Moving charges in a moving frame of reference
Thread starter olgerm
elecrtomagnetism
olgerm
If 2 bodies with charge q are in rest then both have electric force ##F_1=\frac{q*q*k_q}{|\vec{r}|^2}##.
But in another frame of reference, that is moving with velocity v relative to first frame of reference, they feel both magnetic and electric force ##F_2=|\vec{F_{electric}}+\vec{F_{magnetic}}|=|\vec{F_{electric}}|+|\vec{F_{magnetic}}|=|q*(\vec{E}+\vec{v}\times \vec{B})|=|q*(\frac{q*k_q}{r^2}+\vec{v}\times \frac{q*\mu_0*\vec{v}}{4*\pi*|\vec{v}|^2})|=q*(\frac{q*k_q}{|\vec{r}|^2}+\frac{q*\mu_0*|\vec{v}|^2}{4*\pi*|\vec{v}|^2})=\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2}##
So forces are not same in both frames of reference. Is it correct?
##\vec{v}## is crosswise to ##\vec{r}##.
Some source said that I should use relation
##E_2=E_1+v \times B_1##
##B_2=B_1-v/c^2\times E_1##
but that would mean that ##E_2\not=\frac{q*k_q}{r^2}## Is it correct? Does that mean that there is free electromagnetic field(electromagnetic wave(s)) in second frame of reference?
Likes Parth
olgerm said:
Yes, that is correct.
However, there is a relativistic generalization of force called the four-force which is the same in all frames of reference.
Likes olgerm
Dale said:
Thanks, where should I use the formulas
Are these correct formulas?
I don't know of the top of my head. I would have to look them up. I am assuming you can do that as well as I can
For what it's worth, I almost never bother transforming the fields like that. If I am using multiple frames then I will use a covariant formulation of Maxwell's equations
If there were dynamometer between the charges: would it show reading ##F_1## or ##F_2##?
jartsa
##E_2## can be calculated from just ##E_1## and v. The closer the v is to c, the more the electric field is distorted.
Here is a formula:
https://en.wikipedia.org/wiki/Biot–Savart_law#Point_charge_at_constant_velocity
By the way, isn't ##B_1## zero?
If it's at rest in frame 1 then it will measure ##F_1##. If it is at rest in frame 2 then it will measure ##F_2##.
Luxucs
One quick question I would appreciate clarification on about this - does this include non-inertial frames as well, or is the magnitude of the four-force only invariant in inertial frames?
Nugatory
Luxucs said:
The four-force is frame-independent so is the same in all frames whether inertial or not.
You'll sometimes see threads here in which someone states some variant of "you need general relativity to work with non-inertial frames" and being correctly told that that is not true, that SR works just fine with non-inertial frames, you just need more mathematical tools than for inertial frames. The four-vector formalism (including four-force) is one of those tools.
Likes Dale
Nugatory said:
Excellent, thank you!
I do not understand that. if dynamometer is tied between 2 moving bodies then it shows one result, but if it is in rest(for example 2 rods towards which the bodies are sliding) it gives another result. It does not seem right.
It does not seem right.
I can't help you there.
Likes davenn
You have created a surprisingly tricky problem by adding a force-measuring instrument. Before we consider that one, let's try answering the question I think you were trying to ask when you asked about connecting a "dynamometer" between the charges: What force does the particle 'feel'? To answer this, we use the four-vector formalism: we calculate the frame-independent four-force from the Faraday tensor (which combines the E and B three-vectors into a single frame-independent object that describes the electromagnetic field as a single entity). The four-force gives us the frame-independent proper acceleration through the four-vector equivalent of ##\vec{F}=m\vec{a}## and that proper acceleration is what the particle 'feels'; it's a measure of how much the worldline of the particle deviates from the free-fall inertial worldline it would follow if it weren't being pushed around by electromagnetic forces.
But if we attach a dynamometer to the particles to see what it reads? One way or another, all force-measuring devices are equivalent to attaching a spring to the object and then measuring how much the spring deflects when it is applying a force that exactly cancels the force we're looking at. That means that it changes the trajectory and speed of the object in question - so it can't tell us about the velocity-dependent Lorentz force. In effect, we now have a completely different problem: the motion of two charged particles connected by an ideal spring.
Likes PeroK
I guess those two rods must be the dynamometer. So it's actually two parallel rods that are connected by many springs.
So this dynamometer (or an observer strapped to the dynamometer) should say that the two bodies:
1: exert a small force on each other (because of their motion)
2: exert a small force on the rails (because of 1)
3: every spring experiences a large force deformation at some moment of time
So this is just another "relativity paradox" which is solved by the relativity of simultaneity.
(If there was just one spring connecting the rods, then the the dynamometer would not function as intended, because of various wave-phenomenons - except if we assumed perfectly rigid rods, in which case there would be a real paradox)
Which part of dynamometer must be moving fordynamomenter to show ##F_2##?
Lets say we have a string that breaks if ##F=\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2}## but does not break if ##F=\frac{q^2*k_q}{|\vec{r}|^2}##for dynamometer.
this seems to lead to paradox.
As i understoof if string is tied between the charged bodies it would not break.
if the bodies would move WITH pipes and string where attached between pipes it would not break.
if the bodies would move IN pipes and string where attached between pipes it would break.
if the bodies would move IN pipes and string where attached between pipes but would move as fast as the charged bodies with sines that connect it to the pipes with it would break(?).
maybe it about in which frame of reference we are not about in which frame of reference the dynamometer is in rest.
jartsa said:
it is. It is in formula, because formula is generally for EM-field not for only this setup.
Which part of dynamometer must be moving fordynamomenter to show F2
The part which the dynamometer's manufacturer indicates should be fixed and immobile. Often a base plate or a housing or frame.
Lets say we have a string that breaks if F=q2∗(kq+μ0∗|→v|2)|→r|2F=q2∗(kq+μ0∗|v→|2)|r→|2F=\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2} but does not break if F=q2∗kq|→r|2F=q2∗kq|r→|2F=\frac{q^2*k_q}{|\vec{r}|^2}for dynamometer.
There is no paradox here; this is just an incomplete specification of the breaking condition. Forces are frame variant so you have to specify which frame is the breaking condition defined in. This is similar to indicating a distance or a time without identifying the frame.
Okay so the equation simplifies to ##E_2=E_1##, and that is not the truth. Electric field is not the same in all frames. It changes depending on how close to c the speed of the electric field is, as I said.
First, second and fourth are the same. String moves with bodies.
The third one is the interesting one. The moving bodies can be thought to be a moving dynamometer that is measuring how much force the string is generating. Or the string can be thought to be a dynamometer that is measuring how much force the bodies are genarating.
There is a "paradox": How can both dynamometers consider the other dynamometer to be the dynamometer that is moving?
There is a similar problem with devices that measure time. I mean in special relativity.
(If two clocks move relative to each other, how can both clocks consider the other clock to be time dilated?)
But the rule is that: The faster a pair of repelling charges move, the larger their magnetic attraction is.
The faster a pair of repelling charges move, the larger their magnetic attraction is.
They would move with same speed in in the setup with pipes as in setup without pipes. Pipes are just for transferring radial force from moving bodies to the string.
You said the string breaks in the third scenario. What is the reasoning behind that? J.C.Maxwell would say that the net force between the bodies approaches zero as the speed of the bodies approaches c.
PeroK
What if you replace the dynamometer with a clock? The clock might ultimately keep time using EM forces. The moving clock must feel exactly the same forces as the stationary clock and must tick at the same rate. Perhaps the idea of time dilation does not seem right either?
There is no paradox here; this is just an incomplete specification of the breaking condition. Forces are frame variant so you have to specify which frame is the breaking condition defined in.
I think strings and other bodies break under some ultimate mechanical pullingforce irrespective of whether they are moving or not. The force needed to break can be measured measured anywhere.
Can this question be answered with non-relativistic physics?
I thought that because string is moving in frame of chraged bodies, then in feels force ##F_2##, that is sufficient to break it.
This is incorrect. Force is frame variant. That is simply a fact of physics. The same force measured in different frames gives different results. You cannot simply wish that fact away.
Yes that is quite reasonable. As the speed of the string approaches c the forces between charges in the string approach zero.
Well, I guess that when these two things that wrestle with each other have a zero relative speed, then they exert equal forces on each other, and when their relative speed increases, then they of course feel their own strength to remain the same, while the force from the other thing is felt to decrease or increases or stay constant. But because of the symmetry of this scenario the change must be the same for both things, I mean the transformation of the force from the other thing must be the same for both things.
So if the string does break, then the two bodies are pulled together by the string.
If that thing above is hard to follow, I mean that if the string says "I won this wrestling match thanks to the weakening of the opponent caused by motion", then the two bodies will say "I won this wrestling match thanks to the weakening of the opponent caused by motion". Because of symmetry.
If that sounds impossible, then maybe it never happens. So in that case it must be so that the thing made of two charged bodies will say: "the string is getting weaker as it accelerates, but for some reason I can not cause the string to break by pushing on these rods."
So with typical ultimate stress test we can calculate the stress in needed to break body if it is in rest(not moving)?
How to calculate how strong foce is needed to break the same body if it is moving with speed v?
When F is perpendicular to v:
##F' = \frac { F } { \gamma } ##
## \gamma = \frac {1} { \sqrt {1-v^2/c^2} } ##
When F and v are aligned:
##F' = F ##
http://www.sciencebits.com/Transformation-Forces-Relativity
Thanks, but should it not be ##F´=F*\gamma##, because in frame where the object moves the force is greater (at least in the setup I described in original post).
Thanks, but should it not be ##F´=F*\gamma##, because in frame where the object moves the force is greater
Well to me it seems that in frame where the object moves the force is smaller.
What force exactly is larger in the frame where the object moves?
##\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2}>\frac{q*q*k_q}{|\vec{r}|^2}##
The calculation is wrong then, because it's a wrong result that the force between the bodies increases as speed increases.
Let's assume those charge bodies move side by side, that simplifies calculations.
The electric field strength increases by gamma, so electric force increases by gamma. So
##F_{electric} = \gamma * \frac {k* q_1*q_2}{r^2}##
And now the magnetic field. ##B=\frac {1}{v^2}v \times E ##
That E there is the increased field: ##\gamma * E_{rest}##
So the magnetic force is ##q*v \times B##
Total force = ## |F_{electric}| - |F_{magnetic}| ##
Now I can see that this calculation will result in decrease of force and I think that this the right way to calculate.
I got my formulas from here:
That complicated equation for E simplifies to multiplying Erest by gamma in this case.
OK, let's work this through explicitly with covariant notation using the (+---) sign convention and units where c=1.
The electromagnetic field tensor is: $$
F_{\mu\nu}=\left(
\begin{array}{cccc}
0 & -E_x & -E_y & -E_z \\
E_x & 0 & -B_z & B_y \\
E_y & B_z & 0 & -B_x \\
E_z & -B_y & B_x & 0 \\
\right)$$ and the Lorentz four-force is $$f_{\mu}=q F_{\mu\nu} u^{\nu}$$For a charge, q, at rest in a pure E field we have$$f_{\mu}=q F_{\mu\nu} u^{\nu} = q \left(
E_x & 0 & 0 & 0 \\
E_y & 0 & 0 & 0 \\
E_z & 0 & 0 & 0 \\
\right) (1,0,0,0) = (0, q E_x, q E_y, q E_z)$$
If we boost to a frame where the charge is moving with velocity v in the x direction then we can write the Lorentz transform matrix as $$\Lambda = \left(
\gamma & v \gamma & 0 & 0 \\
v \gamma & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
\right)$$ so we can immediately write that the four-force in the other frame is $$f'_{\mu}=\Lambda f_{\mu} = (\gamma v q E_x, \gamma q E_x, q E_y, q E_z)$$
If we are masochists and don't want to do things the easy way then we can boost the field tensor and the four-velocity and get $$f'_{\mu} = q F'_{\mu\nu} u'^{\nu} $$ $$= q \left(
0 & -E_x & -\gamma E_y & -\gamma E_z \\
E_x & 0 & \gamma v E_y & \gamma v E_z \\
\gamma E_y & -\gamma v E_y & 0 & 0 \\
\gamma E_z & -\gamma v E_z & 0 & 0 \\
\right) (\gamma,\gamma v,0,0) $$ $$= (\gamma v q E_x, \gamma q E_x, q E_y, q E_z)$$
Now from this tedious exercise we notice a few things. First, the field tensor which has no B field in the unprimed frame has a mixture of E and B fields in the primed frame. Second, the Lorentz force law holds in both frames with the modified field and modified velocity. Third, since for small forces the spacelike part of the four-force is ##\gamma^2## times the three-force, we have that the component of the three force in the direction of the boost is smaller by a factor of ##1/\gamma## and the transverse components are smaller by a factor of ##1/\gamma^2##. Finally, if you have a breakage force in a rest frame, then you can simply boost that four-force to a different frame to get the breakage condition in that other frame.
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