\phantom{\rule{0.2em}{0ex}}\text{A}.$ One way to check that the solutions are consistent is to check the power supplied by the voltage sources and the power dissipated by the resistors:
$\begin{array}{c}{P}_{\text{in}}={I}_{1}{V}_{1}+{I}_{3}{V}_{2}=130\phantom{\rule{0.2em}{0ex}}\text{W},\hfill \\ {P}_{\text{out}}={I}_{1}^{2}{R}_{1}+{I}_{2}^{2}{R}_{2}+{I}_{3}^{2}{R}_{3}+{I}_{3}^{2}{R}_{4}=130\phantom{\rule{0.2em}{0ex}}\text{W}.\hfill \end{array}$
Note that the solution for the current ${I}_{3}$ is negative. This is the correct answer, but suggests that the arrow originally drawn in the junction analysis is the direction opposite of conventional current flow. The power supplied by the second voltage source is 58 W and not −58 W.
### Example
#### Calculating Current by Using Kirchhoff’s Rules
Find the currents flowing in the circuit in Figure 10.29.
#### Strategy
This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the series-parallel techniques—it is necessary to use Kirchhoff’s rules. Currents have been labeled ${I}_{1},$${I}_{2},$ and ${I}_{3}$ in the figure, and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h. In the solution, we apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents.
#### Solution
Applying the junction and loop rules yields the following three equations. We have three unknowns, so three equations are required.
$\begin{array}{c}\text{Junction}\phantom{\rule{0.2em}{0ex}}c:\phantom{\rule{0.2em}{0ex}}{I}_{1}+{I}_{2}={I}_{3}.\hfill \\ \text{Loop}\phantom{\rule{0.2em}{0ex}}abcdefa:\phantom{\rule{0.2em}{0ex}}{I}_{1}\left({R}_{1}+{R}_{4}\right)-{I}_{2}\left({R}_{2}+{R}_{5}+{R}_{6}\right)={V}_{1}-{V}_{3}.\hfill \\ \text{Loop}\phantom{\rule{0.2em}{0ex}}cdefc:\phantom{\rule{0.2em}{0ex}}{I}_{2}\left({R}_{2}+{R}_{5}+{R}_{6}\right)+{I}_{3}{R}_{3}={V}_{2}+{V}_{3}.\hfill \end{array}$
Simplify the equations by placing the unknowns on one side of the equations.
$\begin{array}{c}\text{Junction}\phantom{\rule{0.2em}{0ex}}c:\phantom{\rule{0.2em}{0ex}}{I}_{1}+{I}_{2}-{I}_{3}=0.\hfill \\ \text{Loop}\phantom{\rule{0.2em}{0ex}}abcdefa:\phantom{\rule{0.2em}{0ex}}{I}_{1}\left(3\phantom{\rule{0.2em}{0ex}}\text{Ω}\right)-{I}_{2}\left(8\phantom{\rule{0.2em}{0ex}}\text{Ω}\right)=0.5\phantom{\rule{0.2em}{0ex}}\text{V}-2.30\phantom{\rule{0.2em}{0ex}}\text{V}.\hfill \\ \text{Loop}\phantom{\rule{0.2em}{0ex}}cdefc:\phantom{\rule{0.2em}{0ex}}{I}_{2}\left(8\phantom{\rule{0.2em}{0ex}}\text{Ω}\right)+{I}_{3}\left(1\phantom{\rule{0.2em}{0ex}}\text{Ω}\right)=0.6\phantom{\rule{0.2em}{0ex}}\text{V}+2.30\phantom{\rule{0.2em}{0ex}}\text{V}.\hfill \end{array}$
Simplify the equations. The first loop equation can be simplified by dividing both sides by 3.00. The second loop equation can be simplified by dividing both sides by 6.00.
$\begin{array}{c}\text{Junction}\phantom{\rule{0.2em}{0ex}}c:\phantom{\rule{0.2em}{0ex}}{I}_{1}+{I}_{2}-{I}_{3}=0.\hfill \\ \text{Loop}\phantom{\rule{0.2em}{0ex}}abcdefa:\phantom{\rule{0.2em}{0ex}}{I}_{1}\left(3\phantom{\rule{0.2em}{0ex}}\text{Ω}\right)-{I}_{2}\left(8\phantom{\rule{0.2em}{0ex}}\text{Ω}\right)=\text{−}1.8\phantom{\rule{0.2em}{0ex}}\text{V}.\hfill \\ \text{Loop}\phantom{\rule{0.2em}{0ex}}cdefc:\phantom{\rule{0.2em}{0ex}}{I}_{2}\left(8\phantom{\rule{0.2em}{0ex}}\text{Ω}\right)+{I}_{3}\left(1\phantom{\rule{0.2em}{0ex}}\text{Ω}\right)=2.9\phantom{\rule{0.2em}{0ex}}\text{V}.\hfill \end{array}$
The results are
${I}_{1}=0.20\phantom{\rule{0.2em}{0ex}}\text{A},\phantom{\rule{0.5em}{0ex}}{I}_{2}=0.30\phantom{\rule{0.2em}{0ex}}\text{A},\phantom{\rule{0.5em}{0ex}}{I}_{3}=0.50\phantom{\rule{0.2em}{0ex}}\text{A}.$
#### Significance
A method to check the calculations is to compute the power dissipated by the resistors and the power supplied by the voltage sources:
$\begin{array}{}\\ {P}_{{R}_{1}}={I}_{1}^{2}{R}_{1}=0.04\phantom{\rule{0.2em}{0ex}}\text{W}.\hfill \\ {P}_{{R}_{2}}={I}_{2}^{2}{R}_{2}=0.45\phantom{\rule{0.2em}{0ex}}\text{W}.\hfill \\ {P}_{{R}_{3}}={I}_{3}^{2}{R}_{3}=0.25\phantom{\rule{0.2em}{0ex}}\text{W}.\hfill \\ {P}_{{R}_{4}}={I}_{1}^{2}{R}_{4}=0.08\phantom{\rule{0.2em}{0ex}}\text{W}.\hfill \\ {P}_{{R}_{5}}={I}_{2}^{2}{R}_{5}=0.09\phantom{\rule{0.2em}{0ex}}\text{W}.\hfill \\ {P}_{{R}_{6}}={I}_{2}^{2}{R}_{6}=0.18\phantom{\rule{0.2em}{0ex}}\text{W}.\hfill \\ {P}_{\text{dissipated}}=1.09\phantom{\rule{0.2em}{0ex}}\text{W}.\hfill \\ {P}_{\text{source}}={I}_{1}{V}_{1}+{I}_{2}{V}_{3}+{I}_{3}{V}_{2}=0.10\phantom{\rule{0.2em}{0ex}}\text{W}+0.69\phantom{\rule{0.2em}{0ex}}\text{W}+0.30\phantom{\rule{0.2em}{0ex}}\text{W}=1.09\phantom{\rule{0.2em}{0ex}}\text{W}.\hfill \end{array}$
The power supplied equals the power dissipated by the resistors.
In considering the following schematic and the power supplied and consumed by a circuit, will a voltage source always provide power to the circuit, or can a voltage source consume power?
Show Solution
The circuit can be analyzed using Kirchhoff’s loop rule. The first voltage source supplies power: ${P}_{\text{in}}=I{V}_{1}=7.20\phantom{\rule{0.2em}{0ex}}\text{mW.}$ The second voltage source consumes power: ${P}_{\text{out}}=I{V}_{2}+{I}^{2}{R}_{1}+{I}^{2}{R}_{2}=7.2\phantom{\rule{0.2em}{0ex}}\text{mW}.$
### Example
#### Calculating Current by Using Kirchhoff’s Rules
Find the current flowing in the circuit in Figure 10.30.
#### Strategy
This circuit can be analyzed using Kirchhoff’s rules. There is only one loop and no nodes. Choose the direction of current flow. For this example, we will use the clockwise direction from point a to point b. Consider Loop abcda and use Figure 10.23 to write the loop equation. Note that according to Figure 10.23, battery ${V}_{1}$ will be added and battery ${V}_{2}$ will be subtracted.
#### Solution
Applying the junction rule yields the following three equations. We have one unknown, so one equation is required:
$\text{Loop}\phantom{\rule{0.2em}{0ex}}abcda:\phantom{\rule{0.2em}{0ex}}-I{R}_{1}-{V}_{1}-I{R}_{2}+{V}_{2}-I{R}_{3}=0.$
Simplify the equations by placing the unknowns on one side of the equations. Use the values given in the figure.
$\begin{array}{c}I\left({R}_{1}+{R}_{2}+{R}_{3}\right)={V}_{2}-{V}_{1}.\hfill \\ \\ I=\frac{{V}_{2}-{V}_{1}}{{R}_{1}+{R}_{2}+{R}_{3}}=\frac{24\phantom{\rule{0.2em}{0ex}}\text{V}-12\phantom{\rule{0.2em}{0ex}}\text{V}}{10.0\phantom{\rule{0.2em}{0ex}}\text{Ω}+30.0\phantom{\rule{0.2em}{0ex}}\text{Ω}+10.0\phantom{\rule{0.2em}{0ex}}\text{Ω}}=0.20\phantom{\rule{0.2em}{0ex}}\text{A}.\hfill \end{array}$
#### Significance
The power dissipated or consumed by the circuit equals the power supplied to the circuit, but notice that the current in the battery ${V}_{1}$ is flowing through the battery from the positive terminal to the negative terminal and consumes power.
$\begin{array}{}\\ \\ {P}_{{R}_{1}}={I}^{2}{R}_{1}=0.40\phantom{\rule{0.2em}{0ex}}\text{W}\hfill \\ {P}_{{R}_{2}}={I}^{2}{R}_{2}=1.20\phantom{\rule{0.2em}{0ex}}\text{W}\hfill \\ {P}_{{R}_{3}}={I}^{2}{R}_{3}=0.80\phantom{\rule{0.2em}{0ex}}\text{W}\hfill \\ {P}_{{V}_{1}}=I{V}_{1}=2.40\phantom{\rule{0.2em}{0ex}}\text{W}\hfill \\ {P}_{\text{dissipated}}=4.80\phantom{\rule{0.2em}{0ex}}\text{W}\hfill \\ {P}_{\text{source}}=I{V}_{2}=4.80\phantom{\rule{0.2em}{0ex}}\text{W}\hfill \end{array}$
The power supplied equals the power dissipated by the resistors and consumed by the battery ${V}_{1}.$
When using Kirchhoff’s laws, you need to decide which loops to use and the direction of current flow through each loop. In analyzing the circuit in Example 10.7, the direction of current flow was chosen to be clockwise, from point a to point b. How would the results change if the direction of the current was chosen to be counterclockwise, from point b to point a?
Show Solution
The current calculated would be equal to $I=-0.20\phantom{\rule{0.2em}{0ex}}\text{A}$ instead of $I=0.20\phantom{\rule{0.2em}{0ex}}\text{A}.$ The sum of the power dissipated and the power consumed would still equal the power supplied.
### Multiple Voltage Sources
Many devices require more than one battery. Multiple voltage sources, such as batteries, can be connected in series configurations, parallel configurations, or a combination of the two.
In series, the positive terminal of one battery is connected to the negative terminal of another battery. Any number of voltage sources, including batteries, can be connected in series. Two batteries connected in series are shown in Figure 10.31. Using Kirchhoff’s loop rule for the circuit in part (b) gives the result
$\begin{array}{c}{\epsilon }_{1}-I{r}_{1}+{\epsilon }_{2}-I{r}_{2}-IR=0,\hfill \\ \left[\left({\epsilon }_{1}+{\epsilon }_{2}\right)-I\left({r}_{1}+{r}_{2}\right)\right]-IR=0.\hfill \end{array}$
When voltage sources are in series, their internal resistances can be added together and their emfs can be added together to get the total values. Series connections of voltage sources are common—for example, in flashlights, toys, and other appliances. Usually, the cells are in series in order to produce a larger total emf. In Figure 10.31, the terminal voltage is
${V}_{\text{terminal}}=\left({\epsilon }_{1}-I{r}_{1}\right)+\left({\epsilon }_{2}-I{r}_{2}\right)=\left[\left({\epsilon }_{1}+{\epsilon }_{2}\right)-I\left({r}_{1}+{r}_{2}\right)\right]=\left({\epsilon }_{1}+{\epsilon }_{2}\right)+I{r}_{\text{eq}}.$
Note that the same current I is found in each battery because they are connected in series. The disadvantage of series connections of cells is that their internal resistances are additive.
Batteries are connected in series to increase the voltage supplied to the circuit. For instance, an LED flashlight may have two AAA cell batteries, each with a terminal voltage of 1.5 V, to provide 3.0 V to the flashlight.
Any number of batteries can be connected in series. For N batteries in series, the terminal voltage is equal to
${V}_{\text{terminal}}=\left({\epsilon }_{1}+{\epsilon }_{2}+\text{⋯}+{\epsilon }_{N-1}+{\epsilon }_{N}\right)-I\left({r}_{1}+{r}_{2}+\text{⋯}+{r}_{N-1}+{r}_{N}\right)=\sum _{i=1}^{N}{\epsilon }_{i}-I{r}_{\text{eq}}$
where the equivalent resistance is ${r}_{\text{eq}}=\sum _{i=1}^{N}{r}_{i}$.
When a load is placed across voltage sources in series, as in Figure 10.32, we can find the current:
$\begin{array}{c}\left({\epsilon }_{1}-I{r}_{1}\right)+\left({\epsilon }_{2}-I{r}_{2}\right)=IR,\hfill \\ I{r}_{1}+I{r}_{2}+IR={\epsilon }_{1}+{\epsilon }_{2},\hfill \\ I=\frac{{\epsilon }_{1}+{\epsilon }_{2}}{{r}_{1}+{r}_{2}+R}.\hfill \end{array}$
As expected, the internal resistances increase the equivalent resistance.
Voltage sources, such as batteries, can also be connected in parallel. Figure 10.33 shows two batteries with identical emfs in parallel and connected to a load resistance. When the batteries are connect in parallel, the positive terminals are connected together and the negative terminals are connected together, and the load resistance is connected to the positive and negative terminals. Normally, voltage sources in parallel have identical emfs. In this simple case, since the voltage sources are in parallel, the total emf is the same as the individual emfs of each battery.
Consider the Kirchhoff analysis of the circuit in Figure 10.33(b). There are two loops and a node at point b and $\epsilon ={\epsilon }_{1}={\epsilon }_{2}$.
Node b: ${I}_{1}+{I}_{2}-I=0$.
Loop abcfa: $\begin{array}{ccc}\hfill \epsilon -{I}_{1}{r}_{1}+{I}_{2}{r}_{2}-\epsilon & =\hfill & 0,\hfill \\ \hfill {I}_{1}{r}_{1}& =\hfill & {I}_{2}{r}_{2}.\hfill \end{array}$
Loop fcdef: $\begin{array}{ccc}\hfill {\epsilon }_{2}-{I}_{2}{r}_{2}-IR& =\hfill & 0,\hfill \\ \hfill \epsilon -{I}_{2}{r}_{2}-IR& =\hfill & 0.\hfill \end{array}$
Solving for the current through the load resistor results in $I=\frac{\epsilon }{{r}_{\text{eq}}+R}$, where ${r}_{\text{eq}}={\left(\frac{1}{{r}_{1}}+\frac{1}{{r}_{2}}\right)}^{-1}$. The terminal voltage is equal to the potential drop across the load resistor $IR=\left(\frac{\epsilon }{{r}_{\text{eq}}+R}\right)$. The parallel connection reduces the internal resistance and thus can produce a larger current.
Any number of batteries can be connected in parallel. For N batteries in parallel, the terminal voltage is equal to
${V}_{\text{terminal}}=\epsilon -I{\left(\frac{1}{{r}_{1}}+\frac{1}{{r}_{2}}+\text{⋯}+\frac{1}{{r}_{N-1}}+\frac{1}{{r}_{N}}\right)}^{-1}=\epsilon -I{r}_{\text{eq}}$
where the equivalent resistance is ${r}_{\text{eq}}={\left(\sum _{i=1}^{N}\frac{1}{{r}_{i}}\right)}^{-1}$.
As an example, some diesel trucks use two 12-V batteries in parallel; they produce a total emf of 12 V but can deliver the larger current needed to start a diesel engine.
In summary, the terminal voltage of batteries in series is equal to the sum of the individual emfs minus the sum of the internal resistances times the current. When batteries are connected in parallel, they usually have equal emfs and the terminal voltage is equal to the emf minus the equivalent internal resistance times the current, where the equivalent internal resistance is smaller than the individual internal resistances. Batteries are connected in series to increase the terminal voltage to the load. Batteries are connected in parallel to increase the current to the load.
### Solar Cell Arrays
Another example dealing with multiple voltage sources is that of combinations of solar cells—wired in both series and parallel combinations to yield a desired voltage and current. Photovoltaic generation, which is the conversion of sunlight directly into electricity, is based upon the photoelectric effect. The photoelectric effect is beyond the scope of this chapter and is covered in Photons and Matter Waves, but in general, photons hitting the surface of a solar cell create an electric current in the cell.
Most solar cells are made from pure silicon. Most single cells have a voltage output of about 0.5 V, while the current output is a function of the amount of sunlight falling on the cell (the incident solar radiation known as the insolation). Under bright noon sunlight, a current per unit area of about $100{\phantom{\rule{0.2em}{0ex}}\text{mA/cm}}^{2}$ of cell surface area is produced by typical single-crystal cells.
Individual solar cells are connected electrically in modules to meet electrical energy needs. They can be wired together in series or in parallel—connected like the batteries discussed earlier. A solar-cell array or module usually consists of between 36 and 72 cells, with a power output of 50 W to 140 W.
Solar cells, like batteries, provide a direct current (dc) voltage. Current from a dc voltage source is unidirectional. Most household appliances need an alternating current (ac) voltage.
### Summary
• Kirchhoff’s rules can be used to analyze any circuit, simple or complex. The simpler series and parallel connection rules are special cases of Kirchhoff’s rules.
• Kirchhoff’s first rule, also known as the junction rule, applies to the charge to a junction. Current is the flow of charge; thus, whatever charge flows into the junction must flow out.
• Kirchhoff’s second rule, also known as the loop rule, states that the voltage drop around a loop is zero.
• When calculating potential and current using Kirchhoff’s rules, a set of conventions must be followed for determining the correct signs of various terms.
• When multiple voltage sources are in series, their internal resistances add together and their emfs add together to get the total values.
• When multiple voltage sources are in parallel, their internal resistances combine to an equivalent resistance that is less than the individual resistance and provides a higher current than a single cell.
• Solar cells can be wired in series or parallel to provide increased voltage or current, respectively.
### Conceptual Questions
Can all of the currents going into the junction shown below be positive? Explain.
Consider the circuit shown below. Does the analysis of the circuit require Kirchhoff’s method, or can it be redrawn to simplify the circuit? If it is a circuit of series and parallel connections, what is the equivalent resistance?
Show Solution
It can be redrawn.
${R}_{\text{eq}}={\left[\frac{1}{{R}_{6}}+\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}+{\left(\frac{1}{{R}_{4}}+\frac{1}{{R}_{3}+{R}_{5}}\right)}^{-1}}\right]}^{-1}$
Do batteries in a circuit always supply power to a circuit, or can they absorb power in a circuit? Give an example.
What are the advantages and disadvantages of connecting batteries in series? In parallel?
Show Solution
In series the voltages add, but so do the internal resistances, because the internal resistances are in series. In parallel, the terminal voltage is the same, but the equivalent internal resistance is smaller than the smallest individual internal resistance and a higher current can be provided.
Semi-tractor trucks use four large 12-V batteries. The starter system requires 24 V, while normal operation of the truck’s other electrical components utilizes 12 V. How could the four batteries be connected to produce 24 V? To produce 12 V? Why is 24 V better than 12 V for starting the truck’s engine (a very heavy load)?
### Problems
Consider the circuit shown below. (a) Find the voltage across each resistor. (b)What is the power supplied to the circuit and the power dissipated or consumed by the circuit?
Consider the circuits shown below. (a) What is the current through each resistor in part (a)? (b) What is the current through each resistor in part (b)? (c) What is the power dissipated or consumed by each circuit? (d) What is the power supplied to each circuit?
Show Solution
a. ${I}_{1}=0.6\phantom{\rule{0.2em}{0ex}}\text{mA},\phantom{\rule{0.5em}{0ex}}{I}_{2}=0.4\phantom{\rule{0.2em}{0ex}}\text{mA},\phantom{\rule{0.5em}{0ex}}{I}_{3}=0.2\phantom{\rule{0.2em}{0ex}}\text{mA}$;
b. ${I}_{1}=0.04\phantom{\rule{0.2em}{0ex}}\text{mA},\phantom{\rule{0.5em}{0ex}}{I}_{2}=1.52\phantom{\rule{0.2em}{0ex}}\text{mA},\phantom{\rule{0.5em}{0ex}}{I}_{3}=-1.48\phantom{\rule{0.2em}{0ex}}\text{mA}$; c. ${P}_{\text{out}}=0.92\phantom{\rule{0.2em}{0ex}}\text{mW},\phantom{\rule{0.5em}{0ex}}{P}_{\text{out}}=4.50\phantom{\rule{0.2em}{0ex}}\text{mW}$;
d. ${P}_{\text{in}}=0.92\phantom{\rule{0.2em}{0ex}}\text{mW},\phantom{\rule{0.5em}{0ex}}{P}_{\text{in}}=4.50\phantom{\rule{0.2em}{0ex}}\text{mW}$
Consider the circuit shown below. Find ${V}_{1},{I}_{2},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{I}_{3}.$
Consider the circuit shown below. Find ${V}_{1},{V}_{2},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{R}_{4}.$
Show Solution
${V}_{1}=42\phantom{\rule{0.2em}{0ex}}\text{V},{V}_{2}=6\phantom{\rule{0.2em}{0ex}}\text{V},{R}_{4}=18\phantom{\rule{0.2em}{0ex}}\text{Ω}$
Consider the circuit shown below. Find ${I}_{1},{I}_{2},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{I}_{3}.$
Consider the circuit shown below. (a) Find ${I}_{1},{I}_{2},{I}_{3},{I}_{4},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{I}_{5}.$ (b) Find the power supplied by the voltage sources. (c) Find the power dissipated by the resistors.
Show Solution
a. ${I}_{1}=1.5\phantom{\rule{0.2em}{0ex}}\text{A},{I}_{2}=2\phantom{\rule{0.2em}{0ex}}\text{A},{I}_{3}=0.5\phantom{\rule{0.2em}{0ex}}\text{A},\phantom{\rule{0.2em}{0ex}}{I}_{4}=2.5\phantom{\rule{0.2em}{0ex}}\text{A},{I}_{5}=2\phantom{\rule{0.2em}{0ex}}\text{A}$; b. ${P}_{\text{in}}={I}_{2}{V}_{1}+{I}_{5}{V}_{5}=34\phantom{\rule{0.2em}{0ex}}\text{W}$;
c. ${P}_{\text{out}}={I}_{1}^{2}{R}_{1}+{I}_{2}^{2}{R}_{2}+{I}_{3}^{2}{R}_{3}+{I}_{4}^{2}{R}_{4}=34\phantom{\rule{0.2em}{0ex}}\text{W}$
Consider the circuit shown below. Write the three loop equations for the loops shown.
Consider the circuit shown below. Write equations for the three currents in terms of R and V.
Show Solution
${I}_{1}=\frac{2}{3}\frac{V}{R},{I}_{2}=\frac{V}{3R},{I}_{3}=\frac{V}{3R}$
Consider the circuit shown in the preceding problem. Write equations for the power supplied by the voltage sources and the power dissipated by the resistors in terms of R and V.
A child’s electronic toy is supplied by three 1.58-V alkaline cells having internal resistances of $0.0200\phantom{\rule{0.2em}{0ex}}\text{Ω}$ in series with a 1.53-V carbon-zinc dry cell having a $0.100\text{-}\text{Ω}$ internal resistance. The load resistance is $10.0\phantom{\rule{0.2em}{0ex}}\text{Ω}$. (a) Draw a circuit diagram of the toy and its batteries. (b) What current flows? (c) How much power is supplied to the load? (d) What is the internal resistance of the dry cell if it goes bad, resulting in only 0.500 W being supplied to the load?
Show Solution
a.
;
b. 0.617 A; c. 3.81 W; d. $18.0\phantom{\rule{0.2em}{0ex}}\text{Ω}$
Apply the junction rule to Junction b shown below. Is any new information gained by applying the junction rule at e?
Apply the loop rule to Loop afedcba in the preceding problem.
Show Solution
${I}_{1}{r}_{1}-{\epsilon }_{1}+{I}_{1}{R}_{4}+{\epsilon }_{4}+{I}_{2}{r}_{4}+{I}_{4}{r}_{3}-{\epsilon }_{3}+{I}_{2}{R}_{3}+{I}_{1}{R}_{1}=0$
### Glossary
junction rule
sum of all currents entering a junction must equal the sum of all currents leaving the junction
Kirchhoff’s rules
set of two rules governing current and changes in potential in an electric circuit
loop rule
algebraic sum of changes in potential around any closed circuit path (loop) must be zero | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9925146102905273, "perplexity": 625.2846105018505}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950383.8/warc/CC-MAIN-20230402043600-20230402073600-00657.warc.gz"} |
https://medcraveonline.com/MOJES/structuralndashparametric-model-electroelastic-actuator-nanondash-and-microdisplacement-of-mechatronics-systems-for-nanotechnology-and-ecology-research.html | # Ecology & Environmental Sciences
Research Article Volume 3 Issue 5
# Structural–parametric model electroelastic actuator nano– and microdisplacement of mechatronics systems for nanotechnology and ecology research
#### Sergey M Afonin function clickButton(){ var name=document.getElementById('name').value; var descr=document.getElementById('descr').value; var unCopyslNo=document.getElementById('unCopyslNo').value; document.getElementById("mydiv").style.display = "none"; $.ajax({ type:"post", url:"https://medcraveonline.com/captchaCode/server_action", data: { 'name' :name, 'descr' :descr, 'unCopyslNo': unCopyslNo }, cache:false, success: function (html) { //alert('Data Send');$('#msg').html(html); } }); return false; } Verify Captcha × Regret for the inconvenience: we are taking measures to prevent fraudulent form submissions by extractors and page crawlers. Please type the correct Captcha word to see email ID. function refreshPage(){ $("#mydiv").load(location.href + " #mydiv"); }$(document).ready(function () { //Disable cut copy paste $('#msg').bind('cut copy paste', function (e) { e.preventDefault(); }); //Disable mouse right click$("#msg").on("contextmenu",function(e){ return false; }); }); .noselect { -webkit-touch-callout: none; /* iOS Safari */ -webkit-user-select: none; /* Safari */ -khtml-user-select: none; /* Konqueror HTML */ -moz-user-select: none; /* Firefox */ -ms-user-select: none; /* Internet Explorer/Edge */ user-select: none; /* Non-prefixed version, currently supported by Chrome and Opera */ cursor: none; }
Department of Intellectual Technical Systems, National Research University of Electronic Technology (MIET), Russia
Correspondence: Sergey M Afonin Department of Intellectual Technical Systems, National Research University of Electronic Technology (MIET), Moscow, 124498 Moscow, Russia
Received: April 21, 2017 | Published: September 4, 2018
Citation: Afonin SM. Structural–parametric model electroelastic actuator nano– and microdisplacement of mechatronics systems for nanotechnology and ecology research. MOJ Eco Environ Sci. 2018;3(5):306-309 DOI: 10.15406/mojes.2018.03.00104
# Abstract
The structural–parametric model, the decision of the wave equation, the parametric structural schematic diagram, the transfer functions of the electroelastic actuator of the mechatronics system for the nanotechnology and the ecology research are obtained. Effects of geometric and physical parameters of the piezoactuator and the external load on its dynamic characteristics are determined. The parametric structural schematic diagram and the transfer functions of the piezoactuator for the transverse, longitudinal, shift piezoelectric effects are obtained from the structural–parametric model of the piezoactuator. For calculation of the mechatronics systems for the nanotechnology with the piezoactuator it’s the parametric structural schematic diagram and the transfer functions are determined. The generalized parametric structural schematic diagram of the electroelastic actuator is constructed.
Keywords: electroelastic actuator, piezoactuator, deformation, structural–parametric model, parametric structural schematic diagram, decision wave equations, transfer functions
# Introduction
For the nanotechnology, the ecology research, the nanobiology, the power engineering, the microelectronics, the astronomy for the large compound telescopes, the antennas satellite telescopes and the adaptive optics equipment is promising for use the mechatronics system with the actuator based on the electroelasticity for the piezoelectric or the electrostriction effects. The piezoactuator is the piezomechanical device intended for the actuation of mechanisms, systems or the management based on the piezoelectric effect, the converts electrical signals into the mechanical movement or the force.1–5 In the present work is solving the problem of building the structural parametric model of the electroelastic actuator in contrast Cady and Mason electrical equivalent circuits for calculation of piezoelectric transmitter and receiver.6–9 The structural–parametric model of the piezoactuator describes the structure and conversion the energy electric field into the mechanical energy and the corresponding displacements and forces at its the faces. The structural–parametric model of the electroelastic actuator of the mechatronics system is determined by using the method of the mathematical physics. The transfer functions and the parametric structural schematic diagrams of the electroelastic actuator are obtained from its structural–parametric model.3–14 The piezoactuator for the nano– and microdisplacement of the mechatronics system operates based on the inverse piezoeffect. The displacement is achieved due to deformation of the piezoactuator when the external electric voltage is applied to it. The piezoactuator for the drives of nano– and micrometric movements provide a movement range from several nanometers to tens of micrometers, a sensitivity of up to 10 nm/V, a loading capacity of up to 1000 N, a transmission band of up to 100 Hz. The piezoactuator provides high speed and force, its return to the initial state when switched off. The use of the piezoactuator solves the problems of the precise alignment and the compensation of the temperature and gravitational deformations. The piezoactuator is used in the majority mechatronic systems for the nanotechnology, the ecology research in the scanning tunneling microscopes and the atomic force microscopes.11–16
# Decision wave equation and structural parametric model of electroelastic actuator
The deformation of the electroelastic actuator corresponds to its stressed state. In the piezoactuator there are six stress components ${T}_{1}$ , ${T}_{2}$ ${T}_{3}$ , ${T}_{4}$ , ${T}_{5}$ , ${T}_{6}$ where the components ${T}_{1}-{T}_{3}$ are related to extension–compression stresses and the components ${T}_{4}-{T}_{6}$ to shear stresses. The matrix state equations8,11 connecting the electric and elastic variables for the polarized piezoceramics have the following form:
$D=dT+{\epsilon }^{T}E$ , (1)
$S={s}^{E}T+{d}^{t}E$ , (2)
where the first equation describes the direct piezoelectric effect, and the second - the inverse piezoelectric effect; $D$ is the column matrix of electric induction along the coordinate axes; $S$ is the column matrix of relative deformations; $T$ is the column matrix of mechanical stresses; $E\text{\hspace{0.17em}}$ is the column matrix of electric field strength along the coordinate axes; ${s}^{E}\text{\hspace{0.17em}}$ is the elastic compliance matrix for $E=\text{const}$ ; ${\epsilon }^{T}$ is the matrix of dielectric constants for $T=\text{const}$ ; ${d}^{t}\text{\hspace{0.17em}}$ is the transposed matrix of the piezoelectric modules.
In polarized piezoceramics from lead zirconate titanate PZT for the piezoactuator on Figure 1 there are five independent components ${s}_{11}^{E}$ ${s}_{12}^{E}$ , ${s}_{13}^{E}$ ,${s}_{33}^{E}$ , ${s}_{11}^{E}$ in the elastic compliance matrix, three independent components ${d}_{33}$ ,${d}_{31}$ , ${d}_{15}$ in the transposed matrix of the piezoelectric modules and three independent components, ${\epsilon }_{11}^{T}$ ,${\epsilon }_{33}^{T}$ ${\epsilon }_{22}^{T}$ in the matrix of dielectric constants.
Figure 1 Piezoactuator.
Let us consider the piezoactuator for the longitudinal piezoelectric effect, where $\delta$ is thickness and the electrodes deposited on its faces perpendicular to axis 3, the area of which is equal${S}_{0}$ . The direction of the polarization axis Р, i.e., the direction along which polarization was performed, is usually taken as the direction of axis 3. The equation of the inverse longitudinal piezoelectric effect8,11 has the form:
${S}_{3}={d}_{33}{E}_{3}\left(t\right)+{s}_{33}^{E}{T}_{3}\left(x,t\right)$ , (3)
where ${S}_{3}=\partial \xi \left(x,t\right)/\partial x$ is the relative displacement of the cross section of the piezoactuator, ${d}_{33}$ is the piezomodule for the longitudinal piezoeffect, ${E}_{3}\left(t\right)=U\left(t\right)/\delta$ is the electric field strength, $U\left(t\right)$ c is the voltage between the electrodes of actuator, $\delta$ is the thickness, ${s}_{33}^{E}$ is the elastic compliance along axis 3, and ${T}_{3}$ is the mechanical stress along axis 3.
The equation of equilibrium for the force acting on the piezoactuator on Figure 1 can be written as
${T}_{3}{S}_{0}=F+M\frac{{\partial }^{2}\xi \left(x,t\right)}{\partial {t}^{2}}$ , (4)
Where F is the external force applied to the piezoactuator, ${S}_{0}$ is the cross section area and M is the displaced mass.
the equation of the inverse longitudinal piezoeffect, the wave equation using Laplace transform, the equations of the forces acting on the faces of the piezoactuator. The calculations of the piezoactuators are performed using the wave equation8,11,12 describing the wave propagation in the long line with damping but without distortions in the following form:
$\frac{1}{{\left({с}^{E}\right)}^{2}}\frac{{\partial }^{2}\xi \left(x,t\right)}{\partial {t}^{2}}+\frac{2\alpha }{{c}^{E}}\frac{\partial \xi \left(x,t\right)}{\partial t}+{\alpha }^{2}\xi \left(x,t\right)=\frac{{\partial }^{2}\xi \left(x,t\right)}{\partial {x}^{2}}$, (5)
where $\xi \left(x,t\right)$ is the displacement of the section, x is the coordinate, t is the time, ${c}^{E}$ is the sound speed for $E=\text{const}$, $\alpha$ is the damping coefficient. We can reduce the original problem for the partial differential hyperbolic equation of type (5) using Laplace transform to a simpler problem for the linear ordinary differential equation[10,12]. Applying the Laplace transform to the wave equation (5)
$\Xi \left(x,p\right)=L\left\{\xi \left(x,t\right)\right\}=\underset{0}{\overset{\infty }{\int }}\xi \left(x,t\right){e}^{-pt}dt$, (6)
Setting the zero initial conditions we obtain the linear ordinary second–order differential equation with the parameter p in the form
$\frac{{d}^{2}\Xi \left(x,p\right)}{d{x}^{2}}-{\gamma }^{2}\Xi \left(x,p\right)=0$, (7)
With its solution being the function
$\Xi \left(x,p\right)=C{e}^{-x\gamma }+B{e}^{x\gamma }$, (8)
Where $\Xi \left(x,p\right)$ is the Laplace transform of the displacement of the section of the piezoelectric actuator, $\gamma =p/{c}^{E}+\alpha$ is the propagation coefficient.
We denote for the faces of the piezoactuator
$\Xi \left(0,p\right)={\Xi }_{1}\left(p\right)$ for $x=0$,(9)
$\Xi \left(\delta ,p\right)={\Xi }_{2}\left(p\right)$ for $x=\delta$.
Then we get the coefficients C and B
$C=\left({\Xi }_{1}{e}^{\delta \gamma }-{\Xi }_{2}\right)/\left[2\text{sh}\left(\delta \gamma \right)\right]$ ,$B=\left({\Xi }_{2}-{\Xi }_{1}{e}^{-\delta \gamma }\right)/\left[2\text{sh}\left(\delta \gamma \right)\right]$ , (10)
The solution (7) can be written as
$\Xi \left(x,p\right)=\left\{{\Xi }_{1}\left(p\right)\text{sh}\left[\left(\delta -x\right)\gamma \right]+{\Xi }_{2}\left(p\right)\text{sh}\left(x\gamma \right)\right\}/\text{sh}\left(\delta \gamma \right)$ , (11)
The equations for the forces on the faces of the piezoactuator
${T}_{3}\left(0,p\right){S}_{0}={F}_{1}\left(p\right)+{M}_{1}{p}^{2}{\Xi }_{1}\left(p\right)$ for $x=0$ , (12)
${T}_{3}\left(\delta ,p\right){S}_{0}=-{F}_{2}\left(p\right)-{M}_{2}{p}_{2}{\Xi }_{1}\left(p\right)$ for $x=\delta$ ,
Where ${T}_{3}\left(0,p\right)$ and ${T}_{3}\left(\delta ,p\right)$ are determined from the equation of the inverse piezoelectric effect. For $x=0$ and ${T}_{3}\left(\delta ,p\right)$ , we obtain the set of equations for determining stresses in the piezoactuator:11−14
${T}_{3}\left(0,p\right)=\frac{1}{{s}_{33}^{E}}{\frac{d\Xi \left(x,p\right)}{dx}|}_{x=0}-\frac{{d}_{33}}{{s}_{33}^{E}}{E}_{3}\left(p\right)$ , (13)
${T}_{3}\left(\delta ,p\right)=\frac{1}{{s}_{33}^{E}}{\frac{d\Xi \left(x,p\right)}{dx}|}_{x=\delta }-\frac{{d}_{33}}{{s}_{33}^{E}}{E}_{3}\left(p\right)$ .
The set of equations (13) yield the set of the equations for the structural–parametric model of the piezoactuator and the parametric structural schematic diagram of the voltage–controlled piezoactuator for the longitudinal piezoelectric effect on Figure 2.
${\Xi }_{1}\left(p\right)=\left[1/\left({M}_{1}{p}^{2}\right)\right]\cdot \left\{-{F}_{1}\left(p\right)+\left(1/{\chi }_{33}^{E}\right)\text{\hspace{0.17em}}\left[{d}_{33}{E}_{3}\left(p\right)-\left[\gamma /\text{sh}\left(\delta \gamma \right)\right]\text{\hspace{0.17em}}\left[\text{ch}\left(\delta \gamma \right){\Xi }_{1}\left(p\right)-{\Xi }_{2}\left(p\right)\right]\right]\right\},$ (14)
$\begin{array}{l}{\Xi }_{2}\left(p\right)=\left[1/\left({M}_{2}{p}^{2}\right)\right]\cdot \left\{-{F}_{2}\left(p\right)+\left(1/{\chi }_{33}^{E}\right)\text{\hspace{0.17em}}\left[{d}_{33}{E}_{3}\left(p\right)-\left[\gamma /\text{sh}\left(\delta \gamma \right)\right]\text{\hspace{0.17em}}\left[\text{ch}\left(\delta \gamma \right){\Xi }_{2}\left(p\right)-{\Xi }_{1}\left(p\right)\right]\right]\right\},\\ \end{array}$
Where${\chi }_{33}^{E}={s}_{33}^{E}/{S}_{0}$ .
From (2), (3), (14) we obtain the system of the equations describing the generalized structural–parametric model of the electroelastic actuator
${\Xi }_{1}\left(p\right)=\left[1/\left({M}_{1}{p}^{2}\right)\right]\cdot \left\{{}^{}-{F}_{1}\left(p\right)+{\left(1/{\chi }_{ij}^{\Psi }\right)}_{}^{}\left[{d}_{mi}{\Psi }_{m}\left(p\right)-{\left[\gamma /\text{sh}\left(l\gamma \right)\right]}_{}\left[\text{ch}\left(l\gamma \right){\Xi }_{1}\left(p\right)-{\Xi }_{2}\left(p\right)\right]\right]\right\},$ (15)
$\begin{array}{l}{\Xi }_{2}\left(p\right)=\left[1/\left({M}_{2}{p}^{2}\right)\right]\cdot \left\{{}^{}-{F}_{2}\left(p\right)+{\left(1/{\chi }_{ij}^{\Psi }\right)}_{}^{}\left[{d}_{mi}{\Psi }_{m}\left(p\right)-{\left[\gamma /\text{sh}\left(l\gamma \right)\right]}_{}\left[\text{ch}\left(l\gamma \right){\Xi }_{2}\left(p\right)-{\Xi }_{1}\left(p\right)\right]\right]\right\},\\ \end{array}$
where ${d}_{mi}=\left\{\begin{array}{c}{d}_{33},{d}_{31},{d}_{15}\\ {g}_{33},{g}_{31},{g}_{15}\end{array}$ , ${\Psi }_{m}=\left\{\begin{array}{c}{E}_{3},{E}_{3},{E}_{1}\\ {D}_{3},{D}_{3},{D}_{1}\end{array}$ , ${s}_{ij}^{\Psi }=\left\{\begin{array}{c}{s}_{33}^{E},{s}_{11}^{E},{s}_{55}^{E}\\ {s}_{33}^{D},{s}_{11}^{D},{s}_{55}^{D}\end{array}$ , $l=\left\{\text{\hspace{0.17em}}\delta ,h,b$ , ${c}^{\Psi }=\left\{\text{\hspace{0.17em}}{c}^{E},{c}^{D}$ , ${\gamma }^{\Psi }=\left\{\text{\hspace{0.17em}}{\gamma }^{E},{\gamma }^{D}$ ,
${\chi }_{ij}^{\Psi }={s}_{ij}^{\Psi }/{S}_{0}$ , i = 1, 2…, 6, j = 1, 2, … , 6, m = 1, 2, 3,
Then the parameter $\Psi$ of the control parameter for the electroelastic actuator: E for the voltage control, D for the current control. On Figure 3 is shown the generalized parametric structural schematic diagram of the electroelastic actuator corresponding to the set (15) of the equations.
Figure 2 Parametric structural schematic diagram of a voltage-controlled piezoactuator for longitudinal piezoelectric effect.
Figure 3 Generalized parametric structural schematic diagram of the electroetoelastic actuator.
# Transfer functions of electroelastic actuator
From the generalized structural–parametric model (15) of the electroelastic actuator after the algebraic transformations we obtain the transfer functions in matrix form.11−14 The transfer functions are the ratio of the Laplace transform of the displacement of the face for the electroelastic actuator and the Laplace transform of the corresponding control parameter or force at zero initial conditions.
${\Xi }_{1}\left(p\right)={W}_{11}\left(p\right){\Psi }_{m}\left(p\right)+{W}_{12}\left(p\right){F}_{1}\left(p\right)+{W}_{13}\left(p\right){F}_{2}\left(p\right)$ , (16)
${\Xi }_{2}\left(p\right)={W}_{21}\left(p\right){\Psi }_{m}\left(p\right)+{W}_{22}\left(p\right){F}_{1}\left(p\right)+{W}_{23}\left(p\right){F}_{2}\left(p\right)$ ,
where the generalized transfer functions
${W}_{11}\left(p\right)={\Xi }_{1}\left(p\right)/{\Psi }_{m}\left(p\right)={{d}_{mi}\text{\hspace{0.17em}}\left[{M}_{2}{\chi }_{ij}^{\Psi }{p}^{2}+\gamma \text{th}\left(l\gamma /2\right)\right]/A}_{ij}{}_{}^{}$ ,
$\begin{array}{l}{A}_{ij}={M}_{1}{M}_{2}{\left({\chi }_{ij}^{\Psi }\right)}^{2}{p}^{4}+\left\{\text{\hspace{0.17em}}\left({M}_{1}+{M}_{2}\right){\chi }_{ij}^{\Psi }/\left[{c}^{\Psi }\text{th}\left(l\gamma \right)\right]\text{\hspace{0.17em}}\right\}{p}^{3}+\left[\left({M}_{1}+{M}_{2}\right){\chi }_{ij}^{\Psi }\alpha /\text{th}\left(l\gamma \right)+1/{\left({c}^{\Psi }\right)}^{2}\right]{p}^{2}+2\alpha p/{c}^{\Psi }+{\alpha }^{2},\\ \end{array}$
${W}_{21}\left(p\right)={\Xi }_{2}\left(p\right)/{\Psi }_{m}\left(p\right)={{d}_{mi}\text{\hspace{0.17em}}\left[{M}_{1}{\chi }_{ij}^{\Psi }{p}^{2}+\gamma \text{th}\left(l\gamma /2\right)\right]/A}_{ij}{}_{}^{}$ ,
${W}_{12}\left(p\right)={\Xi }_{1}\left(p\right)/{F}_{1}\left(p\right)=-{{\chi }_{ij}^{\Psi }{}_{}^{}\left[{M}_{2}{\chi }_{ij}^{\Psi }{p}^{2}+\gamma /\text{th}\left(l\gamma \right)\right]/A}_{ij}$ ,
${W}_{13}\left(p\right)={\Xi }_{1}\left(p\right)/{F}_{2}\left(p\right)={W}_{22}\left(p\right)={\Xi }_{2}\left(p\right)/{F}_{1}\left(p\right)={\left[{\chi }_{ij}^{\Psi }\gamma /\text{sh}\left(l\gamma \right)\right]/A}_{ij},$
${W}_{23}\left(p\right)={\Xi }_{2}\left(p\right)/{F}_{2}\left(p\right)=-{{\chi }_{ij}^{\Psi }\left[{M}_{1}{\chi }_{ij}^{\Psi }{p}^{2}+\gamma /\text{th}\left(l\gamma \right)\right]/A}_{ij}$ .
From the set (15) of the equations we obtain the generalized matrix equation for the electroelastic actuator
$\left(\begin{array}{c}{\Xi }_{1}\left(p\right)\\ {\Xi }_{2}\left(p\right)\end{array}\right)=\left(\begin{array}{c}\begin{array}{ccc}{W}_{11}\left(p\right)& {W}_{12}\left(p\right)& {W}_{13}\left(p\right)\end{array}\\ \begin{array}{ccc}{W}_{21}\left(p\right)& {W}_{22}\left(p\right)& {W}_{23}\left(p\right)\end{array}\end{array}\right)\text{\hspace{0.17em}}\left(\begin{array}{c}{\Psi }_{m}\left(p\right)\\ {F}_{1}\left(p\right)\\ {F}_{2}\left(p\right)\end{array}\right)$ . (17)
Let us find the displacement of the faces for the electroelastic actuator in the stationary regime for the inertial load at ${\Psi }_{m}\left(t\right)={\Psi }_{m0}×1\left(t\right)$ , ${F}_{1}\left(t\right)={F}_{2}\left(t\right)=0$ .
Then we get the static displacement of the faces for the electroelastic actuator
$\begin{array}{l}{\xi }_{1}\left(\infty \right)=\underset{t\to \infty }{\text{lim}}{\xi }_{1}\left(t\right)=\underset{\begin{array}{c}p\to 0\\ \alpha \to 0\end{array}}{\text{lim}}p{W}_{11}\left(p\right){\Psi }_{m0}/p={d}_{mi}l{\Psi }_{m0}\left({M}_{2}+m/2\right)/\left({M}_{1}+{M}_{2}+m\right),\\ \end{array}$ (18)
${\xi }_{2}\left(\infty \right)=\underset{t\to \infty }{\text{lim}}{\xi }_{2}\left(t\right)=\underset{\begin{array}{c}p\to 0\\ \alpha \to 0\end{array}}{\text{lim}}p{W}_{21}\left(p\right){\Psi }_{m0}/p={d}_{mi}l{\Psi }_{m0}\left({M}_{1}+m/2\right)/\left({M}_{1}+{M}_{2}+m\right),$ (19)
${\xi }_{1}\left(\infty \right)+{\xi }_{2}\left(\infty \right)=\underset{t\to \infty }{\text{lim}}\left({\xi }_{1}\left(t\right)+{\xi }_{2}\left(t\right)\right)={d}_{mi}l{\Psi }_{m0}$ , (20)
where $m$ is the mass of the electroelastic actuator, ${M}_{1},\text{\hspace{0.17em}}{M}_{2}$ are the load masses.
Let us consider the static characteristics of the piezoactuator from the piezoceramics PZT under the longitudinal piezoelectric effect at $m<<{M}_{1}$ and$m<<{M}_{2}$ . For ${d}_{33}=4\cdot {10}^{-10}$ m/V, $U=50$ V, ${M}_{1}=2$ kg and ${M}_{2}=8$ kg we obtain the static displacement of the faces of the piezoactuator${\xi }_{1}\left(\infty \right)=16$ nm,${\xi }_{2}\left(\infty \right)=4$ nm, ${\xi }_{1}\left(\infty \right)+{\xi }_{2}\left(\infty \right)=20$ nm. The displacements in the stationary regime of the faces for the piezoactuator under the transverse piezoelectric effect and the inertial load at $U\left(t\right)={U}_{0}\cdot 1\left(t\right)$ ,${E}_{3}\left(t\right)={E}_{30}\cdot 1\left(t\right)=\left({U}_{0}/\delta \right)\cdot 1\left(t\right)$ , ${F}_{1}\left(t\right)={F}_{2}\left(t\right)=0$ can be written in the following form
$\begin{array}{l}{\xi }_{1}\left(\infty \right)=\underset{t\to \infty }{\text{lim}}{\xi }_{1}\left(t\right)=\underset{\begin{array}{c}p\to 0\\ \alpha \to 0\end{array}}{\text{lim}}p{W}_{11}\left(p\right)\left({U}_{0}/\delta \right)/p=d{}_{31}\left(h/\delta \right){U}_{0}\left({M}_{2}+m/2\right)/\left({M}_{1}+{M}_{2}+m\right),\\ \end{array}$ (21)
$\begin{array}{l}{\xi }_{2}\left(\infty \right)=\underset{t\to \infty }{\text{lim}}{\xi }_{2}\left(t\right)=\underset{\begin{array}{c}p\to 0\\ \alpha \to 0\end{array}}{\text{lim}}p{W}_{21}\left(p\right)\left({U}_{0}/\delta \right)/p={d}_{31}\left(h/\delta \right){U}_{0}\left({M}_{1}+m/2\right)/\left({M}_{1}+{M}_{2}+m\right)\\ \end{array}$ (22)
${\xi }_{1}\left(\infty \right)+{\xi }_{2}\left(\infty \right)=\underset{t\to \infty }{\text{lim}}\left({\xi }_{1}\left(t\right)+{\xi }_{2}\left(t\right)\right)={d}_{31}\left(h/\delta \right){U}_{0}$ . (23)
From (21), (22) we obtain the static displacements of the faces of the piezoactuator under the transverse piezoeffect at $m<<{M}_{1}$ , $m<<{M}_{2}$ in the form
${\xi }_{1}\left(\infty \right)=\underset{t\to \infty }{\text{lim}}{\xi }_{1}\left(t\right)=\underset{\begin{array}{c}p\to 0\\ \alpha \to 0\end{array}}{\text{lim}}p{W}_{11}\left(p\right)\left({U}_{0}/\delta \right)/p={d}_{31}\left(h/\delta \right){U}_{0}{M}_{2}/\left({M}_{1}+{M}_{2}\right),$ (24)
$\begin{array}{l}{\xi }_{2}\left(\infty \right)=\underset{t\to \infty }{\text{lim}}{\xi }_{2}\left(t\right)=\underset{\begin{array}{c}p\to 0\\ \alpha \to 0\end{array}}{\text{lim}}p{W}_{21}\left(p\right)\left({U}_{0}/\delta \right)/p={d}_{31}\left(h/\delta \right){U}_{0}{M}_{1}/\left({M}_{1}+{M}_{2}\right).\\ \end{array}$ (25)
Let us consider the static characteristics of the piezoactuator from piezoceramics PZT under the transverse piezoelectric effect at$m<<{M}_{1}$ and $m<<{M}_{2}$ . For ${d}_{31}=2\cdot {10}^{-10}$ m/V, $h=4\cdot {10}^{-2}$ m, $\delta =2\cdot {10}^{-3}$ m, $U=50$ V, ${M}_{1}=2$ kg and ${M}_{2}=8$ ${M}_{2}=8$ kg we obtain the static displacement of the faces of the piezoelectric actuator ${\xi }_{1}\left(\infty \right)=160$ nm, ${\xi }_{2}\left(\infty \right)=40$ nm, ${\xi }_{1}\left(\infty \right)+{\text{ξ}}_{2}\left(\infty \right)=200$ nm.
From (16) we obtain the transfer functions of the piezoactuator with the fixed end and the elastic inertial load so that${M}_{1}\to \infty$ and $m<<{M}_{2}$ in the following form
${W}_{2}\left(p\right)=\frac{{\Xi }_{2}\left(p\right)}{U\left(p\right)}=\frac{{d}_{33}}{\left(1+{C}_{e}/{C}_{33}^{E}\right)\text{\hspace{0.17em}}{\left({T}_{t}^{2}{p}^{2}+2{T}_{t}{\xi }_{t}p+1\right)}_{}^{}}$ , (26)
where the time constant ${T}_{t}$ and the damping coefficient ${\xi }_{t}$ are determined by the formulas
${T}_{t}=\sqrt{{M}_{2}/\left(C{}_{e}+{C}_{33}^{E}\right)}$ , ${\xi }_{t}=\alpha {\delta }^{2}{C}_{33}^{E}/\left(3{c}^{E}\sqrt{M\left({C}_{e}+{C}_{33}^{E}\right)}\right)$ .
Let us consider the operation of the piezoactuator from piezoceramics PZT with one face rigidly fixed and the elastic inertial load so that ${M}_{1}\to \infty$ and $m<<{M}_{2}$ for ${M}_{2}=10$ kg,${C}_{33}=2.1\cdot {10}^{6}$ N/m, ${C}_{e}=0.4\cdot {10}^{6}$ N/m we obtain ${T}_{t}=2\cdot {10}^{-3}$ c. The experimental and calculated values for the piezoactuator are in agreement to an accuracy of 5%.
# Conclusion
The structural–parametric models, the decision of the wave equation, the parametric structural schematic diagram, the transfer functions of the electroelastic actuator are obtained using Laplace transform. The parametric structural schematic diagram and the transfer functions of the piezoactuator for the transverse, longitudinal, shift piezoelectric effects are determined from the structural–parametric model of the electroelastic actuator. The transfer functions in matrix form are describes deformations of the piezoactuator during its operation as part of the mechatronics system for the nanotechnology and the ecology research. From the decision of the electroelasticity equation, the wave equation and the features of the deformations along the coordinate axes we obtain the generalized structural–parametric model and the parametric structural schematic diagram of the electroelastic actuator for the mechatronics system and its dynamic and static properties.
None.
# Conflict of interest
The author declares there is no conflict of interest.
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©2018 Afonin. This is an open access article distributed under the terms of the, which permits unrestricted use, distribution, and build upon your work non-commercially. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 142, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9506349563598633, "perplexity": 3392.864647193645}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711360.27/warc/CC-MAIN-20221208183130-20221208213130-00660.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=2007_Indonesia_MO_Problems/Problem_6&diff=prev&oldid=119769 | # Difference between revisions of "2007 Indonesia MO Problems/Problem 6"
## Problem
Find all triples of real numbers which satisfy the simultaneous equations
## Solution
To start, since all three equations have a similar form, we can let to see if there are any solutions. Doing so results in Note that has complex solutions, so the solution where is .
Additionally, note that and are monotonically increasing functions, so is a monotonically increasing function. Thus, we can suspect that is the only solution. To prove this, we can use proof by contradiction.
Assume that . Because is a monotonically increasing function, , so and , so . By doing the same steps, we can show that and . However, that would mean that , which does not work, so there are no solutions where .
Similarly, assume that . Because is a monotonically increasing function, , so and , so . By doing the same steps, we can show that and . However, that would mean that , which does not work, so there are no solutions where .
Thus, we proved that is the only solution, and by substituting the value into the original equations, we get the only solution of . | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9925717711448669, "perplexity": 244.5951078324554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178360107.7/warc/CC-MAIN-20210228024418-20210228054418-00040.warc.gz"} |
https://www.arxiv-vanity.com/papers/1411.5517/ | # Observational studies of transiting extrasolar planets
John Southworth \affilAstrophysics Group, Keele University, Staffordshire, ST5 5BG, UK\email
###### Abstract
The study of transiting extrasolar planets is only 15 years old, but has matured into a rich area of research. I review the observational aspects of this work, concentrating on the discovery of transits, the characterisation of planets from photometry and spectroscopy, the Homogeneous Studies project, starspots, orbital obliquities, and the atmospheric properties of the known planets. I begin with historical context and conclude with a glance to a future of TESS, CHEOPS, Gaia and PLATO.
\aspSuppressVolSlug\resetcounters\paperauthor
John SKeele UniversityAstrophysics GroupNewcastle-under-LymeStaffordshireST5 5BGUK
## 1 History and context
The first widely accepted detection of an extrasolar planet orbiting a normal star was made by Mayor & Queloz (1995), using high-precision radial velocity (RV) measurements. They found an object with a minimum mass of M orbiting the solar-like star 51 Peg every 4.2 days. Earlier discoveries had been made, but were either treated with caution, had a significantly larger mass, or were orbiting pulsars (see Wright & Gaudi 2013 for an historical account). The second 51 Peg-type planetary system followed quickly afterwards (Marcy & Butler, 1996) and by the start of the year 2000 a total of 25 planets had been detected, all by the RV method. Whilst valuable discoveries, only their minimum mass, orbital period, eccentricity and semimajor axis could be measured; their radius and thus density were unattainable.
One of the early RV planets was HD 209458, and in late 1999 it was found to transit its host star (Henry et al., 2000; Charbonneau et al., 2000). Transiting extrasolar planets (TEPs) are intrinsically more useful because the depth of the transit depends on the planetary radius, ultimately allowing measurement of its density, surface gravity and true mass. The second known TEP was unveiled three years later and in a very different way, by RV follow-up of a star showing transits (Konacki et al., 2003).
Whilst the initial rate of discovery of exoplanets was slow, it has shown exponential growth and now exceeds 1800 objects of which over 1150 are transiting111Data from TEPCat (Southworth, 2011) at: http://www.astro.keele.ac.uk/jkt/tepcat/. Fig. 1 shows the discovery rate of the known TEPs and breaks this down into the contributions from different consortia. The roughly exponential discovery rate gives a constant slope in this logarithmic plot, with the exception of the 851 planets in 340 multiple systems which were statistically validated by Rowe et al. (2014) in early 2014. The greatest number of discoveries have come from the Kepler satellite (Borucki et al., 2010), whose large aperture and space-based location yielded data of extremely high precision, duty cycle, and time coverage. The second most productive consortium is SuperWASP (Pollacco et al., 2006), followed by HAT (Bakos et al., 2002); these groups rely on small ground-based robotic telescopes equipped with telephoto lenses.
Fig. 2 shows the sky positions of the known TEPs, again colour-coded according to discovery consortium. The stand-out feature is the agglomeration of Kepler discoveries (green points at RA 19–20 h and Dec = 40–50). The smaller brown groupings near the two intersections of the celestial equator and Galactic plane are due to the CoRoT satellite (Moutou et al., 2013), and the spread of blue in the Southern hemisphere come from the SuperWASP-South installation in South Africa.
Fig. 3 shows the masses and radii of the known TEPs (main part of the diagram) and their host stars (dense assembly of points at the top-right). The fractional scatter in the properties of the planets is much more than that in the properties of their well-behaved FGK dwarf hosts, an indicator of the complexity of the physical effects which affect giant planets. The huge scatter in the properties of low-mass planets is due partly to the difficulty in characterising these small and low-mass objects, and partly to their extreme and poorly understood diversity (e.g. Masuda, 2014).
## 2 Discovering and characterising transiting extrasolar planets
Early work on the identification of TEPs concentrated mostly on the ‘hot Jupiters’, which I consider to be gaseous planets of mass greater than 0.3 M and orbital period less than 10 d. These are the most easily identifiable planets because their relatively large radii lead to deep transits, and their masses and short orbital periods cause a comparatively large reflex velocity in the host stars. Eight TEPs were first identified using RV measurements and subsequently found to transit, including the two most-studied examples [HD 209458 and HD 189733 (Bouchy et al., 2005)].
Planet detection via RV measurements is inherently expensive, requiring large telescopes and ultra-stable spectrographs, which are capable of observing only one target at once. As only a small fraction of stars host TEPs, this approach is an inefficient method of detection. The great majority of TEPs have therefore been found from large-scale photometric surveys, such as OGLE (Udalski et al., 2002), Kepler, WASP and HAT, which have the advantage of monitoring thousands of stars simultaneously. A major disadvantage of finding TEPs from photometric surveys is that not all transit events are due to planets. False positives can be caused by low-mass stars (late-M dwarfs have radii close to that of Jupiter; recall Fig. 3), faint eclipsing binaries whose light contaminates that of the target star, and instrumental effects. Planet candidates therefore have to be studied in detail to confirm their planetary nature.
Kepler’s space location and relatively high spatial resolution result in it having a low rate of false positives (see Morton & Johnson 2011 but also Santerne et al. 2012 and Coughlin et al. 2014). For the CoRoT satellite, which has an inferior spatial resolution, % of candidates are false positives and only 6% are confirmed planets (Moutou et al., 2013), with the remainder being unsolved. The estimated false-positive rate for WASP-South is representative of a typical ground-based survey: roughly 1 in 14 candidates turns out to be of planetary mass (Hellier et al., 2011a).
### 2.1 Spectroscopic radial velocity measurements
Once a transit event has been found, the planetary nature of the transiting object needs to be proved by measuring its mass. This can be done by obtaining multiple RV measurements using one of the current generation of high-resolution spectrographs such as Keck/HIRES, CORALIE or HARPS (see Pepe et al., 2014, for a recent review). The extremely high RV quality of which these instruments are capable allows the orbital motion of the host star to be measured. With some knowledge of the mass of the star, its orbital velocity amplitude () indicates the mass of the transiting object222Subscripted letters ‘A’ and ‘b’ indicate properties of the host star and planet, respectively.. The RVs also yield the planet’s orbital eccentricity () and argument of periastron ().
A bonus feature of the high-resolution spectra is that they can be used to determine the atmospheric parameters of the host star: its effective temperature (), surface gravity () and metallicity ([M/H] or [Fe/H]). This process is typically achived by comparing the observed spectra to synthetic spectra either directly or via the measured equivalent widths of spectral lines (e.g. Torres et al., 2012). These quantities, especially , are vital for determining the mass of the star and thus the mass of the planet.
An alternative approach to RV measurements has been pursued for most of the Kepler planet candidates, necessitated by the faintness of most of these objects which makes high-resolution spectroscopy prohibitively expensive (often completely impossible) with current facilities. A large number of Kepler candidates have been ‘validated’ by demonstrating the low probability of them being a false positive, instead of proving their planetary nature with a mass determination. The Kepler candidates are well suited to this approach because they are relatively small (too small to be a low-mass star) and very unlikely to be a result of contamination by a third object. The contamination can be investigated by high-resolution imaging and checking for apparent shifts in the position of the star during transit, effectively shrinking the sky area where contaminating objects can plausibly be located to a very small – and therefore unlikely – solid angle.
### 2.2 Follow-up light curves
Once a transiting object has been identified and proven to be of planetary origin via RV measurements, the next step is to obtain a high-quality light curve. The shape of the transit is a crucial piece of information for deducing the physical properties of the system, but discovery light curves from ground-based surveys are typically very scattered (see Fig. 4). A method of obtaining high-precision photometry which is now widely used is that of telescope defocussing (e.g. Alonso et al., 2008; Southworth et al., 2009), whereby the point spread function (PSF) is broadened to cover hundreds or even thousands of pixels. There are two main advantages of this method. Firstly, flat-fielding noise is averaged down by the square-root of the number of pixels, i.e. several orders of magnitude. Secondly, longer exposure times are possible without saturating individual pixels, so less time is lost to reading out the CCD and more time is available to observe, thus decreasing the photon and scintillation noise.
As an example of telescope defocussing, Fig. 4 shows three light curves of the transit of WASP-2. The first panel shows the data used to detect the transit – this was obtained using the SuperWASP-North installation which consists of 200 mm telephoto lenses with a plate scale of 14 px. The second panel shows an example of a follow-up light curve from a 1.2 m telescope operated in focus (Charbonneau et al., 2007), reaching a very creditable scatter of 1.9 mmag per point. The third panel displays a light curve obtained with a defocussed 1.5 m telescope (Southworth et al., 2010), which achieves a scatter of only 0.46 mmag per point. Fig. 5 shows an example PSF and the resulting light curve of a transit of WASP-50 obtained with NTT/EFOSC2.
Once the shape of the transit has been observed, several important pieces of information can be extracted from it. Firstly, the depth of the transit is a strong indicator of the ratio of the radius of the planet to that of the star (a quantity called ), as the flux deficit indicates what fraction of the stellar surface is blocked by the dark planet. Secondly, the duration of the transit indicates how long it took the planet to pass in front of the star. This is closely related to the size of the star: the actual quantity measured is the fractional radius where is the true radius of the stars and is the orbital semimajor axis. This quantity is often inverted and labelled . Thirdly, the duration of the partial phases of the transit (when only part of the planet is in front of the star) is a gauge for which part of the stellar disc the planet transits, i.e. the orbital inclination of the system (). The orbital inclination is related to the impact parameter () by:
b=1−e21±esinωcosirA
where the is ‘’ for the transit and ‘’ for the occultation (secondary eclipse).
An important attribute of the fractional radius of the star is that it is very closely related to the stellar density, (Seager & Mallén-Ornelas, 2003). From Kepler’s third law and the definition of density we can derive the relation:
rA=R3Aa3=3πGP21ρA(MAMA+Mb)
where is the Newtonian gravitational constant, is the orbital period, and and are the masses of the star and planet. As , the quantity in brackets can be ignored. An alternative formulation well suited to light curve analysis is:
ρA+(RARb)3ρb=3πGP2(aRA)3⇒ρA+k3ρb=3πGP2r3A
where is usually small, so is negligible,and the term can be ignored.
The photometric parameters (, and ) can be obtained by fitting transit light curves with a simple geometric model such as the jktebop program (Southworth, 2008, 2013) or the occultsmall subroutine (Mandel & Agol, 2002). The orbital ephemeris (period and reference time of mid-transit ) is easily obtained in the same way. There do, however, exist several complications.
Limb darkening is one nuisance parameter which must be included in the model, and theoretically-derived coefficients are available for several approximation ‘laws’ (e.g. Claret & Bloemen, 2011). The use of theoretical coefficients is generally fine for data of ground-based but not of space-based quality (Southworth, 2008).
Orbital eccentricity affects the transit durations, because the orbital speed of the planet is no longer constant. It is essentially impossible to fit for this effect using only transit light curves (Kipping, 2008). One must use the information provided by the RVs of the host star, either directly or by applying constraints to the light curve fit.
Cadence. Some space-based light curves have a poor time sampling; most egregiously the Kepler long-cadence data with effective interagration times of 1765 s (Jenkins et al., 2010). In these cases one must integrate the model to match the nature of the data or suffer potentially large errors in the results. Southworth (2011) showed that ignoring this problem gave photometric parameters wrong by 30% for a typical case.
Contaminating light. Faint stars close to TEP systems may contaminate light curves, causing the transit to be diluted and the planetary radius to be underestimated (Daemgen et al., 2009). This cannot be fitted for directly in the transit light curve (Southworth, 2010) as it is completely correlated with other parameters. But if faint stars can be detected using high-resolution imaging (e.g. Southworth et al., 2010; Lillo-Box et al., 2014), their light can be accounted for in the model fit.
I finish this section by discussing the terminology for the different types of eclipses seen in TEP systems. The correct terminology (see Fig. 6) has been established for many years for for eclipsing binary systems (e.g. Hilditch, 2001), and of course for solar and lunar eclipses. A ‘transit’ is when a smaller object (e.g. planet) passes completely in front of a larger object (e.g. star). An ‘occultation’ is when the planet passes behind the star. ‘Partial eclipses’ can occur when part of one object never eclipses or is eclipsed by the other object. Eclipsing systems can have only one transit per orbit, so references to a ‘primary transit’, ‘anti-transit’ or ‘secondary transit’ are incorrect.
### 2.3 Determining the physical properties of transiting planets
Fitting the RVs of the host star gives the parameters of the spectroscopic orbit: , and . The combination terms and are often used instead of and themselves because they are less strongly correlated and not biased to higher values of . Fitting the transit light curve gives the photometric parameters , and . We also have extra information from the spectra of the host star: its , and [M/H].
This situation is essentially that of an eclipsing binary system where only one star is seen in the spectra. The lack of RVs for the secondary component means cannot be measured, so we are one piece of information short of being able to determine the physical properties of the system. Thankfully, an additional constraint can been obtained using the spectroscopic properties of the host star and either empirical calibrations of stellar properties or theoretical stellar evolutionary models. An elegant way to do this is to guess a value of and use the other quantities (, , , , and ) to determine the mass and radius of the star and planet using standard formulae (e.g. Hilditch, 2001). , and can then be checked for consistency with the additional constraint, and iteratively adjusted to maximise this consistency.
Most studies of TEPs have used theoretical stellar models to provide the additional constraint, in which case the advantage of conceptual simplicity is offset by the fact that it is not trivial to interpolate to arbitrary values within a tabulated grid of theoretical predictions. The reliance on stellar theory is worrying, as it is difficult to assess the effect of this on the results. One option is to try multiple sets of models and see how well they agree: Southworth (2010) found a minimum scatter of 1% for , 0.6% for and less for other quantities. However, this only provides a lower limit on the true uncertainties because different sets of theoretical models have many areas of commonality such as computational approach, opacities and parameterisation of mixing.
An alternative to stellar theory is to construct (semi-)empirical calibrations of stellar properties based on the values measured for detached eclipsing binary systems (dEBs). This approach has its own advantange and disadvantage: the continuous nature of the calibrations means interpolation is not required, but it is not clear if the properties of low-mass stars are well-represented by dEBs (Torres, 2013). Calibrations were first used by Southworth (2009), based on a simple mass-radius relation for late-type dwarfs. The problem with this approach is that the neglect of stellar evolution meant the results were not very reliable. A better approach was proposed by Torres et al. (2010), who calculated calibrations for stellar mass and radius as a function of , and [Fe/H]. Enoch et al. (2010) further improved this approach by using instead of , motivated by the fact that is directly obtained from transit light curves whereas can be inferred to only a lower precision by spectral analysis. Finally, Southworth (2011) followed the approach of Enoch et al. (2010) but based it on many more objects (180 versus 38 stars, sourced from the DEBCat333DEBCat (Southworth, 2014b) can be found at: http://www.astro.keele.ac.uk/jkt/debcat/ catalogue of measured physical properties of well-studied dEBs).
Several quantities can be measured without requiring the additional constraint. The stellar density, , was already discussed in Sect. 2.2. The surface gravity of the planet can be obtained using only measured quantities (Southworth et al., 2007):
gb=2πP√1−e2KAr 2bsini
where is the fractional radius of the planet.
The planetary equilibrium temperature is also independent of the scale of the system:
Teq=Teff√RAa[f(1−AB)]1/4=Teff√rA[f(1−AB)]1/4
where is the Bond albedo and is the heat redistribution parameter (e.g. Sheets & Deming, 2014). A common approach is to assume in which case the equation becomes very simple: .
## 3 Homogeneous Studies of Transiting Extrasolar Planets (HSTEP)
Back in 2006-7 it became clear that the number of known TEPs was increasing quickly, and that studies of these objects were done in a variety of different ways, especially concerning the additional constraint. This variety of approaches led inexorably to inhomogeneous results, so the properties of different TEPs were not directly comparable. The obvious solution was an homogeneous analysis. For this, I select good published light curves and model them using the jktebop code. Careful attention is paid to the inclusion of limb darkening, numerical integration to account for long exposure times, correction for contaminating ‘third’ light, and in accounting for eccentric orbits. Four error analysis methods are implemented: Monte Carlo simulations, residual-permutation, multiple analyses of the same data using different choices of limb darkening, and separate analyses of different datasets for the same TEP (Southworth, 2008).
Once the photometric parameters have been obtained, I add published spectroscopic results (, , [Fe/H]) and calculate the physical properties of the systems. Statistical errors are prepagated from all input values by a perturbation analysis which yields a full error budget for each output value (Southworth et al., 2005; Southworth, 2009). This process is done using each of five sets of theoretical stellar models, allowing a systematic error to be assigned to each output parameter based on the variation between the five results. Further details can be found in the original papers, and a summary has been given by Southworth (2014a).
At this point, a total of 89 planetary systems have been studied in the course of HSTEP (Southworth, 2008, 2009, 2010, 2011, 2012), most based on published data but some on new light curves obtained for the project (see Southworth et al., 2014, and references therein). A paper in preparation will push this number up to 120 systems.
One feature of the HSTEP results is that the error estimates for the calculated parameters often are much larger than those for published works; in many cases the results agree according to the HSTEP errorbars but not according to the published errorbars. This implies that published errorbars can be rather too small: particular offenders are CoRoT-5, CoRoT-8, CoRoT-13, Kepler-5 and Kepler-7 from Paper IV (Southworth, 2011), and CoRoT-19, CoRoT-20, Kepler-15, Kepler-40 (KOI-428) and OGLE-TR-56 from Paper (Southworth, 2012). Three of these systems deserve special mention.
CoRoT-8. I found a planet radius of R (Southworth, 2011, sect. 6.8) versus R from Bordé et al. (2010). The orbital ephemeris in the discovery paper is incorrect, predicting the transits in the CoRoT data to occur 0.06 d too early.
CoRoT-13. The CoRoT satellite obtained two light curves for this object, which strongly disagree on the transit shape. I adopted the results from the better of the two, finding a planet radius of R (Southworth, 2011, sect. 6.13) versus R in the discovery paper (Cabrera et al., 2010). Whilst CoRoT-13 was thought to be an extremely dense planet with a massive core of heavy elements, my results are consistent with a typical gas giant slightly less dense than Jupiter.
OGLE-TR-56. This was the second known TEP (Konacki et al., 2003) and its faintness means large telescopes are required to obtain good transit light curves. Adams et al. (2011) obtained many excellent light curves, and determined the physical properties of the system based on these and on an assumed and . The problem was that the chosen and (from Torres et al., 2008) were inconsistent with the from the light curve. The HSTEP analysis changed the measured planetary radius from R to R (Southworth, 2012, sect. 5.21).
By far the most common method of obtaining errorbars on measured parameters of TEPs and their host stars is that of MCMC (Markov chain Monte Carlo), a very powerful technique for both model optimisation and calculation of the posterior probability density for parameter values. A common feature of the HSTEP reanalysis of published data is agreement with published results within the HSTEP errorbars but not with the often very small errorbars calculated using MCMC in these publications. This suggests that the error analysis methods using the HSTEP project are robust, but that those arising from MCMC analysis sometimes are not. Like any other statistical tool, MCMC has to be used carefully to ensure good results.
### 3.1 TEPCat: the catalogue of physical properties of transiting extrasolar planets
By Paper IV (Southworth, 2011) it was obvious that readers could not reasonably be expected to trawl through all four papers to compile the full results from the HSTEP project. I therefore created the TEPCat catalogue444TEPCat (Southworth, 2011) can be found at: http://www.astro.keele.ac.uk/jkt/tepcat/ to make these results available in convenient formats (Fig. 7). It was also a good site for placing a compilation of the physical properties of all known TEPs and their host stars, a database which I was already keeping for my own use. At this point TEPCat contains tables in html, ascii and csv formats with the best available values for the stellar properties (, [Fe/H], , , , ), planet characteristics (, , , , ), orbital parameters (, , , ) and references for all confirmed TEPs. A catalogue of orbital obliquities from the Rossiter-McLaughlin effect is also maintained, along with various goodies such as plots, links, explanation, and the set of physical constants used in the HSTEP project.
## 4 Rossiter-McLaughlin effect
Rossiter (1924) and McLaughlin (1924) contemporaneously discovered an RV anomaly during primary eclipse, in the eclipsing binaries Lyrae and Persei. This is caused by the eclipsing object blocking out part of the rotating surface of its companion, removing flux from part of its spectral line profiles and thus biasing measured RVs away from the Keplerian value. The effect is much smaller in TEPs (typically less than 50 m s versus 13 km s for Lyrae) but easier to study because the spectral line profiles come from only one object (the host star). The Rossiter-McLaughlin (RM) effect has now been observed in a total of 91 TEPs (e.g. Triaud et al., 2010; Albrecht et al., 2012), mostly by RV measurements. This approach can only give the sky-projected value () of the true orbital obliquity ().
Whilst early RM measurements (the first being Queloz et al., 2000) indicated aligned orbits, a significant number of misaligned and even retrograde planets are now known (the first being WASP-17; Anderson et al., 2010). Winn et al. (2010) found that misaligned orbits occur mostly for hotter host stars ( K), although Triaud (2011) asserted that this was caused by the younger age of such systems. Tidal dissipation is a critical part of interpreting measurements (see Albrecht et al., 2012).
## 5 Starspots
An alternative way to measure the RM effect is via transits with starspot anomalies. If a planet transits a dark spot on the stellar surface, it temporarily blocks slightly less of the overall starlight. The overall brightness of the system blips upwards, by an amount which depends on the size of the spot and its brightness relative to the rest of the stellar surface. Multi-band photometry of this effect allows the spot temperature to be obtained (e.g. Mancini et al., 2014) and the spot position to be measured precisely.
If several transits are observed over a short period of time, the change in position of a starspot could be tracked. This directly yields the motion of the spot as the star rotates, relative to the planet’s orbit, allowing to be measured (Nutzman et al., 2011; Sanchis-Ojeda et al., 2011) as well as the rotation period of the star (Silva-Valio, 2008). Tregloan-Reed et al. (2013) constructed a physically realistic model of this situation (prism) and used it to measure from two transits of WASP-19 (Fig. 9), a much more precise value than the RM alternative of (Hellier et al., 2011b). Having three or more observations of the same starspot at different positions would allow as well as to be obtained.
## 6 Occultations
Although planets are much fainter than their host stars, it is possible to detect the dips in brightness as they are eclipsed by their star. These miniscule occultations can only be measured using very high-precision photometry, but are nevertheless a valuable source of two types of information.
Firstly, the time of mid-occultation constrains and for eccentric systems. Specifically, the difference in orbital phase between the occultation and the halfway point between the two adjacent transits gives the combination term independently of RV measurements:
Δϕ=(toccult−ttransitP)−0.5=(1+csc2iπ)ecosω
Here, means orbital phase, and and indicate the eclipse midpoints
Secondly, the depth of the occultation gives the brightness of the planet (at a given wavelength or passband) relative to that of the star. This means that the spectrum of the planet can be constructed from occultation observations at a range of wavelengths. The spectrum is that of the irradiated ‘dayside’ of the planet. As a rough approximation, reflected light from the star dominates at optical wavelengths (e.g. Angerhausen et al., 2014) and thermal emission dominates in the infrared (e.g. Charbonneau et al., 2008). Planetary spectra can be used to investigate the chemical composition and structure of their atmospheres (e.g. Ranjan et al., 2014; Madhusudhan & Seager, 2010).
## 7 Transmission spectroscopy
Whilst occultations can be used to measure the flux emitted by the planet, observations of the depth of the transit as a function of wavelength allow the opacity spectrum of a planetary atmosphere to be measured. Occultations probe the dayside of the planet whereas transmission spectroscopy is sensitive to the properties of the atmosphere at the terminator (the limb of the planet). This effect is difficult to observe, but is helped by the very extended atmospheres of some planets. The best example is WASP-17 (Anderson et al., 2010), which is the largest known planet at R (Southworth et al., 2012). Its low surface gravity of m s yields a huge atmospheric scale height of 2000 km (0.028 R). The largest features in the optical and near-infrared spectrum of a hot Jupiter can be 5–10 atmospheric scale heights (Sing et al., 2011, 2014), which are detectable using ground-based large telescopes.
Theoretical spectra of irradiated giant planets show characteristic features at optical wavelengths due to sodium and potassium (Fortney et al., 2008), possibly sulphur compounds (Zahnle et al., 2009), and Rayleigh and Mie scattering in the blue. Infrared wavelengths are predicted to show features due to molecules such as HO, CO, CO and CH, depending on the atmospheric temperature. This is an active area of research which has generated a wide variety of results: some planets show flat transmission spectra indicative of high-altitude clouds (Kreidberg et al., 2014; Knutson et al., 2014), some show signatures of molecules (Tinetti et al., 2010; Wakeford et al., 2013), some show Rayleigh or Mie scattering (Pont et al., 2013; Sing et al., 2013), and at least one planet shows all of these features (Sing et al., 2014).
## 8 Future
We have passed through the initial stages of development of the study of transiting planets and are now in the early characterisation phase. The easy-to-find TEPs are being identified in bulk by ground-based surveys (e.g. Bakos et al., 2012; Hellier et al., 2012) and our boundaries of ignorance are being gradually pushed back by discoveries in new areas of parameter space (e.g. Doyle et al., 2011; Sanchis-Ojeda et al., 2013; Ciceri et al., 2014). Exhaustive examinations of a small subset of TEPs have established them as tracers of the formation, structure and evolution of giant planets. The best-studied TEPs have mass and radius measurements to a few percent precision, projected or true orbital obliquity measurements, and atmospheric abundances of some atoms and molecules through transit and occultation spectroscopy. Whilst Kepler has truly revolutionised the study of TEPs, ground-based surveys remain relevant as they observe many more targets so can find rarer types of planet.
In the near future, the Gaia satellite will fill an important hole in our understanding of TEPs and their host stars. The Gaia parallax measurements will give direct distance and thus luminosity estimates. As , these parallaxes can replace the additional constraint which troubles existing mass and radius measurements of TEPs (see Sect. 3). The high photometric precision of Gaia will also enable it to be used to discover TEPs (see Dzigan & Zucker, 2012), although it is likely that huge observational resources would be needed to follow up the identified planet candidates.
Although the main missions of Kepler and CoRoT have been terminated by technical problems, their archives remain rich in untapped results. Kepler has been reincarnated as the K2 mission, with lower photometric precision but still much better than achievable from the ground. CHEOPS (Broeg et al., 2013) is slated for launch in 2017, for a 3.5 yr mission to detect transits in low-mass planets discovered by the RV method.
The next landmark mission is TESS (Ricker et al., 2014), also due for launch in 2017. TESS will photometrically observe 26 fields covering most of the sky, concentrating on bright stars but for much shorter time intervals (27 days near the ecliptic ranging to one year around the celestial poles). Further ahead, the PLATO mission (Rauer et al., 2014) is planned for launch in 2024 as a precision photometry survey instrument. PLATO will have a much larger field of view than Kepler: it will observe brighter stars which makes follow-up observations much easier. It will also observe patches of sky for several years, thereby avoiding the low sensitivity to long-period planets suffered by TESS. Our knowledge of transiting planets is set to improve immensely. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8682780265808105, "perplexity": 1804.3322244900382}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039550330.88/warc/CC-MAIN-20210421191857-20210421221857-00245.warc.gz"} |
https://physics.stackexchange.com/questions/192742/what-does-it-mean-by-saying-the-generators-of-translations-transform-as-vectors | # What does it mean by saying the generators of translations transform as vectors under the Lorentz Group?
The commutator of generators of Lorentz transformation and translation is as follow: $$[M^{\mu\nu},P^\sigma]=i(P^\mu\eta^{\nu\sigma}-P^\nu\eta^{\mu\sigma} ).$$ Then from this we usually say that the generators of translations transform as vectors under the Lorentz group.
I don't quit understand this statement, could anyone explain it?
Notation: I will write a Poincaré transformation as ${x'}^\mu = {\Lambda^\mu}_\nu x^\nu + a^\mu$, the operator representing this transformation on the Hilbert space is $U(\Lambda, a)$. An infinitesimal transformation with ${\Lambda^\mu}_\nu = \delta^\mu_\nu + {\omega^\mu}_\nu$ and $a^\mu = \epsilon^\mu$ can be expanded as $$U(\delta + \omega, \epsilon) = 1 + \frac{i}{2} \omega_{\rho\sigma} M^{\rho\sigma} - i \epsilon_\rho P^\rho + \cdots \tag{1}$$
Definition: Operators $V^\mu$ transform as a vector under Poincaré transformations iff $$U(\Lambda, a) V^\rho U^{-1}(\Lambda, a) = {\Lambda_\mu}^\rho V^\mu \tag{2}$$ which I would say makes sense, intuitively.
Theorem: The condition (2) is equivalent to
• $[M^{\mu\nu}, V^\sigma] = i \left(V^\mu \eta^{\nu\sigma} - V^\nu \eta^{\mu\sigma} \right)$ (what you wrote)
• and $[P^\mu, V^\sigma] = 0$.
That should answer your question. One direction of the equivalence is easy to show by inserting the expansion (1) into (2) and equation coefficients of $\omega$ and $\epsilon$.
(For the other direction, one could probably take the derivative of (2) with respect to $\omega$ and $\epsilon$ and show that both sides satisfy the same differential equation, but I don't think I've ever seen anyone actually do this proof.)
Notes:
• I've taken this more or less from Weinberg QFT 1, Chapters 2.3 - 2.4.
• The situation is pretty similar in quantum mechanics, just with SO(3) instead of the Poincaré group. Link
• For equation (2), isn't that a vector should transform as $\Lambda^\rho_{\ \mu}V^{\mu}$, instead of $\Lambda^{\ \rho}_{\mu}V^{\mu}$? I mean the positions of the indexs on $\Lambda$ seem not right for me, could you explain again? Thank you! – Nahc Jul 6 '15 at 15:34
• You could equivalently write $U^{-1} V^\rho U = {\Lambda^\rho}_\mu V^\mu$. I agree that this would be a bit nicer (I was following Weinberg). – Noiralef Jul 6 '15 at 17:04
A representation of the Lorentz group implies a set of matrices $(S_{\mu\nu})_a{}^b$ (one for each $\mu,\nu$) that satisfy the Lorentz algebra, i.e. satisfies a relation of the form (I am not keeping track of signs) $$[S_{\mu\nu} , S_{\rho\sigma} ] = i ( \eta_{\mu\rho} S_{\nu\sigma} + \cdots )$$ What this means is that there is a vector space, with vectors $\psi^a$ so that $$[ \psi_a , M_{\mu\nu} ] = (S_{\mu\nu})_a{}^b \psi_b$$ A particular representation of the Lorentz algebra is the vector representation, for which $a=\mu$ and $$(S_{\mu\nu})_\rho{}^\sigma = i (\eta_{\mu\rho} \delta_\nu^\sigma - \delta^\sigma_\mu \eta_{\nu\rho} )$$ The vectors in this representation are therefore of the form $A_\mu$. Then $$[ A_\rho , M_{\mu\nu} ] = (S_{\mu\nu})_\rho{}^\sigma A_\sigma = - i (\eta_{\mu\rho} \delta_\nu^\sigma - \delta^\sigma_\mu \eta_{\nu\rho} ) A_\sigma = - i ( \eta_{\mu\rho} A_\nu - \eta_{\nu\rho} A_\mu )$$ This is precisely the algebra that $M_{\mu\nu}$ satisfies with $P_\mu$. This therefore indicates that $P_\mu$ transforms in the vector representation of the Lorentz algebra (as it should) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.981070876121521, "perplexity": 180.72775895768194}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371886991.92/warc/CC-MAIN-20200410043735-20200410074235-00385.warc.gz"} |
http://mathhelpforum.com/trigonometry/166428-sinusoidal-function-print.html | # sinusoidal function
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• Dec 16th 2010, 04:00 PM
terminator
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sinusoidal function
the amplitude is (max -min)/2 = 13-1/2=6
is b=2pi/period , the period seems to be about 6, so b=pi/3
c = horizontal shift(I doesn't seem to be a horizontal shift). How do I calculate it?
d= (max+min)/2 = 13 + 1/2=14/2=7
• Dec 16th 2010, 05:04 PM
Sudharaka
Quote:
Originally Posted by terminator
the amplitude is (max -min)/2 = 13-1/2=6
is b=2pi/period , the period seems to be about 6, so b=pi/3
c = horizontal shift(I doesn't seem to be a horizontal shift). How do I calculate it?
d= (max+min)/2 = 13 + 1/2=14/2=7
Dear terminator,
According to the details you have found,
$y=6\sin\{\frac{\pi}{3}\left(x+c\right)\}+7$
When $x=0\Rightarrow y=7$
$7=6\sin\{\frac{\pi}{3}\left(c\right)\}+7\Rightarro w \dfrac{\pi c}{3}=n\pi~where~n\in Z$
$c=3n~where~n\in Z$
Hope this will help you.
• Dec 16th 2010, 09:00 PM
SammyS
Quote:
Originally Posted by Sudharaka
Dear terminator,
According to the details you have found,
$y=6\sin\{\frac{\pi}{3}\left(x+c\right)\}+7$
When $x=0\Rightarrow y=7$
$7=6\sin\{\frac{\pi}{3}\left(c\right)\}+7\Rightarro w \dfrac{\pi c}{3}=n\pi~where~n\in Z$
$c=3n~where~n\in Z$
Hope this will help you.
It's also true that $\displaystyle y'>0$ at $\displaystyle x=0$
so, $\displaystyle {{\pi c}\over{3}}\approx 2\pi n$.
Thus: $\displaystyle c\approx 6 n$.
Since the simplest case of $\displaystyle c\approx 6 n$ is when $\displaystyle n=0$, I would choose $\displaystyle c\approx 0\ .$
Also, it looks like the period is closer to 6.2 than it is to 6, (BTW: That's very close to $\displaystyle 2\pi\ .$) so I would say that $\displaystyle b\approx {{\pi}\over{3.1}}\ ,$ maybe even $\displaystyle b\approx 1\ .$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9573000073432922, "perplexity": 3567.4536750020925}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560279368.44/warc/CC-MAIN-20170116095119-00542-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2661366/showing-closure-of-open-interval-is-the-closed-interval | # Showing Closure of Open Interval is the Closed Interval
Show that, in $\mathbb{R}$ with the usual metric, $\overline{(a, b)}$ = [$a, b$] for all $a < b$.
In order to show the equality, you need to show that LHS $\subseteq$ RHS and RHS $\subseteq$ LHS, right? This is what I was thinking:
To show $\overline{(a, b)}$ $\subseteq$ [$a, b$]: we know [$a, b$] is closed. And, since the closure of a set is the smallest closed set containing the set, we must have $\overline{(a, b)}$ $\subseteq$ [$a, b$].
To show [$a, b$] $\subseteq \overline{(a, b)}$: this is where I got stuck. Since $\overline{(a, b)}$ = $(a, b) \bigcup$ {accumulation points of $(a, b)$}, we need to show that [$a, b$] consists of only those elements, and everything outside of it is not an accumulation point, right? How would I go about doing that? Should I assume $x > b$ and $x < a$ is an accumulation point and show $x$ cannot be an accumulation point through contradiction? How should I go about doing that? I considered $x$ infinitesimally bigger than $b$, but don't know how to there is a neighborhood that doesn't contain an element from [$a, b$]. Or is there a much simpler way to prove this statement? Thank you.
While it is true that every $x>b$ and every $x<a$ is not an accumulation point of $(a,b)$, it is not necessary to prove this claim in order to prove that $[a,b]\subset\overline{(a,b)}$. The fact that every $x>b$ and every $x<a$ is not an accumulation point of $(a,b)$ follows from the direction you've already proved.
Instead, to show that $[a,b]\subset\overline{(a,b)}$, it is enough to prove that $a$ and $b$ are accumulation points of $(a,b)$.
• Thank you. How should I proceed to show $a$ and $b$ are accumulation points of $(a, b)$? Is it through contradiction? I'm a bit stuck on this part. – Max Feb 22 '18 at 5:57
• Well, if $U$ is an open set containing $b$, then there is some $\varepsilon>0$ such that $(b-\varepsilon,b+\varepsilon)\subset U$, and so the intersection of $U$ and $(a,b)$ is non-empty. The same argument works for $a$. – carmichael561 Feb 22 '18 at 6:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9618364572525024, "perplexity": 72.38274462458793}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00516.warc.gz"} |
https://www.physicsforums.com/threads/chiral-gauge-theory-and-c-symmetry.766593/ | # Chiral gauge theory and C-symmetry
1. Aug 19, 2014
### mkgsec
Hi, I have a question in Srednicki's QFT textbook.
In p.460 section 75(about Chiral gauge theory), it says
"In spinor electrodynamics, the fact that the vector potential is odd under charge conjugation implies that the sum of these diagrams(exact 3photon vertex at one-loop) must vanish."
That's good, because it's just the Furry's theorem for odd number of external photons. But I find the subsequent statement confusing.
"For the present case of a single Weyl field(coupled to U(1) field), there is no charge conjugation symmetry, and so we must evaluate these diagrams."
But shouldn't the transformation rule of $A^\mu (x)$ under C,P or T be universal regardless of specific theory?
2. Aug 19, 2014
### The_Duck
In spinor electrodynamics, the Lagrangian is *invariant* under the C operation, which among other things takes $A \to -A$. One consequence of this is that there should be an equality between the following three-point functions:
$\langle A(x) A(y) A(z) \rangle = \langle(-A(x))(-A(y))(-A(z))\rangle = -\langle A(x) A(y) A(z) \rangle$
The only way this equality can be satisfied is if $\langle A(x) A(y) A(z) \rangle$ vanishes. This is Furry's theorem. This result relies crucially on the fact that the C operation, under which $A \to -A$ and also $\psi$ transforms nontrivially, is a *symmetry of the Lagrangian*. That's why we have the equality above.
In the chiral case, you can still define a transformation under which $A \to -A$ but it is no longer a symmetry of the Lagrangian, no matter what transformation rule you choose for $\psi$. Therefore you cannot make the same argument and Furry's theorem does not go through.
3. Aug 19, 2014
### mkgsec
But doesn't Furry's theorem still hold even if the langrangian has no C-symmetry? I thought it is valid as long as the vacuum is invariant under C. What am I missing here?
4. Aug 19, 2014
### The_Duck
Why do you think the vacuum is invariant under C?
5. Aug 19, 2014
### mkgsec
Well... Maybe because the vacuum contains no particle? Are the invariance of the vacuum and invariance of lagrangian related to each other?
6. Aug 19, 2014
### The_Duck
Right. If your theory has a symmetry and you have a unique vacuum state, then the vacuum is invariant under that symmetry. If you have some operation that is not a symmetry of your theory, there's no reason to expect the vacuum to be invariant under that symmetry.
7. Aug 19, 2014
### mkgsec
It's very clear now! I always assumed that the vacuum is invariant under any operation. Thank you, it was very helpful :) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.971065104007721, "perplexity": 699.5571738513975}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257822598.11/warc/CC-MAIN-20160723071022-00054-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/23173/what-makes-the-stars-that-are-farther-from-the-nucleus-of-the-galaxy-go-faster-t | # What makes the stars that are farther from the nucleus of the galaxy go faster than those in the middle?
It has no sense that stars that have a bigger radius and apparently less angular speed($\omega$) goes faster than the ones near the center.
-
It's caused by the Jones force. This is the most common but most misunderstood force in the universe. You're subject to it, I'm subject to it, little crystal-armored arachnoids on a planet orbiting a neutron star in the vicinity of Ganymede are subject to it. It's the Jones force - which makes everyone try to keep up with everyone else, even the ones that are trying to keep up with THEM, resulting in recursive up-keeping and the downfall of market-based economies. Coincidentally, it's now thought that this may be evidence for quantum loop gravity. AND it's no crazier than "dark energy"... – Bob Jarvis Aug 28 at 3:15
The dark matter or the dark energy.
We don't know at all what are they. We just know that they exist, but as I say, we don't have idea.
-
well dark energy isn't required to explain the flat rotation curves. And while we don't what DM is, we know what DM isn't, cf MACHOs – Nic Apr 2 '12 at 14:58
DM isnt that mysterious. Also, this doesn't really answer the question--you need to include the explanation as well. – Manishearth Apr 2 '12 at 15:02
The wikipedia page on dark matter gives some details en.wikipedia.org/wiki/Dark_matter – Frédéric Grosshans Apr 2 '12 at 15:12
The question is a bit ambiguous. If the question is
why do star velocity increase with distance close to the galactic centre ?
because their orbit encompass more mass, and this corresponds to a stronger gravity pull.
If the question is
why does their velocity stays constant and does not decrease at big radii, where the star density decreases ?
We have to add dark matter to observed stars to explain that.
The curve B below (taken from wikipedia) plots the the observed star velocity as function of the distance from the galactic centre.
The curve A corresponds to the expected curve without dark matter. As you can see, beyond a given distance, the velocity is expected to decrease, but it actually stays roughly constant. Dark matter has been initially postulated as a solution to this discrepancy.
The increase of velocity close to the centre is independent to the presence of dark-matter or not. The velocity of a star on a circular orbit of radius $r$ in the galactic plane is given by a balance of the centrifugal acceleration and the gravity it feels: $$\begin{gather} \frac{v^2}r= G \frac{m(r)}{r^2}\\ v=\sqrt{ G\frac{m(r)}{r}} \end{gather}$$ where $m(r)$ is the mass of stars contained in a spheroid centred on the galactic centre of radius $r$ (see e.g. here for more details. And then adapt it to the geometry of a galaxy). If $r$ is smaller than the galaxy thickness, the number of star is proportional to the volume of the sphere, and we expect \begin{align} m(r)&\propto r^3& v&\propto r \end{align} which is consistent with the initial increase.
When $r$ is bigger than the thickness, if the star density is constant, we have then \begin{align} m(r)&\propto r^3& v&\propto \sqrt r, \end{align} and this still corresponds to a velocity increase.
When $r$ is big enough, the density of star decreases with $r$ upto a point where $m(r)< C r$ and this should give the decreasing curve A. On the other hand, the observed curve B is essentially constant, and this can bee seen as a measurement of $m(r)\propto r$. This is not consistent with the observed star repartition, but it is consistent with the presence of dark (i.e. not seen) matter with a radial density $\propto\frac1{r^2}$.
-
+1 for using mathcal :D. And for the interesting, detailed, answer. – Manishearth Apr 2 '12 at 16:30
@Manishearth : Apparently, \mathcal was not standard (even if I learned it this way ;-) ), so Garmen1778 removed it. I hope the +1 stays anyway ! – Frédéric Grosshans Jun 4 '12 at 15:36
since no one mentioned it, i think its only fair to provide at least one answer about MOND (Modified Newtonian Dynamics).
Basically the galaxy rotation curve is the reason dark matter was proposed in the first instance. However, the dark matter explanation, putting aside for a moment other considerations as supersymmetric weakly coupled partners, is not very satisfying from a scientific point of view, since its an instance of adjusting parameters (i.e: unseen matter) in order to preserve a model. Think of Ptolomeus model of the solar system, with the epicyclic orbits postulated in order to preserve the earth in the center of the model.
MOND as an alternative explanation to dark matter is widely discredited, specially after the observation of gravitational lensing in the middle regions of the Bullet cluster, which suggests to some that a transparent source of gravity is causing the lensing effect.
However, despite this, the MOND hypothesis, at least as an heuristic to retrodict the galaxy rotation curves works extremely well. The hypothesis basically says that the gravitational mass coupling becomes weaker when accelerations drop below $a_0 \approx 10^{-10} m/s^2$. From that simple assumption, it is able to predict most of the galaxy rotation curves that are currently observed
-
More on MOND theory: physics.stackexchange.com/q/5762/2451 – Qmechanic Apr 3 '12 at 22:08
The striking similarity of a galaxy and an hurricane .
Do you need DM or SMBH in the hurricane ? Their force field is similar. The hurricane is governed by a vortex, then how can a galaxy be subjected to such a field that is more important than the central force at the centre of the galaxy ? I'm sure that there is a way.
I will try to explain why there is a vortex :
All images were taken from outra Física CC.
In this PSE-link (anti gravity) I show with graphs and equations how the voids have to grow due to the gravitational field. All the mass that was previously inside them is now in the exterior shell of the voids.
When two voids intersect a we have :
and the field is like
To see the field of a vortex you need to remove the mean field
As you see there is no need of DM neither SMBH. The only think that is needed is to follow the rules of physics (and a lot of imagination of my friend Alfredo that allow s me to see only physics as the true nature of Nature, when everybody else can only explain with a magical Dark Matter).
There are recent news (arxiv and cosmiclog) that DM probably has to have a two different composition/properties and the astronomers dont now how to justify the observed behaviour.
Of course one can continue the pursue of DM quest. I dont.
The resistence that the readers have to this kind of model, that only follows the laws of gravity, has to have with the fact that this asks for a different kind of beginning of the Universe. Why not? Ahh the BB and the Dark Energy, you say!
Then, the beginning of the Universe can be an infinite universe (or almost), homogeneous, and with temperature = 0. Temperature will grow, contradicting our deepest convictions that this is impossible.
If you want to understand clearer read that blog from the first to the last post (it is written, in portugueese, in such a way that any person can understand), or to focus only the present question you should start at O nascimento de uma Bolha until you arrive at As espirais Galácticas.
And the Dark Energy? You got rid of it, again with the help of my friend Alfredo, with his paper: A self-similar model of the Universe unveils the nature of dark energy. I posted a short math proof of his argument in this PSE link (are-the-rest-masses-of-fundamental-particles-certainly-constants). Why sould the particles be shrinking giving us the impression of a space expansion ? Their associated fields expand in space, the fields have energy, and are sourced by the particles. To preserve the total energy budget they must 'shrink'. Hey, in the lab I dont see that shrinking, how come? It is impossible to detect the phenomena locally because the lab, the instrumentation and yourself are shrinking, and above all no one likes to be shrinking.
I do realize that when DM was 'invented' was only a data fit. They have no model to justify the observations and instead of thinking harder they postulated its existence. Do we need GR in the cosmology if the space is flat arxiv-BAO-survey? No. That is why I invite the readers to pay attention to my friends paper, and be the first to say 'I spot an error'.
When I went to school I already 'knew' that the space is expanding, and the same with you. It is shocking if it is 'on the contrary'.
## The bigest enemy we face in the search of the knowledge is our 'rooted beliefs'.
Now, that I took my precious time to write this long answer I kindly ask the downvoters to say in what particular points they beleive that I am mistaken.
-
-1 : It is not an answer about the velocity distribution. If you explicitely link you hurricane intuition with the increase of velocity, I may remove the -1 – Frédéric Grosshans Apr 3 '12 at 12:39
@Frédéric - the galacic vortex is in place. The outer regions have bigger speeds as you can see. – Helder Velez Apr 3 '12 at 19:14
I don't think this place is for original research. Your answer seems to be far from the scientific consensus. Maybe the consensus is wrong, but if you manage to convince a significant part of the astrophysics community that your explanation is the good one, I promise to remove my downvote, even if you will probably not care at this step. – Frédéric Grosshans Apr 10 '12 at 13:25
And I will not read your explanation on your blog beacause : 1. Even if I'm a professional physicist, I do not have enough knowledge in astrophysics to judge the relevance of arguments about the dark matter model. 2. I'm not any person, since I cannot understand Portuguese. Essayez en Français, pour que je puisse comprendre ;-) – Frédéric Grosshans Apr 10 '12 at 13:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8203828930854797, "perplexity": 838.2063521175945}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119645898.42/warc/CC-MAIN-20141024030045-00295-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://1260d.com/2019/08/25/rod-of-pi-update/ | # Rod of Pi π (Update)
The equation is arranged as “A time, times, and a half-time” (“d, G, g, y”), Dan 9:27, 12:7, Rev 12:14. The two largest numbers (“G” and “g”) refer to One Year of Precession, but also, reversely, to the smallest part of a day. The “182” is half an Enochian year of 364 days
##### The Rod of π (Pi)
Accurate to the diameter of π to the 25th decimal, the formula incorporates all biblical timeframes and is the frame upon which all events occur — especially Biblical history.
The full formula can similarly be expressed as:
The rest is explained here.
The following was written before the above formula (completed by Sept 8, 2019) had fully been worked out. It is still accurate, just not complete.
The below formula is true to Pi π by an astonishing 1/1,000,000,000,000,000,000
This is nearly 1000 times more accurate than what NASA uses for all its space endeavors. Yet it unfolds Enoch’s calendar (and all calendars) and is the numerical sum of much that is found in nature and the Bible.
As discussed in previous documents, (written before this formula was discovered)…
• 360 is the prophetic year (see 360calendar.com)
• 364 is a year on Enoch’s calendar
• 365 is the rounded solar Year (Egyptian Cal.)
• 25920 and 25800 years are the “Great Year“, that is, Precession of the Equinox. And as the 1/25920th part of a day (and thus 1/25800th, too), it is the smallest unit according to Babylonian and Rabinnic reckoning (= 3.33… seconds)
## The Logical Next Step Towards true Pi
Increases accuracy several decimals to about 7/1,000,000,000,000,000,000,000.
While the addition of “1/12^6t-P” to the formula works well, it does not increase accuracy towards Pi as dramatically as the simpler formula.
The rod of God is One. Revelation 11:1-14 (and 12:6, 14) measures time as “1260 days”, or “42 months”, or 2/7 (“3.5 years”), or as “a time” (365), “times” (364 + 360), “and the dividing of time” (180 or apart thereof, i.,e 1/25920 or 1/25800). All this, in turn, is based upon Daniel’s 70 x 7, and Enoch’s 70 x 70. Compare the red to the formula.
Main Formula, accurate to Pi to the 20th decimal ( = 3.141592653589793238456 — 10,000 times more accurate than what NASA uses for all its space explorations. To verify, copy: (11/14y – (42 + 2/7P – 1/70^2p + 1/(12^6*t)P)/Y)/(y/4) where Y=365, y=364, t=360, P=1/25920, p=1/25800) , and paste here. The “y/4” at bottom of formula divides Enoch’s solar year into its four seasons (364/4=91 days)
Therefore, this addition to the formula is an extremely flexible choice. It is, therefore, the logical further division of the smallest part of a day, (3.333… seconds; i.e., 1/25920th of a day). Moreover, “t = 360” can represent exactly one degree of the circle (π), as well as to units of measurement of time and space.
The final fraction for this formula is large:
Circumference 153148817550530282861117/diameter 48748782683691417600000
The importance of 360 in the diameter is obvious:
Therefore, the various orbital lengths of the earth and moon in relation to the sun and other stars appear related to Pi, as well as the other planets in our solar system due to orbital resonance, with earth, apparently the fulcrum.
### The continual fractions best display the connection of Pi to the Rod of Enoch (and his calendar)
I will elaborate later on how the rationalization of π using continual fractions unravels the above formula in stunning stages.
Example, take the sixth fraction of 104348/33215. This means that the perimeter is 33215*4, which is 364*365. This is just one example.
Fractions: Approximate fractions include (in order of increasing accuracy)
22/7,
333/106,
355/113,
52163/16604,
103993/33102,
104348/33215,
and… 245850922/78256779.[30] Wiki
Click here for the signs on earth and in the heavens in regard to the revealing of this formula: “Hurricane Dorian, the Eye of Pi & Sun/Moon” | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9031907320022583, "perplexity": 4104.065372006793}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574058.75/warc/CC-MAIN-20190920175834-20190920201834-00024.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/41308-find-probability-print.html | # find probability
• June 11th 2008, 08:54 AM
cowgirl123
find probability
I have a probability problem where probability of neither A nor B is 35%. 30% is A and 40% is B. How do i find the probability of both A and B?
Also, where can i review how to do probability and stats like this?
Please dont leave the answer.. I need to try to figure this out, thanks!
• June 11th 2008, 08:59 AM
Isomorphism
Quote:
Originally Posted by cowgirl123
I have a probability problem where probability of neither A nor B is 35%. 30% is A and 40% is B. How do i find the probability of both A and B?
Also, where can i review how to do probability and stats like this?
Please dont leave the answer.. I need to try to figure this out, thanks!
Hint: $P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$
• June 11th 2008, 09:06 AM
cowgirl123
Is the answer .12? Or is this wrong?
• June 11th 2008, 09:12 AM
Isomorphism
Quote:
Originally Posted by cowgirl123
Is the answer .12? Or is this wrong?
Its wrong... Show us your work and we will tell your mistake. I can give you the solution but it may not help you a lot. I will give you a new hint.
Hint:
$P(\overline{A} \cap \overline{B}) = 1 - P(A \cup B) = 1 - (P(A) + P(B) - P(A \cap B))$
• June 11th 2008, 09:24 AM
cowgirl123
Quote:
Originally Posted by Isomorphism
Its wrong... Show us your work and we will tell your mistake. I can give you the solution but it may not help you a lot. I will give you a new hint.
Hint:
$P(\overline{A} \cap \overline{B}) = 1 - P(A \cup B) = 1 - (P(A) + P(B) - P(A \cap B))$
I think my problem is not knowing what exactly to do with the unions and intersections. I thought the intersection sign meant to multiply the numbers.. But i dont think this is right. Bc i end up with
$1 - P(A \cup B) = 1 - (.3 + .4 - (.3 * .4)) = 1 - (.7 - .12) = 1 - .58 = .42$
• June 11th 2008, 09:30 AM
Isomorphism
Quote:
Originally Posted by cowgirl123
I think my problem is not knowing what exactly to do with the unions and intersections. I thought the intersection sign meant to multiply the numbers.. But i dont think this is right. Bc i end up with
$1 - P(A \cup B) = 1 - (.3 + .4 - (.3 * .4)) = 1 - (.7 - .12) = 1 - .58 = .42$
What you are thinking is wrong generally. Only for the specific case of A and B being independent events, we can multiply probabilities...
"probability of neither A nor B is 35%" $\Rightarrow P(\overline{A} \cap \overline{B})$
We want the "probability of both A and B" $\Rightarrow P(A \cap B)$
$P(\overline{A} \cap \overline{B}) = 1 - (P(A) + P(B) - P(A \cap B)) \Rightarrow P(A \cap B) = P(\overline{A} \cap \overline{B}) + P(A) + P(B) - 1$
$P(A \cap B) = P(\overline{A} \cap \overline{B}) + P(A) + P(B) - 1 \Rightarrow P(A \cap B) = 0.35 + 0.3 + 0.4 - 1 = 0.05$
• June 11th 2008, 09:36 AM
cowgirl123
.05?
Could you give me the list of...rules.. for things like this? I have a lot to learn.. and i cant find a set of things to do for certain problems.
$.35 + .3 + .4 - 1 = .35 + .7 - 1 = 1.05 - 1 = .05$
• June 11th 2008, 10:01 AM
Isomorphism
Quote:
Originally Posted by cowgirl123
.05?
Could you give me the list of...rules.. for things like this? I have a lot to learn.. and i cant find a set of things to do for certain problems.
$.35 + .3 + .4 - 1 = .35 + .7 - 1 = 1.05 - 1 = .05$
Dont you have a text book? I can suggest a plan though... First learn basic set theory. Intersection, Compliments, DeMorgans Laws, Unions, Venn Diagrams etc. Then try learning probability :)
• June 11th 2008, 03:40 PM
mr fantastic
Quote:
Originally Posted by cowgirl123
I have a probability problem where probability of neither A nor B is 35%. 30% is A and 40% is B. How do i find the probability of both A and B?
Also, where can i review how to do probability and stats like this?
Please dont leave the answer.. I need to try to figure this out, thanks!
I'd suggest drawing a Karnaugh Table:
$\, \begin{tabular}{l | c | c | c} & A & A\, ' & \\ \hline B & & & 0.4 \\ \hline B\, ' & & 0.35 & \\ \hline & 0.3 & & 1 \\ \end{tabular}
$
Fill in the blank cells (just like a sudoko puzzle). Then add up the probabilities in the cells that relate to your question.
• June 16th 2008, 07:26 AM
cowgirl123
Quote:
Originally Posted by mr fantastic
I'd suggest drawing a Karnaugh Table:
$\, \begin{tabular}{l | c | c | c} & A & A\, ' & \\ \hline B & & & 0.4 \\ \hline B\, ' & & 0.35 & \\ \hline & 0.3 & & 1 \\ \end{tabular}
$
Fill in the blank cells (just like a sudoko puzzle). Then add up the probabilities in the cells that relate to your question.
I've never done sudoko.. are the columns and rows supposed to add up to the same number? And do you just have to start guessing numbers until it works?
• June 16th 2008, 07:17 PM
mr fantastic
Quote:
Originally Posted by cowgirl123
I've never done sudoko.. are the columns and rows supposed to add up to the same number? And do you just have to start guessing numbers until it works? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.841752290725708, "perplexity": 700.6573565423591}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430454040787.28/warc/CC-MAIN-20150501042040-00055-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://mathhelpforum.com/pre-calculus/83903-using-regression-equation.html | # Math Help - Using a Regression Equation
1. ## Using a Regression Equation
I have to find the half-life of Iodine-129 using the regression equatioon
y=100e^-0.153x
The chart I was given is divided into age (billions of years) and % of original II-129:
2.0 (74%), 3.5 (59), 4.2 (53%), 4.3 (52%)
I know that to find the half-life I have to find the value of x. However, I am not sure how to do this. Could you help? Thanks.
2. To solve for half-life,
you see that 100 is starting value when x=0.
So, $100 \cdot e^{-0.153x} = 50$
$x = \frac{ln(\frac{1}{2})}{-0.153} \approx 4.53$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8983725905418396, "perplexity": 1170.899660233701}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500836108.12/warc/CC-MAIN-20140820021356-00238-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=48941 | ## Question on 2A. 17 part c
Alvaro Chumpitaz 4D
Posts: 56
Joined: Wed Sep 11, 2019 12:17 am
### Question on 2A. 17 part c
Predict the number of valence electrons present for each of the following ions
c) Co (3+)
How can you predict the number of valence electrons for a transition metal?
JChen_2I
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am
### Re: Question on 2A. 17 part c
Cobalt is in group 9 so it has 9 valence electrons in its neutral state. Then because the charge the 3+, it means there are 3 less electrons than it does during its neutral state so there are 6 valence electrons. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8413094878196716, "perplexity": 2511.169424605339}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141652107.52/warc/CC-MAIN-20201201043603-20201201073603-00240.warc.gz"} |
http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/statug_introreg_sect029.htm | # Introduction to Regression Procedures
### Linear Regression Models
In matrix notation, a linear model is written as
where is the design matrix (rows are observations and columns are the regressors), is the vector of unknown parameters, and is the vector of unobservable model errors. The first column of is usually a vector of 1s and is used to estimate the intercept term.
The statistical theory of linear models is based on strict classical assumptions. Ideally, you measure the response after controlling all factors in an experimentally determined environment. If you cannot control the factors experimentally, some tests must be interpreted as being conditional on the observed values of the regressors.
Other assumptions are as follows:
• The form of the model is correct (all important explanatory variables have been included). This assumption is reflected mathematically in the assumption of a zero mean of the model errors, .
• Regressor variables are measured without error.
• The expected value of the errors is 0.
• The variance of the error (and thus the dependent variable) for the ith observation is , where is a known weight factor. Usually, for all i and thus is the common, constant variance.
• The errors are uncorrelated across observations.
When hypotheses are tested, or when confidence and prediction intervals are computed, an additional assumption is made that the errors are normally distributed. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9936248064041138, "perplexity": 703.5611535852898}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891791.95/warc/CC-MAIN-20180123072105-20180123092105-00094.warc.gz"} |
https://www.elementascience.org/articles/10.1525/elementa.225/ | A- A+
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# The influence of environmental drivers on the enrichment of organic carbon in the sea surface microlayer and in submicron aerosol particles – measurements from the Atlantic Ocean
## Abstract
The export of organic matter from ocean to atmosphere represents a substantial carbon flux in the Earth system, yet the impact of environmental drivers on this transfer is not fully understood. This work presents dissolved and particulate organic carbon (DOC, POC) concentrations, their enrichment factors in the sea surface microlayer (SML), and equivalent measurements in marine aerosol particles across the Atlantic Ocean. DOC concentrations averaged 161 ± 139 μmol L–1 (n = 78) in bulk seawater and 225 ± 175 μmol L–1 (n = 79) in the SML; POC concentrations averaged 13 ± 11 μmol L–1 (n = 80) and 17 ± 10 μmol L–1 (n = 80), respectively. High DOC and POC enrichment factors were observed when samples had low concentrations, and lower enrichments when concentrations were high. The impacts of wind speed and chlorophyll-a levels on concentrations and enrichment of DOC and POC in seawater were insignificant. In ambient submicron marine aerosol particles the concentration of water-soluble organic carbon was approximately 0.2 μg m–3. Water-insoluble organic carbon concentrations varied between 0.01 and 0.9 μg m–3, with highest concentrations observed when chlorophyll-a concentrations were high. Concerted measurements of bulk seawater, the SML and aerosol particles enabled calculation of enrichment factors of organic carbon in submicron marine ambient aerosols, which ranged from 103 to 104 during periods of low chlorophyll-a concentrations and up to 105 when chlorophyll-a levels were high. The results suggest that elevated local biological activity enhances the enrichment of marine-sourced organic carbon on aerosol particles. However, implementation of the results in source functions based on wind speed and chlorophyll-a concentrations underestimated the organic fraction at low biological activity by about 30%. There may be additional atmospheric and oceanic parameters to consider for accurately predicting organic fractions on aerosol particles.
##### Knowledge Domain: Ocean Science
How to Cite: van Pinxteren, M., Barthel, S., Fomba, K.W., Müller, K., von Tümpling, W. and Herrmann, H., 2017. The influence of environmental drivers on the enrichment of organic carbon in the sea surface microlayer and in submicron aerosol particles – measurements from the Atlantic Ocean. Elem Sci Anth, 5, p.35. DOI: http://doi.org/10.1525/elementa.225
Published on 30 Jun 2017
Accepted on 29 May 2017 Submitted on 27 Feb 2017
Domain Editor-in-Chief: Jody W. Deming; University of Washington, US
Associate Editor: Laurenz Thomsen; Jacobs University Bremen, GE
## 1 Introduction
The oceans are a significant source of marine aerosol particles which may impact the radiation budget of the earth directly by scattering and absorbing solar radiation and indirectly by affecting cloud formation processes. Marine aerosol particles are produced at the sea surface primarily due to interaction between wind and waves (bubble bursting) and secondarily by gas-to-particle conversion (secondary organic aerosol, SOA; e.g., de Leeuw et al. 2014. These naturally produced aerosol particles can then be influenced by various environmental factors. In particular, submicron aerosol particles (PM1) that have an atmospheric lifetime of several days (Madry et al., 2011) may be affected as a result of processing in the atmosphere and by anthropogenic impacts. Frossard et al. (2014) measured significant contributions of continental sources and ship pollution to aerosol particles sampled in different regions of the Atlantic and Pacific Ocean. In other regions (North Atlantic and Arctic Ocean), however, these authors could identify the aerosol particles as largely marine in nature. Using isotopic carbon analysis, Ceburnis et al. (2011) found a strong marine organic aerosol source in the North Atlantic under bloom conditions and showed that over 80% of the detected carbon originated directly from plankton emissions. Recently, Miyazaki et al. (2016) identified a significant contribution of oceanic dissolved organic carbon to submicron particles in the marine boundary layer, regardless of the oceanic area. They demonstrated that water soluble organic carbon in seawater is closely linked with the submicron water soluble organic carbon on aerosol particles. McCoy et al. (2015) reported that natural aerosol particles are linked to seasonal and spatial patterns of Southern Ocean cloud albedo. In conclusion, despite possible anthropogenic influences, naturally produced aerosol particles may play a dominant role in the remote marine boundary layer, depending on the region and meteorological condition.
The chemical composition of aerosol particles determines their role in cloud processes. Research of the last two decades has shown that marine aerosol particles contain a large quantity of organic material (e.g., Quinn and Bates, 2011, and references therein). As enrichment increases with particle size (O’Dowd et al., 2004), the sub-micrometer aerosol particles that play a critical role in the formation of cloud condensation nuclei are often dominated by organic matter (OM) (Quinn and Bates, 2011). Yet, the environmental drivers and mechanisms for the OM enrichment are not very clear. Biological activity in the surface ocean has been suggested to be one important parameter. Concentrations of marine organic aerosols were found to be dependent on seasonal cycles of phytoplankton activity as reflected in correlations between chlorophyll-a (chl-a) and organic aerosol mass (O’Dowd et al., 2004; Facchini et al., 2008a; Ovadnevaite et al., 2011; Rinaldi et al., 2013; Schwier et al., 2015). On the other hand, there are several studies in which a direct link between chl-a concentration and organic matter on aerosol particles could not be detected. For example, Quinn et al. (2014) suggested that the high reservoir of dissolved organic carbon in the ocean is responsible for the organic enrichment in freshly emitted sea spray aerosol, thus dominating over any influence of recent local biological activity based on chlorophyll concentrations. In a mesocosm experiment, a negative correlation between organic particles (of a specific size range) and chl-a was identified, although a positive correlation between these particles and bacterial abundance was observed (Prather et al., 2013). These authors concluded that the response of aerosol composition and its mixing state to changes in seawater is size-dependent. Such findings demonstrate the high variability of characteristics observed in organic aerosol particles in the marine environment, while also reflecting the limitations of current knowledge on the effects of biogeochemical drivers on marine aerosol particle organic composition.
When regarding the upwards export of organic material from the oceans, not only the bulk ocean water has to be considered, but also the sea surface microlayer (SML), representing the uppermost layer of the ocean and therefore the interface for all gaseous, liquid and particulate mass transfer processes between the ocean and the atmosphere (Cunliffe et al., 2013). The SML has often been reported to be enriched with both inorganic and organic matter (e.g., Liss and Duce, 1997; Wurl et al., 2011b; Engel and Galgani, 2016) and to enable interfacial photochemical processes leading to the production of important aerosol precursors such as volatile organic compounds (Liss and Duce, 1997; Ciuraru et al., 2015). Specific organic compound groups (e.g., gel-like organic material) originating from the SML have been detected on submicron aerosol particles. These biopolymers are understood to influence the cloud condensation and ice nucleating properties of the particles (Matrai et al., 2008; Orellana et al., 2011; Wilson et al., 2015). Russell et al. (2010) have demonstrated that the enrichment of OM in the submicron marine aerosol particles is several orders of magnitude higher than OM enrichment measured in the SML (in comparison to bulk seawater).
In a conceptual relationship Gantt et al. (2011) proposed that, besides aerosol size, wind speed and biological activity (as reflected in chl-a concentrations) are the most important parameters regulating the organic matter content of primary aerosol particles (particles that are generated at the ocean surface, often referred to as sea spray aerosol or SSA). Based on this model it is predicted that highest organic enrichment in SSA particles occurs at low wind conditions when marine aerosol production is minimal and the SML covers the ocean surface. At higher wind speeds the sea spray production is increased; however, the organic matter in the sea spray aerosol (OMSSA) is reduced due to the stronger mixing of the SML with bulk seawater and the consequently smaller SML coverage. Rinaldi et al. (2013) confirmed an inverse correlation between the organic aerosol fraction and wind speed and presented a function of OMSSA including wind speed and chl-a as driving factors from a two-dimensional regression analysis. In the context of the depicted source function the best fit was obtained using chl-a data with a time resolution of about one week, thus averaging over fast temporal changes.
Despite the progress in parameterizations, currently there are few comprehensive field studies on organic matter in the marine environment that take into account biogeochemical parameters in order to understand processes and to evaluate and develop concepts outlining air-sea exchange of organic matter, especially involving the SML.
In this study, measurements of organic carbon in seawater (both in the SML and the bulk water), as well as in submicron marine aerosol particles, were conducted as concerted measurements throughout different regions of the Atlantic Ocean. A large dataset was obtained during various field campaigns covering large parts of the Atlantic Ocean. Specifically, concentrations of dissolved organic carbon (DOC) and particulate organic carbon (POC) both in seawater and in the SML, as well as water-soluble organic carbon (WSOC) and water-insoluble organic carbon (WISOC) in the submicron aerosol particles, were determined. Enrichment factors of organic carbon in the submicron aerosol particles (EFaer) were calculated, representing the first EFaer for ambient marine aerosol particles based on simultaneous measurements. In order to investigate the influence of biogeochemical parameters, the organic carbon concentrations (in the SML, bulk water and aerosol particles) were analyzed as functions of chl-a concentration and wind speed to determine if and to what extend wind stress and biological activity in the surface ocean have an impact on the concentration and enrichment of organic carbon in the marine environment. Finally, the WISOC results were applied to existing (Rinaldi et al., 2013), as well as newly developed, source functions based on chl-a concentration and wind speed.
## 2 Experimental
### 2.1 Sampling areas
The sampling regions for the present study were located in the tropical and the temperate climate zones of the Atlantic Ocean (Table 1, Figure 1). Samples were taken during campaigns conducted at the Cape Verde Atmospheric Observatory (CVAO), a remote marine station described in Carpenter et al. (2010) and Fomba et al. (2014) during May 2011 and November 2011 and 2013. Samples were also taken at the location of Raune Fjord in Bergen, Norway, following a spring bloom during May–June 2011, as well as during two transects with the RV Polarstern from Punta Arenas, Chile, to Bremerhaven (ANT-XXVIII/5) during spring 2012 and from Cape Town to Bremerhaven (ANT-XXIX/10) during spring 2014. Within these two transects regions in the tropical climate zone (CVAO) and the temperate climate zone (Bergen), as well as a large part of the Atlantic Ocean, are covered.
Table 1
Details of the sampling campaigns included in this study. DOI: https://doi.org/10.1525/elementa.225.t1
Campaign Sampling area Sampling date No. of paired samples of SMLa + bulk seawater No. of marine aerosol samples (filters)
Cape Verde SOPRAN 2011-I North Atlantic (tropical zone) offshore May 2011 6 6
Cape Verde SOPRAN 2011-II North Atlantic (tropical zone) offshore November 2011 15 7
Cape Verde SOPRAN 2013 North Atlantic (tropical zone) offshore/onshore November 2013 11 8
Bergen 2011 North Atlantic onshore May–June 2011 10 15
Polarstern ANT-XXVIII/5 2012 North–South Atlantic (transect) offshore April–May 2012 22 20
Polarstern ANT-XXIX/10 2014 North/South Atlantic (transect) offshore March–April 2014 17 8
a Sea surface microlayer.
Figure 1
Illustration of generalized cruise tracks of the different campaigns included in this study. DOI: https://doi.org/10.1525/elementa.225.f1
#### 2.1.1 Aerosol particle sampling
A high volume Digitel sampler DHA-80 (Walter Riemer Messtechnik, Germany) was used for aerosol particle sampling. At the CVAO measurement station, the aerosol sampler was installed at a height of 30 m on top of a tower located directly at the seaside. During the Bergen campaign, the aerosol sampler was placed on a hill, at a height of approximately 10 m, facing the Fjord. During the Polarstern cruises, the sampler was mounted on top of the observation deck at a height of approximately 30 m. Submicron aerosol particles (PM1) were collected on preheated 150 mm quartz fiber filters (Munktell, MK 360) at a flow rate of 500 L min–1. A sampling time of approximately 24 hours, from midnight to midnight (UTC), was applied. Following the sampling, filters were stored in aluminum boxes that were kept in a freezer at –20 °C. At the end of the campaigns, the samples were transported on dry ice to the laboratories located in the TROPOS Institute, Leipzig.
#### 2.1.2 Water sampling: SML and bulk water
At the CVAO, the water samples were taken by use of a fishing boat either in front of the station (onshore samples) or within a distance of at least 5 km from the station (offshore samples). During the Polarstern campaigns, water sampling was performed using a rubber boat that was launched from the vessel; samples were taken beyond a distance of approximately 500 m from the vessel in order to avoid contamination. In Bergen, the samples were taken in the Fjord from a rubber boat while at a maximum distance of 500 m to the coast (onshore samples).
For SML sampling, the glass plate technique as one typical SML sampling strategy was applied (Cunliffe and Wurl, 2014). A glass plate with a sampling area of 2000 cm2 was vertically immersed into the water and then slowly drawn upwards. The surface film adheres to the surface of the glass and is removed using framed Teflon wipers (Stolle et al., 2010; van Pinxteren et al., 2012). Great care was taken to ensure that during all field campaigns a uniform withdrawal rate of approximately 10 cm s–1 was applied. Bulk seawater was collected from a depth of 1 m using a specially designed device consisting of a glass bottle mounted on a telescopic rod used to monitor sampling depth. The bottle was opened underwater at the intended sampling depth with a specifically conceived seal-opener.
A portion of each water sample was filtered over Whatman GF-F filters (pore size: 0.7 µm) for POC analysis. Also, additional bulk water samples were taken at each sampling day and filtered over Whatman GF-F filters for chl-a analysis (volumes of 0.5–5 L). All water samples were stored in glass bottles at –20 °C until the time of analysis. Following each sampling procedure, all materials were cleaned extensively using reagent water.
Blank samples were collected at the beginning, middle and end of the campaign during all campaigns. To this end, reagent water was loaded into bottles (the same kind used for the purpose of bulk water/SML sampling), which were taken to the field and subjected to the same procedures applied to the field samples, such as filtering and freezing. Blank values were generally lower than 10% of the DOC and POC concentrations values identified in the field samples. Average blank values were subtracted from the measured DOC and POC concentrations.
### 2.2 Analytics
#### 2.2.1 Aerosol particle filter analytics
Prior to their application, aerosol filters were preheated at a temperature of 105 °C for a duration of 24 hours. Organic carbon (OC) and elemental carbon (EC) determination was carried out by means of a thermal-optical method using the Sunset Laboratory Dual-Optical Carbonaceous Analyzer (Sunset Laboratory Inc., U.S.A.) from a filter piece with an area of 1.5 cm2. The EUSAAR 2 temperature protocol was utilized, and a charring correction was applied (Cavalli et al., 2010). The correction value for pyrolytic carbon was determined based on measurements of a sample transmission using a 678 nm laser. Samples were thermally desorbed from the filter medium under an inert He-atmosphere followed by an oxidizing O2/He-atmosphere while applying carefully controlled heating ramps. A flame ionization detector was used to quantify methane following a catalytic methanation of CO2.
The WSOC content of the collected aerosol samples was measured by analyzing filtered aqueous particle extract (from a filter piece of approximately 21 cm2 extracted in 25 mL water) using a 0.45 µm syringe filter and, subsequently, a TOC-VCPH analyzer (Shimadzu, Japan). To convert inorganic carbonates to carbon dioxide (CO2), 1.5 vol% of a hydrochloric acid solution (2 M) was added to the aqueous extract, resulting in a pH value of 2. Afterwards, synthetic particle-free air was channeled through the acidified sample in order to remove the formed CO2. The oxidation of the carbon present within the sample was induced by adding 300 µL of the carbonate-free solutions to a catalyst (aluminum oxide/platinum on quartz wool) maintained at a temperature of 720 °C. The formed carbon dioxide was analyzed by means of non-dispersive infrared (NDIR) detection. For quantification, an external calibration with potassium hydrogen phthalate was applied. WISOC concentrations were calculated as the difference between OC and WSOC (Cavalli et al., 2004).
Concentrations of sodium and other inorganic ions (chloride, sulfate, nitrate, ammonium, potassium, magnesium, calcium) in filtered (0.45 µm syringe filter) aqueous extracts (25% of the filter in 30 mL of water) were determined using ion chromatography (ICS3000, Dionex, Sunnyvale, CA, USA), as described by Müller et al., (2010).
Field blanks were obtained by inserting the filters into the Digitel sampler for a duration of 24 hours without loading them. Three field blanks were collected during the course of each campaign. The mean values of the obtained blanks were subtracted from the WSOC concentrations and OC concentrations, whereas sodium blanks always fell below the detection limit.
#### 2.2.2 Seawater analytics
For seawater DOC, the organic carbon was quantified in filtered seawater (GF-F Whatman filters) using the TOC-VCPH analyzer (Shimadzu, Japan), the same instrument applied to aerosol particles. Repetitive analysis of frozen samples revealed that no artifacts were caused as a result of freezing and storing procedures, as the values of the re-analyzed samples showed relative standard deviations below 10%.
For the purpose of POC analysis, different quantities of seawater were filtered over preheated Whatman GF-F filters kept at 450 °C for 5 hours, following the procedure reported by Taucher et al. (2012). From these filters, POC concentrations were measured directly by applying a two-step thermographic method using a commercial carbon analyzer C-mat 5500 (Ströhlein, Germany). The applied method is a variation of the method described in the VDI Guideline (VDI 2465, Part 2) issued in Germany. As a first step, nitrogen was employed to act as a carrier gas, kept at a temperature of 650 °C, for the purpose of OC volatilization. Furthermore, as a second step, EC was combusted at a temperature of 650 °C in an oxygen atmosphere. The EUSAAR 2 temperature protocol could not be applied, as the final temperature (850 °C) would have caused the GF-F filters to melt.
Salinity in the SML and the bulk water was measured occasionally, using a refractometer (OCS.tec GmbH & Co. KG, Neuching, Germany). The detected salinities varied between 35 and 39 ppt (formerly given as PSU) in all investigated regions.
#### 2.2.3 Back trajectories and wind speed
Information regarding the origins of air masses was obtained on the basis of back trajectory analyses. Seven-day back trajectories were calculated on an hourly basis within the sampling intervals, using the NOAA HYSPLIT model (HYbrid Single-Particle Lagrangian Integrated Trajectory, http://www.arl.noaa.gov/ready/hysplit4.html, 26.11.16) in the ensemble mode at an arrival height of 500 m ± 200 m (van Pinxteren et al., 2010).
The wind speeds reported here represent either the averaged wind speed over the aerosol sampling period (typically 24 hours) or the wind speed at the sampling time in case of seawater sample collection. The wind speed values were achieved on the basis of on-board measurements during the Polarstern cruises (König-Langlo, 2012; König-Langlo, 2014) and from local weather stations at the CVAO and the Bergen campaign. The registered values were converted to a uniform wind speed at a height of 10 m according to Equation 1 provided by Kleemann and Meliß (1993):
${v}_{h}={v}_{10}{\left(\frac{h}{10}\right)}^{g}$
(1)
where vh indicates wind speed at the measured height while v10 indicates the wind speed at a height of 10 m. For open areas such as oceans or deserts, the exponent g is set to 0.16 (Kleemann and Meliß, 1993). Wurl et al. (2011b) have categorized the wind speeds into three groups: low (0–2 m s–1, referring to state 0), moderate (2–5 m s–1, referring to states 1–2) and high (5–10 m s–1, referring to states above 2.5) conditions according to Pierson and Moskowitz (1964). The data obtained in the present study consistently showed wind speeds at higher than 2 m s–1. Therefore, we slightly modified the respective ranges suggested by Wurl et al. (2011b) and classified our data into two categories designating either “low wind”, at values ranging between 2 and 5 m s–1, or “high wind” conditions at wind speeds of higher than 5 m s–1.
#### 2.2.4 Chlorophyll-a (Chl-a)
Chl-a is a regularly applied and well accessible indicator for the purpose of describing biological activity of the surface ocean (O’Dowd et al., 2004; de Leeuw et al., 2011). Chl-a values may be derived from satellite retrievals or directly from seawater measurements. For this study, chl-a concentrations were measured by means of HPLC and fluorescence detection (Dionex, Sunnyvale, CA, USA). GF-F filters were extracted in 5 mL ethanol, and 20 µL of the extract were injected into the HPLC system under gradient elution using methanol/acetonitrile/water systems as eluents. Chl-a concentrations were also obtained from satellite retrievals provided by the Ocean Color Web operated by the NASA (http://oceancolor.gsfc.nasa.gov/, 26.07.16). The concentrations were obtained using MODIS-AQUA and MODIS-TERRA within a radius of 1° from the sampling location and then averaged based on results from both instruments in order to fill data gaps due to cloud coverage. The resulting values from both approaches were categorized into “low chl-a” for chl-a values between 0.1 and 0.6 µg L–1 or “high chl-a” for values starting at 0.9 µg L–1. Chl-a concentration values below the detection limit were set to the detection limit value and classified as within the “low chl-a” category.
#### 2.2.5 Data analysis
The enrichment factor (EF) in the SML was calculated by dividing the concentration (c) of DOC or POC in the SML by the respective concentration measured in the bulk water (BW) using the following equations:
(2)
(3)
Enrichment of DOC and POC in the SML is indicated by EF > 1, while depletion is indicated by EF < 1.
The thickness of the SML (Df) was determined based on the quotient between the sampled volume of the SML and the sampling area of the glass plate:
(4)
In order to calculate the enrichment factors for the organic carbon on submicron aerosol particles, the WSOC (on aerosol particles) was assigned to DOC (in seawater) and the WISOC (on aerosol particles) was assigned to POC (in seawater). The abbreviations EFaer (DOC) and EFaer (POC) were applied uniformly:
(5)
(6)
The EFaer values were calculated combining the data points sampled on a daily basis. To this end, aerosol concentrations, typically sampled at a 24-hour interval, were combined with SML concentrations (spot samples) that had been collected in the middle of the aerosol sampling period. During the Bergen campaign, however, the aerosol sampling did not occur simultaneously with the water sampling, as the aerosol filters collected during SML sampling did not meet the criterion of clean marine air (i.e., low EC concentrations, air masses of marine origin, similar and overall low concentrations of inorganic ions across campaigns; more details can be found in Section 3.2). Therefore, for the Bergen campaign, aerosol and SML concentrations were clustered according to the respective chl-a and wind speed categories. Subsequently, median DOC and POC concentrations as well as median WSOC, WISOC and sodium concentrations were calculated to then determine EFaer (DOC) and EFaer (POC) according to Equations 5 and 6.
According to the applied analytical standard protocol, the separation of WSOC/DOC and WISOC/POC is obtained via filtering. As previously described, filters with a pore size of 0.7 µm are utilized routinely for DOC analysis of seawater (GF-F glass fiber filter, Whatman). However, in order to separate WSOC and WISOC in aerosol particle analysis, pore sizes of 0.45 µm are applied as a standard procedure, which is necessary in order to be able to compare data with values reported in the literature. The usage of different filter sizes may cause a higher ratio of DOC to POC compared to the ratio of WSOC to WISOC.
A Pearson correlation matrix was compiled for all data (XLSTAT software). To this end, a Box-Cox transformation of the data was applied when necessary to reduce skewness in the variables. Also, T-tests were performed to test statistical significance using the analysis of variance (ANOVA). Finally, two-dimensional regression approaches were calculated using the software MATLAB.
## 3 Results and discussion
### 3.1 Seawater
#### 3.1.1 Overview of concentration and enrichment of DOC and POC in seawater and the SML
The measured concentrations of DOC averaged 161 ± 139 (standard deviation, SD; n = 78) with a median value of 122 µmol L–1 in bulk seawater, and averaged 225 ± 175 (n = 79) with a median value of 164 µmol L–1 in the SML (for the complete data set see Table S1). These concentration values lie within the range of DOC concentrations for coastal and open ocean water samples reported in the literature (e.g., Carlson, 1983; Mostofa et al., 2013). DOC concentrations in the SML differed significantly from DOC concentrations in bulk water (ANOVA, oneway, p = 0.024 at a 0.05 level). POC concentrations averaged 13 ± 11 µmol L–1 (n = 80) (median: 7.4 µmol L–1) in bulk water, and 17 ± 10 µmol L–1 (n = 80) (median: 13.2 µmol L–1) in the SML. POC concentrations in the SML also differed significantly from bulk water concentrations (p = 0.029). Romankevich (1984) summarized the results of POC measurements obtained in a number of studies and found average POC concentrations in the Atlantic Ocean to be approximately 5 µmol L–1, with maximum values up to 52 µmol L–1. However, in most of the studies presented by Romankevich (1984), GF-C filters with a pore size of 1.2 µm were applied for the purpose of POC analysis. In the present study we applied GF-F filters with a pore size of 0.7 µm in accordance with the Practical Guidelines for the Analysis of Seawater (Wurl, 2009). GF-F filters are the classical filter material due to ease of cleaning and low blanks (Wurl, 2009). The smaller pore size applied in the present study might explain the slightly higher average POC values when compared to Romankevich (1984), as a larger amount of POC is retained on the filters. Overall differences in analytical protocols and determination methods applied during DOC and POC analysis need to be taken into account when comparing DOC and POC concentrations to literature values. The average enrichment of POC in the SML is slightly higher (EF = 2.3 ± 1.9, n = 76) than enrichment of DOC (EF = 1.8 ± 1.5, n = 75).
There were no clear patterns observable between wind speed and SML enrichment or concentration of DOC or POC. Had particular subgroups of DOC and POC been measured, they might have shown a different enrichment behavior depending on wind stress. For example, surfactants are consistently enriched in the SML at wind speeds above 5 m s–1 (Wurl et al., 2011b), while carbohydrates and gel-particles show depletion in the SML and their concentrations often correlate negatively with wind speed (Wurl et al., 2011a, and references therein). Rather than a consistent enrichment of DOC or POC in the SML, our data sets indicate that high enrichment generally occurs when bulk water samples have low DOC concentrations, while low to no enrichment (or even depletion) is observed when bulk water samples have high DOC concentrations (Figure 2a). This observation is confirmed by the negative Pearson coefficient (–0.60) determined for bulk water concentration and enrichment factor (Table 2). It seems to be a general trend within various data sets, as summarized by Burrows et al. (2014), and is consistent with a surface saturation effect. At higher bulk water concentrations, there are more molecules competing for a smaller available surface area (Burrows et al., 2014). Lower DOC concentrations are typically observed in open ocean waters, whereas coastal seawaters contain a higher amount as well as a higher variance in DOC concentrations, partly as a result of terrestrial inputs (Mostofa et al., 2013). DOC concentrations determined in the present study are consistent with this regional difference: generally lower concentrations were observed in the open ocean while higher concentrations were detected in onshore regions (Figure 2a). These findings are also in agreement with a study conducted by Carlson (1983), who reported DOC as often enriched in the SML, while differences between DOC concentrations measured in SML and bulk water were said to be generally small. He concluded that differences in enrichment between costal and oceanic samples may reflect differences in microlayer accumulation and removal processes depending on the region. For POC concentrations in this study, the same patterns were observed as for DOC; i.e., high enrichment at low bulk water concentrations and vice versa (Pearson coefficient of –0.8; Table 2). However, unlike the case for DOC – where low concentrations were mainly found in oceanic samples – low POC concentrations also occurred in coastal samples, and high POC concentrations in oceanic samples (Figure 2b). We conclude that POC is more evenly distributed among oceanic and coastal water than is DOC.
Figure 2
Bulk water concentrations versus enrichment. Bulk water concentrations of DOC (a) and POC (b) versus the enrichment in the SML (EFSML) grouped according to different sampling regions. The dashed line represents an enrichment factor of one. DOI: https://doi.org/10.1525/elementa.225.f2
Table 2
Pearson correlation matrix on Box-Cox transformed data.a DOI: https://doi.org/10.1525/elementa.225.t2
Variablesb DOC BW DOC SML POC BW POC SML EFSML (DOC) EFSML (POC) Df WSOC WISOC EC Sodium Chl-a Wind speed EFaer (DOC) EFaer(POC) SST
DOC BW 1 0.271 0.094 –0.147 –0.595 –0.143 0.089 –0.195 0.370 0.062 –0.017 0.029 0.061 –0.149 0.465 0.235
DOC SML 0.271 1 –0.053 0.165 0.499 0.162 0.192 –0.242 0.383 0.146 0.090 0.049 0.072 –0.423 0.286 0.179
POC BW 0.094 –0.053 1 0.103 –0.121 –0.813 0.128 0.264 0.074 0.018 –0.082 0.112 –0.043 0.194 0.106 0.020
POC SML –0.147 0.165 0.103 1 0.231 0.358 0.137 –0.002 –0.065 –0.034 –0.084 0.132 –0.113 0.034 –0.262 –0.047
EFSML (DOC) –0.595 0.499 –0.121 0.231 1 0.261 –0.011 –0.141 0.023 0.014 0.165 0.023 0.036 –0.324 –0.157 –0.048
EFSML (POC) –0.143 0.162 –0.813 0.358 0.261 1 –0.033 –0.250 –0.177 –0.064 0.032 0.017 –0.067 –0.165 –0.261 –0.051
Df 0.089 0.192 0.128 0.137 –0.011 –0.033 1 0.038 0.606 0.094 –0.172 0.610 –0.099 0.063 0.280 –0.332
WSOC –0.195 –0.242 0.264 –0.002 –0.141 –0.250 0.038 1 –0.256 0.106 –0.059 –0.025 –0.352 0.543 –0.093 –0.305
WISOC 0.370 0.383 0.074 –0.065 0.023 –0.177 0.606 –0.256 1 0.465 0.087 0.643 0.055 –0.319 0.780 0.370
EC 0.062 0.146 0.018 –0.034 0.014 –0.064 0.094 0.106 0.465 1 0.166 0.155 –0.063 –0.171 0.206 0.078
Sodium –0.017 0.090 –0.082 –0.084 0.165 0.032 –0.172 –0.059 0.087 0.166 1 0.304 0.152 –0.780 –0.131 –0.239
Chl-a 0.029 0.049 0.112 0.132 0.023 0.017 0.610 –0.025 0.643 0.155 0.304 1 0.081 –0.227 0.221 –0.589
Wind speed 0.061 0.072 –0.043 –0.113 0.036 –0.067 –0.099 –0.352 0.055 –0.063 0.152 0.081 1 –0.338 0.115 –0.038
EFaer (DOC) –0.149 –0.423 0.194 0.034 –0.324 –0.165 0.063 0.543 –0.319 –0.171 –0.780 –0.227 –0.338 1 –0.069 –0.054
EFaer (POC) 0.465 0.286 0.106 –0.262 –0.157 –0.261 0.280 –0.093 0.780 0.206 –0.131 0.221 0.115 –0.069 1 0.320
SST 0.235 0.179 0.020 –0.047 –0.048 –0.051 –0.332 –0.305 0.370 0.078 –0.239 –0.589 –0.038 –0.054 0.320 1
a Pearson coefficients higher than 0.6 in bold.
b BW = bulk seawater, SML = sea surface microlayer, EF = enrichment factor, Df = thickness of SML, WSOC = water-soluble organic carbon, WISOC = water-insoluble organic carbon, EC = elemental carbon, aer = aerosol particles, SST = sea surface temperature.
#### 3.1.2 Concentration and enrichment of DOC and POC in relation to wind and chl-a conditions
As a next step, the DOC and POC concentrations were divided into four groups according to their chl-a and wind speed profile, because biological activity and wind speed have been suggested to be two important drivers for the concentration and enrichment of organic carbon into sea spray aerosol (Gantt et al., 2011). The results of a comparison between measured and satellite-derived chl-a concentrations (for details see Section 2.2.4) revealed a reasonably good correlation between the two approaches (R2 = 0.62, slope: 0.39, intercept: 0.08; Figure S1). Both datasets show the same trend of highly elevated chl-a levels during the Bergen campaign. However, measured values and satellite values do not agree at times, and the measured values are generally higher than those derived from satellite data. This difference has to be taken into account when evaluating the single campaigns in detail. Both of these chl-a approaches have their limitations; i.e., satellite data are sensitive to cloud coverage, and the measured values represent single spot samples. A diurnal variation of chl-a was also visible in our data; i.e., when chl-a samples were taken twice a day, the afternoon values were mostly higher (by about 30%) than those observed in the morning. We cannot account for such variations here, as usually only one data point per day was obtained; however, sample collection was generally performed around midday. As only a limited number of chl-a values based on satellite retrievals were available for some campaigns (e.g., only two values could be obtained for the Bergen campaign for the time span of aerosol sampling; see Table S2), the measured chl-a values were applied consistently in the subsequent analyses of this work.
Box plots of the DOC and POC concentrations are shown in Figure 3, and summary statistics of the data (medians, means, maxima and minima) are presented in Table 3. Most of the data can be seen (Figure 3) to lie within the category “high wind and low chl-a” (n = 56), followed by “low wind and low chl-a” (n = 11). During the Bergen campaign, which was conducted during a bloom period, chl-a was significantly elevated (about 1 µg L–1); the Bergen data can be referred to as encompassing both of the categories, “high wind and high chl-a” (n = 8) and “low wind and high chl-a” (n = 4). DOC concentrations detected in the SML and in bulk water did not differ significantly at low chl-a concentrations. However, during high wind and periods characterized by high chl-a values, the concentrations in the SML were significantly higher than those in bulk water (p < 0.05).
Table 3
Concentrations of dissolved and particulate organic carbon (DOC, POC) in the sea surface microlayer (SML) and in bulk water (BW) and their enrichment factors in the SML (EFSML) according to different wind speed and chl-a categories. DOI: https://doi.org/10.1525/elementa.225.t3
Statistic DOC (µmol L–1) in SML DOC (µmol L–1) in BW
median 153 163 177 234 107 122 140 94.0
mean 164 240 172 258 118 173 142 131
max 314 946 255 459 192 856 199 233
min 64.1 74.0 109 120 64.8 44.4 75.5 79.5
n 11 56 4 8 11 56 4 7
Statistic POC (µmol L–1) in SML POC (µmol L–1) in BW
median 13.2 12.8 16.4 13.3 18.0 6.3 6.1 14.9
mean 20.4 16.1 15.8 16.9 18.6 11.8 6.6 17.3
max 59.6 44.5 19.3 39.3 32.9 43.2 11.4 39.8
min 5.4 2.0 10.2 5.4 4.8 2.7 2.7 2.0
n 11 57 4 8 11 57 4 8
Statistic DOC EFSML POC EFSML
mediana 1.35 1.53 1.37 1.51 1.83 2.25 1.52 1.98
meana 1.36 1.93 1.30 1.99 1.88 2.51 1.42 1.95
min 0.99 0.20 0.76 0.80 0.44 0.20 0.20 0.20
max 1.68 9.09 1.79 5.25 3.96 12.9 2.53 5.06
n 10 54 4 7 10 54 4 8
Variable Wind speed and chl-a categories
wind low high low high low high low high
chl-a low low high high low low high high
a The median and mean enrichment factors were calculated from the data pairs listed in Table S1.
Figure 3
Concentrations of DOC and POC. Box and whisker plot of the concentrations of DOC (a) and POC (b) found in SML and bulk water. Each box encloses 50% of the data, the mean value is represented as an open square and the median value as a line. The bottom of the box marks the 25% limit of the data, while the top marks the 75% limit. The lines extending from the top and bottom of each box are the 5% and 95% percentiles within the data set, while the asterisks indicate the data points lying outside of this range (“outliners”). Boxes are color-coded to indicate paired samplings of DOC and POC. The light green dashed lines encompass data from locations with low chl-a concentrations; the dark green dashed lines, those from locations with high chl-a concentrations. The solid blue fields highlight data from locations with high wind speeds. DOI: https://doi.org/10.1525/elementa.225.f3
For POC, during time periods characterized by high chl-a concentrations and low wind speeds, concentrations detected in the SML are slightly higher than those observed in the bulk water (p = 0.02). Despite the high variability of the data, no further significant differences between POC concentrations in the SML and bulk water were found for the other respective conditions. Occasionally, however, high POC and DOC concentrations occurred at low chl-a concentrations (Figures 2 and 3). These high concentrations point to an organic matter source not represented by chl-a concentration.
A comparison of the enrichment factors (EFSML) for DOC with POC based on different wind and chl-a conditions shows that enrichment was consistently higher than 1 (Table 3 and Figure S2). However, the enrichment did not differ significantly among the investigated categories. For high wind speeds, however, maximum EFs were consistently higher than 5, indicating a higher dynamic at high wind stress.
#### 3.1.3 SML thickness
The thickness of the SML (Df) was determined according to Equation 4 and is shown in Table 4 and Figure 4. To date, no uniform standard procedure exists for SML sampling, and the SML thickness data in this study are based on an operational definition that is certainly dependent on the applied sampling strategy. A valuable contribution to allow representative and standardized SML sampling has been achieved by Cunliffe and Wurl (2014). They published a “guide to best practices to study the ocean’s surface” and reviewed current SML sampling strategies in order to provide best sampling protocols and create a consensus definition of the SML. In all campaigns of the present work, the same glass plate device and a uniform withdrawal rate of 10 cm s–1 were applied. Hence, among the campaigns, data are comparable. The median thickness of the SML at low chl-a concentrations was determined to be approximately 80 µm, which closely corresponds to findings reported by Zhang et al. (2003), who demonstrated in the context of experiments that the typical SML thickness lies at approximately 60 µm. During the Bergen campaign, characterized by a time period of high chl-a conditions, a much thicker SML was observed, with a mean thickness of 230 µm. The sea surface temperature (SST) in Bergen was much lower (11 °C) when compared to the other investigated regions (Table S1). Despite the fact that no clear correlation between SST and SML thickness was found for the different campaigns (Figure S3), the low SST values coincided with the high chl-a levels detected in Bergen (Pearson coefficient = –0.6; Table 2). The cold waters tend to be more nutrient-rich, and phytoplankton more abundant, compared to warm waters (see, e.g., global maps of chl-a concentration and SST on http://earthobservatory.nasa.gov/, 29.11.16). The greater thickness of the SML sampled during the Bergen campaign can be taken to indicate an influence of oceanic biological activity on SML thickness, consistent with Galgani and Engel (2013) who demonstrated that enrichment of organic matter can enhance the SML thickness. However, in the present data set this difference in thickness is not reflected in either the EF or the concentration of the (bulk) organic parameters. A much thicker surface film was observed without detecting, correspondingly, a higher enrichment of DOC or POC in the sampled film when compared to the other regions. Surface-active compounds are likely to be enriched in the SML at high chl-a conditions, as previous work has shown that marine gel particles such as transparent exoploymer particles (TEP) and surfactants such as amino acids and fatty acids are strongly enriched there (Wurl et al., 2011a; Engel and Galgani, 2016). Such compounds, however, represent only a portion of DOC and POC; hence, variations in their concentrations and enrichment may not be reflected in the bulk parameters (e.g., van Pinxteren et al., 2012). A more detailed study on the nature of DOC and POC appears to be required in order to further investigate the impact of specific compound groups on the SML thickness.
Table 4
Film thickness (Df) of the SML, according to different wind speed and chl-a categories. DOI: https://doi.org/10.1525/elementa.225.t4
Statistic Df (µm)
median 89.4 82.1 243 237
mean 91.0 83.0 242 209
min 50 52.0 241 91.8
max 114 111 243 243
n 11 57 3 8
Variable Wind speed and chl-a categories
wind low high low high
chl-a low low high high
Figure 4
Film thickness of the SML. Box and whisker plot of the film thickness (Df) of the SML. Each box encloses 50% of the data with the mean value represented as an open square and the median value represented as a line. The bottom of the box marks the 25% limit of the data, while the top marks the 75% limit. The lines extending from the top and bottom of each box are the 5% and 95% percentiles within the data set, while the asterisks indicate the data points lying outside of this range (“outliners”). The light green dashed lines encompass data from locations with low chl-a concentrations; the dark green dashed lines, those from locations with high chl-a concentrations. The solid blue fields highlight high wind speeds. DOI: https://doi.org/10.1525/elementa.225.f4
Overall, the parameters POC and DOC were often found to be enriched, while concentration differences between SML and bulk seawater were relatively small; i.e., the enrichment factor was not high. POC was not significantly more enriched than DOC. Less enrichment was observed when concentrations were higher, which may be due to a sea-surface saturation effect. At high wind speeds, a higher variability of EFs occurred: i.e., higher maximum values were observed. At high chl-a conditions only slight differences in DOC and POC concentrations between SML and bulk water were detected. Although the SML thickness was significantly increased at high chl-a conditions, strong dependencies of DOC and POC concentrations and enrichment on wind conditions and chl-a concentrations were not observed.
### 3.2 Aerosol particles
#### 3.2.1 Organic parameters WSOC and WISOC
Submicron aerosol particles (PM1) were sampled at all areas listed in Table 1. Similar to seawater and SML measurements, OC was distinguished into a water-soluble part (WSOC) and a water-insoluble part (WISOC). To ensure that measured aerosol particles would be of marine origin, only samples showing EC concentrations ≤ 70 ng m–3, and therefore below the maximal EC concentration reported for clean marine air (Cavalli et al., 2004), as well as those from locations with air masses predominantly originating from the ocean (residence time index over water/ice > 0.7; van Pinxteren et al., 2010), were considered in this study. Therefore, approximately 90% of the sampled aerosol filters utilized at the CVAO, but only 20% (Polarstern ANT-XXIX/10 2014) and 50% (Polarstern ANT-XXVIII/5 2012) of the filters applied during the ship cruises and 40% of the filters obtained during the Bergen campaign could be included in the interpretation. For the selected aerosol particles, no correlation between EC and WSOC or WISOC was observed. This lack of correlation served as a further confirmation that the influence of anthropogenic sources is of minor importance (Rinaldi et al., 2013). The observed concentrations of the main inorganic ions were quite similar across the different campaigns, falling within the ranges reported to indicate clean marine aerosol particles. Sulfate concentrations were observed to be between 0.42 and 0.73 µg m–3, which is in good agreement with concentration values reported for the remote tropical region (e.g., Müller et al., 2010). Sodium and magnesium correlated very well (R2 = 0.78) and nitrate concentrations were between 0.02 and 0.05 µg m–3. The molar ratio of Cl to Na+ was between 0.34 and 0.76 for the different campaigns and a lower Cl/Na+ ratio appeared to be mostly connected to a higher sulfate concentration (Figure S4). This connection could point to gas phase reactions (“aging”) of the particles resulting in a depletion of chloride and acidification of the particles. However, acidification of sea spray particles occurs in the timescale of seconds (e.g., Keene et al., 2007). Recently, Miyazaki et al. (2016) found a molar ratio of Cl/Na+ between 1.06 and 0.32 during different parts of a cruise in the Pacific Ocean. Nevertheless, based on isotopic carbon measurements, they found that the natural sources of the aerosol particles (regarding WSOC) were dominant during the entire cruise. Further detailed chemical investigations (e.g., products of gas phase reactions) would be required to study atmospheric processing (“aging”) of the particles. Based on the qualities of low EC, air masses of marine origin, and similar and overall low concentrations of inorganic ions across campaigns, the aerosol data reported in this study were considered to represent natural marine aerosol particles with minor contribution of anthropogenic sources. In the future, however, we plan to further confirm the “clean marine aerosol” classification by additional methods such as isotopic carbon measurements (e.g., Ceburnis et al., 2011).
WSOC concentrations of approximately 0.2 (±0.12, n = 64) µg m–3 were detected, therefore falling within the range of previously reported marine aerosol concentrations (e.g., by Cavalli et al., 2004). WISOC concentrations showed a higher variance with concentrations between 0.01 and 0.9 µg m–3. During some periods, the high WISOC concentrations appeared to correspond to low EC concentrations between 10 and 30 ng m–3. An overview about organic aerosol concentrations in clean marine air has been published by Gantt and Meskhidze (2013). The concentrations of the present work fit well with the compilation of Gantt and Meskhidze (2013); however, in most of the studies no differentiation was made between WSOC and WISOC. Cavalli et al. (2004) summarized WISOC concentrations of the Mace Head region and reported WISOC values that averaged 0.66 µg m–3. At the same location, very high organic marine aerosol concentrations, peaking at 3.8 µg m–3 during moderate wind speeds, have been reported by Ovadnaite et al. (2011) using aerosol mass spectrometry (measuring PM1 aerosol particles). These concentrations are much higher than determined with offline techniques, and they comprised 77% of the total submicron non-refractory mass. The pattern of the organic mass also revealed a unique marine fingerprint (Ovadnaite et al., 2011). O’Dowd et al. (2015) had reported previously high ambient submicron organic matter concentration (measured with aerosol mass spectrometry) exceeding 4 µg m–3 at the Mace Head station during blooms. These aerosol organic matter concentrations measured with online techniques are not directly comparable to the offline method applied in the present work because with time-averaged offline sampling such high plumes cannot be captured. Nevertheless, the measured concentrations suggest a far stronger source of primary organic matter in marine air than assumed.
Regarding all aerosol data in this study, no significant correlation between WSOC and WISOC concentrations was found. In addition, no correlation of WSOC or WISOC concentrations to wind speed could be observed over the sampling time of 24 hours, nor was a significant correlation apparent between chl-a concentration, serving as an indicator of seawater biological activity, and WSOC concentration. However, a connection between measured chl-a and WISOC concentrations was observed (Pearson coefficient = 0.64; Table 2).
#### 3.2.2 WSOC and WISOC in relation to wind and chl-a categories
As for procedures applied to seawater, the obtained WSOC and WISOC data were classified into groups according to differing biological and meteorological conditions. The results are presented in Table 5 and Figure 5. Compared to the results obtained for DOC and POC concentrations in seawater, wind and chl-a conditions generally had a stronger impact on WSOC and WISOC concentrations sampled on aerosol particles. Differences in concentration values at low chl-a conditions were minor.
Table 5
Concentrations of water-soluble organic carbon (WSOC) and water-insoluble organic carbon (WISOC) in submicron aerosol particles (PM1) and enrichment factors of DOC and POC in PM1 (EFaer), according to different wind speed and chl-a categories. DOI: https://doi.org/10.1525/elementa.225.t5
Statistic WSOC (µg m–3) in PM1 WISOC (µg m–3) in PM1
median 0.18 0.16 0.33 0.21 0.04 0.10 0.83 0.68
mean 0.19 0.17 0.35 0.19 0.07 0.14 0.75 0.62
max 0.26 0.64 0.50 0.30 0.17 0.66 0.92 0.90
min 0.16 0.01 0.24 0.05 0.01 0.01 0.44 0.14
n 10 39 4 11 8 37 4 11
Statistic EFaer (DOC)a EFaer (POC)a
median 1.1 × 104 1.5 × 104 3.4 × 104 1.2 × 104 6.4 × 103 9.0 × 104 6.6 × 105 5.7 × 105
mean 1.5 × 104 1.8 × 104 2.1 × 104 1.8 × 105
max 4.0 × 104 6.4 × 104 8.6 × 104 9.8 × 105
min 4.4 × 103 6.4 × 101 1.8 × 103 3.3 × 103
n 6 35 1 1 5 31 1 1
Variable Wind speed and chl-a categories
wind low high low high low high low high
chl-a low low high high low low high high
a EFaer were calculated based on combinations of daily samples of SML and aerosol sampling events (see text for details).
Figure 5
Concentrations of WSOC and WISOC. Box and whisker plot of the concentrations of water-soluble organic carbon (WSOC) and water-insoluble organic carbon (WISOC) found on aerosol particles. Each box encloses 50% of the data with the mean value represented as an open square and the median value represented as a line. The bottom of the box marks the 25% limit of the data, while the top marks the 75% limit. The lines extending from the top and bottom of each box are the 5% and 95% percentiles within the data set, while the asterisks indicate the data points lying outside of this range (“outliners”). Boxes are color-coded to indicate paired samplings of WSOC and WISOC. The light green dashed lines encompass data from locations with low chl-a concentrations; the dark green dashed lines, those from locations with high chl-a concentrations. The solid blue fields highlight data from locations with high wind speeds. DOI: https://doi.org/10.1525/elementa.225.f5
During the Bergen campaign that was shaped by high chl-a concentrations, concentrations of both WSOC and WISOC were higher and showed greater variance, and WISOC concentrations were significantly higher than WSOC (p < 0.05; compare box plots in the high chl-a section of Figure 5). On average, WSOC was higher by 30% and WISOC by 85% compared to periods shaped by low chl-a concentrations. Furthermore, approximately 60% more WISOC than WSOC was present on the aerosol particles. That contamination from terrestrial sources may have biased the data cannot be ruled out completely. However, the application of the “clean marine air” criteria suggested that only a minor contribution may have derived from anthropogenic sources. Besides chl-a concentrations, the WISOC was also connected to SML thickness (Pearson coefficient = 0.61; Table 2). The high WISOC concentrations are likely caused by an oceanic bubble bursting source, as also previously reported (Ceburnis et al., 2008; Facchini et al., 2008b), and seem to be additionally affected by biological activity in the ocean. High WISOC concentrations occurring even at low wind speeds suggest that wind intensity above the oceans is consistently sufficient for organic matter to be transferred from the oceans to the atmosphere. However, besides wind, other sources have been identified that may be responsible for the transfer of organic matter from the ocean to the atmosphere, as previously observed, for example, in the Arctic at low wind speeds (Leck et al., 2002). Bubbles can be formed as a result of processes such as microbial respiration (Johnson and Wangersky, 1987), presenting an additional transfer possibility for organic matter – predominantly at time periods characterized by high biological activity, such as during the Bergen campaign. WSOC concentrations can in part be attributed to other (secondary) sources and may therefore not be as strongly affected by oceanic biological activity as WISOC.
### 3.3 Enrichment factors in submicron aerosol particles (EFaer)
As a next step, the organic carbon concentrations detected in the SML and on the aerosol particles were considered in relation to each other. Seawater and aerosol samples were taken simultaneously, and only clean marine air masses were considered (see Section 3.2.1). At the CVAO, for example, the air masses were demonstrated to follow the water current (Figure S5), enhancing the organic carbon link between the SML and the aerosol particles. As mainly winds drive the ocean currents in the upper 100 m of the ocean, there is a likely connection between organic carbon on aerosol particles and organic carbon observed in the SML.
To compare organic carbon concentrations found in seawater and aerosol particles, enrichment factors of the organic carbon on the submicron aerosol particles (EFaer) were determined by relating the WSOC (WISOC) observed in the aerosol particles and the DOC (POC) found in the SML to sodium as a conservative tracer for sea salt (e.g., Sander et al., 2003; Keene et al., 2007; Russell et al., 2010) applying Equations 5 and 6. Overall, the EFaer (DOC) is about 104 and the EFaer (POC) is about 105 (Table 5, Figure S6). These data represent the first EFaer (DOC) and EFaer (POC) obtained on the basis of concerted measurements, i.e., from simultaneous measurements of the SML and submicron aerosol particles in the marine environment. Figure 6 illustrates that the enrichment of DOC and POC detected in the aerosol particles is orders of magnitude higher than the enrichment of DOC and POC in the SML. The concentrations of DOC and POC detected in the SML, however, are strongly dependent on the thickness of the SML sampled and thus the sampling technique. By use of the described sampling method involving a glass plate, SML thicknesses of 83–242 µm were determined (Table 4). It is likely that the thinner the sampled SML, the higher the DOC/Na and POC/Na ratios, so that methods yielding thinner SML samples would result in smaller EFaer for DOC and POC.
Figure 6
The enrichment of DOC and POC in the SML and in aerosol particles. Comparison of the seawater and aerosol ratios of DOC (WSOC) to sodium (Na+) (a) and POC (WISOC) to Na+(b) to illustrate their enrichments in SML and aerosol particles at different wind and chl-a conditions. The secondary organic aerosol (SOA) arrow in (a) illustrates that additional DOC originating from the gas phase is likely to contribute to the concentrations measured on aerosol particles. DOI: https://doi.org/10.1525/elementa.225.f6
Data reported in the literature for enrichment factors of organic carbon in aerosol particles have been obtained mostly in the context of laboratory experiments, often utilizing natural seawater with an artificial bubbling apparatus. As the enrichment factors for organic carbon are often reported as a total sum (without distinguishing between DOC and POC), they are designated as EFaer (OC) in the following discussion. Based on artificial bubbling experiments, EFaer (OC) values ranging between 103 and 104 but also values up to 105 have been reported (Quinn et al., 2015, and references therein). Only one study reports EFaer (OC) values for ambient marine aerosol particles, with maximal values of about 103 (Russell et al., 2010). These authors measured organic carbon on aerosol particles and combined them with data on organic carbon in seawater from the literature. Given the range of published EFaer (OC) values, the EFaer (DOC) and the EFaer (POC) values in the present study (Table 5), based on ambient marine measurements, are located at the upper end.
The results presented in Figure 6 and Table 5 demonstrate that the EFaer (DOC) values are quite uniform (104), independent of the different wind and chl-a conditions. As mentioned in the introduction, DOC can be transferred to aerosol particles via bubble bursting and via secondary production by gas-to-particle conversion (secondary organic aerosol, SOA). The contribution of SOA in the marine atmosphere, however, is highly uncertain according to a recent review, and probably significantly smaller than the direct or primary transfer mechanism via bubble bursting (Quinn et al., 2015, and references therein). From the data here, we cannot differentiate between primary and secondary sources for the WSOC on the aerosol particles and therefore indicated the additional SOA processes in Figure 6. However, the transfer of POC to WISOC is naturally related to primary, bubble bursting transfer processes. The EFaer (POC) values, as a result of POC and WISOC concentrations, exhibit a greater variability and are significantly higher at high chl-a conditions. Highest enrichment values of organic matter in aerosol particles occurring when biological activity is also high has been reported previously (O’Dowd et al., 2004; Facchini et al., 2008b). However, in other studies of mainly oligotrophic areas, such a correlation between organic matter on aerosol particles and chl-a concentration could not be observed (Russell et al., 2010; Quinn et al., 2014). Some studies report high organic aerosol concentrations in connection to dimethylsulfide (Bates et al., 2012) or to heterotrophic bacteria (Prather et al., 2013) rather than to chl-a conditions. The EFaer (POC) values obtained during this study support the hypothesis that organic carbon enrichment in marine aerosol particles is coupled to oceanic biological activity represented by chl-a conditions.
In general, on the basis of the results reported here, no strong fluctuations in bulk water and SML concentrations or in the SML enrichment of DOC and POC at different wind speeds and chl-a conditions were observed. WSOC and WISOC concentrations in the aerosol particles, however, were significantly affected by elevated chl-a concentrations. These findings, along with very high and highly variable values for EFaer (DOC) and EFaer (POC), suggest that processes within the water column selectively transport organic carbon from the bulk water to the SML and subsequently from the ocean surface into the atmosphere. Such transfer processes have been studied in detail by Schmitt-Kopplin et al. (2012), who reported a vertical enrichment of biomolecules (such as functionalized fatty acids, monoterpenes and sugars) from the surface ocean into the atmosphere by bubble-mediated processes. The surface-active molecules are selectively transported from the surface water to the bubble interface and then into the aerosol phase via sea spray. The authors concluded that there is a chemo-selective transfer of natural organic compounds from seawater to atmosphere.
More specifically, Frossard et al. (2014) found that the organic composition of aerosol particles changed while bubbling oligotrophic versus eutrophic seawater, although the average composition of seawater OM remained fairly similar. These authors suggested that the presence of a thick SML at high biological activity stabilizes the bubble film and increases bubble persistence at the seawater surface. The longer the bubbles persist, the more they potentially drain. This increase in bubble persistence time leads to a drainage of more soluble compounds, leaving the less soluble compounds enriched on the bubbles. The end result is the presence of bubble films and primary aerosol particles with a larger fraction of non-soluble compounds (Frossard et al., 2014).
The findings of the present study – that more WISOC (less soluble compounds) are present on aerosol particles at high biological activity while the seawater concentrations do not differ significantly – agree with the observations described by Frossard et al. (2014). They indicate the involvement of different transport mechanisms depending on the level of oceanic biological activity. Furthermore, the low SST registered in Bergen (compared to the other investigated regions) may also affect the organic transfer processes, as bubble bursting processes are suggested to be influenced by the SST (Lewis and Schwartz, 2004). In the future, more systematic laboratory studies are needed in order to elucidate the transfer of organic carbon with its different chemical compound classes with emphasis on clarifying the role of bubble interfaces against direct transfer from the SML.
### 3.4 Implementation of OMSSA in a source function based on wind speed and chl-a conditions
Rinaldi et al. (2013) presented a source function for the prediction of sea spray organic matter (OMSSA) in PM1 based on chl-a conditions and wind speed. This source function was developed for use at the Mace Head station at the easterly rim of the North Atlantic Ocean. As the present measurements were conducted in the Atlantic Ocean, with the majority located in the Northern hemisphere, these measurements were compared with the source function. First, the OMSSA fraction was calculated on the basis of the WISOC measurements (as shown in Figure S7), based on the assumption that primary marine organic matter is primarily water-insoluble (Rinaldi et al., 2013). The OMSSA fractions are highly variable with fractions up to 90% in the productive region of Bergen. These fractions are much higher than the low and constant (around 10%) OMSSA fractions reported by Quinn et al. (2015) in northwestern Atlantic and Pacific waters. Model calculations report overall organic fractions of submicron sea spray in the range of 0.5 to 0.7 over blooming regions (Burrows et al., 2016). However, at the Mace Head station in the northeastern Atlantic Ocean (regionally close to Bergen), Rinaldi et al. (2013) observed a maximum OMSSA fraction of 78%, and O’Dowd et al. (2015) reported that OMSSA factions can reach up to 95% of submicron mass. The comparison of the measured and the predicted OMSSA fractions is illustrated in Figure 7. It shows an apparent large scattering for low OMSSA fractions, while better agreement is achieved for high OMSSA fractions at elevated chl-a values. A strong underestimation of calculated OMSSA (approximately 30%) is found under conditions characterized by low chl-a concentrations and high wind speeds. Rinaldi et al. (2013) were able to improve the source function by using chl-a data with a time lag of one week between the chl-a and the particle OMSSA determinations. O’Dowd et al. (2015) also recently suggested that large quantities of transferable organic carbon are released during the decay phase of the bloom. However, the application of chl-a data with a time delay of seven days to the present measurements led to an even greater underestimation of OMSSA in the high OMSSA range (Figure S8). A time delay was therefore not applied to the present data. Altogether, it appears that the source function yields good results for time periods characterized by high degrees of biological activity, but fails to predict OMSSA in oligotrophic regions, especially at high wind speeds.
Figure 7
Calculated and measured organic mass fractions (OMSSA). Scatter plot showing the relation between measured WISOC concentrations (transferred to OMSSA) and the calculated OMSSA according to the parameterization by Rinaldi et al. (2013) using chl-a concentrations and wind speed conditions as input parameters. The applied equation from Rinaldi et al. (2013) is (R = 0.82): OMSSA = (56.9 × chl-a) + (–4.64 × wind speed) + 40.9. DOI: https://doi.org/10.1525/elementa.225.f7
As a next step, the issue of whether a variation of the parameters of the source function would result in a better description of the OMSSA in relation to wind speed and chl-a concentrations was investigated. To this end, several two-dimensional regression approaches were tested. The large scatter of the data, however, especially in the low chl-a concentration range, could not be represented properly in the different applied source functions. The best and quite simple function was based on a linear regression, with R2 = 0.48:
$\text{O}{\text{M}}_{\text{SSA}}=23.64+\left(35.25×\text{chl - a}\right)+\left(0.11×\text{wind speed}\right).$
(7)
Once again, the results (Figure 8) clearly reveal that OMSSA levels are at their highest at elevated chl-a conditions (OMSSA median: 78%), which is in good agreement with Rinaldi et al. (2013). However, at low chl-a conditions, the OMSSA levels occur in two distinct fractions: a medium (OMSSA median: 42%) and a low (OMSSA median: 7%) fraction. The low OMSSA fractions are again in good agreement with the results of the source function from Rinaldi et al. (2013). However, the medium OMSSA fractions are not in good agreement with the source function and apparently cannot be represented by chl-a and wind speed exclusively. This result suggests the presence of an additional OMSSA source during oligotrophic conditions. This additional source could either be biological – where chl-a, however, does not serve as an indicator – or possibly of a different nature. As mentioned in Section 3.3, other authors have observed high organic aerosol concentrations in connection to dimethylsulfide (Bates et al., 2012) or to heterotrophic bacteria (Prather et al., 2013). The medium OMSSA fractions were mainly found in the region of the Cape Verde islands (campaigns: November 2011 and 2013) under remote marine conditions, with chl-a concentrations around 0.2 µg L–1 and wind speeds of 7.6 m s–1 on average. Of particular interest to our future investigations will be a closer inspection of the nature of the medium OMSSA fraction in oligotrophic regions. Regarding the impact of wind speed, Rinaldi et al. (2013) and Gantt et al. (2011) predicted that the highest organic enrichment in sea spray aerosol particles will occur during calm winds when the ocean surface is strongly covered by the SML. Gantt et al. (2011) demonstrated that at higher wind speeds (up to 8 m s–1), the sea spray production is elevated, whereas OMSSA in aerosol particles is reduced due to the stronger mixing of SML with bulk water and the corresponding lower SML coverage. Based on the WSOC and WISOC measurements as well as the OMSSA fraction (Figure S7) obtained in this study, no inverse relation between OMSSA fraction and wind speed could be observed. Generally, the biological activity of the ocean, as reflected by chl-a conditions, had a significantly higher impact on OC transfer from seawater onto aerosol particles compared to wind speed. Source functions aimed at predicting the occurrence of organic matter on marine aerosol particles based solely on wind speed and chl-a concentration levels, however, should be taken with caution when investigating different regions of the Atlantic Ocean. Certainly, other meteorological parameters besides wind speed, as well as biological parameters beyond chl-a concentration, affect organic matter on marine aerosol particles, especially in oligotrophic regions.
Figure 8
Plot of the OMSSA source function. Plot of the two-dimensional regression analysis for OMSSA based on chl-a concentrations and wind speed. Function of best fit (R2 = 0.48): OMSSA = 23.64 + (35.25 × chl-a) + (0.11 × wind speed). The upper limit of OMSSA in the source function was set to 91% (corresponding to the maximum OMSSA values observed), while the lower limit was set to 0%, as some combinations of chl-a levels and wind speed lead to negative OMSSA values. The color bar indicates % OMSSA; the graph clearly illustrates a threefold division of the OMSSA fractions: high OMSSA (OMSSA median: 78%) at high chl-a conditions, and medium OMSSA (OMSSA median: 42%) and low OMSSA (OMSSA median: 7%) at low chl-a conditions. DOI: https://doi.org/10.1525/elementa.225.f8
## 4 Summary and conclusions
The results of measurements of DOC and POC concentrations and enrichment factors in the SML, as well as WSOC and WISOC measurements in marine aerosol particles, were presented. The concentrations of DOC and POC in seawater (bulk water and the SML) tended to be relatively uniform across the particular regions of the North and Central Atlantic Ocean. Organic carbon was enriched in the SML by a factor of two on average, following a pattern that may be explained by a surface saturation effect (higher enrichment at lower concentrations). The impact of different wind speeds and varying degrees of biological activity, as indicated by chl-a concentration levels, on DOC and POC concentrations as well as enrichment factors were observed to be minor. However, a thicker surface film was observed at elevated chl-a concentration levels, which may be due to surface-active species assumed to represent only a small fraction of DOC and POC. In contrast to organic carbon measured in seawater, the organic carbon levels observed in aerosol particles were significantly affected by varying levels of chl-a.
On the basis of the present ambient concerted measurements, enrichment factors for DOC and POC in ambient marine aerosol particles, hitherto not reported in the literature, were obtained. These enrichment factors fall at the upper end of the range of values reported in the framework of experimental studies using artificial bubbling devices in the laboratory and in the field. This finding indicates that the transfer of organic carbon from the SML into the aerosol phase in the real-world marine environment is consistent with processes described in laboratory studies.
The results obtained in the present study support the thesis that elevated local biological activity (as indicated by chl-a concentrations) enhances (primary) organic carbon concentration on aerosol particles. These field results furthermore support the thesis of a chemo-selective transfer of organic compound groups, which has been reported in the framework of studies using artificial bubbling devices. However, a prediction of the WISOC measurement results (transferred to OMSSA) by source functions based on wind speed and chl-a levels showed a strong underestimation of OMSSA in oligotrophic regions, especially at high wind speeds. For the North and Central Atlantic Ocean there may be additional parameters – either connected to biological factors not served by chl-a level as an indicator or to factors of a different nature – that must be taken into consideration in order to accurately predict the organic fractions on aerosol particles.
In the future, a more detailed investigation of the chemical nature of DOC and POC in seawater and aerosol particles is required in order to identify suitable tracers for primary and secondary organic aerosols. Moreover, a more comprehensive classification of “clean marine aerosol particles” is needed. Besides characterization of the chemical nature, isotopic analysis can help to differentiate marine and continental sources. In addition, further studies on suitable proxies for oceanic biological activity are needed to describe the organic composition of seawater. Burrows et al. (2014) recently introduced a new concept aimed at predicting the mass of organic matter in aerosol particles based on their different physical and chemical properties. Representative model compounds as suggested by Burrows et al. (2014) should be measured in ambient marine samples.
## Data Accessibility Statement
The datasets will be available in the PANGAEA database.
## Acknowledgements
The authors would like to acknowledge the captain and crew of the Polarstern ANT-XXVIII/5 2012 (BSH20120081) and Polarstern ANT-XXIX/10 2014 (BSH20120320) cruises as well as the organizers of the Bergen 2011 campaign at the Raune Fjord; Kerstin Lerche for chlorophyll-a analysis, René Rabe, Marina Voyevoda, Susanne Fuchs, Anett Dietze and Anke Roedger for technical assistance. Finally, the authors thank the anonymous reviewers for their valuable comments and suggestions.
## Funding information
The authors would like to thank the Bundesministerium fuer Bildung und Forschung (BMBF) (German Federal Ministry of Education and Research) for funding within the SOPRAN project (FKZ03F0662-J) and the Leibniz Association (OCEANET project in the framework of PAKT).
## Competing interests
The authors have no competing interests to declare.
## Author contributions
• MvP and HH designed and organized the field campaigns.
• MvP, KWF and KM performed the field campaigns and contributed to the acquisition of the data.
• SB contributed to the regression analysis.
• WvT organized and conducted analysis of biological tracers.
• MvP and HH performed the data interpretation and wrote the paper with input from all authors.
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http://physics.stackexchange.com/questions/48543/do-some-half-lives-change-over-time/48544 | Do some half-lives change over time?
I was recently doing some physics tuition on radioactivity and the student claimed her chemistry teacher had said that radioactive substances can be grouped into two divisions: those whose half-life is constant and those whose half-life changes over time.
I had never heard of this before and can't think of any reason why a half-life should change, so does anyone else know anything about this?
(I know some half-lives can be altered under certain conditions, but I'm talking about a natural change over time).
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The short answer is no: halflives are constant.
However, let's discuss a situation in which that comment might have some kind of truth behind it. If you have a parent nucleus that decays to a radioactive daughter so that there will be two (or more) decays before stability. In general there are two possibilities for this:
• The daughter has a shorter halflife than the parent. In this case the concentration of the daughter is always $\displaystyle\frac{\tau_\text{daughter}}{\tau_\text{parent}}$ of the parent concentration. This means that the concentration of the daughter actually decays on the parent's halflife (because the daughter is constantly refreshed from the parent).
• The daughter has a longer half life than the parent. In this case the daughter will accumulate steadily as the parent decays away.
The latter case is interesting to us here because at the start the sample will register an activity that decays with the parent's (short) halflife, but after a number of those halflifes have passed the activity of the sample will be dominated by the daughter and exhibit a longer halflife.
That is something that your instructor could have meant which would not be wrong. However, the halflife of each isotope remains the same: it is only the halflife of the sample (which contains more than one isotope) that varies.
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Thanks, OP here (sorry, couldn't remember password, had to sign in under a different name). I did consider what you mention, but it doesn't seem to fit particularly well with having two distinct categories of radionuclide. I suspect the answer is one or both of a) the student's misremembering what the teacher said and/or b) the teacher talking nonsense. Wouldn't be the first time... – James Jan 9 '13 at 14:48
There have been reports of annual modulation of radioactive decay rates in certain elements. That is, a change in decay rates that depends (apparently) on the position of the Earth around the sun. Here is a fairly recent example (disclaimer: I can't get behind the paywall at the moment). The effect is very small and at the moment there is no consensus on the cause. It is probably a systematic error in the measurements themselves. A more exotic (and unlikely) possibility is new physics.
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The following may have little practical use (at least so far), but it seems very interesting to me that, strictly speaking, the exponential law (and, therefore, constant half-life) is incompatible with quantum mechanics (this is an old result by Khalfin). It is very difficult to observe the deviations from the exponential law though. Some details and a reference to Khalfin's work can be found in Nature vol. 335, p. 298 (22 September 1988).
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Is this the same work refernced in Long time deviations from exponential decay in radioactivity? – dmckee Jan 9 '13 at 11:04
Yes. There is also a somewhat earlier work - L. A. Khalfin (1957, Doklady Akad. Nauk USSR, 115, 277) - the original paper in Russian (L.A.Khalfin, Sov.phys.Doklady 2, 340 (1957) - the English translation). – akhmeteli Jan 9 '13 at 11:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8503449559211731, "perplexity": 670.7858029977699}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644064590.32/warc/CC-MAIN-20150827025424-00310-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://carlbrannen.wordpress.com/category/gravity/ | # Category Archives: gravity
## Quantization of event horizon radius and Quasar Redshifts
I’m getting ready for the FFP10 meeting later this month. In reading the abstracts of those who will be giving talks or posters, I came upon “Analyses of the 2dF deep field” by Chris Fulton, Halton Arp and John G. Hartnett. The abstract is about the relationship between low redshift and high redshift astronomical objects. The claim is that some quasars have redshifts that do not give their true distance; instead, they are much closer. Looking on arXiv finds: The 2dF Redshift Survey II: UGC 8584 – Redshift Periodicity and Rings by Arp and Fulton.
If these high and low redshift objects actually are related, this places doubt on the Hubble relation. In addition, when low and high redshift objects appear to be related, their redshifts are related by quantum values . From observations, Arp has proposed that quasars evolve from high to low redshift, and finally become regular galaxies.
Now for quasars to have redshifts that differ from their true distances implies that their redshifts are determined gravitationally; that is, what we are seeing is partly the redshift of light climbing out of a gravitational potential. And if these redshifts are quantized, this gives a clue that the structure inside the event horizon of a black hole is not a simple central singularity but instead there must be repetitive structure.
In a classical black hole, the region inside the event horizon can only be temporarily visited by regular matter. Even light cannot be directed so as to increase its radius in this region. Let’s refer to this region as the “forbidden region” of the black hole as it is near the central singularity. For the classical black hole, this includes everything inside the event horizon. We will be considering the possibility that the forbidden regions of a black hole occur as infinitesimally thin shells, and that between these shells, light can still propagate outwards:
Forbidden regions shown in red.
Event Horizons as Quantum amplitudes
If we were looking for a quantum mechanical definition of the inside of a black hole, we could define it as the region where particles have a zero probability of moving outwards. We could say that the transition probability for the particle moving outwards is zero. However, in quantum mechanics probabilities are defined as the squared magnitudes of complex amplitudes. The way we compute transition probabilities is from complex transition amplitudes. If the transition amplitude between two states is zero, we say that they are “orthogonal”. Zero transition amplitudes correspond to points where a sine wave is zero; at these points, deviations to either side give nonzero transition amplitudes:
How to get zero probabilities from nonzero in QM.
Filed under gravity, particle physics, physics
## My Gravity paper accepted for publication
I’ve just got notice that my gravity paper, titled The force of gravity in Schwarzschild and Gullstrand-Painleve coordinates has been accepted for publication in the International Journal of Modern Physics D, with only a very minor modification.
I’m kind of surprised by this, given that the paper proposes a new theory of gravity. I was expecting to have that portion excised.
And to help make a week more perfect, my paper for Foundations of Physics, titled Spin Path Integrals and Generations, got a good review along with a nasty one (and much good advice from both), and the editor has asked for me to revise the manuscript and resubmit. So I suppose this paper will also eventually be published. I’m a little over half finished with the rewrite. This paper is, if anything, even more radical than the gravity paper.
Finally, the Frontiers of Fundamental and Computational Physics conference organizers have chosen my abstract (based on the Foundations of Physics paper) for a 15 minute talk. The title is Position, Momentum, and the Standard Model Fermions. Marni Sheppeard (my coauthor for a third paper, “The discrete Fourier transform and the particle mixing matrices” which so far is having some difficulty getting published), is giving a related talk, Ternary logic in lepton mass quantum numbers immediately following mine.
So all in all, I am a very lucky amateur physicist
Filed under gravity, heresy, particle physics, physics
## Dark Flow, the Speed of Gravity, and the CMB
Kashlinsky, Atrio-Barandela, Kocevski, and Ebeling have just put out a preprint on the peculiar motions of galactic clusters: A measurement of large-scale peculiar velocities of clusters of galaxies: results and cosmological implications. In short, they claim that all galactic clusters appear to have a motion with respect to the cosmic microwave background (CMB). The motion of a galactic cluster slightly effects the energy of the microwave radiation that travels through it, so they use the temperature map of the CMB to determine the velocity of those galactic clusters.
And the result is that the whole (observable) universe appears to be moving with respect to the CMB. This was not expected because the observable universe is approximately isotropic and so shouldn’t be going anywhere. They write (in the abstract):
This flow is difficult to explain by gravitational evolution within the framework of the concordance LCDM model and may be indicative of the tilt exerted across the entire current horizon by far-away pre-inflationary inhomogeneities.
However, the tilt is easy to explain when you assume that the speed of gravity is larger than C: If gravitational interactions travel faster than light, you will automatically be able to feel the gravitational attraction of matter even if it is too far away for you to see.
Filed under gravity, physics
## Lorentz Violation and Feynman’s Checkerboard Model
Lubos Motl brings to our attention a paper by Ted Jacobson and Aron C. Wall on black hole theremodynamics and Lorentz invariance, hep-ph/0804.2720 and claims that theories that violate Lorentz invariance are ruled out because they will also violate the second law of thermodynamics, the law that requires that entropy never decreases. Lubos concludes, “At any rate, this is another example showing that the “anything goes” approach does not apply to quantum gravity and if someone rapes some basic principles such as the Lorentz symmetry or any other law that is implied by string theory, she will likely end up not only with an uninteresting, ugly, and umotivated theory but with an inconsistent theory.” I disagree with this.
First, the abstract of the article:
Recent developments point to a breakdown in the generalized second law of thermodynamics for theories with Lorentz symmetry violation. It appears possible to construct a perpetual motion machine of the second kind in such theories, using a black hole to catalyze the conversion of heat to work. Here we describe the arguments leading to that conclusion. We suggest the implication that Lorentz symmetry should be viewed as an emergent property of the macroscopic world, required by the second law of black hole thermodynamics.
From the abstract, we see that Lubos has put the cart in front of the horse. Rather than proving that Lorentz symmetry has to be exact “all the way down”, the authors instead say that Lorentz symmetry does not have to be present at the foundations of elementary particles because it will automatically emerge macroscopically as a result of requiring that the second law of thermodynamics apply to black holes. And I agree wholeheartedly with this.
Filed under gravity, heresy, physics
## The Painleve Equations of Motion
In the general theory of relativity, the orbits are given by geodesics. A geodesic is a path that extremizes the path length. The path length is defined as the integral of $ds$ over the path, where $ds^2$ is the metric. For the case of Painleve coordinates on the Schwarzschild metric, $ds^2$ is given by:
. Let’s let our path start at time t=0 and end at time t=1. For the path to be a geodesic, we must extremize the following integral (I’ll quickly sneak in a minus sign to make the path be timelike instead of spacelike):
To make life easier for us, we will make the assumption that the orbital motion is in the $\theta = \pi/2$ plane so there’s no $\theta$ dependence. That turns the angular part of the square root into $r^2\;(d\phi/dt)^2$. Furthermore, since the simulation is going to use Cartesian, (x,y) coordinates, we might as well replace $r^2\;(d\phi/dt)^2$ with $(x\;dy/dt-y\;dx/dt)^2$, and $dr/dt$ with $x\;dx/dt + y\;dy/dt$, their Cartesian equivalents. And put M=1, we can always fix it later by dimensional analysis.
Filed under gravity, physics
## Painleve Coordinates
In the previous post, we took a tour through the literature and found that when general relativity is translated into the elegant mathematical language of the geometric algebra, the natural coordinates for a black hole turn out to be Painleve or Gullstrand-Painleve coordinates instead of the more common Schwarzschild coordinates. Our next post will derive the equations of motion for orbits in this coordinate system, but before we get into the difficult mathematics, we should take a quick look at the Painleve coordinates.
First of all, most of my readers will know that in general relativity, the choice of coordinates is quite arbitrary. Both Schwarzschild and Painleve coordinates describe the same object, the gravity field of a gravitating object which is spherically symmetric (and therefore non rotating), i.e. they are all descriptions of the black hole. In this sense they correspond to the same solution to Einstein’s field equations, which is sometimes called “Schwarzschild’s Solution”, or the “Schwarzschild Metric”. This is a little confusing, “Schwarzschild” was the person who found the Schwarzschild metric and he found it using the Schwarzschild coordinates, so his name is used twice here.
I guess I should put a pretty picture from the gravity simulation that resulted from all this here so it will show above the fold. This is a set of “knife edge” orbits, that is, orbits that quite nearly fall into the black hole but do not. Due to the time spent near the black hole, whose event horizon is marked in gray, the test masses get huge precession:
We will be discussing the less pretty, but more mathematical subject of Painleve and Schwarzschild coordinates in this post.
1 Comment
Filed under gravity, physics
## General Relativity, Painleve and QFT
The first problem in writing gravitation as a particle interaction is the fact that QFT works best on flat space, while general relativity is almost always written in arbitrary coordinates. One of the underlying principles of general relativity is that coordinates shouldn’t matter (background independence), so this problem appears to be a deep one. The point of view we will take here is that of the “new physics” .
That is, we will treat the equations of the old physics with more respect than we treat their theories. Consequently, instead of chasing after will-o-the-wisps like background independence, we will instead search for a method of writing general relativity using the mathematical tools of quantum field theory. Very fortunately for us, that method has already been found; it is the gauge theory of gravity discovered by the Cambridge Geometry Algebra Research Group. The purpose of this post is to introduce the theory to those who have not yet been exposed to it, and to note that this gravity theory (which is identical to GR so long as you restrict your attention to stuff that happens outside of the event horizons of black holes) picks out Painleve coordinates as a natural flat space (and therefore QFT compatible) coordinate system for a non rotating black hole.
Those with a graduate education in physics are already familiar with the Geometric Algebra (GA) in that it is equivalent to the Gamma matrices used throughout quantum field theory. So a gravitation theory that is equivalent to general relativity, but is written with gamma matrices, is a natural starting point for a search for a unified field theory.
The primary proponent for the use of GA in physics (outside of QFT) is David Hestenes, who applied it to classical and quantum mechanics. As the introduction to GA article at the Cambridge Geometry group’s website puts it:
We believe that there are two aspects of Hestenes’ work which physicists should take particularly seriously. The first is that the geometric algebra of spacetime is the best available mathematical tool for theoretical physics, classical or quantum. Related to this part of the programme is the claim that complex numbers arising in physical applications usually have a natural geometric interpretation that is hidden in conventional formulations. David’s second major idea is that the Dirac theory of the electron contains important geometric information, which is disguised in the conventional matrix based approaches.
Now that’s a pretty big claim: that geometric algebra is the best mathematical tool for all physics. I will spend the rest of this post exploring this claim in the case of general relativity, and then tracing the consequences for a unified field theory. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 10, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8582367897033691, "perplexity": 454.77983864262274}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891817523.0/warc/CC-MAIN-20180225225657-20180226005657-00147.warc.gz"} |
https://mathoverflow.net/questions/104645/descartes-rule-of-signs-for-a-noncommutative-polynomial-in-matrix-variables | # Descartes rule of signs for a noncommutative polynomial in matrix variables
Recently, while studying certain notions of "averaging" a set of input matrices, I obtained a nonlinear polynomial in matrix variables. A simple example is
\begin{equation*} \mathcal{G}(X) := X^n - \sum_{i=0}^{n-1} (C_iX^i + X^iC_i), \end{equation*}
where the $C_i$ are symmetric positive definite matrices.
If all the terms above were scalars, then Descartes' Rule of Signs tells us that $\mathcal G$ has exactly one positive root. This led me to wonder if a similar "rule" is also known for the above case.
Does there exist a unique symmetric positive definite solution to $\mathcal{G}(X)=0$?
EDIT If the $C_i$ commute with each other, we can simplify matters as follows. Since the $C_i$ commute, they can be simultaneously diagonalized by the same orthogonal matrix $U$; thus, let $C_i=U\Lambda_i U^T$. Suppose for a moment that $\mathcal{G}(X)=0$ does have a solution. Then, pre and post multiplying by $U^T$ and $U$, respectively, we see that this solution must satisfy
\begin{eqnarray*} U^TX^nU &=& \sum_{i=0}^{n-1} (U^TC_iUU^TX^iU + U^TX^iUU^TC_iU),\\\\ Y^n &=& \sum_{i=0}^{n-1} (\Lambda_iY^i + Y^i\Lambda_i), \end{eqnarray*} where $Y=U^TXU$. Now, if we pick $Y$ to be diagonal, then we see that indeed, for each diagonal entry we have a separate polynomial that has a unique positive root. Hence, we have a unique diagonal matrix $Y$. But as Mark Sapir alerted me in a comment below, it seems that having a unique diagonal $Y$ (and thus possibly non-diagonal $X=UYU^T$), does not yet rule out the possibility of other solutions.
Update After some hours of struggle due to the pressure of having posted my question on MO, under the additional assumption that each term in the sum is positive definite, the answer is "yes". But this assumption seems to be way too strong...
• This may be a stupid question, but while it is obvious that $\mathcal{G}(X)$ is symmetric when $X = I,$ it is not so clear it is ever symmetric otherwise, unless the system is very special. Am I missing something? – Igor Rivin Aug 13 '12 at 21:53
• @Igor: I think $X$ is supposed to be symmetric too. Then, since $C_i$ are symmetric, the sum is also symmetric. – user6976 Aug 13 '12 at 21:59
• Aha, I see, I was careless reading the equation! – Igor Rivin Aug 13 '12 at 23:15
• @Suvrit: Could you explain why if all $C_i$ pairwise commute, you can diagonalize everything? You cannot assume that $X$ commutes with $C_i$. – user6976 Aug 14 '12 at 1:14
• @Mark: Hmm...if all the $C_i$ commute, we diagonalize them, and then we see that there exists a unique diagonal matrix that satisfies the equation. But you are right, just because we have a unique diagonal matrix, does not mean that the solution to the original equation is unique. I'll edit the question to work in these details. Thanks. – Suvrit Aug 14 '12 at 7:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9667109251022339, "perplexity": 248.6716062750115}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046152236.64/warc/CC-MAIN-20210727041254-20210727071254-00236.warc.gz"} |
https://www.kybernetika.cz/content/2005/6/713 | František Rublík
# Abstract:
The two-sample Lepage test, devised for testing equality of the location and scale parameters against the alternative that at least for one of the parameters the equality does not hold, is extended to the general case of $k>1$ sampled populations. It is shown that its limiting distribution is the chi-square distribution with $2(k-1)$ degrees of freedom. This $k$-sample statistic is shown to yield consistent test and a formula for its noncentrality parameter under Pitman alternatives is derived. For some particular alternatives, the power of the $k$-sample test is compared with the power of the Kruskal-Wallis test or with the power of the Ansari-Bradley test by means of simulation estimates. Multiple comparison methods for detecting differing populations, based on this multisample version of the Lepage test or on the multisample version of the Ansari-Bradley test, are also constructed. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9179932475090027, "perplexity": 527.7666803413639}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00368.warc.gz"} |
https://chemistry.stackexchange.com/questions/7705/explaining-the-oxidation-states-of-palladium | # Explaining the oxidation states of palladium
Palladium follows an exception to normal electron filling-up rules and so has the electron configuration of $\ce{[Kr] 4d^{10} 5s^0}$
The oxidation states of palladium are $\ce{+II}$ and $\ce{+IV}$. How can these oxidation states be explained by looking at the special electron configuration of $\ce{Pd}$?
Ions of the transition metals do not often occur "naked" the way one might think of a sodium ion. Of course, sodium ions cannot occur by themselves either, but let's compare the two for a moment.
Sodium nitrate, $\ce{NaNO3}$
When $\ce{NaNO3}$ dissolves in water, the two ions dissociate.
$$\ce{NaNO3 -> Na+(aq) + NO3- (aq)}$$
Both ions are solvated by the solvent. The water molecules (which are polar) surround the ions. The partial charges due to the dipoles in the water molecules stabilize the charges on the ions. However, the interactions between the sodium ions and the water molecules are fleeting. These interactions are not strong enough to be consider "bonds" (there is no covalent electron sharing between water and sodium ion). Thus, we often consider sodium ions in solution (despite their solvation) to be "just" sodium ions.
Palladium nitrate, $\ce{Pd(NO3)2}$
Transition metal ions are different. They tend to form complex ions by coordinating solvent molecules. These dative interactions are far stronger than the ion-dipole interactions in solvated sodium ions.
Palladium nitrate does dissociate, but the palladium ion picks up water molecules that it holds tightly. Thus, we do not worry about the electron configuration of $\ce{Pd^{2+}}$, since it never exists on its own. It is always part of a complex ion or coordination compound, and we care about the electron configuration of that complex ion.
$$\ce{Pd(NO3)2 + nH2O -> Pd(H2O)6^{2+}(aq) + 2NO3- (aq)}$$
• But he didn't say solvated in water- Of course the ion in salt going to have some covalent nature. – user2617804 Jan 9 '14 at 6:29
• The OP did not say solvated in water, but then there will always be a counterion. – Ben Norris Jan 9 '14 at 13:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8250565528869629, "perplexity": 1386.0904820430767}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250592261.1/warc/CC-MAIN-20200118052321-20200118080321-00133.warc.gz"} |
http://www.distributome.org/js/exp/FExperiment.html | X Distribution graph Y Distribution graph F Distribution graph
#### Description
The experiment simulates a random sample $$\boldsymbol{X} = (X_1, X_2, \ldots, X_m)$$ of size $$m$$ from the normal distribution with mean $$\mu$$ and variance $$\sigma^2$$, and a random sample $$\boldsymbol{Y} = (Y_1, Y_2, \ldots, Y_n)$$ of size $$n$$ from the normal distribution with mean $$\nu$$ and standard deviation $$\tau$$. The samples are independent. Random variable $$V$$ is
$V = \frac{S_X^2 / \sigma^2}{S_Y^2 / \tau^2}$
where $$S_X^2$$ and $$S_Y^2$$ are the sample variances of $$\boldsymbol{X}$$ and $$\boldsymbol{Y}$$, repsectively. Random variable $$V$$ has the $$F$$ distribution with $$m$$ degrees of freedom in the numerator and $$n$$ degrees of freedom in the denominator. On each run, the applet shows the normal samples in the first two graphs. The $$F$$ distribution is shown in the last graph and in the second table. As the experiment runs, the empricial density and moments are shown in this graph and recorded in this table. The parameters $$\mu$$, $$\sigma$$, $$\nu$$, $$\tau$$, $$m$$ and $$n$$ can be varied with the input controls. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9771977066993713, "perplexity": 92.82277287099458}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917123270.78/warc/CC-MAIN-20170423031203-00293-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://www.cs.technion.ac.il/users/wwwb/cgi-bin/tr-info.cgi/1994/CS/CS0799 | # Technical Report CS0799
TR#: CS0799 Class: CS Title: A PARALLELIZABLE INCOMPLETE LU-TYPE PRECONDITIONER FOR THE SOLUTION OF SPARSE LINEAR SYSTEMS. Authors: Y. Shapira, A. Sidi and M. Israeli PDF Not Available Abstract: An algebraic condition ensuring stability of the ILU (1,1) decomposition of sparse matrices is given. A stability and convergence theory for some multi-dimensional recursions that are relevant to the ILU preconditioning method for the solution of sparse linear systems is presented. Relying on this theory, a parallelizable truncated ILU (PTILU) preconditioning method is developed. Numerical experiments show that for grids of size up to 160 \times 160, with 8 \times 8 subdomains, the amount of arithmetic operations of PTILU is very similar to that of standard ILU, no more than 3 times larger when implemented as a modification of Row-Sum ILU (RSILU) and no more that twice larger when implemented as a modification of Alternating Direction Implicit (ADI) methods. In addition, PTILU is applicable as a smoother in multigrid methods. Copyright The above paper is copyright by the Technion, Author(s), or others. Please contact the author(s) for more information
Remark: Any link to this technical report should be to this page (http://www.cs.technion.ac.il/users/wwwb/cgi-bin/tr-info.cgi/1994/CS/CS0799), rather than to the URL of the PDF or PS files directly. The latter URLs may change without notice. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8614874482154846, "perplexity": 1598.5857035053061}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637900029.49/warc/CC-MAIN-20141030025820-00090-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://www.techtimes.com/articles/145249/20160330/scientists-detect-first-possible-signature-of-dark-matter-annihilation-in-the-milky-way.htm | # Scientists Detect First Possible Signature Of Dark Matter Annihilation In The Milky Way
Close
This decade has been fascinating for astrophysics because of dramatic discoveries such as cosmic acceleration, exoplanets and the detection of gravitational waves from merging black holes. However, none of these breakthroughs have been as challenging and surprising as the identification of dark matter.
Most people think that the entire universe is comprised of ordinary tangible matter, or those that are made up of atoms. Scientists say that more than 80 percent of the universe is comprised of dark matter, which cannot be observed directly, but has a mass and can affect normal matter.
Various theories have been formulated about what comprises dark matter, and the common belief is that besides gravity, it shares another property with general matter.
Dark matter also comes in two forms - regular matter and anti-matter. When these two collide, they annihilate, or destroy each other. In the process, energy is conserved and a new high energy particle like a photon or gamma-ray is formed.
In a recent study, published in the journal Physics of the Dark Universe, researchers have recognized one such signature of dark matter annihilation.
The physicists studied the spatial distribution of gamma-ray emission, particularly in the Milky Way's Galactic Center. This region has higher matter density and is relatively nearby.
Should dark matter annihilation occur in this area, the location is expected to become bright. Indeed, scientists have observed a large gamma-ray signature that extends beyond hundreds of light years.
There are other possible sources of gamma-ray, such as, a large number of rapidly spinning pulsars emitting electromagnetic radiation. Hence, the scientists revisited some earlier observations and applied new data reduction methods in order to better assimilate the location of gamma-ray emissions.
The findings demonstrated that the distribution of the radiation works better with models of dark matter annihilation than with pulsar models. If confirmed, the study can be used as evidence about the mysterious dark matter.
TAG
#### Most Popular
Earth/Environment
Public Health
Space
Animals
Animals | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9514099955558777, "perplexity": 984.9775763245913}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863109.60/warc/CC-MAIN-20180619173519-20180619193519-00603.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/bent-wire-shown-figure-lies-uniform-magnetic-field-straight-section-24-m-long-makes-angle--q973357 | ## Magnetic Force on a Current-Carrying Wire
The bent wire shown in the figure below lies in a uniform magnetic field. Each straight section is 2.4 m long and makes an angle of ? = 60° with the x axis, and the wire carries a current of 2.0 A.
(a) What is the net magnetic force on the wire in unit-vector notation if the magnetic field is 4.0T?
magnitude ____N direction ______ (x, y, z, -x, -y, -z, or Force is zero).
(b) What is the net magnetic force on the wire in unit-vector notation if the magnetic field is 4.0 T?
magnitude ___N direction _____ (x, y, z, -x, -y, -z, or Force is zero). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9503851532936096, "perplexity": 680.3081299525428}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368697843948/warc/CC-MAIN-20130516095043-00058-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://piping-designer.com/index.php/disciplines/electrical/1015-current-electrical | # Current (Electrical)
Written by Jerry Ratzlaff on . Posted in Electrical
Current is the rate of flow of electricity in a circuit, measured in amperes.
Amp is a unit of current. One ampere (amp) is the current flowing through one ohm of resistance at one volt potential.
## formula
$$Amps = \frac{Volts}{Ohms}\;\; = \;\;I = \frac{V}{R}$$
$$Amps = \frac{Watts}{Volts}\;\; = \;\;I = \frac{P}{V}$$
$$Amps = \sqrt{\frac{Watts}{Ohms}}\;\; = \;\;I = \sqrt{\frac{P}{R}}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9809167981147766, "perplexity": 3160.399085362914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823229.49/warc/CC-MAIN-20171019050401-20171019070401-00422.warc.gz"} |
https://www.physicsforums.com/threads/spivaks-calculus-5-x-use-ix-backwards.263468/ | # Spivak's Calculus, 5(x) - Use (ix) backwards
1. Oct 11, 2008
### Tasaio
Spivak's Calculus, 5(x) -- "Use (ix) backwards..."
1. The problem statement, all variables and given/known data
Prove the following:
(x) If $$a,b\geq0$$ and $$a^{2}<b^{2}$$, then $$a<b$$. (Use (ix), backwards.)
2. Relevant equations
(ix) If $$0 \leq a<b$$, then $$a^{2}<b^{2}$$.
3. The attempt at a solution
Suppose $$a,b\geq0$$ and $$a^{2}<b^{2}$$.
Here's my problem. What does "Use (ix) backwards" mean? I'll assume he means to use the converse of (ix). In that case...
The converse of (ix):
$$\neg(a^{2}<b^{2})\rightarrow\neg(0\leq a<b)$$
Hence $$(a^{2}\geq b^{2})\rightarrow\neg(0\leq a\&\&a<b)$$;
hence $$(a^{2}\geq b^{2})\rightarrow(0>a)\Vert(a\geq b).$$ $$(\star)$$
Since $$a^{2}<b^{2}$$, then $$a^{2} \leq b^{2}$$. So $$b^{2} \geq a^{2}$$.
Then by $$(\star)$$, $$(0>b)\Vert(b \geq a)$$.
Since $$b \geq 0$$, then we know $$0>b$$ cannot be true.
This means that $$b\geq$$ a must be true.
But if $$b=a$$, then $$b^{2}=a^{2}$$; this is a contradiction since we are given that $$a^{2}<b^{2}$$.
Hence b>a must be true.
Hence a<b.
2. Oct 11, 2008
### HallsofIvy
Re: Spivak's Calculus, 5(x) -- "Use (ix) backwards..."
What you have done is perfectly good. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.993353009223938, "perplexity": 2091.4736183226055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647707.33/warc/CC-MAIN-20180321234947-20180322014947-00708.warc.gz"} |
https://www.physicsforums.com/threads/probability-measure-on-smooth-functions.435404/ | # Probability measure on smooth functions
1. ### Tac-Tics
810
Is there a "standard" probability measure one would use for the set of smooth real-valued functions on [a, b]?
My intuition is picturing a setup where you cut out shapes in the x-y plane, and then the set of functions whose graphs are contained in that shape have a measure proportional to the Euclidean area of the shape. But I can't quite make that intuition exact.
2. ### Eynstone
336
Do you have the Borel measure ( under the sup metric ) in mind?
3. ### simeonsen_bg
9
I suppose, you have to consider functions uniformely bounded by some constant M (or even vith uniformely bounded variation?), otherwise the whole set gets infinite measure, not 1, the way you described the measure. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9906584620475769, "perplexity": 676.4671663412092}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398447729.93/warc/CC-MAIN-20151124205407-00303-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.physics.wisc.edu/events/?date=2017-09-26&name=HE%2FCosmo | # Theory Seminar (High Energy/Cosmology)
<< Summer 2017 Fall 2017 Spring 2018 >>
Subscribe your calendar or receive email announcements of events
### Events on Tuesday, September 26th, 2017
Weak gravity conjecture, Multiple point principle and standard model landscape
Time: 3:30 pm
Place: 5280 Chamberlin
Speaker: Yuta Hamada, University of Wisconsin-Madison & KEK Theory Center
Abstract: The requirement for an ultraviolet completable theory to be well-behaved upon compacti cation has been suggested as a guiding principle for distinguishing the landscape from the swampland. Motivated by the weak gravity conjecture and the multiple point principle, we investigate the vacuum structure of the standard model compacti ed on S^1 and T^2. The measured value of the Higgs mass implies, in addition to the electroweak vacuum, the existence of a new vacuum where the Higgs eld value is around the Planck scale. We explore two- and three-dimensional critical points of the moduli potential arising from compacti cations of the electroweak vacuum as well as this high scale vacuum, in the presence of Majorana/Dirac neutrinos and/or axions. We point out potential sources of instability for these lower dimensional critical points in the standard model landscape. We also point out that a high scale AdS_4 vacuum of the Standard Model, if exists, would be at odd with the conjecture that all non-supersymmetric AdS vacua are unstable. We argue that, if we require a degeneracy between three- and four-dimensional vacua as suggested by the multiple point principle, the neutrinos are predicted to be Dirac, with the mass of the lightest neutrino O(1-10) meV, which may be tested by future CMB, large scale structure and 21cm line observations. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.824855625629425, "perplexity": 1640.6035546321784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00359.warc.gz"} |
http://lmfa.ec-lyon.fr/~lmfa/spip.php?article1156&lang=fr | # Laboratoire de Mécanique des Fluides et d’Acoustique - UMR 5509
LMFA - UMR 5509
Laboratoire de Mécanique des Fluides et d’Acoustique
Lyon
France
## Nos partenaires
Article dans J. Non-Newtonian Fluid Mech. (2015)
## Experimental determination of the viscosity at very low shear rate for shear thinning fluids by electrocapillarity
Mohamed Hatem Allouche, Valéry Botton, Daniel Henry, Séverine Millet, R. Usha, Hamda Ben Hadid
Non-Newtonian fluids can present a complex rheological behaviour involving shear-thinning, viscoelastic or thixotropic effects. We focus here on the characterization of generalized Newtonian fluids as shear-thinning fluids. Rotational rheometers are torque-sensitive, which makes the measurement of viscosity at low shear-rates very difficult, in particular for low viscosity fluids. An accurate knowledge of this value is, however, particularly important for the characterization of several types of flows, in particular those featuring a free surface, a symmetry axis or a symmetry plane. The experimental determination of viscosity and surface tension is enabled from the propagation of attenuated capillary waves. An optical technique has been implemented in order to determine the shear-thinning viscosity at values of the shear-rate as small as $10^{−3} \mathrm{s}^{−1}$ from measurements of the spatial attenuation and wavelength. We have performed measurements on two different polymer solutions ; we show that this technique is complementary to the classical rotational rheometer techniques since it gives access to the zero shear viscosity, which could not be measured by using rotational rheometers. The present technique is thus used to characterize the Newtonian plateau, while a rotational rheometer is used to characterize the fluid behaviour at moderate to high shear rates. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9321399927139282, "perplexity": 2090.8933253267237}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00063.warc.gz"} |
https://studycorgi.com/changes-in-circulation-ventricular-septal-defect/ | # Changes in Circulation: Ventricular Septal Defect
## Changes in Circulation
Certain changes in circulation occur at birth, and they are related to the gas exchange in the lungs and the work of the heart. Furthermore, one of the main causes of these changes is the fact that the umbilical cord becomes clamped (Hooper et al., 2015). As a result, the lungs can be viewed as prepared to breathing actively while becoming larger, and the capillary network also becomes wide to decrease resistance and increase the flow of blood.
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These changes lead to decreases in the pressure observed in the right atrium (Hooper et al., 2015). Changes in the pressure also lead to closing shunts located between the left and right atrium. Furthermore, the pressure in the aorta also increases, and a right-to-left shunt begins to function as a left-to-right shunt. The increase in the pressure in the aorta leads to the creation of a functional seal because of the contractions of the musculature (Hooper et al., 2015). In newborns, this shunt typically disappears in several weeks.
There are two forms of shunts which are left-to-right and right-to-left shunts. Thus, left-to-right shunts are associated with processes when blood recirculates in the lungs. A ventricular septal defect is related to this type of shunts (Hooper et al., 2015). Right-to-left shunts are observed when venous blood bypasses the lungs. This situation can be associated with intracardiac or intrapulmonary defects (Hooper et al., 2015). The difference in these two shunts is that the right-to-left shunts affect the cardiac output and inhalational anesthetics more significantly in comparison to the effects of the left-to-right shunts.
## Ventricular Septal Defect
The ventricular septal defect is a specific type of congenital heart defect which is observed when the communication between the ventricles is problematic. While discussing the pathophysiology of this condition, it is important to note that the shunting is usually directed from the left side with high pressure to the right side with low pressure (McCance & Huether, 2014). Pulmonary vascular resistance can significantly affect the degree of the problem. When ventricular septal defects are small, the resistance to shunting is high, and the flow of blood is limited.
Changes in the amount of shunted blood can be observed in several weeks after birth when the resistance associated with the pulmonary vascular function decreases (McCance & Huether, 2014). However, the left-to-right shunting associated with increases in the pulmonary flow of blood leads to decreasing pulmonary vessels’ diameters, and in this case, it is possible to observe the resistance to the flow of blood which can potentially lead to irreversible changes associated with the ventricular septal defect.
It is necessary to state that the occurrence of the ventricular septal defect is frequent while comparing it to the occurrence of other congenital heart defects. Thus, about 30% of these heart defects are classified as ventricular septal defects (McCance & Huether, 2014). Furthermore, it is important to pay attention to the fact that ventricular septal defects are divided into perimembranous, muscular, and supracristal ones, among which perimembranous ventricular septal defects are viewed as most typical. The treatment of the ventricular septal defect includes the use of the patch closure, sternotomy, a specific transarterial method, and the use of cardiopulmonary bypass (McCance & Huether, 2014). All these methods can be viewed as effective when they are applied during the first weeks after an infant’s birth.
## References
Hooper, S. B., Te Pas, A. B., Lang, J., Van Vonderen, J. J., Roehr, C. C., Kluckow, M.,… Polglase, G. R. (2015). Cardiovascular transition at birth: A physiological sequence. Pediatric Research, 77(5), 608-614.
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McCance, K. L., & Huether, S. E. (2014). Pathophysiology: The biologic basis for disease in adults and children (7th ed.). St. Louis, MO: Elsevier.
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If you are the original creator of this paper and no longer wish to have it published on StudyCorgi, request the removal. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8191755414009094, "perplexity": 4562.542856059985}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488559139.95/warc/CC-MAIN-20210624202437-20210624232437-00023.warc.gz"} |
http://math.stackexchange.com/questions/325815/conditional-expectation-in-the-case-of-mathcala-emptyset-omega-a-ac/325849 | # Conditional expectation in the case of $\mathcal{A}=\{\emptyset,\Omega,A,A^c\}$
I have a small computation to do and I am not able to prove it:
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probabiltiy space. Let $X$ be an integrable random variable and $A\in\mathcal{F}$ an event. Let $\mathcal{A}=\{\emptyset,\Omega,A,A^c\}$.
What is $\mathbb{E}[X \mid \mathcal{A}]$? I do not how to procede, is there anybody who can give a detailed proof of the expression of $\mathbb{E}[X \mid \mathcal{A}]$ in this particular situation?
-
By the definition of conditional expectation, $Y:= \mathbb{E}(X \mid \mathcal{A})$ is $\mathcal{A}$-measurable. So, it's a good start to think about the question how $\mathcal{A}$-measurable random variables look like. In this case, you can easily show that they are of the form $$c_1 \cdot 1_A + c_2 \cdot 1_{A^c}$$ where $c_1, c_2$ are constants. Thus we conclude $$Y= c_1 \cdot 1_A + c_2 \cdot 1_{A^c} \tag{1}$$ for some suitable constants $c_1$, $c_2$. Obviously, we still have to determine $c_1, c_2$.
Again by the definition of conditional expectation, we know that $Y$ has to fulfill $$\forall B \in \mathcal{A}: \int_B X \, d\mathbb{P} = \int_B Y \, d\mathbb{P} \stackrel{(1)}{=} \int_B (c_1 \cdot 1_A + c_2 \cdot 1_{A^c}) \, d\mathbb{P}$$
First, we choose $B=A$ and obtain $$\int_A X \, d\mathbb{P} = \int_A (c_1 \cdot 1_A + c_2 \cdot 1_{A^c}) \, d\mathbb{P} = \int (c_1 \cdot 1_A + c_2 \cdot \underbrace{1_A \cdot 1_{A^c}}_{0}) \, d\mathbb{P} = c_1 \cdot \mathbb{P}(A) \\ \Rightarrow c_1 = \begin{cases} \frac{\mathbb{E}(X \cdot 1_A)}{\mathbb{P}(A)} & \mathbb{P}(A) \not= 0 \\ 0 & \mathbb{P}(A) = 0 \end{cases}$$
A similar calculation for $B=A^c$ yields $$c_2 = \begin{cases} \frac{\mathbb{E}(X \cdot 1_{A^c})}{\mathbb{P}(A^c)} & \mathbb{P}(A^c) \not= 0 \\ 0 & \mathbb{P}(A^c) = 0 \end{cases}$$
-
Let $Y=\mathbb E[X\mid \mathcal A]$. There are two properties of $Y$.
1. $Y\in\mathcal A$
2. $\int_{A\in\mathcal A} XdP=\int_{A\in\mathcal A} YdP$
This is the definition of conditional expectation.
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While I know what you mean, $\int_{\mathcal A}$ is not very standard notation. – cardinal Mar 9 '13 at 21:07
@cardinal: Thanks, edited. – Bravo Mar 9 '13 at 21:38
Sorry, but I know what its definition is, in this case it is an exercise, we have to give its form in the case of the particular $\mathcal{A}$. – Mathoman Mar 9 '13 at 21:39
Ok thank you, I see what to do now! – Mathoman Mar 10 '13 at 8:56
Prove that the following quantities satisfy the definition of conditional expectation for $A \in \mathcal{A}$ and $A^c \in \mathcal{A}$ (the other two are obvious): $$\frac{\int_A X \,\mathrm{d}P}{P(A)} \quad\textrm{and}\quad \frac{\int_{A^c} X \,\mathrm{d}P}{P(A^c)}$$ This is actually a special case of a more general result about expectation conditional on the sigma-field generated by a (possibly infinite) measurable partition of $\Omega$.
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https://www.cofault.com/2009/08/trivial-exercise.html | ## 2009-08-21
### a trivial exercise.
Let's find a sum
$$\sum_{n=1}^\infty {1\over{n(n+2)}}$$
There is a well-know standard way, that I managed to recall eventually. Given that
$${1 \over n(n+2)} = {1 \over 2}\cdot \left({1\over n} - {1\over n+2}\right)$$
the sum can be re-written as
$$\sum_{n=1}^\infty {1\over{n(n+2)}} = \sum_{n=1}^\infty {1 \over 2}\left({1\over n} - {1\over n+2}\right) = {1\over 2}\left({1\over 1} - {1\over 3} + {1\over 2} - {1\over 4} + {1\over 3} - {1\over 5} + {1\over 4} - {1\over 6} \cdots\right)$$
with almost all terms canceling each other, leaving
$$\sum_{n=1}^\infty {1\over{n(n+2)}} = {1\over 2}\left(1 + {1\over 2}\right) = {3\over 4}$$
While this is easy to check, very little help is given on understanding how to arrive to the solution in the first place. Indeed, the first (and crucial) step is a rabbit pulled sans motif out of a conjurer hat. The solution, fortunately, can be found in a more systematic fashion, by a relatively generic method. Enter generating functions.
First, introduce a function
$$f(t) = \sum_{n=1}^\infty {t^{n + 1}\over n}$$
The series on the right converge absolutely when |t| < 1, so one can define
$$g(t) = \int f(t) dt = \int \sum_{n=1}^\infty {t^{n + 1}\over n} = \sum_{n=1}^\infty \int {t^{n + 1}\over n} = \sum_{n=1}^\infty {t^{n + 2}\over {n(n+2)}} + C$$
with the sum in question being
$$\sum_{n=1}^\infty {1\over{n(n+2)}} = g(1) - C = g(1) - g(0)$$
Definition of the g function follows immediately from the form of the original sum, and there is a limited set of operations (integration, differentiation, etc.) applicable to g to produce f.
The rest is more or less automatic. Note that
$$- ln(1 - t) = t + {t^2\over 2} + {t^3\over 3} + \cdots$$
so that
$$f(t) = t^2 + {t^3\over 2} + {t^4\over 3} + \cdots = - t \cdot ln(1-t)$$
therefore
$$g(t) = - \int t \cdot ln(1-t) dt = \cdots = {1\over 4} (1 - t)^2 - {1\over 2} (1 - t)^2 ln(1 - t) + (1 - t) ln(1 - t) + t + C$$
where the integral is standard. Now,
$$g(1) - g(0) = 1 - {1\over 4} = {3\over 4}$$
Voilà!
And just to check that things are not too far askew, a sub-exercise in a pointless programming:
scala> (1 to 10000).map(x => 1.0/(x*(x+2))).reduceLeft(_+_)
res0: Double = 0.749900014997506
PS: of course this post is an exercise in tex2img usage.
PPS: Ed. 2022: tex2img is gone, switch to mathjax.
#### 1 comment:
1. that's not pointless because of the xs | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.859818160533905, "perplexity": 1484.3090701263125}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950422.77/warc/CC-MAIN-20230402074255-20230402104255-00753.warc.gz"} |
https://www.esdemc.com/zh/%E6%8A%80%E6%9C%AF%E6%96%87%E6%A1%A3/pb2021-10-js-002%E6%A0%87%E5%87%86%E7%9A%841%CF%89%E7%9B%98%E5%BC%8F%E7%94%B5%E9%98%BB%E5%99%A8%E5%85%A8%E6%B3%A2%E5%BB%BA%E6%A8%A1/ | X
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PB2021.10 JS-002标准的1Ω盘式电阻器全波建模
1 Ω DISK RESISTOR FULL-WAVE MODELING FOR JS-002 STANDARD
Abstract –A 1 Ω disk resistor is specified in the CDM standard as the current sensing element. However, its transfer impedance is frequency dependent which is not considered in the standard. In this work, a full-wave model and a simple equivalent circuit model is provided to explain the root cause of the variation the transfer impedance of the 1 Ω disk resistor.
I. Introduction
When the pin or pad of a charged IC approaches an external metal object, and the breakdown voltage is exceeded a Charged Device Model (CDM) event occurs. During CDM testing, the discharge current of the CDM event is measured by a 1 Ω disk resistor sensing element that is located at the top of a pogo pin probe as described in the industry standard ANSI/ESDA/JEDEC JS-002 [1]-[3]. The resistance of this element is specified to have a value of 1.0 Ω ± 10% and a transfer impedance that does not have a deviation relative to the DC value greater than 3 dB up to 9 GHz [3]. The standard does not consider frequency variations of the transfer impedance when calculating the current from the measured voltage. This work investigates this assumption by characterizing the sensing element (1 Ω disk resistor) of a field induced CDM tester in the frequency domain. Additionally, a simple circuit model and a full-wave model are presented to explain the variation of the transfer impedance up to 27 GHz.
II. CDM tester
A. Discharge Circuit
The cross section of a CDM tester [3] is shown in Figure 1.
To charge the device, the field plate is brought to the specified charge voltage, then the pogo pin is lowered to contact the DUT pin. A spark is initiated and the current flows via the pogo pin to the 1 Ω disk resistor (Figure 2). The transfer impedance, properties of the oscilloscope and cable losses will determine the voltage that is displayed at the scope. The voltage is measured across the 1 Ω disk resistor from which the discharge current waveform is then calculated. However, the transfer impedance of the 1 Ω disk resistor is not constant over the frequency range and deviates from the DC value of the 1 Ω resistor [5].
B. 1 Ω Disk Resistor Measurement
As shown in Figure 2, the disk resistor consists of a resistive sheet on one side and a ceramic substrate made from beryllium oxide (BeO) on the other side that provides mechanical strength. In a CDM test head, the resistive sheet can be mounted downward (Figure 3a, resistive sheet towards pogo tip) or upward (Figure 3b, resistive sheet towards coaxial line).
Figure 4a shows a fixture to measure the disk resistor. The fixture is composed of two 50 Ω surface mountable connectors and a plate that aligns the disk resistor. The thickness of the plate was chosen to prevent any gap between the 50 Ω connectors and the surface of the disk resistor. The surface mount connectors were connected to ports 1 and 2 of a VNA and port extensions were performed up to the connector surfaces (Figure 4a). Figure 4b shows the definition of current and voltage of the two-port measurement setup (disk resistor setup measured with VNA). Knowing the S-parameter across the disk resistor, the transfer impedance of the disk can be calculated as in [6]:
As shown in Figure 5, the S21 and S12 of the disk are nearly identical which gives evidence for the accuracy of the measurement. However, the S22 and S11 are different (Figure 6). Port 1 of the measurement fixture (Figure 4) is toward the resistive sheet of the disk resistor, and port 2 is connected to the ceramic side of the disk. The reflection coefficient S11 is nearly flat but S22 has lower value at higher frequencies. The difference between S11 and S22 is important in understanding the CDM system behavior at high frequencies. Ringing was observed in the time domain discharge waveform if the substrate is mounted toward the pogo-pin but it reduced if the disk was flipped. The ringing will be discussed in Section IV of this paper. At low frequencies, the transfer impedance equals the disk resistance. Thus, the impedance measured with VNA should match the measurements obtained by an LCR meter (1 Ω @ 1kHz).
As shown in Table I, both LCR Meter (@ 1kHz) and disk measurements (Figure 4) show about 1 Ω for different orientations of disk (up to 1GHz). For each sample, both methods showed nearly identical low frequency values (Figure 7). At higher frequencies, the transfer impedance obtained using (1) is increasing with frequency (Figure 8) for all disks. Additionally, manufacturing tolerances lead to variations between disk samples. The behavior of the S21 can be explained using a simple circuit model in the next section.
III. Device Modeling
A. Simple Circuit Model
Although different disk resistors showed slightly different transfer impedance values versus frequency, they all behave similar (Figure 8). The transfer impedance of the disks increased with frequency from 1 Ω @ 1 MHz to about 3 Ω @ 20 GHz. This behavior can be explained by the influence of the ceramic substrate. It forms a short, 13 ps delay transmission line (see Figure 2) which acts as transmission line transformer and changes the match of the 1 Ω disk side to the 50 Ω system. The substrate has a relative permittivity of around 7. However, the structure is not easy to model in 3D due to possibility of higher order modes if they are excited. Only considering TEM modes, a simple circuit model is created in Advanced Design System (ADS) [7] to evaluate the influence of the short ceramic transmission line (Figure 9).
Based on the geometry and permittivity of the Beryllium Oxide (BeO) substrate, the characteristic impedance of the ceramic portion was calculated to be roughly 17 Ω. A 13 ps long 17 Ω lambda/4 transmission line transformer converts 50 Ω to transfer impedance of about 3 Ω at 20 GHz which can be calculated with (1).
This transmission line behavior explains the observed increase of S21 and transfer impedance. As shown in Figure 10, S11 is almost flat over the entire frequency range whereas S22 decreases with frequency as the short transmission line changes the match to 50 Ω. A comparison between measured impedance and calculated from the simple circuit model will be discussed in the next section.
B. Full-wave Model
The simple circuit model gives a qualitative insight into only the dominating effects, excluding the influence of the skin effect, higher order modes, and details of the geometry. A full-wave model can reveal additional details about the frequency-dependent impedance. Figure 11 shows the core elements of the full-wave model: the 50 Ω connectors, the geometry of the disk resistor, and two waveguide ports which are placed across the 50 Ω connectors. Two short 50 Ω connectors are placed on both sides of the disk. As shown in Figure 2, the 1 Ω disk resistor has a resistive sheet on one side and a ceramic carrier made of beryllium oxide on the other side. A thin resistive sheet material (this does not model skin effect) and BeO were imported from the library of CST Studio Suite [8].
C. Comparison Between Measurement and Simulation
Figure 12 and Figure 13 compares the magnitude and phase variation data for the measured, circuit model, and full-wave model of the 1 Ω disk resistor. Both models and the data agree on the increase of the impedance above 1 GHz and the peak around 20 GHz. This increase and the peaking may cause some error in the peak current measurement for CDM if the actual current contains relevant spectral content in this frequency range.
The fact that the simple circuit model and the measured data match up to 20 GHz can be seen as indicator that the short transmission line is the dominating reason for the observed increase in transfer impedance and the behavior of S11 and S22. The full-wave model predicted a higher peak value, 3.3 Ω relative to 2.6 Ω in the measurements (Figure 12 for resistive sheet down). The reason is not known, but as the frequency matches the other data one can be assured that the dielectric constant of the BeO was correctly estimated from literature data. The full-wave simulation shows additional resonant behavior around 25 GHz which is also seen in the measurements in the same frequency range. We have not investigated the field distribution at these frequencies within the full-wave results to identify the nature of these resonances. Furthermore, the transfer impedance of the disk may decrease at higher frequencies when skin effect starts to decouple the front and the back side of the very thin resistive sheet. As the authors do not know the exact thickness and material of the resistive layer, it is not known above which frequency the decoupling effect of the skin effect would reduce S21. The data indicates that this is not the case below 20 GHz, as the simple ADS model matches the measurement in its principle behavior. Our full-wave model is also not able to simulate skin effect as an infinitely thin electrical layer was used to model the resistive sheet. The metallization of the resistive layer may not be fully homogeneous. This would cause a current flow that is not radially symmetric. As known from current shunts, this would increase the mutual inductance between both sides of the resistive disk. Such a behavior is not observed which leads to a tentative conclusion that the resistive sheet is homogeneous within the boundaries of the analysis.
IV. Effect of Disk Orientation on CDM Event
To investigate the effect of disk orientation, both measured and simulated results have been compared in a CDM test setup.
A. Effect of Disk Orientation on Measured Discharge Current
CDM classification levels have been reduced [9] and that further reductions are to be expected. This will lead to a faster rise time in CDM. This, paired with faster I/O on ICs may lead to measurement problems in CDM testing due to the mounting direction of the disk resistor. To investigate the effect of disk orientation, CDM discharge tests have been measured using a 23 GHz bandwidth oscilloscope [10] with different orientations of disk resistor as shown in Figure 3. One disk resistor was used within one test head by flipping the orientation between tests to prevent any unwanted effects or variation in the test setup. Discharge data from multiple pogo-pins with different length and discharge currents have been captured for charge voltage of 500 V (Figure 14 through Figure 17).
As shown, all plots have a low frequency component around 1 GHz which is the main CDM discharge current. However, there are some high frequency components as well which create ringing waveform over the low frequency waveform. As shown, high frequency ringing was observed in discharge current when the resistive sheet of disk resistor was mounted upward (Figure 3b).
However, the ringing is weaker if the resistive sheet of disk resistor is mounted downward (substrate toward oscilloscope and the resistive sheet toward DUT as shown in Figure 3a). To isolate the ringing from the familiar low frequency response, a Maximum Overlap Discrete Wavelet Transform based Multiresolution Analysis (MODWT MRA) was used [11]. This method yields excellent decomposition and reconstruction while maintaining sharp edge definition and minimizing non-causality introduced by traditional high pass filtering. The high frequency ringing signal was found to be well isolated from the rest of the signal by using the db7 wavelet with a scaling factor of 2. Furthermore, the Wigner-Ville distribution [12] of ringing is used to visualize the time dependent frequency composition of the time dependent current.
The time domain signal, the power spectral density and time-frequency scalogram of ringing for different pogo-pins and for resistive sheet up and down is shown in Figure 18 and Figure 19. As indicated in timefrequency spectra of Figure 18, when the disk resistor is mounted upward for pogo pin length of 8.25 mm, 9.4 mm and 10.5 mm, there are two main high frequency component which make up the ringing with the corresponding interference term between the two main components.
However, in all time-frequency spectra of Figure 19, there is only one frequency component. Similarly, when the resistive sheet of disk resistor is mounted upward the power spectrum has two main frequency components for pogo pin length of 8.25 mm, 9.4 mm and 10.5 mm. Table II summarizes the two observed frequencies if the resistive sheet of disk is mounted upward (Figure 18).
Two sinusoidal signals are used in Figure 20 to reconstruct the ringing for pogo pin of 10.5 mm (blue curve in Figure 20) and is compared with the original ringing (red curve in Figure 20).
Therefore, it is possible to reconstruct the ringing by summing two sinusoidal signals e.g., f1 and f2 as shown in Figure 20. This motivates us to consider the nature of ringing and determine the physical agents which correspond to these responses.
Figure 21 shows the half wavelength versus frequency (blue curve) and is compared with the length of the pogo pins versus the first sinusoidal signal (f1) from Table II (black curve). As shown, the first sinusoidal signal (f1) is directly related to the length of pogo pin and can be calculated relative to the length of the pogo pin. The second sinusoidal component is related to the disk orientation. As shown in Figure 18 and 19, the second sinusoidal signal (f2) exist for resistive sheet upward for pogo pin 8.25 mm, 9.4 mm and 10.5 mm. However, this signal disappears when the resistive sheet is mounted downward. For the pogo pin 6.6 mm, two sinusoidal signals cannot be distinguished since f1 is very close to f2. When the resistive sheet of disk resistor in mounted downward (resistive sheet toward DUT), only one frequency can be observed in time-frequency spectrum of ringing signal (Figure 19) which indicate the effect of disk orientation.
B. Effect of Disk Orientation in Simulation
As shown in Figure 18 through Figure 20, two sinusoidal signals contribute to the high-frequency ringing of the CDM discharge current. A full-wave model is created for CDM test setup (Figure 22) to obtain a qualitative insight into the dominating effects up to 30 GHz. Two discrete ports are placed on both sides of the pogo pin providing the corresponding connection for co-simulation simulation in ADS which is shown in Figure 23. Measured S-parameters of the 1 Ω disk resistor from Section II or simulated S-parameter file from Section III can be imported into ADS model of Figure 23. It is also possible to use the simple circuit model of the disk resistor (Figure 9) into the circuit model of Figure 23. As shown in Figure 24 and Figure 25, if the orientation of the disk is changed, the high frequency ringing also changes. The first ringing in the discharge current relates to the length of the pogo pin which exists in the discharge current regardless of disk orientation. However, the second ringing corresponds to the disk orientation and will disappear if the resistive sheet of the disk is mounted toward the DUT (red curve of Figure 24 and Figure 25). If the substrate of the disk is mounted toward the DUT (blue curve in Figure 24 and Figure 25), the waveform has more ringing (high frequency contents).
The effect of disk orientation in simulated data (Figure 24 and Figure 25) is not as strong as the measured waveform (Figure 14 through Figure 17), but in principle they follow the same behavior, i.e., ringing is stronger if the ceramic substrate is mounted toward the DUT and gets weak if the resistive sheet is mounted toward the DUT.
It is known that the current measured at disk is not necessarily equal to the current at the DUT [13]. Ultimately, the current at the DUT is the stress that the device experienced. For accurate comparison between measured and simulated data in this paper, only current at the disk resistor was studied. Future work should incorporate current at the DUT to get a more accurate measurement of the stress that the device experiences.
A. Discussion
The fundamental question is how important is the frequency response of the transfer impedance of the disk resistor above 10 GHz?
Present ICs have data rates up to 50 GHz and more. Thus, their I/O can be damaged by high frequency content of every strong signal [14]. The charge voltages of CDM will be further reduced such that the rise times will further reduce [9]. Thus, the importance of the larger than 10 GHz spectral content will increase. Right now, CC-TLP testers can be based on a 40 ps or less transmission line pulser [14]. A 30 ps rise time equates to 10 GHz. To avoid problems in testing of future ICs and for comparing test methods, the analysis > 10 GHz is suggested.
V. Conclusion and Further Investigations
The frequency response of the transfer impedance of a 1 Ω disk resistor has been investigated through measurement, full-wave modeling, and a simplified equivalent circuit. The transfer impedance increases with frequency and shows a maximum of about 3 Ω at 20 GHz for disks mounted downward (resistive sheet toward DUT). However, the transfer impedance increases with frequency and shows a maximum of about 15-20 Ω around 20 GHz for a disk mounted upward (resistive sheet toward oscilloscope).
This is explained by considering the inner structure of the 1 Ω disk resistor. Only one side of the resistor’s ceramic carrier contains the resistive sheet material. Thus, it is asymmetric. The thin ceramic carrier creates a short transmission line. The effect of this short transmission line section is clearly visible in measurement, simplified and full-wave simulation both in S11, S22 and S21 data. It cannot match the 1 Ω but it strongly changes the match and causes an increase of the transfer impedance. The CDM current has been captured with different pogo pin lengths. High frequency ringing was observed. It can be explained as the sum of two sinusoids. The first sinusoidal signal was directly related to the resonance frequency of the pogo pin structure, i.e., its length and the second one was created due to the transfer impedance of the disk orientation.
VI. Acknowledgements
We would like to thank Dr. David Johnnsson and Dr. Timothy Maloney for their useful discussions and comments. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8810781240463257, "perplexity": 1184.4004159553842}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00515.warc.gz"} |
http://garden.irmacs.sfu.ca/op/invariant_subspace_problem | # Invariant subspace problem
Importance: High ✭✭✭
Author(s):
Subject: Analysis
Keywords: subspace
Recomm. for undergrads: no
Posted by: tchow on: February 9th, 2009
Problem Does every bounded linear operator on an infinite-dimensional separable Hilbert space have a non-trivial closed invariant subspace?
Let be a Hilbert space. The subspaces and are trivially invariant under any linear operator on , and so these are referred to as the trivial invariant subspaces. The problem is concerned with determining whether bounded operators necessarily have non-trivial invariant subspaces.
This is one of the most famous open problems in functional analysis. Enflo [1] constructed Banach spaces for which the corresponding question has a negative answer, and recently Argyros and Haydon constructed a Banach space for which the corresponding question has a positive answer [4].
For a nice overview to the problem see [2], [3] or [5].
## Bibliography
[1] P. Enflo, On the invariant subspace problem for Banach spaces, Acta Math. 158 (1987), 213-313. MathSciNet
[2] B. S. Yadav, The Present State and Heritages of the Invariant Subspace Problem, Milan J. Math. 73 (2005), 289-316. MathSciNet another link
[3] H. Radjavi and P. Rosenthal, The Invariant Subspace Problem, The Mathematics Intelligencer 4 (1982), no. 1, 33-37. MathSciNet
[4] S. A. Argyros and R. G. Haydon, A hereditarily indecomposable -space that solves the scalar-plus-compact problem, arXiv:0903.3921 (2009).
[5] J. Noel. The Invariant Subspace Problem. Honours Thesis, Thompson Rivers University. Link to pdf.
* indicates original appearance(s) of problem.
### closed
Does every bounded linear operator on an infinite-dimensional separable Hilbert space have a non-trivial closed invariant subspace?
### ستار اكاديمي 8
ستار اكاديمي 8
thank you man, u'r good :)
## Comment viewing options
Select your preferred way to display the comments and click "Save settings" to activate your changes. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8867349624633789, "perplexity": 1611.8806235055222}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039744750.80/warc/CC-MAIN-20181118221818-20181119003818-00247.warc.gz"} |
http://mathhelpforum.com/trigonometry/8084-trig-question-finding-all-solutions-print.html | # Trig question finding all solutions
• November 27th 2006, 03:06 PM
kcsteven
Trig question finding all solutions
what is the best way to find all the solutions of sin x-1=0, with the answer in A+Bkpi, where A=? with 0<A<pi
where B=? and k is any integer
Thankx
Keith Stevens
• November 27th 2006, 04:52 PM
topsquark
Quote:
Originally Posted by kcsteven
what is the best way to find all the solutions of sin x-1=0, with the answer in A+Bkpi, where A=? with 0<A<pi
where B=? and k is any integer
Thankx
Keith Stevens
There is no such x, $0 < x < \pi$.
$sin(x) - 1 = 0$
$sin(x) = 1$
For what values of x is sin(x) = 1?
$x = 0, \pi$
Neither of which is in your indicated domain.
-Dan | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9058868288993835, "perplexity": 4508.762488510829}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982293195.16/warc/CC-MAIN-20160823195813-00175-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/3075705/integral-evaluation-with-delta-dirac | # Integral evaluation with delta Dirac
I am having doubts about the following integral:$$\int \limits _{0}^{10} \int \limits _{0}^{10} \frac{x^2y^2}{(x^2+y^2)^{5/2}}\ \delta(x)\ \mathrm{d}x\mathrm{d}y$$ If we apply the definition of the Delta Dirac function we should get: $$\int \limits _{0}^{10}\int \limits _{0}^{10} \frac{x^2y^2}{(x^2+y^2)^{5/2}}\ \delta(x)\ \mathrm{d}x\mathrm{d}y=\int \limits _{0}^{10} 0\ \mathrm{d}y=0$$ Nevertheless, when one plots the 3D function: $$z(x,y)=\frac{x^2y^2}{(x^2+y^2)^{5/2}}$$ and intersect it with the plane $$x=0$$, what he gets is:
1) $$z(0,y)=0$$ for $$y\neq 0$$
2) $$z(0,0)=\lim \limits _{x,y\rightarrow0}z(x,y)=- \infty$$
so basically $$z(0,y)=-\delta\ (y)$$ and therefore: $$\int \limits _{0}^{10}\int \limits _{0}^{10} \frac{x^2y^2}{(x^2+y^2)^{5/2}}\ \delta(x)\ \mathrm{d}x\mathrm{d}y=-\int \limits _{0}^{10} \delta\ (y)\ \mathrm{d}y=-H(0)$$ N.B. If I choose any negative number as the lower integration extremum for the variable $$y$$ the final result is obviously $$-H(a)=1\neq 0$$. I do not know which of the two approaches is right, or if they are both wrong and in this case, what is the correct one?
Let us re-name the inner integral in terms of a mapping $$g:[0,10]\to\mathbb R\qquad y\mapsto \int_0^{10}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\delta(x)\,dx=\int_0^{10}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\,\delta(dx)$$ with $$\delta$$ being the Dirac measure (which is not all too relevant for this question, I just wanted to clarify that whatever we do with the so-called "delta function" is rigorously defined when one considers measure theory). Anyways, we get $$g(y)=0$$ for all $$y\in[0,10]$$ (actually for all $$y\in\mathbb R$$) as is readily verified. Then $$\int_0^{10} \int_0^{10}\frac{x^2y^2}{(x^2+y^2)^{5/2}}\,\delta(dx) \,dy=\int_0^{10} g(y)\,dy=0$$ as you correctly claimed in the beginning.
1. $$\lim_{(x,y)\to (0,0)} z(x,y)$$ does not exist as $$\lim_{n\to\infty}z\Big(\frac1n,0\Big)=\lim_{n\to\infty}0=0$$ but $$\lim_{n\to\infty} z\Big(\frac1n,\frac1n\Big)=\lim_{n\to\infty} \frac{n}{2^{5/2}}=\infty$$ so the path on which you approach the origin does matter for the limit, which would not happen if the limit in question existed in the usual sense. (One would say that $$z$$ is discontinuous at the origin or that $$z$$ cannot be extended continuously into the origin).
2. The idea that $$\delta(0)=\infty$$, $$\delta(x)=0$$ if $$x\neq 0$$ and $$\int_{\mathbb R}\delta(x)\,dx=1$$ ís nothing more than a heuristic characterization simply to get an intuition what the dirac measure does - because no mapping on the real numbers satisfies the three above conditions at once. Therefore, one should refrain from using this for anything technical beyond the standard property $$\int_{\mathbb R} f(x)\delta(x)\,dx=f(0)$$ for $$f$$ continuous. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9446170330047607, "perplexity": 96.47247331876348}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371858664.82/warc/CC-MAIN-20200409122719-20200409153219-00266.warc.gz"} |
https://math.stackexchange.com/questions/2281262/find-solution-of-maximum-of-a-function-with-conditions | # Find solution of maximum of a function with conditions
\label{eqgen} \begin{aligned} & \underset{x}{\text{maximize}} & & F(x)\\ & \text{subject to} & & x \geq 0, \\ & & &x + f(x)\leq d, \end{aligned} where $F(x), f(x)$ are continuous functions from $\mathbb{R}^+ \to \mathbb{R}^+$.
Here is my approach. Intuitively, $x+f(x)\leq d, x\geq 0$ is equivalent to $x \in [a_1, b_1] \cup [a_2, b_2]\cup \cdots \cup [a_{n}, b_{n}]$, where $a_i\leq b_i$, $a_1$ is either 0 or solution of $x+f(x) = d$, $b_n$ is either d or solution of $x+f(x) = d$, and $a_i, b_i$ are solutions of $x+f(x) = d$ for other cases.
Hence, the optimization problem becomes \begin{aligned} & \underset{x}{\text{maximize}} & & F(x)\\ & \text{subject to} & & x \in [a_1, b_1] \cup [a_2, b_2]\cup \cdots \cup [a_{n}, b_{n}]. \end{aligned}
max $F(x)$ on $[a_i, b_i]$ is attained at either $a_i, b_i$ or maximum of $F(x)$ on $[a_i, b_i]$. We can conclude that the optimal solution is either solutions of $x+f(x) = d, or x=0, x= d$ or local maximum of function $F(x)$. This is also what I want to prove.
How can I formally write down the solution of the above approach? I wrote this and my professor does not accept my solution. Thank you in advance!
• Have you tried: en.wikipedia.org/wiki/… May 15 '17 at 17:14
• Thanks for your hint. But it does not work May 23 '17 at 19:54
• Can you tell me what you mean by it does not work? You would not find any closed form solutions in any case. Under differentiability, you would get first order conditions. May 24 '17 at 1:19
There are a lot of intuitive tricks that one can use if one knows exactly the values of $f(x),\ d$, but given it is not given, I will lay out a general way of Kuhn-Tucker conditions.
Write: $L=F(x)+\lambda_1 x-\lambda_2 (x+f(x)-d)$ with $\lambda_1, \lambda_2 \geq0$
Think of the intuition behind the sign of the lambdas. If $x$ becomes negative that is bad for our purpose given the constraint, and that is getting reflected by our objective function getting a negative "shock" through the positive $\lambda_1$. You can reason through similarly for the other lambda.
Now, take derivative w.r.t $x$ to get:
$F'(x)-\lambda_1+\lambda_2(1+f'(x))=0$
Couple this equation with the constraints:
$\lambda_1 x=0$
$\lambda_2 (x+f(x)-d)=0$
And these conditions have complementary slackness.
Case 1: Suppose both constraints are binding, then $\lambda_1, \lambda_2>0$. And, $x=0$, and, $f(0)=d$. If, $f(0) \ne d$, we can immediately move on from this case to the second case.
Otherwise, plug in $x=0$ in the derivative condition, solve for lambdas and check if they are actually positive, otherwise, as before, we move on from this case. Check here:
$F'(0)-\lambda_1+\lambda_2(1+f'(0))=0$
Case 2: Constraint 1 binds and 2 does not. So, $\lambda_1>0$ and $\lambda_2=0$ and $x=0$.
Check back into the derivative condition and check if the values match up with it and with the condition
$f(0)\leq d$
Case 3: Constraint 2 binds and 1 does not. So, $\lambda_1=0$ and $\lambda_2>0$ and $x=0$.
Solve for $x$ from $f(x)=d$, given we are not in case 1, $x$ should not be zero. Use that value of the solution to check back if $x\geq 0$ holds, and if the derivative condition is satisfied for positive $\lambda_2$.
Case 4: Both constraints bind So, $\lambda_1=0$ and $\lambda_2=0$
Solve it like an unconstrained problem, find $x$ and check back to confirm that the inequality conditions are satisfied.
Finally at any case when you find a solution, you check for the second derivative of the maximand (Lagrangian function L) to be negative at the optimal point ($x^*, \lambda^*$). Also, if you found multiple solutions from the different cases, choose among them by plugging them directly into the actual function $F$ to see which one gives the maximum value.
• $d$ is given and plays the role of constant. May 24 '17 at 5:46
• I mean, do you know how $f(0)$ and $d$ are exactly related? Also does the above answer help you? May 24 '17 at 5:58
• Many thanks. I am checking. If it is correct I will mark it. May 24 '17 at 6:22
• I think it's correct and make it as completed answer. But how can we deal with the case when $F, f$ are continuous but not differentiable. From my picture I think it still holds. May 24 '17 at 8:54
• Without differentiability, your best hope would be to take the same four cases as above, and find the optimal in each case by plotting+eye inspection or by logical reasoning. May 24 '17 at 15:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.8911601305007935, "perplexity": 226.22801295659994}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587963.12/warc/CC-MAIN-20211026231833-20211027021833-00300.warc.gz"} |
http://math.stackexchange.com/questions/215315/noetherian-ring-and-primary-decomposition-result | # Noetherian ring and primary decomposition result
I'm struggling with the following problem and I would appreciate some help if possible
Let $R$ be Noetherian and let $I,J$ be ideals. Define $(I:J^{\infty}) = \bigcup_{n}(I:J^{n})$.
(a) If $Q$ is primary, prove that $(Q:J^{\infty}) = Q$ for any $J \subset R$ with $J$ not contained in the radical of $Q$.
(b) If $P$ is prime and $I = Q_{1} \cap \ldots \cap Q_{k}$ is a finite intersection of primary ideals, then show that $(I : P^{\infty})$ is the intersections of the $Q_{i}$ for which $P \not \subset P_{i}$.
Thank you!
-
(a) If $aJ^n\subseteq Q$, $Q$ primary and $J\nsubseteq \sqrt{Q}$ (therefore $J^n\nsubseteq \sqrt{Q}$), then... – user26857 Oct 17 '12 at 9:11
(b) Apply (a) using that $(Q_1\cap\cdots\cap Q_k:P^{\infty})=(Q_1:P^{\infty})\cap\cdots\cap(Q_k:P^{\infty})$. – user26857 Oct 17 '12 at 9:20
Thanks a lot! I had (b), but I don't know why I couldn't see a) :)); this is what happens when you don't sleep enough – Dquik Oct 18 '12 at 1:46
An alternative definition of primary ideals is the following: $Q$ is primary iff for every ideals $I,J$ with $IJ\subseteq Q$ we have $I\subseteq Q$ or $J\subseteq\sqrt{Q}$ (or viceversa). IN my answer to (a) take $I=(a)$. – user26857 Oct 18 '12 at 8:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8201568722724915, "perplexity": 244.283144283857}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802775085.124/warc/CC-MAIN-20141217075255-00089-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://www.groundai.com/project/phase-entanglement-complementarity-with-time-energy-uncertainty/ | Phase-Entanglement Complementarity with Time-Energy Uncertainty
# Phase-Entanglement Complementarity with Time-Energy Uncertainty
Fu-Lin Zhang Department of Physics, School of Science, Tianjin University, Tianjin 300072, P. R. China Mai-Lin Liang Department of Physics, School of Science, Tianjin University, Tianjin 300072, P. R. China
July 15, 2019
###### Abstract
We present a unified view of the Berry phase of a quantum system and its entanglement with surroundings. The former reflects the nonseparability between a system and a classical environment as the latter for a quantum environment, and the concept of geometric time-energy uncertainty can be adopted as a signature of the nonseparability. Based on this viewpoint, we study their relationship in the quantum-classical transition of the environment, with the aid of a spin-half particle (qubit) model exposed to a quantum-classical hybrid field. In the quantum-classical transition, the Berry phase has a similar connection with the time-energy uncertainty as the case with only a classical field, whereas the geometric phase for the mixed state of the qubit exhibits a complementary relationship with the entanglement. Namely, for a fixed time-energy uncertainty, the entanglement is gradually replaced by the mixed geometric phase as the quantum field vanishes. And the mixed geometric phase becomes the Berry phase in the classical limit. The same results can be draw out from a displaced harmonic oscillator model.
###### pacs:
03.65.Vf; 03.65.Ud
## I Introduction
A quantum system always need to be isolated from its surroundings, which is referred to an open quantum system. Otherwise, we would have to treat the system and the rest of the universe Berry (1984) as a whole. An significant obstacle to this separation is the entanglement Horodecki et al. (2009) between the open system and its environment caused by their quantum nature. The system is exactly described by its reduced density matrix defined as the partial trace of a global state over the Hilbert space of environment. In an ideal case, the decoherence of an open system due to entanglement with its environment is negligible. The effects of the environment represents as parameters, or say classical fields, which are time-dependent in most cases, in the local Hamiltonian of the open system. If the field is slowly altered, the open system will behave like a closed one, namely staying adiabatically in an instantaneous eigenstate of the time-dependent Hamiltonian. The difference is that the open system undergoing a cyclic adiabatic evolution gains a Berry phase Berry (1984). In Berry’s original paper Berry (1984), the phase is said to be geometric because it results from the geometric properties of the parameter space of the local Hamiltonian. In other words, it depends on the properties of the external field in evolutionary process, and thus reveals a nonseparable relation between the system and environment.
In the two aforementioned cases, the nonseparability between a system and an environment exhibits either entanglement or Berry phase as the environment is either quantum or classical. This naturally leads to an interesting question: What is their relationship in the quantum-classical transition of the environment? To answer this question, we introduce a model of a spin-half particle (qubit) coupled to an adiabatically rotating quantum-classical hybrid field. And before that, pointed out that the Berry phase is caused by a neglected Hamiltonian and accompanied by a geometric time-energy uncertainty Anandan and Aharonov (1990); Anandan (1991); Uhlmann (1992). This uncertainty is shown to be a signature of the total nonseparability between the system and the quantum-classical hybrid field. Because of the entanglement between the principal qubit and the field, the original definition of the Berry phase is no longer applicable. We study its two extensions, one of which is presented in this work based on the neglected Hamiltonian and the other is the geometric phase for mixed state Sjöqvist et al. (2000); Singh et al. (2003); Tong et al. (2004). In this work, we call the former the Berry phase and the latter the mixed geometric phase. When the classical part of the field vanishes, one can find the Berry phase leads to the definition in the works of vacuum induced Berry phase Fuentes-Guridi et al. (2002); Liu et al. (2010, 2011); Larson (2012) in the Jaynes-Cummings (JC) model Jaynes and Cummings (1963), where they devise the phase with the aid of a phase shift operator. In the quantum-classical transition of external field and for a fixed time-energy uncertainty, the mixed geometric phase replaces the entanglement gradually, and becomes the Berry phase in the classical limit. This shows a complementary relationship between the mixed geometric phase and the entanglement to reflect the nonseparability.
## Ii Neglected Hamiltonian
Before introducing our model, we first carefully examine the connection between the Berry phase of an open system and its nonseparability with a classical environment. The connection will become apparent if we consider the Berry phase as the adiabatic limit of the Aharonov-Anandan (AA) phase Aharonov and Anandan (1987); Zeng and Lei (1995, 1996). The latter is an extension of the Berry phase without adiabatic approximation. It has been shown that only for a nonstationary state after a cyclic evolution might the AA phase appear Zeng and Lei (1995). That is, if an open quantum system stays in an instantaneous eigenstate of its Hamiltonian and acquires an AA phase after a cyclic evolution, there should exist a Hamiltonian which is neglected but affects the system in addition to , and Berry (2009). Therefore, the uncertainty in energy when the system is isolated from its environment causes the AA phase. In the adiabatic limit, , the expectation value of the Hamiltonian becomes the instantaneous eigenvalue of , and the AA phase returns the Berry phase. The Berry phase comes from the fact that even the uncertainty of energy tends to zero in the adiabatic limit, its effect in a cyclic evolution is finite and nonzero. The accumulating result of the uncertanty of energy can be represented as the geometric quantum uncertainty relation Anandan and Aharonov (1990); Anandan (1991); Uhlmann (1992)
S=2∫τ0ΔEdt, (1)
where is the energy fluctuation, defined by
ΔE2=⟨ψ|H2|ψ⟩−⟨ψ|H|ψ⟩2. (2)
It is the distance that the system traverses during its evolution in the projective Hilbert space measured by Fubini-Study metric Anandan and Aharonov (1990). In the following section, we will extend the Berry phase to a system in a quantum-classical hybrid field with the aid of the neglected Hamiltonian, and show the connection between the time-energy uncertainty and its entanglement with the field.
## Iii Qubit Model
Let us now consider the model of a qubit coupled to a rotating quantum-classical hybrid field with the Hamiltonian
H0=−μJ⋅σ−B⋅σ (3)
where are the Pauli operators of the qubit, is the classical part of the field, and denotes the quantum part with being the angular momentum operators of a spin- particle. Experimental realization of the interaction between two spins and observation of the Berry phase for such a system is feasible by current technology Jones et al. (2000); Du et al. (2003). In the frame of quantum optics, the physical meaning of the pure quantum term can be easily understood if we consider the spin operator as the Schwinger representation Fuentes-Guridi et al. (2002) of two modes of a quantized optical field. The Hamiltonian denotes a interacting process between the two modes of optical field and a two level system conserving the total photon number. The whole Hamiltonian (3) could also be regarded as a semiclassical spin star model Calvani et al. (2013a, b); Richter and Voigt (1994); Deng and Fang (2008). As shown in Fig. 1, the system qubit interacts with two Heisenberg chains, i.e. two sets of environmental spins. One of the chains has a total spin , and the other is treated as a classical field as its entanglement with the system qubit is negligible Calvani et al. (2013a, b). The quantum-classical transition of the environment can be shown by tuning the parameter from infinity to zero. Thus the experimental realization of the Hamiltonian (3) provides a simulation of the crossover from a quantum to a classical environment.
Classical Field.– For the simplest case in the absence of quantum field, where , we can treat the principal qubit in the space of itself. The instantaneous eigenstates of corresponding to the eigenvalues are
|ψ+⟩=U(t)|↑⟩, |ψ−⟩=U(t)|↓⟩, (4)
with . To keep the system in the eigenstates , the neglected Hamiltonian should be , and the coefficients . After a period , the total phases are determined by the eigenvalues and the property of the spin-half rotation operation. For the dynamic phase, which is defined by Aharonov and Anandan (1987), we get Hence the Berry phase, as the AA phase in the limit of , can be written as
γ±=∓π(1−cosθ). (5)
They equal the half of the solid angle subtended by the path followed by in the parameter space, and the sign depends on whether the spin was aligned or against the direction of the field. Although the fluctuation of energy defined by (2) approaches zero in the adiabatic limit, the time-energy uncertainty (1) in a period is a constant independent of the frequency as
S±=2πsinθ. (6)
which is nothing but the shortest perimeter surrounding the solid angle in an unit sphere. In this work, we confine ourselves to the case of rotating fields not only for simplicity but also to gain as large the geometric phases as possible under a fixed time-energy uncertainty.
Furthermore, if we consider the geometric phase for noncyclic evolution Samuel and Bhandari (1988) and require corresponding to a reversal of the interference fringes, we can easily obtain a time-energy uncertainty relation in the present model not (a)
⟨ΔE⟩Δt≥h2, (7)
where is the time-averaged uncertainty in energy during the time to gain the geometric phase and is the Planck constant. Equality in (7) holds when the classical field is perpendicular to z axis, i.e. .
Quantum-Classical Hybrid Field.– We will now derive the results for the full Hamiltonian (3). One can diagonalize the Hamiltonian in the subspace of , where is the eigenvector of with and In the unitary transformation, a term is introduced to eliminate the influence of the phase brought by a odd not (b). The instantaneous eigenstates of corresponding to eigenvalues are
|ψ+m⟩=U(t)(cosαm2|m⟩|↑⟩+sinαm2|m+1⟩|↓⟩), |ψ−m⟩=U(t)(sinαm2|m⟩|↑⟩−cosαm2|m+1⟩|↓⟩), (8)
with . Here, the values of index range from to . For or , only two states and are physically possible. They are direct products of two spin-coherent states aligned or against the classical field, and the states of spin- are the closest to the classical states Layton et al. (1990). The concurrence Wootters (1998) of states (III) can be written as
C=sinαm (9)
which is the degree of entanglement between the system qubit and its external field.
In the case of the classical field discussed above, the dynamics of the field is not described by the Hamiltonian (3), but is considered as a time-dependent variable. A quantum field cannot be treated in this way anymore, as its state can be influenced by the interaction with the principal qubit. We have to take into account a Hamiltonian driving the quantum field to rotate with the same frequency as the classical field. The most natural choice of the Hamiltonian is . Staying in the instantaneous eigenstates (III) requires the whole system to satisfy the Schrödinger equation
i|˙Ψ±m⟩=(H0+Hj+HΔ)|Ψ±m⟩, (10)
with . Consequently, the neglected Hamiltonian can be found as , and the coefficients . After a period, the total phase is and the dynamic phase is with . Hence we obtain the Berry phase
γq = π(1−cosθcosαm) (11) = π(1−cosθ)+π(1−cosαm)cosθ,
where we omit the sign for clarity. The second term in (11) shows a correction in Berry phase aroused by the quantum part of the field. We call this term quantum field induced Berry phase (QBP), as it is caused by quantum fluctuation and entanglement. In addition, when the classical field vanishes, one can choose due to the symmetry of the Hamiltonian (3) and find the QBP has the same physical meaning as the vacuum induced Berry phase Fuentes-Guridi et al. (2002); Liu et al. (2010, 2011); Larson (2012). A remarkable quantum nature of QBP can be understood by its sharp contrast to the classical one. Namely, the Berry phase is robust against the perturbations by classical environment, particularly is invariant under a changing of the strength of the field. But it is sensitive to the quantum perturbation even though the polarization direction of the spin- particle in the eigenstates remains parallel to the classical field. From the form of QBP in (11) one can surmise it revelent to a solid angle in a space corresponding to the quantum fluctuation of spin-. The QBP reaches its maximum for the most entangled states with or and vanishes for the two separable eigenstates and . And in the former case the states give rise to the maximum dispersion for square of the spin angular momentum, and the latter two states lead to the minimum Layton et al. (1990).
For further analysis of the physical meaning of QBP and the whole Berry phase in (11) we calculate the geometric time-uncertainty relation (1) during a period of rotation . Substituting the expressions of and into (1) leads to
Sq=2π√1−cos2θcos2αm. (12)
Similar to QBP, it is increased by the entanglement between the qubit and field. It is interesting to note that if one defines an angle , there also exists a relationship of solid angle and perimeter between and as the case of the classical field. That is, the classical part and quantum part of the external field parallel to each other behave as an effective classical field in another direction. This deflection caused by the fluctuation of the quantum field is most visible when , where only the quantum field contributes to and , and vanishes for with the classical filed making the most contribution. Moreover, for a fixed value of geometric phase , the inequality (7) still holds for the qubit system with a quantum-classical hybrid field. And the entanglement between the principal qubit and its external field reduces the value of for a given .
In the above discussion about QBP, we treat the system qubit and the quantum field as a composite system when we consider their evolution. We actually remove terms irrelevant to the qubit in the total phase of the whole system and obtain the Berry phase (11). That is, the Berry phase (11) depends on the evolution of the quantum field entangled with the qubit. To study the phase determined only by the geometry of the path of the qubit, we derive the time-dependent mixed state
ρ=sin2αm2|ψ+⟩⟨ψ+|+cos2αm2|ψ−⟩⟨ψ−|, (13)
by tracing out the quantum filed in the eigenstates (III), where is defined in (4). Here, we only give the case for the pure state , the mixed geometric phase for can be easily obtained by changing into . We can calculate the mixed geometric phase by using the definition with a kinematic description Sjöqvist et al. (2000); Singh et al. (2003); Tong et al. (2004) and get
γmix = arg(sin2αm2eiγ++cos2αm2eiγ−) (14) = arctan(cosαmtanγ−),
where are the Berry phases (5) resulting from the classical field. The phase is manifestly gauge invariant and can be experimentally tested in interferometry.
Let us give a further discussion about the relationship among these quantities related to the nonseparability between the system and the external field. For a fixed value of , , and the entanglement widens their gaps. They are equal to the Berry phase for a separable eigenstate. It is important to note that the time-energy uncertainty leads to an upper bound of the entanglement . The equality holds when the mixed geometric phase vanishes. These reveal a complementary relationship between the mixed geometric phase and the entanglement to reflect the nonseparability between the system and the external field, while and can be considered to be their sum. For a fixed amount of , the entanglement is gradually replaced by the geometric phase in the quantum-classical transition of external field, and becomes the Berry phase when the quantum field vanishes. In a figurative sense, Berry phase is semiclassical entanglement between a quantum system and a classical environment.
## Iv Harmonic Oscillator
We also study a displaced harmonic oscillator which is another canonical example of the Berry phase Chaturvedi et al. (1987) to verify our above discussions. Here we add a qubit as the part of the its environment, which is described in terms of the Pauli operators . Suppose the qubit-oscillator interaction is characterized by a JC term Jaynes and Cummings (1963), we get
H0=νb†b+g(σ−b†+σ+b), (15)
where are displaced creation and annihilation operators with the frequency , is the coupling constant. We take and to be slowly rotating parameters. One can find its instantaneous eigenstates in the subspace , where , and satisfies . The instantaneous eigenstates with eigenvalues are derived as
|ψ+n⟩=V(t)(cosαn2|n⟩|↑⟩−sinαn2|n+1⟩|↓⟩), |ψ−n⟩=V(t)(sinαn2|n⟩|↑⟩+cosαn2|n+1⟩|↓⟩), (16)
where . The concurrence of states is
Ch=sinαn, (17)
which is zero for the state , corresponding to the eigenstate in (IV) with . We also choose a Hamiltonian to drive the rotation of the qubit with the frequency of , which is . The neglected Hamiltonian in this case is . Follow the same steps of the qubit case in the above section, we obtain the time-energy uncertainty
Sqh=2π√4(2n+1)|β|2+sin2αn, (18)
the Berry phase
γqh=2π|β|2±π(1−cosαn), (19)
and the mixed geometric phase
γmixh=2π|β|2. (20)
Here, both the phases are defined modulo . Obviously there exist an entanglement induced term in each of the expression of the time-energy uncertainty and the Berry phase. And the entanglement and the mixed geometric phase are complementary for a fixed value of .
## V Summary
In this work, we present a viewpoint of the Berry phase that it reflects the nonseparability between an open system and its classical environment just like the entanglement between a system and a quantum environment. This unified view of the two concepts inspires us to explore their properties and connection in the quantum-classical transition of the environment. The viewpoint is supported by the fact that the Berry phase can be considered as the effect of a neglected Hamiltonian which affects the system but has no effect on the eigenvalues and eigenstates in the adiabatic limit. This understanding can be obtained by checking the relation between the Berry phase and the AA phase, and provides an approach to extend the Berry phase to the system in a quantum-classical hybird environment. In addition, the geometric time-energy uncertainty as an accumulating result of the neglected Hamiltonian in a cyclic evolution is found to be a signature of the nonseparability not only for a classical environment but also for a quantum-classical hybird one.
Based on foregoing considerations we investigate a qubit under an adiabatically rotating quantum-classical hybrid field. The entanglement between the principal qubit and the field introduces a correction to the time-energy uncertainty and a corresponding effect on the Berry phase. For a fixed time-energy uncertainty, the entanglement is gradually replaced by the mixed geometric phase, determined solely by the geometry of the the qubit, in the quantum-classical transition of external field. That is, the geometric phase and entanglement have complementary relationship to reflect the nonseparability between the system and the external field. We also make similar calculations for a model of displaced harmonic oscillator to verify these conclusions, in which a qubit acts as the quantum part of the field.
We have been very careful to introduce a quantum part of the environment in our models without giving up the conditions of adiabaticity and cyclicity in Berry’s original definition of geometric phase. These models allow us to treat the geometric phases and quantum entanglement uniformly. We believe that, by removing the restrictions in this work, one may uncover more general connections among geometric aspects of quantum mechanics and different quantum correlations Modi et al. (2010).
###### Acknowledgements.
The idea was initiated in our discussions with Wu-Sheng Dai, Mi Xie and Da-Bao Yang. We are very indebted to Sir Michael Berry for his remarks and encouragement. F.L.Z. also thanks Jing-Ling Chen, Mang Feng, Paola Verrucchi, and Kamal Berrada for their valuable comments. This work was supported by NSF of China (Grant No. 11105097).
## References
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• Jaynes and Cummings (1963) E. T. Jaynes and F. W. Cummings, Proc. IEEE 51, 89 (1963).
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• not (a) In the full text but this equation we set the reduced Planck constant .
• not (b) This shows the Berry phases induced by quantum field we obtained below violate the gauge invariance, as it can be affected by the choice of instantaneous eigenstates. However the difference of the Berry phases between two eigenstates is an invariant under the gauge transformations. This remark is also applicable to the results in Fuentes-Guridi et al. (2002) as well. One can change the Berry phases in Fuentes-Guridi et al. (2002) by replacing the phase shift operator by without influence on the rotation of the field.
• Layton et al. (1990) E. Layton, Y. Huang, and S.-I. Chu, Phys. Rev. A 41, 42 (1990).
• Wootters (1998) W. K. Wootters, Phys. Rev. Lett. 80, 2245 (1998).
• Chaturvedi et al. (1987) S. Chaturvedi, M. S. Sriram, and V. Srinivasan, J. Phys. A: Math. Gen. 20, L1071 (1987).
• Modi et al. (2010) K. Modi, T. Paterek, W. Son, V. Vedral, and M. Williamson, Phys. Rev. Lett. 104, 080501 (2010).
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The feedback must be of minimum 40 characters and the title a minimum of 5 characters | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9285941123962402, "perplexity": 669.0424218780158}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703517559.41/warc/CC-MAIN-20210119011203-20210119041203-00478.warc.gz"} |
https://physics.stackexchange.com/questions/176717/ground-state-for-interacting-field-thoeries | # Ground state for interacting field thoeries
Are there references where the ground state of an interacting quantum field theory is explicitly written in terms of states of the underlying free theory?
For example, let us suppose to have a self interacting scalar field theory (with a potential $\phi^4$). Are there references expressing its ground state in terms of free states of the underlying free scalar field theory (without the potential $\phi^4$)?
In fact, there are an many references about perturbation theory in field theory but I do not seem to find one addressing this problem. For example, I guess it might be possible to use some standard time-independent perturbation theory but it would be nice to have a reference as guidance to correctly deal with the infinities.
Here I constructed perturbation-like approximants converging to the vacuum in $\phi^4_2g(x)$ (technically an interacting QFT, although not translation invariant, so Haag's theorem does not apply). There are no "infinities" in this case. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8961877226829529, "perplexity": 217.74046202827938}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146004.9/warc/CC-MAIN-20200225014941-20200225044941-00481.warc.gz"} |
https://overbrace.com/bernardparent/viewtopic.php?f=17&t=307&start=10 | Intermediate Thermodynamics Assignment 4 — Conservation of Energy
There was a typo in the formulation of Question 2, and this is why the answers for Q2 seemed wrong. Check the problem formulation again.
04.23.19
Previous 1 , 2 • PDF 1✕1 2✕1 2✕2
$\pi$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8104126453399658, "perplexity": 909.1673408167268}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195527396.78/warc/CC-MAIN-20190721225759-20190722011759-00367.warc.gz"} |
http://mathoverflow.net/questions/64029/if-a-manifold-suspends-to-a-sphere?sort=newest | If a manifold suspends to a sphere…
I have a topological manifold whose suspension is homeomorphic to the sphere $S^{k+1}$. Is it necessarily itself homeomorphic to $S^k$?
I know that this is not true if I replace "suspension" with "double suspension", because I found the helpfully named http://en.wikipedia.org/wiki/Double_suspension_theorem.
-
This is way outside my area of expertise, so perhaps someone can explain why the answer does not follow from the double suspension theorem: start with the Poincare dodecahedral space $M$ (a homology 3-sphere with nontrivial fundamenatal group) and suspend it once. If you get something homeomorphic to $S^4$, then $M$ is a counterexample. If not, then by DST $SM$ is homeomorphic to $S^5$ so $SM$ is a counterexample. – Pete L. Clark May 5 '11 at 18:29
Interesting plan! But it is not obvious to me that the suspension of M (or any other space obtained by a similar method) is a topological manifold. – James Cranch May 5 '11 at 18:54
Okay, so that's what I was missing: that the suspension of a manifold might or might not be a manifold. Like I said: not my area of expertise. (I guess the upvotes on my previous comment mean: "yes, I was wondering that too...") – Pete L. Clark May 5 '11 at 20:15
Yes, it's pretty easy that the suspension of a space $X$ cannot possibly be an $n+1$-manifold unless $X$ is homotopy equivalent to $S^n$. – Tom Goodwillie May 6 '11 at 0:37
Suppose $M$ is a closed $n$-manifold whose suspension is homeomorphic to $S^{n+1}$. Removing the two "singular" points from the suspension gives $M\times \mathbb R$, while removing two points from $S^{n+1}$ gives $S^n\times\mathbb R$. Thus $M\times \mathbb R$ and $S^n\times\mathbb R$ are homeomorphic, which easily implies that $M$ and $S^n$ are h-cobordant, and hence $M$ and $S^n$ are homeomorphic.
You need $n>4$, though, don't you? – Benoît Kloeckner May 5 '11 at 19:10
To see that $M$ and $S^n$ are h-cobordant consider a homeomorphism $h$ of their products with $\mathbb R$, and use excision in homology to show that the submanifolds $S^n\times 0$ and $h(M\times t)$ bound an h-cobordism, where $t$ need to be sufficiently large to ensure that the submanifolds are disjoint. – Igor Belegradek May 5 '11 at 19:38
In fact, one need not involve h-cobordisms at all: just note that $M$ and $S^n$ are homotopy equivalent and use Poincare's conjecture. I guess, I just like to advertize that fact that if two closed manifolds become homeomorphic after multiplying by $\mathbb R$, then they are $h$-cobordant. :) – Igor Belegradek May 5 '11 at 20:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8759718537330627, "perplexity": 285.15429110758544}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824756.90/warc/CC-MAIN-20160723071024-00128-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://dojiboy9.atspace.cc/contest/view-post.php?p=camelot | # Camelot
We are given an $$r \times c$$ chessboard $$(1 \leq R \leq 30, \ 1 \leq C \leq 26)$$ where some cells have knights and one cell has a king.The knights and king can move freely, and there can be more than one piece in any cell at any time (the cells are very "large"). The goal is to select a cell for all of the pieces to meet in, in such a way that the sum of the number of moves taken by all pieces is minimal. At any point, if a king and a knight occupy the same cell, the king may join with the knight, and the two pieces become one knight piece.
1. Euler tour?
2. Shortest paths
3. Floyd-Warshall algorithm
4. Pick a knight to pick-up the king
5. Greedy (always meet on a king, or a knight?)
###### Algorithm 1.1
Use floyd-warshall to compute all shortest paths by knight and by king.
For each (meeting-point s, pick-up-point p, knight to pick up the king k):
cost(s,p,k) = dist(k,p,by_knight) + dist(king,p,by_king) + dist(p,s,by_knight)
For each knight h != k:
cost(s,p,k) += dist(h,s)
return min(cost(s,p,k)) over all (s,p,k)
Basically the main idea is to use Floyd-Warshall to compute all the distances. We then pick a knight to go pick up the king, and also a cell where they should meet. Given a meeting cell, a pick-up cell, and a knight, we can use the previously computed distances to calculate the total sum of distances / moves. Take the minimum over all such choices, and this must be our answer. This approach is correct. However, it is far too slow. The Floyd-Warshall algorithm alone runs in $$O(N^3)$$ time where $$N = R \times C$$ is the number of cells, which can be around 800. So, this approach got TLE.
Since the number of nodes is too high, an $$O(N^3)$$ algorithm is too slow. Therefore, I need an algorithm that is based on some greedy observation or on classic "shortest path" algorithms, if I hope to solve this problem.
The conditions are complicated and not independent. How does the problem vary when we drop some combination of the conditions?
How do we solve this problem if we were not given any king at all?
If we have no king, we simply have to find the s which minimizes the sum of distances to all knights. Doing Floyd-Warshall is still too slow here, so we can't use it at all. However, we can use Dijkstra's algorithm, once from each possible "meeting-point", to compute the distances, and then take the sum. There is probably no better way to solve this modified problem anyway. With Dijkstra's algorithm, we can solve this in $$O(NlogN)$$ time per "meeting-point". So we get $$O(N^2logN)$$ in total, which is better than $$O(N^3)$$.
What if we only have a king (no knights)?
If we just have a king, we should always just place the meeting point on the king (obviously). If we (for some reason) still had to go through all meeting points, the answer is just $$\max(dx,dy)$$ where $$dx$$ and $$dy$$ are the absolute differences of $$x$$ and $$y$$ coordinate from the king to the meeting point; we would then try all meeting points (this solution would likely be $$O(N)$$).
What if we have a king, but we need not "pick-up" the king?
Then these problems are independent. We simply proceed with dijkstra's, computing the distances, and also computing the king distance. We then sum over these and take the minimum meeting point.
In attempting to generalize to the "pick-up-the-king" version, the above simplifications reminded me of something which I've heard called "2D-Dijkstra". In essence, we use a modified version of Dijkstra's algorithm, maintaining more than one value (i.e.: a pair, triple, tuple, or list of values) at each node; the values will be modified similarly to with Dijkstra's algorithm, but may not directly correspond to a "shortest" path (or may correspond to a shortest path on some auxiliary graph).
Given a fixed meeting-point $$s$$, can we do some kind of modified Dijkstra's algorithm (or possibly a 2D-Dijkstra) to solve the main problem? (It seems like this is the right approach based on the simplified problems; we have to try each source anyway, it seems.)
Assume that we have a fixed meeting cell $$s$$, and a king starting at some fixed cell $$k$$. Take any cell $$(r,c)$$; Take any path from $$(r,c)$$ to $$s$$. We define the detour distance of this path to be the total number of steps that a knight would take on this path, plus the minimum number of steps that it would take for the king $$k$$ to move to some cell on this path.
That is, the detour distance of this path is the total number of moves needed if a knight at $$(r,c)$$ were to go pick up the king somewhere on this path, and the king were to move to that cell, and then they would travel together (as one knight) to $$s$$.
With a fixed meeting cell $$s$$ and a king $$k$$, we define the detour distance of the cell $$(r,c)$$ to be the minimum detour distance over all paths from $$(r,c)$$ to $$s$$. The detour distance of cell $$(r,c)$$ will be denoted $$f(r,c)$$.
The goal is that, hopefully, these "detour distances", defined as above, can be computed via Dijkstra's algorithm in some way.
Two cells $$(r,c)$$ and $$(r',c')$$ are said to be neighbors if a knight on cell $$(r,c)$$ can move to cell $$(r',c')$$ in one step.
How do detour distances of neighbors relate? In particular, can we generate formulas and inequalities that relate detour distances of neighbors similarly to regular distances?
After fiddling around with some formulas, we get the following lemma.
$f(r,c) \leq \min_{\text{neighbors} (r',c')}{f(r', c') + 1}$
Take any optimal solution (i.e.: a path which generates a shortest detour distance) for $$(r,c)$$. There are two cases: either the king comes to this cell, or it does not. If the king must come to that cell, then the "detour distance" of any knight starting on that cell is the sum of the number of moves it takes for the king to get to that cell and for a knight to go from that cell to the meeting-point $$s$$. If the king will not go to that cell, then any knight starting on $$(r,c)$$ which is to pick up the king must travel at least one unit away. Hence, it must travel through some neighboring cell $$(r',c')$$ (which takes 1 step). After that, we can take the shortest path to pick up the king and go to the end from there (which takes $$f(r',c')$$ steps by definition). And this can be extended to any pair of neighbors. $$(r,c)$$ can always take a path through $$(r',c')$$. So, $$f(r,c) \leq (r',c') + 1$$. Furthermore, there must be some neighbor for which this applies (or else the first case applies, and the king must travel to the cell $$(r,c)$$). This proves the statement.
We can apply a modified version of Dijkstra's algorithm to solve this problem.
See analyses of the Algorithm below. Also, see my tutorial on shortest paths. Generally speaking, we have a graph where the nodes are the cells, and there are edges of length 1 between neighbors, and the weight of the "empty" path is some number dependent on the vertex it is on.
Let $$d(r,c)$$ be the distance (by knight) from cell $$(r,c)$$ to point $$s$$. Let $$f(r,c)$$ be the detour distance from cell $$(r,c)$$ to point $$s$$. MEETING-POINT(R, C, king_r, king_c, knights):
ans = infinity
For each possible meeting-point s:
Compute d(r,c) over all cells (r,c) by Dijkstra from s.
For each cell (r,c):
f(r,c) = d(r,c) + (distance from king to (r,c) by king-steps)
while(f(r,c) > f(r',c') + 1) for some neighbors (r,c) and (r',c'):
Choose the (r',c') with the minimum f(r',c') value
Set f(r,c) = f(r',c') + 1
The above algorithm is equivalent to running Dijkstra's algorithm on an auxiliary graph which would return the correct detour distances.
For each cell $$(r,c)$$ we define two auxiliary nodes: $$(r,c,no)$$ and $$(r,c,yes)$$ to represent a knight being on cell $$(r,c)$$ without the king, or with the king (respectively). We construct a graph on these auxiliary nodes so that an edge exists between $$(r,c,no)$$ and $$(r',c',no)$$ of length 1 if $$(r,c)$$ and $$(r',c')$$ are neighbors by knight. Similarly, an edge exists between $$(r,c,yes)$$ and $$(r',c',yes)$$ of length 1 for neighbors $$(r,c)$$ and $$(r',c')$$ by knight. Now, an edge exists between $$(r,c,no)$$ and $$(r,c,yes)$$ of length equal to the minimum steps it would take to get from the king to the cell $$(r,c)$$ by king-steps. Informally, we can model this problem by knights either "taking a step" or "calling the king". The king does not move until some knight specifically calls it to the knight's cell. Then, the king will travel directly to that cell, and the knight will now "have" the king; and the knight can then move on, taking steps again. Once the king has been "called" to the knight at cell $$(r,c)$$, the knight moves from state $$(r,c,no)$$ to state $$(r,c,yes)$$. So, the total number of "steps" needed to transition from $$(r,c,no)$$ to $$(r,c,yes)$$ was exactly the number of steps it took the king to get to cell $$(r,c)$$. Then, the detour distance $$f(r,c)$$ to a fixed cell $$s = (sr,sc)$$ can actually be seen as the shortest path distance from $$(r,c,no)$$ to $$(sr,sc,yes)$$. That is, we must eventually end up at cell $$s$$, and we must eventually have the king. This corresponds to eventually ending up in cell $$(sr,sc,yes)$$. And the minimum weight path over all these is the solution. (Note: This also gives us an alternative algorithm for solving this problem.) To show the reduction from our algorithm above to the shortest path algorithm described in this proof, we notice that, for any cell $$(r,c)$$, the distance from $$(r,c,yes)$$ to $$(sr,sc,yes)$$ is just the length of the shortest direct path (i.e.: minimum number of steps a knight must take to get there, $$d(r,c)$$). Once a knight has the king, the knight can simply move directly to the sink cell $$s$$. So, for a node $$(r,c,no)$$, the length of the path is no more than the length of the path $$(r,c,yes)$$ plus the cost of achieving/calling the king. This is equal to $$d(r,c) + \text{(distance from king to (r,c) by king-steps)}$$. And this is exactly what we initialize $$f(r,c)$$ to for each node $$(r,c)$$. Then, after some iterations, we update $$f(r,c)$$ based on the edges, which is equivalent to updating the distances of nodes $$(r,c,no)$$ based on neighbors $$(r',c',no)$$. Hence, our above algorithm is equivalent to running Dijkstra's algorithm on this auxiliary graph; but our algorithm automatically fills in the distances for the $$(r,c,yes)$$ nodes to be $$d(r,c)$$ (which they would be by the end of Dijkstra's algorithm anyway) and skips directly to the point in the algorithm where the $$(r,c,no)$$ nodes need to filled in. So, in the end $$f(r,c)$$ will be set exactly equal to the proper detour distance according to a reduction to this auxiliary graph.
Upon termination of the algorithm above, $$f(r,c)$$ will be equal to the detour distance of cell $$(r,c)$$.
We could also solve this problem by doing Dijkstra's algorithm directly on the auxiliary graph. This would be an alternative approach
The above algorithm can be made to run in $$O(N \log N)$$ time using the priority queue version of Dijkstra's algorithm.
In the end, this problem was reducible to solving shortest paths on multiple sources. Whenever one has multiple-source shortest-paths, the immediate idea is to use Floyd-Warshall's algorithm for this, due to its simplicity to code. However, the Floyd-Warshall algorithm runs in $$O(N^3)$$ (where N is the number of edges) whereas (at least for sparse graphs with small edges, in this case), simply calling Dijkstra's algorithm once for each source is $$O(N^2 \log N)$$ or something similar, which is much faster. In fact, in this problem, with $$N=800$$, Floyd-Warshall was utterly impossible. Once having decided that we want to use some variant of Dijkstra's algorithm once for each "source" (where I am using a "source" to mean "final meeting place"; it might be better to call it a "sink", but hopefully the reader understands), it is not easy to figure out "how" to use Dijkstra's algorithm here. The key realization is that, for a knight on cell $$(r,c)$$ either the king will go there directly, or the cell will have to make at least one step before meeting the king. This "at least one step" property is the key property that allows this problem to be solvable by Dijkstra's algorithm. Specifically, that means that the "detour distance" of a cell $$(r,c)$$ is no more than the detour distance of any of its knightly-neighbors, plus 1. This yields an inequality that smells a lot like the shortest-path inequalities: $$f(r,c) \leq f(r',c') + 1$$ over all neighbors $$(r',c')$$ (where neighbors are those which can be reached in one knight-step). After working this out, one comes to either a reduction to an auxiliary graph (where the states are the cells, along with an additional bit for whether the knight has the king or not), or a modified Dijkstra's algorithm that computes these $$f(r,c)$$ directly. This problem was neat because it was a nice demonstration of modified shortest paths, especially non-trivial reductions. It also showed that, anytime we can get a system of inequalities of the form: $$f(x) \leq f(x') + w(x,x')$$, then we can likely solve the problem via shortest paths in some form or another.
1. $$f(r,c) \leq \min_{\text{neighbors} (r',c')}{f(r', c') + 1}$$ (the detour distance of a cell is no more than 1 plus the detour distance of its neighbor). This was the crucial observation. It took me all of my mental capabilities to get to this observation (probably because I was too busy thinking about optimizing the Floyd-Warshall brute-force approach). But the way I realized this was by applying some problem solving processes (namely, "drop and/or vary the conditions", or "simplify the problem") and realizing that, even under a simplified version of this problem, Dijkstra's algorithm would be required in some capacity.
2. This problem is reducible to Dijkstra's algorithm on an auxiliary graph. This became immediately intuitive once we realized the inequalities. It was much harder to prove this though. In the end I simply wrote the code for a "modified dijkstra's algorithm" that appeared correct; and it took me several attempts at this analysis before I could formally prove its correctness. In the end, my formal proof of correctness relied on reducing it to another algorithm (which was more literally an application of Dijkstra's algorithm); I would have liked to use a more direct proof. But nonetheless, sometimes it is better to code first and prove later, for problems that are "intuitive but hard to prove".
1. Recognizing more quickly that an $$O(N^3)$$ algorithm is too slow would have saved me a lot of time on this problem. Being able to recognize the necessity for a greedy or classic shortest-path algorithm would have been an asset.
2. The conditions / constraints (meet up at a place, pick up king, shortest path, etc.) were far to interrelated and dependent. A good problem-solving technique is to vary the constraints, drop different combinations of the constraints, and solve simpler forms of the problem, in order to determine what underlying structures there are inherent to the problem. For me, this meant asking the questions:
1. What happens if there is no king?
2. What happens if there is only a king and no knights? (With variations on this)
3. What about if there are kings and knights, but they can't "combine"?
These lead to some insights about the structure of this problem. For example, without a king, we still have to perform some kind of dijkstra's algorithm from each source, since Floyd-Warshall is too slow anyway.
3. I remembered the buzz-words: "2D-Dijkstra" and "modified Dijkstra". Once I decided that I wanted to use shortest-paths, I focused my attention on these kinds of algorithms (variants of Dijkstra's algorithm), which eventually became the final solution.
• Shortest Paths / Dijkstra's Algorithm / Modified Dijkstra
• Shortest Paths / All-Pairs Shortest Paths / Sparse Graphs
• Checkerboard / Chess Board / Knights and Kings
• Intuition / Intuitive but Hard to Prove problems
• Graph Theory / Shortest Paths
• Steiner Trees / Minimum Spanning Trees?
• I need to ask good questions (i.e.: the "problem-solving questions"). By asking "How does this problem change when I vary / drop some conditions", I was able to better understand the structure of the problem and then solve it.
• Anytime I have a set of objects, and a function over that set of objects that preserves some inequalities of the form: $$f(x) \leq f(y) + w(y,x)$$, then we can apply shortest path algorithms. We can apply shortest path algorithms in more general (but very similary) settings, depending on our definitions of $$+", \leq",$$ and $$w(y,x)$$. As such, I decided to write a tutorial / post analyzing the general shortest-path problem. See the tutorial here.
• Try FIXING one of the parameters (in this case, the source/sink cell). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8079427480697632, "perplexity": 422.12740875752485}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986673250.23/warc/CC-MAIN-20191017073050-20191017100550-00024.warc.gz"} |
https://repository.uantwerpen.be/link/irua/111815 | Publication
Title
Modeling ultrashort laser-induced emission from a negatively biased metal
Author
Abstract
A theoretical study of ultrashort laser-induced electron emission from a negatively biased metallic cathode has been performed. Classical as well as tunneling electron emission mechanisms are considered. It was found that electron emission is governed by an interplay of processes inside as well as above the cathode. A hybrid model is proposed, where the electron distribution within the target is retrieved from Boltzmann scattering integrals, while the charge distribution above it is studied by a Particle-In-Cell simulation. The results indicate that non-equilibrium effects determine the initial emission process, whereas the space charge above the target suppresses the effectively emitted charge.
Language
English
Source (journal)
Applied physics letters / American Institute of Physics. - New York, N.Y., 1962, currens
Publication
New York, N.Y. : American Institute of Physics, 2013
ISSN
0003-6951 [print]
1077-3118 [online]
Volume/pages
103:22(2013), p. 1-4
Article Reference
221603
ISI
000327696300020
Medium
E-only publicatie
Full text (Publisher's DOI)
Full text (publisher's version - intranet only)
UAntwerpen
Faculty/Department Research group [E?say:metaLocaldata.cgzprojectinf] Publication type Subject Affiliation Publications with a UAntwerp address | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8532037734985352, "perplexity": 4620.026272679862}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806066.5/warc/CC-MAIN-20171120130647-20171120150647-00073.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/152054-probability-involving-z-score-print.html | # probability involving z-score
Printable View
• July 26th 2010, 07:45 PM
Jskid
[RESOLVED]probability involving z-score
A bottling company uses a machine to fill plastic bottles with pop. Due to variations in the filling process, the pop amounts in all bottles are normally distributed with a mean of 300 ml and a standard deviation of 3 ml. One bottle is randomly selected, what is the probability that bottle contains less than 299.5 ml?
$z=\frac{299.5-300}{\frac{3}{\sqrt{n}}}$ I'm not sure what value n is, or if it's needed.
I think I'm suppose to take n=1.
For this question how I'm calculating the probability do I write $P(\bar{x}<299.5)$ or $P(x<299.5)$?
• July 26th 2010, 09:20 PM
mr fantastic
Quote:
Originally Posted by Jskid
A bottling company uses a machine to fill plastic bottles with pop. Due to variations in the filling process, the pop amounts in all bottles are normally distributed with a mean of 300 ml and a standard deviation of 3 ml. One bottle is randomly selected, what is the probability that bottle contains less than 299.5 ml?
$z=\frac{299.5-300}{\frac{3}{\sqrt{n}}}$ I'm not sure what value n is, or if it's needed.
Where has n come from? $\displaystyle z = \frac{x - \mu}{\sigma}$ and you have been told the values of mean, sd and x.
• July 27th 2010, 06:26 PM
Jskid
$\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$ where n is sample size.
What I don't understand is a remeber doing similar questions but just used the S.D. $\sigma$
• July 27th 2010, 06:42 PM
pickslides
Quote:
Originally Posted by Jskid
$\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$ where n is sample size.
What I don't understand is a remeber doing similar questions but just used the S.D. $\sigma$
$P(\bar{X}<299.5) \implies P\left(Z<\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$
$P(X<299.5) \implies P(Z<\frac{X-\mu}{\sigma})$
Can you see the difference?
• July 27th 2010, 08:22 PM
Jskid
Quote:
Originally Posted by pickslides
$P(\bar{X}<299.5) \implies P\left(Z<\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$
$P(X<299.5) \implies P(Z<\frac{X-\mu}{\sigma})$
Can you see the difference?
I think I understand. The top equation is for the probability of a sample mean and the lower one is for the probability of getting a sample of a certain value? Does that mean the two formula are the same because $\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{1}}}=\frac{X-\mu}{\frac{\sigma}{1}}$
• July 27th 2010, 10:18 PM
pickslides
Quote:
Originally Posted by Jskid
I think I understand. The top equation is for the probability of a sample mean and the lower one is for the probability of getting a sample of a certain value? Does that mean the two formula are the same because $\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{1}}}=\frac{X-\mu}{\frac{\sigma}{1}}$
I have never thought of it like that. But I am inclinded to say this is not true as you can be given a sample size $n>1$ for $X$ and the standard tranformation would apply given it was normal. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9506937265396118, "perplexity": 612.784676385238}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802771716.117/warc/CC-MAIN-20141217075251-00123-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/143233-complex-integration-print.html | # Complex Integration
• May 5th 2010, 12:18 PM
zizou1089
Complex Integration
How would I compute:
$I=\int_{2\pi }^{0}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}$
I have tried finding the poles by changing the values into complex values
where $z=e^{i\theta }$
$2cos\theta =z+\frac{1}{z}$
$2sin\theta =-i({z-\frac{1}{z}})$
I have concluded
$I=\oint \frac{\frac{4dz}{iz}}{(z+\frac{1}{z})-\frac{3i}{2}(z-\frac{1}{z})+2i}$
Could someone tell me if this correct and I cant find the poles cause I cant find anyway of factorising! Thanks in advance!
• May 5th 2010, 02:18 PM
ques
Quote:
Originally Posted by zizou1089
How would I compute:
$I=\int_{2\pi }^{0}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}$
$=> I= - \int_{0}^{2\pi}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}$
I have tried finding the poles by changing the values into complex values
where $z=e^{i\theta }$
$2cos\theta =z+\frac{1}{z}$
$2sin\theta =-i({z-\frac{1}{z}})$
I have concluded
$I=-\oint \frac{\frac{4dz}{iz}}{(z+\frac{1}{z})-\frac{3i}{2}(z-\frac{1}{z})+2i}$
Could someone tell me if this correct and I cant find the poles cause I cant find anyway of factorising! Thanks in advance!
Multiply iz with the denominator.
Simplifying will give us the denominator's expression as,
( 3 + 2i ) z^2 - 4 z + ( 2i - 3 )
Now you should be able to figure out the compex roots quite easily.(Happy)
• May 5th 2010, 03:19 PM
zizou1089
Quote:
Originally Posted by ques
Multiply iz with the denominator.
Simplifying will give us the denominator's expression as,
( 3 + 2i ) z^2 - 4 z + ( 2i - 3 ) = 0
Now you should be able to figure out the compex roots quite easily.(Happy)
Thats the bit Im stuck on (Worried)..
• May 5th 2010, 03:21 PM
Bruno J.
Quote:
Originally Posted by zizou1089
Thats the bit Im stuck on (Worried)..
Do you know the quadratic formula?
• May 5th 2010, 03:25 PM
ques
Consider ( 3 + 2i ) z^2 - 4 z + ( 2i - 3 ) = 0
For a(z^2) + bz + c = 0 the roots are
z = {-b +- (b^2 - 4ac)^(1/2) }/2a
Here z = [ 4 +- {16 - ( 2i - 3 ) (2i + 3) }]/ {2(2i + 3)}
You should be able to take it on from here.
• May 5th 2010, 03:27 PM
ques
Quote:
Originally Posted by Bruno J.
Do you know the quadratic formula?
I am new here and I am sorry that I gave out way too much!
• May 5th 2010, 03:48 PM
zizou1089
Yep got it with the quadratic formula, problem is sometimes using the formula i get a complex number within the square root, and the function cannot be factorised either, so is there a more consistent way of getting the roots??
• May 5th 2010, 03:51 PM
ques
Quote:
Originally Posted by zizou1089
Yep got it with the quadratic formula, problem is sometimes using the formula i get a complex number within the square root, and the function cannot be factorised either, so is there a more consistent way of getting the roots??
Why do you think that i^(1/5) or that ( i + 1)^(1/35) does not exist ?
• May 5th 2010, 04:15 PM
Bruno J.
Quote:
Originally Posted by ques
I am new here and I am sorry that I gave out way too much!
No worries mate, welcome aboard! (Clapping) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8990272879600525, "perplexity": 1709.0577122542127}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721355.10/warc/CC-MAIN-20161020183841-00073-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://encyclopediaofmath.org/wiki/Hyperbolic_partial_differential_equation,_numerical_methods | Hyperbolic partial differential equation, numerical methods
Methods for solving hyperbolic partial differential equations using numerical algorithms.
Various mathematical models frequently lead to hyperbolic partial differential equations. Only very infrequently such equations can be exactly solved by analytic methods. The most widely used methods are numerical methods. They find extensive use in solving problems of the mechanics of a continuous medium, in particular for the equations of gas dynamics (cf. Gas dynamics, numerical methods of), which are quasi-linear.
Numerical methods for solving hyperbolic partial differential equations may be subdivided into two groups: 1) methods involving an explicit separation of the singularities of the solution; 2) indirect computation methods, in which the singularities are not directly separated but are obtained in the course of the computation procedure as domains with sharp changes in the solutions.
The first group includes, for instance, the method of characteristics, which is only used for solving hyperbolic partial differential equations; it has found extensive use in solving the problems of gas dynamics.
The methods in the second group yield non-singular difference schemes (cf. Difference scheme). Let there be given, for example, the hyperbolic equation
$$\tag{1 } \frac{\partial w }{\partial t } = A \frac{\partial w }{\partial x } ,$$
where $A$ is an $( m \times m )$- matrix with $m$ distinct real eigen values and $w = w( x, t)$ is a vector function with $m$ components. The matrix $A$ may either be a function in $x$, $t$( in such a case (1) is a linear hyperbolic system), or may depend on $w = w ( x, t )$( a quasi-linear system). In the latter case, let the system of equations (1) be reduced to divergence form:
$$\frac{\partial w }{\partial t } = \ \frac{\partial F }{\partial x } + \psi ,$$
where $F$ is a vector function in $w$, $x$, $t$ such that $A = dF/dw$ and $\psi$ is a vector function in $w$, $x$, $t$. In the most important case $A$, $F$ and $\psi$ depend only on $w$. For (1) one may pose the Cauchy problem
$$w ( x, 0) = w _ {0} ( x)$$
with suitable boundary conditions.
As a rule, the basis on which finite-difference schemes are constructed is an approximation by the integral conservation law corresponding to the differential equation, using certain quadrature formulas on the contour of integration of the differential cell. If the solutions are smooth, approximation by the integral conservation law is equivalent to direct approximation of the corresponding differential equation. Finite-difference schemes must satisfy the approximation and stability requirements. These requirements are mutually independent and, in a sense, contradict one another. In divergence systems of differential equations it is the divergence (or the conservation) condition of the finite-difference scheme which is fundamental. Moreover, the finite-difference schemes must also satisfy a number of other necessary conditions, like dissipativeness, economy, etc. A two-level explicit finite-difference scheme for a linear equation of type (1) has the form
$$w _ {j} ^ {n + 1 } = \Lambda w _ {j} ^ {n} ,$$
where $\Lambda$ is a finite operator, i.e. can be represented in the form
$$\Lambda = \sum _ { \alpha = - q _ {1} } ^ { {q _ 2 } } B _ \alpha T _ {1} ^ \alpha ,$$
where $B _ \alpha$ are $( m \times m)$- matrices with coefficients that depend on $\tau$, $h$, $x$, $t$; $t = n \tau$, $x = jh$; $\tau$, $h$ are the steps of the finite-difference grid along the axis $t$ and $x$, respectively; the numbers $q _ {1}$ and $q _ {2}$ do not depend on $\tau$, $h$; $\kappa = \tau / h$; and $T _ {1}$ is a displacement operator with respect to $x$.
The consistency conditions lead to the relationships
$$\sum _ { \alpha = - q _ {1} } ^ { {q _ 2 } } B _ \alpha = I,$$
$$\sum _ { \alpha = - q _ {1} } ^ { {q _ 2 } } \alpha B _ \alpha = \kappa A,$$
where $I$ is the identity matrix.
The implicit finite-difference scheme may be written as follows:
$$\Lambda _ {1} w _ {j} ^ {n + 1 } = \Lambda _ {0} w _ {j} ^ {n} ,$$
where $\Lambda _ {1}$ and $\Lambda _ {0}$ are finite operators,
$$\Lambda _ {k} = \ \sum _ { \alpha = - q _ {1} } ^ { {q _ 2 } } B _ \alpha ^ {k} T _ {1} ^ \alpha ,\ \ k = 0, 1,$$
where $B _ \alpha ^ {k}$ are $( m \times m )$- matrices which depend on $\tau$, $h$, $x$, $t$, and the operator $\Lambda _ {1}$ contains at least two non-zero matrices $B _ \alpha ^ {1}$. The operator $\Lambda _ {1}$ is assumed to be invertible, but its inverse need not be finite.
The various schemes may be subdivided into two classes by their approximation properties: the conditionally approximating schemes and the absolutely-approximating schemes. The conditionally-approximating finite-difference schemes approximate the initial differential equation for $\tau$, $h$ tending to zero with a certain interdependence between them: $\tau = \phi ( h)$. Absolutely-approximating finite-difference schemes approximate the initial differential equation as $\tau$, $h$ tend to zero in accordance with an arbitrary law.
In the case of conditional approximation the finite-difference equation may approximate various differential equations for various conditions of limit transition. For instance, for the equation
$$\tag{2 } \frac{\partial u }{\partial t } + a \frac{\partial u }{\partial x } = 0,\ \ a = \textrm{ const } > 0,$$
one may consider, e.g., the two finite-difference schemes
$$\tag{3 } \left . \begin{array}{c} { \frac{u _ {j} ^ {n + 1 } - \overline{u}\; {} _ {j} ^ {n} } \tau } + a { \frac{u _ {j + 1 } ^ {n} - u _ {j - 1 } ^ {n} }{2h} } = 0, \\ \overline{u}\; {} _ {j} ^ {n} = { \frac{1}{2} } ( u _ {j + 1 } ^ {n} + u _ {j - 1 } ^ {n} ), \\ \end{array} \right \}$$
$$\tag{4 } { \frac{u _ {j} ^ {n + 1 } - u _ {j} ^ {n} } \tau } + a { \frac{u _ {j} ^ {n} - u _ {j - 1 } ^ {n} }{h} } = 0.$$
If the law governing the limit transition is
$${ \frac \tau {h} } = \textrm{ const } ,$$
then the finite-difference scheme (3) approximates equation (2), while if that law is
$${ \frac \tau {h} ^ {2} } = \textrm{ const } ,$$
it approximates the equation
$$\frac{\partial u }{\partial t } + a \frac{\partial u }{\partial x } = \mu \frac{\partial ^ {2} u }{\partial x ^ {2} } ,\ \ \mu = \frac{h ^ {2} }{2 \tau } .$$
The finite-difference scheme (4) absolutely approximates equation (2).
In a similar manner, finite-difference schemes may be subdivided into conditionally-stable schemes and absolutely-stable schemes. Thus, the finite-difference scheme (4) will be stable if the following condition (Courant's condition) is met:
$$\frac{\tau a }{h } \leq 1,$$
i.e. if it is conditionally stable. On the other hand, the implicit finite-difference scheme
$$\frac{u _ {j} ^ {n + 1 } - u _ {j} ^ {n} } \tau + a \frac{u _ {j + 1 } ^ {n + 1 } - u _ {j - 1 } ^ {n + 1 } }{2h } = 0$$
is stable for all relations between $\tau$ and $h$, i.e. it is absolutely stable.
Explicit finite-difference schemes are simple in realization, but are either conditionally stable or conditionally approximating. In the case of an absolutely-approximating finite-difference scheme the condition of stability of the explicit scheme usually has the form
$$\tau \leq \textrm{ const } \cdot h ^ \beta \ \ ( \beta \geq 1),$$
which results in too small a step $\tau$ and involves unjustifiable labour in computations. All absolutely-stable absolutely-approximating schemes are contained in the class of implicit schemes.
Implicit finite-difference schemes are more complex in realization on passing from one time level to another, but then the step size $\tau$ may often be chosen arbitrary large, and is thus determined by considerations of accuracy only.
The convergence theorems for finite-difference schemes approximating linear differential equations make it possible to reduce the study of the convergence of such schemes to a study of their stability.
The study of the approximation of a finite-difference scheme corresponding to a hyperbolic equation is rather simple in the case of smooth solutions, has a local character, and in fact amounts to expansion into Taylor series; if the solution is discontinuous, the problem becomes difficult and consists of the verification of integral conservation laws. The study of stability is much more complicated.
The stability of finite-difference schemes which approximate hyperbolic equations with constant coefficients is studied by the Fourier method — i.e. an estimate is made of the norm of the Fourier transform of the step operator of the finite-difference scheme. Since the spectral radius of the matrix of the Fourier transform of the step operator does not exceed the norm of the matrix, a necessary criterion for stability follows: For a finite-difference scheme to be stable it is necessary that the spectral radius of the Fourier transform of the step operator does not to exceed $1 + O ( \tau )$, where $\tau$ is the step of the scheme along the $t$- axis. This is also a necessary condition for finite-difference schemes with variable coefficients and, if certain supplementary restrictions are imposed, this is also a sufficient condition. The following methods are used in the study of finite-difference schemes with variable coefficients, as well as for certain non-linear equations: the method of majorizing or of a priori estimates, and a local algebraic method.
The method of a priori estimates is analogous to the corresponding method for differential equations, but in the case of finite differences its realization involves major difficulties, owing to the specific features of finite-difference analysis in which — unlike in the method of a priori estimates in the theory of differential equations — many relationships take a tedious form.
The simplest majorizing estimate is the estimate for finite-difference schemes with positive coefficients.
For instance, consider the following finite-difference scheme for equation (2) with $a = a( x)$:
$$\tag{5 } { \frac{u _ {j} ^ {n + 1 } - u _ {j} ^ {n} } \tau } + a _ {j} { \frac{u _ {j} ^ {n} - u _ {j - 1 } ^ {n} }{h} } = 0,\ \ a _ {j} = a ( jh).$$
Then, if
$$0 \leq 1 - \kappa _ {j} \leq 1,\ \ \kappa _ {j} = { \frac{\tau a _ {j} }{h} } ,$$
the following estimate holds:
$$\| u ^ {n + 1 } \| \leq \| u ^ {n} \| ,$$
where
$$\| u ^ {n} \| = \max _ { j } | u _ {j} ^ {n} | .$$
It follows that the scheme (5) is uniformly stable in the space $C$. This estimate can be applied to finite-difference schemes which approximate hyperbolic systems of equations in invariants.
A very important, though limited, class of finite-difference schemes is represented by schemes with positive coefficients and matrices (so-called majorant schemes). If the coefficients of such schemes are symmetric, positive and Lipschitz continuous with respect to $x$, then they are stable in the space $L _ {2}$. As a rule, these are finite-difference schemes of first-order approximation in which the derivatives are approximated by one-sided differences. In higher-order approximations, when central differences are taken, one need not obtain positive coefficients. In such cases a priori estimates of a more general type in the spaces $W _ {2} ^ {p}$ are used.
For instance, let (1) be an acoustic scheme where
$$w = \left ( \begin{array}{c} u \\ v \end{array} \right ) ,\ \ A = \left ( \begin{array}{cc} 0 & a \\ 1 & 0 \\ \end{array} \right ) ;$$
and let the functions $a( x, t)$, $u( x, t)$ and $v( x, t)$ be periodic with respect to $x$ with period $2 \pi$. The a priori estimate for the finite-difference scheme
$$\tag{6 } \left . \begin{array}{c} { \frac{u ^ {n + 1 } ( x) - u ^ {n} ( x) } \tau } = a { \frac{v ^ {n + 1 } ( x + h) - v ^ {n + 1 } ( x) }{h} } , \\ { \frac{v ^ {n + 1 } ( x) - v ^ {n} ( x) } \tau } = \ { \frac{u ^ {n + 1 } ( x) - u ^ {n + 1 } ( x - h) }{h} } \\ \end{array} \right \}$$
has the form
$$\| E ^ {n + 1 } \| ^ {2} \leq \ \frac{1 + b _ {1} \tau }{1 + b _ {2} \tau } \| E ^ {n} \| ^ {2} ,$$
where
$$\| E \| ^ {2} = \ \int\limits _ {- \pi } ^ \pi ( u ^ {2} + av ^ {2} ) dx,$$
$$b _ {1} = \sup _ {x, n, \tau } \ \left | \frac{a ^ {n + 1 } ( x) - a ^ {n} ( x) }{\tau a ^ {n} ( x) } \right | ,$$
$$b _ {2} = \sup _ {x, n, h } \left ( { \frac{1}{\sqrt {a ^ {n + 1 } ( x + h) }} } \left | \frac{a ^ {n + 1 } ( x + h) - a ^ {n + 1 } ( x) }{h} \right | \right ) .$$
The given estimate proves the stability of the finite-difference scheme (6) and is analogous to the energy inequality for the system of acoustic equations.
The local algebraic method is based on the study of the properties of the local finite-difference operator obtained from the respective finite-difference operator with variable coefficients by "freezing" the coefficients. In this way the analysis of the stability of a finite-difference operator with variable coefficients is replaced by the analysis of an entire family of operators with constant coefficients. The local stability criterion is a generalization of the method of "freezing" the coefficients employed in the theory of differential equations.
The local stability criterion is closely connected with the dissipative stability criterion. A finite-difference scheme is called dissipative of order $\nu$, where $\nu$ is an even number, if there exists a $\delta > 0$ such that
$$| \rho | \leq 1 - \delta | \xi | ^ \nu ,$$
$$| \xi | = | kh | \leq \pi ,$$
where $\rho$ is the eigen value of maximum modulus of the transition matrix (the Fourier transform of the step operator) of the finite-difference scheme, and $k$ is a dual variable. Then if the order of approximation of the scheme is $2r + 1$( $2r + 2$), $r = 0, 1 \dots$ and if it is dissipative of order $2r + 2$( $2r + 4$), the scheme will be stable in $L _ {2}$ for hyperbolic systems of first-order differential equations with a Hermitian matrix.
In studying stability of finite-difference schemes for non-linear hyperbolic equations (in particular, for the equations of gas dynamics), the differential approximation method, in which the analysis of the finite-difference scheme is replaced by the analysis of its differential approximation, is employed.
For instance, the differential approximation of the finite-difference scheme (4) for equation (2) is constructed as follows. The expansion in (4) of the functions
$$u _ {j} ^ {n + 1 } = u ( jh, ( n + 1) \tau ),$$
$$u _ {j - 1 } ^ {n} = u (( j - 1) h, n \tau )$$
into Taylor series with respect to the point $x = jh$, $t = n \tau$ by the parameters $\tau$ and $h$ yields the $\Gamma$- form of the differential representation of the finite-difference scheme:
$$\tag{7 } \frac{\partial u }{\partial t } + \sum _ {l = 2 } ^ \infty { \frac{\tau ^ {l - 1 } }{l!} } \frac{\partial ^ {l} u }{\partial t ^ {l} } + a \frac{\partial u }{\partial x } + a \sum _ {l = 2 } ^ \infty { \frac{(- 1) ^ {l - 1 } h ^ {l} }{l!} } \frac{\partial ^ {l} u }{\partial x ^ {l} } = 0.$$
The elimination of the derivatives
$$\frac{\partial ^ {2} u }{\partial t ^ {2} } ,\ \frac{\partial ^ {3} u }{\partial t ^ {3} } \dots$$
from (7) yields the $\Pi$- form of the differential representation of the scheme (4):
$$\tag{8 } \frac{\partial u }{\partial t } + a \frac{\partial u }{\partial x } = \ \sum _ {l = 2 } ^ \infty C _ {l} \frac{\partial ^ {l} u }{\partial x ^ {l} } ,$$
where $C _ {l}$ are certain coefficients which depend on $\tau$, $h$, $a$, and $C _ {l} = O ( \tau ^ {l-} 1 , h ^ {l-} 1 )$. Elimination of the terms of order $O ( \tau ^ {2} , h ^ {2} )$ from (7) and (8) yields, respectively, the $\Gamma$- form of the first differential approximation of the scheme (4):
$${ \frac \tau {2} } \frac{\partial ^ {2} u }{\partial t ^ {2} } + \frac{\partial u }{\partial t } + a \frac{\partial u }{\partial x } - { \frac{ah}{2} } \frac{\partial ^ {2} u }{\partial x ^ {2} } = 0 ,$$
and the $\Pi$- form of the first differential approximation of the scheme (4):
$$\tag{9 } \frac{\partial u }{\partial t } + a \frac{\partial u }{\partial x } = \ C _ {2} \frac{\partial ^ {2} u }{\partial x ^ {2} } ,$$
$$C _ {2} = { \frac{ah}{2} } \left ( 1 - \frac{a \tau }{h} \right ) .$$
It has been proved in the linear case, for several finite-difference schemes, that if the first differential approximation is correct, then the respective finite-difference scheme is stable. Thus, in the case of scheme (4) above, the fact that (9) is correct means that $C _ {2} \geq 0$, i.e. that a necessary and sufficient condition, $a \tau /h \leq 1$, for the stability of the scheme is satisfied. The terms with even derivatives in equation (8) ensure the dissipative properties of the finite-difference scheme, while those with odd derivatives are responsible for its dispersive properties.
The dissipation of the finite-difference scheme (4) is the magnitude
$$d = 1 - | \rho | ^ {2} = \ 1 - \mathop{\rm exp} \left [ \sum _ {l = 1 } ^ \infty (- 1) ^ {l} \frac \tau {h ^ {2l} } C _ {2l} \xi ^ {2l} \right ] ,$$
where
$$\rho = 1 - \frac{a \tau }{h} ( 1 - e ^ {- i \xi } )$$
is the amplification factor of the scheme; the dispersion of the scheme (4) is the magnitude
$$\kappa = \kappa a \xi - \mathop{\rm arg} \rho = \ \sum _ {l = 1 } ^ \infty (- 1) ^ {l + 1 } \frac \tau {h ^ {2l + 1 } } C _ {2l + 1 } \xi ^ {2l + 1 } .$$
The dissipative terms in (8) determine the properties of approximate viscosity of the scheme (i.e. a certain mechanism of smoothing in the finite-difference scheme). The form of the dissipative terms is determined both by artificial terms introduced into the initial differential equation and by the structure of the finite-difference scheme itself. The first differential approximation yields the principal term of the approximate viscosity. The method of differential approximation is widely employed in the study of differential schemes for non-linear equations and makes it possible to explain the instability effects of various finite-difference schemes which can be encountered in specific computations and which are not locally revealed by the Fourier method.
The construction of finite-difference schemes in multi-dimensional cases is based on splitting methods (weak approximation) and fractional-step methods, by which the integration of the initial multi-dimensional equation is reduced to the integration of equations of a simpler structure (cf. Fractional steps, method of).
The method of solving hyperbolic equations may be developed by using the finite-element method, which may be considered as a finite-difference method on a special irregular grid.
References
[1] S.K. Godunov, V.S. Ryaben'kii, "The theory of difference schemes" , North-Holland (1964) (Translated from Russian) MR0163421 [2] R.D. Richtmeyer, K.W. Morton, "Difference methods for initial value problems" , Wiley (1967) MR220455 [3] B.L. Rozhdestvenskii, N.N. Yanenko, "Systems of quasilinear equations and their applications to gas dynamics" , Amer. Math. Soc. (1983) (Translated from Russian) MR0694243 [4] A.A. Samarskii, A.V. Gulin, "Stability of difference schemes" , Moscow (1973) (In Russian) MR2263771 MR1800050 MR1473135 MR1441771 MR0405838 MR0431671 MR0314262 MR0268694 MR0264870 MR0232229 MR0146975 MR0156471 Zbl 1002.65101 Zbl 1189.34107 Zbl 0939.65082 Zbl 0814.65092 Zbl 0368.65032 Zbl 0368.65031 Zbl 0334.65072 Zbl 0303.65079 Zbl 0313.65081 [5] N.N. Yanenko, Yu.I. Shokin, "The first differential approximation to finite-difference schemes for hyperbolic systems of equations" Siberian Math. J. , 10 : 5 (1969) pp. 868–880 Sibirsk. Mat. Zh. , 10 : 5 (1969) pp. 1173–1187 Zbl 0196.11701 [6] S.I. Serdyukova, "A necessary and sufficient condition for the stability of a class of difference boundary-value problems" Soviet Math. Dokl. , 14 : 1 (1973) pp. 50–54 Dokl. Akad. Nauk SSSR , 208 : 1 (1973) pp. 52–55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9660999178886414, "perplexity": 340.12355351797913}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154214.36/warc/CC-MAIN-20210801123745-20210801153745-00227.warc.gz"} |
http://www.computer.org/csdl/trans/tp/1999/01/i0058-abs.html | Subscribe
Issue No.01 - January (1999 vol.21)
pp: 58-65
ABSTRACT
<p><b>Abstract</b>—This paper studies the computation of projective invariants in pairs of images from uncalibrated cameras and presents a detailed study of the projective and permutation invariants for configurations of points and/or lines. Two basic computational approaches are given, one algebraic and one geometric. In each case, invariants are computed in projective space or directly from image measurements. Finally, we develop combinations of those projective invariants which are insensitive to permutations of the geometric primitives of each of the basic configurations.</p>
INDEX TERMS
Uncalibrated stereo, projective and permutation invariants, indexation, projective reconstruction, cross ratio, Grassmann-Cayley algebra.
CITATION
Gabriella Csurka, Olivier Faugeras, "Algebraic and Geometric Tools to Compute Projective and Permutation Invariants", IEEE Transactions on Pattern Analysis & Machine Intelligence, vol.21, no. 1, pp. 58-65, January 1999, doi:10.1109/34.745735 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9577596187591553, "perplexity": 2368.9062313557583}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398452560.13/warc/CC-MAIN-20151124205412-00120-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://mathhelpforum.com/discrete-math/125452-proof-series-binomial-coefficients.html | Thread: Proof of a series with binomial coefficients
1. Proof of a series with binomial coefficients
I don't have any experience dealing with binomial coefficients, so I'm really confused on how to do this following problem. I wrote the binomial coefficients in parenthesis with some space between where the top and bottom part should be (not sure how to format it otherwise
Problem: Prove that
n
Sigma (n i) = 2^n
i=0
and
n
Sigma (-1)^n * (n i) = 0
i=0
I'm not sure if I need to use induction or not to prove either of these and I'm not sure where I'd start even the base of induction if I did have to.
2. Originally Posted by uberbandgeek6
I don't have any experience dealing with binomial coefficients, so I'm really confused on how to do this following problem. I wrote the binomial coefficients in parenthesis with some space between where the top and bottom part should be (not sure how to format it otherwise
Problem: Prove that
n
Sigma (n i) = 2^n
i=0
and
n
Sigma (-1)^n * (n i) = 0
i=0
I'm not sure if I need to use induction or not to prove either of these and I'm not sure where I'd start even the base of induction if I did have to.
The first can be done in a plethora of ways. I guess the easiest is to note that both are forms of the binomial expansion. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9676059484481812, "perplexity": 240.77010812072436}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163986869/warc/CC-MAIN-20131204133306-00030-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://homework.cpm.org/category/CON_FOUND/textbook/ac/chapter/14/lesson/14.2.1.7/problem/2-104 | ### Home > AC > Chapter 14 > Lesson 14.2.1.7 > Problem2-104
2-104.
The general equation for an aritmetic sequence is t(n) = mn + b.
Use t(8) = 1056 to write an equation.
t(8) = m(8) + b or 1056 = 8m + b
Use t(13) = 116 to write another equation.
Solve the system of equations you wrote for m and b.
t(n) = −188n + 2560
To find t(5), substitute 5 for n in the equation that you wrote. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8068653345108032, "perplexity": 3159.777272880476}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400202007.15/warc/CC-MAIN-20200921175057-20200921205057-00611.warc.gz"} |
https://getrevising.co.uk/revision-tests/kinetics-9 | Kinetics
HideShow resource information
• Created by: r98
• Created on: 19-03-16 12:34
What are the main factors that affect the rate of reactions?
Temperature, concentration, pressure, surface area and catalysts.
1 of 17
What's the definition of rate of reaction?
The change in concentration (of any of the reactants or products) with unit time.
2 of 17
How do you find the rate at a particular instant?
Draw a tangent to the curve at that time and then find its gradient (slope).
3 of 17
Why do we have to keep the temperature constant when measuring rate of reaction?
Because rate varies with temperature.
4 of 17
What are the units of reaction rates?
mol dm-3 s-1
5 of 17
Can a species thats not in the chemical equation appear in the rate equation?
Yes, for example a catalyst.
6 of 17
Is the rate constant, k, different for every reaction?
Yes.
7 of 17
What does k vary with?
Temperature.
8 of 17
What happens when all the species in the rate equation are 1 mol dm-3?
The rate of reaction is equal to the value of k.
9 of 17
What's the order of reaction?
The order of reaction, with respect to one species, is the power to which the concentration of that species is raised in the rate expression.
10 of 17
What does the order of reaction tell you?
How the rate depends on the concentration of that species.
11 of 17
Do all species in the chemical equation appear in the rate equation?
No, not necessarily.
12 of 17
Do the coefficience of species in the chemical equation have any importance in the rate expression?
No, they have no relevance.
13 of 17
What are the units of the rate constant, k?
The units of the rate constant vary depending on the overall order of the reaction.
14 of 17
Increasing temperature always increases what two things?
The rate of reaction & the value of the rate constant, k.
15 of 17
Do steps that occur after the rate-determining step affect the rate?
No, provided that it's fast compared with the rate-determining step.
16 of 17
Do species that are involved in the mechanism after the rate-determining step appear in the rate expression?
No, only species involved in or before the rate-determining step could affect the overall rate and therefore appear in the rate expression.
17 of 17
Other cards in this set
Card 2
Front
What's the definition of rate of reaction?
Back
The change in concentration (of any of the reactants or products) with unit time.
Card 3
Front
How do you find the rate at a particular instant?
Card 4
Front
Why do we have to keep the temperature constant when measuring rate of reaction?
Card 5
Front
What are the units of reaction rates? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9113836288452148, "perplexity": 1945.970915466473}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719155.26/warc/CC-MAIN-20161020183839-00103-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://nbviewer.org/github/jump-dev/SumOfSquares.jl/blob/gh-pages/v0.6.2/generated/Systems%20and%20Control/lyapunov_function_search.ipynb | Lyapunov Function Search¶
Adapted from: SOSTOOLS' SOSDEMO2 (See Section 4.2 of SOSTOOLS User's Manual)
In [1]:
using DynamicPolynomials
@polyvar x[1:3]
Out[1]:
(DynamicPolynomials.PolyVar{true}[x₁, x₂, x₃],)
We define below the vector field $\text{d}x/\text{d}t = f$
In [2]:
f = [-x[1]^3 - x[1] * x[3]^2,
-x[2] - x[1]^2 * x[2],
-x[3] - 3x[3] / (x[3]^2 + 1) + 3x[1]^2 * x[3]]
Out[2]:
3-element Vector{MultivariatePolynomials.RationalPoly{DynamicPolynomials.Polynomial{true, Int64}, DynamicPolynomials.Polynomial{true, Int64}}}:
(-x₁³ - x₁x₃²) / (1)
(-x₁²x₂ - x₂) / (1)
(3x₁²x₃³ + 3x₁²x₃ - x₃³ - 4x₃) / (x₃² + 1)
We need to pick an SDP solver, see here for a list of the available choices. We use SOSModel instead of Model to be able to use the >= syntax for Sum-of-Squares constraints.
In [3]:
using SumOfSquares
using CSDP
solver = optimizer_with_attributes(CSDP.Optimizer, MOI.Silent() => true)
model = SOSModel(solver);
We are searching for a Lyapunov function $V(x)$ with monomials $x_1^2$, $x_2^2$ and $x_3^2$. We first define the monomials to be used for the Lyapunov function:
In [4]:
monos = x.^2
Out[4]:
3-element Vector{DynamicPolynomials.Monomial{true}}:
x₁²
x₂²
x₃²
We now define the Lyapunov function as a polynomial decision variable with these monomials:
In [5]:
@variable(model, V, Poly(monos))
Out[5]:
$$(_[1])x_{1}^{2} + (_[2])x_{2}^{2} + (_[3])x_{3}^{2}$$
We need to make sure that the Lyapunov function is strictly positive. We can do this with a constraint $V(x) \ge \epsilon (x_1^2 + x_2^2 + x_3^2)$, let's pick $\epsilon = 1$:
In [6]:
@constraint(model, V >= sum(x.^2))
Out[6]:
$$(_[1] - 1)x_{1}^{2} + (_[2] - 1)x_{2}^{2} + (_[3] - 1)x_{3}^{2} \text{ is SOS}$$
We now compute $\text{d}V/\text{d}x \cdot f$.
In [7]:
using LinearAlgebra # Needed for dot
dVdt = dot(differentiate(V, x), f)
Out[7]:
$$\frac{(-2 _[1])x_{1}^{4}x_{3}^{2} + (-2 _[2])x_{1}^{2}x_{2}^{2}x_{3}^{2} + (-2 _[1] + 6 _[3])x_{1}^{2}x_{3}^{4} + (-2 _[1])x_{1}^{4} + (-2 _[2])x_{1}^{2}x_{2}^{2} + (-2 _[1] + 6 _[3])x_{1}^{2}x_{3}^{2} + (-2 _[2])x_{2}^{2}x_{3}^{2} + (-2 _[3])x_{3}^{4} + (-2 _[2])x_{2}^{2} + (-8 _[3])x_{3}^{2}}{x_{3}^{2} + 1}$$
The denominator is $x[3]^2 + 1$ is strictly positive so the sign of dVdt is the same as the sign of its numerator.
In [8]:
P = dVdt.num
Out[8]:
$$(-2 _[1])x_{1}^{4}x_{3}^{2} + (-2 _[2])x_{1}^{2}x_{2}^{2}x_{3}^{2} + (-2 _[1] + 6 _[3])x_{1}^{2}x_{3}^{4} + (-2 _[1])x_{1}^{4} + (-2 _[2])x_{1}^{2}x_{2}^{2} + (-2 _[1] + 6 _[3])x_{1}^{2}x_{3}^{2} + (-2 _[2])x_{2}^{2}x_{3}^{2} + (-2 _[3])x_{3}^{4} + (-2 _[2])x_{2}^{2} + (-8 _[3])x_{3}^{2}$$
Hence, we constrain this numerator to be nonnegative:
In [9]:
@constraint(model, P <= 0)
Out[9]:
$$(2 _[1])x_{1}^{4}x_{3}^{2} + (2 _[2])x_{1}^{2}x_{2}^{2}x_{3}^{2} + (2 _[1] - 6 _[3])x_{1}^{2}x_{3}^{4} + (2 _[1])x_{1}^{4} + (2 _[2])x_{1}^{2}x_{2}^{2} + (2 _[1] - 6 _[3])x_{1}^{2}x_{3}^{2} + (2 _[2])x_{2}^{2}x_{3}^{2} + (2 _[3])x_{3}^{4} + (2 _[2])x_{2}^{2} + (8 _[3])x_{3}^{2} \text{ is SOS}$$
The model is ready to be optimized by the solver:
In [10]:
JuMP.optimize!(model)
We verify that the solver has found a feasible solution:
In [11]:
JuMP.primal_status(model)
Out[11]:
FEASIBLE_POINT::ResultStatusCode = 1
We can now obtain this feasible solution with:
In [12]:
value(V)
Out[12]:
$$15.718362431619164x_{1}^{2} + 12.28610732110516x_{2}^{2} + 2.995845325502817x_{3}^{2}$$
This notebook was generated using Literate.jl. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8563966751098633, "perplexity": 2792.518124719459}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570827.41/warc/CC-MAIN-20220808122331-20220808152331-00642.warc.gz"} |
https://aviation.stackexchange.com/questions/66199/are-rhumb-lines-ever-used-as-leg-paths-for-rnp-rnav/66201 | # Are rhumb lines ever used as leg paths for RNP/RNAV?
Are rhumb lines ever used by an FMS as leg paths for RNP/RNAV? If so, what arinc 424 leg type would use them?
No. The FMS does not use rhumb lines.
All RNP/RNAV flight path legs are geodesics. The only exceptions are RF legs (a constant radius circular path about a turn center) and hold legs (a closed racetrack pattern).
To quote RTCA DO-236C, Minimum Aviation System Performance Standards: Required Navigation Performance for Area Navigation:
RNP Routes and user-preferred trajectories are assumed to use a series of fixes. The desired path is defined by a series of geodesic tracks joining successive fixes.
It also includes the following definition:
GEODESIC LINE [Bowditch]
A line of shortest distance between any two points on a mathematically defined surface. A geodesic line is a line of double curvature and usually lies between the two normal section lines which the two points determine. If the two terminal points are nearly in the same latitude, the geodesic line may cross one of the normal section lines. It should be noted that, except along the equator and along the meridians, the geodesic line is not a plane curve and cannot be sighted over directly.
• thank you very much – Daniel Ogburn Jul 3 '19 at 16:53 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8523069620132446, "perplexity": 2079.078109847539}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00284.warc.gz"} |
https://www.physicsforums.com/threads/question-on-uniform-convergence.180995/ | # Question on uniform convergence
1. Aug 19, 2007
### Simfish
5: Let p > 0. Let $$f_n (x) = x n^p e^{-nx}$$
(i) For what values of p is$$f_n (x) -> 0$$ uniformly on [0,1]?
(ii) For what values of p is it true that $$lim_{n->\infty} \int_0^1 f_n (x) dx = 0$$
For (i), we know that uniform convergence implies that the convergence to 0 depends ONLY on the value of N, so it should converge to 0 for all x. Now, we can make our function arbitrarily small for any p, if p < 1,since then for n -> infinity, both functions converge to 0.
Let's try (i) first.
We want to find the maximum value of $$f_n (x)$$. So...
\begin{align} \left[ f_n (x) \right]' = \left[- x n^{p+1} e^{-nx} + n^p e^{-nx}\right] = 0 \\ \left[ e^{nx} n^{p} (-nx + 1) \right] = 0 \\ -nx + 1 = 0 \\ nx = 1 \\ x = 1/n,\\ \end{align}
plug back into equation. the value of the maximum is... $$\left[ n^{p-1} * \frac{1}{e} \right]$$
which can only imply uniform convergence for all x for p $$<$$ 1.
But what of (ii)? Integration by parts.
$$x n^p \left[\frac{-x e^{-nx}}{n} - \frac{e^{-nx}}{n^2} \right]_{0}^{1} = \left[\frac{-e^{-n}}{n} - \frac{e^{-n}}{n^2} + \frac{1}{n^2} \right] x n^p$$
...which would imply convergence of the integral for all p < 2. Am I wrong anywhere? Someone else said that p > 0, but this relation seems to be completely relevant for all negative values of p as well.
Last edited: Aug 19, 2007
Can you offer guidance or do you also need help?
Similar Discussions: Question on uniform convergence | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.999562680721283, "perplexity": 661.0660690787881}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320362.97/warc/CC-MAIN-20170624221310-20170625001310-00070.warc.gz"} |
https://www.physicsforums.com/threads/two-easy-circuit-analysis-problems.191394/ | # Two easy circuit-analysis problems
1. Oct 15, 2007
### Molecular
First problem:
1. The problem statement, all variables and given/known data
I'm supposed to find the voltage across the 8k resistor in the following circuit:
3. The attempt at a solution
This just seems so mindboggling easy. My first thought is to use kirckhoffs current laws on all three nodes, giving me three equations with three unknowns.
The problem is this always ends up giving me a 0 = 0 equation. I'm guessing this might be because the equation for the upper left node = 1, the down left node = 2 and the node on the right = 3, so you could say left node equations = right node equation. I don't know, at any rate I always get a 0 = 0 equation so there's no help there either.
I've also tried using the node-voltage method, but this also turns out wrong every time. Admittedly this may be calculation errors on my side, but to some extent since I can't see the paths back to whatever node I chose as the reference node I've got a feeling it won't be correct anyway.
The problem seems so easy, and I know for a fact it's supposed to be easy, but i'm somehow missing something essential here, any help would be greatly appreciated.
Problem two:
Nevermind this one, I figured it out =\
1. The problem statement, all variables and given/known data
I'm supposed to find the Thevénin equivalent voltage of the following circuit:
Again this is a somewhat easy problem.
I start by simplifying the voltage source to a power source:
230v / 20 = 11.5 A in paralell with a 20 ohm resistance and a 30 ohm resistor, which together becomes a 12 ohm resistor.
So now I've got this 11.5 A power source in paralell with a 12 ohm resistor in paralell with a 12 A power source.
First question: Considering I can now splice these two power sources together, how come I need to subtract them from eachother when their current goes the same way? I know that's what I'm supposed to do, I'd just think it would be 12 + 11.5 instead of 12 - 11.5.
Second question: Aren't I correct in my observation that since the 12A current is stronger than the 11.5 one, the current after I put these together as one current source will be 0.5 but pointed downwards on the left side?
Anyway at the end I get a Tvh of -3.75, but it's not supposed to be negative. But I can't see how it's supposed to not become negative considering i'm almost positive that will be the direction of the current after adding the two sources up.
This is probably because I can't figure out the exact mechanics on adding two current sources together. All I know is I'm supposed to use kirckhoff's current law to do it, but I've really got no idea how, I just take it for granted as a rule that I can.
Last edited: Oct 15, 2007
2. Oct 15, 2007
### Molecular
Ok never mind problem two, I realize why now. 11.5 ampere goes down, 12 goes up, this means that the current across the 12 ohm resistor and the other 12 ohm resistor is a total of 0.5, this has to go upwards from the left current source because the 12 ampere current is headed upwards. Phew.
Still stumped at the first question though. I'm gonna try the node voltage method one more time ;o.
Edit: And don't ya know it, 5 minutes after I made this thread this one was right aswell. I get -12 V. I'd still like to ask though wether this is solvable in any easier way than using the node-voltage method? As I'm fairly sure it's supposed to be.
Last edited: Oct 15, 2007 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8969414830207825, "perplexity": 481.71352103305844}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988718278.43/warc/CC-MAIN-20161020183838-00498-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/quick-boolean-logic-question.370717/ | # Homework Help: Quick Boolean Logic Question
1. Jan 18, 2010
### rokimomi
1. The problem statement, all variables and given/known data
let p, q, and r be the following propositions
p: You get an A on the final exam.
q: You do every exercise in this book
r: You get an A in this class
translate: You get an A on the final, but you don't do every exercise in this book; nevertheless, you get an A in this class.
2. Relevant equations
3. The attempt at a solution
Do I have to include the part about doing exercises at all? Since it's sufficient enough to have p$$\rightarrow$$r to convey the message? What I'm worried about is if they want us to include it anyways so someone can go from this logic to English again. How would I include q then?
I would assume that
2. Jan 18, 2010
### story645
Yes, because it has a truth value and therefore affects the truth value of the whole sentence.
But is a conjunction, so logically/grammatically it works the same way as an and.
3. Jan 18, 2010
### rokimomi
How about ((p$$\wedge$$q)$$\vee$$(p$$\wedge$$$$\neg$$q))$$\rightarrow$$r
Is there any way to convey this simpler?
4. Jan 19, 2010
### story645
Using distrbutive properties, you end up with :
(p$$\wedge$$($$\neg$$q$$\vee$$q))$$\rightarrow$$r, which is back to p$$\rightarrow$$r, which again means a loss of the but clause.
5. Jan 19, 2010
### rokimomi
Wait, wasn't that my goal though? Something that simplifies to "if p then q".
Hm, im rereading it again, and im getting the feeling that I should just word for word put it into logic. So
(p$$\wedge$$$$\neg$$q) $$\rightarrow$$ r
So is their use of "but" just to confuse me?
Last edited: Jan 19, 2010
6. Jan 19, 2010
### story645
Probably.
That's my usual assumption with these types of problems.
7. Jan 19, 2010
### rokimomi
Oh wow, I over-read your comment about "but" the first time through. Sorry bout that and thanks for the help.
8. Jan 19, 2010
### cepheid
Staff Emeritus
Yeah, exactly right.
No, it's standard english. 'But' is the right conjunction to use, because the clause that comes after it is a negative, and tends to have the effect of lessening the impact of the first. It was your job to figure out that this sentence given in proper english is logically equivalent to:
"You get an A in the final exam and you do NOT do every exercise in the book..."
and you did figure it out. If somebody had said either wording to you, you would have understood what he meant. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9012282490730286, "perplexity": 1327.4694568804732}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267868135.87/warc/CC-MAIN-20180625150537-20180625170537-00378.warc.gz"} |
https://www.physicsforums.com/threads/implicitly-differentiating-pde-multivariable-calculus.271249/ | # Homework Help: Implicitly differentiating PDE (multivariable calculus)
1. Nov 11, 2008
### Legion81
The problem:
Find the value of dz/dx at the point (1,1,1) if the equation xy+z3x-2yz=0 defines z as a function of the two independent variables x and y and the partial derivative exists.
I don't know how to approach the z3x part. I thought you would use the product rule and get 3(dz/dx)2x + z3. But if that is right, the final equation looks something like
y + 3x(dz/dx)2 + z3 - 2y(dz/dx) = 0
And I don't think that is right. The only way I know to solve that would be with the quadratic equation and that gives a complex value. Am I forgeting the chain rule somewhere or just way off on approaching this problem?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Nov 11, 2008
### Frillth
I have done very little multivariable so this could easily be wrong, but if you implicitly differentiated z^3x shouldn't you get 3z^2x*dz/dx + z^3?
3. Nov 11, 2008
### Legion81
That's right because you have the 3z2x(dz/dx) + z3(dx/dx). I don't know what I was thinking... Thank You!
4. Nov 11, 2008
### ducnguyen2000
You didn't apply the chain rule, remember df(u)/dx = f(u)'*du/dx | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9468915462493896, "perplexity": 1090.9615593422616}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746528.84/warc/CC-MAIN-20181120171153-20181120193153-00557.warc.gz"} |
https://math.answers.com/Q/Can_the_product_of_two_numbers_be_less_than_either_of_the_numbers | 0
# Can the product of two numbers be less than either of the numbers?
Wiki User
2010-10-04 20:00:52
If one of the numbers is negative, but the other is positive, then the product is negative - and therefore smaller than both numbers in the question. For example, 2 x -4 = -8.
=====================================
Also, whenever the absolute magnitude of both factors is less than ' 1 ',
the absolute magnitude of the product is less than either factor.
Wiki User
2010-10-04 20:00:52
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https://www.physicsforums.com/threads/find-the-minimal-distance-between-any-point-on-the-sphere.270042/ | # Find the minimal distance between any point on the sphere
1. Nov 7, 2008
### helix999
How can we find the minimal distance between any point on the sphere whose centre is 2,1,3 and radius is 1 and any point on the sphere centred at -3,2,4 and radius is 4?
is there any formula for finding this minimum distance ?
2. Nov 7, 2008
### aniketp
Re: spheres
find the distance between the centres, and subtract the sum of radii of the spheres....
3. Nov 7, 2008
### helix999
Re: spheres
thnx aniketp...that was helpful...i got my answer
4. Nov 7, 2008
### aniketp
Re: spheres
no problem, but i hope you understood WHY this works.
5. Nov 7, 2008
### helix999
Re: spheres
well...yes...i have a math exam tomorrow....have to finish a LOT of things
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https://destevez.net/2019/04/diffraction-in-dslwp-b-lunar-occultations/ | # Diffraction in DSLWP-B lunar occultations
In February this year, the orientation of the orbit of DSLWP-B around the Moon was such that, when viewed from the Earth, it passed behind the Moon on every orbit. This opened up the possibility for recording the signal of DSLWP-B as it hid behind the Moon, thus blocking the line of sight path. The physical effect that can be observed in such events is that of diffraction. The power of the received signal doesn’t drop down to zero in a brick-wall fashion just after the line of sight is blocked, but rather behaves in an oscillatory fashion, forming the so called diffraction fringes.
The signal from DSLWP-B was observed and recorded at the Dwingeloo 25m radiotelescope for three days in February: 4th, 13th and 15th. During the first two days, an SSDV transmission was commanded several minutes before DSLWP-B hid behind the Moon, so as to guarantee a continuous signal at 436.4MHz to observe the variations in signal power as DSLWP-B went behind the Moon. On the 15th, the occultation was especially brief, lasting only 28 minutes. Thus, DSLWP-B was commanded to transmit continuously before hiding behind the Moon. This enabled us to also observe the end of the occultation, since DSLWP-B continued transmitting when it exited from behind the Moon. This is an analysis of the recordings made at Dwingeloo.
The recordings can be found here. I have used my Moonbounce processing scripts to compute the signal power from the recordings. These scripts were made to study the power of the Moonbounce signal, but can be used here as well. Essentially, the script performs Doppler correction using the tracking files published in dslwp_dev as orbit state vector and GMAT as propagator, detects the frequency offset of the GMSK signal from DSLWP-B, lowpass filters it, and takes averaged power readings in windows of 102.4ms.
The occultations were planned as shown in this post. After computing the expected times of the occultations, the relevant segments of the power readings are plotted, obtaining the figures shown below.
The first two figures correspond to the start of the occultations on February 4 and 13. The last two figures correspond to the start and end of the occultation on Februrary 15. The first figure looks somewhat different from the rest because the occultation was much slower. Take note at the time axis.
We can clearly see the diffraction fringes. Before dropping down to zero, the power of the received signal oscillates. The diffraction pattern that happens during a lunar occultation is explained in this document. There, the case of study is the occultation of a star by the Moon at optical wavelengths. In the derivation done in that document, it is implicitly assumed that the star is very far, so the wave arriving the Moon can be assumed to be a plane wave. This is not the case for DSLWP-B, which is much more near to the Moon. In this case, we need to consider a point source instead of a plane wave. The same kind of diffraction pattern is obtained in both cases, but the temporal frequency at which the diffraction fringes occur (the fringe rate) can be quite different, especially because the distance from DSLWP to the surface of the Moon is changing quickly.
The typical diffraction pattern is given by the Fresnel integral$D(\nu) = \frac{1}{2}\left|\int_{-\infty}^\nu e^{i\pi s^2/2}\,ds\right|^2.$A plot of this function can be seen in the figure below.
The variable $$\nu$$, depends on the geometry of the diffraction, and is computed differently in the case of a plane wave or a point source. The point $$\nu=0$$ represents the case of a grazing diffraction, which means that the line of sight path is tangent to the obstacle. The case $$\nu < 0$$ corresponds to the situation when the direct line of sight is blocked, and $$\nu > 0$$ corresponds to the situation when there is a direct line of sight path. As the obstacle gets away from the line of sight and $$\nu \to +\infty$$, we have $$D(\nu) \to 1$$, so we recover the power we would have if the obstacle wasn’t present.
Without discussing for now much about the geometry of the problem and how to compute $$\nu$$ with respect to time $$t$$, we just assume that it is reasonable to approximate $$\nu$$ by its first order Taylor expansion about the instant $$t_0$$ when the grazing diffraction happens:$\nu \approx v(t-t_0).$
The next step in the analysis is to fit a function of the form$P_0D(v(t-t_0))$to the observed data, where $$P_0$$, $$v$$, and $$t_0$$ are unknown parameters representing the unobstructed signal power, the fringe rate, and the instant of grazing respectively.
To that end, we use a non-linear least squares fit. It is important to choose the initial values for the unknown parameters correctly to avoid falling in a local minimum. We do the following educated guesses. For $$t_0$$ we take the instant when the power achieves its average value. To estimate $$v$$ we try to locate the first local minimum and maximum of the diffraction fringes and compute $$v$$ in terms of their temporal spacing. Finally, $$P_0$$ is estimated as the mean value of the power after the first local minimum. The time interval over which the fit is performed is selected by hand. Also, averaging of the data is performed for the first occultation.
After taking these precautions, the fit is rather good, as shown in the figures below.
The fit gives us empirical values for $$t_0$$ and $$v$$. Let us compare those with the theoretical values obtained by using the tracking files from dslwp_dev as ephemeris and propagating the orbit in GMAT. This requires us to speak about how $$\nu$$ is computed in terms of the geometry, for the case of a point source.
This is treated in this text very succinctly. It states that$\nu = \pm2\sqrt{\frac{\Delta}{\lambda}},$where $$\lambda$$ is the wavelength and $$\Delta$$ is the difference in distances between a certain “reflected” path and the direct line of sight path. The sign $$\pm$$ is chosen according as to whether the direct line of path is blocked or not, as explained above.
In more detail, assume that our transmitter and receiver are at points $$A$$ and $$B$$ respectively. Consider the point $$R$$ in the obstacle that is closest to the line segment joining $$A$$ and $$B$$. Then $\Delta = \|A-R\| + \|R-B\| – \|A-B\|,$ where $$\|P-Q\|$$ denotes the distance between the points $$P$$ and $$Q$$. Thus, $$\Delta$$ is the extra distance traveled by a ray that goes from $$A$$ to $$B$$ reflecting off $$R$$.
As an interesting fact, we have that the time derivative $$d\Delta/dt$$ is equal to the difference in Dopplers between the direct signal and the Moonbounce signal (see the Appendix in this post). However, we are not really interested in $$d\Delta/dt$$ but rather in the time derivative of $$\nu$$, which involves $$\sqrt{\Delta}$$, so it is not easy to relate the fringe rate and the Doppler directly.
In the case of the lunar occultation of DSLWP-B, we compute $$\Delta$$ as follows. From GMAT we get the coordinates of the points $$A$$ and $$B$$ corresponding to DSLWP-B and Dwingeloo, and $$M$$, which corresponds to the centre of the Moon. We compute the distance $$\delta$$ between $$M$$ and the line passing through $$A$$ and $$B$$ by computing $l = \frac{\langle M – B, A – B\rangle}{\|A-B\|},$the projection of the vector joining Dwingeloo and the Moon onto the unit line of sight vector pointing from Dwingeloo to DSLWP-B. Then$\delta = \sqrt{|M-B|^2 – l^2}.$We note that$\Delta = |\delta – r|,$where $$r$$ is the lunar radius, and that the sign of $$\nu$$ coincides with the sign of $$\delta – r$$.
The instant of grazing diffraction is found by locating the time $$t_0$$ for which $$\Delta = 0$$. Then, the time derivative $$\frac{d\nu}{dt}(t_0)$$ is evaluated numerically as a difference quotient.
For each of the four diffractions studied in this post, we obtain the following time differences in seconds between the instant of grazing $$t_0$$ obtained from the fit to the measurements and the GMAT simulation:
[-116.05367306500001, 13.311127206, 31.371431057000002, 41.360679158]
We see that there are errors on the order tens of seconds. This can be explained by errors in the ephemeris. Indeed, it is difficult to estimate the mean anomaly of the orbit accurately, and it is the first parameter that becomes inaccurate with time, since orbital perturbations always build up in the mean anomaly. An offset in mean anomaly translates directly into a time offset in the instant of grazing. Determining accurately the instant of grazing is actually a good way to measure the mean anomaly precisely.
It is also interesting to observe the last two time differences, which correspond to the entry and exit of the same occultation on February 15. They have a difference of 10.01s, meaning that the duration of the occultation had such a difference between the observations and the GMAT prediction. This could also be explained by ephemeris errors, but the situation is not as simple as an offset in the mean anomaly.
The relative errors in the calculation of the fringe rate $$\frac{d\nu}{dt}(t_0)$$ between the observations and the GMAT prediction are shown below in parts per one:
[0.012628611461853678, -0.14849525967852584, 0.03838583537440288, 0.021923209485447237]
We observe errors between 1% and 15%. This means that our model for the geometry of the diffraction explains the observations very well.
The grazing instants determined from the observations are
2019-02-04T10:06:58.3454239352019-02-13T21:31:11.2576072062019-02-15T14:23:36.4204000572019-02-15T14:51:46.844356158
The fringe rates, in units of 1/s determined from the observations are
[-0.1682628343896966, -1.2532957532402556, -1.7370407299575834 1.6895344971498276]
The computations and plots shown in this post have been done in this Jupyter notebook. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9089514017105103, "perplexity": 357.0454088069032}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00774.warc.gz"} |
https://www.physicsforums.com/threads/wilson-theorem-and-formulas-for-pi-x.118199/ | # Wilson theorem and formulas for pi(x)
1. Apr 20, 2006
### eljose
Hello..my question is from wilson theorem:
$$(p-1)!=-1mod(p)$$
how could you derive the formulae for $$pi(x)$$?..this should be impossible as you can,t solve the congruence exactly a proposed method would be to find the roots of:
$$f(x)=[\gamma(x)+1/x]-(\gamma(x)+1)/x$$
that precisely x=p for every prime and intege...
2. Apr 20, 2006
### matt grime
Do you not feel the slightest desire to define gamma?
3. Apr 20, 2006
Nor to use proper English sentences?
I mean what the hell is a "root of [equation] such that precisely x=p for every prime and intege..."
Last edited: Apr 20, 2006
4. Apr 21, 2006
### eljose
-Umm..sometimes i don,t know if you,re speaking seriously or if this some kind of British-American sarcastic humour:
$$\Gamma(x)=\int_{0}^{\infty}dxt^{x-1}e^{-t}$$
a "root" is just a root in the sense that is a number that satisfies for some function that f(c)=0 then c is a root of the function. for the primes and using Wilson,s theorem we would have that
$$Sin(\pi[\Gamma(x)+1]/x)=0$$ iff x is p for p prime.
and using this "Gamma" function Wilson,s theorem becomes
$$\Gamma(p)+1=0mod(p)$$ but how can you obtain the Prime number counting function from wilson,s theore?..thanks.
Last edited: Apr 21, 2006
5. Apr 21, 2006
### matt grime
it is weariness at seeing you repeatedly post articles that contain contradictory notation (using n for two different things in one equation for instance), symbols that you simply do not explain and careless abuse of mathematics: notice how your gamma is now an uppercase gamma when it started off as a lower case one.
notice how this produces a real number, for real x...
Does it? In what sense does it become this, since mod p is not something that makes sense for the real numbers? What is the point of introducing the gamma function if you can only use it for certain values in the domain where we already have a better description of its value (at these points)? So, what good does introducing the gamma function do? You can't use it, can you? No arguments, that (p-1)! is indeed Gamma(p), and that Wilson's theorem is that p-1!=-1 mod p, but why use Gamma instead of p-1!? The proof of Wilson's theorem is not one that uses real analysis, so why do this?
Last edited: Apr 21, 2006
6. Apr 21, 2006
### shmoe
$$\gamma$$ and $$\Gamma$$ look different to me, we aren't mind readers.
see http://mathworld.wolfram.com/PrimeFormulas.html (3)-(5) show how to get pi(x) using Wilson's theorem. Wilson's theorem is not at all practical for computing pi(x) like this, nor for locating primes by looking at zeros of that sin function. n! grows too fast. Besides, the version you have with real numbers for x in $$\sin (\pi(\Gamma(x)+1)/x)$$ has zeros at many real numbers that aren't prime integers.
Last edited: Apr 21, 2006
7. Apr 22, 2006
### eljose
the main question is not computing a prime this was not my intention..but perhaps to throw some of light into the problem of time primes in fact for m and m+2 twin primes:
$$4((m-1)!+1)+m=0mod(m(m+2)$$ so we could compute the density of twin primes in the form:
$$\pi_{2}(m)=\sum_{2}^{m}cos^{2}[\pi([4((m-1)!+1)+m/(m+2)m]-4((m-1)!+1)+m/(m+2)m)$$
if somehow we could invert this twin prime counting function t obtai the k-th twin prime in the form:
$$p_{n}=C+\sum_{k=2}^{f(n)}F(k,\pi_{2}(k))$$ we could prove that p-n is not bounded for p a twin prime so it must be an oo number of them.
8. Apr 22, 2006
### shmoe
If you mean something essentially like (3)-(5) from that link, then sure. What you have here has a few errors in the parenthesis so it's hard to tell what you mean. You would also want the floor of the cos terms at some point, otherwise this is most likely not even an integer.
You can do essentially the same as (10)-(11) from the mathworld link. As in the case of primes, this isn't a very enlightening expression, it's mostly a novelty.
9. Apr 22, 2006
### eljose
-I think that the "cos" term is enclosed by a floor function [cos] o this only can be 1 or 0.
-If wemanaged in the same case that for ordinary primes to prove that the k-th twin prime is:
$$p_{k}=\sum_{n=2}^{f(k)}G(n,\pi_{2}(n)+C$$
we would have proved "twin prime conjecture" as there would be the (k+1)-th prime so:
$$p_{k+1}-p_{k}=\sum_{f(k)}^{f(k+1)}G(n,\pi_{2}(n))+C$$
this formula somehow would use the [] function as the difference would be an integer so if this sum is $$p_{k+1}-p_{k}>0$$ then for every k exist a higher prime (k+1) wich is the next twin prime in the series.
10. Apr 22, 2006
### shmoe
Which we can do. If you have the counting function of a sequence, (10)-(11) can be used to give you the nth term of the sequence. However:
This isn't true. Using that expression to show there are in fact infinitely many twin primes will rely on already knowing that $$\pi_2(x)$$ goes to infinity as x does, i.e. you already need to know there are infinitely many twin primes.
Try using the formula from Hardy&Wright ((10)-(11) in the link) to prove there are infinitely many primes without already knowing that pi(x)->infinity as x->infinity (edit-more specifically, try proving that the p_n defined by this formula are in fact prime without already knowing there are infinitely many primes)
Last edited: Apr 22, 2006
11. Apr 23, 2006
### eljose
-umm...i see in this form (perhaps wrong) if we have the formula:
$$p_{k+1}-p_{k}=\sum_{f(k)}^{f(k+1)}G(n,\pi_{2}(n))+C$$
this means that for every k-th twin prime there is another consecutive (k+1)-th prime no matter what k is of course if the number of twin primes is finite you have that the maximum value of $$\pi_{2}(n)=a$$ your only chance for having a finite number of primes is that somehow the function inside the sum including the floor function would be <1 and form tis we could perhaps calculate the last twin prime so either there are infinite number of twin primes or the last number can be calculated. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9474522471427917, "perplexity": 928.4706416454557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542668.98/warc/CC-MAIN-20161202170902-00132-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/conditional-expectation.636636/ | # Conditional Expectation
1. Sep 17, 2012
### Scootertaj
1. Let the joint pdf be f(x,y) = 2 ; 0<x<y<1 ; 0<y<1
Find E(Y|x) and E(X|y)
2. Relevant equations
E(Y|x) = $\int Y*f(y|x)dy$
f(y|x) = f(x,y) / f(x)
3. The attempt at a solution
f(x) = $\int 2dy$ from 0 to y = 2y
f(y|x) = f(x,y)/f(x) = 1/2y
E(Y|x) = $\int Y/2Y dy$ from x to 1 = $\int 1/2 dy$ from x to 1
= -(x-1)/2
= (1-x)/2
The answer is supposed to be (1+x)/2
2. Sep 17, 2012
### jbunniii
Your expression for f(x) is wrong. It should be a function of x, not of y. Try drawing a picture of the region where f(x,y) is nonzero. Then answer this question: for a fixed value of x, what values of y will give you a nonzero f(x,y)?
3. Sep 17, 2012
### Scootertaj
The only other way I can think of doing f(x) would be to integrate from 0 to 1 instead. f(x) is defined as the integral of the joint pdf in terms of y.
So, we could get integral(2dy) from x to 1?
4. Sep 17, 2012
### jbunniii
Correct, from x to 1 (not as 0 to 1 as you wrote in the previous paragraph). Also be sure to state which values of x this is valid for.
5. Sep 17, 2012
### Scootertaj
That will give us 2(1-x) so f(y|x) = 1/(1-x)
I'm confused how this will give (1+x)/2 for E(y|x)
6. Sep 17, 2012
### Scootertaj
Just kidding I worked it out, thanks.
7. Sep 17, 2012
### jbunniii
That's the answer I got. What integral are you calculating for E(y|x)?
Cool, I see you got it.
Similar Discussions: Conditional Expectation | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9012023210525513, "perplexity": 1728.6392483458558}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812758.43/warc/CC-MAIN-20180219171550-20180219191550-00166.warc.gz"} |
http://mathhelpforum.com/geometry/205148-ladder-problem.html | # Math Help - ladder problem
There are 2 ladders, one is 25 ft long the other is 20 ft long, they are crossing eachther in an alley between two buildings, find the distance between the two buildings. The point where the ladders intersect to the ground is 10 ft.
I would let the width of the alley be x, where $0. Drop a vertical line down from the point where the ladders meet to the ground. Let the horizontal distance from the left wall to the vertical line be $x_1$ and the horizontal distance from the vertical line to the right wall be $x_2$, hence:
$x_1+x_2=x$
Now, by similarity (and the Pythagorean theorem), we find:
$x_1=\frac{10x}{\sqrt{20^2-x^2}}$
$x_2=\frac{10x}{\sqrt{25^2-x^2}}$
Now solve for x. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.95272296667099, "perplexity": 1371.949302010825}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131297622.30/warc/CC-MAIN-20150323172137-00234-ip-10-168-14-71.ec2.internal.warc.gz"} |
http://papers.nips.cc/paper/6545-consistent-kernel-mean-estimation-for-functions-of-random-variables | # NIPS Proceedingsβ
## Consistent Kernel Mean Estimation for Functions of Random Variables
[PDF] [BibTeX] [Supplemental] [Reviews]
### Abstract
We provide a theoretical foundation for non-parametric estimation of functions of random variables using kernel mean embeddings. We show that for any continuous function f, consistent estimators of the mean embedding of a random variable X lead to consistent estimators of the mean embedding of f(X). For Matern kernels and sufficiently smooth functions we also provide rates of convergence. Our results extend to functions of multiple random variables. If the variables are dependent, we require an estimator of the mean embedding of their joint distribution as a starting point; if they are independent, it is sufficient to have separate estimators of the mean embeddings of their marginal distributions. In either case, our results cover both mean embeddings based on i.i.d. samples as well as "reduced set" expansions in terms of dependent expansion points. The latter serves as a justification for using such expansions to limit memory resources when applying the approach as a basis for probabilistic programming. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8968904614448547, "perplexity": 311.1427355586108}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549423723.8/warc/CC-MAIN-20170721062230-20170721082230-00025.warc.gz"} |
https://mathhelpboards.com/threads/proof-of-a-linear-transformation-not-being-onto.6580/ | # Proof of a linear transformation not being onto
#### baseball3030
##### New member
Sep 18, 2013
9
proof onto
Prove: A linear Map T:Rn->Rm is an onto function :
The only way I have thought about doing this problem is by proving the contrapositive:
Last edited:
#### ModusPonens
##### Well-known member
Jun 26, 2012
45
Prove: A linear Map T:Rn->Rm is not onto if m>n.
The only way I have thought about doing this problem is by proving the contrapositive:
If m<=n then T:Rn->Rm is onto.
I would start by letting there be a transformation
matrix with dimension mxn.
Then the only thing I can think of doing is using the rank nullity thm to show that the dimension of the range=dimension of v. Does anyone know of any other ways to go about this or if my way would be correct? Thank you so much
The counterpositive is "If T is onto, then m<=n" It's easier to do it by the theorem that says that the dimensions of the kernel and of the image of T add to n.
#### Deveno
##### Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
You can do a direct proof, and I do not think one needs to invoke rank-nullity.
let $B = \{v_1,\dots,v_n\}$ be a basis for $\Bbb R^n$.
Consider the set $T(B) \subset \Bbb R^m$.
If $u \in T(\Bbb R^n) = \text{im}(T)$ we have:
$u = T(x)$ for some $x = c_1v_1 + \cdots + c_nv_n \in \Bbb R^n$.
Thus:
$u = c_1T(v_1) + \cdots c_nT(v_n)$, which shows $T(B)$ spans $\text{im}(T)$.
Since $|T(B)| \leq n < m$, we have that the dimension of $\text{im}(T) \leq n < m$.
However, if $\text{im}(T) = \Bbb R^m$, we have that $T(B)$ spans $\Bbb R^m$ leading to:
$m \leq |T(B)| \leq n < m$, a contradiction.
As a general rule, functions can, at best, only "preserve" the "size" of their domain, they cannot enlarge it. Dimension, for vector spaces, is one way of measuring "size".
While it is technically possible to have a function from $\Bbb R^n \to \Bbb R^m$ where $m > n$ (so called "space-filling functions") that is onto, such functions turn out to be rather bizarre and cannot be linear (they do not preserve linear combinations). Linear maps cannot "grow in dimension", for the same reason the column rank cannot exceed the number of rows in a matrix (even if we have more columns than rows).
#### baseball3030
##### New member
Sep 18, 2013
9
How are you able to say that
Since |T(B)|≤n<m , we have that the dimension of im(T)≤n<m ?
I understand the T(B)≤n but don't know how you know that n<m?
I understand everything up to that point. Thank you very much, I appreciate it.
#### Deveno
##### Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
$n < m$ is given by the problem, yes?
#### baseball3030
##### New member
Sep 18, 2013
9
Yeah but the problem states that the linear map is not onto if m<n so are you assuming that it is onto and then arriving at a contradiction? Thanks,
#### Deveno
##### Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Yes, I am doing a proof by contradiction (or reductio ad absurdum). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.923481285572052, "perplexity": 707.8655430475877}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046152085.13/warc/CC-MAIN-20210805224801-20210806014801-00597.warc.gz"} |
https://phys.libretexts.org/TextBooks_and_TextMaps/Astronomy_and_Cosmology_TextMaps/Map%3A_Stellar_Atmospheres_(Tatum)/11%3A_Curve_of_Growth/11.5%3A_Curve_of_Growth_for_Lorentzian_Profiles | $$\require{cancel}$$
# 11.5: Curve of Growth for Lorentzian Profiles
The optical depth of a line broadened by radiation damping is given, as a function of wavelength, by
$\label{11.5.1}\tau (x)=\tau(0)\frac{l^2}{x^2+l^2},$
where the HWHM is
$\label{11.5.2} l=\frac{\Gamma \lambda_0^2}{4\pi c}$
and the optical thickness at the line center is
$\tau(0) = \frac{e^2\mathcal{N}_1 f_{12}}{m\epsilon_0 c\Gamma}.$
Here $$x = \lambda − \lambda_0$$, and the damping constant $$\Gamma$$ may include a contribution from pressure broadening. As in Section 11.4, in figure XI.4 I draw line profiles for optical thicknesses at the line center $$\tau(0)=\frac{1}{16},\frac{1}{8},\frac{1}{4},\frac{1}{2}, 1 , 2 , 4 , 8$$.
We see that the wings continue to add to the equivalent width as soon as, and indeed before, the central depth has reached unity. On combining equations 11.3.4, \ref{11.5.1},2 and 3, we obtain the following expression for the equivalent width:
$\label{11.5.4}W=\frac{\Gamma \lambda_0^2}{2\pi c}\int_0^\infty \left ( 1-\text{exp}\left \{-\frac{\tau(0)}{y^2+1}\right \}\right )\,dy,$
in which $$y = x/l$$ - i.e. distance from the line center in units of the HWHM.
If we now substitute $$y = \tan \theta$$, the expression for the equivalent width becomes
$\label{11.5.5}W=\frac{\Gamma \lambda_0^2}{2\pi c} \int_0^{\pi/2} \frac{1-\text{exp}\left [-\tau(0)\cos^2 \theta \right ]}{\cos^2 \theta }\,d\theta .$
Now that we have a finite upper limit, the expression can be integrated numerically without artificial and unjustified truncation. As described in the Appendix to Chapter 10, calculation of the trigonometric function cos can be avoided, and hence the integration much speeded up, by the substitution of $$t=\tan (\frac{1}{2}\theta)$$. Although the denominator of the integrand is obviously zero at the upper limit, so is the numerator, and the value of the integrand at the upper limit is finite and equal to $$\tau (0)$$. Figure XI.5 shows the equivalent width, in units of $$\frac{\Gamma \lambda_0^2}{2\pi c}$$ as a function of $$\tau(0)$$.
$$\text{FIGURE XI.4}$$ $$\text{FIGURE XI.5}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9859660863876343, "perplexity": 541.7479886651271}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221219197.89/warc/CC-MAIN-20180821230258-20180822010258-00076.warc.gz"} |
https://www.omnicalculator.com/other/note-frequency | Note name
A
Octave
4 (1-line)
Frequency
440
Hz
# Note Frequency Calculator
By Rita Rain
Last updated: Sep 07, 2020
Table of contents:
This note frequency calculator is a convenient tool to determine the frequency of musical notes tuned in the twelve-tone equal temperament. In the text, you'll find the note frequency chart and a guide on how to use the note frequency converter.
We'll also cover the basics of the physics of sound - if you want to learn more about it, check out our relevant calculators:
## The frequency of musical notes
Sound is a wave which travels through a medium, such as air or water. All sounds consist of sine waves with different frequencies and amplitudes. Amplitude determines volume, and frequency determines pitch.
Technically, frequency is the number of completed wave cycles per second, and is expressed in Hertz. For example the frequency of the musical note C4 is approximately 262 Hz. This means the sound wave has 262 cycles per second.
Musical notes have determined frequencies. For example, in the twelve-tone equal temperament (the most common tuning system since the 18th century), note A4 has a frequency of 420 Hz.
If one note is double the frequency of another, they sound similar, so we designate them with the same letter. We then add a number to represent the octave - that is to say whether we mean a higher or lower D.
This way of representing musical notes if called the scientific pitch notation.
## How to use the note frequency calculator
1. Choose the name of the note. Note pairs like F♯ and G♭, or C♯ and D♭, are enharmonic equivalents, which means they denote the same sound.
2. Choose an octave.
3. By default, the calculator rounds the frequency of the musical note to one decimal place. You can change it in the `advanced mode` (below the note frequency calculator).
## The note frequency chart
Below you can see a table with the rounded frequencies of musical notes in Hertz:
Octave / Note
0
1
2
3
4
5
6
7
8
C
16
33
65
131
262
523
1047
2093
4186
C♯
17
35
69
139
277
554
1109
2217
4435
D
18
37
73
147
294
587
1175
2349
4699
D♯
19
39
78
156
311
622
1245
2489
4978
E
21
41
82
165
330
659
1319
2637
5274
F
22
44
87
175
349
698
1397
2794
5588
F♯
23
46
93
185
370
740
1480
2960
5920
G
25
49
98
196
392
784
1568
3136
6272
G♯
26
52
104
208
415
831
1661
3322
6645
A
28
55
110
220
440
880
1760
3520
7040
A♯
29
58
117
233
466
932
1865
3729
7459
B
31
62
123
247
494
988
1976
3951
7902
Rita Rain | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9263936877250671, "perplexity": 1099.1665016864636}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038061562.11/warc/CC-MAIN-20210411055903-20210411085903-00506.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=146&t=62067&p=237997 | ## Negative overall order
Rida Ismail 2E
Posts: 139
Joined: Sat Sep 07, 2019 12:16 am
### Negative overall order
Does anyone understand how an overall order can be negative and if so can you please explain it to me?
Sarah Zhari 1D
Posts: 103
Joined: Sat Sep 14, 2019 12:16 am
### Re: Negative overall order
I think that an order can be negative if as you increase the concentration of a particular reactant, the overall rate of reaction decreases.
Anisha Chandra 1K
Posts: 118
Joined: Thu Jul 11, 2019 12:17 am
### Re: Negative overall order
Can the whole reaction have a negative overall order or only certain steps?
William Chan 1D
Posts: 102
Joined: Sat Sep 14, 2019 12:15 am
### Re: Negative overall order
I think it is rare, but if you increase the concentration of certain reactants, then the overall reaction rate will decrease.
I found an example of the conversion of ozone to oxygen here: https://www.quora.com/What-are-negative-order-reactions-And-what-are-the-examples
KBELTRAMI_1E
Posts: 108
Joined: Sat Jul 20, 2019 12:17 am
### Re: Negative overall order
Like first vs second order? Negative as in less than a zero order? What would that be called and can you give an example if you found a problem with it? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8974067568778992, "perplexity": 3963.012837505463}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00419.warc.gz"} |
http://www.mathcaptain.com/geometry/vertical-angles.html | Vertical angles can be defined as the pairs of non-adjacent angles in opposite positions formed by the intersection of two straight lines are known as vertical angles.
In the figure, $\angle$1 and $\angle$3 are a pair of vertical angles. Another pair of vertical angles is $\angle$2 and $\angle$4.
## Vertical Angles Theorem
If two angles are vertical angles, then they have equal measures.
Construction: Let $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ be two intersecting lines in a plane, intersecting at the point P, as shown in the figure. Here, the angles $\angle$APD and $\angle$BPC are vertical angles.
To Prove:
$$\angle APD \cong \angle BPC$$
Proof:
S. No.StatementReason
1∠APD+∠APC = 180oAngles on a straight line (Supplementary angles)
2∠BPC+∠APC = 180oAngles on a straight line
3∠APD+∠APC=∠BPC+∠APCFrom (1) & (2)
4∠APC≅∠BPC(3), Subtracted ∠APC from both sides
Hence, we proved the theorem.
### Complementary Angles
A Vertical Line Icosahedron Vertices What are Angles Find the Area of a Parallelogram with Vertices Pentagonal Prism Vertices Rational Function Vertical Asymptote Rectangular Prism Vertices Triangular Pyramid Vertices Vertical Distance Formula Angle Relationship Angle of Depression and Angle of Elevation A Acute Angle
Find Vertical Asymptote Calculator Angle Solver Coterminal Angles Calculator | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9117703437805176, "perplexity": 1360.474096724979}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507443869.1/warc/CC-MAIN-20141017005723-00064-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-6-section-6-6-solving-quadratic-equations-by-factoring-exercise-set-page-459/77 | ## Algebra: A Combined Approach (4th Edition)
$(x-6)(x+1)=0$
The answer is $(x-6)(x+1)=0$ since the values $x=6$ and $x=-1$ are the only ones that satisfy the equation, i.e., these values are the only solutions | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9783880114555359, "perplexity": 221.26808338042704}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583509996.54/warc/CC-MAIN-20181016030831-20181016052331-00211.warc.gz"} |
http://theinfolist.com/php/SummaryGet.php?FindGo=Albedo | TheInfoList
The percentage of diffusely reflected sunlight relative to various surface conditions
Albedo (/ælˈbd/) (Latin: albedo, meaning 'whiteness') is the measure of the diffuse reflection of solar radiation out of the total solar radiation and measured on a scale from 0, corresponding to a black body that absorbs all incident radiation, to 1, corresponding to a body that reflects all incident radiation.
Surface albedo is defined as the ratio of radiosity to the irradiance (flux per unit area) received by a surface.[1] The proportion reflected is not only determined by properties of the surface itself, but also by the spectral and angular distribution of solar radiation reaching the Earth's surface.[2] These factors vary with atmospheric composition, geographic location and time (see position of the Sun). While bi-hemispherical reflectance is calculated for a single angle of incidence (i.e., for a given position of the Sun), albedo is the directional integration of reflectance over all solar angles in a given period. The temporal resolution may range from seconds (as obtained from flux measurements) to daily, monthly, or annual averages.
Unless given for a specific wavelength (spectral albedo), albedo refers to the entire spectrum of solar radiation.[3] Due to measurement constraints, it is often given for the spectrum in which most solar energy reaches the surface (between 0.3 and 3 μm). This spectrum includes visible light (0.4–0.7 μm), which explains why surfaces with a low albedo appear dark (e.g., trees absorb most radiation), whereas surfaces with a high albedo appear bright (e.g., snow reflects most radiation).
Albedo is an important concept in climatology, astronomy, and environmental management (e.g., as part of the Leadership in Energy and Environmental Design (LEED) program for sustainable rating of buildings). The average albedo of the Earth from the upper atmosphere, its planetary albedo, is 30–35% because of cloud cover, but widely varies locally across the surface because of different geological and environmental features.[4]
The term albedo was introduced into optics by Johann Heinrich Lambert in his 1760 work Photometria. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 2, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8628647923469543, "perplexity": 1685.4145187630756}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178358064.34/warc/CC-MAIN-20210227024823-20210227054823-00183.warc.gz"} |
https://www.physicsforums.com/threads/can-you-help-me-please.232258/ | # Can you help me , please?
1. Apr 30, 2008
### vip89
Can you help me , please??
Can you solve this problem by using Calculus l "" Volumes by Slicing and Rotation About an Axis"" ??
Develop a formula for the volume of an ellipsoid of the form
( x^2/a^2)+(y^2/b^2)+(z^2/c^2)=1, a, b, c > 0.
I want the steps , because I know the final answer .
2. Apr 30, 2008
### tiny-tim
Welcome to PF!
Hi vip89! Welcome to PF!
Show us what you've tried, and where you're stuck, and then we'll know how to help you!
3. Apr 30, 2008
### vip89
I know that I must sketch,find the equation of the cross section , find the area from A=piab and x^2/a^2+y^2/b^2=1
Finally,I must find the definit integral of the area
the final answer will be 4pi/3 abc
4. Apr 30, 2008
### vip89
I dont know how to do this , and how to apply my steps?!
5. Apr 30, 2008
### tiny-tim
Hi vip89!
ok … we usually take horizontal cross-sections (don't have to) …
in other words, the intersection with a plane z = constant.
So what is the equation (in x and y) if you put z = w (a constant) … ?
… and then what is the area?
6. May 2, 2008
### vip89
thnkx very much
7. May 6, 2008
### vip89
I trid to solve it by this way,but strang things will appear,the equation be more complicated!!
8. May 6, 2008
### Tedjn
Show us your work. You can either take a picture, scan it, and upload it, or you can try using LaTeX in between [noparse]$$and$$[/noparse] tags.
9. May 7, 2008
### tiny-tim
Hi vip89!
Show us how far you got …
What equation (in x and y) did you get for the horizontal cross-sections when you put z = w (a constant)?
10. May 7, 2008
### rocomath
lol ... funny kid we're not here to do your hw
11. May 7, 2008
### tiny-tim
oi … rocomath! … that was vip89's very first post (seven days ago)!
… but he's getting the hang of it now!
If you can't help, just say "welcome!"
12. May 7, 2008
### rocomath
sorry :( i've grown less tolerant towards ppl just wanting us to do their hw.
13. May 9, 2008
### vip89
These are my trials
I have the idea but cant get the right answer
THNKX
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• ###### pass3 (2).JPG
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14. May 9, 2008
### vip89
Continue:
#### Attached Files:
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15. May 10, 2008
### HallsofIvy
Staff Emeritus
In your first sheet, you start with $$\int A(x)dx$$ and then immediately switch to $$\int x y[/itex] without a "dx" at all. How did that happen? I also cannot see any good reason for writing x as a function of y and z and writing y as a function of x and z. You are correct that [tex]y= b\sqrt{1- x^/a^2- z^2/c^2}$$ and that $$z= c\sqrt{1- x^2/a^2- y^2/b^2$$. The first tells you that when z= 0, $$y= b\sqrt{1- x^2/a^2}$$ and the second tells you that when y= 0, $$z= c\sqrt{1- x^2/a^2}$$. In other words, at each x, the cross section is an ellipse with semi-axes $$b\sqrt{1-x^2/a^2}$$ and $$c\sqrt{1- x^2/a^2}$$. Do you know that the area of an ellipse with semi-axes a and b is $$\pi ab$$?
16. May 11, 2008
### tiny-tim
hmm … let's start at the top of your page 3 …
which I take it is following my suggestion …
And then you try to use the formula A = πab for the area of an ellipse.
BUT … that formula only applies if the equation for the ellipse is in the standard form, with nothing but x² and y² on the left and "= 1" on the right.
You must put w² = (1 - x²/a² - y²/b²)c² into that form first.
Then you will get a "new a and b" that are not the same as the original a and b.
ok … rearrange w² = (1 - x²/a² - y²/b²)c² into the standard form … and then find the area.
Similar Discussions: Can you help me , please? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8718137741088867, "perplexity": 2524.4964378532018}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463612018.97/warc/CC-MAIN-20170529053338-20170529073338-00299.warc.gz"} |
https://everything.explained.today/Poincar%C3%A9_group/ | # Poincaré group explained
The Poincaré group, named after Henri Poincaré (1906),[1] was first defined by Hermann Minkowski (1908) as the group of Minkowski spacetime isometries.[2] It is a ten-dimensional non-abelian Lie group, which is of importance as a model in our understanding the most basic fundamentals of physics.
## Overview
A Minkowski spacetime isometry has the property that the interval between events is left invariant. For example, if everything were postponed by two hours, including the two events and the path you took to go from one to the other, then the time interval between the events recorded by a stop-watch you carried with you would be the same. Or if everything were shifted five kilometres to the west, or turned 60 degrees to the right, you would also see no change in the interval. It turns out that the proper length of an object is also unaffected by such a shift. A time or space reversal (a reflection) is also an isometry of this group.
In Minkowski space (i.e. ignoring the effects of gravity), there are ten degrees of freedom of the isometries, which may be thought of as translation through time or space (four degrees, one per dimension); reflection through a plane (three degrees, the freedom in orientation of this plane); or a "boost" in any of the three spatial directions (three degrees). Composition of transformations is the operation of the Poincaré group, with proper rotations being produced as the composition of an even number of reflections.
In classical physics, the Galilean group is a comparable ten-parameter group that acts on absolute time and space. Instead of boosts, it features shear mappings to relate co-moving frames of reference.
## Poincaré symmetry
Poincaré symmetry is the full symmetry of special relativity. It includes:
The last two symmetries, J and K, together make the Lorentz group (see also Lorentz invariance); the semi-direct product of the translations group and the Lorentz group then produce the Poincaré group. Objects that are invariant under this group are then said to possess Poincaré invariance or relativistic invariance.
10 generators (in four spacetime dimensions) associated with the Poincaré symmetry, by Noether's theorem, imply 10 conservation laws: 1 for the energy, 3 for the momentum, 3 for the angular momentum and 3 for the velocity of the center of mass.[3] [4]
## Poincaré group
The Poincaré group is the group of Minkowski spacetime isometries. It is a ten-dimensional noncompact Lie group. The abelian group of translations is a normal subgroup, while the Lorentz group is also a subgroup, the stabilizer of the origin. The Poincaré group itself is the minimal subgroup of the affine group which includes all translations and Lorentz transformations. More precisely, it is a semidirect product of the translations and the Lorentz group,
R1,3\rtimesO(1,3),
with group multiplication
(\alpha,f)(\beta,g)=(\alpha+f\beta,fg)
.[5]
Another way of putting this is that the Poincaré group is a group extension of the Lorentz group by a vector representation of it; it is sometimes dubbed, informally, as the inhomogeneous Lorentz group. In turn, it can also be obtained as a group contraction of the de Sitter group SO(4,1) ~ Sp(2,2), as the de Sitter radius goes to infinity.
Its positive energy unitary irreducible representations are indexed by mass (nonnegative number) and spin (integer or half integer) and are associated with particles in quantum mechanics (see Wigner's classification).
In accordance with the Erlangen program, the geometry of Minkowski space is defined by the Poincaré group: Minkowski space is considered as a homogeneous space for the group.
In quantum field theory, the universal cover of the Poincaré group
R1,3\rtimesSL(2,C),
which may be identified with the double cover
R1,3\rtimesSpin(1,3),
is more important, because representations of
SO(1,3)
are not able to describe fields with spin 1/2, i.e. fermions. Here
SL(2,C)
is the group of complex
2 x 2
matrices with unit determinant, isomorphic to the Lorentz-signature spin group
Spin(1,3)
.
## Poincaré algebra
The Poincaré algebra is the Lie algebra of the Poincaré group. It is a Lie algebra extension of the Lie algebra of the Lorentz group. More specifically, the proper ($\det\Lambda =1$), orthochronous ($\Lambda^0_ \geq 1$) part of the Lorentz subgroup (its identity component), $SO(1,3)_+^$, is connected to the identity and is thus provided by the exponentiation $\exp(ia_ P^)\exp\left(\frac\omega_ M^\right)$ of this Lie algebra. In component form, the Poincaré algebra is given by the commutation relations:[6] [7] where
P
is the generator of translations,
M
is the generator of Lorentz transformations, and
η
is the
(+,-,-,-)
Minkowski metric (see Sign convention).
The bottom commutation relation is the ("homogeneous") Lorentz group, consisting of rotations, $J_i = \frac\epsilon_ M^$, and boosts, $K_i = M_$. In this notation, the entire Poincaré algebra is expressible in noncovariant (but more practical) language as
[Jm,Pn]=i\epsilonmnkPk~,
[Ji,P0]=0~,
[Ki,Pk]=iηikP0~,
[Ki,P0]=-iPi~,
[Jm,Jn]=i\epsilonmnkJk~,
[Jm,Kn]=i\epsilonmnkKk~,
[Km,Kn]=-i\epsilonmnkJk~,
where the bottom line commutator of two boosts is often referred to as a "Wigner rotation". The simplification $[J_m + iK_m,\, J_n -iK_n] = 0$ permits reduction of the Lorentz subalgebra to $\mathfrak(2) \oplus \mathfrak(2)$ and efficient treatment of its associated representations. In terms of the physical parameters, we have
\left[lH,pi\right]=0
\left[lH,Li\right]=0
\left[lH,Ki\right]=i\hbarcpi
\left[pi,pj\right]=0
\left[pi,Lj\right]=i\hbar\epsilonijkpk
\left[pi,K
j\right]= i\hbar clH\delta
ij
\left[Li,Lj\right]=i\hbar\epsilonijkLk
\left[Li,Kj\right]=i\hbar\epsilonijkKk
\left[Ki,Kj\right]=-i\hbar\epsilonijkLk
The Casimir invariants of this algebra are $P_ P^$ and $W_W^$ where $W_$ is the Pauli–Lubanski pseudovector; they serve as labels for the representations of the group. The Poincaré group is the full symmetry group of any relativistic field theory. As a result, all elementary particles fall in representations of this group. These are usually specified by the four-momentum squared of each particle (i.e. its mass squared) and the intrinsic quantum numbers $J^$, where
J
is the spin quantum number,
P
is the parity and
C
is the charge-conjugation quantum number. In practice, charge conjugation and parity are violated by many quantum field theories; where this occurs,
P
and
C
are forfeited. Since CPT symmetry is invariant in quantum field theory, a time-reversal quantum number may be constructed from those given.
As a topological space, the group has four connected components: the component of the identity; the time reversed component; the spatial inversion component; and the component which is both time-reversed and spatially inverted.[8]
## Other dimensions
The definitions above can be generalized to arbitrary dimensions in a straightforward manner. The d-dimensional Poincaré group is analogously defined by the semi-direct product
IO(1,d-1):=R1,d-1\rtimesO(1,d-1)
with the analogous multiplication
(\alpha,f)(\beta,g)=(\alpha+f\beta,fg)
.
The Lie algebra retains its form, with indices and now taking values between and . The alternative representation in terms of and has no analogue in higher dimensions.
## Super-Poincaré algebra
A related observation is that the representations of the Lorentz group include a pair of inequivalent two-dimensional complex spinor representations
2
and
\overline{2}
whose tensor product
2 ⊗ \overline{2}=3 ⊕ 1
is the adjoint representation. One may identify this last bit with four-dimensional Minkowski space itself (as opposed to identifying it with a spin-1 particle, as would normally be done for a pair of fermions, e.g. a pion being composed of a quark-anti-quark pair). This strongly suggests that it might be possible to extend the Poincaré algebra to also include spinors. This leads directly to the notion of the super-Poincaré algebra. The mathematical appeal of this idea is that one is working with the fundamental representations, instead of the adjoint representations. The physical appeal of this idea is that the fundamental representations correspond to fermions, which are seen in nature. So far, however, the implied supersymmetry here, of a symmetry between spatial and fermionic directions, cannot be seen experimentally in nature. The experimental issue can roughly be stated as the question: if we live in the adjoint representation (Minkowski spacetime), then where is the fundamental representation hiding?
## References
• Book: Group Theory in Physics . Wu-Ki Tung . 1985 . World Scientific Publishing . 9971-966-57-3.
• Book: Weinberg, Steven . The Quantum Theory of Fields . 1 . 1995 . Cambridge University press . Cambridge . 978-0-521-55001-7 . registration .
• Book: Quantum Field Theory . L.H. Ryder . Cambridge University Press . 2nd . 0-52147-8146 . 1996 . 62 .
## Notes and References
1. (Wikisource translation: On the Dynamics of the Electron). The group defined in this paper would now be described as the homogeneous Lorentz group with scalar multipliers.
2. (Wikisource translation: The Fundamental Equations for Electromagnetic Processes in Moving Bodies).
3. Web site: Survey of Symmetry and Conservation Laws: More Poincare. 2021-02-14. frankwilczek.com.
4. Barnett. Stephen M. 2011-06-01. On the six components of optical angular momentum. Journal of Optics. 13. 6. 064010. 10.1088/2040-8978/13/6/064010. 2040-8978.
5. Book: Oblak, Blagoje. BMS Particles in Three Dimensions. 2017-08-01. Springer. 9783319618784. 80. en.
6. Book: General Principles of Quantum Field Theory . N.N. Bogolubov. Springer . 2nd . 0-7923-0540-X . 1989 . 272 .
7. Book: 978-1-13950-4324 . T. Ohlsson. Tommy Ohlsson . Relativistic Quantum Physics: From Advanced Quantum Mechanics to Introductory Quantum Field Theory . Cambridge University Press . 2011 . 10 .
8. Web site: Topics: Poincaré Group. 2021-07-18. www.phy.olemiss.edu. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9347200989723206, "perplexity": 748.1902005597507}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304835.96/warc/CC-MAIN-20220125130117-20220125160117-00424.warc.gz"} |
https://converter.ninja/length/millimeters-to-feet/652-mm-to-ft/ | # 652 millimeters in feet
## Conversion
652 millimeters is equivalent to 2.13910761154856 feet.[1]
## Conversion formula How to convert 652 millimeters to feet?
We know (by definition) that: $1\mathrm{mm}\approx 0.0032808399\mathrm{ft}$
We can set up a proportion to solve for the number of feet.
$1 mm 652 mm ≈ 0.0032808399 ft x ft$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{ft}\approx \frac{652\mathrm{mm}}{1\mathrm{mm}}*0.0032808399\mathrm{ft}\to x\mathrm{ft}\approx 2.1391076148\mathrm{ft}$
Conclusion: $652 mm ≈ 2.1391076148 ft$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 foot is equal to 0.467484662576687 times 652 millimeters.
It can also be expressed as: 652 millimeters is equal to $\frac{1}{\mathrm{0.467484662576687}}$ feet.
## Approximation
An approximate numerical result would be: six hundred and fifty-two millimeters is about two point one four feet, or alternatively, a foot is about zero point four seven times six hundred and fifty-two millimeters.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9254350662231445, "perplexity": 1593.7441883188017}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00727.warc.gz"} |
https://www.physicsforums.com/threads/line-integral-dl-in-spherical-polar-coordinates.302970/ | # Line Integral dl in spherical polar coordinates
1. Mar 27, 2009
### wam_mi
1. The problem statement, all variables and given/known data
Hi guys,
I'm trying to evaluate a line integral, Integration of Vector A dot dL
The vector A was given to be a function of r, theta and fi in spherical polar coordinates.
The question states that an arbitrary closed loop C is the circle parametrised by fi at some arbitrary values of (r, theta).
I would like to ask the following questions:
(i) How do I rewrite dl, the line element, in terms of spherical polar co-ordinates? What does it mean by the closed curve C is the circle parametrised by fi at some arbitrary values of (r, theta). How does it change my dl if C wasn't a circle?
(ii) What are the limits of the line integral?
Thanks guys!
2. Relevant equations
3. The attempt at a solution
Can you offer guidance or do you also need help?
Similar Discussions: Line Integral dl in spherical polar coordinates | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.984839141368866, "perplexity": 875.0728747177222}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886110573.77/warc/CC-MAIN-20170822085147-20170822105147-00485.warc.gz"} |
http://physics.stackexchange.com/questions/65782/does-the-existence-of-higgs-imply-the-existence-of-magnetic-monopoles | # Does the existence of Higgs imply the existence of Magnetic Monopoles?
I am aware that in theories with spontaneous symmetry breaking, Magnetic Monopoles can exist as topological solitons. Can the same be done with the Standard Model gauge group. I am familiar with the contents of 't Hooft's paper Magnetic Monopoles in Unified Gauge theories. But the analysis in that paper is done for the $\operatorname{SO}(3)$ gauge group.
Is there a similar analysis for the standard model gauge group? Does the discovery of Higgs particle imply the existence of magnetic monopoles as topological solitons, and magnetic charge being treated as a topological charge?
-
No, I believe the Standard Model does not predict monopoles as a result of symmetry breaking. This is because the symmetry breaking $\mathrm{SU(2)} \times \mathrm{U(1)} \rightarrow \mathrm{U(1)}$ does not allow for topological solitons to exist.
Edit: $\pi_2(\mathrm{SU(2)} \times \mathrm{U(1)}/\mathrm{U(1))}=\pi_2(S^3)=0$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9584017395973206, "perplexity": 128.635276805986}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783393093.59/warc/CC-MAIN-20160624154953-00170-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://pdfkul.com/information-acquisition-and-strategic-sequencing-in-_59bbfecc1723dd99e8792bbe.html | Strategic Ignorance in Sequential Procurement∗ Silvana Krasteva Department of Economics Texas A&M University Allen 3054 College Station, TX 77843 E-mail: [email protected]
Huseyin Yildirim Department of Economics Duke University Box 90097 Durham, NC 27708 E-mail: [email protected]
March 17, 2017
1
Introduction
The procurement of complementary goods and services often entails dealing with independent sellers. Examples include: a real estate developer buying up adjacent parcels from different landowners; an employer recruiting a team of employees; a lobbyist securing bipartisan support; and a vaccine manufacturer obtaining required antigens from patent holders. In many cases, the buyer needs to deal with the sellers bilaterally – perhaps, convening multiple sellers is infeasible, or the sellers fear leaking business plans to the rivals. Given the ∗ Previously entitled “Information Acquisition and Strategic Sequencing in Bilateral Trading: Is Ignorance Bliss?” We thank seminar participants at the Duke Theory Lunch, Econometric Society Summer Meetings, 12th International Industrial Organization Conference, 25th International Conference on Game Theory, Texas Theory Camp, UC-Berkeley, UC-San Diego, and University of Toronto for comments. All remaining errors are ours.
1
an interesting discussion and further applications of sequencing in bilateral trading, see Sebenius (1996) and Wheeler (2005). 2 In particular, the buyer’s stand-alone valuations are assumed to be more uncertain than her joint valuation. For instance, a developer may be less sure about the success of a smaller shopping mall built on a single parcel; a lobbyist may be more worried about passage of the legislation through only one-party endorsement; or a vaccine manufacturer may be more uncertain about the effectiveness of the vaccine that uses only a subset of the required antigens. 3 For instance, an employer can assign scheduling of job interviews to an (uninformed) administrative assistant or ask job candidates to pick an interview slot from available ones. In some applications, the buyer’s sheer concern
2
3
4
The strategic value of being uninformed has also been indicated in other contexts. For instance, Carrillo and Mariotti (2000) argue that a decision-maker with time-inconsistent preferences may choose to remain ignorant of the state to control future consumption. In a principal-agent framework, Riordan (1990), Cremer (1995), Dewatripont and Maskin (1995) and Taylor and Yildirim (2011), among others, show that an uninformed principal may better motivate an agent while Kessler (1998) makes a similar point for the agent who may stay ignorant to obtain a more favorable contract. Perhaps, in this vein, papers closest in spirit to ours are those that incorporate signaling. Among them, Kaya (2010) examines a repeated contracting model without commitment and finds that the principal may delay information acquisition to avoid costly signaling through contracts. In a duopoly setting with role choice, Mailath (1993) and Daughety and Reinganum (1994) show that the choice of production period (as well as production level) may have signaling value and dampen incentives to acquire information. The issue of signaling in our setting is very different from these models, and the value of information critically depends on the prior belief in a non-monotonic way. The rest of the paper is organized as follows. The next section sets up the base model, followed by the equilibrium characterization with exogenous information in Section 3. Section 4 endogenizes information. We explore several extensions and variations in Section 5 and the case of substitutes in Section 6. Section 7 concludes. The proofs of formal results are relegated to an appendix.
2
Base model
A risk-neutral buyer (b) aims to purchase two complementary goods such as adjacent land parcels from two risk-neutral sellers (si , i = 1, 2). It is commonly known that the buyer’s joint value is 1, while her stand-alone value for good i, vi , is an independent draw from a nondegenerate Bernoulli distribution:11 Pr{vi = 0} = q ∈ (0, 1) and Pr{vi =
1 2}
= 1 − q.
We say that as q increases, goods become stronger complements for the buyer. In particular, with probability q2 goods are believed to be perfect complements. The outside option of each player is normalized to 0. The buyer meets with the sellers only once and in the sequence of her choice: s1 → s2 or acquisition (hence the signaling and strategic ignorance issues here) and explore instead the optimal sequencing of the sellers with ex ante heterogenous bargaining powers – i.e., the probability of making the offer. 11 For expositional purposes, we take v ∈ {0, 1 } throughout, but our results, especially that on the negative i 2 value of information, would generalize to vi ∈ {v L , v H } where 0 ≤ v L < v H ≤ 12 ; see Appendix B.
5
rule out c = 0 in the analysis to avoid a trivial equilibrium multiplicity when the value of information is exactly zero, though some of our key results will hold even for c = 0.
6
tial for an uninformed buyer. Finally, we restrict attention to one-time bilateral interactions; ¨ see Horner and Vieille (2009) for a similar restriction. This greatly simplifies the analysis with multiple sellers and is reasonable if the buyer has a limited time to undertake the project or an employer is in urgent need of filling vacancies. In Section 5, we relax many of our modeling assumptions. We begin our analysis with an exogenous information structure and then determine the value of being informed for the buyer. Without loss of generality, we re-label the sellers so that seller 1 refers to the first or leading seller in the sequence unless stated otherwise.
3
Informed vs. uninformed sequencing
Suppose that it is commonly known whether the buyer sequences informed (I) or uninformed (U). Given ex ante identical sellers, sequencing is inconsequential for an uninformed buyer. For an informed buyer, let θ1 (vi , v−i ) be the probability that the first (-place) seller has standalone value vi . To ease our analysis, we restrict attention to symmetric equilibria, in which equal sellers are treated equally: θ1 (v, v) = 12 , which reduces sequencing decision to choosing θ1 ( 21 , 0). Let h ∈ {0, 1} indicate the buyer’s purchase history and ( p1z , p2z (h)) denote the corresponding pair of prices where z = I, U. Our first result shows that under weak complements, equilibrium prices do not respond to informed sequencing. Lemma 1 Suppose q ≤ 12 . In equilibrium, (a) ( p1z , p2z (h)) = ( 12 , 12 ) for all z and h, and (b) the buyer purchases the bundle with certainty. If goods were independent, i.e., q = 0, each seller would post his monopoly price of 21 , inducing a joint purchase irrespective of the buyer’s information. Lemma 1 implies that the same applies to weak complements, q ≤ 12 . Lemma 1 is, however, uninteresting for our purposes as it trivially rules out information acquisition. For q > 12 , Proposition 1 characterizes the equilibrium in which prices do respond to the buyer’s information and thus the focus of our ensuing analysis.13
13 As with most signaling games, for q > 1 , there are also trivial equilibria in which the buyer always ignores 2 her private information when sequencing and has no incentive to acquire information as a result.
7
1 I Proposition 1 Suppose q > 12 . In equilibrium, pU 2 ( h = 0) = p2 ( h = 0) = 2 . Moreover, for
(a) an uninformed buyer: 1− q with prob. 2 U p1 = 1 with prob. 2
1− q q
and 2q−1 q
(b) an informed buyer: p1I = p2I (h = 1) =
p1I =
1− q2 2 1 2
with prob. with prob.
and θ1 ( 12 , 0) = 1 for q >
1− q2 q2 2q2 −1
1 2
pU 2 (h
= 1) =
and θ1 ( 21 , 0) >
and p2I (h = 1) =
q2
1 2
1 2
1 with prob.
with prob. 1 − q
for q ≤
√1 ; 2
q;
and
1 2
1 with prob.
with prob. 1 − q2 q2 ,
√1 . 2
(c) Demand: A buyer with v1 = 0 accepts only the low p1z but all p2z (h = 1) whereas a buyer with v1 =
1 2
accepts all p1z but only the low p2z (h = 1).
To understand Proposition 1, notice that with sunk payments, a key strategic concern for the buyer is being held up by the second seller. Consider an uninformed buyer. Upon observing the purchase history, the second seller optimally charges the buyer’s marginal value from the bundle, which is either
1 2
or 1. He must strictly mix between these prices; otherwise,
a sure price of 1 would strictly discourage a low value buyer from acquiring the first good and lead the second seller to reduce his price to
1 2
whereas a sure price of
1 2
would guarantee
the sale of the first good and encourage the second seller to raise his price to 1 given that the prior strictly favors a low value buyer, q > 12 . Not surprisingly, seller 2 mixes according to the prior on the first good and thus stochastically increases his price with the probability of a low value buyer, q. Note that a low value buyer demands the first good in the hope of paying less than the full surplus for the second. In particular, in equilibrium, such a buyer expects to pay
1+ q 2
for the second good and is therefore willing to pay
1− q 2
for the first, which
is exactly what seller 1 might offer. Seller 1 might, however, also offer a high price of
1 2
to
target a high value buyer. Seller 1’s mixing between these two prices accommodates that of 2’s by keeping his posterior “unbiased” at 12 . As q increases, seller 1 drops his discount price, 1− q 2 ,
to (partially) subsidize a low value buyer for the future holdup, but interestingly he also
drops the frequency,
1− q q ,
of this enticing offer so that his subsidy is not captured by seller 8
2.14 The uninformed prices in part (a) also explain the equilibrium demand in part (c): a low value buyer purchases the first good only at the discount price, upon which she proceeds to purchase the second with certainty, while the opposite is true for a high value buyer. Note that because each seller prices at the buyer’s marginal value, uninformed prices trend upward: the first seller charges no higher and the second seller charges no lower than the stand-alone value. Hence, an informed buyer is more likely to sequence the sellers from high to low value. If this sequencing is strict, namely θ1 ( 12 , 0) = 1, then the informed buyer has low value for the first good only in the case of perfect complements, occurring with probability q2 . Substituting this posterior for the prior q in the uninformed prices yields informed prices in part (b) so long as q >
√1 . 2
That is, an informed buyer begins with the high value
seller if goods are strong complements. For moderate complements,
1 2
< q ≤
√1 , 2
the in-
formed buyer might mix over the sequence although due to rising prices, she is still strictly more likely to begin with the high value seller, θ1 ( 21 , 0) ∈ ( 21 , 1]. Such mixing over the sequence requires equal prices, which can only be at 21 . Inspecting Proposition 1, we can determine the buyer’s payoff and identify the two key effects of being informed: sequencing and pricing. Recall that the first seller offers the discount price to entice a low value buyer, leaving her with no expected surplus. This means that despite a joint purchase, a low value buyer incurs a loss if she receives a high price from the second seller. Such holdup does not apply to a high value buyer because she can opt to purchase only the first good. Corollary 1 records this useful observation about the payoffs. Corollary 1 A low value buyer of the first good (v1 = 0) obtains an expected payoff of 0 while a high value buyer (v1 = 12 ) obtains a positive expected payoff equal to her expected payoff from the first purchase. From Corollary 1 and Proposition 1, the expected payoff of an uninformed buyer is found to be 1−q B ( q ) = (1 − q ) q U
=
1 1−q − 2 2
(1 − q )2 1 if q > , 2 2
where 1 − q is the probability that v1 = 1− q 2 ,
1 2
and
1− q q
(1)
is the probability of the discount price,
by the first seller. For strong complements, the expected payoff of an informed buyer is
14 Indeed,
1− q
with probability q q = 1 − q, a low value buyer acquires both goods but ends up with a loss of illustrating the holdup problem.
9
1− q 2 ,
analogously found by replacing 1 − q in (1) with 1 − q2 – the probability that v1 =
1 2
under
strategic sequencing. For moderate complements, the expected informed payoff is zero since the first seller targets the high value buyer; hence, (1− q2 )2 if 2 I B (q) = 0 if
q> 1 2
√1 2
(2)
√1 . 2
To identify the two effects of being informed, we also compute a counterfactual payoff for the buyer in which she sequences informed but the sellers are “nonstrategic” in that they keep their uninformed prices. Substituting the probability 1 − q2 for 1 − q in the first term of (1), we find the expected informed payoff with nonstrategic sellers: I
B (q) =
(1 − q2 )(1 − q) 1 if q > . 2 2
(3)
I
Evidently, B (q) > BU (q), implying a positive sequencing effect of being informed: given uninformed prices, the buyer strictly benefits from the ability to match a high value good I
with a low price seller. Moreover, B I (q) < B (q) for q >
√1 ; 2
1 2
< q ≤
√1 , 2
I
and B I (q) > B (q) for
so the pricing effect of being informed is negative for moderate complements and
positive for strong complements. As indicated by Corollary 1, the direction of the pricing effect depends on the first seller. Note from Proposition 1 that the first seller offers an expected price of
q 2
to an uninformed buyer while he offers a higher price of
and a lower expected price of
q2 2
1 2
for moderate complements
for strong complements to an informed buyer. Intuitively,
informed sequencing increases the probability that the first seller faces a high value buyer. For moderate complements, this probability is significant enough that the second seller chooses a low price, ruling out the holdup and in turn, inducing aggressive pricing by the first seller. For strong complements, the probability of a high value buyer is less significant and thus the second seller also puts weight on the full – surplus extracting – price of 1, leading the first seller to decrease his average price for a low value buyer. An interesting implication of the pricing effect is that for strong complements, an informed buyer prefers strategic sellers who read into her sequencing to those who do not while for moderate complements, she prefers nonstrategic sellers. From Corollary 1, it is clear that an informed buyer sequences to reduce the risk of holdup by the last seller.15 We therefore predict that an informed buyer is more likely to purchase the 15 This
strategy is consistent with the evidence on land assembly alluded to in the Introduction.
10
bundle than the uninformed. To confirm, we calculate from Proposition 1 that an uninformed buyer purchases the bundle with probability q
1−q + (1 − q)(1 − q) = (1 − q)(2 − q), q
whereas an informed buyer purchases the bundle with certainty for moderate complements and with probability (1 − q2 )(2 − q2 ) for strong complements. Since q2 < q, we have Corollary 2 An informed buyer is strictly more likely to purchase the bundle than an uninformed buyer. Note that both informed and uninformed buyers are less likely to purchase the bundle of stronger complements, i.e., a greater q, due to the increased chance of holdup. Nevertheless, given complementarity, the buyer should be less inclined to purchase a single good. Corollary 3 An informed buyer is strictly more likely to purchase the bundle than a single good. The same is true for an uninformed buyer if and only if q < 34 . Corollary 3 follows because unable to sequence optimally, the uninformed buyer guards against the holdup by acquiring only one unit when the holdup is sufficiently likely. Together with Corollary 2, this result points to the social value of information. The value of information to the buyer, however, depends on the sequencing and pricing effects, as we study next.
4
Information acquisition
Before examining information acquisition when it is unobservable to the sellers, we establish two benchmarks, one in which the buyer can publicly commit to visiting the sellers informed or uninformed, and the other in which a social planner dictates such commitment. Optimal information acquisition. By definition, the buyer’s value of information is the difference between her informed and uninformed payoffs: ∆(q) ≡ B I (q) − BU (q). Using (1) and (2), we have ∆(q) =
(1−q)2 (q2 +2q) 2
2 − (1−2q)
q>
if
√1 2
(4) if
1 2
√1 . 2
Eq.(4) implies that for moderate complements, the buyer is strictly worse off being informed! As discussed above, informed sequencing causes the first seller to set the high price 11
in this case, leaving no surplus to the buyer. Put differently, for moderate complements, the negative pricing effect of being informed dominates the positive sequencing effect. For strong complements, both effects are positive and so is the value of information, which the buyer weighs against the cost of information, c. Proposition 2 If goods are strong complements, q >
√1 , 2
and the information cost is low enough,
c < ∆(q), then the buyer optimally acquires information. If, on the other hand, goods are moderate complements,
1 2
√1 , 2
she optimally stays uninformed.
Hence, the buyer prefers informed sequencing if and only if goods are strong complements and the information cost is low. Otherwise, even with no information cost, the buyer prefers to sequence uninformed. The buyer can credibly remain uninformed by: (1) significantly raising her own cost, perhaps through overloading with multiple tasks (Aghion and Tirole, 1997); (2) delegating her sequencing decision to an uninformed third party; or (3) letting the sellers self-sequence. Note that if the sellers were nonstrategic, the value of information would be positive for all q > 12 . To see this, we subtract (1) from (3): ∆(q) ≡ Interestingly, ∆(q) < ∆(q) for q >
√1 . 2
(1 − q )2 q . 2
(5)
That is, for strong complements, the buyer has a
greater incentive to be informed when the sellers are strategic and read into her sequence, which simply follows from the positive pricing effect identified above. Since informed sequencing increases the probability of a joint sale, the buyer’s optimal information strategy is unlikely to be (socially) efficient, which we demonstrate next. Efficient information acquisition. Suppose that a social planner who maximizes the expected welfare can publicly instruct the buyer whether or not to acquire information. Consider an uninformed buyer. From Proposition 1, the expected welfare defined as the expected total surplus is computed to be 1−q 2q − 1 1 1 U W (q) = q (1) + (1 − q)( ) + (1 − q) q( ) + (1 − q)(1) q q 2 2 1 = (1 − q)(3 + q). 2 Similarly, the expected welfare under an informed buyer is W I (q) = q >
√1 , 2
1 2 (1
− q2 )(4 − q2 ) if
and W I (q) = 1 if q ∈ ( 21 , √1 ] since in the latter case, the bundle is purchased with 2
certainty. Hence, the social value of information is ∆W (q) ≡ W I (q) − W U (q) or 12
∆W ( q ) =
∆(q) +
1− q2 2
q>
if
√1 2
(6)
q2 +2q−1 2
if
1 2
√1 . 2
the sellers believe that with φ∗ , the buyer performs informed sequencing.
13
much on informed sequencing while the opposite holds for strong complements under which the pricing effect is positive. It is intuitive that by restricting her ability to commit, the unobservability of information acquisition cannot make the buyer better off than her optimal strategy. It may, however, strictly increase the welfare by encouraging informed sequencing for moderate complements. As mentioned above, although the buyer would want to sequence the sellers of moderate complements uninformed, this is not credible. She would not sequence them informed either because the value of information, ∆(q), is negative in this region, establishing strict mixing ¯ (q). The unobservability may also lower the welfare: for strong in equilibrium for c < ∆ complements, the buyer acquires information even less frequently than efficient.
5
Extensions and Variations
In this section, we consider several extensions of the model pertaining to the bargaining protocol and information structure.
5.1
Sequential procurement vs. auction
Up to now, we have assumed sequential procurement of goods and services. This is natural if, as with job interviews, the buyer has a capacity or privacy concern to deal with both sellers. Absent such concerns, the buyer could alternatively hold an auction in which she receives simultaneous price offers from the sellers and decides which good(s) to acquire after being informed of all prices and valuations. The obvious advantage of an auction over sequential procurement is that the buyer avoids the holdup problem and will incur no ex post loss. Its potential disadvantage is that having no sequence, both sellers are likely to target the buyer’s extra surplus from complementarity. Therefore, in the auction, the sellers are expected to coordinate prices to avoid exceeding the buyer’s joint valuation, but this makes them less generous in their price discounts. Lemma 2 confirms this conjecture. A = 1 , for all q. For Lemma 2 In the auction, there is a symmetric-price equilibrium, piA = p− 2 h i i √ 1− q 5−1 1 A A = q ≥ 2 , there is also a continuum of asymmetric-price equilibria: pi ∈ 2 , 1 − 2q and p− i
1 − piA . The multiplicity of equilibria is not surprising because the sellers play a simultaneous game of price coordination in the auction. In equilibrium, prices sum to the joint valuation of 14
1, with each being no lower than the discount price,
1− q 2 ,
offered by the leading seller under
uninformed sequencing (see Proposition 1(a)). Note that the buyer efficiently purchases the bundle in the symmetric equilibrium but enjoys no surplus – i.e., B A (q) = 0 whereas in the h i (1− q )2 (1− q ) q asymmetric equilibria, she receives a positive expected payoff, B A (q) ∈ , , as she 2q 2 may realize a high value on the lower price item. To understand the buyer’s choice between sequential procurement and auction, consider uninformed sequencing. This comparison is the most meaningful because the auction is strategically equivalent to uninformed sequencing except that price offers are “nonexploding” – i.e., all purchases are decided after visiting both sellers.17 Using (1), it is evident that the buyer will inefficiently choose sequential procurement if she anticipates symmetric pricing in the auction and Lemma 2 indicates that such an equilibrium always exists. On the other hand, the buyer may also choose an auction if she anticipates asymmetric pricing. Proposition 5 records these observations.
Proposition 5 For all q, there is an equilibrium in which an uninformed buyer chooses sequential √ procurement over an auction. This equilibrium is unique if q ∈ 12 , 52−1 ; otherwise, there is also an equilibrium in which she holds an auction. Hence, the buyer may adopt sequential procurement as assumed in the base model. While the multiplicity of equilibria in the auction prevents a clear prediction of this choice for all q, it is worth noting that the symmetric-price equilibrium maximizes the sellers’ joint payoff and is therefore likely to be their “focal point”. Based on the strategic equivalence alluded to above, an alternative interpretation of Proposition 5 is that the buyer may prefer exploding offers to nonexploding offers in sequential procurement. Again, while the former expose the buyer to holdup, they also compel the first seller to significantly cut price to entice the initial purchase.
5.2
Correlated values
Up to now, we have also assumed that stand-alone values, vi , are independent. Yet, in many applications, they may be (positively) correlated.18 For instance, a developer who is unable to acquire the desired land parcels for a shopping mall may appraise each similarly for a smaller 17 Moreover, in either setting, the buyer reveals no price information interim and eventually learns all her valuations. 18 The argument for negatively correlated goods is symmetric.
15
project. Here we show that correlation reduces the incentive for informed sequencing. To this end, consider the following joint distribution of stand-alone values: Pr(v1 , v2 ) 0 1 2
0 2 q + rq(1 − q) (1 − r ) q (1 − q )
1 2
(1 − r ) q (1 − q ) (1 − q)2 + rq(1 − q)
where r ∈ [0, 1] denotes the correlation coefficient, with r = 0 and 1 referring to the base model and ex post homogenous goods, respectively. The equilibrium characterization with correlation closely mimics Proposition 1 (see Proposition A2). In particular, the expected uninformed payoff in (1) remains intact since, as in the base model, the buyer’s equilibrium payoff depends on the first deal. The expected informed payoff in (2) is, however, slightly modified by replacing the posterior q2 with Pr(0, 0): 2 [1−Pr2(0,0)] if q > q (r ) BC,I (q; r ) = 0 if 12 < q ≤ q(r ) where q(r ) ≥
1 2
uniquely solves Pr(0, 0) =
1 2
such that q0 (r ) < 0, q(0) =
√1 , 2
(7)
and q(1) = 12 . By
subtracting (1) from (7), we obtain the value of information under correlation:
∆ (q; r ) = C
[1−Pr(0,0)]2 −(1−q)2 2
2 − (1−2q)
q > q (r )
if
(8) if
1 2
< q ≤ q (r ).
As expected, ∆C (q; 0) = ∆(q). Moreover, ∆C (q; 1) = 0. This makes sense because when goods are ex post homogeneous, the buyer’s ability to match a high value good with a low price seller under informed sequencing is inconsequential. More generally, informed sequencing becomes less consequential when goods are more correlated and thus less heterogeneous: formally, ∆C (q; r ) is strictly decreasing in r for q > q(r ). It is, however, worth noting that since Pr(0, 0) is increasing in r, q0 (r ) < 0; that is, correlation reduces the incentive to remain uninformed by increasing the likelihood of perfect complements.
5.3
Partial information
In the base model, the buyer can discover both valuations ex ante by paying a fixed cost; that is, information is all-or-nothing. If the marginal cost of information is significant, however, the buyer may choose to learn only one valuation. We argue that the buyer is unlikely to
16
gain from such “partial” information. Suppose that prior to meeting with the sellers, the buyer privately discovers only vi . If she approaches seller i second, then she engenders the uninformed equilibrium described in Proposition 1 and obtains a positive expected payoff for q > 21 . If, instead, the buyer approaches seller i first, she receives an expected payoff of 0 irrespective of vi . For a low value buyer, this follows from Corollary 1. For a high value buyer, this follows because seller i would infer the buyer’s valuation from sequencing and charge a sure price of 12 , leaving no surplus to the buyer. We therefore obtain Proposition 6. Proposition 6 Suppose q >
1 2
and that the buyer is privately informed of vi only. Then, she optimally
sequences seller i second and receives her uninformed payoff in (1). Proposition 6 justifies our focus on all-or-nothing information. Intuitively, the buyer cannot exploit partial information as it leaks through her sequencing; to avoid this, the buyer begins with the seller of the uncertain good, effectively committing to behaving uninformed. This contrasts with a fully informed buyer whose sequencing leaves a significant probability that the first seller has a low value item.
5.4
To identify strategic sequencing as a source of bargaining power for the buyer, we have assumed that sellers make the price offers – i.e., each operates in a seller’s market. We predict that the buyer will value sequencing less if she expects a buyer’s market. To confirm, let mi ∈ {si , b} denote the state of market i, which favors either seller i or the buyer as the pricesetter. We assume that sellers already know their respective market conditions but the buyer needs to find out.19 Specifically, the buyer is assumed to learn m1 and m2 at an interim stage between information acquisition and meeting with the sellers.20 Letting Pr(m1 , m2 ) be the joint probability distribution over the states of the markets, the following proposition shows that the buyer discounts the value of information by the likelihood of facing a seller’s market in each meeting.
19 In the real estate market, the buyer can discover the market condition from the stock of listings or expert opin-
ions while in the labor market, the employer can ascertain it from the initial screening of candidates, anticipating that peer institutions receive similar applications. 20 Though convenient, this timing of events is not crucial. Our conclusion in Proposition 7 would not change if the buyer learned m1 and m2 before her information decision.
17
m
5.5
Uncertain joint value
In the base model, we have also maintained that the buyer’s joint value is commonly known. This fits well applications where the buyer has a large winning project: e.g., a retail chain opening a large enough store that ensures monopolizing a local market; a vaccine company acquiring all the necessary antigens to guarantee an effective vaccine; and a lobbyist seeking bipartisan support that secures the favorable legislation. In other applications, the buyer’s joint value may be uncertain – at least initially. For instance, an academic department may be unsure of the synergy level between potential faculty hires. Here we show that consistent with the base model, the buyer has an incentive to learn her joint value to avoid the future holdup. To distinguish it from the information incentive due to sequencing, suppose that stand-alone values are equal and commonly known, v1 = v2 = v ∈ [0, 12 ]. The joint value, however, is uncertain: V = 1 or V > 1 where Pr{V = V } = α ∈ (0, 1). As in the base model, the buyer can privately discover V at a cost prior to meeting with the sellers or wait until she meets with both (so the second purchase is always informed). Proposition 8 characterizes the value of information in this setting.
21 That is, sequential procurement would essentially turn into a one-market problem where the seller makes the offer to a buyer with a privately known outside option, 0 or 12 .
18
Proposition 8 Consider the setting with an uncertain joint value as described above. In equilibrium, the buyer’s uninformed payoff is B J,U (α) = 0 whereas her informed payoff and thus her value of information is
∆ J (α) = B J,I (α) =
α (V − 1) v 1− v α ( V
if
− 1) if
0
if
α≤ v v +V −1
v v +V −1
<α≤
α>
1− v V −v
1− v . V −v
An uninformed buyer receives no expected surplus because the first seller offers the highest price acceptable in expectation. This means that an uninformed buyer may realize a loss after the second purchase if her joint value turns out to be low. To minimize such holdup, the buyer therefore has an incentive to approach the sellers informed. An informed purchase from the first seller, however, leads the second seller to be more optimistic about a high joint value and raise price, fully extracting the buyer’s surplus when a high joint value is sufficiently likely, i.e., α >
1− v . V −v
Similar to the base model, Proposition 8 indicates that a negative
pricing effect may completely outweigh the benefit of informed purchases. Together, ∆ J (α) and ∆(q) imply that the buyer is likely to discover her stand-alone values for strong complements and her joint value for moderate complements.
6
Substitutes
Sequential procurement and thus the issues of information acquisition and sequencing can also be pertinent to substitutes – e.g., parcels at rival locations and job candidates with comparable skills. We, however, argue that with substitutes, sequential procurement is undesirable for the buyer as it forecloses competition between the sellers; instead, the buyer is likely to hold an auction with simultaneous price offers. To make the point, let the buyer’s joint value be 1 (as in the base model) but her stand-alone values be independently distributed such that Pr{vi = 1} = qu and Pr{vi =
1 2}
= 1 − qu . Clearly, with probability q2u , goods are perfect
substitutes whereas with probability (1 − qu )2 , they are independent. We assume qu >
1 2
so
that perfect substitutes are more likely.22 Proposition 9 Consider substitute goods with qu > 12 . Then, an uninformed buyer strictly prefers auction to sequential procurement. 22 Again, the comparison is for an uninformed buyer because as mentioned in Section 5.1, the auction is strategically equivalent to uninformed sequencing except that price offers are nonexploding.
19
Proposition 9 is easily understood for (almost) perfect substitutes, qu ≈ 1. Unsurprisingly, the auction engenders the most competitive prices of 0 and in turn, the highest expected payoff of 1 for the buyer. In contrast, sequential procurement results in the monopoly prices of 1 and yields the lowest payoff of 0. The latter follows because with no previous purchase, the last seller sets his monopoly price and anticipating this, so does the first seller, leaving no surplus to the buyer. The buyer continues to receive monopoly prices under sequential procurement for imperfect substitutes, qu > 12 , but due to competition, lower prices are likely in the auction. In particular, the proof of Proposition 9 establishes that there is no pure strategy equilibrium in the auction: the sellers trade off pricing for perfect substitutes and pricing for independent units.
7
Conclusion
In this paper, we have explored the optimal sequencing of complementary negotiations with privately known values. Our analysis has produced three main observations. First, an informed buyer begins with the high value seller to mitigate the future holdup. Second, because of the sellers’ pricing response, the buyer may be strictly worse off with informed sequencing; that is, ignorance may be bliss. And third, the buyer underinvests in information from a social stand point: informed sequencing increases the likelihood of an efficient (joint) purchase but also the risk of holdup. As mentioned in the Introduction, the empirical evidence on land assembly corroborates our first observation in that real estate developers are estimated to assemble land parcels with a flexible design in mind to dissuade possible holdouts. On the other hand, the evidence on labor market supports our second observation in that job candidates are often advised by career consultants against reading into the interview sequence. Nonetheless, job candidates’ interest in the sequence is consistent with their potential uncertainty about who actually schedules interviews – the employer or (uninformed) staff member.
20
Appendix A As in the text, we re-label the sellers so that the sequence is s1 → s2 unless stated otherwise. For future reference, Proposition A1 characterizes the equilibrium with the following information structure: the buyer privately knows z ∈ { I, U } but the sellers commonly believe that Pr{z = I } = φ ∈ [0, 1]. Conditional on this information structure, let q1 (φ) = Pr{v1 = 0|φ} be the posterior belief that s1 is of low value. 1 2
Proposition A1. In equilibrium, p2 (h = 0) = (a) if q1 (φ) < 21 , then p1 = p2 (h = 1) = (b) if q1 (φ) = 21 , then p1 =
β 2
and
1 2
and the buyer purchases the bundle with certainty; 1 β 2 with prob. , and p2 (h = 1) = 1 with prob. 1 − β
where β ≥ 12 . The buyer purchases from s1 with certainty and s2 only if v1 = 0 or p2 (h = 1) = 21 ; (c) if q1 (φ) > 12 , then
p1 =
1− q1 ( φ ) 2
1 2
with prob.
1− q1 ( φ ) q1 ( φ )
and p2 (h = 1) = with prob.
2q1 (φ)−1 q1 ( φ )
1 2
1 with prob.
with prob. 1 − q1 (φ) . q1 ( φ )
Moreover, a buyer with v1 = 0 accepts only low p1 but all p2 (h = 1) whereas a buyer with v1 =
1 2
accepts all p1 but only the low p2 (h = 1). Proof. Consider pricing by s2 . Clearly p2 (h = 0) =
1 2
since s2 realizes a positive payoff
only if v2 = 12 . Let h = 1. Then a buyer with v1 = 0 accepts any offer p2 (h = 1) ≤ 1 whereas a buyer with v1 =
1 2
accepts only p2 (h = 1) ≤ 12 . Thus, s2 ’s optimal price is
p2 ( h = 1) =
1 2
1 if qb1 (φ, 1) ≥ 12 ,
if
qb1 (φ, 1) ≤
1 2
(A-1)
where qb1 (φ, h) = Pr{v1 = 0|φ, h} is the posterior conditional on the buyer’s information and purchase history. Anticipating p2 (h = 1), a buyer with v1 is willing to pay s1 up to p1 (v1 ) such that max{1 − p1 (v1 ) − p2 (h = 1), v1 − p1 (v1 )} = 0, 21
or simplifying, p1 (v1 ) = max{1 − p2 (h = 1), v1 }. Next we show that qb1 (φ, 1) ≤
1 2
(A-2)
in equilibrium. Suppose, to the contrary, that qb1 (φ, 1) > 21 .
Then p2 (h = 1) = 1, p1 (v1 = 0) = 0, and p1 (v1 = 12 ) = 12 . But this would imply p1 =
1 2
and
in turn qb1 (φ, 1) = 0 – a contradiction. We exhaust two possibilities for qb1 (φ, 1). qb1 (φ, 1) <
1 2
: Then p2 (h = 1) =
1 2
from (A-1), and p1 (v1 = 0) = p1 (v1 = 21 ) =
2). This implies qb1 (φ, h) = q1 (φ) and thus q1 (φ) <
1 2,
1 2
from (A-
which reveals that the buyer purchases
the bundle with certainty, proving part (a). qb1 (φ, 1) =
1 2
: By (A-1), s2 is indifferent between the prices
with probability β. Then, by (A-2), p1 (v1 = 12 ) = the prices 12 and accepts
β 2
β 2
1 2
1 2
and 1. Suppose s2 offers
1 2
β
and p1 (v1 = 0) = 2 . Let s1 mix between
by offering the latter with probability γ ∈ [0, 1]. Evidently, the buyer always
whereas only the buyer with v1 =
1 2
accepts 12 . Using Bayes’ rule, we therefore have
1− q1 ( φ ) . For q1 (φ) = 12 , γ = 1 q1 ( φ ) β β or p1 = 2 . By the buyer’s optimal purchasing decision, this means 2 (1) ≥ 12 (1 − q1 (φ)) or equivalently β ≥ 12 , resulting in the equilibrium multiplicity in part (b). Finally, for q1 (φ) > 12 , β 1− q1 ( φ ) γ ∈ (0, 1). Such strict mixing by s1 requires 2 = or β = 1 − q1 (φ), proving part (c). 2 1 Proof of Lemma 1. Suppose q ≤ 2 . For z = U, φ = 0 and q1 (0) = q. For z = I or φ = 1, it must be that q1 (1) ≤ 21 ; otherwise, if q1 (1) > 21 , equilibrium prices in Proposition A1 would imply an informed sequence strictly from high to low value – i.e., θ1 ( 12 , 0) = 1, and in turn, q1 (1) = q2 < 12 – a contradiction. Given that q1 (0) ≤ 12 and q1 (1) ≤ 12 , Proposition A1 further reveals that ( p1z , p2z (h)) = ( 21 , 21 ) for z = I, U and h = 0, 1, inducing a joint purchase, where the sellers’ indifference at q1 (φ) = 12 is broken in favor of efficient pricing. Proof of Proposition 1. Suppose q > 12 . For z = U, parts (a) and (c) are immediate from Proposition A1 since q1 (0) = q. Next, consider z = I. If q > √1 , the proof of Lemma 1 has 2 established that θ1 ( 12 , 0) = 1 and q1 (1) = q2 > 12 . Therefore, q1 (1) ≤ 12 if q ∈ ( 12 , √1 ], where 2 sellers break indifference in favor of efficient pricing at q1 (1) = 21 . This implies θ1 ( 21 , 0) > 12
qb1 (φ, 1) =
γq1 (φ) , γq1 (φ)+1−q1 (φ)
which, given qb1 (φ, 1) = 21 , implies γ =
given that by Bayesian updating, q1 (1) =
q2 21 + q(1 − q)[1 − θ1 ( 12 , 0)] 1 2
1 = q2 + 2q(1 − q)[1 − θ1 ( , 0)]. 2 Applying Proposition A1, we obtain parts (b) and (c) for z = I. 22
(A-3)
Proof of Corollary 1. Obvious from Proposition 1. Proof of Corollary 2. Directly follows from the probability of a joint purchase found in the text. Proof of Corollary 3. From Proposition 1, an informed buyer obtains the bundle of moderate complements with probability 1. She obtains a single unit of strong complements if and only if she has at least one high value and receives a high second price, whose probability is
(1 − q2 )q2 and strictly less than (1 − q2 )(2 − q2 ). The probability of a single purchase by an uninformed buyer is
(1 − q ) q + q
2q − 1 (1 − q) = (1 − q)(3q − 1), q
where the first term is the probability that v1 = is the probability of rejecting pU 1 =
1 2
1 2
and pU 2 ( h = 1) = 1, while the second term
1 due to v1 = 0 and accepting pU 2 ( h = 0) due to v2 = 2 .
Since the probability of a joint purchase is (1 − q)(2 − q), the buyer is more likely to purchase both if and only if q < 34 . Proof of Proposition 2. Directly follows from (4). Proof of Proposition 3. Directly follows from (4) and (6). Proof of Proposition 4. Consider first q >
√1 . 2
Then θ1 ( 12 , 0) = 1 and q1 (1) = q2 (by
(A-3)), which imply q1 (φ) = φq1 (1) + (1 − φ)q ≥ φq2 + (1 − φ)q > 21 . Moreover, the value of ˆ (φ) = q(1 − q) 1−q1 (φ) . Comparing with (4), information under unobservable acquisition is ∆ 2
ˆ (φ) ≤ ∆(q) for all φ. To determine when φ∗ < 1, note that for c < ∆(q), it is optimal for ∆ ˆ (1) = the buyer to acquire information. With unobservability, however, φ∗ = 1 requires c ≤ ∆ 1− q2 2 = 1 consider 2 < q
q (1 − q )
(1 + q)∆(q). Therefore, φ∗ < 1 for (1 + q)∆(q) < c < ∆(q), as claimed. Next, ≤ √12 . For φ∗ = 0, q1 (0) = q and ∆ˆ (0) = q(1 − q) 1−2 q = ∆(q). Hence, φ∗ > 0
for c < ∆(q), as desired.
Proof of Lemma 2. In an auction, the sellers play a simultaneous-move pricing game and thus the equilibrium occurs at the intersection of their best responses. Consider seller i’s best response Pi ( p−i ) to price p−i by the other seller. Note that if p−i ≤ generates a sale for si only if v−i = 0, while pi = payoffs, (1 − p−i )q and if 1 −
1 2q
1 2,
1 2
1 2,
then pi = 1 − p−i
guarantees a sale. Comparing si ’s resulting
it follows that Pi ( p−i ) = 1 − p−i if p−i ≤ 1 −
1 2q ,
and Pi ( p−i ) =
1 2
≤ p−i ≤ 21 . If, on the other hand, p−i > 12 , then since v−i ≤ 12 , seller s−i realizes a
sale only if the buyer acquires the bundle. Given this, the price pi = 1 − p−i ensures a sale for si whereas pi =
1 2
is accepted only if vi = 21 , leading to the respective payoffs: 1 − p−i and
(1 − q) 12 . From here, Pi ( p−i ) = 1 − p−i if
1 2
< p −i ≤ 23
1+ q 2 ,
and Pi ( p−i ) =
1 2
if p−i ≥
1+ q 2 .
To
sum up,
Pi ( p−i ) =
1 1 − p−i if 0 ≤ p−i ≤ 1 − 2q 1 1 if 1 − 2q ≤ p−i ≤ 12 2 1 2
1 − p−i if 1 if 2
≤ p −i ≤ p −i ≥
(A-4)
1+ q 2
1+ q 2 .
A = 1. In equilibrium, Pi ( P−i ( piA )) = piA for all i, which, given (A-4), implies that piA + p− i A = Clearly, piA = p− i
1 2
satisfies this condition for all q. Moreover, the only asymmetric prices h i 1− q 1 1 1+ q A = 1 − pA ∈ that satisfy this condition are: piA ∈ [ 2 , 1 − 2q ] and p− , i i 2q 2 . The interval √
for piA is nonempty if and only if q ≥
5−1 2 .
A = Proof of Proposition 5. By Lemma 2, piA = p− i
1 2
is an equilibrium for all q when the
buyer holds an auction, resulting in the expected payoff B A = 0. Under sequential procurement with an uninformed buyer, BU = 0 for q ≤
1 2
(by Lemma 1) and BU =
(1− q )2 2
for q >
1 2
(by (1)). Therefore, there is an equilibrium, in which the buyer chooses sequential procurement with an off-equilibrium belief that symmetric pricing would occur under the auction. √ For q ∈ 21 , 52−1 , B A = 0 is the unique equilibrium payoff and BU > B A , indicating a √
unique equilibrium with sequential procurement. For q ≥ 52−1 , given the equilibrium prich i 2 (1− q )2 (1− q ) q ing in Lemma 2, B A ∈ , ∪ {0}. Note that BU < (1−2qq) , implying that in this 2q 2 region there is also an equilibrium in which the buyer chooses to hold an auction and the sellers charge asymmetric prices piA ∈ [
1− q 1 2 , 1 − 2q ]
A = 1 − pA. and p− i i
Proposition A2. (Informed prices with correlation) As defined in Section 5.2, let q(r ) be the unique solution to Pr(0, 0) = 1) =
p1I
1 2
=
1 2,
where Pr(0, 0) = q2 + rq(1 − q). In equilibrium, p1I = p2I (h =
for q ≤ q(r ) with θ1 ( 21 , 0) >
1−Pr(0,0) 2
with prob.
1 2
with prob.
1 2
for q > 12 ; and
1−Pr(0,0) Pr(0,0)
and 2 Pr(0,0)−1 Pr(0,0)
p2I (h
= 1) =
1 2
1 with prob.
with prob. 1 − Pr(0, 0) Pr(0, 0)
and θ1 ( 12 , 0) = 1 for q > q(r ). Proof. Using the joint distribution Pr(v1 , v2 ) in Section 5.2, the posterior belief in (A-3)
24
generalizes to: q1 (1) =
Pr(0, 0) × 21 + Pr( 21 , 0) 1 − θ1 ( 12 , 0) 1 2
1 1 = Pr(0, 0) + 2 Pr( , 0) 1 − θ1 ( , 0) . 2 2 By Proposition A1, if q1 (1) > 12 , then θ1 ( 12 , 0) = 1 and q1 (1) = Pr(0, 0). Therefore q1 (1) > 1 2 , or equivalently q > q (r ). On the other hand, for q ≤ q (r ), q1 (1) ≤ 21 , which requires θ1 ( 12 , 0) > 12 . Equilibrium prices follow from Proposition A1. Proof of Proposition 6. Suppose q > 12 and that the buyer is privately informed of vi only. If si is second in the sequence, the buyer receives the uninformed payoff in (1), BU (q) > 0, 1 2
if and only if Pr(0, 0) >
because she is uninformed of v−i and the second seller’s pricing depends only on the prior q in this case. Suppose, instead, that si is first and let qe1 = Pr{vi = 0|si is first}. If vi = 0, the buyer receives an expected payoff of 0 because, by Proposition A1, for qe1 ≤ 12 , each seller charges
1 2
whereas for qe1 >
1 2,
si sets his low price to leave no expected surplus. Hence, a
buyer with vi = 0 strictly prefers to sequence si second and obtain BU (q) > 0. This implies qe1 = 0 and by Proposition A1, an expected payoff of 0 for the buyer when approaching si first. Therefore, in equilibrium, si is sequenced second, yielding the buyer her uninformed payoff in (1). Proof of Proposition 7. Let m = (m1 , m2 ). By definition, the buyer’s expected value of information is m
2
∆ (q) = Pr(b, b)∆(b,b) (q) + ∑ Pr(si , b)∆(si ,b) (q) + Pr(s1 , s2 )∆(s1 ,s2 ) (q). i =1
If m1 = m2 = b, the buyer optimally offers 0 to each seller, implying B I (q) = BU (q) = 1 and in turn, ∆(b,b) (q) = 0. If, on the other hand, mi = si for i = 1, 2, the setting reduces to our base model, implying ∆(s1 ,s2 ) (q) = ∆(q) where ∆(q) is as stated in (4). It therefore remains to prove that if mi = si and m−i = b, then ∆(si ,b) (q) = 0. Suppose m = (si , b). We consider uninformed and informed buyers in turn. Uninformed buyer: If the sequence is s−i → si , the buyer always purchases from s−i (at price 0) and thus the optimal price by si is given by (A-1) where φ ( = 0 and qb1 (0, 1) = q. 1 i f q ≤ 21 2 This means that the buyer’s uninformed payoff is: BU (s−i → si ) = . If, 1− q i f q > 21 2 however, the sequence is si → s−i , the buyer’s expected payoff from rejecting si ’s offer is
25
1− q 2 ,
which is simply the expected payoff from acquiring good i only. Therefore, the high-
est acceptable price by si in the first period satisfies max{1 − p1 , v−i − p1 } = ing p1 =
1+ q 2
and an expected payoff:
BU ( s
i
1− q 2 .
→ s −i ) =
1− q 2 ,
reveal-
Comparing the two payoffs,
BU = BU ( s −i → s i ). Informed buyer: If the sequence is s−i → si , the optimal price by si is given by (A-1) where φ = 1 and qb1 (1, 1) = q1 (1) since the buyer always purchases from s−i . If the sequence is si → s−i , the highest price acceptable to the buyer in the first meeting satisfies max{1 − p1 , v−i − p1 } = vi . Therefore, the optimal price by si is 1 2 i f q2 (1) ≤ p1 = 1 i f q2 (1) ≥
1 2
(A-5)
1 2
where q2 (1) = Pr{v−i = 0|s−i is second}. The buyer with v−i = 0 accepts p1 for sure whereas a buyer with v−i = 1 accepts only the low p1 . Let θbk (v−i ) = Pr{s−i is kth |v−i } and qk (1) = 2
Pr{v−i = 0|s−i is kth}. Then, by Bayes’ rule, q k (1) =
qθbk (0) qθbk (0) + (1 − q)θbk ( 12 )
.
(A-6)
We show that there is no equilibrium in which B I 6= BU . If, in equilibrium, θbk (0) = θbk ( 12 ) = 1 for some k = 1, 2, then qk (1) = q and, by (A-1) and (A-5), B I = BU . Suppose θbk (0) ∈ (0, 1). Then qk (1) is uniquely pinned down for k = 1, 2 using (A-6). We consider three cases for q k (1). 1 2
for k = 1, 2 : Since θb−k (v−i ) = 1 − θbk (v−i ), by (A-6), such an equilibrium belief requires 2q − 1 ≤ qθbk (0) − (1 − q)θbk ( 1 ) ≤ 0 and in turn, q ≤ 1 . From (A-1) and
• q k (1) ≤
2
2
(A-5), we therefore have that the informed and uninformed prices by si is 21 , resulting in B I = BU . • q k (1) >
1 2
for k = 1, 2 : By (A-6), this requires q > 21 , which, by (A-1) and (A-5), induces
the informed and uninformed prices of 1 by si . Therefore, B I = BU . • q k (1) ≤
1 2
< q−k (1) : By (A-1) and (A-5), approaching s−i in kth place results in a price
1 2
by si , while approaching s−i in −kth place results in a price of 1 by si . Therefore, a buyer with v−i = 0 has a strict preference to approach s−i kth, i.e. θbk (0) = 1. By of
(A-6), however, this implies that q−k (1) = 0 < qk (1), contradicting the existence of an equilibrium with qk (1) ≤
1 2
< q − k (1). 26
Consequently, there is no equilibrium with B I 6= BU . An equilibrium with B I = BU obtains by setting θb1 (0) = θb1 ( 21 ) ∈ (0, 1), resulting in qk (1) = q for k = 1, 2, which, by (A-1) and (A-5), yields identical informed and uninformed pricing by si . Proof of Proposition 8. Suppose that v1 = v2 = v ∈ [0, 12 ] and the joint value is V = 1 or V > 1 where Pr{V = V } = α ∈ (0, 1). Let αˆ (h) = Pr(V = V |h). Then, the optimal price by the second seller upon observing a prior purchase is
p2 ( h = 1) =
1−v
if αˆ (1) ≤
1− v V −v
V − v if αˆ (1) ≥
1− v V −v
.
(A-7)
Without a prior purchase, the second seller trivially offers p2z (h = 0) = v. The pricing by the first seller depends on whether the buyer is informed or uninformed. Uninformed buyer: Let p1 be the first seller’s maximum price acceptable to the buyer. Denoting by E[.] the usual expectation operator, p1 satisfies: max { E[V ] − p1 − E [ p2 (h = 1)] , v − p1 } = 0, or p1 = max{ E[V ] − E[ p2 (h = 1)], v}. ˆ (1) = α. Therefore, Since any higher price is rejected for sure, pU 1 = p1 and by Bayes’ rule, α 1− v U U for α ≤ V −1 , p1 = α V − 1 + v and p2 (h = 1) = 1 − v while for α > V1−−v1 , pU 1 = v and J,U ( v ) = 0. pU 2 ( h = 1) = V − v. The resulting expected payoff for the buyer is B
Informed buyer: In this case, p1 satisfies p1 = max{V − E[ p2 (h = 1)], v}. We consider three possibilities for αˆ (1). • αˆ (1) <
1− v V −v
: Then, p2 (h = 1) = 1 − v by (A-7), implying that p1L = v is accepted
for sure, whereas p1H = (V − 1) + v is accepted only if V = V. Therefore, for α ≤ v 1− v < , the first seller optimally sets p1 = v (breaking the indifference at α = v +V −1 V −v v v +V −1
in favor of efficiency), which reveals αˆ (1) = α. As a result, for α ≤
v , v +V −1
the
price pair p1I = v and p2I (h = 1) = 1 − v constitute an equilibrium, resulting in the payoff: B J,I (α) = α(V − 1). For α ∈ v+Vv −1 , V1−−vv , the first seller sets p1 = (V − 1) + v, which implies αˆ (1) = 1 and a profitable deviation for the second seller to p2 (h = 1) = V − v. Hence, αˆ (1) < • αˆ (1) =
1− v V −v
1− v V −v
only if α ≤
v v +V −1
resulting in ∆ J (α) = B J,I (α).
: Then, the second seller is indifferent between V − v and 1 − v. Suppose
that he offers 1 − v with probability σ. Then, E[ p2 (h = 1)] = V − v − σ(V − 1), implying 27
that p1L = v is accepted for sure by the buyer while p1H = v + σ(V − 1) is accepted only if V = V. The first seller is indifferent between p1L and p1H if σ = first seller’s mixing β = Pr( p1 = v) = αˆ (1) =
α β+(1− β)α
1− v . V −v
α≤
=
1− v . V −v
Then,
p1H
α (V −1) (1−α)(1−v)
(1− α ) v , α (V −1)
in which case the
≤ 1 engenders an equilibrium belief
= αv . Note that σ ≤ 1 for α ≥
v v +V −1
and β ≤ 1 for
Therefore, the price pair
1 − v with prob. σ v with prob. β I and p ( h = 1 ) = v 2 with prob. 1 − β V − v with prob. 1 − σ α i is an equilibrium for α ∈ v+Vv −1 , V1−−vv . In such an equilibrium, ∆ J (α) = B J,I (v) = p1I =
v 1− v α ( V
− 1). Note that for v = 0, σ = 0 and in turn, p2I (h = 1) = V. Then, p1 = 0.
Using the efficient tie-breaking rule, the first seller offers p1I = 0. Let η (V ) denote the probability that the buyer accepts the first offer given V. Then, by Bayes’ rule, αˆ (1) = η (V ) α . Since efficiency is maximized for η (V ) = 1, αˆ (1) = V1 , and η (1) η (V )α+η (1)(1−α) α (V −1) . Therefore, for v = 0, the price pair p1I = 0 and p2I (h = 1) = V is supported (1− α ) α (V −1) η (1) = (1−α)(1−v) and αˆ (1) = V1−−vv . The buyer’s payoff is B J,I (v = 0) = 0 = ∆ J (α).
• αˆ (1) >
1− v V −v
= by
: Then, p2 (h = 1) = V − v and p1I = p1 = v. The first price is always
accepted by the buyer, implying that αˆ (1) = α >
1− v V −v
and B J,I = 0 = ∆ J (α).
Proof of Proposition 9. Let qu > 12 . Consider sequential procurement with uninformed buyer. If s2 observes no prior purchase, he offers p2 (h = 0) = 1 since it is accepted with probability qu , resulting in a payoff of qu , whereas the alternative price of tainty, resulting in a payoff of
1 2.
1 2
is accepted with cer-
If s2 observes a prior purchase, he offers p2 (h = 1) =
the buyer’s marginal value for his good is
1 2
1 2
since
or 0. Anticipating such pricing, the highest price,
p1 , acceptable to the buyer in the first meeting satisfies: max {1 − p2 (h = 1) − p1 , v1 − p1 } ≥ 0, or simplifying p1 ≤ max {1 − p2 (h = 1), v1 } . This implies p1 = 1 since p1 = 1 is accepted with probability qu and p1 =
1 2
is accepted for
sure. Given the equilibrium prices, sequential procurement yields a payoff of 0 to the buyer. To prove that the auction yields a positive payoff, it suffices to show that in equilibrium, the sellers choose prices lower than 1 with a positive probability. The following two claims make this point. Claim 1 In the auction, there is no pure strategy equilibrium. 28
Proof of Claim 1. As in the standard Bertrand competition, p1 = p2 = p > 0 cannot arise in equilibrium because with probability q2u , goods are perfect substitutes and a slightly lower price would guarantee a sale in this realization. Without loss of generality, suppose p1 < p2 . If 21 < p1 < p2 , then s1 receives an expected profit π1 = qu (1 − qu ) + q2u p1 , p2 + p1 2 .
implying a profitable deviation to p˜ 1 = p1 < p2 ≤
1 2,
The same profitable deviation also exists if
because in this case, p1 is accepted unless v2 = 1 and v1 =
π1 = [1 − qu (1 − qu )] p1 . Finally, if p1 ≤
( π1 , π2 ) =
1 2
1 2,
resulting in
< p2 , the sellers’ expected profits are if p1 < p2 +
1 2
([1 − (1 − σ1 )qu (1 − qu )] p1 , qu (1 − qu )(1 − σ1 ) p2 ) if p1 = p2 +
1 2
( p1 , 0)
([1 − qu (1 − qu )] p1 , qu (1 − qu ) p2 )
if p1 > p2 +
1 2
where σ1 ∈ [0, 1] is an arbitrary tie-breaking rule when the buyer is indifferent. For p1 < p2 + 12 , s2 clearly has a strict incentive to lower his price. The same is true for p1 = p2 + 12 , in which case the buyer is indifferent. For p1 > p2 + 12 , s2 would deviate to p˜ 2 =
p2 + p1 − 12 2
. In
sum, there is no pure strategy equilibrium. Claim 2 In the auction, the following c.d.f. constitutes a symmetric mixed strategy equilibrium: 1 − qu 1 1 − qu 1 F ( p ) = 2 1 − q u (1 − q u ) − for p ∈ , . qu 2p 2(1 − qu (1 − qu )) 2 Proof of Claim 2. Consider a symmetric mixed strategy equilibrium with a continuous support p < p ≤
1 2
and no mass points. Then,
π ( p) = [ F ( p)(1 − qu ) + (1 − F ( p))(1 − qu (1 − qu ))] p = π, where π is the indifference profit across p ∈ [ p, p]. Note that π ( p) = (1 − qu ) p is increasing in p, implying a profitable deviation to p ∈ ( p, 12 ]. Therefore, p = 12 . Re-writing, 1 π F ( p ) = 2 1 − q u (1 − q u ) − . qu p Since F ( 12 ) = 1, π = Thus, F ( p) =
1− q u 2 .
Given this and the fact that F ( p) = 0, we find that p =
1− q u . 2(1−qu (1−qu ))
1 1 − qu 1 − qu 1 , 1 − q ( 1 − q ) − for p ∈ , u u q2u 2p 2(1 − qu (1 − qu )) 2
29
h i as claimed. It remains to show that there is no unilateral deviation incentive to p ∈ / p, 12 . Without loss of generality, consider a deviation by s1 . Clearly, p1 < p is not profitable, because π ( p1 ) = (1 − qu (1 − qu )) p1 < (1 − qu (1 − qu )) p = π ( p). Next consider a deviation to p1 > 12 . Since p1 > 1 is rejected with probability 1, we restrict attention to p1 ∈ ( 12 , 1] . In this case, s1 realizes a sale only if v1 = 1, v2 = 21 , and 1 − p1 >
1 2
− p2 , or equivalently p2 > p1 − 12 . We
exhaust two cases: • p1 −
1 2
≤ p : Then, π ( p1 ) = qu (1 − qu ) p1 . Since this deviation profit is increasing in p1 ,
the maximum deviation profit in this region is 1 1 − qu 1 1 − qu < + = π. π ( p + ) = q u (1 − q u ) 2 2 2(1 − qu (1 − qu )) 2 Therefore, there is no incentive to deviate to p1 ∈ ( 12 , 12 + p]. •
1 2
+ p < p1 ≤ 1 : Then, s1 ’s probability of a sale is qu (1 − qu ) Pr( p2 > p1 − 12 ) and his
deviation profit is
"
1 − qu 1 π ( p1 ) = qu (1 − qu ) 1 − 2 1 − qu + q2u − qu 2( p1 − 12 ) # " 1 (1 − q u )2 = − 1 p1 . qu 2( p˜ i − 21 ) Simple algebra shows that π ( p1 ) <
1− q u 2
!# p1
= π.
Together Claims 1 and 2 prove Proposition 9.
Appendix B Here, we extend our analysis to a more general Bernoulli distribution with valuations vi ∈
{v L , v H } where 0 ≤ v L < v H ≤
1 2
and Pr {vi = v L } = q ∈ (0, 1). The main difference from
the special case in the text (v L = 0 and v H =
1 2)
is that with v L > 0, the second seller may
not always charge the high price v H upon observing no purchase from the first seller – i.e. h = 0. In particular, by charging a low price of v L > 0, the second seller now leaves a positive surplus to the buyer who did not acquire the first object. As in the base model, let z = I, U refer to informed and uninformed sequencing, and
( p1z ,
p2z (h)) be the corresponding pair of prices. Analogous to the base model, for q ≤ 30
1− v H 1−v L , the
value of information is 0, since p1z = p2z (h = 0) = v H and p2z (h = 1) = 1 − v H in equilibrium. Thus, our analysis here focuses on q >
1− v H 1− v L .
Moreover, for z = I, similar to the base model,
we restrict attention to symmetric equilibria that satisfy the Intuitive Criterion. The following proposition characterizes equilibrium for the intermediate values of q and shows that the value of information is negative. nq vH 1− v H Proposition B1. Let q ∈ 11− , min −v L 1− v L , 1 −
vL vH
o
. In equilibrium, pU 2 ( h = 0) =
p2I (h = 0) = v H . Moreover, for (a) an uninformed buyer:
pU 1 =
with prob. 1 −
vH
(1 − q ) v H with prob. and pU 2 ( h = 1) =
1 − v H with prob.
1 − vL
1− q 1− v H q v H −v L
1− q 1− v H q v H −v L v H (1−q)−v L v H −v L
with prob. 1 −
v H (1−q)−v L ; v H −v L
(b) an informed buyer: p1I = v H and p2I (h = 1) = 1 − v H . (c) Demand: An informed buyer accepts both sellers’ offers, while an uninformed buyer with U v1 = v L accepts only the low pU 1 but all p2 ( h = 1) whereas an uninformed buyer with v1 = v H z accepts all pU 1 but only the low p2 ( h = 1).
Proof. Let qb1 (z, 1) = Pr {v1 = 0|h = 1} and qb2 (z, 0) = Pr {v2 = 0|h = 0} denote the sellers’ posterior beliefs given the buyer’s information and purchase history. The optimal pricing by s2 is p2z (h = 1) =
1 − v H if
qb1 (z , 1) ≤
1− v H 1− v L
1−v L
qb1 (z, 1) ≥
1− v H 1− v L
if
since 1 − v H is accepted for sure and 1 − v L only if v1 = v L ; and p2z (h = 0) =
v H if
qb2 (z , 0) ≤ 1 −
vL vH
vL
qb2 (z, 0) ≥ 1 −
vL vH
if
since v L is accepted for sure and v H only if v2 = v H .
31
For z = U, we have that qb2 (z , 0) = q, since the buyer is uninformed about v2 when making a purchasing decision about good 1. Therefore, for q < 1 −
vL vH ,
pU 2 ( h = 0) = v H .
Then, the equilibrium derivation under z = U is analogous to the proof of Proposition 1 and thus omitted here. Next, consider z = I. Clearly, p1 = v H is a best response to p2 (h = 1) = 1 − v H . Given this pricing, the buyer accepts both offers with probability 1. Therefore, upon observing h = 1, vH s2 has no incentives to deviate as long as qb1 (1, 1) = q2 + 2q(1 − q)θ1 (v L , v H ) ≤ 11− −v L . Since q 1− v L the buyer is indifferent in the order, θ1 (v L , v H ) = 0 and q < 1−v H ensure that s2 has no
incentive to deviate. Finally, p2 (h = 0) = v H is supported by an off-equilibrium belief that qb2 (1, 0) ≤ 1 −
vL vH .
This belief passes the Intuitive Criterion since the buyer’s equilibrium
payoff is 0, while rejecting the first offer would result in a payoff of at least 0 for the buyer under the most favorable beliefs regarding v2 (corresponding to p2 (h = 0) = v L ). Thus, rejecting the first offer is not an equilibrium dominated for any realization of v2 . Note that for v L = 0 and v H = 12 , the pricing in Proposition B1 coincides with Proposition 1. Moreover, it is readily verified that the value of information is
(1 − v H ) v H < 0, vH − vL which, given a payoff of 0 for the informed buyer, is simply the negative of the buyer’s unin nq oi 1− v H vH vL formed payoff. Therefore, for moderate complements, i.e., q ∈ 11− , min , 1 − , −v L 1− v L vH ∆(q) = −(1 − q)2
it is optimal for the buyer to stay uninformed, extending Proposition 2. Analogous to the base model, it can also be shown that the buyer’s optimal strategy to remain uninformed is not credible if it is unobservable to the sellers. In particular, if the cost is not too high, the buyer would acquire information with some positive probability in equilibrium, extending Proposition 4.
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[15] Gul, F., and H. Sonnenschein (1988). “On delay in bargaining with one-sided uncertainty.”Econometrica, 601-611. [16] Horn, H., and A. Wolinsky (1988). “Worker substitutability and patterns of unionization.” Economic Journal, 98, 484-497. ¨ [17] Horner, J., and N. Vieille (2009). “Public vs. private offers in the market for lemons.” Econometrica, 77.1, 29-69. [18] Hwang, I., and F. Li (2017). “Transparency of outside options in bargaining.“ Journal of Economic Theory, 167, 116-147. [19] Kaya, A. (2010). “When does it pay to get informed?” International Economic Review, 51(2), 533-51. [20] Kelsky, K. “The Professor Is In: On Interview Order,” https://chroniclevitae.com/ news, Date accessed: November 6, 2015. [21] Kessler, A. (1998). “The Value of Ignorance,” RAND Journal of Economics 29, 339–54. [22] Krasteva, S., and Yildirim, H. (2012a). “On the role of confidentiality and deadlines in bilateral negotiations.” Games and Economic Behavior, 75(2), 714-730. [23] Krasteva, S., and Yildirim, H. (2012b). “Payoff uncertainty, bargaining power, and the strategic sequencing of bilateral negotiations.” RAND Journal of Economics, 43(3), 514-536. [24] Lippman, S., and J. Mamer. (2012) “Exploding offers.” Decision Analysis 9(1): 6-21. [25] Mailath, G. (1993) “Endogenous Sequencing of Firm Decisions,” Journal of Economic Theory, 59, 169-82. [26] Marx, L., and G. Shaffer (2007). “Rent Shifting and the Order of Negotiations,” International Journal of Industrial Organization, 25(5), 1109-25. [27] Moresi, S., S. Salop, and Y. Sarafidis (2008), “A Model of Ordered Bargaining with Applications.” Available at SSRN: http://ssrn.com/abstract=1287224 [28] Niederle, M., and A. E. Roth. (2009). “Market culture: How rules governing exploding offers affect market performance.” American Economic Journal: Microeconomics, 199-219.
34
[29] Riordan, M. “What Is Vertical Integration?” In M. Aoki, B. Gustafsson, and O.E. Williamson, eds., The Firm as a Nexus of Treaties. London: Sage Publications, 1990. [30] Sebenius, J.K. 1996. Sequencing to build coalitions: With whom should I talk first? In Wise choices: Decisions, games, and negotiations, edited by R. J. Zeckhauser, R. L. Keeney, and J. K. Sebenius. Cambridge, MA: Harvard Business School Press. [31] Taylor, C., and H. Yildirim. (2011) “Subjective Performance and the Value of Blind Evaluation.” Review of Economic Studies, 78, 762-94. [32] Vincent, D. (1989) “Bargaining with common values.” Journal of Economic Theory, 48.1, 47-62. [33] Wheeler, M. (2005) “Which Comes First? How to Handle Linked Negotiations.” Negotiation 8, no.1. [34] Willihnganz, M., and L. Meyers. (1993) “Effects of time of day on interview performance.” Public Personnel Management 22(4): 545-550. [35] Xiao, J. (2015) “Bargaining Order in a Multi-Person Bargaining Game,” University of Melbourne Working Paper. Available at SSRN: http://ssrn.com/abstract=2577142
35
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https://www.groundai.com/project/design-of-a-surface-trap-for-freely-rotating-ion-ring-crystals/ | Design of a Surface Trap for Freely Rotating Ion Ring Crystals
# Design of a Surface Trap for Freely Rotating Ion Ring Crystals
Po-Jen Wang Department of Physics, University of California, Berkeley, California 94720, USA Tongcang Li NSF Nanoscale Science and Engineering Center, 3112 Etcheverry Hall, University of California, Berkeley, California 94720, USA Crystal Noel Department of Physics, University of California, Berkeley, California 94720, USA Xiang Zhang NSF Nanoscale Science and Engineering Center, 3112 Etcheverry Hall, University of California, Berkeley, California 94720, USA Hartmut Häffner Department of Physics, University of California, Berkeley, California 94720, USA
July 15, 2019
###### Abstract
We present a design of an r.f. trap using planar electrodes with the goal to trap on the order of 100 ions in a small ring structure of diameters ranging between 100 m and 200 m. In order to minimize the influence of trap electrode imperfections due to the fabrication, we aim at trapping the ions around 400 m above the trap electrodes. In view of experiments to create freely rotating crystals near the ground state, we numerically study factors breaking the rotational symmetry such as external stray electric fields, local charging of the trap electrodes, and fabrication imperfections. We conclude that these imperfections can be controlled sufficiently well under state-of-the-art experimental conditions to allow for freely rotating ion rings even at energies comparable to the ground state energy of the rotational degree-of-freedom.
###### pacs:
preprint: APS/123-QED
## I Introduction
The electronic and motional degrees-of-freedom of ions trapped with electromagnetic fields are extremely well decoupled from their environment. In addition, lasers and electromagnetic fields allow for excellent control of both degrees-of-freedom on the single quantum level Leibfried et al. (2003); Häffner et al. (2008). Both of those properties make ion crystals nearly perfect systems to study many-body physics in closed systems Friedenauer et al. (2008); Richerme et al. (2014); Jurcevic et al. (2014); Ramm et al. (2014). While most experiments are carried out with linear ion strings, a particularly interesting structure is a ring of trapped ions. Proposals include mini-accelerators Blümel and Smaldino (1999); Schätz et al. (2001), dynamics of KinksLanda et al. (2010), quantum emulation of ring molecules, and the acoustic analog of Hawking radiation Horstmann et al. (2010). Recently, also rings of trapped ions have been suggested to realize the concept of so-called time crystals Wilczek (2012); Li et al. (2012).
However, starting from experiments on how to implement rings of trapped ions by Schätz et. al. Waki et al. (1992); Schätz et al. (2001), it has become clear that imperfections and charging will make it very hard to implement such experiments. Thus, a few design improvements have been proposed Lammert et al. (2006); Austin et al. (2007); Madsen and Gorman (2010). Furthermore, the Sandia group has implemented a ring trap on surface trap technology Tabakov et al. (2012). Common to those designs and experiments is that the resulting ring potential has relatively large diameters, making it difficult to compensate for imperfections.
Inspired by Ref. Clark (2013), we study a novel design, deviating from the idea of bending a conventional linear trap into a ring. In addition, with a planar electrode design amenable to microfabrication, we seek to reduce inevitable imperfections from the geometry as well as local charging of trap electrodes. The main feature of our geometry is to trap the ion ring far away from the trapping electrodes as compared to the typical ion-ion distance and the ring diameter itself. Thus local imperfections from stray charges affect the rotational symmetry of the ion ring much less as if the ions were trapped close to the trap electrodes.
Our design is composed of concentric planar ring electrodes (see Fig. 1). Trapping is accomplished by applying suitable radio frequency (rf) voltage to those rings. Work by Clark suggest that the multipoles of such a trapping potential can be adjusted over a wide range by changing the rf voltage on the various rings Clark (2013). However, it is experimentally difficult to keep several rf high-voltage sources in phase. We thus design a trap requiring only one rf high-voltage source. Fixing this parameter, we aim to find a trap geometry yielding a rotationally symmetric potential minimum at the desired ring diameter and height.
For the study of the physics of symmetrization of the wavefunction of bosonic and fermionic ions discussed in the context of time crystalsWilczek (2012); Li et al. (2012), it will be important that the ion ring can freely rotate at the level of single rotational quanta. Keeping that in mind, we consider identical ions with mass and charge in a ring trap with diameter and a uniform magnetic field. The energy scales of its internal vibration modes are much larger than the energy scale of its collective rotation Burmeister and Maschke (2002). Thus, we focus on the collective rotation. Because of the required symmetrization of identical ions, the canonical angular momentum of the ion ring will be quantized according to for identical bosons, where is the quantum number of the rotation. The kinetic angular momentum is and the eigenenergy of the collective rotation is for identical bosonic ions, where is the normalized magnetic flux. This provides an energy scale of
Egap=Nℏ2Md2. (1)
For identical fermions, the energy scale is the same, although its dependence on can be different. For an ion ring of 100 Ca ions with a diameter of 100 m, the energy scale of the collective rotation is nK. We seek to create a ring potential with sufficiently small imperfections such that a classical ion ring with rotational energy corresponding to the ground state energy would not be pinned by the imperfections.
In view of these considerations, it is important to trap the ion ring far away from trapping electrodes, while at the same time keeping the ion ring as compact as possible. Thus, heating is reduced while still maintaining a reasonable energy scale of the ground state. Additional design constraints are ease of symmetric trap fabrication as well as reasonable trapping voltages while maintaining an appreciable trap depth. To this end we target an ion ring with a diameter of 100 m trapped 400 m above the trap electrodes.
The remainder of the paper is organized as follows. In Sec. II, we briefly summarize the methods outlined in Ref. Clark (2013) on how to efficiently calculate rotationally symmetric potentials. Armed with the potential, we study the structure of ion crystals forming in such ring shaped potentials in Sec. III. We then analyze various imperfections breaking the rotational symmetry in Sec. IV, most notably external stray fields, electrode edge irregularities, and local charging of the trap electrodes. Sec. V addresses the process of cooling and pinning such a ring of ions.
## Ii Calculation of the trapping potential
We start with the trap design proposed in Ref.Clark (2013) composed of planar ring electrodes of different radii and applied voltages. Given this cylindrically symmetric boundary condition, the analytic solution to Laplace’s equation is given by Kim et al. (2010)
Φ(z,r)=∫∞0J0(kr)e−kzA0(k)dk. (2)
where is the Bessel function of order. can be expressed as , and is given by
Ai(k)=Vi(biJ1(kbi)−aiJ1(kai)). (3)
where and are the outer and inner radius of each ring electrode and is the amplitude of the rf voltage applied to the each electrode. In order to study ions in this oscillating trapping potential, we approximate the potential by the time-averaged pseudopotential
Ψ(z,r)=Q24MΩ2rf|→E(z,r)|2. (4)
This approximation is valid when the oscillation frequency of the trapped ion is much smaller than the rf frequency.
### Trap design
Ref.Clark (2013) showed that multipole surface traps can be built from concentric rings with particular sets of ring diameters and applied voltages. However, it is experimentally difficult to keep several rf high-voltage sources in phase. We thus design a trap targeting only one rf high-voltage source. Our design, shown in Fig. 2, is composed of three concentric ring electrodes with outer radius 126 m, 600 m, 1100 m, with the second ring grounded and the other two connected to a fixed rf driving source of amplitude = 1000 V. As the rf driving frequency, we choose MHz. In what follows, we also assume Calcium ions with mass amu. The design leads to a Mexican-hat-shaped pseudopotential in the radial direction and a confining pseudopotential in the axial direction, as shown in Fig. 3. The trap potential has minimum at radius m, height 385 m, and a trap depth of 0.134 eV, leading to single-ion trap frequencies MHz and MHz in the radial and vertical direction, respectively.
### Design variation compensation
In view of fabrication imperfections, we study the effects of small deviation in the size of the center electrode. Our simulation Fig. 4 shows that the position of the potential minimum is very sensitive to the change in the size of the center electrode. In particular, it shows that changes of 1 m in radius will shift the radius of the minimum by 10 m. While we expect that microfabrication allows fabrication with tolerances below the micrometer range, we also can tune the potential by adding a small variable rf voltage with the same driving frequency on the second ring, but with the phase exactly opposite to that. Simulations show that the minimum position is shifted radially inward by about 2.5 m/V, while the trap depth changes by 0.005 eV/V. This small compensation rf voltage provides a powerful tool for fine-tuning the potential in situ.
## Iii Structure of ring crystals
Of particular interest are the conditions under which ultra-cold ions form a ring in this potential. For this, we carry out molecular dynamics simulation to analyze the structure of laser-cooled ions in the surface trap Okada et al. (2007). We calculate the trajectories and velocities of the trapped ions by solving Newtonian equations of motion including the Coulomb interaction, the pseudoforce from the rf potential, and a hypothetical damping force term Meyrath and James (1998). The extra damping term serves as a friction term that will gradually reduce the energy of the ions, thereby simulating laser cooling. Thus, the ions will eventually reach a steady state, which represents the expected structure of the cold ion crystal. The equations of motion of the ion can be written
mid2→xidt2=−γd→xidt+→FT+→FC. (5)
where is the damping coefficient, is the pseudoforce and the Coulomb force is given by
→FC=q24πϵ0N∑i≠j→ri−→rjR3ij, (6)
where is the number of ions in the trap.
Subsequently, the equation of motion is numerically solved by fourth-order Runge-Kutta method with a time step of 20 ns. For a reasonable run-time of the algorithm of a few hours, we choose the damping coefficient kg/s. As a result, our simulation shows that a ring crystal can be formed with up to 92 ions with the parameters and geometry discussed above (outer radius of the inner electrode 126 m). As shown in Fig. 5(a), the 92-ion ring has diameter 116 m and height 385 m,
Keeping all parameters fixed, but adding one more ion yields a 93-ion ring crystal of two layers with about 1 micrometer separation in the plane perpendicular to the trap surface, as shown in Fig. 5(b). We can study this phase transition from single-layer ion rings to double-layer ion rings by fine-tuning the trapping potential. This can be done by adjusting the compensation rf voltage on the second ring, as we have discussed in Sec. II. In the double layer regime, we also find meta-stable kinks as shown in Fig. 5(c). The kink dynamics in a ring might be an interesting subject in its own right Landa et al. (2010). Contrary to studies in linear traps Mielenz et al. (2013); Ulm et al. (2013); Pyka et al. (2013), the kinks are in a homogeneous environment and cannot escape by just traveling to the edge of the ion crystal. Furthermore, working with an odd number of ions enforces the presence of an odd number of kinks and thus of at least one, while working with an even number of ions would lead to an even number of kinks.
## Iv Analysis of imperfections
Of particular interest in our work is to create ion crystals freely rotating even if their rotational energy is comparable to the groundstate energy Li et al. (2012); Wilczek (2012). With with the criterion established in Eq. 1, i.e the energy barrier created by the imperfections should be be smaller than nK, we calculate the energy as a function of the angle when rotating the crystal around the symmetry axis. We study three sources of imperfections: a homogeneous electric field, irregularities on the edge of the electrodes, and the effect of a local charging (Fig. 6).
### Homogeneous Electric Field
First we calculate the energy of the ion ring as a function of rotation angle in presence of an homogeneous electric field. The result will be a sinusoidal periodic function whose amplitude represents the classical energy barrier, which we denote (c.f. Fig. 8). Fig. 7 shows the energy barrier as a function of the applied field for 10, 20, 30, 40, and 50 ions. Fig. 7 illustrates that the energy barrier is drastically suppressed when the number of ions is increased. Recall that the energy gap is also proportional to the number of ions, . This suggests that freely rotating crystals with larger ion numbers are easier to observe, and suffer less from the imperfection of the trap. Intuitively, this can be understood in the following manner: for an increased ion number, the ion-ion spacing is much reduced approaching a more homogeneous and continuous charge distribution. For such an homogeneous charge distribution external imperfections cannot exert a torque on the charge distribution. Thus, the rotational barrier caused by the imperfection drastically reduces with an increasing number of ions.
### Electrode Edge Irregularity
Next we add a square electrode of width 1 m (2 m) adjacent to the center electrode imitating a fabrication imperfection. The electric potential of a square electrode can be calculated from solving Laplace’s equation analytically Gotoh and Yagi (1971). We obtain the full potential by superposing this solution to the one obtained earlier. Fig 8 shows the energy of a 25-ion ring as a function of rotation angle with the square electrode of width 1 m and 2 m.
We concluded that already for 25 ions, the energy barrier is sufficiently small. Increasing the ion number reduces the ion-ion spacing which is expected to reduce the energy barrier further.
### Local Charge
In surface traps with ion-electrode distances on the order of 100 m, we typically find electric fields on the order of 100 V/m before performing micromotion compensation. These fields come potentially from local charges on the trap surface as caused for instance by small charged dust particles on the electrodes. To study the effects of this, we assume that a square of size 10 m 10 m at a position 200 m from the center, which carries a different voltage than the rest of the electrodes. By applying 1V to 50 V DC to the square electrode, we create electric field of strengths from 2 V/m to 100 V/m at the center of the ion ring 385 m above the surface. The result is presented in Fig 9. We find that already a 25-ion crystal is nearly insensitive to charging of a 10 m 10 m surface to tens of volts. Again, increasing the ion number will reduce the energy barrier further, and thus allow for even stronger local charge imperfections.
## V Laser Cooling
Finally, we study the laser cooling dynamics for freely rotating ion rings. In ion trapping experiments, usually a single cooling beam is preferred avoiding complications with interference effects. If this beam is centered perfectly on the ion ring, the radiation pressure will be balanced on each side of the nearly freely rotating ring. If, however, the beam is displaced from center, then a net torque from radiation pressure will result. Treating the ring as a rigid rotating body, we can apply the force equations for Doppler cooling Leibfried et al. (2003) and determine the net torque on the ring. In the limit of small velocities, at temperatures close to the Doppler cooling limit, we linearize the forces and the average torque on a single ion can be written as:
τa=R×F0(1+κv). (7)
where is a radial vector of each ion. is averaged radiation pressure in the direction of laser beam propagation,
|F0|=ℏkΓs/21+s+(2Δ/Γ)2. (8)
The drag coefficient for cooling,
κ=8kΔΓ2cosθ1+s+(2Δ/Γ)2. (9)
is negative when the detuning is negative. The arises from the projection of the laser onto the rotational degree-of-freedom of each ion, , where is the velocity of the ion. The decay rate is given as ns) for Calcium, the wavenumber is defined as nm), and is the saturation parameter Leibfried et al. (2003). In what follows, we assume a saturation parameter of and a gaussian beam profile with a beam waist of 200 m. We chose a large beam waist to minimize the differences in intensity across the ring of ions and assume a detuning of MHz for optimal cooling.
The Doppler cooling action of the laser on the moving ions will counterbalance the torque from the radiation pressure. If the torque is sufficiently small, the forces will cancel at some finite frequency of rotation. For 30 ions confined to a m ring, we calculate the equilibrium frequency and find a rotation frequency of approximately 1.04 kHz per 1 m displacement of the cooling beam.
To counteract the rotation, we turn to the previous discussion of the effects of electric fields on the ring. By applying a strong electric field, we aim at creating a sufficiently strong energy barrier, thus stopping the ion ring from rotating. The maximum slope of the energy of the ion crystal as function of the rotation angle represents the torque needed to overcome the energy barrier and cause the ions to rotate around the ring
τE=∣∣∣dEdθ∣∣∣max. (10)
Aiming at a static ion crystal, the torque from the cooling laser on a single ion is . Next, we look to find an applied electric field that will cancel the torque from the laser for this displacement. This condition can be written as
N∑i=1τiL<τE. (11)
where represents the sum over the torque of all ions by a given electric field, with the field orientation chosen to maximize the torque, c.f. Fig. 10b).
For ion positions in the trap given an electric field of 75 V/m optimally aligned against the propagation of the laser, the total sum of the torque from the laser alone is approximately linear for small laser displacements d. Analyzing the situation for 30 ions and taking into account their calculated positions in the ion ring, we obtain N. The torque from the energy barrier for 75 V/m, given by the maximum slope of the curve in Figure 11, is Nm, allowing for a displacement d of up to 6.0 m to achieve a static ion crystal.
## Vi Conclusions
We have studied a design of a planar trap providing trapping of a 92 ion-ring of diameter 116 m at height 385 m above its surface. This design can be fabricated using micro-fabrication methods with high precision. We also studied the rotational motion under three symmetry-breaking imperfections: homogeneous electric fields, irregularities of electrode edges from fabrication imperfections, and local charges placed on the trap electrodes. We have shown that the rotational energy barrier induced by these imperfections drastically reduces with an increasing number of ions in the ring. We thereby expect that the energy barrier from the imperfections can be reduced below the rotational ground state energy of large ion crystals. In addition, we have shown that laser alignment and strong homogeneous electric field of 75 V/m can be utilized to pin and cool the ion ring for trapping and imaging.
## Acknowledgements
This work is supported by the W.M. Keck Foundation. We acknowledge the contributions of Anthony Ransford and Hao-kun Li to the discussions related to this work.
## References
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https://www.physicsforums.com/threads/lagrangian-dynamics-problem-need-help-with-setup.100378/ | # Lagrangian Dynamics problem - need help with setup
1. Nov 17, 2005
### don_anon25
Lagrangian Dynamics problem -- need help with setup
Here's the problem:
A simple pendulum of length b and bob with mass m is attached to a massless support moving horizontally with constant acceleration a. Determine the equations of motion.
For the pendulum, x = b sin theta and y = b cos theta (Which of these equations should I use? y = b cos theta?)
For the support, x = (v0)t +.5a*t^2 - b sin theta? Is this correct?
Kinetic energy is 1/2 m (x'^2+y'^2), correct?
Potential energy is U=mgy.
If my equations are wrong, could someone tell me why and how to correct them? Your help is greatly appreciated!!!
2. Nov 17, 2005
### Physics Monkey
Remember that the support is massless so it has no kinetic or potential energy; you should just think of the support as being accelerated externally. To find the Lagrangian, you need only know the kinetic and potential energy of the pendulum, right? Start by writing the position of the pendulum relative to the support (hint: this is your usual pendulum position), and then write the position of the support relative to fixed lab (hint: this term should involve the acceleration). Once you have these two pieces, add them to obtain the position of the pendulum relative to the inertial frame. From this you can calculate the kinetic and potential energies and thus the Lagrangian. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.90152907371521, "perplexity": 509.5257724226312}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583656530.8/warc/CC-MAIN-20190115225438-20190116011438-00527.warc.gz"} |
https://blog.minicircuits.com/rf-microwave-equalizers-an-essential-ingredient-for-the-modern-system-designer/ | Select Page
The need for equalization has become commonplace throughout the RF/microwave/millimeter wave frequency ranges. Modern military, satellite and wireless communications systems transmit and receive signals with relatively high instantaneous bandwidths. Gain variation across the signal bandwidth induces distortion in the transmitted or received signal because not all frequency components are amplified equally1. In other words, for multi-phase, multi-amplitude-level-modulated systems, unintended amplitude differences (i.e. due to gain changes) over the occupied bandwidth introduce errors, leading to incorrect interpretation of the signal and an increase in system bit error rate (BER)2. A 5G system, for example, may operate in FR1 (3.3-4.2 GHz) with a bandwidth of 100 MHz or in FR2 (24-54 GHz) with an aggregate contiguous bandwidth of 1.2 GHz3. 5G signals utilize quadrature amplitude modulation (QAM)4, a multi-phase, multi-amplitude-level modulation scheme that is susceptible to amplitude (gain) variation over its occupied bandwidth. Consequently, gain flatness specifications have become more demanding for receiver and transmitter chains operating across wider bandwidths.
Most transmitter and receiver systems experience decreasing gain and increasing losses at progressively higher frequencies. The equalizer is a component designed specifically to compensate for this natural tendency toward roll-off exhibited by these systems. Mini-Circuits’ equalizers extend from DC to 45 GHz. In combination with Mini-Circuits’ family of ZEQ and VEQY connectorized components, the EQY-series SMT and die devices provide the designer great latitude in the selection of an equalizer to fit any application. These equalizers are offered in many different attenuation levels and in bandwidths ranging from DC to 6 GHz, 20 GHz, 28 GHz, and 45 GHz.
## What is an Equalizer?
An equalizer, sometimes referred to as a “lossy equalizer,” is essentially an attenuator placed in a circuit or system that exhibits an insertion loss that is at its maximum at the lowest operating frequency and at its minimum at the highest operating frequency. A complementary statement is that the equalizer exhibits minimum gain at the lowest operating frequency and maximum gain at the highest operating frequency. Placing the equalizer in a system cascade “tilts” the gain response in a positive direction. The magnitude by which the equalizer tilts the system gain response from minimum to maximum operating frequency is called its equalization value or “slope.”
Although far less common, some equalizers tilt gain in a negative direction. This type of equalizer is suitable in applications requiring waveguide components, for example. Additionally, parabolic equalizers have long established a niche in the area of traveling-wave tube (TWT) amplifiers. Regardless of the shape of the equalizer response, the end goal is to flatten or “equalize” the gain of a system over a relatively wide bandwidth.
Figures 1 and 2 show frequency response curves for three SMT devices in Mini-Circuits’ line of 45 GHz equalizers, the EQY-3-453+, EQY-6-453+ and EQY-10-453+. Each equalizer in Figure 1 has a different slope, or magnitude of monotonic decrease in attenuation from DC to 45 GHz. The slope of each model may also be interpreted as the magnitude of monotonic increase in gain from DC to 45 GHz, as shown in Figure 2. It is easier to visualize how the equalizer compensates for gain roll-off when the frequency response is viewed from a gain perspective, as in Figure 2. Mini-Circuits’ family of 45 GHz equalizers includes a unique model for each 1 dB increment in slope from 3 to 10 dB. Additionally, every equalizer in the 45 GHz line is available in die form.
Figure 1: EQY-3-453+, EQY-6-453+ and EQY-10-453+ MMIC SMT equalizer IL vs. frequency.
Figure 2: EQY-3-453+, EQY-6-453+ and EQY-10-453+ MMIC SMT equalizer gain vs. frequency.
## When is an Equalizer Required?
Generally, equalizers are necessary in any wideband system in which elements of gain and insertion loss are cascaded in series. Examples of systems that require equalizers are wideband receivers and transmitters, where typically quite a number of gain and insertion loss elements are cascaded together, and where flat system frequency response is often desired. In most cases, the gain of the amplification stages declines with increasing frequency, and the insertion loss of the lossy elements increases, leading to significant roll-off in overall cascaded system gain. The equalizer provides a negative insertion loss slope to the system (less insertion loss at higher frequencies), and when chosen properly, may reduce or even eliminate frequency-dependent system gain roll-off.
## How Much Equalization is Necessary?
It is possible to determine the correct amount of equalization by examining the characteristics of the discrete system components for a small system. However, even cables, circuit board traces and the mismatch of connector transitions constitute additional losses with frequency that are difficult to predict. So in reality, it is therefore most often necessary to perform simulations on a well-modeled system to determine the correct amount of equalization. Even with precise modelling, designers often need to evaluate multiple slope values at the prototyping stage to achieve measured performance that matches the requirements. Mini-Circuits’ designer kits avail the designer of a multitude of slope values in small quantities for evaluation and prototyping, ensuring that optimum performance can be achieved in practice.
If the system consists of just a handful of components, however, it is quite possible to achieve a reasonable estimate of the required equalization using system calculations associated with simple cascade analysis, which we utilize for the following hypothetical system for illustration.
## Equalization Yields Flat Gain in a Two-Stage Wideband LNA
An example of a two-stage, 400 to 6000 MHz LNA for which equalization is required is shown in the block diagram in Figure 3 below. The aggressive performance goals for the design are:
• Frequency Range: 400-6000 MHz
• Gain: 30 dB min.
• NF: 1dB max.
• Output P1dB: +15 dBm min.
• Output IP3: +25 dBm min.
Figure 3: Two-stage 400-6000 MHz LNA block diagram.
The input stage is the Mini-Circuits TAV2-14LN+. This choice was made based on its very low noise figure (NF). When operated at Vd = 2V, this device maintains a NF of 0.69 to 0.75 dB over the 400-6000 MHz band. The TAV2-14LN+ also exhibits relatively high gain (16-22 dB) over the band of interest, although its gain slope will necessitate equalization. The amplifier for the final stage is the PHA-83W+, selected predominantly since it performs quite well in the linearity category, boasting 25 dBm OP1dB and 36 dBm OIP3 at 400 MHz and Vd = 9V. Not only do OP1dB and OIP3 of the PHA-83+ remain strong to 6000 MHz and beyond, but its total gain flatness across the 400-6000 MHz band is only ±1 dB.
In order to determine the required equalization in this example, it is first necessary to determine the frequency response of the system gain without the use of an equalizer. Since the gain is in dB, the cascaded gain is simply the sum of the gain of the two stages. Figure 4 shows that the gain without equalization exhibits a significant roll-off of 7 dB, 6 dB from the TAV2-14LN+ plus 1 dB from the PHA-83W+. The biggest contributor to negative gain slope, the TAV2-14LN+ input stage, will be equalized, and the results examined.
Figure 4: Gain and NF with and without equalization for two-stage LNA system.
The system is an LNA requiring very low NF (< 1 dB). Therefore, no attenuation should be added to the input. Integrating a single equalizer that has steep slope into the cascade is often not the best way to correct system gain roll-off. It is generally better to distribute equalization throughout the cascade. An arbitrary decision was made to split the required amount of equalization between two identical SMT equalizers, the EQY-3-63+, for which the frequency response is shown in Figure 5. The first of the two 3-dB equalizers is placed on the output of the first stage amplifier which greatly reduces its effect on overall NF (since a gain element precedes it). A collateral benefit of splitting the equalization in the LNA of Figure 3 is that the first equalizer serves as a low value attenuator, improving the interstage 50Ω match and mitigating VSWR interactions between stages.
The additional equalization needed is provided by a second EQY-3-63+ equalizer placed at the output of the final stage of the LNA. The cascaded system gain both with and without equalization is shown in Figure 4. The overall LNA with equalization has a gain vs. frequency slope that is now just slightly negative. The choice of equalization was sufficient to compensate for the vast majority of the gain roll-off for the LNA’s 400-6000 MHz operating range. Additionally, the absolute gain is just slightly lower than the design goal of 30 dB min. in the center of the band, which is a rather successful outcome, given this gain flatness was achieved over more than a decade of bandwidth.
Figure 5: EQY-3-63+ equalizer gain and insertion loss vs. frequency.
## The Equalizer and its Influence on Noise Figure
When an equalizer is introduced into a system, it is important to examine its effect on system noise figure (NF) over frequency since the equalizer is essentially a frequency-dependent attenuator. The effects of equalization on system performance parameters NF, P1dB, and IP3 can often be quite involved, particularly as the complexity of the system grows.
In the relatively simple example discussed above, it is first necessary to take the inverse log of each of the components’ respective noise figures (NF1, NF2, NF3, …(dB)) and Gains (dB) to determine individual noise factors (F1, F2, F3, …) and linear gains (G1, G2, G3, …). Next, the two-stage LNA requires Friis’ formula to cascade the respective, individual noise factors and to compute system noise factor (F). System noise factor is then converted back to system noise figure (NF) in dB. Sample calculations are shown below for cascaded system F and NF at 400 MHz, with equalization:
Fn = 10^(NFn/10) = 10(NFn/10)
F1 = 10^(0.69/10) = 1.17, F2 = 10^(3.77/10) = 2.38, F3 = 10^(2.96/10) = 1.98, F4 = 10^(3.77/10) = 2.38
Gn = 10^(Gn(dB)/10) = 10(Gn(dB)/10)
G1 = 10^(22.0/10) = 158.9, G2 = 10^(-3.77/10) = 0.42, G3 = 10^(6.59/10) = 45.6, G4 = 10^(-3.77/10) = 0.42
F = F1 + (F2 – 1)/G1 + (F3 – 1)/G1G2 + (F4 – 1)/G1G2G3 + …
F = 1.17 + (2.38 – 1)/158.9 + (1.98 – 1)/(158.9)(0.42) + (2.38 – 1)/( 158.9)(0.42)(45.6) = 1.196
NF = 10*log(F) = 10*log(1.196) = 0.78 dB
The results of these calculations for cascaded system NF are shown for the entire frequency range in Figure 4. Without the additional attenuation of the equalizers, the NF of the system is naturally lower, and with equalization it is higher. Even with equalization, however, system NF remains below 1 dB, meeting one of the key design goals for the LNA.
## LNA System Linearity Parameters: Output P1dB and Output IP3
The results of performing a cascade analysis for OP1dB and OIP3 are shown in Figure 6. Both parameters meet their respective design goals of +15 dBm and +25 dBm with a good amount of margin. Naturally, the amount of performance margin at the low end of the band (400 MHz), where the equalizer exhibits maximum attenuation, is lower than the margin at the high end of the band. This particular LNA has nearly flat gain and is capable of providing increasing OP1dB and OIP3 with frequency. Generally, most system components will exhibit a subtle roll-off in linearity with increasing frequency, making these positively-sloping linearity characteristics potentially useful were this LNA to be designed into a larger, more complex system.
Figure 6: Output P1dB and output IP3 with and without equalization for two-stage LNA system.
## Equalizers Ready for System Integration
Equalizers are an essential part of any wideband system for which flat gain response is required. Passive or “lossy” equalizers tend to reduce overall gain but may often have minimal effect on NF as shown in the two-stage LNA example. Additionally, when utilized between amplifier stages, VSWR interaction will be mitigated and 50Ω matching will improve along with gain flatness. While linearity experiences a slight reduction with equalization, it is possible to achieve increasing linearity with frequency. Mini-Circuits has over 70 unique equalizer models in stock with slope values from 1 to 15 dB and operating frequency ranges spanning DC to 45 GHz, including voltage variable equalizers.
Browse our full selection of RF / microwave equalizers >
## Related Blog Posts
Positive Gain Slope Amplifiers Compensate for Gain Roll-Off in Wideband Systems
Flattening Amplifier Gain Slope with MMIC Fixed Equalizers
## References
1. Keysight Support – VNA Help, “Why Measure Small-signal Gain and Flatness” Gain and Flatness (keysight.com)
2. Electronic Design October 10, 2013 “Understanding Error Vector Magnitude” Lou Frenzel Understanding Error Vector Magnitude | Electronic Design
3. Keysight Blogs, RF + Microwave, April 20, 2021 “What You Need to Know about Wideband Signal Analysis” Eric Hsu What You Need to Know about Wideband Signal Analysis | Keysight Blogs
4. 5G Technology World April 16, 2020 “The Basics of 5G’s modulation, OFDM Bob Witte The basics of 5G’s modulation, OFDM – 5G Technology World | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8022664785385132, "perplexity": 2859.3695581046495}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662675072.99/warc/CC-MAIN-20220527174336-20220527204336-00085.warc.gz"} |
https://math.stackexchange.com/questions/3265576/associative-law-for-boolean-logic | # Associative Law for Boolean Logic
The associative law states that for the logic formula:
$$(A \wedge B) \wedge C = A \wedge (B \wedge C)$$
$$(A \vee B) \vee C = A \vee (B \vee C)$$
I asked myself would the associative law hold for multiple operators, so I tested it out on $$(A \wedge B) \vee C$$ vs $$A \wedge (B \vee C)$$. This turned out to not be true once I did I truth table. For the first formula as long as C is true the entire thing is true even if A is not true, while for the second, A must be true in order for the logic formula to hold true.
My question is are there times when the associative law still works when I have multiple operators and not just two (e.g. $$A \wedge B \wedge C \vee D\vee E$$). How can I tell immediately whether or not the associative law is applicable? Are there operators (such as $$\downarrow$$) that work with the associative law even when I have different operators in the same formula?
• @PeterForeman : Disjunction is associative... – Eric Towers Jun 17 '19 at 17:29
Apart from conjunction and disjunction, other associative operations are the biconditional $$\leftrightarrow$$ and the exclusive-or $$\oplus$$, where $$a\oplus b$$ is defined as $$(a\land\lnot b)\lor (\lnot a\land b)$$, i.e., $$a\oplus b$$ is true if and only if exactly one of $$a$$ and $$b$$ is true. The conditional $$\rightarrow$$ is not associative.
To your broader question, I'd say it's not entirely clear what it means to say that the associative law applies to 'multiple operators': typically, it's only applied to iterations of the same operator.
In your example with $$(a\land b)\lor c$$ and $$a\land (b\lor c)$$, you can replace $$a\land b$$ with $$f(a,b)$$, and $$a\lor b$$ with $$g(a,b)$$. Your suggestion is then that associativity would require that $$g(f(a,b), c)$$ is equivalent to $$f(a, g(b,c))$$. However, it's not entirely clear that this is what associativity should mean in this context; and even if it does, you can't expect it to hold in general.
This is analogous to the situation with addition and multiplication. Both are associative by themselves. However, multiplication distributes over addition. You can't have combined associativity with both of them together. That is, for example, it is not true in general that $$(a+b)*c = a+(b*c).$$ In fact, what is true is that $$(a+b)*c = (a*c)+(b*c).$$ Same thing happens with Boolean logic $$\wedge$$ and $$\vee.$$
Associativity only holds if you have an expression with just one kind of binary operator in the expression. You can not combine two kinds of associative operators together in general, as demonstrated by the addition ($$+$$) and multiplication ($$*$$) example.
Of course, there can be special examples where you can combine operators and still have associativity. For example, suppose we define the operator $$a\circ b:=0$$ which is clearly associative (check). Then $$(a*b)\circ c = a*(b\circ c) = 0$$ for all $$a,b,c.$$
• So it's true that, in general, the associative law does not hold? Are there cases where they do hold? – BigBear Jun 17 '19 at 17:43
• @WilsonGuo: It would be more right to say that what you're calling "the associative law" is not the same as what most other people use that name about. Your (vastly extended) "associative law" does not hold in general. – hmakholm left over Monica Jun 17 '19 at 22:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8999279737472534, "perplexity": 141.03706523880928}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00588.warc.gz"} |
http://sourceforge.net/p/gnuplot/bugs/1197/ | ## #1197 postscript terminal: bad vertical centering of text
open
nobody
None
5
2013-06-29
2012-12-17
Ethan Merritt
No
The postscript terminal currently sets an offset used for vertical centering of text only once, in the prolog:
\vshift -<foo> def
This shift is only approximate to begin with and is not adjusted later when the current font changes.
Possible fix (quick-and-dirty): overwrite the vshift definition every time the font is changed using the same approximation used currently.
Better fix: re-write the Lshow/Rshow/Cshow commands to query the current font from inside the PostScript engine rather than using vshift.
## Discussion
The two possibilities I see for querying the current font are:
1. Use the font bounding box for centering.
This might not fit the current string.
1. Extract the metrics of all glyphs in the string.
How should the vertical centering be done? Using the lowest height, or the largest height, the median, ...?
For Latin fonts I would correct by the half of the lowest height (with respect to the base line). But I don't know if this would also fit cyrillic, japanese and other fonts.
If I find some time, I could come up with a proposal patch.
Christoph | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9796760678291321, "perplexity": 3693.94478780617}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375094629.80/warc/CC-MAIN-20150627031814-00243-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.zbmath.org/?q=an%3A1333.76003 | ×
# zbMATH — the first resource for mathematics
Mathematics of two-dimensional turbulence. (English) Zbl 1333.76003
Cambridge Tracts in Mathematics 194. Cambridge: Cambridge University Press (ISBN 978-1-107-02282-9/hbk; 978-1-139-57519-5/ebook). xvi, 320 p. (2012).
Publisher’s description: This book is dedicated to the mathematical study of two-dimensional statistical hydrodynamics and turbulence, described by the 2D Navier–Stokes system with a random force. The authors’ main goal is to justify the statistical properties of a fluid’s velocity field $$u(t,x)$$ that physicists assume in their work. They rigorously prove that $$u(t,x)$$ converges, as time grows, to a statistical equilibrium, independent of initial data. They use this to study ergodic properties of $$u(t,x)$$ – proving, in particular, that observables $$f(u(t,.))$$ satisfy the strong law of large numbers and central limit theorem. They also discuss the inviscid limit when viscosity goes to zero, normalising the force so that the energy of solutions stays constant, while their Reynolds numbers grow to infinity. They show that then the statistical equilibria converge to invariant measures of the 2D Euler equation and study these measures. The methods apply to other nonlinear PDEs perturbed by random forces.
##### MSC:
76-02 Research exposition (monographs, survey articles) pertaining to fluid mechanics 76F02 Fundamentals of turbulence 76D05 Navier-Stokes equations for incompressible viscous fluids
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https://www.physicsforums.com/threads/would-baryons-and-their-antiparticles-interact-via-the-strong-interaction.239730/ | # Would baryons and their antiparticles interact via the strong interaction?
1. Jun 11, 2008
### K.J.Healey
And would the potential be equal in magnitude yet opposite in sign?
If you were to approximate a yukawa potential for some baryon and had it "near" its antiparticle, what would the potential look like. The same for a baryon and another baryon but opposite?
This would just be like a residual strong interaction right? A quick approximation maybe being a light meson exchange?
Thanks!
2. Jun 11, 2008
### humanino
It took me a while to realize what was the question about ! In hadronic physics, we are not used to considering particles vs antiparticles, but more in terms of multiplets of similar particles, where it so happens that the multiplets connect particles with their antiparticles. One of the reasons QCD was quickly realized to be an excellent candidate for strong interactions is that there is an attractive force in the quark-antiquark and three-quark sectors, but repulsive in the quark-quark sector. So things are not as simple as "if you turn a particle into its antiparticle, just reverse the force". However, particles and antiparticles do interact via strong interaction.
The very notion of potential works only in non-relativistic physics, so bearing that in mind, we need to do something like fixing our baryons in static positions at a given distance from one another and the potential would be the resulting energy of the system. It may or may not be relevant to the dynamics, but we can do that. Then most likely the potential will remain essentially the same. Reason is partly answered
we would indeed have meson exchanges between our static colorless sources. The quantum numbers of those mesons will determine whether the interaction is repulsive or attractive, their mass will determine the range of static interaction. So if you take one of your baryons and change it into another one in your multiplet (like its antibaryon) you will need to exchange other mesons but they will still be in the same multiplet of mesons (not the same multiplet as your baryons). Say at short distances you will have a similar more massive vector repulsion and at long distance you will have a less massive scalar attraction (masses are almost degenerate in a given multiplet).
Last edited: Jun 11, 2008
3. Jun 11, 2008
### K.J.Healey
Well, I'm not sure how detailed I should be since I'm supposed to be researching this on my own, but I consider asking questions here part of research.
Now, some of you may have the answer right away, but that not what I want, I really want to make sure I understand this and how to go about figuring it out. Heres the scenario:
Non relativistic is fine.
I'm trying calculate an oscillation suppression of a nnbar in the presence of other neutrons via the strong interaction with a very naive model. I've read a few articles that give a decent method of approximating the residual-strong interaction between nucleons in a nucleus by using a sum of a heavy and a light meson exchange. They have all the data fitted so I have actual values for the coupling contants + masses.
While I'm not doing these calculations for nuclei, they are of the same interaction distance.
I guess I'm asking, if I have this fit of a sum of two meson exchange potentials for a neutron-neutron interaction (approx), what can I do to it (the potential) to make it reflect the interaction (approx) of a neutron-antineutron?
Is it equal in magnitude, yet opposite in sign? If so then I know what I have to do, just toss it in the Hamiltonian, and calculate he oscillation freq. {{E+V, dm},{dm,E-V}} or something similar.
Or is it completely different?
I'm trying to figure out how the strong interaction behaves for the antiparticle part of a nnbar state.
Last edited: Jun 11, 2008
4. Jun 11, 2008
### meopemuk
I also thought about this question. I found a couple of references that might be useful:
[1] M.-L. Yan, S. Li, B. Wu, B.-Q. Ma, Baryonium with a phenomenological
skyrmion-type potential. http://www.arxiv.org/abs/hep-ph/0405087v4
[2] J.M. Richard, Historical Survey of the Quasi-Nuclear Baryonium,
http://arxiv.org/abs/nucl-th/9906006v1
For example, if you change the sign of the nucleon-antinucleon potential shown in fig 1. of [1] you'll get something similar to the nucleon-nucleon potential as usually drawn in textbooks. Experimental studies of baryon-antibaryon potentials are difficult, because they tend to annihilate.
Another evidence for the opposite character of baryon-baryon and baryon-antibaryon potentials may come from the Sakata model, in which mesons are represented as nucleon-antinucleon bound states and baryons are composites of two nucleons and one antinucleon. This model was quite popular in the end of 1950's. Then it was replaced by the quark model and (almost) forgotten. In the paper
K. Matumoto, S. Sawada, Y. Sumi, M. Yonezawa, "Mass formula in the
Sakata model" Progr. Theor. Phys. Suppl. 19 (1961), 66
masses of (then known) mesons and baryons were fitted in the Sakata model, and the conclusion was that interaction energy nucleon-antinucleon has equal magnitude but opposite sign wrt the interaction energy nucleon-nucleon.
Eugene.
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## Section11.2Algebraic Systems
An algebraic system is a mathematical system consisting of a set called the domain and one or more operations on the domain. If $V$ is the domain and $*_1, *_2, \ldots , *_n$ are the operations, $\left[V;*_1, *_2, \ldots , *_n\right]$ denotes the mathematical system. If the context is clear, this notation is abbreviated to $V\text{.}$
### Subsection11.2.1Monoids at Two Levels
Consider the following two examples of algebraic systems.
1. Let $B^*$ be the set of all finite strings of 0's and 1's including the null (or empty) string, $\lambda$. An algebraic system is obtained by adding the operation of concatenation. The concatenation of two strings is simply the linking of the two strings together in the order indicated. The concatenation of strings $a$ with $b$ is denoted $a+b$. For example, $01101+101 =01101101$ and $\lambda +100 = 100$. Note that concatenation is an associative operation and that $\lambda$ is the identity for concatenation.
A note on notation: There isn't a standard symbol for concatenation. We have chosen $+$ to be consistent with the notation used in Python and Sage for the concatenation.
2. Let $M$ be any nonempty set and let * be any operation on $M$ that is associative and has an identity in $M\text{.}$
Our second example might seem strange, but we include it to illustrate a point. The algebraic system $\left[B^*;+\right]$ is a special case of $[M;*]$. Most of us are much more comfortable with $B^*$ than with $M\text{.}$ No doubt, the reason is that the elements in $B^*$ are more concrete. We know what they look like and exactly how they are combined. The description of $M$ is so vague that we don't even know what the elements are, much less how they are combined. Why would anyone want to study $M\text{?}$ The reason is related to this question: What theorems are of interest in an algebraic system? Answering this question is one of our main objectives in this chapter. Certain properties of algebraic systems are called algebraic properties, and any theorem that says something about the algebraic properties of a system would be of interest. The ability to identify what is algebraic and what isn't is one of the skills that you should learn from this chapter.
Now, back to the question of why we study $M\text{.}$ Our answer is to illustrate the usefulness of $M$ with a theorem about $M\text{.}$
\begin{equation*} \begin{split} (a*b)*(a*b) &=a*(b*(a*b))\quad \textrm{ Why?} \\ &=a* ((b*a)*b)\quad \textrm{ Why?}\\ &= a*((a*b)*b)\quad \textrm{ Why?}\\ &= a*(a*(b*b))\quad \textrm{ Why?}\\ &= (a*a)*(b*b)\quad \textrm{ Why?} \end{split} \end{equation*}
The power of this theorem is that it can be applied to any algebraic system that $M$ describes. Since $B^*$ is one such system, we can apply Theorem 11.2.1 to any two strings that commute. For example, 01 and 0101. Although a special case of this theorem could have been proven for $B^*$, it would not have been any easier to prove, and it would not have given us any insight into other special cases of $M\text{.}$
Consider the set of $2\times 2$ real matrices, $M_{2\times 2}(\mathbb{R})$, with the operation of matrix multiplication. In this context, Theorem 11.2.1 can be interpreted as saying that if $A B = B A$, then $(A B)^2= A^2B^2$. One pair of matrices that this theorem applies to is $\left( \begin{array}{cc} 2 & 1 \\ 1 & 2 \\ \end{array} \right)$ and $\left( \begin{array}{cc} 3 & -4 \\ -4 & 3 \\ \end{array} \right)$.
### Subsection11.2.2Levels of Abstraction
One of the fundamental tools in mathematics is abstraction. There are three levels of abstraction that we will identify for algebraic systems: concrete, axiomatic, and universal.
#### Subsubsection11.2.2.1The Concrete Level
Almost all of the mathematics that you have done in the past was at the concrete level. As a rule, if you can give examples of a few typical elements of the domain and describe how the operations act on them, you are describing a concrete algebraic system. Two examples of concrete systems are $B^*$ and $M_{2\times 2}(\mathbb{R})$. A few others are:
1. The integers with addition. Of course, addition isn't the only standard operation that we could include. Technically, if we were to add multiplication, we would have a different system.
2. The subsets of the natural numbers, with union, intersection, and complementation.
3. The complex numbers with addition and multiplication.
#### Subsubsection11.2.2.2The Axiomatic Level
The next level of abstraction is the axiomatic level. At this level, the elements of the domain are not specified, but certain axioms are stated about the number of operations and their properties. The system that we called $M$ is an axiomatic system. Some combinations of axioms are so common that a name is given to any algebraic system to which they apply. Any system with the properties of $M$ is called a monoid. The study of $M$ would be called monoid theory. The assumptions that we made about $M\text{,}$ associativity and the existence of an identity, are called the monoid axioms. One of your few brushes with the axiomatic level may have been in your elementary algebra course. Many algebra texts identify the properties of the real numbers with addition and multiplication as the field axioms. As we will see in Chapter 16, “Rings and Fields,” the real numbers share these axioms with other concrete systems, all of which are called fields.
#### Subsubsection11.2.2.3The Universal Level
The final level of abstraction is the universal level. There are certain concepts, called universal algebra concepts, that can be applied to the study of all algebraic systems. Although a purely universal approach to algebra would be much too abstract for our purposes, defining concepts at this level should make it easier to organize the various algebraic theories in your own mind. In this chapter, we will consider the concepts of isomorphism, subsystem, and direct product.
### Subsection11.2.3Groups
To illustrate the axiomatic level and the universal concepts, we will consider yet another kind of axiomatic system, the group. In Chapter 5 we noted that the simplest equation in matrix algebra that we are often called upon to solve is $A X = B$, where $A$ and $B$ are known square matrices and $X$ is an unknown matrix. To solve this equation, we need the associative, identity, and inverse laws. We call the systems that have these properties groups.
###### Definition11.2.3Group
A group consists of a nonempty set $G$ and a binary operation $*$ on $G$ satisfying the properties
1. $*$ is associative on $G$: $(a*b)*c=a*(b*c)$ for all $a, b, c \in G$.
2. There exists an identity element, $e \in G$ such that $a*e=e*a=a$ for all $a \in G$.
3. For all $a \in G$, there exists an inverse, there exist $b\in G$ such that $a *b = b*a=e$.
A group is usually denoted by its set's name, $G\text{,}$ or occasionally by $[G; * ]$ to emphasize the operation. At the concrete level, most sets have a standard operation associated with them that will form a group. As we will see below, the integers with addition is a group. Therefore, in group theory $\mathbb{Z}$ always stands for $[\mathbb{Z}; +]$.
###### Note11.2.4Generic Symbols
At the axiomatic and universal levels, there are often symbols that have a special meaning attached to them. In group theory, the letter $e$ is used to denote the identity element of whatever group is being discussed. A little later, we will prove that the inverse of a group element, $a\text{,}$ is unique and its inverse is usually denoted $a^{-1}$ and is read “$a$ inverse.” When a concrete group is discussed, these symbols are dropped in favor of concrete symbols. These concrete symbols may or may not be similar to the generic symbols. For example, the identity element of the group of integers is 0, and the inverse of $n$ is denoted by $-n$, the additive inverse of $n\text{.}$
The asterisk could also be considered a generic symbol since it is used to denote operations on the axiomatic level.
1. The integers with addition is a group. We know that addition is associative. Zero is the identity for addition: $0 + n = n + 0 = n$ for all integers $n\text{.}$ The additive inverse of any integer is obtained by negating it. Thus the inverse of $n$ is $-n$.
2. The integers with multiplication is not a group. Although multiplication is associative and 1 is the identity for multiplication, not all integers have a multiplicative inverse in $\mathbb{Z}$. For example, the multiplicative inverse of 10 is $\frac{1}{10}$, but $\frac{1}{10}$ is not an integer.
3. The power set of any set $U$ with the operation of symmetric difference, $\oplus$, is a group. If $A$ and $B$ are sets, then $A\oplus B=(A\cup B)-(A\cap B)$. We will leave it to the reader to prove that $\oplus$ is associative over $\mathcal{P}(U)$. The identity of the group is the empty set: $A\oplus \emptyset = A$. Every set is its own inverse since $A \oplus A = \emptyset$. Note that $\mathcal{P}(U)$ is not a group with union or intersection.
###### Definition11.2.6Abelian Group
A group is abelian if its operation is commutative.
### SubsectionExercises for Section 11.2
###### 1
Discuss the analogy between the terms generic and concrete for algebraic systems and the terms generic and trade for prescription drugs.
Answer
The terms “generic” and “trade” for prescription drugs are analogous to “generic” and “concrete” algebraic systems. Generic aspirin, for example, has no name, whereas Bayer, Tylenol, Bufferin, and Anacin are all trade or specific types of aspirins. The same can be said of a generic group $[G; *]$ where $G$ is a nonempty set and $*$ is a binary operation on $G$, When examples of typical domain elements can be given along with descriptions of how operations act on them, such as $\mathbb{Q}$* or $M_{2\times 2}(\mathbb{R})$, then the system is concrete (has a specific name, as with the aspirin). Generic is a way to describe a general algebraic system, whereas a concrete system has a name or symbols making it distinguishable from other systems.
###### 2
Discuss the connection between groups and monoids. Is every monoid a group? Is every group a monoid?
###### 3
Which of the following are groups?
1. $B^*$ with concatenation (see Subsection 11.2.1).
2. $M_{2\times 3}(\mathbb{R})$ with matrix addition.
3. $M_{2\times 3}(\mathbb{R})$ with matrix multiplication.
4. The positive real numbers, $\mathbb{R}^+,$with multiplication.
5. The nonzero real numbers, $\mathbb{R}^*$, with multiplication.
6. $\{1, -1\}$ with multiplication.
7. The positive integers with the operation $M$ defined by $a M b = \textrm{ the larger of } a \textrm{ and } b$.
Answer
The systems in parts b, d, e, and f are groups.
###### 4
Prove that, $\oplus$, defined by $A \oplus B = (A \cup B) - (A \cap B)$ is an associative operation on $\mathcal{P}(U)$.
###### 5
The following problem supplies an example of a non-abelian group. A rook matrix is a matrix that has only 0's and 1's as entries such that each row has exactly one 1 and each column has exactly one 1. The term rook matrix is derived from the fact that each rook matrix represents the placement of $n$ rooks on an $n\times n$ chessboard such that none of the rooks can attack one another. A rook in chess can move only vertically or horizontally, but not diagonally. Let $R_n$ be the set of $n\times n$ rook matrices. There are six $3\times 3$ rook matrices: $\begin{array}{ccc} I=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) & R_1=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right) & R_2=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right) \\ F_1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) & F_2=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{array} \right) & F_3=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) \\ \end{array}$
1. List the $2\times 2$ rook matrices. They form a group, $R_2,$ under matrix multiplication. Write out the multiplication table. Is the group abelian?
2. Write out the multiplication table for $R_3$ . This is another group. Is it abelian?
3. How many $4\times 4$ rook matrices are there? How many $n\times n$ rook matrices are there?
Answer
1. Elements are $I=\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$, and $T=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right)$, the group is abelian. Operation table is $\begin{array}{c|cc} \cdot & I & T\\ \hline I & I & T\\ T & T & I\\ \end{array}$
2. $\begin{array}{c|c} & \begin{array}{cccccc} I & R_1 & R_2 & F_1 & F_2 & F_3 \\ \end{array} \\ \hline \begin{array}{c} I \\ R_1 \\ R_2 \\ F_1 \\ F_2 \\ F_3 \\ \end{array} & \begin{array}{cccccc} I & R_1 & R_2 & F_1 & F_2 & F_3 \\ R_1 & R_2 & I & F_2 & F_3 & F_1 \\ R_2 & I & R_1 & F_3 & F_1 & F_2 \\ F_1 & F & F_2 & I & R_2 & R_1 \\ F_2 & F_1 & F_3 & R_1 & I & R_2 \\ F_3 & F_2 & F_1 & R_2 & R_1 & I \\ \end{array} \\ \end{array}$ This group is non-abelian since, for example, $F_1 F_2=R_2$ and $F_2 F_1=R_2$.
3. 4! = 24, $n!$.
###### 6
For each of the following sets, identify the standard operation that results in a group. What is the identity of each group?
1. The set of all $2\times 2$ matrices with real entries and nonzero determinants.
2. The set of $2 \times 3$ matrices with rational entries.
###### 7
Let $V = \{e,a,b, c\}$. Let $*$ be defined (partially) by $x * x = e$ for all $x \in V$. Write a complete table for $*$ so that $[V; * ]$ is a group.
Answer
The identity is $e\text{.}$ $a*b = c$, $a*c= b$, $b*c = a$, and $[V; *]$ is abelian. (This group is commonly called the Klein-4 group.) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.940157949924469, "perplexity": 249.8522201395846}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583509170.2/warc/CC-MAIN-20181015100606-20181015122106-00087.warc.gz"} |
http://math.stackexchange.com/questions/759237/criterion-to-decide-the-invertibility-of-polynomial-maps | Criterion to decide the invertibility of polynomial maps
Consider a polynomial map $f:\mathbb{R}^{n-1}\to V\subset\mathbb{R}^n$ where $V$ is $n-1$-dimensional variety in $\mathbb{R}^n$.
Are there any conditions on $f$ to determine whether it defines bi-rational equivalence between $V$ and $\mathbb{R}^{n-1}$?
Are you asking for a rational inverse of $f$ or a regular inverse of $f$ (the latter means that $f$ is an isomorphism to a subvariety of $\mathbb R^n$) ? – user143488 Apr 18 '14 at 14:40
I am asking about rational inverse of $f$. Are there any criterion/conditions on $f$? – Deepak Apr 18 '14 at 14:58
So you want $f$ to induce an immersion on an open subset of $\mathbb R^{n-1}$. A necessary condition is that the tangent map of $f$ is injective at some point. This would be sufficient if moreover $f$ is injective on an open subset. – user143488 Apr 18 '14 at 15:18
Actually bi-rational equivalence is weaker than isomorphism. For example consider the question asked here: math.stackexchange.com/questions/756322/…. In this example rational inverse was given by $$z_1=\frac{a^3-c}{2(a^2-b)}, z_2=\frac{a^3-3ab+2c}{3b-3a^2}$$ However for the points (a,b,c) satisfying $a^2-b=0$ there do not exist any finite $z_1$ and $z_2$. – Deepak Apr 18 '14 at 15:58
I didn't say isomorphism from $\mathbb R^{n-1}$ to $V$. – user143488 Apr 18 '14 at 15:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.879071056842804, "perplexity": 307.5062901900889}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246639674.12/warc/CC-MAIN-20150417045719-00121-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/calculus-word-problem.117949/ | # Calculus word problem
1. Apr 18, 2006
### preet
"An object moves so that its velocity, v is related to its position. s according to v = (b^2 + 2gs) ^1/2 where b and g are constants. Show that the acceleration of the object is constant."
I typed out the question exactly as it is. I'm confused because I don't really get what to do. To show that acceleration is constant, I need to get rid of that "s" variable in the question. Acceleration is = to d(velocity)/dt... but from the given function, you can only get d(v) / ds.
So dv/dt = dv/ds * ds/dt
But how do I find d(s) / dt? Don't I need a function that has position in terms of time?
TiA
Preet
2. Apr 18, 2006
### e(ho0n3
Isn't ds/dt just v?
3. Apr 18, 2006
### Hootenanny
Staff Emeritus
Yes but v isn't a function of t in this case, it is a function of s
4. Apr 18, 2006
### e(ho0n3
Does that matter? I mean, a = dv/ds * ds/dt = v dv/ds. If it can be shown that this is constant, then the problem is solved. There is no need to get rid of s.
5. Apr 18, 2006
### Hootenanny
Staff Emeritus
Could you please show me how;
$$\frac{dv}{ds} \cdot \frac{ds}{dt} = v\frac{dv}{ds}$$
Perhaps I'm missing something? It is late after all and I've run out of coffee :grumpy:
~H
6. Apr 18, 2006
### e(ho0n3
dv/ds * ds/dt = ds/dt * dv/ds by commutativity and substituting ds/dt for v gives the result I gave.
Is there a flaw in my reasoning here?
7. Apr 18, 2006
### Hootenanny
Staff Emeritus
Ahhhh It was so simple I missed it . My frantic scribblings on paper seem so stupid now.
$$\frac{ds}{dt} = v$$
You reasoning is perfect e(ho0n3.
To clarify for the OP;
You were right by using the chain rule to obtain;
$$\frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt}$$
What you (and I) didn't spot is that;
$$\frac{ds}{dt} = v \Rightarrow a = \frac{dv}{dt} = v\frac{dv}{ds}$$
As e(ho0n3 correctly stated.
~H
Last edited: Apr 18, 2006
8. Apr 18, 2006
### preet
Yeah, I feel bad not catching that... it was so obvious =/. Thanks a lot for the help!
-Preet
Similar Discussions: Calculus word problem | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8741639852523804, "perplexity": 2258.5433471011806}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189462.47/warc/CC-MAIN-20170322212949-00024-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/304394/high-power-congruences-finding-x | # high power congruences finding $x$
trying to solve:
$$x^{13} \equiv 11 \pmod{135}$$
I came to the fact that $x = 11^{59}$ but its in mod $72$ and needs to be converted to mod $135$
any suggestions? I'm not sure how to change it to mod $135$ with such a large number
-
$135 = 3^3 \cdot 5$, so $\varphi(135) = 72$. Using Euclid, you find that $$1 = 13 \cdot (-11) + 72 \cdot 2 = 13 \cdot 61 + 72 \cdot (-11),$$ so $$x \equiv (x^{13})^{61} \equiv 11^{61} \pmod{135}.$$
PS Once again, apologies for the initial mistake in this late-night post (never again!), and thanks to Gerry Myerson and user62340 for calling my attention to it.
To finish the calculation by hand, it is probably safer to compute first $$11^{11} = 11 \cdot 11^2 \cdot 11^{8} \equiv 41 \pmod{135},$$ and then the inverse $56$ of $41$ modulo $135$, which is the required solution.
-
$\phi(135)=72$. – Gerry Myerson Feb 14 '13 at 23:24
@GerryMyerson, thank you so much, I shouldn't post this late in the evening, I'll fix it. – Andreas Caranti Feb 14 '13 at 23:27
I thought ϕ(135) = 72 not 24... How did you get it Andreas? ϕ(135) is 2*3*3*4 ? – user62340 Feb 14 '13 at 23:32
@user62340, it's late in the evening, I cannot count anymore. I have fixed it, thanks to you and Gerry Myerson for notifying me. – Andreas Caranti Feb 14 '13 at 23:34
If $\rm\: x^{13}\equiv 11\,\ (mod\ 5\cdot27)\:$ then the same congruence holds mod $5$ and $27$, and we can use CRT (Chinese Remainder Theorem) to recombine the two solutions. It turns out to be very simple:
$\rm\quad\ mod\,\ 5\!:\ x^4\equiv 1,\:$ so $\rm\,x \equiv x (x^4)^3 \equiv x^{13}\equiv 11\equiv \color{#C00}1.\ \$ Next $\ \phi(27) = 18,\:$ hence
$\rm\quad\ mod\ 27\!:\ 1 \equiv x^{18}\equiv x^{13} x^5 \equiv 11 x^5\:$ so $\rm\,x^5 \equiv \dfrac{1}{11}\equiv \dfrac{55}{11}\equiv 5\equiv 2^5\Rightarrow\:x\equiv \color{#0A0}2\$ (unique, see Note)
$\rm\Rightarrow mod\ 5\cdot 27\!:\,\ x \equiv \color{#0A0}2 + 27\,\left[\dfrac{\color{#C00}1\!-\!\color{#0A0}2}{27}\ mod\ 5\right]\! \equiv 56,\$ by $\rm\ mod\ 5\!:\, \dfrac{-1}{27}\equiv \dfrac{4}2 \equiv 2,\$ using Easy CRT.
Note $\,\ 5$'th roots are unique $\rm\,mod\ 27\:$ since $\rm\, (5,\phi(27)) = (5,18) = 1,\:$ so $\rm\:n\equiv 1/5\ mod\ 18\:$ exists, hence $\rm\,x^5 = y^5\:\Rightarrow x\equiv x^{5n}\equiv y^{5n}\equiv y\,$ since $\rm\,5n\equiv 1\,\ (mod\ 18).$
-
$x^{13}\equiv11\pmod{135}\implies x^{13}\equiv11\pmod5$ and $x^{13}\equiv11\pmod{27}$
Using Euler's Totient Theorem, $$a^{\phi(m)}\equiv1\pmod m \text{ where }(a,m)=1$$
So, if $\phi(m)\mid(r-s),a^r\equiv a^s\pmod m$
$\implies x^{13}\equiv x\pmod5$ as $\phi(5)=5-1=4\implies x^{13}\equiv11\pmod5\iff x\equiv1\pmod5--->(1)$
Using this, if $a$ is primitive roots of prime $p$ and $p^2,a$ will be primitive roots of $p^n$ for $n\ge1$
$2^1\equiv-1\pmod3,2^2\equiv1\implies2$ is a primitive root of $3$
Now, $2^1\equiv2\pmod9,2^2\equiv4,2^3\equiv-1\implies ord_92=6=\phi(9)\implies2$ is a primitive root of $9$
So, $2$ is a primitive root of $3^n$ for $n\ge1$
Applying Discrete Logarithm with respect to base $2$ on $x^{13}\equiv11\pmod{27},$
$13\cdot ind_2x\equiv ind_211\pmod{18}$ as $\phi(27)=3^2(3-1)=18$
Now, $2^3\equiv8\pmod{27},2^4\equiv16\equiv-11\pmod{27}$
Again as $2$ is a primitive root of $27,2^{\frac{\phi(27)}2}\equiv-1\pmod{27}$ i.e., $2^9\equiv-1$
So, $11\equiv2^4\cdot2^9\equiv2^{13}\implies ind_211=13$
So, $13\cdot ind_2x\equiv 13\pmod{18}\implies ind_2x\equiv1\pmod{18}$ as $(13,18)=1$
So, $x\equiv2^1\pmod{27}\equiv2--->(2)$
Applying well known CRT on $(1),(2)$ we get $x\equiv2b_1\frac{27\cdot5}{27}+1\cdot b_2\frac{27\cdot5}5\pmod{27\cdot5}\equiv 10b_1+27b_2\pmod{135}$
where $b_1\frac{27\cdot5}{27}\equiv1\pmod {27}\iff 5b_1\equiv1$
and $b_2\frac{27\cdot5}5\equiv1\pmod{27}\iff 27b_2\equiv1$
Expressing $\frac{27}5$ as continued fraction, $\frac{27}5=5+\frac25=5+\frac1{\frac52}=5+\frac1{2+\frac12}$
So, the last convergent of $\frac{27}5$ is $5+\frac12=\frac{11}2$
Using convergent property of continued fraction, $27\cdot2-5\cdot11=-1$
$\implies 5\cdot11\equiv1\pmod{27}\implies 5^{-1}\equiv11\implies b_1\equiv11$
and $27\cdot2\equiv-1\pmod5\implies (27)^{-1}\equiv-2\equiv3\implies b_2\equiv3$
So, $x\equiv27\cdot3+10\cdot11\pmod{135}\equiv191\equiv56$
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9829698801040649, "perplexity": 476.66836540714877}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00172-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.tis-gdv.de/tis_e/verpack/verpackungshandbuch/04verpackungshandbuch_0142/ | 1.4.2 Tipping loads / risk of tipping [German version]
Horizontal acceleration may also cause packages to tip.
A package is at risk of tipping if the area of contact with the floor is small and the center of gravity is high after due consideration of the acceleration to be expected. This is the case, for instance, if the center of gravity is above half the height of the package.
To determine whether packages are at risk of tipping, the ratio between the height of the center of gravity and the distance to the tilting edge must be determined. The tilting edge is always the edge perpendicular to the direction of acceleration (see Figure 2).
FG = weight force FV = acceleration force hs = tipping moment arm bs = resting moment arm
Figure 2: Checking the risk of tipping
Whenever the ratio of the height of the center of gravity over the base to the lateral distance between the tilting edge and the center of gravity is larger than the expected acceleration forces, the box is at risk of tipping and must be secured accordingly. Expressed in terms of ratios, this means:
to the front if bs / hs < 1.0* to the side if bs / hs < 0.5* to the rear if bs / hs < 0.5*
* Example acceleration values during transportation by truck
(CTU Packing Guidelines, see Table 1 in section 1.3.1). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9523763060569763, "perplexity": 795.3050701061863}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103617931.31/warc/CC-MAIN-20220628203615-20220628233615-00717.warc.gz"} |
https://brilliant.org/discussions/thread/opinionsresponsesreport/ | # Opinions/Responses/Report
Dear Brilliantians ,
I am Nihar Mahajan , a schooler studying in $10^{th}$. Most of you have observed that I have created a set named Jee Novices and till date , I have posted $\color{#D61F06}{32}$ problems gathered from various sources. The intention of creating this set was to help the students who have just started their preparation for the $IIT-JEE$ exam , thats why I have named the the set as 'Novices' .The questions are intended to be Level $2-3$ so that even the students who are a bit weak in mathematics will able to give their best and solve those questions which will eventually increase their confidence level.This will give them the potential to fight against the tough Level $4-5$ problems to their best. Of course , I am not stopping to add the problems and will continue to add more.
So , This note is created to discuss your opinions , experience and what do you feel about the formation of this set.
Nihar
Note by Nihar Mahajan
4 years, 2 months ago
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It's a great step Bro.It would help many of us.👏👏👏👏👏👏
- 4 years, 2 months ago
Thanks a lot bro!!! I am happy to help others. :)
- 4 years, 2 months ago
According to me this set is full of multilevel problems . ☺nice job.
- 4 years, 2 months ago
Nihar,your step is significant.As you have mentioned that it is for IIT-JEE Novices,I am also included in this Novices list so thanks a lot.APPRECIATION AND CHEERS!
- 4 years, 2 months ago
Welcome. :)
- 4 years, 2 months ago
I appreciate your efforts a lot.Thanks bro for posting it.
- 4 years, 2 months ago
Thanks a lot!!!!
- 4 years, 2 months ago
Nice job Nihar. It would help a lot of us for the IIT JEE
- 4 years, 2 months ago
Thanks a lot!!!
- 4 years, 2 months ago
Good exercises.
- 4 years, 2 months ago
Thanks!!!!! :)
- 4 years, 2 months ago
Good exercises! @Nihar Mahajan
- 4 years, 2 months ago
Thanks!!!
- 4 years, 2 months ago
Nice work bro .... a helping hand for jee aspirants..........
- 4 years, 2 months ago
Thanks!!!
- 4 years, 2 months ago
Are there any chemistry problems that you can add to the set ? Thank you,
- 3 years, 11 months ago
A good collection of number theory problems for jee aspirants.
- 3 years, 5 months ago
Nice job.........liked the questions
- 4 years, 2 months ago
Thanks!!!
- 4 years, 2 months ago
@Nihar Mahajan Every step taken has its significance and should be appreciated..!! Keep up the good work ppl will come and join you..!!!
- 4 years, 2 months ago
Thanks a lot!!!
- 4 years, 2 months ago
Yo bro, nihar i am also of your same class ( class 10) and what your have posted is really appreciable man. Keep on doing like so . By the way are you preparing for RMO. I am doing so, so that why i was thinking to create some note and probs and share our opinions ...................................................................
- 4 years, 2 months ago
Thanks!!! Go on!
- 4 years, 2 months ago
Hey Nihar your probs are AWESOME!!!
BTW plz add some more physics prob in the set.Thnx!😅😅
- 4 years, 2 months ago
Ok , I will add Physics probs too , thanks!!
- 4 years, 2 months ago
Interesting problems,Please keep posting like this.Waiting for more good sets by you.
- 3 years, 12 months ago
Thanks! And I will post many problems. Stay tuned. :)
- 3 years, 12 months ago
Nihar I've seen some of your Geometry problems and considering you're a 13 year old, I've always been in awe about the level of imagination you have. Plus you have your own Geometry theorem. I know I'm not really qualified to appreciate the beauty of your problems because Geometry has never really been my cup of tea( I solved one today after ages that too trigonometry) but all I can say is keep up the good work. c: Btw talking about Algebra, Bashing Telescope was BEAST. I'll admit I cheated my way through it by programming it out, just for the 400 points. :3
- 3 years, 8 months ago
Lol dude , I am 15 years old. Don't worry , geometry is nothing but a game of detectives. You just have to seek various hints in the figures and thereby progress. Even I was once a beginner in geometry , but as I moved ahead , it felt like I was in heaven while learning that subject!
- 3 years, 8 months ago
I was okay with Geometry till the end of 10th grade. But then, one of my math teachers who was really bad at Geometry really undermined my interest in the subject. So by the end of 10th grade, I have a great "Geometrophobia".
But on the positive side, he developed my interest in Algebra and Number Theory ( though my area is Combinatorics, the subject of the useless) ^.^
- 3 years, 8 months ago | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9741208553314209, "perplexity": 3802.6751119168184}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027330800.17/warc/CC-MAIN-20190825194252-20190825220252-00185.warc.gz"} |
http://rspa.royalsocietypublishing.org/content/455/1984/1235 | On the spectrum of second-order differential operators with complex coefficients
B. M. Brown, D. K. R. McCormack, W. D. Evans, M. Plum
Abstract
The main objective of this paper is to extend the pioneering work of Sims on second-order linear differential equations with a complex coefficient, in which he obtains an analogue of the Titchmarsh–Weyl theory and classification. The generalization considered exposes interesting features not visible in the special case in Sims paper from 1957. An m-function is constructed (which is either unique or a point on a ‘limit-circle’), and the relationship between its properties and the spectrum of underlying m-accretive differential operators analysed. The paper is a contribution to the study of non–self–adjoint operators; in general, the spectral theory of such operators is rather fragmentary, and further study is being driven by important physical applications, to hydrodynamics, electro–magnetic theory and nuclear physics, for instance. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9677591323852539, "perplexity": 510.7979660354554}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207924799.9/warc/CC-MAIN-20150521113204-00248-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://enumeration.askdefine.com/ | # Dictionary Definition
enumeration
### Noun
1 a numbered list [syn: numbering]
2 the act of counting; "the counting continued for several hours" [syn: count, counting, numeration, reckoning, tally]
# User Contributed Dictionary
## English
### Noun
1. The act of enumerating, making separate mention, or recounting.
2. A detailed account, in which each thing is specially noticed.
3. A recapitulation, in the peroration, of the heads of an argument.
4. For example - Monday, Tuesday, Wednesday is an enumeration of days.
# Extensive Definition
In mathematics and theoretical computer science, the broadest and most abstract definition of an enumeration of a set is an exact listing of all of its elements (perhaps with repetition). The restrictions imposed on the type of list used depend on the branch of mathematics and the context in which one is working. In more specific settings, this notion of enumeration encompasses the two different types of listing: one where there is a natural ordering and one where the ordering is more nebulous. These two different kinds of enumerations correspond to a procedure for listing all members of the set in some definite sequence, or a count of objects of a specified kind, respectively. While the two kinds of enumeration often overlap in most natural situations, they can assume very different meanings in certain contexts.
## Enumeration as counting
Formally, the most inclusive definition of an enumeration of a set S is any surjection from an arbitrary index set I onto S. In this broad context, every set S can be trivially enumerated by the identity function from S onto itself. If one does not assume the Axiom of Choice or one of its variants, S need not have any well-ordering. Even if one does assume Choice, S need not have any natural well-ordering.
This general definition therefore lends itself to a counting notion where we are interested in "how many" rather than "in what order." In practice, this broad meaning of enumerable is often used to compare the relative sizes or cardinalities of different sets. If one works in ZF (Zermelo-Fraenkel set theory without choice), one may want to impose the additional restriction that an enumeration must also be injective (without repetition) since in this theory, the existence of a surjection from I onto S need not imply the existence of an injection from S into I.
## Enumeration as listing
When an enumeration is used in an ordered list context, we impose some sort of ordering structure requirement on the index set. While we can make the requirements on the ordering quite lax in order to allow for great generality, the most natural and common prerequisite is that the index set be well-ordered. According to this characterization, an ordered enumeration is defined to be a surjection with a well-ordered domain. This definition is natural in the sense that a given well-ordering on the index set provides a unique way to list the next element given a partial enumeration.
## Enumeration in countable vs. uncountable context
The most common use of enumeration occurs in the context where infinite sets are separated into those that are countable and those that are not. In this case, an enumeration is merely an enumeration with domain ω. This definition can also be stated as follows:
We may also define it differently when working with finite sets. In this case an enumeration may be defined as follows:
• As a bijective mapping from S to an initial segment of the natural numbers. This definition is especially suitable to combinatorial questions and finite sets; then the initial segment is for some n which is the cardinality of S.
In the first definition it varies whether the mapping is also required to be injective (i.e., every element of S is the image of exactly one natural number), and/or allowed to be partial (i.e., the mapping is defined only for some natural numbers). In some applications (especially those concerned with computability of the set S), these differences are of little importance, because one is concerned only with the mere existence of some enumeration, and an enumeration according to a liberal definition will generally imply that enumerations satisfying stricter requirements also exist.
Enumeration of finite sets obviously requires that either non-injectivity or partiality is accepted, and in contexts where finite sets may appear one or both of these are inevitably present.
### Examples
• The natural numbers are enumerable by the function f(x) = x. In this case f: \mathbb \to \mathbb is simply the identity function.
• \mathbb, the set of integers is enumerable by
f(x):= \begin -(x+1)/2, & \mbox x \mbox \\ x/2, & \mbox x \mbox. \end
f: \mathbb \to \mathbb is a bijection since every natural number corresponds to exactly one integer. The following table gives the first few values of this enumeration: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9259456396102905, "perplexity": 520.8316031984513}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934809229.69/warc/CC-MAIN-20171125013040-20171125033040-00639.warc.gz"} |
http://mathoverflow.net/questions/2991/over-which-schemes-can-there-exist-non-trivial-g-a-bundles?sort=oldest | Over which schemes can there exist non-trivial G_a bundles?
The group scheme G_a here is the one-dimensional additive group.
-
You can escape the underscores with a backslash so that they don't cause a problem (i.e. type \\_ to get _). Better yet, instead of typing H^1(X,O\\_X), you can type H<sup>1</sup>(X,O<sub>X</sub>) to get really pretty-looking output. – Anton Geraschenko Oct 28 '09 at 5:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.928361177444458, "perplexity": 2888.2519538287283}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936463093.76/warc/CC-MAIN-20150226074103-00153-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/751539/proof-that-a-product-of-two-quasi-compact-spaces-is-quasi-compact-without-axiom | Proof that a product of two quasi-compact spaces is quasi-compact without Axiom of Choice
A topological space is called quasi-compact if every open cover of it has a finite subcover. Let $X, Y$ be quasi-compact spaces, $Z = X\times Y$. The usual proof that $Z$ is quasi-compact uses a maximal filter, hence Axiom of Choice. Can we prove it without using Axiom of Choice?
Edit(Apr. 14, 2014) If I am not mistaken, I have come up with a proof without using Axiom of Choice. I would like to post it as an answer.
Edit(Apr. 15, 2014) I was mistaken. As Andres Caicedo pointed out, I used AC without noticing it in my "proof".
• @Arthur: Because it will incite another long circle of "why was this question closed", "why was this meta thread closed", "is it wrong to ask questions?", "why is there a group of users that votes against everything I do?" and so on and so forth. – Asaf Karagila Apr 13 '14 at 8:43
• @AsafKaragila Stop it. You are abusing the thread by an off topic matter. – Makoto Kato Apr 13 '14 at 8:48
• @Makoto: Stop it. You are abusing the thread by an off topic matter. I see that I have to start quoting your comments too, in case that you delete them. For example, now it seems that my last comment is "out of the blue" and not at all in reply of you being all naive and innocent about your behavior. – Asaf Karagila Apr 13 '14 at 8:52
• Please improve the question - as it stands, it shows no effort. For example, you could examine the usual proof that the product of two quasicompact spaces is compact, and point out where the axiom of choice is actually used. @OP – Carl Mummert Apr 13 '14 at 12:23
• @Makoto Kato: [[the issue is that you didn't do enough before asking the question.] How do you know that?] You asked already - see math.stackexchange.com/questions/753072/… and math.stackexchange.com/questions/753072/… . You appear to ignore responses such as that, while reading others in a way that makes them more favorable to you than they actually are (e.g. the fact that it's sometimes acceptable to answer one's own question). – Carl Mummert Apr 14 '14 at 15:11
Here is a proof which is choice free. I should also mention that the proof using ultrafilters appears in Herrlich's The Axiom of Choice, where later he says that it can be modified to work without the axiom of choice using some ideas from other proofs. I did not check that claim in details, though.
First, let us prove the following lemma:
Lemma. Let $$X$$ be a topological space and $$\cal B$$ a basis for the topology. $$X$$ is quasi-compact if and only if every cover with elements of $$\cal B$$ has a finite subcover.
Proof. One direction is trivial, if $$X$$ is quasi-compact, certainly every cover with elements of $$\cal B$$ has a finite subcover. In the other direction, suppose that $$\mathcal U=\{U_i\mid i\in I\}$$ is an open cover, consider $$\cal U'$$ to be the refined cover, $$\{V\in\mathcal B\mid\exists i\in I: V\subseteq U_i\}$$. Then $$\cal U'$$ is an open cover as well, since every point in $$U_i$$ lies within some element of $$\cal B$$. Let $$V_1,\ldots,V_n\in\cal U'$$ be a finite subcover, then we can choose $$U_1,\ldots,U_n\in\cal U$$ such that $$V_i\subseteq U_i$$ for all $$i\leq n$$, and this is a finite subcover as wanted. $$\ \square$$
Now we can prove our theorem.
Let $$X,Y$$ be two quasi-compact spaces, and let $$\cal U$$ be an open covering of $$X\times Y$$ using rectangles (that is, sets of the form $$U\times V$$ where $$U$$ is open in $$X$$ and $$V$$ open in $$Y$$). We say that $$A\subseteq X$$ is adequate [for $$\cal U$$] if $$A\times Y$$ has a finite subcover in $$\cal U$$. Our goal is to show that $$X$$ is adequate, then $$X\times Y$$ can be covered by a finite subcover of $$\cal U$$.
First we show that if $$x\in X$$, then there is some $$U\subseteq X$$ which is open and $$x\in U$$ such that $$U$$ is adequate. Let $$U_1\times V_1,\ldots,U_n\times V_n$$ be a finite subcover such that $$\{x\}\times Y\subseteq U_1\times V_1\cup\ldots\cup U_n\times V_n$$ (such finite cover exists since $$\{x\}\times Y$$ is quasi-compact). Now consider $$U=\bigcap_{i=1}^n U_i$$, then $$U$$ is open as a finite intersection of open sets, and non-empty since $$\{x\}\in U$$, as wanted. $$U$$ is adequate since given any $$(u,y)\in U\times Y$$ we have that for some $$i\leq n$$ it is true that $$(x,y)\in U_i\times V_i$$, so $$(u,y)\in U_i\times V_i$$ as well (since $$u\in U_i$$). Therefore $$U_1\times V_1,\ldots,U_n\times V_n$$ is a cover of $$U\times Y$$.
Next we note that the finite union of adequate sets is adequate (as it is covered by the [finite] union of the [finite] subcovers of the adequate sets).
Now $$\{U\subseteq X\mid U\text{ is open and adequate}\}$$ is an open cover of $$X$$, by the above fact that every $$x\in X$$ has an adequate neighborhood, and by quasi-compactness of $$X$$ it has a finite subcover. And therefore $$X$$ is the finite union of adequate sets and it is adequate as well.
Finally, since rectangles form a basis for the product topology, from the lemma above we have that $$X\times Y$$ is indeed quasi-compact, as wanted. $$\quad\square$$
(I found the proof on ProofWiki sometime in the past.)
• It looks like you are claiming that the $U$ you construct in the second paragraph is adequate, but I don't see it very clearly. In particular, there seems to be an implicit claim that $U \times X \subset W_1 \cup \cdots \cup W_n$. – Scott Carnahan Apr 23 '14 at 14:12
• @Scott: Yes, I think you're right. What I should have here is that without loss of generality $W_i=U_i\times V_i$ (and we can make that assumption, even without choice). Then $U=\bigcap U_i$, and given some $(u,y)\in U\times Y$ there is some $V_i$ such that $y\in V_i$ and therefore $(u,y)\in U_i\times V_i$, so $U\times Y$ is adequate. – Asaf Karagila Apr 23 '14 at 14:24
• I'll think about it for a few minutes to see if I can dispense of that assumption of rectangles, and I'll edit. (I also caught a use of "compact" instead of "quasi-compact" which I should correct while I'm at it.) – Asaf Karagila Apr 23 '14 at 14:24
• @Scott: Yes, the use of rectangles is crucial here. Thanks, I'll edit it in. – Asaf Karagila Apr 23 '14 at 14:32
Let $X, Y$ be quasi-compact spaces, $Z = X\times Y$. We will prove that $Z$ is quasi-compact without using Axiom of Choice. Suppose $(W_\alpha)_{\alpha\in A}$ is an open cover of $Z$. Then $W_\alpha = \bigcup_{\beta\in B_\alpha} U_{\alpha\beta} \times V_{\alpha\beta}$, where $U_{\alpha\beta}$ is an open subset of $X$ and $V_{\alpha\beta}$ is an open subset of $Y$. Let $x \in X$. Then $\{x\}\times Y$ is quasi-compact. Since $(U_{\alpha\beta} \times V_{\alpha\beta})_{(\alpha,\beta)\in A\times B_\alpha}$ is an open cover of $\{x\}\times Y$, there exists a finite subcover $U_i(x) \times V_i(x), i = 1, \cdots , n_x$ of $\{x\}\times Y$, where each $U_i(x) \times V_i(x)$ is of the form $U_{\alpha\beta} \times V_{\alpha\beta}$ for some $(\alpha, \beta) \in A\times B_\alpha$. Let $U(x) = U_1(x) \cap \cdots \cap U_{n_x}$. Since $(U(x))_{x\in X}$ is an open cover of $X$, there exists a finite subcover $U(x_1),\cdots, U(x_m)$ of $X$. Then $U(x_i) \times V_k(x_i)$, $i = 1, \cdots, m, k = 1,\cdots, n_{x_i}$ is a finite cover of $X\times Y$. Then for every pair $(i, k)$, there exist $\alpha \in A$ and $\beta \in B_\alpha$ such that $U(x_i) \times V_k(x_i) \subset U_{\alpha\beta} \times V_{\alpha\beta} \subset W_\alpha$. Hence there exists a finite subcover of $(W_\alpha)_{\alpha\in A}$. QED
• A problem with your write-up: For each $x$ there is a finite subcover of $\{x\}\times Y$, but there may be more than one, so you need to specify (if at all possible) how to pick the $U_i(x)\times V_i(x)$, $1\le i\le n_x$. – Andrés E. Caicedo Apr 14 '14 at 16:32
• @AndresCaicedo A good point. By the way, why did you vote to close this question? – Makoto Kato Apr 14 '14 at 16:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 57, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9678136110305786, "perplexity": 137.07646327582452}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703561996.72/warc/CC-MAIN-20210124235054-20210125025054-00530.warc.gz"} |
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