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https://www.bartleby.com/solution-answer/chapter-171-problem-19e-calculus-mindtap-course-list-8th-edition/9781285740621/solve-the-initial-value-problem-9y12y4y0y01y00/0f4faf0b-940a-11e9-8385-02ee952b546e | # Solve the initial-value problem. 9 y ″ + 12 y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0
### Calculus (MindTap Course List)
8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285740621
Chapter
Section
### Calculus (MindTap Course List)
8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285740621
Chapter 17.1, Problem 19E
Textbook Problem
1 views
## Solve the initial-value problem. 9 y ″ + 12 y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0
To determine
To solve:
The initial value problem 9y''+12y'+4y=0 , y0=1, y'0=0
Solution:
y=e-23x+23xe-23x
Explanation:
1) Concept:
i) Theorem: If y1 and y2 are linearly independent solutions of the second order Homogeneous linear differential equation ay''+by'+cy=0 on an interval, and a0, then the general solution is given by yx=c1y1x+c2y2(x) where c1 and c2 are constants.
ii) In general solution of ay''+by'+cy=0
Roots of ar2+br+c=0 General solution r1, r2 are real and distinct y=c1er1x+c2er2x r1=r2=r y=c1erx+c2xerx r1 , r2 complex :α±iβ y=eαx(c1cosβx+c2sinβx)
2) Given:
The initial value problem 9y''+12y'+4y=0 , y0=1, y'0=0
3) Calculations:
The auxiliary equation of the differential equation 9y''+12y'+4y=0 is 9r2+12r+4=0
Solving the auxiliary equation 9r2+12r+4=0
9r2+12r+4=0
9r2+6r+6r+4=0
3r(3r+2)+2(3r+2)=0
(3r+2)(3r+2)=0
r=-23,-23
By given concept general solution of differential equation 9y''+12y'+4y=0 is
y=c1e-23x+c2xe-23x
Differentiate this equation with respect to x
y=-23c1e-23x+c2e-23x-23c2xe-23x
Applying initial conditions y0=1, y'0=0
y0=c1 Therefore , c1=1
y'0=-23c1+c2 Therefore, -23c1+c2=0 this imply c2=23
Thus, the solution of the initial value problem y''-2y'-3y=0 , y0=1, y'0=0 is y=e-23x+23xe-23x
Conclusion:
The solution of the initial value problem y''-2y'-3y=0 , y0=2, y'0=2 is y=e-23x+23xe-23x
...
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Study Guide for Stewart's Multivariable Calculus, 8th | {"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": 0.828567385673523, "perplexity": 4129.01182316555}, "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-2020-05/segments/1579250628549.43/warc/CC-MAIN-20200125011232-20200125040232-00489.warc.gz"} |
https://scoop.eduncle.com/iot-w-be-the-three-dimensional-region-under-the-graph-of-f-x-y-exp-x-y-and-over-the-region-7-in-the-plane | IIT JAM Follow
April 12, 2021 11:52 am 30 pts
Iot W be the three dimensional region under the graph of f(x,y) = exp(x + y) and over the region 7. in the plane defined by 1s x+ y2 < 2. (A) The volume of W is te(e - 1). (B) The volume of W is r(et - e). (C) the flux of the vector field F = (2x - xy) i - yj + yzk out of the region W is z(e' -e). the flux of the vector field F = (2x - xy) i - yj + yzk out of the region W is nele-1). (D)
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I didn't get this can you??
Alka gupta
in last step when we solve by GD theorem we get triple integration dxdydz....which is volume of W ....but they written wrong ans.... because volume is we have already found πe(e-1)
but in solution it's different | {"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": 0.898491621017456, "perplexity": 2174.2032499739976}, "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-2021-21/segments/1620243992516.56/warc/CC-MAIN-20210516075201-20210516105201-00228.warc.gz"} |
http://math.stackexchange.com/questions/235843/find-rotation-that-maps-a-point-to-its-target | # Find rotation that maps a point to its target
I have a 3D point that is rotated about the $x$-axis and after that about the $y$-axis. I know the result of this transformation. Is there an analytical way to compute the rotation angles?
$$v'=R_y(\beta)*R_x(\alpha)*v$$ Here, $v$ and $v'$ are known and I want to compute $\alpha$ and $\beta$. $R_x$ and $R_y$ are the rotation matrices about the x and y axis respectively. The overall matrix will then be:
$$R=\begin{pmatrix} \cos \beta & \sin\alpha * \sin\beta & \cos\alpha * \sin\beta \\ 0 & \cos \alpha & -\sin\alpha \\ -\sin\beta & \cos\beta * \sin\alpha & \cos\alpha * \cos\beta \end{pmatrix}$$
-
You can use law of cosines. Say $v=(x,y,z)$ and $R_x(\alpha)v=(x,y^\prime,z^\prime)$. Draw the triangle with vertices $A=(x,y,z)$, $B=(x,y^\prime,z^\prime)$, and $C=(x,0,0)$. Then $\angle C$ is $\alpha$. You can do the same process for $\beta$. – icurays1 Nov 12 '12 at 18:27
Correct me if I am wrong, but this cannot work because the second rotation (about the y axis) will change the z-coordinate. Therefore, the angle you described is not the correct one. – Nico Schertler Nov 12 '12 at 19:36
True, but all you need to get the first angle is the final $y$ coordinate. The intermediate $z$ coordinate $z^\prime$ just helps establish $\alpha$. Your second triangle would have vertices $(x,y^\prime,z^\prime)$ (the location after the $x$ rotation), $(0,y^\prime,0)$, and $(x^\prime,y^\prime,z^{\prime\prime})$. This last coordinate is the final location of your point. – icurays1 Nov 12 '12 at 19:42
Yes, but how do I apply law of cosines, if I don't know the intermediate position? – Nico Schertler Nov 12 '12 at 19:58
You know the distance to the $x$-axis though, and the $x$ rotation doesn't change this distance. I.e. $r=\sqrt{z^2+y^2}=\sqrt{y^{\prime^2}+z^{\prime^2}}$. You can get $z^\prime$ from this, up to a sign (the sign doesn't matter - the problem doesn't have a unique solution anyway). – icurays1 Nov 12 '12 at 20:03
Rotation about the $y$ axis won't change the $y$ coordinate anymore, so the first rotation has to get that coordinate right straight away. There are in general two angles that do the job, but there might be none, namely if $\sqrt{v_y^2+v_z^2}<|v'_y|$. If there are such angles, then for each there will be a unique rotation about the $y$ axis that gets the vector to the destination $v'$, provided we had $|v|=|v'|$ to begin with, which is of course a necessary condition to succeed with any number of rotations fixing the origin. You can easily write down the required angles using inverse trigonometric functions.
I also thought about beginning with the y-coordinate. But I am somehow stuck at solving $v'_y=cos\alpha*v_y-sin\alpha*v_z$. – Nico Schertler Nov 12 '12 at 19:42
@NicoSchertler Put $v_y+v_z\mathbf i=r\exp(\beta\mathbf i)$ with $r,\beta\in\mathbf R$, then you want $v'_y=\operatorname{Re}(\exp(\alpha\mathbf i)(v_y+v_z\mathbf i))$ which gives $\frac{v'_y}r=\cos(\alpha+\beta)$, and this is solved by $\alpha=\pm\arccos(\frac{v'_y}r)-\beta$. – Marc van Leeuwen Nov 13 '12 at 9:04 | {"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": 0.8405832052230835, "perplexity": 160.17199285098098}, "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-2015-35/segments/1440645298781.72/warc/CC-MAIN-20150827031458-00283-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://paperswithcode.com/conference/ieee-transactions-on-pattern-analysis-and-9 | # Robust Point Set Registration Using Gaussian Mixture Models
Then, the problem of point set registration is reformulated as the problem of aligning two Gaussian mixtures such that a statistical discrepancy measure between the two corresponding mixtures is minimized.
127 | {"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": 0.9621191024780273, "perplexity": 1228.2234016376487}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2020-16/segments/1585371656216.67/warc/CC-MAIN-20200406164846-20200406195346-00357.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=450424 | # eigenvalues for integral operator
by margaret37
Tags: eigenvalues, integral, operator
P: 12 1. The problem statement, all variables and given/known data Find all non-zero eignvalues and eigenvectors for the following integral operator $Kx := \int^{\ell}_0 (t-s)x(s) ds$ in $C[0,\ell]$ 2. Relevant equations $\lambda x= Kx$ 3. The attempt at a solution $\int^{\ell}_0 (t-s)x(s) ds = \lambda * x(t)$ $t\int^{\ell}_0 x(s) ds - s\int^{\ell}_0x(s) ds = \lambda * x(t)$ Am I even going the right direction? I think I need function(s) of x(t) and scalar $\lambda$, when I am finished is that right? And should $\lambda$ be a complex (possibly real) number? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
HW Helper
PF Gold
P: 2,890
Quote by margaret37 1. The problem statement, all variables and given/known data Find all non-zero eignvalues and eigenvectors for the following integral operator $Kx := \int^{\ell}_0 (t-s)x(s) ds$ in $C[0,\ell]$ 2. Relevant equations $\lambda x= Kx$ 3. The attempt at a solution $\int^{\ell}_0 (t-s)x(s) ds = \lambda * x(t)$ $t\int^{\ell}_0 x(s) ds - s\int^{\ell}_0x(s) ds = \lambda * x(t)$ Am I even going the right direction? I think I need function(s) of x(t) and scalar $\lambda$, when I am finished is that right? And should $\lambda$ be a complex (possibly real) number?
Well, you are already on the wrong track because you cannot simply pull the "s" out of this integral when it's the variable of integration!
$$\int^{\ell}_0 sx(s) ds \neq s\int^{\ell}_0x(s) ds$$
P: 12 Oops. :( But is this the right way to do it?
HW Helper
PF Gold
P: 2,890
## eigenvalues for integral operator
Quote by margaret37 Oops. :( But is this the right way to do it?
Well, you haven't got that far yet so I can't say for sure.
But the basic idea is that $\lambda$ will be a complex number (there might be more than one that work!), and corresponding to each such $\lambda$ there will be a family of specific functions $x(t)$ (the eigenvalues) which satisfy
$$\int_0^\ell (t-s) x(s) ds = \lambda x(t)$$
P: 1,412 First of all you should correctly state the problem. It should read: $$(Kx)(t) := \int^{\ell}_0 (t-s)x(s) ds$$ Can you see the difference?
P: 12 I do see the difference. (Which is not how it is written on my assignment.) So does K operate on x? Are t and s scalars? So for a trivial example... $$\int {e^{nx}} = ne^nx$$ So is this a solution to some integral equation similar to the problem? Thank you for your answer
P: 1,412 K takes a function x. It produces a new function Kx. The value of the new function at t is calculated from the values of the old function by the process of integration. You operator could be as well written as $$(Ky)(s) := \int^{\ell}_0 (s-t)y(s) dt$$ That would be exactly the same operator. Think about it! Anyway, you want to solve the equation $$Kx=\lambda x$$ Now two functions are equal when they are equal at all points. So the above means $$(Kx)(t)=\lambda x(t)$$ for all t. So you substitute and get as much as possible from the equation $$\int^{\ell}_0 (t-s)x(s) ds =\lambda x(t)$$ for all t. Now, if $\lambda\neq 0$ can you see from this equation that x(t) is necessarily a linear function of t, that is that it must be of the form x(t)=at+b ???
P: 12 Thank you very much. I think the light MAY be starting to dawn.
P: 1,412 One more thing. You should get now something like $$t\int_0^l(as+b)ds -\int_0^l s(as+b)=\lambda (at+b)$$ This is supposed to hold for all t, so, in particular, for t=0. So you will get two equations from the one above. Two equations for three unknowns: a,b,$\lambda$. But do not worry. The eigenfunction are determined only up to a multiplicative constant.
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https://cstheory.stackexchange.com/questions/38595/fourth-moment-method-for-minimum-value | # Fourth(?) moment method for minimum value
I would like to lower bound the quantity $\Pr[X\ge t, Y\ge t]=\Pr[\min(X,Y)\ge t]$ using the small moments of $X$, $Y$ and $XY$. In particular I am interested in the case where $E[X]=E[Y]=0$, but $E[X Y]>0$.
In "The Fourth Moment Method" Berger shows that if $\frac{E[X^4]}{E[X^2]^2}\le b$, then $\Pr[X\ge \sqrt{E[X^2]}/(4\sqrt{t})]\ge1/(4^{4/3}t)$. Alternative formulations of this principle include the Paley–Zygmund inequality.
Using the fact that $\min(X,Y)=\frac12(X+Y-|X-Y|)$, we get that that $$E[\min(X,Y)^2] = E[\min(X^2,Y^2)] = \tfrac12\left(E[X^2+Y^2]-E[|X^2-Y^2|]\right),\\ E[\min(X,Y)^4] = E[\min(X^4,Y^4)] = \tfrac12\left(E[X^4+Y^4]-E[|X^4-Y^4|]\right).$$ Now using $\frac{E[X^2]^{3/2}}{E[X^4]^{1/2}} \le E[|X|] \le \sqrt{E[X^2]}$ we can get a bound: $$\frac{E[\min(X,Y)^4]}{E[\min(X,Y)^2]^2} = 2\frac{E[X^4+Y^4]-E[|X^4-Y^4|]}{(E[X^2+Y^2]-E[|X^2-Y^2|])^2} \le 2\frac{E[X^4+Y^4]-\sqrt{E[(X^4-Y^4)^2]}}{\left(E[X^2+Y^2]-\frac{E[(X^2-Y^2)^2]^{3/2}}{E[(X^2-Y^2)^4]^{1/2}}\right)^2}.$$
Besides being ugly, this has ended up using the eighth moment of $X$ and $Y$, rather than just the fourth. Is this necessary? Or is there a nicer way taking better advantage of the $\min$ function?
Update: We may wonder what the ideal result would be. If we let $\hat{X}=[X,Y]^T$, $\Sigma=\text{Cov}(\hat X)$ and $Z=\hat X ^T\Sigma^{-1}\hat X$, then Markov's (or the multivariate Chebyshev) inequality tells us $\Pr[Z \ge \epsilon] \le E[Z]/\epsilon=2/\epsilon$. If we assume $\text{Var}[X]=\text{Var}[Y]=1$ then $\min(X,Y)\ge t$ implies $Z\ge t^T\Sigma^{-1}t=\frac{2t^2}{1+\text{Cov}(X,Y)}$; and so $\Pr[X,Y\ge t]\le\frac{1+\text{Cov}(X,Y)}{t^2}$.
With Paley–Zygmund we then get $\Pr[Z\ge 2t] \ge 4(1-t)^2/E[Z^2]$, which is a fourth moment bound. Of course this is a lower bound on the entire space outside the ellipse, however we might(?) hope that it is also correct up to a constant factor for any convex region at distance t, if $X$ and $Y$ are nice enough?
• I would suggest asking this on math.stackexchange.com or mathoverflow.net. – Ryan O'Donnell Jul 27 '17 at 1:13 | {"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": 0.9950101375579834, "perplexity": 244.4668886747767}, "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-2020-45/segments/1603107880038.27/warc/CC-MAIN-20201022195658-20201022225658-00651.warc.gz"} |
http://chemistryadda.blogspot.com/2011/12/class-ix-chemistry-notes-atomic.html | Subscribe:
## Wednesday, December 28, 2011
### Class IX Chemistry E-Notes Chapter Atomic Structure
Dalton’s Atomic Theory
The important postulates of Dalton’s atomic theory are:
1. All elements are composed of atoms. Atom is too small so that it could not be divided into further simpler components.
2. Atom cannot be destroyed or produced.
3. Atoms of an element are similar in all respects. They have same mass and properties.
4. Atoms of different elements combine in a definite simple ratio to produce compounds.
Mass Number
The total number of the protons and neutrons present in the nucleus of an atom is called mass number. The protons and neutrons together are called nucleon. Hence it is also known as nucleon number. It is denoted by A. the number of neutrons present in the nucleus of an atom is rperesented by N.
Discovery of Electron
A discharge tube is a glass tube. It has two electrode, a source of electric current and a vacuum pump.
(Diagram)
Sir William Crooks (1895 performed experiments by passing electric current through gas in the discharge tube at very low pressure. He observed that at 10-4 (-4 is power to 10) atmosphere pressure, shining rays are emitted from cathode. These rays were named cathode rays. Cathode rays are material particles as they have mass and momentum.
Properties of Cathode Rays
The properties of these particles are given below:
1. These particles are emitted from cathode surface and move in straight line.
2. The temperature of the object rises on which they fall.
3. They produce shadow of opaque object placed in their path.
4. These particles are deflected in electric and magnetic fields.
5. These particles are deflected towards positive plate of electric field.
Discovery of Proton
Gold Stein (1886) observed that in addition to the cathode rays, another type of rays were present in the discharge tube. These rays travel in a direction opposite to cathode rays. These rays were named positive rays. By using perforated cathode in the discharge tube the properties of these rays can be studied. Positive rays are also composed of metered particles. The positive rays are not emitted from anode. They are produced by the ionization of residual gas molecules in the discharge tube. When cathode rays strike with gas molecule, electrons are removed and positive particles are produced.
Properties of Positive Rays
1. They are deflected towards negative plate of electric field. Therefore these rays carry positive charge.
2. The mass of positive rays is equal to the mass of the gas enclosed in the discharge tube.
3. The minimum mass of positive particles is equal to the mass of hydrogen ion (H+). These positive ions are called Protons.
4. The charge on proton is equal to +1.602×10^-19 Coulomb. (-19 is power of 10)
The phenomenon in which certain elements emit radiation which can cause fogging of photographic plate is called natural radioactivity. The elements which omit these rays are called radioactive elements like Uranium, Thorium, Radium etc. There are about 40 radioactive elements.
Henri Bequrel (1896) discovered radioactivity.Madam Curei also has valuable contribution in this field. In natural radioactivity nuclei of elements are broken and element converted to other elements. Natural radioactivity is nuclear property of the elements.
Alpha Rays
1. They are helium nuclei. They are doubly positively charged, He2+.
2. They move with speed equal to the 1/10th of the velocity of the light.
3. They cannot pass through thick-metal foil.
4. They are very good ionizer of a gas.
5. They affect the photographic plate.
Beta Rays
1. They are negatively charged.
2. They move with the speed equal to the velocity of light.
3. They can pass through a few millimeter thick metal sheets.
4. They are good ionizer of a gas.
5. They can affect the photographic plate.
Gamma Rays
2. They travel with speed equal to velocity of light.
3. They carry no charge.
4. They have high penetration power than alpha and beta rays.
5. They are weak ionizer of gas.
Rutherford Experiment and Discovery of Nucleus
Lord Rutherford (1911) and his coworkers performed an experiment. They bombarded a very thin, gold fail with Alpha particles from a radioactive source. They observed that most of the particles passed straight through the foil undeflected. But a few particles were deflected at different angles. One out of 4000 Alpha particles was deflected at an angle greater than 150.
Following conclusions were drawn from the Rutherford’s Alpha Particles scattering experiment.
1. The fact that majority of the particles went through the foil undeflected shows that most of the space occupied by an atom is empty.
2. The deflection of a few particles over a wide angle of 150 degrees shows that these particles strike with heavy body having positive charge.
3. The heavy positively charged central part of the atom is called nucleus.
4. Nearly all of the mass of atom is concentrated in the nucleus.
5. The size of the nucleus is very small as compared with the size of atom.
Defects of Rutherford Model
Rutherford model of an atom resembles our solar system. It has following defects:
1. According to classical electromagnetic theory, electron being charged body will emit energy continuously. Thus the orbit of the revolving electron becomes smaller and smaller until it would fall into the nucleus and atomic structure would collapse.
2. If revolving electron emits energy continuously then there should be a continuous spectrum but a line spectrum is obtained.
Bohr’s Atomic Model
Neil Bohr (1913) presented a model of atom which has removed the defects of Rutherford Model. This model was developed for hydrogen atom which has only proton in the nucleus and one electron is revolving around it.
Postulates of Bohr’s Atomic Model
The main postulates of Bohr’s Model are given below:
1. Electrons revolve around the nucleus in a fixed orbit.
2. As long as electron revolves in a fixed orbit it does not emit and absorb energy. Hence energy of electron remains constant.
3. The orbit nearest to the nucleus is the first orbit and has lowest energy. When an electron absorbs energy it jumps from lower energy orbit to higher energy orbit. Energy is emitted in the form of radiations, when an electron jumps from higher energy orbit to lower energy orbit. The unit of energy emitted in the form of radiations is called quantum. It explains the formation of atomic spectrum.
4. The change in energy is related with the quantum of radiation by the equation :
E2 – E1 = hv where,
E1 = Energy of first orbit
E2 = Energy of the second orbit
h = Planck’s constant
Atomic Number
The number of protons present in the nucleus of an atom is called atomic number or proton number.
It is denoted by z. The proton in the nucleus of an atom is equal to number of electrons revolving around its nucleus.
Mass Number = No of Protons + No of neutrons
A = Z + N
Isotopes
The atoms of same elements which have same atomic number but different mas number are called Isotopes.
The number of protons present in the nucleus of an atom remains the same but number of neutrons may differ.
Isotopes of Different Elements
Isotopes of Hydrogen
Hydrogen has three isotopes:
1. Ordinary Hydrogen or Protium, H.
2. Heavy Hydrogen or Deutrium, D.
3. Radioactive Hydrogen or Tritium, T.
Protium
Ordinary naturally occurring hydrogen contains the largest percentage of protium. It is denoted by symbol H. It has one proton in its nucleus and one electron revolve around the nucleus.
Number of Protons = 1
Number of Electrons = 1
Number of Neutrons = 0
Atomic Number = 1
Mass Number = 1
Deutrium
Deutrium is called heavy hydrogen. The percentage of deutrium in naturally occuring hydrogen is about 0.0015%. It has one proton and one neutron in its nucleus. It has one electron revolving around its nucleus. It is denoted by symbol D.
Number of Proton = 1
Number of Electron = 1
Number of Neutrons = 1
Atomic Number = 1
Mass Number = 2
Tritium
Radioactive hydrogen is called tritium. It is denoted by symbol T. The number of tritium isotope is one in ten millions. It has one proton and 2 neutrons in its nucleus. It has one electron revolving around its nucleus.
Number of Proton = 1
Number of Electron = 1
Number of Neutron = 2
Atomic Number = 1
Mass Number = 3
1. Plum pudding Model of atom was discovered by _______________
2. Combining capacity of an atom is called _______________
3. Alpha - particle scattering experiment of Rutherford led to discovery of _________
4. __________ are atoms having the same mass number but different atomic number.
5. The charge on the electron is found to be ________ coulombs.
6. What do you understand by valency of an element? What is valency of boron?
7. State the valencies of the following elements
(a) Magnesium
(b) Phosphorous
(c) Argon
(d) Flurine
8. Draw the atomic diagram of calcium.
9.Calculate the mass of an electron.
11.What is mean by valency of an atom?
12.What is octel rule?
14. List the features of Rutherford's nuclear model of atom.
15. What are the postulates of Bohr Model of an atom? | {"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": 0.8174580335617065, "perplexity": 1262.127833588201}, "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/1429246655962.81/warc/CC-MAIN-20150417045735-00303-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.hepdata.net/record/ins1504058 | • Browse all
Measurement of the $b$-quark production cross-section in 7 and 13 TeV $pp$ collisions
The collaboration
Phys.Rev.Lett. 118 (2017) 052002, 2017
Abstract (data abstract)
CERN-LHC. Measurements of the cross-section for producing $b$ quarks in the reaction $pp \to b\bar{b}X$ are reported in 7 and 13 TeV collisions at the LHC as a function of the pseudorapidity $\eta$ in the range $2 < \eta < 5$ covered by the acceptance of the LHCb experiment. The measurements are done using semileptonic decays of $b$-flavored hadrons decaying into a ground-state charmed hadron in association with a muon. The cross-sections in the covered $\eta$ range are $72.0 \pm 0.3 \pm 6.8 \;\mu$b and $144 \pm 1 \pm 21 \;\mu$b for 7 and 13 TeV. The ratio is $2.00 \pm 0.02 \pm 0.26$, where the quoted uncertainties are statistical and systematic, respectively.
• #### Table 1
Figure 2(a), Table 1
10.17182/hepdata.79130.v1/t1
The cross-section as a function of $\eta$ for $pp \to H_b X$, where $H_b$ is a hadron that contains either...
• #### Table 2
Figure 2(b), Table 1
10.17182/hepdata.79130.v1/t2
The cross-section as a function of $\eta$ for $pp \to H_b X$, where $H_b$ is a hadron that contains either...
• #### Table 3
Figure 2(c), Table 1
10.17182/hepdata.79130.v1/t3
The ration of the cross-sections as a function of $\eta$ for $pp \to H_b X$, where $H_b$ is a hadron... | {"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": 0.9713148474693298, "perplexity": 1632.5209042855029}, "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/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00417.warc.gz"} |
https://community.agilent.com/technical/gcms/f/gcms-forum/4103/unexplained-at-least-for-me-instrument-software-shut-downs?pifragment-8136=2 | "Unexplained" (at least for me) Instrument/Software shut-downs
We have an Agilent GC-MS (6890N-5975C) with MSD chemstation ( E02.02.1431) and experience problems with shut-downs that I can't explain. I attached a couple of error messages that I see. The strange thing is that sometimes a sequence runs all weekend without ever stopping and when I then rerun the sequence it stops. Sometimes the software shuts down completely, sometimes the sequence just pauses, allows me to resume and then pauses again after running one sample.
IT added some memory, but that did not solve the problem.
Here iare the error messages and the Instrument and Software Log. I also sometimes
/resized-image/__size/1714x964/__key/communityserver-discussions-components-files/147/Error-Message-Software-shut_2D00_down.png
/resized-image/__size/1280x960/__key/communityserver-discussions-components-files/147/pastedimage1631171985598v1.png
/resized-image/__size/1280x960/__key/communityserver-discussions-components-files/147/Log-for-Sequence-that-kept-pausing.png
/resized-image/__size/1280x960/__key/communityserver-discussions-components-files/147/Sequence-Log.png
If anyone has guidance what to look for. Please let me know.
Thanks. Bettina | {"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": 0.8320398330688477, "perplexity": 2890.8767373669743}, "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-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00060.warc.gz"} |
http://konedreamhouse.blogspot.com/2013/05/iswhat-is-force-of-gravity.html | 2013/05/19
【IS】What is the force of gravity? / 甚麼是重力?
What is the force of gravity?
In the 17th century, Isaac Newton discovered that there is a small attractive force between any two objects. It exists even between you and this book, or between you and your classmates. Usually, this force is so small that we do not notice it. However, if one of the objects is the Earth (of which the mass is very large), the force is noticeable. Wherever we are on Earth, there is a force pulling us towards the centre of the Earth. This force is called the force of gravity. It keeps us on the Earth's surface. It causes objects to fall back to the Earth. It also acts on a spacecraft and tends to pull it back to Earth.
The force of gravity is a non-contact force. It can act on an object at a distance. For example, the Earth exerts a force of gravity on a spacecraft even when the spacecraft travels into space. The force of gravity decreases as distance increases. Therefore, the Earth will exert a smaller force on a spacecraft when it is further away from the Earth.
＊ The force of gravity tends to pull objects towards the centre of the Earth.
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B. G. | {"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": 0.9182724356651306, "perplexity": 248.20040491396463}, "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-2017-43/segments/1508187825812.89/warc/CC-MAIN-20171023073607-20171023093607-00757.warc.gz"} |
https://www.physicsoverflow.org/6373/naive-question-gauge-transformation-electromagnetic-field | # A naive question on the $U(1)$ gauge transformation of electromagnetic field?
+ 2 like - 0 dislike
472 views
For simplicity, in the following we set the electric charge $e=1$ and consider a lattice spinless free electron system in an external static magnetic field $\mathbf{B}=\nabla\times\mathbf{A}$ described by the Hamiltonian $H=\sum_{ij}t_{ij}c_i^\dagger c_j$, where $t_{ij}=\left | t_{ij} \right |e^{iA_{ij}}$ with the corresponding lattice gauge-field $A_{ij}$. As we know the transformation $\mathbf{A}\rightarrow \mathbf{A}+\nabla\theta$ does not change the physical magnetic field $\mathbf{B}$, and the induced transformation in Hamiltonian reads $$H\rightarrow H'=\sum_{ij}t_{ij}'c_i^\dagger c_j$$ with $t_{ij}'=e^{i\theta_i}t_{ij}e^{-i\theta_j}$. Now my confusion point is:
Do these two Hamiltonians $H$ and $H'$ describe the same physics? Or do they describe some same quantum states? Or what common physical properties do they share?
I just know $H$ and $H'$ have the same spectrum, thank you very much.
This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user K-boy
If you do the transformation $c_i^{(\dagger)}\to e^{-(+)i\theta_i}c_i^{(\dagger)}$, you see that $H'\to H$, and the model is indeed gauge invariant. The physics is thus the same.
This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user Adam
@ Adam Yes,you're right. But why we do this transformation? Can it be deduced from the underlying microscopic model or it's just the requirement for gauge invariant to describe the same physics?
This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user K-boy
In the underlying model, it corresponds to the transformation $A_i(x)\to A_i(x)+\partial_i \theta(x)$ and $\hat\psi(x)\to e^{i\theta(x)}\hat\psi(x)$ (with maybe some minus signs) in the microscopic Hamiltonian. Or if you take your lattice Hamiltonian as "fundamental" (as in lattice QCD), you can show that it reproduces the usual continuous Hamiltonian in the limit of vanishing lattice spacing. So maybe you are asking what does the U(1) gauge transformation means in quantum mechanics ?
This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user Adam
A system is not only determined by its Hamiltonian, but also by its Hilbert space. U(1) gauge theory can come from constraint of your Hilbert space.
This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user Shenghan Jiang
@ Shenghan Jiang Thanks for your comment. But what is the constraint of Hilbert space in this example of my question(QHE system)?
This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user K-boy
Remarks: $e^{i\theta _i \hat{n}_i}c_j^\dagger e^{-i\theta _i \hat{n}_i}=\delta _{ij}e^{i\theta _i} c_j^\dagger +(1-\delta _{ij})c_j^\dagger$, and hence $H'=UHU^{-1}$ with $U=\prod _ie^{i\theta _i\hat{n}_i}$.
This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user K-boy
+ 1 like - 0 dislike
For constant $t_{ij}$, the transformation may be considered as a simple redifinition of the quantum state basis.
A natural basis for you quantum states are the $|\psi_j \rangle = c_j^+|0\rangle$. In this basis, you have : $H|\psi_j \rangle = t_{ij}|\psi_i \rangle$, so this means that $H_{ij}=t_{ij}$, so we may write $H = \sum H_{ij}~ c^+_i c_j$.
Now, we may decide to change the basis $|\psi' \rangle = U |\psi \rangle$, with $U = Diag (e^{i\theta_1},e^{i\theta_2}, ....e^{i\theta_n})$, so that $|\psi_j \rangle \to |\psi'_j \rangle = e^{i\theta_j} |\psi_j \rangle$. The matrix $U$ is unitary, and it transforms an orthonormal basis into an other orthonormal basis.
In this new basis, the hamiltonian is simply $H' = U H U^{-1}$, or expressing the elements of the operator $H'$, we get : $H'_{ij} = e^{i\theta_i} H_{ij}e^{-i\theta_j}$
As you know, multiplying a quantum basis state $|\psi_j \rangle$ by a unit phase $e^{i\theta_j}$ does not change the physical state (which is $|\psi_j \rangle \langle|\psi_j|$), so the physics described by $H$ and $H'$ is the same, the eigenvalues $E_k$ of $H$ and $H'$ are the same, etc...
This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user Trimok
answered Oct 9, 2013 by (950 points)
@ Trimok Thanks for your comment. By the way, I think only the global phase is un physical, while the relative phase(here is the local $U(1)$ transformation) is physical.
This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user K-boy
To be fair, I consider only global transformations in my answer (with constants $t_{ij}$), so I realize it does not answer the question... The local $U(1)$ invariance is not really "physical". The $U(1)$ invariance of the EM field is a mathematical redundancy, not a true symmetry. This means that we overcount the total number of states, that is that one physical state is represented by many mathematical states, and it is necessary to reduce this number, to keep only one representant for one physical state.
This post imported from StackExchange Physics at 2014-03-09 08:41 (UCT), posted by SE-user Trimok
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https://www.physicsforums.com/threads/massless-free-field-equation-maxwells-eqn.246996/ | # Massless free field equation -> Maxwell's eqn.
1. Jul 26, 2008
### lark
Massless free field equation --> Maxwell's eqn.
The massless free field equation is supposed to turn into the empty space Maxwell's equations for spin 1 (like a photon).
But, in the book I'm using, Roger Penrose's "The Road to Reality", there seems to be a typo, because it's not quite working out. Almost but not quite.
Can someone tell me what the problem is?
See http://camoo.freeshell.org/33.24quest.pdf
for details.
(sorry if you find it inconvenient to click on the link, but I'm not going to rewrite everything into the forum's Latex).
Laura
2. Jul 26, 2008
### reilly
Re: Massless free field equation --> Maxwell's eqn.
Could you explain the notation, please.
Regards,
Reilly Atkinson
3. Jul 26, 2008
### muppet
Re: Massless free field equation --> Maxwell's eqn.
I can't help directly... but if this is one of his problems have you tried Penrose's solutions online?
4. Jul 27, 2008
### lark
Re: Massless free field equation --> Maxwell's eqn.
I'm the only one who has posted solutions to the problems in the 2nd half of the book!
Laura
5. Jul 27, 2008
### lark
Re: Massless free field equation --> Maxwell's eqn.
It won't help if I explain it, I could be misinterpreting something and that's part of the question. I'm sure my calculations are OK.
I was hoping somebody would know how the massless free field equation translates into Maxwell's equations, concretely.
6. Jul 27, 2008
### Avodyne
Re: Massless free field equation --> Maxwell's eqn.
I think it's a little more complicated than what you're doing. First of all, there are two psi fields, one with two unprimed indices and one with two primed indices; each field must be treated separately.
See sections 34 and 35 of Srednicki's field theory book for an introduction to this notation (but with conventions that probably don't match Penrose's). The Srednicki book is available free online in draft form at his web page.
7. Jul 28, 2008
### lark
Re: Massless free field equation --> Maxwell's eqn.
I tried the one with two unprimed indices. It's supposed to work out to Maxwell's equations. It does almost, but not quite. That's the problem. Can anyone tell me what the typo is, that if fixed, would make it come out to be Maxwell's equations?
The worked out version is in http://camoo.freeshell.org/33.24.pdf and for example, eqn 7 minus eqn 5 in there, is $$\partial E_x/\partial x + \partial E_y/\partial y +\partial B_z/\partial z=0$$. It wants to be $$\nabla\cdot E=0,$$ but it isn't quite. I'm sure I'm not making an algebra mistake, the problem is mis-stated or something.
Laura
Last edited: Jul 28, 2008
8. Jul 29, 2008
### George Jones
Staff Emeritus
Re: Massless free field equation --> Maxwell's eqn.
After looking at page 323 of Spinors and Space-Time V1 by Penrose and Rindler, it looks like it should be $\psi_{01} = -C_3$, not $\psi_{01} = -iC_3$. This seems to give stuff like divergences and components of curls, but I haven't worked through the details.
9. Jul 29, 2008
### lark
Re: Massless free field equation --> Maxwell's eqn.
I made that change. Now I get $$\nabla\cdot E=0$$ and $$\nabla\cdot B=0$$. But the curl equations have the sign exactly reversed! So that if you reverse the sign of $$t$$, it works out right.
I understand now, I think. I got the sign of $$t$$ in $$\nabla^a$$ wrong, it's actually $$-\partial /\partial t+\partial /\partial x+\partial /\partial y+\partial /\partial z$$ not $$+\partial /\partial t+\partial /\partial x+\partial /\partial y+\partial /\partial z$$.
$$Laura$$
10. Jul 29, 2008
### George Jones
Staff Emeritus
Re: Massless free field equation --> Maxwell's eqn.
Maybe the sign of t was okay, and the signs of the spatial derivatives were wrong. I think Penrose uses the + - - - convention for the metric, so raising the standard partials in $$\nabla_a$$ to $$\nabla^a$$ would put minus signs in front of the spatial derivatives.
11. Jul 29, 2008
### lark
Re: Massless free field equation --> Maxwell's eqn.
Well, whatever. It doesn't matter in this case
I put the whole calculation in http:/camoo.freeshell.org/33.24.pdf
Thanks,
$$Laura$$ | {"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": 0.8482447266578674, "perplexity": 1478.2946997879842}, "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-2017-04/segments/1484560280292.50/warc/CC-MAIN-20170116095120-00376-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://www.ams.org/mathscinet-getitem?mr=57:10435 | MathSciNet bibliographic data MR470689 46L10 Connes, Alain; Størmer, Erling Homogeneity of the state space of factors of type ${\rm III}\sb{1}$${\rm III}\sb{1}$. J. Functional Analysis 28 (1978), no. 2, 187–196. Links to the journal or article are not yet available
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews. | {"extraction_info": {"found_math": true, "script_math_tex": 1, "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": 0.9521071314811707, "perplexity": 4110.585033587388}, "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-2017-09/segments/1487501171162.4/warc/CC-MAIN-20170219104611-00442-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/change-in-amplitude-of-wave-on-reflection.122235/ | # Homework Help: Change in amplitude of wave on reflection
1. May 28, 2006
### Amith2006
Sir,
When a wave is reflected from a free end or a rigid end, will there be a change in amplitude always?
2. May 28, 2006
### maverick280857
What do you think?
3. May 29, 2006
### Amith2006
I think Yes.
4. May 29, 2006
### maverick280857
You're correct, as far as the magnitude is concerned, assuming NO power loss at the "end". | {"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": 0.9505397081375122, "perplexity": 4285.515701162941}, "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-2018-39/segments/1537267156471.4/warc/CC-MAIN-20180920120835-20180920141235-00116.warc.gz"} |
https://byjus.com/question-answer/one-end-of-a-massless-spring-of-spring-constant-k-200-text-n-m-and-4/ | Question
# One end of a massless spring of spring constant k=200 N/m and natural length 0.5 m is fixed at the centre of a frictionless horizontal table. The other end is connected to a load of mass 1 kg lying on the table at rest. If the spring remains horizontal and mass is made to rotate at an angular speed of 2 rad/s, find the elongation in the spring.
A
2 cm
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B
1 cm
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C
4 cm
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D
0.1 cm
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Open in App
Solution
## The correct option is B 1 cmWhen the mass m is moving on a circular path around the vertical axis, the spring undergoes elongation to provide the necessary centripetal force for the motion. Body is in equiibrium in vertical direction (N=mg). So we only have to consider the motion on the horizontal plane. Let the elongation produced in the spring be x metres i.e Spring force (F) = kx Radius of circular path r= natural length of spring =0.5 m Since centripetal force is provided by the spring force, kx=mrω2 ⇒x=mrω2k=1×0.5×(2)2200=2200 m=1 cm NOTE: Elongation in the spring is very small with respect to the natural length of the spring i.e x<<r. Hence our assumption of radius of circular path =r, holds good.
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Explore more | {"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": 0.9492030143737793, "perplexity": 1712.241289384542}, "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/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00184.warc.gz"} |
http://mathoverflow.net/questions/332/does-the-space-of-n-x-n-positive-definite-self-adjoint-real-matrices-have-a-be/2153 | ## Does the space of n x n, positive-definite, self-adjoint, real matrices have a better name?
This is also the space of real, symmetric bilinear forms in R^n.
-
The bilinear form should still be positive definite. – S. Carnahan Oct 12 2009 at 2:54
Better in what sense? – Felix Goldberg May 25 at 19:23
• Standard jargon is SPD (for "symmetric positive-definite").
• This isn't exactly a "name," but the n x n symmetric positive-definite matrices are exactly those matrices A such that the bilinear function (x, y) -> yTAx defines an inner product on Rn. Conversely, every bilinear function is of that form for some A, so with some abuse of terminology, you could equate the set of those matrices with the set of inner products on Rn.
There are many other ways to characterize SPD matrices, but that's the only one I can think of at the moment that can be summarized as a single noun phrase.
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Note that this space is not a vector space, but is a convex cone in the vector space of nxn matrices (it is closed under addition and multiplication by positive scalars). Hence people sometimes refer to the "positive semidefinite cone".
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Yes, I was going to post that as well. – Ilya Nikokoshev Oct 23 2009 at 19:18
This is the symmetric space of GL_n(R)
-
How about ? I have seen or used to denote the set of positive linear transformations in a set of linear transformations on an inner product space, but this was in the context of operator algebras.
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It is often usefull to know that this set can be identified with the set of non-singulat covariance matrices of random vectors with values in $\mathbb(R)^n$. | {"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": 0.9580450057983398, "perplexity": 267.4669948742437}, "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-2013-20/segments/1368698693943/warc/CC-MAIN-20130516100453-00012-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://www.thespectrumofriemannium.com/tag/covariant/ | ## LOG#148. Path integral (III).
Round 3! Fight… with path integrals! XD Introduction: generic aspects Consider a particle moving in one dimension (the extension to ND is trivial), the hamiltonian being of the usual form: The fundamental question and problem in the path integral (PI) … Continue reading
## LOG#032. Invariance and relativity.
Invariance, symmetry and invariant quantities are in the essence, heart and core of Physmatics. Let me begin this post with classical physics. Newton’s fundamental law reads: Suppose two different frames obtained by a pure translation in space: … Continue reading | {"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": 0.9974463582038879, "perplexity": 3860.9574378527723}, "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-2020-05/segments/1579250601040.47/warc/CC-MAIN-20200120224950-20200121013950-00055.warc.gz"} |
https://socratic.org/questions/how-do-you-find-the-standard-form-of-the-equation-vertex-1-2-focus-1-0-vertex-2- | Precalculus
Topics
How do you find the standard form of the equation Vertex: ( -1, 2), Focus (-1, 0) Vertex: (-2, 1), Directrix: x = 1?
Oct 2, 2017
Answer:
Use the fact that a parabola is the locus of points equidistant from the focus point and the directrix line.
Explanation:
The distance from the directrix, $x = 1$, to any point, $\left(x , y\right)$, on the parabola is:
$d = x - 1 \text{ [1]}$
The distance from the focus, $\left(- 1 , 0\right)$ to any point, $\left(x , y\right)$, on the parabola is:
$d = \sqrt{{\left(x - \left(- 1\right)\right)}^{2} + {\left(y - 0\right)}^{2}}$
Simplify:
$d = \sqrt{{\left(x + 1\right)}^{2} + {y}^{2}} \text{ [2]}$
Because the distances must be equal, we can set the right side of equation [1] equal to the right side of equation [2]:
$x - 1 = \sqrt{{\left(x + 1\right)}^{2} + {y}^{2}}$
Square both sides:
${\left(x - 1\right)}^{2} = {\left(x + 1\right)}^{2} + {y}^{2}$
Expand the squares:
${x}^{2} - 2 x + 1 = {x}^{2} + 2 x + 1 + {y}^{2}$
Combine like terms:
$- 4 x = {y}^{2}$
Divide both sides by -4:
$x = - \frac{1}{4} {y}^{2} \leftarrow$ standard form for a parabola that opens left.
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https://www.physicsforums.com/threads/how-to-tell-whether-a-reaction-is-an-elementary-reaction.612234/ | # How to tell whether a reaction is an elementary reaction
1. Jun 7, 2012
### BrianC12
How can you tell whether a reaction is an elementary reaction without knowing the reaction mechanism? I had this question on a recent midterm and got it wrong:
A -> Product
Is this an elementary reaction?
I said yes because it has no intermediates. In retrospect there's no way I could know that unless I knew the reaction mechanism. The only other information given is that the reaction is zero order. Thanks for any help!
2. Jun 8, 2012
### That Neuron
BY checking to see if the stochiometric coefficients equal the exponents in a chemical equilibrium equation.
3. Jun 8, 2012
### That Neuron
aA + bB -----> cC + dD, where all lower case letters are coefficients and all capitals are the Chemical formulae of each reactant/product. So in a chemical equilibrium equation, r=k[A]^a (^b)]/[[C]^c ([D]^d)]. If the exponents found to match the chemical reaction differ from this, the reaction is composed from two or more elementary steps. [] is notation for the concentration in molarity mass of solute/volume of solution.
Last edited: Jun 8, 2012
4. Jun 8, 2012
### BrianC12
So that would mean this is NOT an elementary reaction because the coefficient of 1 for A doesn't match the power of 0 for the equilibrium equation? But if that's true, then doesn't that imply that any zero order reaction isn't an elementary reaction since a coefficient > 0 can obviously never equal 0.
5. Jun 8, 2012
### BrianC12
Ok I'm just confused now. The second part of the question asked to write the differential rate law. I wrote -d[A]/dt = k[A]^0 = k and got full credit. If this isn't an elementary reaction then I wouldn't be able to write the rate law. But according to the TA who graded my test, there are intermediates in the reaction, meaning it can't be an elementary reaction.....
6. Jun 9, 2012
### epenguin
Not elementary my dear Brian :tongue:
If the reaction is zero order it cannot be an elementary reaction.
Looks like there is no other participant, just A goes to something.
In an elementary reaction all the A molecules - say they split into something - have the same chance of doing so. So the actual rate of reaction - number of molecules of A that change into something else per second - is proportional to A. I think you can write the rate law and name it.
But with zero order kinetics - that means say you put in 10 times more A, yet only as many as before react per second. Each molecule does not have the same chance of reacting as before. That strongly suggests that in order to react they need to interact with something else, e.g. a catalyst, before they can split. Not elementary.
Zero order kinetics are typical of reactions involving solid catalysts or enzymes or some free radical catalyses when you get above a certain concentration (i.e. saturation).
7. Jun 9, 2012
### BrianC12
Ahh alright that makes sense, thanks. I actually checked with my friend's midterm. He wrote "this is not an elementary reaction because elementary reactions are characterized by collisions with other molecules"(that's the definition in the book) and got full credit. I don't really understand how that explains anything since first of all unimolecular elementary reactions don't involve any collisions and your explanation implies that there must be a collision, which my friend's answer contradicts. I asked him and he said he didn't actually know, he just saw a previous midterm with a similar question where the person got it wrong, so he switched the answer and wrote the definition from the book. Is that an insufficient answer or am I just missing something?
On top of that doesn't the book's definition contradict itself too? Because it states that definition and then goes on to show an example of a unimolecular reaction where an energized molecule of N2O5 dissociates into NO2 and NO3, which is just a "reaction" at high temperature that does not involve any collisions.
I hope I don't come across as arrogant for doubting everything, in my mind it just seems like a lot of the information I've been getting from professors, TA's, and the book contradict each other and I'm trying to figure out what's right and what's wrong.
Again, thanks for the help!
8. Jun 9, 2012
### epenguin
I think your reasoning is OK. I don't think there is this contradiction. When I said "all the A molecules - say they split into something..." your unimolecular N2O5 dissociation is an example of that.
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https://math.stackexchange.com/questions/3784641/proof-of-interior-gradient-estimate-for-laplaces-equation | # Proof of interior gradient estimate for Laplace's equation
I have a question about the proof of the estimate $$|\nabla u(x_0)| \leq \frac{n}{R} \max_{\bar{B}_R(x_0)} |u|$$ where $$u$$ is assumed to be harmonic.
Since $$u_{x_i}$$ is harmonic, by the mean value property and integration by parts, $$u_{x_i}(x_0) = \frac{r}{\omega_n R^n}\int_{B_R(x_0)} u_{x_i}(y) dy = \frac{n}{\omega_n R^n}\int_{\partial B_R(x_0)} u(y) \nu_i dS_y.$$ Taking the absolute value, we obtain $$|u_{x_i}(x_0)| \leq \frac{n}{\omega_n R^n} \int_{\partial B_R(x_0)} |u(y)| dS_y \leq \frac{n}{R}\max_{\bar{B}_R(x_0)} |u|.$$ I understand the preceding steps. What I don't understand is how this obviously proves the desired result. This is my attempt at obtaining the desired result: \begin{align*} |\nabla u(x_0)|^2 &= u_{x_1}^2(x_0) + \cdots + u_{x_n}^2(x_0) \\ &\leq \underbrace{\frac{n^2}{R^2}(\max_{\bar{B}_R(x_0)} |u|)^2 + \cdots + \frac{n^2}{R^2}(\max_{\bar{B}_R(x_0)} |u|)^2}_{\text{n times}}\\ &=\frac{n^3}{R^2}(\max_{\bar{B}_R(x_0)} |u|)^2. \end{align*} Taking the square root, $$|\nabla u(x_0)| \leq \left(\frac{n^3}{R^2}(\max_{\bar{B}_R(x_0)} |u|)^2\right)^{1/2} = \frac{n^{3/2}}{R}\max_{\bar{B}_R(x_0)}.$$
I'm not sure where my logic is wrong and I am aware this must be something simple...
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The line $$u_{x_i}(x_0) =\frac{n}{\omega_n R^n}\int_{\partial B_R(x_0)} u(y) \nu_i \,dS_y$$ would be the components of the equation $$\nabla u(x_0) = \frac{n}{\omega_n R^n}\int_{\partial B_R(x_0)} u(y) \nu\, dS_y$$ And now you do the approximation at this level: \begin{align}|\nabla u(x_0)| &= \frac{n}{\omega_n R^n}\left|\int_{\partial B_R(x_0)} u(y) \nu\, dS_y\right| \\ &\le \frac{n}{\omega_n R^n}\int_{\partial B_R(x_0)} |u(y) \nu|\, dS_y \\ &=\frac{n}{\omega_n R^n}\int_{\partial B_R(x_0)} |u(y) |\, dS_y \\ &\le \frac{n}{R}\|u\|_{L^\infty(\overline{B_R(x_0)})}. \end{align} | {"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": 10, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999808073043823, "perplexity": 321.2652914116948}, "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/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00407.warc.gz"} |
http://clay6.com/qa/41250/a-positively-charged-conductor-carrying-a-charge-q-placed-near-an-uncharged | Browse Questions
# A positively charged conductor carrying a charge +Q placed near an uncharged and grounded conductor, results in the phenomenon of electrostatic induction. An electric charge is induced in the uncharged conductor because of this phenomenon. The electric charge induced in the uncharged conductor is
$\begin{array}{1 1}(A)\;-2Q\\ (B)\;-Q \\ (C)\;+Q \\(D)\; +2Q \end{array}$ | {"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": 0.9962307810783386, "perplexity": 441.7773705954945}, "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/1476988721067.83/warc/CC-MAIN-20161020183841-00283-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/146481/what-symmetry-operation-mixes-states-with-different-ell-in-hydrogen-atom?noredirect=1 | What symmetry operation mixes states with different $\ell$ in hydrogen atom? [duplicate]
We can mix states with different $m$ in hydrogen atom by rotating it around some axis (not coinciding with $z$). Thus rotation is the symmetry operation which mixes states with different $m$.
As hydrogen has hidden symmetry, which leads to degeneracy in $\ell$, there must be some symmetry operation, which would mix different $\ell$'s. What is this operation? I understand that this symmetry is related to Laplace-Runge-Lenz vector conservation, but I don't quite see what symmetry operation corresponds to it. | {"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": 0.932541012763977, "perplexity": 443.35426960935314}, "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-2019-47/segments/1573496669730.38/warc/CC-MAIN-20191118080848-20191118104848-00342.warc.gz"} |
http://gradestack.com/MCAT-Complete-Tutor/The-standard-potential-Eo/2-3316-3365-16562-sf | ## General Chemistry
### Electrochemistry
Question
3 out of 10
The standard potential Eo is related to equilibrium constant K by the following relation:
In a reaction occurring at standard state conditions, if the concentrations of the products are greater than that of the reactants, which of the following is true?
A K is negative B Eo is negative C The reaction is nonspontaneous D None of the above
Ans. D Based on the given information in the question, we can say that Eo is directly related to K. The question also says that the product concentrations are greater than the reactant concentrations. So we can say that the equilibrium constant value is greater than 1. We can rule out Choice A, because there is no such thing as a negative concentration. So K cannot be negative anyway. Since we know that K is greater than 1, the natural logarithm of K will be positive. So from the equation, we can say that Eo cannot be negative, which rules out Choice B. Hence, the Eo value should be positive. Since we have a positive Eo value, the reaction is most likely to be spontaneous. Keep in mind the conditions at which the reaction occurs - standard state (1 atm, 25 oC). | {"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": 0.9607617855072021, "perplexity": 439.25756543918783}, "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/1476988719273.38/warc/CC-MAIN-20161020183839-00107-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.groundai.com/project/orbital-kondo-effect-in-cobalt-benzene-sandwich-molecules/ | Orbital Kondo effect in Cobalt-Benzene sandwich molecules
# Orbital Kondo effect in Cobalt-Benzene sandwich molecules
## Abstract
We study a Co-benzene sandwich molecule bridging the tips of a Cu nanocontact as a realistic model of correlated molecular transport. To this end we employ a recently developed method for calculating the correlated electronic structure and transport properties of nanoscopic conductors. When the molecule is slightly compressed by the tips of the nanocontact the dynamic correlations originating from the strongly interacting Co shell give rise to an orbital Kondo effect while the usual spin Kondo effect is suppressed due to Hund’s rule coupling. This non-trivial Kondo effect produces a sharp and temperature-dependent Abrikosov-Suhl resonance in the spectral function at the Fermi level and a corresponding Fano line shape in the low bias conductance.
###### pacs:
73.63.Rt, 71.27.+a, 72.15.Qm, 31.15.A–
The discovery of ferrocene and bisbenzene chromium Kealy and Pauson (1951); ? over half a century ago was the starting point for experimental and theoretical work both in chemistry and physics concerning the intricate properties of organometallic compounds. The investigation of these and related sandwich complexes is driven by their relevance in various chemical applications (e.g. catalysis) and more recently also because of their prospective nanotechnological applications for example as molecular magnets Gatteschi et al. (2006) or spintronic devices Xiang et al. (2006); ?.
Molecules with a transition metal (TM) center coupled to aromatic groups are also of high interest from a fundamental point of view. The strong electronic correlations in the shell of the TM can modify the ground state and electronic transport properties of such molecules, leading to many-body phenomena like the Kondo effect Hewson (1997) as recently observed in TM-phthalocyanine molecules Zhao et al. (2005); ?. The importance of dynamical correlations in nanoscopic devices in general is further substantiated by the recent observation of the Kondo effect in nanocontacts made from TMs Calvo et al. (2009); Jacob et al. (2009). Also recently a so-called underscreened Kondo effect has been reported for a molecule trapped in a breakjunction Parks et al. (2010). While the Kondo effect is commonly associated with the screening of a local magnetic moment by the conduction electrons, it is also possible that another internal degree of freedom associated with a degeneracy gives rise to a Kondo effect Cox and Zawadowski (1998). One example is the so-called orbital Kondo effect where the pseudo-spin arising from an orbital degeneracy is screened by the conduction electrons Kolesnychenko et al. (2002); ?.
Such complex many-body phenomena call for a theoretical description beyond the standard treatment with Kohn-Sham (KS) density functional theory (DFT) which cannot capture many-body physics beyond the effective mean-field picture. The incorrect behavior of the KS DFT for strongly correlated systems can be remedied by augmenting the DFT with a local Hubbard-like interaction. This DFT++ approach is the de facto standard in the theory of solids Kotliar et al. (2006); ?. Recently this approach has been adapted to the case of nanoscopic conductors Jacob et al. (2009); Jacob and Kotliar (2010); ?; Lucignano et al. (2009); ?; ?; ?.
In this letter we apply this method to investigate the transport properties of cobalt benzene sandwich molecules (CH)Co(CH) (CoBz in the following). We find that dynamical correlations in the Co shell give rise to an orbital Kondo effect in a doubly degenerate level of the Co shell while the usual spin Kondo effect is suppressed due to Hund’s rule coupling.
We perform DFT calculations using the CRYSTAL06 code Dovesi et al. () employing the LDA Kohn and Sham (1965), PW91 Perdew and Wang (1986) and the hybrid functional B3LYP Becke (1993), and using the all electron Gaussian 6-31G basis set. The geometries of the wires were relaxed beforehand and kept fixed during the calculations. The geometry of the molecule in contact with the wires was relaxed employing the B3LYP functional geo ().
In order to capture many-body effects beyond the DFT level, we have applied the DFT+OCA (Density Functional Theory + One-Crossing Approximation) for nanoscopic conductors developed by one of us in earlier work Jacob et al. (2009); Jacob and Kotliar (2010). To this end the system is divided into three parts: The two semi-infinite leads L and R and the device region D containing the molecule and part of the leads. The KS Green’s function (GF) of region D can now be obtained from the DFT electronic structure as where is the KS Hamiltonian of region D and are the so-called lead self-energies which describe the coupling of D to L and R and which are obtained from the DFT electronic structure of the nanowire leads.
Next the mean-field KS Hamiltonian is augmented by a Hubbard-like interaction term that accounts for the strongly interacting electrons of the Co shell. Here we use a simplified interaction which only takes into account the direct Coulomb repulsion and the Hund’s rule coupling . These are different from the bare interactions due to screening processes. Here we assume that the interactions are somewhat increased as compared to their bulk values since the screening should be weaker than in bulk. Hence we use eV and eV; indeed we find that our results are qualitatively stable for a reasonable range of values for and (see below).
The interacting Co shell coupled to the rest of the system (benzene+leads) constitutes a so-called Anderson impurity model (AIM). The AIM is completely defined by the interaction parameters and , the energy levels of the orbitals and the so-called hybridization function . The latter describes the (dynamic) coupling of the Co -shell to the rest of the system and can be obtained from the KS GF as where is the chemical potential, are the KS energy levels of the orbitals and is the KS GF projected onto the subspace. The energy levels are obtained from the static crystal field where as usual in DFT++ approaches a double counting correction (DCC) has to be subtracted to compensate for the overcounting of interaction terms. Here we employ the so-called fully localized or atomic limit DCC Czyżyk and Sawatzky (1994) .
The AIM is solved within the OCA Haule et al. (2001); Kotliar et al. (2006). This yields the electronic self-energy which accounts for the electronic correlations of the -electrons due to strong electron-electron interactions. The correlated GF is then given by . Correspondingly, the correlated GF for D is given by where and only act within the subspace. From we can calculate the transmission function where . For small bias voltages , the transmission yields the conductance: .
We consider a single CoBz molecule in contact with two semi-infinite Cu nanowires as shown in Fig. 1a. The CoBz molecule is the smallest instance of a general class of MBz complexes, where M stands for a metal atom, that have been prepared and investigated Kurikawa et al. (1999); ?; ?; Xiang et al. (2006); ?; ?. The semi-infinite Cu wires exhibit the hexagonal symmetry of the molecule and correspond to the (6,0) wires described in Ref. Tosatti et al., 2001. We investigate the system at three different Cu-tip-Co distances 3.6 Å, 4.0 Å and 4.3 Å (see Fig. 1a,b). As can be seen from Fig. 1b the Bz-Bz distance varies between 3.4 Å and 3.65 Å depending on the distance of the Cu tip to the Co atom in the center of the molecule. At distances Å and Å the molecule is slightly compressed compared to its free height of about Å, whereas it is slightly stretched at Å.
The hexagonal symmetry of the system leads to a lifting of the degeneracy of the five orbitals. The symmetry adapted representations are: the group consisting of the orbital only, the doubly degenerate group consisting of the and orbitals and the group consisting of the and orbitals.
Figs. 1c+d show the imaginary parts of the hybridization functions calculated from the LDA electronic structure. The imaginary part of the hybridization function exhibits a distinct peak close to the Fermi level () in the channel, whose position, width and height depend significantly on the molecular geometry, specifically on the Bz-Co distance. The other channels and show only a negligible hybridization close to . The dominant feature stems, similarly as shown for graphene Wehling et al. (2010); Jacob and Kotliar (2010) from hybridization with the orbital state of the benzene rings. The feature does not depend qualitatively on the DFT functional used, as we have found the same feature within GGA and also in B3LYP calculations. The presence of strong molecular resonances in the hybridization function makes this case different from the case of nanocontacts with magnetic impurities where the hybridization functions are generally much smoother (see Ref. Jacob et al. (2009) for comparison).
Fig. 2a compares the correlated spectral functions of the Co electrons for the three distances considered here at high temperatures on a large energy scale. The spectra vary considerably as the distance changes. Most importantly, for around 3.6 Å when the molecule is slightly compressed, a sharp temperature-dependent peak appears right at , as can be seen from Fig. 2b. The peak is strongly renormalized (i.e. it only carries a small fraction of the spectral weight) due to the strong electron-electron interactions.
The sharp peak in the spectral function at that starts to develop already at temperatures of eV K stems from the channel which is the only channel with appreciable hybridization near , see Figs. 1b+c. Correspondingly, the transmission function (Fig. 2c) shows a Fano-like feature around zero energy. A renormalized, sharp and temperature-dependent resonance in the spectral function at is commonly associated with the Kondo effect. Looking at the orbital occupations, we find that the channel that gives rise to the resonance for Å has an occupation of about while the total occupation of the shell is . The fractional occupation numbers indicate the presence of valence fluctuations where the charges in the individual impurity levels fluctuate in contrast to the pure Kondo regime where these fluctuations are frozen.
Analyzing the atomic states of the Co -shell contributing to the ground state of the system we find that the principal contribution (%) is an atomic state with electrons and a total spin of as shown in Fig. 3a. The total spin 1 stems from holes in the and channels. The charge fluctuations in the channel are mainly due to the contribution (%) of an atomic state. There are considerably weaker contributions (%) from atomic and states. The individual contributions of the remaining atomic states are very small (below 1%) but add up to a total contribution of 34%.
By exclusion of individual atomic states from the calculation of the spectra we can determine which fluctuations are responsible for the different spectral features. We find that the fluctuations between the and the states are primarily responsible for the three spectral features close to including the sharp Kondo-like peak right at , as illustrated in Fig. 3b. Also the broad peak around eV below originates from these fluctuations while the broad peak about eV above arises from fluctuations between the atomic state and the principal atomic state. The two peaks in the spectral function nearest to the Kondo resonance arise from the strong energy dependence of the hybridization function whose real part has poles just below and above roughly at the positions of these two spectral features.
Note that the fluctuations from the to the states that give rise to the Kondo-like peak at actually cannot lead to a spin Kondo effect since the spin of the state is higher than the spin of the principal state. Instead the fluctuations between the and the states which give rise to the Kondo-like resonance at correspond to an orbital Kondo effect in the doubly-degenerate levels of the Co shell as illustrated in the upper panel of Fig. 3c. Here the index labeling the two orbitals with symmetry takes over the role of a pseudo spin. In the principal atomic state the levels are occupied with three electrons and hence have a pseudo spin of . By excitation to the state the electron with minority real spin and with some pseudo spin state is annihilated. By relaxation to the principal electronic state a minority real spin electron can now be created in one of the two pseudo spin states. Those processes that lead to a flip of the pseudo spin then give rise to the orbital Kondo effect and the formation of the Kondo peak at .
The absence of a normal spin Kondo effect where the total spin 1 of the principal atomic state is screened, is easily understood on the following grounds: First, in general the Kondo scale decreases exponentially with increasing spin of the magnetic impurity Nevidomskyy and Coleman (2009). In addition, here the level does not couple at all to the conduction electrons around (no hybridization). Thus the spin associated with it cannot be flipped directly through hopping processes with the conduction electron bath.
On the other hand, an underscreened Kondo effect as reported in Ref. Parks et al. (2010) where only the spin within the shell is screened is also suppressed compared to the orbital Kondo effect due to Hund’s rule coupling: Screening of the spin in the shell can take place by fluctuations to the () state. However, the Hund’s rule coupling favors the high spin () state over the low spin () state as can also be seen from the smaller weight of the latter compared to the former. Hence the Kondo scale is considerably lower for the underscreened Kondo effect than for the orbital Kondo effect found here. At lower temperatures (not accessible with OCA) the two Kondo effects may in fact coexist.
We have also checked the dependence of the LDA+OCA spectra on the DCC as well as on the interaction parameters and (not shown sup ()). In general we find the spectra and also the Kondo peak to be qualitatively robust against shifts of the impurity levels in energy over a range of several electron volts, and changes of between 3 and 7 eV, and of between 0.6 to 1 eV. As expected for the Kondo effect the sharp resonance stays pinned to the Fermi level when shifting the impurity levels in energy, and only height and width somewhat change.
Stretching the molecule by displacing the tips of the Cu nanowires the Kondo resonance and the concomitant Fano line shape in the transmission disappear for distances Å (not shown). This is accompanied by an increase of the occupation of the Co shell. The new regime is characterized by a strong valence mixing between the and atomic state of roughly equal contribution indicating that the system is now in the so-called mixed valence regime (see e.g. Ref. Hewson, 1997, Chap. 5). Hence the orbital Kondo effect and the associated spectral features can be controlled by stretching or compressing the molecule via the tip atoms of the Cu nanocontact. This strong sensitivity on the molecular conformation stems from the sharp features in the hybridization function which change considerably when the molecule is stretched or compressed as can be seen from Fig. 1d. This peculiar behavior is qualitatively different from the case of the nanocontacts containing magnetic impurities where the hybridization functions are much smoother Jacob et al. (2009).
In conclusion, we have shown for a CoBz sandwich molecule coupled to Cu nanowires that the dynamic correlations originating from the strongly interacting Co electrons give rise to an orbital Kondo effect in the doubly degenerate levels while a spin Kondo effect is suppressed by Hund’s rule coupling. Due to the sensitivity of the electronic correlations on the molecular conformation it is possible to control the appearance of the orbital Kondo effect by stretching and compressing the molecule with the tips of the Cu leads. It should be possible to prepare a setup similar to the one considered here in an actual experiment and measure the orbital Kondo effect, e.g. by contacting a CoBz molecule deposited on a Cu(111) surface with a scanning tunneling microscope. Finally, we mention that we have also explored the case of a NiBz sandwich. In this case we do not find a Kondo effect. Instead the system is in the mixed valence regime characterized by a strong peak with weak temperature dependence close to but not at sup ().
We thank K. Haule for providing us with the OCA impurity solver and for helpful discussions. Support from SFB 668, LEXI Hamburg and ETSF are acknowledged. MK gratefully acknowledges the hospitality of the Max-Planck-Institute in Halle (Saale).
### References
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37. See supplemental material for details.
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# Inverse trigonometric functions differentiation Calculator
## Get detailed solutions to your math problems with our Inverse trigonometric functions differentiation step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here!
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### Difficult Problems
1
Solved example of inverse trigonometric functions differentiation
$\frac{d}{dx}\left(\arcsin\left(4x^2\right)\right)$
2
Taking the derivative of arcsine
$\frac{1}{\sqrt{1-\left(4x^2\right)^2}}\frac{d}{dx}\left(4x^2\right)$
3
The power of a product is equal to the product of it's factors raised to the same power
$\frac{1}{\sqrt{1-16x^{4}}}\frac{d}{dx}\left(4x^2\right)$
4
The derivative of a function multiplied by a constant ($4$) is equal to the constant times the derivative of the function
$4\left(\frac{1}{\sqrt{1-16x^{4}}}\right)\frac{d}{dx}\left(x^2\right)$
5
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{8x}{\sqrt{1-16x^{4}}}$
## Final Answer
$\frac{8x}{\sqrt{1-16x^{4}}}$
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http://mathhelpforum.com/math-software/122394-derivative-using-wolfram-alpha.html | # Thread: Derivative using Wolfram Alpha
1. ## Derivative using Wolfram Alpha
Why is Wolfram Alpha calculating this wrong:
$f(u)=(arcsinu)^{-1}$ then what is $f'(\frac{1}{\sqrt2})=$
My calculator says $f'(\frac{1}{\sqrt2})=-2.99547$
Why does Wolfram Alpha say -2.29 instead?
2. Originally Posted by Kataangel
Why is Wolfram Alpha calculating this wrong:
$f(u)=(arcsinu)^{-1}$ then what is $f'(\frac{1}{\sqrt2})=$
My calculator says $f'(\frac{1}{\sqrt2})=-2.99547$
Why does Wolfram Alpha say -2.29 instead?
my calculator says -2.292636673
3. Originally Posted by Kataangel
Why is Wolfram Alpha calculating this wrong:
$f(u)=(arcsinu)^{-1}$ then what is $f'(\frac{1}{\sqrt2})=$
My calculator says $f'(\frac{1}{\sqrt2})=-2.99547$
Why does Wolfram Alpha say -2.29 instead?
Please explain exactly what you think you mean by $f(u)=(arcsinu)^{-1}$.
Also please give the exact command string you have given Wolfram Alpha.
CB | {"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": 10, "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": 0.9454335570335388, "perplexity": 2570.3827927597627}, "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-2017-17/segments/1492917122886.86/warc/CC-MAIN-20170423031202-00450-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://www.ias.ac.in/describe/article/pram/079/04/0875-0878 | • Search for new physics in dijet mass and angular distributions in $pp$ collisions at $\sqrt{s} = 7$ TeV measured with the ATLAS detector
• # Fulltext
https://www.ias.ac.in/article/fulltext/pram/079/04/0875-0878
• # Keywords
ATLAS; dijet distributions.
• # Abstract
We present a search for physics beyond the Standard Model in proton–proton collisions at a centre-of-mass energy of $\sqrt{s} = 7$ TeV, performed with the ATLAS detector at the Large Hadron Collider (LHC). No evidence for new physics is found in dijet mass and angular distributions and stringent limits are set on a variety of models of new physics, including excited quarks, quark contact interactions, axigluons, and quantum black holes.
• # Author Affiliations
1. Kirchhoff Institute for Physics, Heidelberg University, Im Neuenheimer Feld 227, 69120 Heidelberg, Germany
• # Pramana – Journal of Physics
Volume 94, 2020
All articles
Continuous Article Publishing mode
• # Editorial Note on Continuous Article Publication
Posted on July 25, 2019 | {"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": 0.9173325300216675, "perplexity": 3206.0130731836107}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2020-24/segments/1590347406785.66/warc/CC-MAIN-20200529214634-20200530004634-00342.warc.gz"} |
https://www.jiskha.com/questions/1502461/a-circular-disk-of-mass-0-4-kg-and-radius-36-cm-initially-not-rotating-slips-down-a-thin | # Physics
A circular disk of mass 0.4 kg and radius 36 cm, initially not rotating, slips down a thin spindle onto a turntable (disk) of mass 1.7 kg and the same radius, rotating freely at 3.4
a) Find the new angular velocity of the combination;
b) The change in the kinetic energy?
c) If the motor is switched on after the disk has landed, what is the constant torque needed to regain the original speed in 2.2 s?
I figured out a) and b), but I'm not sure what c) is asking or how to solve it.
1. 👍 0
2. 👎 0
3. 👁 229
1. 3.4 WHAT??
but hard to say. I will use that but if RPS or something, convert
I disk = (1/2)m r^2
so
Id = .5*.4*.36^2
and
It = .5 * 1.7 * .36^2
initial angular momentum = It*3.4
final angular momentum (the same of course)
= It omega +Id omega = omega(It+Id)
so
It * 3.4 = (It+Id)omega
but It =1.7/.4 * Id
so It = 4.25 Id
so
3.4 (4.25 Id) = (5.25 Id) omega
omega = 3.4 (4.25/5.25)
omega = 2.75 whatever your units are
----------------
initial Ke = .5 It(3.4)^2
final Ke = .5 (It+Id)(2.75)^2
-------------------
Torque = moment = change in I*omega /time
[just like force =change in m*v/t]
= (It+Id) (3.4-2.75) / 2.2
1. 👍 0
2. 👎 0
posted by Damon
2. Thanks Damon you're the real MVP!
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2. 👎 0
posted by Ana
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https://iwaponline.com/hr/article-abstract/26/2/73/644/Regional-Flood-Relationships-by-Nonparametric | Since some theoretical assumptions needed in linear regression are not always fulfilled in practical applications, nonparametric regression was investigated as an alternative method in regional flood relationship development. Simulation studies were developed to compare the bias, the variance and the root-mean-square-errors of nonparametric and parametric regressions. It was concluded that when an appropriate parametric model can be determined, parametric regression is preferred over nonparametric regression. However, where an appropriate model cannot be determined, nonparametric regression is preferred. It was found that both linear regression and nonparametric regression gave very similar regional relationships for annual maximum floods from New Brunswick, Canada. It was also found that nonparametric regression can be useful as a screening tool able to detect data deficient relationships.
This content is only available as a PDF. | {"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": 0.9177387952804565, "perplexity": 868.8872050175422}, "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-2019-09/segments/1550247495147.61/warc/CC-MAIN-20190220150139-20190220172139-00613.warc.gz"} |
https://experts.illinois.edu/en/publications/costs-of-energy-efficiency-mandates-can-reverse-the-sign-of-rebou | Costs of energy efficiency mandates can reverse the sign of rebound
Don Fullerton, Chi L. Ta
Research output: Contribution to journalArticlepeer-review
Abstract
Improvements in energy efficiency reduce the cost of consuming services from household cars and appliances and can result in a positive rebound effect that offsets part of the direct energy savings. We use a general equilibrium model to derive analytical expressions that allow us to compare rebound effects from a costless technology shock (CTS) to those from a costly energy efficiency standard (EES). We decompose each total effect on the use of energy into a direct efficiency effect, direct rebound effect, and indirect rebound effect. We show which factors determine the sign and magnitude of each. Rebound from a CTS is generally positive, as in prior literature, but we also show how a pre-existing EES can negate the direct energy savings from the CTS – leaving only the positive rebound effect on energy use. Then we analyze increased stringency of an EES, and we show exactly when the increased costs reverse the sign of rebound. Using plausible parameter values in this model, we find that indirect effects can outweigh the direct effects captured in partial equilibrium models, and that the total rebound from a costly EES can be negative.
Original language English (US) 104225 Journal of Public Economics 188 https://doi.org/10.1016/j.jpubeco.2020.104225 Published - Aug 2020
Keywords
• Costless technology improvement
• Energy efficiency standards
• Energy mandates
• General equilibrium
• Rebound
ASJC Scopus subject areas
• Finance
• Economics and Econometrics
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http://theoryapp.com/tag/dictator-function/ | ## Hastad Dictator Test for Boolean Functions
A function $$f\colon \{0,1\}^n\to \{0,1\}$$ is a dictator function if there exists $$i\in [n]$$ such that $$f(x) = x_i$$. That is, the function is completely determined by its $$i$$-th input. A function $$f\colon \{0,1\}^n\to \{ \pm 1\}$$ is a dictator
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https://www.physicsforums.com/threads/analysis-proof-help-s.13249/ | # Analysis Proof Help :S
1. Jan 25, 2004
### gimpy
Hi, im taking my first analysis course and we are studying Limits right now. My prof said they are the most important thing to remember out of this whole semester. Anyways i have two problems im trying to solve that i could do with some help.
1) Show that if $$f(x) \leq 0$$ and $$\lim_{x->a} f(x) = l$$, then $$l \leq 0$$.
2) If $$f(x) \leq g(x)$$ for all x, then $$\lim_{x->a} f(x) \leq \lim_{x->a} g(x)$$.
If those limits exist.
For number one i can see this is obvious but i don't know where to start to try and prove it.
I know the definitions for limits, do i use them somehow?
For number 1) i think that i can use a proof by contradiction somehow.
2. Jan 25, 2004
### Hurkyl
Staff Emeritus
Yah, using the definition of limit will probably be helpful. Proof by contradiction sounds like a good plan, let us know how far you get!
3. Jan 25, 2004
### gimpy
I have been working ont he second question:
2) If $$f(x) \leq g(x)$$ for all x, then $$\lim_{x->a} f(x) \leq \lim_{x->a} g(x)$$.
If those limits exist.
Suppose that $$f(x) \leq g(x)$$. Let $$h(x) = g(x)-f(x)$$. So $$h(x) \geq 0$$.
$$\lim_{x->a} h(x) = \lim_{x->a} g(x) - \lim_{x->a} f(x) \geq 0$$. Therefore $$\lim_{x->a} f(x) \leq \lim_{x->a} g(x)$$
Is this right?
Can someone give me a hint for question 1?
Start me off or something......
Last edited: Jan 25, 2004
4. Jan 25, 2004
### Hurkyl
Staff Emeritus
What is your justification for this?
Your hint for problem (1) is to assume $l > 0$ and rewrite the entire problem in terms of epsilons and deltas; exactly what you said you thought you should do.
5. Jan 26, 2004
### gimpy
Ok i had a good nights sleep went to school and came back tonight with a fresh mind to give it another shot. And i believe it worked :D
1) Show that if $$f(x) \leq 0$$ and $$\lim_{x->a} f(x) = l$$, then $$l \leq 0$$.
$$\forall\epsilon > 0, \exists\delta > 0 \ni$$ for all $$x$$ , $$0< |x - a| < \delta$$, then $$|f(x) - l| < \epsilon$$.
Assume that $$l > 0$$
$$\forall\epsilon > 0, \exists\delta > 0 \ni$$ for all $$x$$ , $$0< |x - a| < \delta$$, then $$l - f(x) < \epsilon$$.
since $$l - f(x) > 0 \implies 0 < l - f(x) < \epsilon$$
add $$f(x)$$ to each side to get $$f(x) < l < \epsilon + f(x)$$.
Choose $$\alpha$$ to be the distance from $$f(x)$$ to $$0$$. Then we choose $$\delta \ni \epsilon = \alpha$$, then $$\epsilon + f(x) = 0$$
therefore, $$f(x) < l < 0$$ which is a contradiction to $$l > 0$$.
Im sure this must be correct.
I will try number 2 now. Thanks for your hint ;)
6. Jan 26, 2004
### Hurkyl
Staff Emeritus
You're almost there: given the assumption $l > 0$, you've correctly proven
$$\forall \epsilon > 0 \exists \delta > 0 \forall x: 0 < |x - a| < \delta \implies f(x) < l < f(x) + \epsilon$$
however, the next steps are wrong. For instance, the distance from $f(x)$ to zero could, in fact, be zero. (e.g. take $f(x) = 0$, or $f(x) = -|x|$)
I'm also not sure what you're trying to do when you say "choose $\delta$ such that $\epsilon = \alpha$"... when deriving the contradiction you do get to choose $\epsilon$, but the contradiction must hold for all $\delta$.
You need another hint on this one?
As for number two, the proof you gave almost works, now that you've proven #1... see if you can rewrite it to take advantage of knowing #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": 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": 0.9632222652435303, "perplexity": 427.2893625796525}, "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-2018-47/segments/1542039746171.27/warc/CC-MAIN-20181119233342-20181120015342-00247.warc.gz"} |
http://stats.stackexchange.com/questions/2932/metric-spaces-and-the-support-of-a-random-variable/20769 | Metric spaces and the support of a random variable
Does the use of metric spaces to describe the support of a random variable provide any greater illumination? I ask this after reading about how metrics spaces have been used to unify the mathematical measure theoretic nature of probability and the physical intuition that most associate with probability. You can read my inspiration here: http://www.arsmathematica.net/archives/2009/02/14/complete-metric-spaces-and-the-interpretation-of-probability/
-
Here are some technical conveniences of separable metric spaces (a) If $X$ and $X'$ take values in a separable metric space $(E,d)$ then the event $\{X=X'\}$ is measurable, and this allows to define random variables in the elegant way: a random variable is the equivalence class of $X$ for the "almost surely equals" relation (note that the normed vector space $L^p$ is a set of equivalence class) (b) The distance $d(X,X')$ between the two $E$-valued r.v. $X, X'$ is measurable; in passing this allows to define the space $L^0$ of random variables equipped with the topology of convergence in probability (c) Simple r.v. (those taking only finitely many values) are dense in $L^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": 0.9515782594680786, "perplexity": 165.33461742574897}, "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-2013-20/segments/1368705305291/warc/CC-MAIN-20130516115505-00010-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/348881/wrong-answer-rewrite-rational-number-as-a-fraction | # Wrong Answer - Rewrite Rational Number as a Fraction.
This number 2.962962 can be rational
$$x=2.962962$$ $$10x=29.62962$$ $$100x=296.2962$$ $$1000x=2962.962$$ $$1000x-10x=\frac{990x}{990}=\frac{2933}{990}$$
why is this wrong? That way of getting the answer is how I was said to do it
Comment: $$1000x-x=\frac{999x}{999}=\frac{2960}{999}=?$$
-
The last line is wrong. If the decimal part is supposed to be repeating, I imagine it should read $1000x - x = 999x = 2960$. – Qiaochu Yuan Apr 2 '13 at 8:54
Your solution makes very less sense. You have made many direct mistakes. Check it once again. – lsp Apr 2 '13 at 8:55
@QiaochuYuan That can't be right because the answer has to be rational – user1838781 Apr 2 '13 at 8:56
@user: Can you explain your reasoning in that last comment? – Hurkyl Apr 2 '13 at 9:02
@Hurkyl it has to be a fraction = no decimals – user1838781 Apr 2 '13 at 9:06
$$1000x -x = 2962.962962\ldots - 2.962962\ldots = 2960.$$ The fractional part cancels out completely.
Then $$x = \frac{2960}{999} = \frac{37\cdot80}{37\cdot27} = \frac{80}{27}.$$
-
I assume you meant a repeating decimal like $2.\overline{962}$ (i.e. $2.9629629\ldots$) rather than one that terminates like you write.
Your work at the end is somewhat confused; I can't tell what you were trying to do. But the calculation of $1000x - 10x$ yields
$$\begin{matrix} 2&9&\not 6^5&{}^1 2&.&9&\not6^5&{}^1 2&9&\not6^5&{}^12&... \\ & & 2&9&.&6&2&9&6&2&9&... \\\hline \\ 2 & 9 & 3 & 3 & . & 3 & 3 & 3 & 3 & 3 & 3 & \cdots \end{matrix}$$
(I hope that is how they still notate subtraction these days) and so you have
$$1000 x - 10 x = 2933 + \frac{1}{3}$$
and
$$1000 x - 10 x = 990 x$$
and so we've derived
$$990 x = 2933 + \frac{1}{3}$$
Of course, it would have been easier to compare $1000x$ to $x$....
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Hint: there's a very easy way to transform periodic into decimals without any calculation. Here's an example: $$0.123\overline{4567}=\frac{1234567-123}{9999000}.$$ In your case it's just $$2\frac{962}{999}.$$
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http://math.stackexchange.com/questions/60945/understanding-the-musical-isomorphisms-in-vector-spaces/70101 | # Understanding the Musical Isomorphisms in Vector Spaces
I am trying to solidify my understanding of the muscial isomorphisms in the context of vector spaces. I believe I understand the definitions but would appreciate corrections if my understanding is not correct. Also, as I have had some difficulty tracking down related material, I would welcome any suggested references that would expand upon this material and related results.
So, let $V$ be a finite-dimensional vector space over $\mathbb{R}$ with inner product $\langle \cdot, \cdot \rangle$ and let $V^*$ denote its dual. For each element $v \in V$, one can define a mapping $V \rightarrow V^*$ by $v \mapsto \langle \cdot, v \rangle$. By the Riesz representation theorem this mapping actually determines an isomorphism that allows us to identify each element in $V$ with a unique functional in $V^*$ The "flat" operator $\flat$ is defined by by $v^{\flat}(u) = \langle u , v \rangle$ for all $u \in V$ and is thus just the Riesz isomorphism in the direction $V \rightarrow V^*$ as defined above. On the other hand, given a linear functional $f_v \in V^*$, we know that there exists a unique $v \in V$ such that $f_v(u) = \langle u , v \rangle$ for all $u \in V$ and the "sharp" operator $\sharp$ is defined by by $f_v^{\sharp} = v$ and this represents the other direction of the Riesz isomorphism.
Is this understanding correct? Can anyone provide a reference to some examples/exercises that would explore these operators in a concrete way? The Wikipedia page on the topic isn't of much help.
Update: I am adding a bounty to this question in hopes that someone will be able to provide examples/exercises (or references to such) that illustrate the use of the musical isomorphisms in the context of vector spaces.
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In finite dimensions, you don't need Riesz representation. By the nondegeneracy of the inner product, the map $v \mapsto \langle v,\cdot\rangle$ from $V\to V^*$ has rank $\dim V$, so is a bijection. Also, you need that you are working over real vector spaces for the map to be an isomorphism. Over $\mathbb{C}$, it is an anti-isomorphism (scalars get changed to complex conjugates). Otherwise, yeah, your understanding in the second paragraph is correct. – Willie Wong Aug 31 '11 at 13:38
In fact, in finite dimensions, given any non-degenerate bilinear form $B$ on a vector space $V$, you can define the musical isomorphisms relative to $B$; whereas the proof of Riesz representation usually uses that for an inner product: $\|u-v\|^2 = 0 \iff u = v$, which maybe false if $B$ has mixed signatures. – Willie Wong Aug 31 '11 at 13:45
@Willie Thanks for the clarifications; I have updated the question to specify that the vector space in question is over the reals. – ItsNotObvious Aug 31 '11 at 13:47
Here's a visual way to imagine $\flat$: let's say the underlying field were $\mathbb{N}$ (even though that's not a field, we'll extend what I say at the end to $\mathbb{R}$). Counting we think of $5$ as 5 stones: ⬤⬤⬤⬤⬤. But one could also think of $5$ as a line through the origin with the slope of 5. $f(x)=5x$ takes some input and does $f(■■■■)=5+5+5+5$ or $f(■■)=5+5$. These two visions biject insofar as any $r \in \mathbb{R}$ could be represented as a "regular number" or a sloped line. – isomorphismes May 13 '14 at 2:22
"Musical isomorphism" sounds like "catastrophe theory". But, irony aside, what you think and write is absolutely correct. (A minor point: You are not "given a functional $f_v\in V^*$", but you are given a functional $f\in V^*$, and the isomorphism in question guarantees you the existence of a unique $v\in V$ such that $f(u)=\langle v,u\rangle$ for all $u\in V$.)
Now it seems that you cannot make the connection with the Wikipedia page about the "musical isomorphism". This has to do with the way a scalar product has to be encoded when the given basis is not orthonormal to start with. In this respect one can say the following:
If $V$ is a real vector space provided with a scalar product $\langle\cdot,\cdot\rangle$, and if $(e_1,\ldots,e_n)$ is an arbitrary (i.e. not necessarily orthonormal) basis of $V$ then (a) any vector $x\in V$ is represented by a $(n\times 1)$ column vector $x\in{\mathbb R}^n$ such that $x=\sum_{k=1}^n x_k\> e_k$, and (b) there is a matrix $G=[g_{ik}]$ such that $$\langle x,y\rangle \ =\ x'\> G \> y\qquad \forall x,\forall y\ .$$ This implies that the functional $v^\flat$ appears in the form $$v^\flat(u)=\langle v,u\rangle=v'\> G\> u\ .$$ Putting $\sum_{i=1}^n v_i g_{ik}=: \bar v_k$ one therefore has $$v^\flat(u)=\sum_{k=1}^n \bar v_k\> u_k\ ,$$ so that one can interpret the coefficients $\bar v_k$ as "coordinates“ of the functional $v^\flat$.
Similar considerations apply to the operator $\sharp$, and going through the motions one finds that the inverse of the matrix $G$ comes into play. The details are as follows: A given functional $f\in V^*$ appears "coordinatewise" as $$f(u)\ =\ \sum_{k=1}^n f_k u_k\quad (u\in V)\ ,$$ where the coefficient vector $f=(f_1,\ldots, f_n)$ is some real $n$-tuple. Now you want to represent $f$ by means of the scalar product as a vector $f^\sharp\in V$ in the following way: We should have $$f(u)=\langle f^\sharp, u\rangle\qquad \forall u\in V\ ,$$ which in terms of matrix products means $$f'\> u\ =\ (f^\sharp)'\> G\>u\qquad\forall u\in{\mathbb R}^n$$ or $$f^\sharp =G^{-1}\>f\$$ (note that the matrix $G$ is symmetric). This formula gives you the coordinates of $f^\sharp$ with respect to the basis $(e_1,\ldots,e_n)$.
In old books you read about "covariant" and "contravariant" coordinates of one and the same vector (or functional) $v$ resp. $f$.
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I do complex differential geometry, so for me these things pop up in the context of finite dimensional linear algebra. I'll address your question in those terms.
For references, I've profited much from Lee's Riemannian Manifolds and Coffman's Trace, metric and reality, which is an amazing book that talks about these things in a coordinate free way, which is almost essential to understanding what is going on.
So, let's take a finite dimensional real vector space $V$, and equip it with an inner product $g$. Here $g(u,v) = \langle u, v \rangle$ in your notation.
Each vector space $V$ is canonically isomorphic to its double dual $V^{**}$, but it is not to its dual $V^*$. Thus if we want to transport vectors from $V$ to $V^*$ (we want to do this a lot, just note that the transpose of a vector is an element of the dual space) we need to choose an isomorphism $V \to V^*$.
The inner product $g$ is exactly the choice of such an isomorphism. More precisely, given $u$ in $V$ we send it to the linear form $v \mapsto g(u,v)$. We could also have sent $u$ to the linear form $v \mapsto g(v,u)$, but as $g$ is symmetric this gives the same form.
Suppose now that we have a vector space of the form $V^{\otimes p} \otimes V^{*\otimes q}$. An element of this space is often called a $(p,q)$ tensor. Pick one of the spaces $V$, say the first one. Then the inner product $g$ induces an isomorphism $V^{\otimes p} \otimes V^{*\otimes q} \to V^{\otimes p-1} \otimes V^{*\otimes q+1}$. In the literature, this map is called "lowering the index", or, in musical notation, the "flat" map.
The classical "raising of index", or the "sharp" map, is now just the inverse of the isomorphism $V \to V^*$ given by $g$. As we are working over finite dimensional spaces, so that $V^{**} = V$, it coincides with the dual inner product $V^* \to V^{**} = V$.
This is far from being (only) abstract nonsense. As one example, this point of view clarifies immensly the definition of positive-definitiveness of a matrix $A$. Recall that a matrix $A$ (in the canonical basis of $\mathbb R^n$) is positive-definite if ${}^t u A u > 0$ for all non-zero vectors $u$. Recalling that we have the standard inner product on $\mathbb R^n$, we re-interpret this condition as saying that $\langle Au, u \rangle > 0$ for all vectors $u$.
But now it is immediatly clear that this property depends on the choice of an inner product, and not a basis. So, let $g$ be an inner product on $V$. Then we say that a linear morphism $f : V \to V$ is $g$-positive-definite if the bilinear form (i.e. map $V \to V^*$) $g \circ f$ is positive-definite, in the sense that $g \circ f (u,u) = g(f(u),u) > 0$ for all non-zero $u$. This does however not necessarily give the same condition as setting $g \circ f(u,u) = g(u,f(u))$.
To get around this, note that usually we only talk about positive-definiteness for symmetric matrices $A$. Such a matrix represents a linear map $f : V \to V$, or in other words, an element of $V^* \otimes V$. This last space is auto-dual, so $f^*$ is again a linear map $V \to V$. Saying that $A$ is symmetric exactly means that $f^* = f$. We are now free to define a $g$-positive-definite linear map $f : V \to V$ as a symmetric map as above.
As another example, this point of view will make clear what "trace with respect to a metric" actually means. (Hint: The trace of a linear map is invariant of basis, and has nothing to do with an inner product. A 2-form however, that is something else. If only there was a way to convert a 2-form into a linear map.)
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Wow, I think Coffman's notes could probably be subtitled "Everything you wanted to know about the tensor product and commutative diagrams in linear algebra but were afraid to ask". I've never seen anything quite like them. – ItsNotObvious Oct 6 '11 at 11:32
Why is there a canonical isomorphism from $V \to V^{**}$ but not from $V \to V^*$? – isomorphismes May 13 '14 at 19:18
My copy of Spivak is in storage, unfortunately, but I think he discusses it in his Comprehensive Intro to Differential Geometry (volume 1, probably). I would also guess that Boothby discusses it in his intro to differentiable manifolds, but I am again not sure. I am almost certain that Spivak has a great exercise working through showing that this isomorphism is not canonical.
As far as I can tell, this question shows up in three places. The first is in Hilbert space theory, where Willie Wong's remark is important -- in particular, you end up with some differences between the Banach space adjoint of an operator and the Hilbert space adjoint. This caught me by surprise when I first studied this. Reed & Simon, Functional Analysis, volume 1, has a discussion of this.
The second context I have seen this in, is in Riemannian geometry. Here, you end up with an isomorphism between TM and T^*M induced by the metric. The third context is in symplectic geometry, where the symplectic form induces this isomorphism (not called a musical isomorphism anymore). This is one of the cases covered by Willie Wong's second comment. (Presumably also this shows up in Lorentzian geometry, but I haven't ever studied that myself.)
This isomorphism induced by the metric is quite important in Riemannian geometry -- it's what allows you to do the "raising and lowering of indices". This is why Riemannian geometry books often have discussions of these isomorphisms.
Another possible source for a discussion of this is in Wendl's notes on connections, Appendix A. http://www.homepages.ucl.ac.uk/~ucahcwe/connections.html
As for problems, I don't think any of these give you problems. However, I think there are some simple things you can compute directly that are worth playing around with.
Let $V$ be an $n$-dimensional real vector space with an inner product on it.
Choose a basis and use this to identify $V$ with $\mathbb{R}^n$. Use the dual basis to identify $V^*$ with $\mathbb{R}^n$.
(a) show that the inner product becomes where <,> denotes the standard inner product on $\mathbb{R}^n$, and $A$ is a matrix. Show $A$ is positive definite and symmetric. Also show that any positive definite symmetric matrix induces an inner product on $V$. (b) what does the "musical" isomorphism induced by the inner product look like as a map from $\mathbb{R}^n \to \mathbb{R}^n$, using the isomorphisms to $V$ and $V^*$ given by the basis/dual-basis isomorphisms? (c) How do these isomorphisms transform if you change basis for $V$ (and thus change dual basis for $V^*$)? (d) How about if you take some arbitrary basis for $V$ and $V^*$? (i.e. not necessarily dual bases) (e) Now consider how to represent a linear map from $W \to V$ and from $V \to W$, composed/precomposed with the musical isomorphism.
(f) The musical isomorphism allows you to put an inner product on $V^*$. What does its corresponding self-adjoint positive definite matrix look like?
Most of these are actually redundant, but once you've fiddled with this, I think you will understand the concept a lot better. I don't think it's the musical isomorphism that you don't understand, but the non-canonical nature of the isomorphism between $V$ and $V^*$.
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Nice exercises and the notes you pointed me to were also helpful – ItsNotObvious Oct 6 '11 at 11:30
We can go back to Euclidean space to get some concrete intuition for what's going on. A motivation for introducing differential forms comes in the context of work being done on a particle along a curve $\gamma$ with some force field present. Now, one learns in a manifolds course that what is essential for integration is only the fact that we have linear functionals on the tangent space at each point; physically, this means we know the amount of work needed to push the particle an infinitesimal amount in some tangent direction.
If we have an inner product, there's a very natural way to associate vector fields with such linear functionals. Thinking about this physically, imagine the particle on our curve with some tangent vector; we want to see how much the force field is pushing the particle in the direction of this tangent vector, and the most obvious thing to do is to look at the angle formed, which is defined using the Euclidean dot product. So given any vector field, we can use the dot product to get a linear way of assigning the amount of work required to push the particle in some direction.
If we want to formulate the physical intuition above in terms of Riemannian manifolds, it would look something like this: for any vector field $X$ on a Riemannian manifold $M$ with metric $\langle\cdot, \cdot\rangle_p$, then $\langle X_p, \cdot \rangle$ defines a linear functional on $T_p M$. This is exactly the canonical identification of a tangent space with it's cotangent space.
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https://www.physicsforums.com/threads/force-on-a-body-due-to-pressure-of-a-fluid.823140/ | # Force on a body due to pressure of a fluid
1. Jul 12, 2015
### better361
• If I placed a anti-symmetric object in water, and if it floats,is it possible for it to move left or right due to a force imbalance in calculating $$\int p * \vec{dA}$$.
• I know that if the object is a closed surface, we can apply the divergence theorem and because we also know how pressure changes with position ($\rho g z$), we can prove that the resultant force can only be in the z direction.
2. Jul 13, 2015
### Svein
Have you tried it in practice?
3. Jul 13, 2015
### better361
I don't think I have something that it very anti-symmetric and floats.
Ohh, and correction to my second statement, "I know I know that if the object is a closed surface, and if it is completely submerged, we can apply the....."
4. Jul 13, 2015
### GlenMedina
So which pressure are you talking here? Are we applying any pressure from above, equivalent to the formula mentioned?? Or is it the pressure inside?? It's that I'm not clear about the procedure here.
5. Jul 13, 2015
### Orodruin
Staff Emeritus
You can still apply the divergence theorem when the object floats. The only difference is that the pressure gradient changes at the water surface.
6. Jul 13, 2015
### A.T.
Don't you think, that after thousands of years of building boats, we would have found that self-propelling hull shape by now?
7. Jul 13, 2015
### better361
@GlenMedina Essentially, I am talking about the buoyant force.
@Orodruin Wait, so divergence theorem still holds even if the vector field is discontinuous in a sense?
@A.T. : I guess not, but I can only put this to rest with a mathematical or very thorough physical explanation . :)
8. Jul 13, 2015
### Orodruin
Staff Emeritus
Yes, you can show it for distributions in general. As long as you are careful with what you do with the discontinuities you will be fine. In this case the discontinuity is in the derivative of the pressure field and the integral is well defined without any further assumptions. It is worse for point charges and other types of singularities where the discontinuities essentially give you delta functions.
9. Jul 13, 2015
### better361
Ok, thanks! Also, can you link me to somewhere/some book that explains how to deal with discontinuities? I suspect it would just be in any regular multi book, but I just remember them saying when stuff isn't "nice", you can't apply the theorem.
10. Jul 13, 2015
### A.T.
How about considering the pressure forces from air and water separately? The two partial object surfaces aren't closed, but the missing parts are planar and horizontal, so they cannot introduce any horizontal force disbalance.
11. Jul 13, 2015
### Orodruin
Staff Emeritus
This is true for the theorem as it is usually presented. If you allow for distributions where things such as $\Delta \phi(\vec x) = - \delta^{(3)}(\vec x)$ are well defined, you will essentially be safe also in these cases.
Edit: The way this would be handled with the "usual" theorem would be to use the fact that the $\delta$ function is zero (and therefore nice) everywhere except at $\vec x = 0$ to rewrite integrals as an integral over a small surface around $\vec x = 0$.
12. Jul 13, 2015
### SteamKing
Staff Emeritus
Floating objects tend to try to find a point of stability, such that the buoyant force and the weight of the object line up to produce zero net moment. Once stability is achieved, the object is in equilibrium and remains motionless until some external force disturbs its equilibrium.
13. Jul 13, 2015
### better361
Do you mean the surface that intersects the horizontal plane at the surface? So if I had a sphere where half of it is submerged, the missing part would be the great circle around the equator?
14. Jul 15, 2015
### Khashishi
If you could have a boat that just accelerated due to equal pressure of the water on two asymmetric sides, then that would violate laws of thermodynamics.
15. Jul 15, 2015
### better361
How so?
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http://www.incompleteideas.net/book/ebook/node74.html | Next: 7.3 The Backward View Up: 7. Eligibility Traces Previous: 7.1 -Step TD Prediction Contents
# 7.2 The Forward View of TD()
Backups can be done not just toward any -step return, but toward any average of -step returns. For example, a backup can be done toward a return that is half of a two-step return and half of a four-step return: . Any set of returns can be averaged in this way, even an infinite set, as long as the weights on the component returns are positive and sum to 1. The overall return possesses an error reduction property similar to that of individual -step returns (7.2) and thus can be used to construct backups with guaranteed convergence properties. Averaging produces a substantial new range of algorithms. For example, one could average one-step and infinite-step backups to obtain another way of interrelating TD and Monte Carlo methods. In principle, one could even average experience-based backups with DP backups to get a simple combination of experience-based and model-based methods (see Chapter 9).
A backup that averages simpler component backups in this way is called a complex backup. The backup diagram for a complex backup consists of the backup diagrams for each of the component backups with a horizontal line above them and the weighting fractions below. For example, the complex backup mentioned above, mixing half of a two-step backup and half of a four-step backup, has the diagram:
The TD() algorithm can be understood as one particular way of averaging -step backups. This average contains all the -step backups, each weighted proportional to , where (Figure 7.3). A normalization factor of ensures that the weights sum to 1. The resulting backup is toward a return, called the -return, defined by
Figure 7.4 illustrates this weighting sequence. The one-step return is given the largest weight, ; the two-step return is given the next largest weight, ; the three-step return is given the weight ; and so on. The weight fades by with each additional step. After a terminal state has been reached, all subsequent -step returns are equal to . If we want, we can separate these terms from the main sum, yielding
(7.3)
This equation makes it clearer what happens when . In this case the main sum goes to zero, and the remaining term reduces to the conventional return, . Thus, for , backing up according to the -return is the same as the Monte Carlo algorithm that we called constant- MC (6.1) in the previous chapter. On the other hand, if , then the -return reduces to , the one-step return. Thus, for , backing up according to the -return is the same as the one-step TD method, TD(0).
We define the -return algorithm as the algorithm that performs backups using the -return. On each step, , it computes an increment, , to the value of the state occurring on that step:
(7.4)
(The increments for other states are of course , for all .) As with the -step TD methods, the updating can be either on-line or off-line.
The approach that we have been taking so far is what we call the theoretical, or forward, view of a learning algorithm. For each state visited, we look forward in time to all the future rewards and decide how best to combine them. We might imagine ourselves riding the stream of states, looking forward from each state to determine its update, as suggested by Figure 7.5. After looking forward from and updating one state, we move on to the next and never have to work with the preceding state again. Future states, on the other hand, are viewed and processed repeatedly, once from each vantage point preceding them.
The -return algorithm is the basis for the forward view of eligibility traces as used in the TD() method. In fact, we show in a later section that, in the off-line case, the -return algorithm is the TD() algorithm. The -return and TD() methods use the parameter to shift from one-step TD methods to Monte Carlo methods. The specific way this shift is done is interesting, but not obviously better or worse than the way it is done with simple -step methods by varying . Ultimately, the most compelling motivation for the way of mixing -step backups is that there is a simple algorithm--TD()--for achieving it. This is a mechanism issue rather than a theoretical one. In the next few sections we develop the mechanistic, or backward, view of eligibility traces as used in TD().
Example 7.2: -return on the Random Walk Task Figure 7.6 shows the performance of the off-line -return algorithm on the 19-state random walk task used with the -step methods in Example 7.1. The experiment was just as in the -step case except that here we varied instead of . Note that we get best performance with an intermediate value of .
Exercise 7.4 The parameter characterizes how fast the exponential weighting in Figure 7.4 falls off, and thus how far into the future the -return algorithm looks in determining its backup. But a rate factor such as is sometimes an awkward way of characterizing the speed of the decay. For some purposes it is better to specify a time constant, or half-life. What is the equation relating and the half-life, , the time by which the weighting sequence will have fallen to half of its initial value?
Next: 7.3 The Backward View Up: 7. Eligibility Traces Previous: 7.1 -Step TD Prediction Contents
Mark Lee 2005-01-04 | {"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": 0.8891461491584778, "perplexity": 1193.8718468860052}, "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/1656104628307.87/warc/CC-MAIN-20220705205356-20220705235356-00221.warc.gz"} |
http://www.show-my-homework.com/2012/10/inertia-moment-homework-3-103.html | # Inertia Moment (Homework 3-103)
1. A hollow sphere (M = 3 kg, R = 60 cm) is rolling down an inclined surface with a height of 2.0 meters and an inclined angle of 25 degrees.a) find the moment of inertia of the sphere.
b) find the speed of the center of the mass when it reaches the bottom of the hill.
c) find the rotational kinetic energy at the bottom
d) find the linear acceleration of the sphere
e) for the sphere to make pure rotation find the magnitude of the static frictional coefficient.
2. A hollow spherical ball is let go from the top of a track and goes around a circular loop (R = 60 cm).
a) What must be the minimum height of the track so that the ball will not fall off the track at the topmost point of the loop?
b) when the above condition is met, what is the tangential speed of the ball when it reaches the topmost point?
3. A bucket (m = 4 kg) is falling down a 25 meter deep well as shown in the figure. The cylindrical drum on which is winded the bucket rope has a mass of 28 kg and radius of 15 cm.
a) find the tension in the rope.
b) find the linear acceleration of the descending bucket.
c) find the angular acceleration of the drum
d) find the angular velocity of the cylinder when the bucket touches the bottom of the well.
1.
a) The moment of inertia of a hollow sphere of mass M and radius R is
$I_sphere = 2*M*R^2/3 = 2*3*0.6^2/3 = 0.72 kg*m^2$
b. The total energy of the sphere at up and at the bottom is the same. Up there is only potential energy, at the bottom there is kinetic energy of rotation +kinetic energy of the center of mass
$M*g*H = I*\omega^2/2 +M*V(cm)^2/2$, $\omega$ is the angular speed of the sphere
the contact point of the spere with the plane has instant speed zero, the diametrical opposite point has instant speed $v = \omega*R,$ hence the instant speed of the center of mass is
$V(cm) = v/2 =omega*R/2$
$\omega = 2*V(cm)/R$
$M*g*H = I*4*V(cm)^2/R^2/2 + M*V(cm)^2/2$
$M*g*H = [I*4/R^2/2+M/2]*V(cm)^2$
$M*g*H = [(2M/3)*2 +M/2]*V(cm)^2$
$9.81*2 =[(2/3)*2 +1/2]*v(cm)^2$
$19.62 =1.83*V(cm)^2$
$V(cm) =3.27 m/s$
c. Rotational kinetic energy is
$E = I*\omega^2/2 = I*4*V(cm)^2/R^2/2 =(2M/3)*2*V(cm)^2 =(2*3/3)*2*3.27^2 =42.77 J$
4. V(cm)^2 =2*a*d, where d is the sliding distance down the inclined plane
$d = H/sin(\alpha) =2/sin(25) =4.73 m$
$a = V(cm)^2/2/d =3.27^2/2/4.73 =1.13 m/s^2$
d. The total friction force is equal to the friction force
$M*a = \mu*M*g$
$\mu = a/g = 1.13/9.81 =0.115$
2.
The condition of not falling is
$m*v^2/R = m*g$ (centrifugal force is equal to the weight)
$V^2/R = g$, or $V =\sqrt{(R*g)} = \sqrt{(0.6*9.81)} = 2.42 m/s$
The energy at the top of the loop (kinetic + potential) is equal to the energy at the maximum height (only potential)
$m*v^2/2 +m*g*(2R) = m*g*H$
$H =(V^2/2/g + 2R) =2.42^2/2/9.81 +2*0.6 = 1.498 m$
3.
considering the drum a solid cylinder it has a moment of inertia
$I = m*R^2/2 =28*0.15^2/2 =0.315 kg*m^2$
for the bucket one can write
$F +m*a = m*g$ (F is the force in the string)
For the drum one can write
$F = I*\epsilon$ (epsion is the angular acceleration of the drum)
$a =\epsilon*R$
Hence
$F = m*(g-a)$
$F= I*a/R$
or $I*a/R =mg-ma$
$(I/R+m)a =m*g$
$(0.315/0.15 +4)*a =4*9.81$
$a =6.43 m/s^2$
the force in the rope is
$F =m(g-a)= 4*(9.81-6.43) =13.52 N$
angular acceleration $\epsilon = a/R =6.43/0.15 =42.86 1/s^2$
tangential velocity of the cylinder is
$v= \sqrt{(2*a*s)} =\sqrt{(2*6.43*25)} = 17.93 m/s$
angular velocity is
$\omega =v/R =17.93/0.15 =119.54 1/s^2$ | {"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": 0.956141471862793, "perplexity": 415.89928760275285}, "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/1529267862929.10/warc/CC-MAIN-20180619115101-20180619135101-00501.warc.gz"} |
http://math.stackexchange.com/users/76798/mare?tab=summary | # Mare
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# 26 Questions
7 Given three integers in $\{0,\ldots,100\}$ which sum up to $100$. What is the probabilty that two of them are the same? 4 Is uncountable subset of separable space separable? 4 Cardinal arithmetic questions 3 Aproximating rational with fraction with “smallest numerator and denumerator possible” 3 When does equality holds in $A\subseteq P(\cup A)$
# 359 Reputation
+5 Is this complete partial order? +5 Proving a set is language generated by given grammar +5 Is $\alpha + \omega$ limit ordinal +5 About described DFA
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# 2 Accounts
Mathematics 359 rep 7 MathOverflow 101 rep 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": 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": 0.804081916809082, "perplexity": 4247.809130564744}, "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-2013-48/segments/1386164034642/warc/CC-MAIN-20131204133354-00067-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://www.nag.com/numeric/MB/manual64_24_1/html/F07/f07cef.html | Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
# NAG Toolbox: nag_lapack_dgttrs (f07ce)
## Purpose
nag_lapack_dgttrs (f07ce) computes the solution to a real system of linear equations AX = B $AX=B$ or ATX = B ${A}^{\mathrm{T}}X=B$, where A $A$ is an n $n$ by n $n$ tridiagonal matrix and X $X$ and B $B$ are n $n$ by r $r$ matrices, using the LU $LU$ factorization returned by nag_lapack_dgttrf (f07cd).
## Syntax
[b, info] = f07ce(trans, dl, d, du, du2, ipiv, b, 'n', n, 'nrhs_p', nrhs_p)
[b, info] = nag_lapack_dgttrs(trans, dl, d, du, du2, ipiv, b, 'n', n, 'nrhs_p', nrhs_p)
## Description
nag_lapack_dgttrs (f07ce) should be preceded by a call to nag_lapack_dgttrf (f07cd), which uses Gaussian elimination with partial pivoting and row interchanges to factorize the matrix A $A$ as
A = PLU , $A=PLU ,$
where P $P$ is a permutation matrix, L $L$ is unit lower triangular with at most one nonzero subdiagonal element in each column, and U $U$ is an upper triangular band matrix, with two superdiagonals. nag_lapack_dgttrs (f07ce) then utilizes the factorization to solve the required equations.
## References
Anderson E, Bai Z, Bischof C, Blackford S, Demmel J, Dongarra J J, Du Croz J J, Greenbaum A, Hammarling S, McKenney A and Sorensen D (1999) LAPACK Users' Guide (3rd Edition) SIAM, Philadelphia http://www.netlib.org/lapack/lug
## Parameters
### Compulsory Input Parameters
1: trans – string (length ≥ 1)
Specifies the equations to be solved as follows:
trans = 'N'${\mathbf{trans}}=\text{'N'}$
Solve AX = B$AX=B$ for X$X$.
trans = 'T'${\mathbf{trans}}=\text{'T'}$ or 'C'$\text{'C'}$
Solve ATX = B${A}^{\mathrm{T}}X=B$ for X$X$.
Constraint: trans = 'N'${\mathbf{trans}}=\text{'N'}$, 'T'$\text{'T'}$ or 'C'$\text{'C'}$.
2: dl( : $:$) – double array
Note: the dimension of the array dl must be at least max (1,n1)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$.
Must contain the (n1)$\left(n-1\right)$ multipliers that define the matrix L$L$ of the LU$LU$ factorization of A$A$.
3: d( : $:$) – double array
Note: the dimension of the array d must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
Must contain the n$n$ diagonal elements of the upper triangular matrix U$U$ from the LU$LU$ factorization of A$A$.
4: du( : $:$) – double array
Note: the dimension of the array du must be at least max (1,n1)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-1\right)$.
Must contain the (n1)$\left(n-1\right)$ elements of the first superdiagonal of U$U$.
5: du2( : $:$) – double array
Note: the dimension of the array du2 must be at least max (1,n2)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}-2\right)$.
Must contain the (n2)$\left(n-2\right)$ elements of the second superdiagonal of U$U$.
6: ipiv( : $:$) – int64int32nag_int array
Note: the dimension of the array ipiv must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
Must contain the n$n$ pivot indices that define the permutation matrix P$P$. At the i$i$th step, row i$i$ of the matrix was interchanged with row ipiv(i)${\mathbf{ipiv}}\left(i\right)$, and ipiv(i)${\mathbf{ipiv}}\left(i\right)$ must always be either i$i$ or (i + 1)$\left(i+1\right)$, ipiv(i) = i${\mathbf{ipiv}}\left(i\right)=i$ indicating that a row interchange was not performed.
7: b(ldb, : $:$) – double array
The first dimension of the array b must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array must be at least max (1,nrhs)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{nrhs}}\right)$
The n$n$ by r$r$ matrix of right-hand sides B$B$.
### Optional Input Parameters
1: n – int64int32nag_int scalar
Default: The first dimension of the array b The dimension of the arrays d, ipiv.
n$n$, the order of the matrix A$A$.
Constraint: n0${\mathbf{n}}\ge 0$.
2: nrhs_p – int64int32nag_int scalar
Default: The second dimension of the array b.
r$r$, the number of right-hand sides, i.e., the number of columns of the matrix B$B$.
Constraint: nrhs0${\mathbf{nrhs}}\ge 0$.
ldb
### Output Parameters
1: b(ldb, : $:$) – double array
The first dimension of the array b will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array will be max (1,nrhs)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{nrhs}}\right)$
ldbmax (1,n)$\mathit{ldb}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
The n$n$ by r$r$ solution matrix X$X$.
2: info – int64int32nag_int scalar
info = 0${\mathbf{info}}=0$ unless the function detects an error (see Section [Error Indicators and Warnings]).
## Error Indicators and Warnings
info = i${\mathbf{info}}=-i$
If info = i${\mathbf{info}}=-i$, parameter i$i$ had an illegal value on entry. The parameters are numbered as follows:
1: trans, 2: n, 3: nrhs_p, 4: dl, 5: d, 6: du, 7: du2, 8: ipiv, 9: b, 10: ldb, 11: info.
It is possible that info refers to a parameter that is omitted from the MATLAB interface. This usually indicates that an error in one of the other input parameters has caused an incorrect value to be inferred.
## Accuracy
The computed solution for a single right-hand side, $\stackrel{^}{x}$, satisfies an equation of the form
(A + E) x̂ = b , $(A+E) x^=b ,$
where
‖E‖1 = O(ε)‖A‖1 $‖E‖1 =O(ε)‖A‖1$
and ε $\epsilon$ is the machine precision. An approximate error bound for the computed solution is given by
(‖x̂ − x‖1)/(‖x‖1) ≤ κ(A) (‖E‖1)/(‖A‖1) , $‖ x^-x ‖1 ‖ x ‖1 ≤ κ(A) ‖E‖1 ‖A‖1 ,$
where κ(A) = A11 A1 $\kappa \left(A\right)={‖{A}^{-1}‖}_{1}{‖A‖}_{1}$, the condition number of A $A$ with respect to the solution of the linear equations. See Section 4.4 of Anderson et al. (1999) for further details.
Following the use of this function nag_lapack_dgtcon (f07cg) can be used to estimate the condition number of A $A$ and nag_lapack_dgtrfs (f07ch) can be used to obtain approximate error bounds.
The total number of floating point operations required to solve the equations AX = B $AX=B$ or ATX = B ${A}^{\mathrm{T}}X=B$ is proportional to nr $nr$.
The complex analogue of this function is nag_lapack_zgttrs (f07cs).
## Example
```function nag_lapack_dgttrs_example
trans = 'No transpose';
dl = [0.8823529411764706;
0.01960784313725495;
0.1400560224089636;
-0.01479925303454714];
d = [3.4;
3.6;
7;
-6;
-1.015373482726424];
du = [2.3;
-5;
-0.9;
7.1];
du2 = [-1;
1.9;
8];
ipiv = [int64(2);3;4;5;5];
b = [2.7, 6.6;
-0.5, 10.8;
2.6, -3.2;
0.6, -11.2;
2.7, 19.1];
[bOut, info] = nag_lapack_dgttrs(trans, dl, d, du, du2, ipiv, b)
```
```
bOut =
-4.0000 5.0000
7.0000 -4.0000
3.0000 -3.0000
-4.0000 -2.0000
-3.0000 1.0000
info =
0
```
```function f07ce_example
trans = 'No transpose';
dl = [0.8823529411764706;
0.01960784313725495;
0.1400560224089636;
-0.01479925303454714];
d = [3.4;
3.6;
7;
-6;
-1.015373482726424];
du = [2.3;
-5;
-0.9;
7.1];
du2 = [-1;
1.9;
8];
ipiv = [int64(2);3;4;5;5];
b = [2.7, 6.6;
-0.5, 10.8;
2.6, -3.2;
0.6, -11.2;
2.7, 19.1];
[bOut, info] = f07ce(trans, dl, d, du, du2, ipiv, b)
```
```
bOut =
-4.0000 5.0000
7.0000 -4.0000
3.0000 -3.0000
-4.0000 -2.0000
-3.0000 1.0000
info =
0
``` | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 90, "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": 0.9880006909370422, "perplexity": 4760.383318884872}, "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-2015-06/segments/1422115857200.13/warc/CC-MAIN-20150124161057-00038-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/cycloid-motion-of-electron-in-perpendicular-e-and-b-field.614936/ | # Homework Help: Cycloid motion of electron in perpendicular E and B field
1. Jun 18, 2012
### bobred
1. The problem statement, all variables and given/known data
An infinite metal plate occupies the xz-plane. The plate is kept at zero potential. Electrons are liberated from the plate at y = 0. The initial velocity of the electrons is negligible. A uniform magnetic field B is maintained parallel to the plate in the positive z-direction and a uniform electric field E is maintained perpendicular to the plate in the negative y-direction. The electric field is produced by a second infinite plate parallel to the first plate, maintained at a constant positive voltage $V_{0}$ with respect to the first plate. The separation of the plates is $d$. Show that the electron will miss the plate at $V_{0}$ if
$d>\sqrt{\frac{2mV_{0}}{eB^2}}$
2. Relevant equations
$v_{x}=\frac{E}{B}\left(1-\cos\left(\frac{qB}{m}t\right)\right)$
$v_{y}=\frac{E}{B}\sin\left(\frac{qB}{m}t\right)$
$v_{z}=0$
3. The attempt at a solution
I know this produces a cycloid travelling in the minus x direction. If $r$ is the radius of a rolling circle then $d>2r$ to miss. I think I should be using conservation of energy but dont know the form of the velocity. I am assuming the perpendicular velocity will be the sum of a transverse and rotational velocity?
2. Jun 18, 2012
### TSny
You might try integrating the expression for vy with respect to time to get an expression for y as a function of time. Choose the constant of integration to match the initial condition for y. Then examine the expression.
I don't see an easy way to use energy conservation.
3. Jun 18, 2012
### collinsmark
Oooh, nice problem.
You can use conservation of energy to solve this problem. Well, that and the work-energy theorem. Conservation of energy makes this problem a lot easier. Here are a few things that are noteworthy (you can call them hints if you like):
1) The magnetic forces always acts in a direction perpendicular to the electron's velocity. In other words, the magnetic force never causes the electron's speed to increase or decrease, it only changes the direction. Still in other words, the magnetic force does no work on the electron.
2) You're going to have to determine the maximum speed of the electron. But there are couple of tricks you can do to make it simpler, if you choose to use them. When the electron is at its maximum speed, which direction is going? What's the maximum value of [1-cos(x)]?
3) You'll need to determine a relationship between E and V0, but that should be pretty simple.
4. Jun 18, 2012
### TSny
Ah, nice. I now see that using conservation of energy is a good way to get the result.
My suggestion of integrating vy to get y as a function of time also gets the answer in short order. But I like the energy approach. Thanks.
5. Jun 19, 2012
### bobred
Hi, thanks for the replies.
Part of the question before asked for the expressions of $x(0)=0$ and $y(0)=0$ giving
$x=\frac{E}{B}t-\frac{Em}{qB^{2}}\sin\left(\frac{qB}{m}t\right)$
$y=\frac{Em}{qB^{2}}\left(1-\cos\left(\frac{qB}{m}t\right)\right)$
The expression $\left(1-\cos\left(\frac{qB}{m}t\right)\right)$ at maximum is 2 so $y$ has a max of
$y=\frac{2Em}{qB^{2}}$ and $E=V_{0}/d$ so
$y=\frac{2mV_{0}}{qdB^{2}}$
I keep going around in circles with this.
6. Jun 19, 2012
### TSny
You're essentially there Just interpret what you got. The electron will barely reach the plate if y-max equals what value? Put this value of y into your result and solve for d.
7. Jun 19, 2012
### collinsmark
I think you're on the right track.
What's the magnitude of the electron's charge q? (As in terms of e)?
The variable y is a measure of length (well, technically displacement in the y direction, but that's still a measure of length). What is the value of y when it is at its maximum? (I.e. what's the significance of ymax = d?)
[Edit: TSny beat me to the hint.]
Last edited: Jun 19, 2012
8. Jun 19, 2012
### bobred
Hi
Sorry, went back to the start and had a look at the Lorentz force equations and worked forward from there and using conservation of energy to get the result.
Thanks again. | {"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": 0.8422085046768188, "perplexity": 392.63703511324957}, "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-2018-39/segments/1537267156423.25/warc/CC-MAIN-20180920081624-20180920101624-00494.warc.gz"} |
https://www.physicsforums.com/threads/need-help-solving-initial-value-differential-equation.253552/ | # Need Help Solving Initial Value Differential Equation
1. Sep 3, 2008
### swooshfactory
1. The problem statement, all variables and given/known data
dy/dx + y*cos(x) = 3*cos(x) . Find the particular solution of the differential equation dy/dx + y*cos(x) = 3*cos(x) satisfying the initial condition y(0)=5.
y(x) = ________________________
2. Relevant equations
3. The attempt at a solution
I thought I had done it the right way, but my computer based homework system disagreed.
Heres my work:
p(t)=cos(x)
g(t)=3cos(x)
m: mu
m(t)=exp [int] cos(x)dx
m(t)= e^sin(x)
then,
e^sinx(y)= [int]3cosx(e^sinx)dx
[int]3cosx(e^sinx)dx= 3(e^sinx)+C
then to solve for y,
y= 3(e^sin(x)+C)/(e^sin(x))
using the intial value to solve for C, [[the initial value was y(0)=5]]
5=3(e+C)/e
5e=3e + C<<<right here is it C or 3c?>>
c=2e
and plugging it back into the y-equation, I get y=(3(e^sinx)+2e)/(e^sinx)
Where have I gone wrong? My homework system says this is wrong... Any help would be greatly appreciated. Thank you for looking.
2. Sep 3, 2008
### Dick
sin(0)=0 and e^(0)=1 not e. | {"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": 0.8998998999595642, "perplexity": 3079.917627398253}, "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/1521257647649.70/warc/CC-MAIN-20180321121805-20180321141805-00066.warc.gz"} |
http://math.stackexchange.com/questions/208976/what-prime-number-generating-algorithms-are-used?answertab=votes | # What prime number generating algorithms are used?
You sometimes hear bout these huge prime numbers (RSA prime number challenge comes to mind) and I was curious about what algorithms or formulas prime-number generators use in practice ? For example in cluster / cloud computing, parallel computing ...etc.
I imagine that everyone has their own "custom" optimized versions of more known prime number generational algorithms but what are the "base methods" used to do so.
Thank you very much in advance !
edit: I hope that I used the right [tags].
-
Suppose we are interested in producing a prime near a very large number $M$. It is convenient to not bother testing for primality the multiples of small primes, like the even numbers! For the remaining numbers, test, one after the other, candidates using a "probable prime" test such as Miller-Rabin.
Miller-Rabin is very fast, and can be used to quickly rule out candidates.
If we are not too fussy, we can accept such a probable prime as prime. There is a possibility of being wrong, but if we make the probability of error less than say $10^{-100}$, effectively we have certainty.
Or else, after locating a number which is "almost certainly prime" via Miller-Rabin, one can use one of the relatively expensive primality tests to make sure.
-
Thank you very much for both of your answers ! – Awake Zoldiek Oct 8 '12 at 0:40
As far as I know, the AKS primality test is the most effective known method to decide if a large number picked at random is prime or not.
If we pick a large number $N$, the probability it is prime is about $\frac{1}{\ln (N)}$. Thus, if you look for a prime number with $t$ digits, if you pick about $t \log(10)$ $t$-digit numbers at random, there is a pretty big chance one will be prime...The time involved is usually reasonable for a computer....
-
AKS is seldom used in practice. When it comes down to it, algorithms that were present before it - especially the Miller-Rabin algorithm and its variants - are much faster, even if AKS is asymptotically better (or just deterministic). – Yoni Rozenshein Oct 7 '12 at 23:55
Neither of the algorithms mentioned in the two previous answers are used to find record large primes. Lucas-Lehmer has the advantage that it is as fast as Miller-Rabin, and deterministic (it guarantees primality/compositeness rather than being probabilistic). However it is severely limited in that it only works on numbers of the form $2^n-1$ (Mersenne numbers). All of the last dozen or so record primes have been Mersenne primes located by massive volunteer effort and verified by Lucas-Lehmer test.
As far as I know, RSA has no prime number challenge, but it does have factoring challenges (which require considerably different algorithms). Are you perhaps thinking of the EFF cooperative computing awards for finding large primes?
-
yes, I kind of mistakenly took the RSA challenge for the EFF one. Sorry about that. – Awake Zoldiek Oct 8 '12 at 10:12
but shouldn't better to generate primes AT RANDOM ??¿ i mean if i were a 'hacker' i first would study all the prime generating algorithms to crack codes :D :D so perhaps a prime-generating algorithm is not quite safe. – Jose Garcia Oct 8 '12 at 10:45
@JoseGarcia My point is that the method for generating primes depends on your intent. It could be for cryptography, for amusement, for record-breaking, etc. And the standards for testing could vary in strictness (probabilistic vs provable primes). – Erick Wong Oct 8 '12 at 14:12 | {"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": 0.8422048687934875, "perplexity": 853.4148025896606}, "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-2014-41/segments/1412037663167.8/warc/CC-MAIN-20140930004103-00371-ip-10-234-18-248.ec2.internal.warc.gz"} |
http://www.reference.com/browse/ADM+formulation | Definitions
The ADM Formalism developed by Arnowitt, Deser and Misner is a Hamiltonian formulation of general relativity.
The formalism supposes that spacetime is foliated into a family of spacelike surfaces $Sigma_t$, labeled by their time coordinate $t$, and with coordinates on each slice given by $x^i$. The dynamic variables of this theory are taken to be the metric of three dimensional spatial slices $gamma_\left\{ij\right\}\left(t,x^k\right)$ and their conjugate momenta $pi^\left\{ij\right\}\left(t,x^k\right)$. Using these variables it is possible to define a Hamiltonian, and thereby write the equations of motion for general relativity in the form of Hamilton's equations.
In addition to the twelve variables $gamma_\left\{ij\right\}$ and $pi^\left\{ij\right\}$, there are four Lagrange multipliers: the lapse function, $N$, and components of shift vector field, $N_i$. These describe how each of the "leaves" $Sigma_t$ of the foliation of spacetime are welded together. These variables are nondynamical, and their "equations of motion" are constraint equations that enforce invariance under time reparameterizations and coordinate changes on the spatial slices, respectively.
Using the ADM formulation, it is possible to attempt to construct a quantum theory of gravity, in the same way that one constructs the Schrödinger equation corresponding to a given Hamiltonian in quantum mechanics. That is, replace the canonical momenta $pi^\left\{ij\right\}\left(t,x^k\right)$ by functional differential operators
$pi^\left\{ij\right\}\left(t,x^k\right) to -i frac\left\{delta\right\}\left\{delta gamma_\left\{ij\right\}\left(t,x^k\right)\right\}$
This leads to the Wheeler-deWitt equation.
Search another word or see ADM formulationon Dictionary | Thesaurus |Spanish | {"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": 0.8391144275665283, "perplexity": 372.14510275991034}, "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-2014-42/segments/1414119646554.36/warc/CC-MAIN-20141024030046-00315-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/90121-del-operator-divergence-curl.html | # Thread: del operator, divergence , and the curl
1. ## del operator, divergence , and the curl
In physics the del operator is fundamental.The basic laws of Electromagnetism are defined using it.I wanted to know the actual significance of the curl ( $\nabla \times$)and divergence.I also intended this post to be a starter for a discussion about mathematical fields,their difference and applications(in physics or other subjects).
2. One reason why $\nabla$ is so important is that it is invariant under rigid motions: rotating or translating the coordinate system does not change the form of $\nabla$. Roughly speaking, the "divergence", $\nabla\cdot \vec{v}$, measures how much a vector field "spreads out" (diverges) while the "curl", $\nabla\times\vec{v}$ measure its tendency to rotate- hence the names "divergence" and "curl". | {"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": 5, "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": 0.9364268183708191, "perplexity": 1293.56320633441}, "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/1500549424884.51/warc/CC-MAIN-20170724142232-20170724162232-00647.warc.gz"} |
https://www.coursehero.com/file/71478352/%D8%AD%D8%B3%D8%A7%D8%A8-%D8%A7%D9%84%D8%AA%D9%83%D8%A7%D9%85%D9%84-%D8%A7%D9%84%D8%BA%D8%A7%D9%85%D8%AF%D9%8Apdf/ | # u062du0633u0627u0628...
• 207
• 100% (1) 1 out of 1 people found this document helpful
This preview shows page 1 out of 207 pages. | {"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": 0.8922795653343201, "perplexity": 3060.2357198404047}, "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-2021-43/segments/1634323585380.70/warc/CC-MAIN-20211021005314-20211021035314-00428.warc.gz"} |
https://brilliant.org/problems/all-even-2/ | # All Even
If $$\dfrac{n}{2}$$ is an even number, then $$n(n^{2}-1)$$ must be divisible by which of the following numbers?
× | {"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": 0.9663919806480408, "perplexity": 150.86317572312913}, "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-2019-18/segments/1555578529962.12/warc/CC-MAIN-20190420180854-20190420202854-00379.warc.gz"} |
https://hiof.brage.unit.no/hiof-xmlui/browse?value=Kvaal,%20Knut&type=author | Viser treff 1-1 av 1
• Evaluating Flood Exposure for Properties in Urban Areas Using a Multivariate Modelling Technique
(Journal article; Peer reviewed, 2017)
Urban flooding caused by heavy rainfall is expected to increase in the future. The main purpose of this study was to investigate the variables characterizing the placement of a house, which seem to have an impact when 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": 0.8331400156021118, "perplexity": 1678.2707308822608}, "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-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00113.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/138212-homomorphism-a4.html | # Math Help - homomorphism from A4
1. ## homomorphism from A4
Show that
(1) there is no homomorphism from A4 onto a group of order 2, 4, or 6
(2) there is a homomorphism from A4 onto a group of order 3.
Can anyone please give me some hints ?
2. Let's see some work for the first one.
Originally Posted by xmcestmoi
(2) there is a homomorphism from A4 onto a group of order 3.
Can anyone please give me some hints ?
How much do you know about $A_4$? A (moderately) common fact is that $A_4$ has only one subgroup of order four (namely $\{e, (12)(34), (13)(24), (14)(23)\}=N$). But, it is fairly easy to prove that since $N$ is the only subgroup of order four and conjugation by any element of that subgroup (i.e. $gNg^{-1}$) is another subgroup of order four that $N$ must be invariant under conjugation. Thus, $N\unlhd A_4$. So, the canonical homomorphism is $\theta:G\to G/N$ given by $g\mapsto gN$. This is a surjective homomorphism and $|G/N|=[G:N]=\frac{12}{4}=3$
3. Thank you !!
4. Now I'm curious. How do you know there's no homorphism onto groups with order 4 or 6? | {"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": 10, "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": 0.9585838317871094, "perplexity": 213.8724112171301}, "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/1448398461529.84/warc/CC-MAIN-20151124205421-00006-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.therightgate.com/line-integration-electromagnetics/ | Line Integration in Electromagnetics | Definition and Significance
# What is Line Integration in Electromagnetics?
Almost all the Science Pupil are well aware of the term Integration. It is a common term in Mathematics. It also plays a very important role in the theory of Electromagnetics. In this article, I am going to discuss the formal meaning of integration and specifically what is Line Integration and its significance in the Electromagnetics.
In Mathematics, formally the Integration gives the area under the curve. Consider the integration shown in the figure below.
The differential or infinitesimal length is “dx” and corresponding to this small length an infinitesimal rectangular strip is shown. The area of this strip can be considered as multiplication of the value of f(x) at that x and length dx. In the process of integration, we add all such infinitesimal strips from A to B to give the area under the curve.
In Electromagnetics, we deal with various vector fields like Electric field, Magnetic field etc. Consider any such vector field E as shown in the following figure. Say, a line or vector AB is present within that field.
If we want the product of the length of the vector and component of the field along this vector (line) at every point of line then we write the integration as follows: –
This integration is termed as a line integration.
## The significance of Line Integration in Electromagnetics
The formula for work in Electromagnetics is based on the concept of line integration. Say we are moving a charge in the electric field, then work done is calculated by
Basically, the line integration is required when we want to calculate the component or value of any vector field along the given vector path as in Ampere’s law.
## Different types of Differential lengths
In Electromagnetics we deal with Cartesian Coordinate System, Cylindrical Coordinate System, and Spherical Coordinate System. So the infinitesimal length “dl” should be defined in each of the coordinate systems. As we deal with vectors and vector fields in Electromagnetics, “dl” consists of infinitesimal change along each axis of the respective coordinate system. | {"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": 0.9746731519699097, "perplexity": 308.70858779601105}, "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-2021-43/segments/1634323585522.78/warc/CC-MAIN-20211022212051-20211023002051-00601.warc.gz"} |
http://mathhelpforum.com/pre-calculus/110639-solving-trig-equation.html | # Math Help - Solving a trig equation?
1. ## Solving a trig equation?
I need to solve the following trig expression:
4cos4θ=0 (reads four cosine four theta)
Its the formula of a graph but i need to find the zeros or the points that it passes through the pole (its a polar graph)
Would i use the double angle formula or find when cos=0 and multiply the period by 4? Im confused
2. I think you are making this more difficult than it really is....
4cos4θ=0
the four is insignificant (because 0/4 is still 0). so now we have:
cos4θ=0.
I think the 4 is confusing you so let x=4θ so now we have:
cosx=0
and the solutions to that is when x=pi/2 or 3pi/2
now substitute 4θ in for x and you have
4θ=pi/2 and 4θ=3pi/2
so dividing by four yields:
θ=pi/8 and θ=3pi/8
Hope that helped!!
3. Also what domain are you looking for solutions in?
The previous post gives the solutions on $x \in \left[0,\frac{\pi}{2}\right]$
You can find all solutions as $x = \frac{(2n+1)\pi}{8}, n\in \mathbb{Z}$ | {"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": 0.9900264143943787, "perplexity": 1644.2391084570995}, "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-2015-32/segments/1438042990112.92/warc/CC-MAIN-20150728002310-00179-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://mathhelpboards.com/threads/sequence-with-recursive-definition.2367/ | # sequence with recursive definition?
#### skatenerd
##### Active member
Oct 3, 2012
114
Sorry to spam my problems all over this forum but series have me struggling somewhat. Last problem on my homework is the sequence an defined recursively by:
a1=1
and
an+1= $$\frac{a_n}{2}$$ + $$\frac{1}{a_n}$$
First part was the only part i know how to do. it was to find an for n=1 through 5.
However this next part has me stumped. Assume that:
The limit as n approaches infinity = alpha > 0
Obtain the value alpha by taking the limit of both sides of (1) [(1) is the info given in the beginning].
Now I can already tell this limit is going to $$\sqrt{2}$$ because it is kind of hinted to in the later parts of this problem. However I am kind of confused as to how to approach doing this limit of both sides of an+1= $$\frac{a_n}{2}$$ + $$\frac{1}{a_n}$$ . How would I evaluate this limit when the right side of the equation is in terms of an?
#### MarkFL
Staff member
Feb 24, 2012
13,775
One way (though not as directed by the problem) to demonstrate the sequence converges to $\displaystyle \sqrt{2}$ is to observe that if we begin with the function:
$\displaystyle f(x)=x^2-2=0$
which we know has the positive root $\displaystyle \sqrt{2}$ and apply Newton's method:
$\displaystyle f'(x)=2x$
$\displaystyle x_{n+1}=x_n-\frac{x_n^2-2}{2x_n}$
$\displaystyle x_{n+1}=\frac{2x_n^2-x_n^2+2}{2x_n}$
$\displaystyle x_{n+1}=\frac{x_n^2+2}{2x_n}$
$\displaystyle x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}$
For any positive initial value, the series will converge to the positive root of the defining function.
#### CaptainBlack
##### Well-known member
Jan 26, 2012
890
Sorry to spam my problems all over this forum but series have me struggling somewhat. Last problem on my homework is the sequence an defined recursively by:
a1=1
and
an+1= $$\frac{a_n}{2}$$ + $$\frac{1}{a_n}$$
First part was the only part i know how to do. it was to find an for n=1 through 5.
However this next part has me stumped. Assume that:
The limit as n approaches infinity = alpha > 0
Obtain the value alpha by taking the limit of both sides of (1) [(1) is the info given in the beginning].
Now I can already tell this limit is going to $$\sqrt{2}$$ because it is kind of hinted to in the later parts of this problem. However I am kind of confused as to how to approach doing this limit of both sides of an+1= $$\frac{a_n}{2}$$ + $$\frac{1}{a_n}$$ . How would I evaluate this limit when the right side of the equation is in terms of an?
If the limit $$\alpha$$ exists and is non-zero then:
$\alpha=(\alpha/2) + (1/\alpha)$
(arrived at by taking limits of both sides of the recursion relation)
CB
#### skatenerd
##### Active member
Oct 3, 2012
114
Oh man thank you to you both. To CaptainBlack, I guess I didn't realize that seeing that if limit of n approaching infinity of an = alpha, then so will the this limit of an+1. Makes sense now! Thanks again.
#### chisigma
##### Well-known member
Feb 13, 2012
1,704
Sorry to spam my problems all over this forum but series have me struggling somewhat. Last problem on my homework is the sequence an defined recursively by:
a1=1
and
an+1= $$\frac{a_n}{2}$$ + $$\frac{1}{a_n}$$
First part was the only part i know how to do. it was to find an for n=1 through 5.
However this next part has me stumped. Assume that:
The limit as n approaches infinity = alpha > 0
Obtain the value alpha by taking the limit of both sides of (1) [(1) is the info given in the beginning].
Now I can already tell this limit is going to $$\sqrt{2}$$ because it is kind of hinted to in the later parts of this problem. However I am kind of confused as to how to approach doing this limit of both sides of an+1= $$\frac{a_n}{2}$$ + $$\frac{1}{a_n}$$ . How would I evaluate this limit when the right side of the equation is in terms of an?
The solving procedure for this type of problems is illustrated in...
http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/
The difference equation can be written as...
$\displaystyle \Delta_{n}= a_{n+1}-a_{n}= \frac{1}{a_{n}}- \frac{a_{n}}{2} = f(a_{n})$ (1)
The function f(x) has two 'attractive fixed points' in $x=\pm \sqrt{2}$ and, because for both the fixed points the theorem 4.2 is satisfied, roughly specking any initial value >0 will generate a sequence which tends to $+\sqrt{2}$ and any initial value <0 will generate a sequence which tends to $-\sqrt{2}$. An important detail in the case $a_{0}>0$ is that the sequence for any n>0 is decreasing...
Kind regards
$\chi$ $\sigma$
#### skatenerd
##### Active member
Oct 3, 2012
114
My teacher went over this problem recently and showed that it was a funny way of doing a proof of The Newtonian Method. Seeing as it was something we had never learned about before, I guess it was just a little something our teacher threw in to see if we could teach ourselves a concept on our own. | {"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": 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": 0.9489923119544983, "perplexity": 412.61818570414493}, "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/1610704804187.81/warc/CC-MAIN-20210126233034-20210127023034-00068.warc.gz"} |
https://www.physicsforums.com/threads/integration-by-parts-and-substitution-help.295186/ | # Integration by parts and substitution help
• Start date
• #1
1
0
Integration by parts and substitution help!!
∫0
-1 e ^√x+1
## The Attempt at a Solution
• #2
HallsofIvy
Homework Helper
41,833
964
I'm sorry, but I simply cannot figure out what that integral is intended to be.
Is that $$\int_{-1}^0 e^{\sqrt{x+1}} dx$$ or $$\int_{-1}^0 e^{\sqrt{x}+ 1} dx$$ or $$\int_{-1}^0 e^{\sqrt{x}}+ 1 dx$$?
If it is the first, take $u= \sqrt{x+1}= (x+1)^{1/2}$ so that $du= (1/2)(x+1)^{-1/2}dx$ so that $(x+1)^{1/2}du= u du= dx$. When x= -1, u= 0, when x= 0, u= 1 The integral becomes
$$\int_0^1 ue^{u}du$$
and can be done by "integration by parts".
If the second, write it as $$e\int_{-1}^0 e^{\sqrt{x}}dx$$ and let $u= \sqrt{x}$.
If the third, write it as $$\int_{-1}^0 e^{\sqrt{x}}dx+ \int_{-1}^0 dx$$.
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https://kluedo.ub.uni-kl.de/frontdoor/index/index/docId/946 | ## Structural relaxation and magnetic anisotropies in Co/Cu(001) films
• The magnetic anisotropy of Co/Cu~001! films has been investigated by the magneto-optical Kerr effect, both in the pseudomorphic growth regime and above the critical thickness where strain relaxation sets in. A clear correlation between the onset of strain relaxation as measured by means of reflection high-energy electron diffraction and changes of the magnetic anisotropy has been found.
$Rev: 13581$ | {"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": 0.9530336260795593, "perplexity": 1983.2686436960546}, "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/1510934807344.89/warc/CC-MAIN-20171124085059-20171124105059-00080.warc.gz"} |
https://in.mathworks.com/help/signal/ref/lp2lp.html | lp2lp
Change cutoff frequency for lowpass analog filter
Syntax
``[bt,at] = lp2lp(b,a,Wo)``
``[At,Bt,Ct,Dt] = lp2lp(A,B,C,D,Wo)``
Description
example
````[bt,at] = lp2lp(b,a,Wo)` transforms an analog lowpass filter prototype given by polynomial coefficients (specified by row vectors `b` and `a`) into a lowpass filter with cutoff angular frequency `Wo`. The input system must be an analog filter prototype.```
````[At,Bt,Ct,Dt] = lp2lp(A,B,C,D,Wo)` converts the continuous-time state-space lowpass filter prototype (specified by matrices `A`, `B`, `C`, and `D`) to a lowpass filter with cutoff angular frequency `Wo`. The input system must be an analog filter prototype.```
Examples
collapse all
Design an 8th-order Chebyshev Type I analog lowpass filter prototype with 3 dB of ripple in the passband.
`[z,p,k] = cheb1ap(8,3);`
Convert the prototype to transfer function form and display its magnitude and frequency responses.
```[b,a] = zp2tf(z,p,k); freqs(b,a)```
Transform the prototype to a lowpass filter with a cutoff frequency of 30 Hz. Specify the cutoff frequency in rad/s. Display the magnitude and frequency responses of the transformed filter.
```Wo = 2*pi*30; [bt,at] = lp2lp(b,a,Wo); freqs(bt,at)```
Input Arguments
collapse all
Prototype numerator and denominator coefficients, specified as row vectors. `b` and `a` specify the coefficients of the numerator and denominator of the prototype in descending powers of s:
`$\frac{B\left(s\right)}{A\left(s\right)}=\frac{b\left(1\right){s}^{n}+\cdots +b\left(n\right)s+b\left(n+1\right)}{a\left(1\right){s}^{m}+\cdots +a\left(m\right)s+a\left(m+1\right)}$`
Data Types: `single` | `double`
Prototype state-space representation, specified as matrices. The state-space matrices relate the state vector x, the input u, and the output y through
`$\begin{array}{l}\stackrel{˙}{x}=Ax+Bu\\ y=Cx+Du\end{array}$`
Data Types: `single` | `double`
Cutoff angular frequency, specified as a scalar. Express `Wo` in units of rad/s.
Data Types: `single` | `double`
Output Arguments
collapse all
Transformed numerator and denominator coefficients, returned as row vectors.
Transformed state-space representation, returned as matrices.
Algorithms
`lp2lp` transforms an analog lowpass filter prototype with a cutoff angular frequency of 1 rad/s into a lowpass filter with any specified cutoff angular frequency. The transformation is one step in the digital filter design process for the `butter`, `cheby1`, `cheby2`, and `ellip` functions.
`lp2lp` is a highly accurate state-space formulation of the classic analog filter frequency transformation. If a lowpass filter has cutoff angular frequency ω0, the standard s-domain transformation is
`$s=p/{\omega }_{0}$`
The state-space version of this transformation is
`$At={\omega }_{0}\cdot A$`
`$Bt={\omega }_{0}\cdot B$`
`$Ct=C$`
`$Dt=D$`
The `lp2lp` function can perform the transformation on two different linear system representations: transfer function form and state-space form. See `lp2bp` for a derivation of the bandpass version of this transformation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "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": 0.990606427192688, "perplexity": 2480.464898537542}, "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-2021-43/segments/1634323585265.67/warc/CC-MAIN-20211019105138-20211019135138-00103.warc.gz"} |
http://mathoverflow.net/questions/125988/questions-of-localization-of-topos | # questions of localization of topos
Let $T$ be a topos, and $F \in T$, $T/F$ a localization of $T$. So we have a natural morphism $i: T/F \longrightarrow T$.
My questions are:
1.What are the definitions of $i_{\ast}$ and $i^{\ast}$ without using site ?
2.Dose there exist a site $S$ such that $Sh(S) \cong T$ and there is an $U \in S$ which $U^{\sim} \cong F$ ?
-
Hello !
$i^*$ is the functor which send an object $X \in T$ to $X \times F$ with the natural projection as map into $F$.
$i_*$ is a little harder to describe, if $p: Y \rightarrow F$ is an object of $T/F$, then $i\_*(Y)$ is the sub-object of $[F,Y]$ (the internal hom object) which corresponds to map $f$ from $F$ to $Y$ such that $p \circ f =Id\_F$ this can be express as an equaliser or with internal language as you prefers)
You also have a $i\_!$ functor (left adjoint to $i^*$ ) which is just the forget functor who send $p:Y \rightarrow F$ to $Y$.
You can chose any generating family $B$ of $T$ (for example the image by the yoneda embeddings of a site of definition of $T$) add $F$ to this familly, the familly $B'$ you obtain seen as a full subcategory of $T$ and endowed with the canonical topology of $T$ is (by the Grothendieck comparison lemma) a site of definition of $T$, and you simply choose $U$ to be $F$. | {"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": 0.927498996257782, "perplexity": 99.30816999522864}, "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-2013-48/segments/1386163043081/warc/CC-MAIN-20131204131723-00024-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://mathhelpforum.com/statistics/189658-find-number-different-order-triples.html | # Thread: find the number of different order triples
1. ## find the number of different order triples
Find the number of different order triples (a,b,c) s.t a+b+c=12 if
a) a,b,c must be nonegetive integers
b) a,b,c must be positive integers
is this permatutaion or combination? Suppose there is 12 choices for a , then b will be depend on a, and c will depend on both a and b
2. ## Re: find the number of different order triples
Hello, wopashui!
Believe it or not this is a Combination problem.
But you'll have see the explanation.
Find the number of different ordered triples $(a,b,c)$
such that $a+b+c\,=\,12$ if:
. . a) $a,b,c$ must be nonnegetive integers.
Place 12 objects in a row, inserting a space before, after and between them.
. . $\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;* \;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_$
Place two "dividers" in any of the 13 spaces.
So that: . $*\;*\;*\;*\;*\,|\,*\;*\;*\;*\,|\,*\;*\;\:*$ .represents: $5 + 4 + 3$
. . and: . $*\;*\;*\;*\,||\,*\;*\;*\;*\;*\;*\;*\;\,*$ .represents: $4 + 0 + 8$
If the dividers are placed in two different spaces, there are: . $_{13}C_2 = 78$ ways.
If the dividers are placed in the same space, there are: $13$ ways.
Hence, there are: . $78 + 13 \,=\,91$ ways to place the dividers.
Therefore, there are $91$ triples of nonnegative integers whose sum is 12.
b) $a,b,c$ must be positive integers.
Place 12 objects in a row, inserting a space between them.
. . $*\;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;* \;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;*$
Choose 2 of the 11 spaces and insert "dividers".
So that: . $*\;*\;*\;*\;*\,|\,*\;*\;*\;*\,|\,*\;*\;\:*$ .represents: $5 + 4 + 3$
. . and: . $*\;*\;*\;*\,|\,*\;*\;*\;*\;*\;*\;*\,|\,*$ .represents: $4 + 7 + 1$
The dividers are placed in two different spaces.
. . There are: . $_{11}C_2 = 55$ ways.
Therefore, there are $55$ ordered triples of positive integers whose sum is 12.
3. ## Re: find the number of different order triples
thanks alot, but i dun understand why is the second part only has 11C2, not 13C2?
4. ## Re: find the number of different order triples
anyone can explain?
5. ## Re: find the number of different order triples
Originally Posted by wopashui
Find the number of different order triples (a,b,c) s.t a+b+c=12 if
b) a,b,c must be positive integers
Originally Posted by wopashui
thanks alot, but i dun understand why is the second part only has 11C2, not 13C2?
Think of the 12 as being twelve 1's. Lets make the variables positive by putting a 1 is each to begin with. Now that leaves us with nine 1's to distribute: $\binom{9+3-1}{9}$.
6. ## Re: find the number of different order triples
thx,, so the first part would be 12+3-1 C 14, which is 91
7. ## Re: find the number of different order triples
Originally Posted by wopashui
thx,, so the first part would be 12+3-1 C 14, which is 91
@wopashui,
You have absolutely no idea about any of this.
You are either playing us for fools or you just don't get it.
I am done with you in any case.
Good luck. | {"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": 21, "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": 0.8258772492408752, "perplexity": 1523.2517862592015}, "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-2017-39/segments/1505818695726.80/warc/CC-MAIN-20170926122822-20170926142822-00310.warc.gz"} |
https://www.hardinscientific.com/blogs/news/if-the-strong-nuclear-force-is-stronger-than-electrostatic-repulsion-why-dont-nuclei-collapse-into-a-point | # If the strong nuclear force is stronger than electrostatic repulsion, why don't nuclei collapse into a point?
Today in class we were discussing the strong nuclear force, and our teacher was explaining about how the strong nuclear force counteracts the repulsion force between protons in a nucleus. When asked about the relative strength of the two forces in question, she said that "The strong nuclear force is the strongest force of nature, and is infinitely stronger than the repulsion force between the protons". Now, if that were true, how would the atom remain in equilibrium, because if I'm correct, Equilibrium is achieved when the net force on a body is zero. However, in this situation, that doesn't seem to be the case. Could someone elaborate on this apparent contradiction? $\begingroup$ I would strongly disagree with that statement (bad puns are great). Obviously the strong force is not infinitely times greater than the repulsion force. In order for that to be true, there would either have to be zero force repelling the protons or infinite strong force holding them together. I'm afraid neither is the case. Even a cursory search on wikipedia should tell you that the strong force is only 137 times greater than the EM force. Go back to your prof and tell her you disagree $\endgroup$ – Jim Jul 26 '16 at 13:11 $\begingroup$ The origin of the strong force is the fact that the electromagnetic theory as it is known now demands the nucleus to fall apart. It isn't the case, so something had to be invented. $\endgroup$ – bright magus Jul 26 '16 at 13:12 $\begingroup$ Are you sure that your teacher said 'strong force isinfinitely stronger than repulsive force between protons? Is she said that, then its wrong. $\endgroup$ – AMS Jul 26 '16 at 13:16 $\begingroup$ @Jim That is a pretty weak argument.The gravity of this question makes me wonder about how strongly you are trying to, em, help, given the bad pun in your answer.Together, unified, we can do better; we can lift up a standard of high quality comments, break the confinement of off topic comments, be grander than ourselves, stop stringing people along, and model better behavior.In theory, QED, but more importantly: OP, does simply pointing out that the ratio isn't actually infinite sufficient to answer your question, or are you wondering how the protons don't collide? $\endgroup$ – Yakk Jul 26 '16 at 15:29 $\begingroup$ @Yakk I sentence you to walk the Planck. $\endgroup$ – Don Branson Jul 26 '16 at 15:41 First, the strong force acts on scales where our classical idea of forces as something that obeys Newton's laws breaks down anyway. The proper description of the strong force is as a quantum field theory. On the level of quarks, this is a theory of gluons, but on scales of the nucleus, only a "residual strong force", the nuclear force remains, which can be thought of as being effectively mediated by pions. Now, a force mediated by pions is very different from one mediated by photons, for the simple reason that pions are massive. Massive forces do not, in their classical limit, follow a pure inverse square law, but yield the more general Yukawa potential, which goes as $\propto \frac{\mathrm{e}^{-mr}}{r^2}$ where $m$ is the mass of the mediating particle. That is, massive forces fall off far faster than electromagnetism. So this makes it already difficult to tell what the "strength" of a force exactly is - it depends on the scale you are looking at, as Wikipedia's table for the strengths of the fundamental forces rightly acknowledges. However, in no sense is the strong force "infinitely stronger" than the electromagnetic force - it is simply much stronger than it, sufficient to keep nuclei together against electromagnetic repulsion. Now, the person who said that it is "infinitely stronger" might have had something different in mind which is not actually related to the strength of the force but to its fundamentally quantum mechanical nature: Confinement, the phenomenon that particles charged under the fundamental (not the residual) strong force cannot freely exist in nature. When you try - electromagnetically or otherwise - to separate two quarks bound by the strong force, then you will never get two free quarks. The force between these two quarks stays constant with increasing distance, it does not obey an inverse square law at all, and in particular the energy to being on of the two quarks to infinity is not finite. At some point, when you have invested enough energy, there will be a spontaneous creation of a new quark-antiquark pair and you will end up with two bound quark systems, but no free quark. In this sense, one might say that the strong force is "infinitely stronger", but crucially this is not the aspect of the strong force that keeps nuclei together; the theory of pions shows no confinement. $\begingroup$ The equilibrium question: shouldn't mentioning exclusion pressure be part of an answer? $\endgroup$ – Yakk Jul 26 '16 at 15:15 $\begingroup$ @Yakk Yes, if one wanted to do a full account of all the forces at work, exclusion would add to the electromagnetic repulsion. But the question appeared to me to be about a misconception that the strong force is "infinitely strong" - I presume OP has no problem with accepting that finitely strong forces (no matter whether there are two, three, four, or even more at work) can achieve equilibrium. $\endgroup$ – ACuriousMind♦ Jul 26 '16 at 15:23 $\begingroup$ Nice answer, I just would like to point out that it is not the force between two quarks that grows linearly with distance. It is the energy of the pair. $\endgroup$ – Diracology Jul 26 '16 at 15:59 $\begingroup$ @Diracology: Ahh, right. The force is constant, I guess. $\endgroup$ – ACuriousMind♦ Jul 26 '16 at 16:01 $\begingroup$ This is still a bit short of the posed question, though. (i) The strong force doesn't need to be infinitely stronger than electromagnetic repulsion, it just needs to be stronger at all positions, and (ii) Yukawa potentials die off past a certain point, but the question concerns what happens in the $r\to0$ limit, and there the Yukawa potential is essentially equivalent to a Coulombic force. When two protons approach, at some point the strong attraction turns off and they must begin to repel past an equilibrium position - that's what the question is asking for. $\endgroup$ – Emilio Pisanty Jul 27 '16 at 0:56 Your existing answer talks about quark confinement, but stable nuclei can't really be described using quark and gluon degrees of freedom.Also your existing answer doesn't answer your title question: why don't nuclei collapse to a point? To first approximation, nuclei do collapse into a point.The diameter of a nucleus is typically about $10^{-5}$ the diameter of an atom, which means the nucleus occupies something like $10^{-15}$ of the atomic volume.If your atom were the size of a house, the nucleus, to scale, would be the size of a grain of salt (and yet containing 99.95% of the atom's mass). In nuclear physics it makes more sense to talk about energy than it does to talk about force.The two concepts are closely related.Two protons separated by a distance $r$ have an "electrical" interaction energy $$U_E = +\alpha\frac{\hbar c}{r}.$$ Here $\hbar c = 200\rm\,MeV\,fm$ relates distance and energy. The fine structure constant, $\alpha \approx 0.0073 \approx 1/137$, represents the strength of the electrical interaction. The positive sign, and the $r$ on the bottom, mean that bringing two protons closer together stores energy in the electric field between them. If you find yourself with two protons at rest some distance apart, they'll tend to evolve to reduce the energy stored in the electric field by moving yet further apart; this is the sense in which the electric interaction is repulsive. The electric field isn't the only place where interacting protons can store energy.There is also the pion field, whose interaction energy is given by a Yukawa potential, $$U_\pi = -\alpha_\pi \frac{\hbar c}r \times e^{-m_\pi r}.$$ This is mostly the same as the electrical potential, but the differences are important: The sign is negative: protons liberate energy from the pion field by approaching each other, so the force is attractive. There's a different coupling constant, $\alpha_\pi$.If I'm converting units correctly, the pion coupling constant is something like $\alpha_\pi\sim100$: the energy associated with the "pion exchange" interaction may be as much as ten thousand times stronger than the energy associated with a "photon exchange" interaction. Critically, there's an exponential factor that depends on the mass of the pion, $m_\pi$.To be proper this should have some factors of $\hbar$ and $c$ to make the argument of the exponential dimensionless; alternatively we can be cleverer about units and not have any distracting cruft.This means there is a length scale $r_0 \propto 1/m_\pi$ beyond which the pion interaction totally dies away, but for very short distances the interaction looks like electricity with a different coupling constant. These differences address a little of your confusion. The strong interaction is much stronger than electromagnetism, but not infinitely stronger.But your title question still remains: why doesn't the nucleus collapse exactly to a point?That's because we still haven't exhausted the forest of different ways that two protons can store energy.The next two that are important are the rho and omega fields, \begin{align} U_\rho &= +\alpha_\rho \frac{\hbar c}r \times e^{-m_\rho r}, \\ U_\omega &= +\alpha_\omega \frac{\hbar c}r \times e^{-m_\omega r}. \end{align} Like the pion field, these meson interactions "turn off" when the nuclei separate from each other, but the distances are shorter than for the pion because the masses are larger.So the interaction between two protons has at least three different regions: Here's a plot using the potentials we've discussed so far: This simple model reproduces several real features of real nuclei: In uniform nuclear matter, nucleons are separated by about 1.2 fm. The energy required to remove one nucleon from a nucleus is typically about 10 MeV. In very heavy nuclei, the length scale for nuclear attraction is shorter than the diameter of the nucleus; such nuclei are unstable. You can't tell from this plot, but the total interaction energy becomes positive --- that is, electrostatic repulsion wins out over strong attraction --- at a separation of about 12.5 fm. Uranium-235, famously unstable, has diameter 14 fm. Uranium is unstable because protons at one end of the nucleus are repelled by, rather than attracted to, protons at the other end of the nucleus. So let me get this correct. Gravity is what binds things together but the strong force is the glue holding everything in place!! Hmm something not right here. The universeis made of 99.9% plasma. And electromagnetism rules our reality. Why concentrate on the front force at scales of Planck length which have no importance on our reality whatsoever.
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Analysis of the SARS-CoV-2 proteome via visual tools | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9021735787391663, "perplexity": 392.5857888446353}, "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-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00413.warc.gz"} |
http://www.cs.columbia.edu/~rocco/papers/stoc03ptfdb.html | New Degree Bounds for Polynomial Threshold Functions.
R. O'Donnell and R. Servedio.
Combinatorica 30(3), 2010, pp. 327-358.
Preliminary version in 35th Annual Symposium on Theory of Computing (STOC), 2003, pp. 325-334.
Abstract:
We give new upper and lower bounds on the degree of real multivariate polynomials which sign-represent Boolean functions. Our upper bounds for Boolean formulas yield the first known subexponential time learning algorithms for formulas of {\em superconstant} depth. Our lower bounds for constant-depth circuits and intersections of halfspaces are the first new degree lower bounds since 1968, improving results of Minsky and Papert. The lower bounds are proved {\em constructively}; we give explicit dual solutions to the necessary linear programs. | {"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": 0.9934104681015015, "perplexity": 1031.473243312575}, "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-2014-42/segments/1414637898141.14/warc/CC-MAIN-20141030025818-00175-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://space.stackexchange.com/tags/performance/hot | # Tag Info
111
Because linear increases in delta-v require exponential increases in mass, small changes to the assumptions you make about fuel tank structural mass and engine thrust-to-weight ratio start to make very large changes in the final size of the rocket. For example, if you're getting off a 3.6g planet with a 7-stage rocket, the difference between 88% fuel ...
26
First, let us look at the rocket equation: $$\Delta v=\ln \left(\frac{m_0}{m_f}\right)v_e$$ That tells how much a rocket can change its velocity (the $\Delta v$). The requirements for reaching a higher velocity for a minimal orbit would increase on your heavier Earth. (For constant density it is proportional to the radius.) How can we increase the $\Delta ... 19 A few different factors contribute to the Raptor's higher thrust: The specific impulse -- force delivered per mass of propellant consumed -- of methane-LOX combustion is generally higher than kerosene-LOX, because the exhaust is composed of simpler and lighter molecules; The Raptor uses the staged combustion cycle, where the hot partially-combusted gases ... 14 First, let's get terminology straight: "Tilt maneuver", or "Gravity turn", sometimes also called "Pitch maneuver". It was called "Roll Program" in case of Space Shuttles, because it was connected with a roll, necessary for technical reasons but not contributing to flight efficiency directly. All rockets (and all flying bodies on Earth for that matter) are ... 13 The delta-v needed from low Earth orbit to a Hohmann transfer orbit with a periapse inside the Sun is actually "just" 21,300 m/s. But there is a better option. A bi-elliptic transfer to just hit a central body is better when the ratio between the orbital radius and the radius of the central body is larger than 4.82. The orbital radius of the Earth divided ... 13 According to Wikipedia, HIBEX employed a star-grain "composite modified double-base propellant", known as FDN-80, created from the mixing of ammonium perchlorate, aluminum, and double-base smokeless powder, with zirconium staples "randomly dispersed" throughout the matrix. APCP (ammonium perchlorate composite propellant) is occasionally ... 12 Launchers generally start at full throttle, so for the most part their immediate performance off the pad is determined solely by the type of launcher and mass of payload. Falcon 9's initial acceleration is quite modest, about 1.5 m/s2 vertically (thrust to weight ratio about 1.15, cancelling and exceeding 9.8m/s2 of surface gravity). Shuttle's initial TWR ... 11 This depends on what "better" is. But let us talk about fluorine. The main point of using it is that turning your hydrogen into$HF$gives more energy than getting$H_2O$for pure hydrogen, that gives a small improvement of$I_{sp}$. However, for fuels containing carbon, like JP1, (or perhaps RP-1 or JP4. JP1 has some unfortunate properties for rocket ... 9 I can answer the second question — engines are by and large fuel specific. There's plenty of complicated stuff that goes on in the combustion and the transition to supersonic flow that means you can't just exchange one fuel for another and expect it to be high-performing. As far as I know, there are no pre-existing methalox-burning engines that SpaceX ... 9 One thing mentioned was that it allowed the vehicle to lift more mass. All things being equal, I fail to see how tilt can increase the ability to increase the potential energy of the payload. All things weren't equal. The Shuttle was not an axially symmetric vehicle. The Shuttle roll program was performed starting about ten seconds after launch and lasted ... 9 There is likely minimal effect. At rocket speeds, there is very little effect of shear stress, the only significant effect is particles hitting the leading surface of the rocket. Also due to how fast hey are going, the effect of the rocket of "pushing air out of the way" does not have time to get far ahead of the rocket, and this drops further behind as ... 7 Almost all the launch vehicles lift off vertically and are designed to reach orbital speed, altitude and orientation as the upper stage completes its injection burn. Consider a launch vehicle lifting off vertically- The vehicle accelerates to overcome two forces- earth's gravity and the atmospheric drag. Image from rocketmime.com If the launch vehicle goes ... 7 note: I've accepted an answer 2.5 years ago. This paper was published recently so I thought I would add this supplemental answer since it may be an interesting reference for future readers. The Space.com article No Way Out? Aliens on 'Super-Earth' Planets May Be Trapped by Gravity links to Michael Hippke's ArXiv preprint Spaceflight from Super-Earths is ... 6 I'd assumed that Isp expresses the axial component of the velocity, so even if there is some transverse flow in the expanding exhaust, that wouldn't have to be accounted for beyond Isp, but is there something else? To be pedantic, Isp is specific impulse, or more fully "mass specific impulse": the impulse delivered per unit of mass flow. Impulse is force ... 6 The final stage of a rocket has to lift not just the payload but also itself. So lets look not at the payload mass alone but at the total mass lifted to the final orbit. According to http://www.spaceflightinsider.com/hangar/falcon/ the empty mass of the second stage is 3,900 kg. Lets add that the numbers in the table. If we add the mass of the final stage ... 6 All else being equal (e.g. propellant choice), higher-energy trajectories favor a rocket with more stages. One thought experiment that can help you understand this is to consider two missions launched on identical rockets, the first to LEO and the second to an interplanetary trajectory. Both missions have payloads sized to "max out" the capacity of the ... 6 This response addresses just the last part of your question "what sort of$I_\text{sp}$and thrust might it have", and does so in the sense of an ideal resistojet. I hope this isn't too basic an approach but its worth stepping through it to develop a more intuitive view of the very high level physics. There are similar equations in the link you gave though, ... 6 There were several factors in the launch date The late John F. Kennedy had promised to go to the moon within the decade. This was a secondary concern, but an important one nevertheless The failure of the Soviet N1 rocket on July 3, 1969. The Soviet program was desperately trying to keep up with the American program. NASA feared the Soviets would overtake ... 6 "Yes", but that would sort of be silly and it would highly depend on where you live, since it may be illegal in your country to mix your own rocket propellant. HIBEX used what is essentially a precursor to modern APCP (Ammonium Perchlorate Composite Propellant), which is widely used in amateur rocketry and commercially available. Its propellant ... 5 Borrowing from Wikipedia's article on modern ICBMs: One particular weapon developed by the Soviet Union (Fractional Orbital Bombardment System) had a partial orbital trajectory, and unlike most ICBMs its target could not be deduced from its orbital flight path. It was decommissioned in compliance with arms control agreements, which address the maximum ... 5 Not a planetological exposition in sight so, I'll add my two cents to this rather theoretical discussion. 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The area$A$is a ... 5 They aren't launched slower but they are launched in different trajectories. This can be to control abort conditions and G-Loads. The most efficient launch is to get as high as possible before going for speed. The STS however was shallower to allow for the abort modes and control the forces. The best example I can think of at the moment is the man rating ... 5 Hall effect thrusters do not use strong magnetic fields and radio frequency power to heat electrons to high energy for ionization of the gas. Instead they rely on a DC current of electrons accelerated to a few hundred volts flowing through the gas volume for ionization. Since Krypton has a higher ionization potential than Xenon (roughly 14.0 versus 12.1 eV), ... 5 The season when the Saturn V launches doesn't matter, especially in Florida where the temperature remains relatively warm year round. The Saturn V was engineered to be very resilient and to withstand almost every weather condition. You can see the launch commit criteria of modern rockets here. As Uwe said, there have been Apollo missions launched in the ... 4 Besides pure oxygen, several other oxidizers in theory have better performance with hydrocarbon fuels, but none of them are used because of stability, toxicity and price issues. But let mention some of them: Ozone O3. It's allotrope of oxygen with enthalpy of formation 142,67 KJ/mol (2.97MJ/kg) which can bust combustion energy up to 30% depending on fuel ... 4 Not sure you'll get much useful data, but here's a useful datum point: Bloodhound SSC - the worlds fastest car is expected to reach 1000mph in 55 seconds from rest (42 seconds would be a perfect run, but 55 is the expected run), with a peak acceleration of 2g. 0 - 300mph (Point a) using the EJ-200 jet engine (generating around 0.5g), where the rocket will ... 4 Ignoring the complicated separation issue, there is a simple relation to calculate the effect on thrust based on mismatch of the exit plane pressure and ambient pressure, namely: $$F = q V_e + (P_e-P_a) A_e$$ Where$P_e$is the exit plane pressure,$P_a$is ambient pressure, and$A_e$is the exit plane area.$qV_e\$ without the correction term gives the ...
4
According to this article, in an overexpanded nozzle, the loss of efficiency is caused by the "pinching" of the exhaust plume by the ambient air pressure. In grossly overexpanded nozzles, there's another, more serious problem, where the exhaust flow separates from one side of the nozzle, adhering to the opposite side, which causes very uneven heating and ...
Only top voted, non community-wiki answers of a minimum length are eligible | {"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": 0.969946026802063, "perplexity": 1280.1328595706339}, "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/1606141733122.72/warc/CC-MAIN-20201204040803-20201204070803-00289.warc.gz"} |
https://www.physicsforums.com/threads/future-pointing-and-past-pointing-time-like-vectors.850151/ | # Future-pointing and past-pointing time-like vectors
1. Dec 30, 2015
### spaghetti3451
1. The problem statement, all variables and given/known data
I need to prove the following:
a) If $P^{a}$ and $Q^{a}$ are time-like and $P^{a}Q_{a}>0$, then either both are future-pointing or both are past-pointing.
b) If $U^{a}$, $V^{a}$ and $W^{a}$ are time-like with $U^{a}V_{a}>0$ and $U^{a}W_{a}>0$, then $V^{a}W_{a}>0$.
2. Relevant equations
Using the 'mostly minus' convention, $A^a$ is time-like, null and space-like if $A^{a}A_{a}$ is $>0,=0,<0$ respectively.
A time-orientation is chosen by taking at will some time-like vector, say $U^{a} = (1,0,0,0)$, and designating it to be future-pointing. Any other time-like or null vector $V^{a}$ such that $g_{ab}U^{a}V^{b}>0$ is also future-pointing, whereas if $g_{ab}U^{a}V^{b}<0$, then $V^{a}$ is past-pointing.
3. The attempt at a solution
a) If $P^{a}$ is time-like future-pointing and $P^{a}Q_{a}>0$, then (by definition) $Q^{a}$ is also future-pointing.
Simialarly, for $P^{a}$ past-pointing.
b) If $U^{a}$ is time-like future-pointing and $U^{a}V_{a}>0$, then (by definition) $V^{a}$ is also future-pointing.
If $U^{a}$ is time-like future-pointing and $U^{a}W_{a}>0$, then (by definition) $W^{a}$ is also future-pointing.
Since both $V^{a}$ and $W^{a}$ are future-pointing, $V^{a}W_{a}>0$.
Similar argument for $U^{a}$ past-pointing.
Are my proofs sound?
2. Dec 30, 2015
### Staff: Mentor
Where is the definition for that?
Not by definition, but via (a).
(b) is fine once you fix (a).
Schwarz Inequality is your friend (from today!)
3. Dec 30, 2015
### spaghetti3451
I can see that the following theorem is useful:
For $U^{a}$ and $V^{a}$ timelike and $U^{0}, V^{0} >0$, we have that $U^{a}V_{a}>0$.
How do I relate the fact that $U^{0}, V^{0} >0$ with future-pointing and past-pointing vectors?
Last edited: Dec 30, 2015
4. Dec 30, 2015
### Staff: Mentor
That you can get from the definition now, if you evaluate the sums there explicitely.
5. Jan 1, 2016
### spaghetti3451
Alright, here's my proof.
We are given that $P^{a}$ and $Q^{a}$ are timelike and that $P^{a}Q_{a}>0$.
If $P^{a}$ is timelike, then $P^{a}P_{a}>0 \implies (P^{0})^{2}>(\vec{P})^{2}$. Similarly, for $Q^{a}$.
Therefore, $(P^{0})^{2}(Q^{0})^{2}>(\vec{P})^{2}(\vec{Q})^{2}$.
So, by the Cauchy-Schwarz inequality, $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$, which implies that either $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$, where the RHS is positive or $P^{0}Q^{0}<\vec{P}\cdot{\vec{Q}}$, where the RHS is negative.
On the other hand, if $P^{a}Q_{a}>0$, then $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$. Therefore, $P^{0}Q^{0}>0$.
So, either both $P^{0}$ and $Q^{0}$ are positive and hence future-pointing, or both $P^{0}$ and $Q^{0}$ are negative and hence past-pointing.
Is my proof correct?
6. Jan 1, 2016
### Staff: Mentor
Correct.
You can add a 1-line proof that future-pointing <=> P0>0.
7. Jan 1, 2016
### spaghetti3451
Isn't it a convention that a four-vector $P^{0}$ is future pointing iff $P^{0}>0$?
8. Jan 1, 2016
### Staff: Mentor
The definition you have in post 1 expresses it in a slightly different way, although you get this result by evaluating the sums there.
9. Jan 1, 2016
### spaghetti3451
So, you mean that a four-vector $P^{0}$ being future pointing iff $P^{0}>0$ is not a convention?
10. Jan 1, 2016
### spaghetti3451
The definition says that we choose an arbitrary time-like vector $P^{a}=(1,0,0,0)$ with $P^{0}>0$ and designate it to be future-pointing. Then, if $g_{ab}P^{a}Q_{a}>0$ for some time-like vector $Q^{a}$, then $Q^{a}$ is also future-pointing.
What strikes me is that the definition arbitrarily assigns a time-like vector $P^{a}$ with $P^{0}>0$ as being future-pointing. That's why I ask if it's a convention that a time-like vector $P^{a}$ with $P^{0}>0$ is called future-pointing.
And if so, I don't see why we need to use the $g_{ab}P^{a}Q_{a}>0$ to evaluate sums.
Last edited: Jan 1, 2016
11. Jan 1, 2016
### Staff: Mentor
It is an arbitrary convention, right - you could also assign positive time-components to past-pointing and negative ones to future-pointing. Would be odd, but it would lead to consistent physics as well.
You need to use it because you are given this definition of future-pointing.
12. Jan 1, 2016
### spaghetti3451
Ah! I see.
So, it is indeed an arbitrary definition. But what I do need to show is that the class of future-pointing vectors $P^{a}$ all have $P^{0}>0$ if at least one of them have $P^{0}>0$.
That is the reason why we choose $P^{a}=(1,0,0,0)$ because then, when we evaluate $P^{a}Q_{a}>0$, all the space-components of $Q_{a}$ are eliminated and we are left with the simple relation that $Q^{a}>0$.
Thus, we have shown basically that the class of future-pointing vectors and the class of past-pointing vectors do not overlap, and that each class can be identified by the sign of the temporal component of the four-vectors.
Am I correct?
13. Jan 1, 2016
### spaghetti3451
Alright, so now let me show the equivalence between statements 1 and 2.
1. A time-orientation is chosen by taking at will some time-like vector, say $U^{a}=(1,0,0,0)$ and designating it to be future-pointing. Any other time-like or null vector $V^{a}$ such that $U^{a}V_{a}>0$ is also future-pointing, whereas if $U^{a}V_{a}<0$, then $V^{a}$ is past-pointing.
2. A four-vector $P^{a}$ is future-pointing if and only if $P^{0}>0$.
If some time-like vector $U^{a}=(1,0,0,0)$ is chosen to be future-pointing, then for any other time-like vector $V^{a}$ with $U^{a}V_{a}>0$, we have
$U^{a}V_{a}>0 \implies U^{0}V_{0}+U^{i}V_{i}>0 \implies U^{0}V^{0}>0 \implies V^{0}>0$.
Therefore, all time-like vectors $V^{a}$ with $V^{0}>0$ are future-pointing.
Similarly, for any other time-like vector $W^{a}$ with $U^{a}W_{a}<0$, we have
$U^{a}W_{a}<0 \implies U^{0}W_{0}+U^{i}W_{i}<0 \implies U^{0}W^{0}<0 \implies W^{0}<0$.
Therefore, all time-like vectors $W^{a}$ with $W^{0}<0$ are future-pointing.
14. Jan 1, 2016
### Staff: Mentor
Correct.
15. Jan 1, 2016
### TSny
I'm not following how you claim that the statement $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$ implies that either $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$, where the RHS is positive or $P^{0}Q^{0}<\vec{P}\cdot{\vec{Q}}$, where the RHS is negative.
Are you saying that the statement $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$ by itself implies that if $\vec{P}\cdot{\vec{Q}}$ is positive then $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$?
Suppose $P^{\alpha} = (2, 1, 0, 0)$ and $Q^{\alpha} = (-2, 1, 0, 0)$. Both vectors are timelike. $P^{0}Q^{0} = -4$ while $\vec{P}\cdot{\vec{Q}} = 1$. Then $(P^{0}Q^{0})^{2}>(\vec{P}\cdot{\vec{Q}})^{2}$ and $\vec{P}\cdot{\vec{Q}} > 0$, but it is not true that $P^{0}Q^{0}>\vec{P}\cdot{\vec{Q}}$.
Hope I'm not overlooking something or misinterpreting your statement.
16. Jan 1, 2016
### spaghetti3451
Yes.
Ah! I see! Thanks for pointing out the mistake.
I thought I could use the fact that $x^{2}>y^{2} \implies x>y$ for $y$ positive or $x<y$ for $y$ negative. But, I realise now that it's not so simple, because $x$, in this case $P^{0}Q^{0}$, is itself a product of two numbers.
So, I have to go back to the drawing board and rethink a new proof, or at least, modify the existing proof.
17. Jan 1, 2016
### TSny
Hmm. Even if $x$ were not the product of two numbers, it would still not be true in general that $x^{2}>y^{2} \implies x>y$ for $y$ positive.
You are close. Along with $x^{2}>y^{2}$ you have $x > y$ (from $P^{\alpha}Q_{\alpha} > 0$). From both of these together you can conclude something important about $x$.
18. Jan 1, 2016
### spaghetti3451
Got it!
$x^{2}>y^{2} \implies (x+y)(x-y)>0$.
Now, $x>y \implies x-y>0 \implies x+y>0 \implies x>-y$.
The only way for $x>y$ and $x>-y$ are both valid is if $x>0$, that is $P^{0}Q^{0}>0$.
19. Jan 1, 2016
### spaghetti3451
Is it correct?
20. Jan 1, 2016
### TSny
The middle $\implies$ is of course using both $x-y>0$ and $(x+y)(x-y)>0$.
Yes. Nice.
Draft saved Draft deleted
Similar Discussions: Future-pointing and past-pointing time-like vectors | {"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": 0.9795055985450745, "perplexity": 711.9539778393254}, "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-2018-09/segments/1518891813322.19/warc/CC-MAIN-20180221024420-20180221044420-00202.warc.gz"} |
https://blogs.princeton.edu/imabandit/2016/05/11/bandit-theory-part-i/?replytocom=368136 | # Bandit theory, part I
This week I’m giving two 90 minutes lectures on bandit theory at MLSS Cadiz. Despite my 2012 survey with Nicolo I thought it would be a good idea to post my lectures notes here. Indeed while much of the material is similar, the style of a mini-course is quite different from the style of a survey. Also, bandit theory has surprisingly progressed since 2012 and many things can now be explained better. Finally in the survey we completely omitted the Bayesian model as we thought that we didn’t have much to add on this topic compared to existing sources (such as the 2011 book by Gittins, Glazebrook, Weber). For a mini-course this concern is irrelevant so I quickly discuss the famous Gittins index and its proof of optimality.
i.i.d. multi-armed bandit, Robbins [1952]
Known parameters: number of arms and (possibly) number of rounds .
Unknown parameters: probability distributions on with mean (notation: ).
Protocol: For each round , the player chooses based on past observations and receives a reward/observation (independently from the past).
Performance measure: The cumulative regret is the difference between the player’s accumulated reward and the maximum the player could have obtained had she known all the parameters,
This problem models the fundamental tension between exploration and exploitation (one wants to pick arms that performed well in the past, yet one needs to make sure that no good option has been missed). Almost every week new applications are found that fit this simple framework and I’m sure you already have some in mind (the most popular one being ad placement on the internet).
i.i.d. multi-armed bandit: fundamental limitations
How small can we expect to be? Consider the -armed case where and where is unknown. Recall from Probability 101 (or perhaps 102) that with expected observations from the second arm there is a probability at least to make the wrong guess on the value of . Now let be the expected number of pulls of arm up to time when . One has
We refer to Bubeck, Perchet and Rigollet [2013] for the details. The important message is that for fixed the lower bound is , while for the worse (which is of order ) it is . In the -armed case this worst-case lower bound becomes (see Auer, Cesa-Bianchi, Freund and Schapire [1995]). The -lower bound is slightly “harder” to generalize to the -armed case (as far as I know there is no known finite-time lower bound of this type), but thankfully it was already all done 30 years ago. First some notation: let and the number of pulls of arm up to time . Note that one has . For let
Theorem [Lai and Robbins [1985]]
Consider a strategy s.t. , we have if . Then for any Bernoulli distributions,
Note that so up to a variance-like term the Lai and Robbins lower bound is . This lower bound holds more generally than just for Bernoulli distributions, see for example Burnetas and Katehakis [1996].
i.i.d. multi-armed bandit: fundamental strategy
Hoeffding’s inequality teaches us that with probability at least , ,
The UCB (Upper Confidence Bound) strategy (Lai and Robbins [1985], Agarwal [1995], Auer, Cesa-Bianchi and Fischer [2002]) is:
The regret analysis is straightforward: on a probability event one has
so that and in fact
i.i.d. multi-armed bandit: going further
• The numerical constant in the UCB regret bound can be replaced by (which is the best one can hope for), and more importantly by slightly modifying the derivation of the UCB one can obtain the Lai and Robbins variance-like term (that is replacing by ): see Cappe, Garivier, Maillard, Munos and Stoltz [2013].
• In many applications one is merely interested in finding the best arm (instead of maximizing cumulative reward): this is the best arm identification problem. For the fundamental strategies see Even-Dar, Mannor and Mansour [2006] for the fixed-confidence setting (see also Jamieson and Nowak [2014] for a recent short survey) and Audibert, Bubeck and Munos [2010] for the fixed budget setting. Key takeaway: one needs of order rounds to find the best arm.
• The UCB analysis extends to sub-Gaussian reward distributions. For heavy-tailed distributions, say with moment for some , one can get a regret that scales with (instead of ) by using a robust mean estimator, see Bubeck, Cesa-Bianchi and Lugosi [2012].
Adversarial multi-armed bandit, Auer, Cesa-Bianchi, Freund and Schapire [1995, 2001]
For , the player chooses based on previous observations, and simultaneously an adversary chooses a loss vector . The player’s loss/observation is . The regret and pseudo-regret are defined as:
Obviously and there is equality in the oblivious case ( adversary’s choices are independent of the player’s choices). The case where is an i.i.d. sequence corresponds to the i.i.d. model we just studied. In particular we already know that is a lower bound on the attainable pseudo-regret.
The exponential weights strategy for the full information case where is observed at the end of round is defined by: play at random from where
In five lines one can show with and a well-chosen learning rate (recall that ):
For the bandit case we replace by in the exponential weights strategy, where
The resulting strategy is called Exp3. The key property of is that it is an unbiased estimator of :
Furthermore with the analysis described above one gets
It only remains to control the variance term, and quite amazingly this is straightforward:
Thus with one gets .
• With the modified loss estimate one can prove high probability bounds on , and by integrating the deviations one can show .
• The extraneous logarithmic factor in the pseudo-regret upper can be removed, see Audibert and Bubeck [2009]. Conjecture: one cannot remove the log factor for the expected regret, that is for any strategy there exists an adaptive adversary such that .
• can be replaced by various measure of “variance” in the loss sequence, see e.g., Hazan and Kale [2009].
• There exist strategies which guarantee simultaneously in the adversarial model and in the i.i.d. model, see Bubeck and Slivkins [2012].
• Many interesting variants: graph feedback structure of Mannor and Shamir [2011] (there is a graph on the set of arms, and when an arm is played one observes the loss for all its neighbors), regret with respect to best sequence of actions with at most switches, switching cost (interestingly in this case the best regret is , see Dekel, Ding, Koren and Peres [2013]), and much more!
Bayesian multi-armed bandit, Thompson [1933]
Here we assume a set of “models” and prior distribution over . The Bayesian regret is defined as
where simply denotes the regret for the i.i.d. model when the underlying reward distributions are . In principle the strategy minimizing the Bayesian regret can be computed by dynamic programming on the potentially huge state space . The celebrated Gittins index theorem gives sufficient condition to dramatically reduce the computational complexity of implementing the optimal Bayesian strategy under a strong product assumption on . Notation: denotes the posterior distribution on at time .
Theorem [Gittins [1979]]
Consider the product and -discounted case: , , , and furthermore one is interested in maximizing . The optimal Bayesian strategy is to pick at time the arm maximizing the Gittins index:
where the expectation is over drawn from with , and the supremum is taken over all stopping times .
Note that the stopping time in the Gittins index definition gives the optimal strategy for a 2-armed game, where one arm’s reward distribution is while the other arm reward’s distribution is with as a prior for .
Proof: The following exquisite proof was discovered by Weber [1992]. Let
be the Gittins index of arm at time , which we interpret as the maximum charge one is willing to pay to play arm given the current information. The prevailing charge is defined as (i.e. whenever the prevailing charge is too high we just drop it to the fair level). We now make three simple observations which together conclude the proof:
• The discounted sum of prevailing charge for played arms is an upper bound (in expectation) on the discounted sum of rewards. Indeed the times at which the prevailing charge are updated are stopping times, and so between two such times the expected sum of discounted reward is smaller than the discounted sum of the fair charge at time which is equal to the prevailing charge at any time in .
• Since the prevailing charge is nonincreasing, the discounted sum of prevailing charge is maximized if we always pick the arm with maximum prevailing charge. Also note that the sequence of prevailing charge does not depend on the algorithm.
• Gittins index does exactly 2. (since we stop playing an arm only at times at which the prevailing charge is updated) and in this case 1. is an equality. Q.E.D.
For much more (implementation for exponential families, interpretation as a multitoken Markov game, …) see Dumitriu, Tetali and Winkler [2003], Gittins, Glazebrook, Weber [2011], Kaufmann [2014].
Bayesian multi-armed bandit, Thompson Sampling (TS)
In machine learning we want (i) strategies that can deal with complicated priors, and (ii) guarantees for misspecified priors. This is why we have to go beyond the Gittins index theory.
In his 1933 paper Thompson proposed the following strategy: sample and play .
Theoretical guarantees for this highly practical strategy have long remained elusive. Recently Agrawal and Goyal [2012] and Kaufmann, Korda and Munos [2012] proved that TS with Bernoulli reward distributions and uniform prior on the parameters achieves (note that this is the frequentist regret!). We also note that Liu and Li [2015] takes some steps in analyzing the effect of misspecification for TS.
Let me also mention a beautiful conjecture of Guha and Munagala [2014]: for product priors, TS is a 2-approximation to the optimal Bayesian strategy for the objective of minimizing the number of pulls on suboptimal arms.
Bayesian multi-armed bandit, Russo and Van Roy [2014] information ratio analysis
Assume a prior in the adversarial model, that is a prior over , and let denote the posterior distribution (given ). We introduce
The key observation is that (recall that )
Indeed, equipped with Pinsker’s inequality and basic information theory concepts one has (we denote for the mutual information conditionally on everything up to time , also denotes the law of conditionally on everything up to time ):
Thus which gives the claim thanks to a telescopic sum. We will use this key observation as follows (we give a sequence of implications leading to a regret bound so that all that is left to do is to check that the first statement in this sequence is true for TS):
Thus writing and we have
For TS the following shows that one can take :
Thus TS always satisfies . Side note: by the minimax theorem this implies the existence of a strategy for the oblivious adversarial model with regret (of course we already proved that such a strategy exist, in fact we even constructed one via exponential weights, but the point is that the proof here does not require any “miracle” –yes exponential weights are kind of a miracle, especially when you consider how the variance of the unbiased estimator gets automatically controlled).
Summary of basic results
• In the i.i.d. model UCB attains a regret of and by Lai and Robbins’ lower bound this is optimal (up to a multiplicative variance-like term).
• In the adversarial model Exp3 attains a regret of and this is optimal up to the logarithmic term.
• In the Bayesian model, Gittins index gives an optimal strategy for the case of product priors. For general priors Thompson Sampling is a more flexible strategy. Its Bayesian regret is controlled by the entropy of the optimal decision. Moreover TS with an uninformative prior has frequentist guarantees comparable to UCB.
This entry was posted in Optimization, Probability theory. Bookmark the permalink. | {"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": 0.9790366888046265, "perplexity": 1099.9326693121482}, "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/1529267864139.22/warc/CC-MAIN-20180621094633-20180621114633-00561.warc.gz"} |
https://www.physicsforums.com/threads/fluid-flow-question-through-hose.185661/ | # Fluid flow question through hose
1. Sep 19, 2007
### greydient
Here's a silly question.
If I'm comparing two different sizes of hose, using the same flow rate in gpm, how do I calculate the pressure drops through each for comparison?
I'm considering using hoses with interior diameter .438 and .688, each with a length of 5 inches.
2. Sep 19, 2007
### FredGarvin
I am assuming that you want to measure the delta P, not calculate it. Calculating it is basic stuff.
If you are flowing to atmosphere on one end, you will have to adjust the pressure side of the pump you are using to get the flowrates to be equal between the two hose sizes. You can rely on a pressure gauge right at the outlet of the pump to give you the required delta P data.
3. Sep 19, 2007
### stewartcs
Pressure drop depends on these items. The fluid density, fluid velocity, pipe (or hose) roughness, length of the pipe (or hose), diameter of pipe (or hose), and any elevation changes between the two ends.
A standard equation for pressure drop is given by:
delta p = ((pf*f*L*v^2)/(144*d*2*g)) + ((z*pf)/144)
where,
delta p = pressure drop in lbs/in^2
pf = density of fluid in lbs/ft^3
f = friction factor
L = length in feet
v = velocity of fluid in ft/sec
d = diameter in ft
g = gravitational acceleration in ft/sec^2
z = elevation change in feet
The friction factor, f, is found depending on whether the flow is laminar or turbulent. If you find that the flow is in the critical region (between laminar and turbulent, 2000 > Re < 4000), then emperical testing is more accurate.
Hope this helps...
PS
Make sure your units are the same as shown above!
4. Sep 22, 2007
### jaap de vries
look up Moody chart the rest is standard and described above. (find Re #) | {"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": 0.8716850876808167, "perplexity": 2115.721696488508}, "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-2017-04/segments/1484560280850.30/warc/CC-MAIN-20170116095120-00008-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://jp.maplesoft.com/support/help/maplesim/view.aspx?path=worksheet/reference/contextmaplein | Context Bar for Math Input - Maple Help
Context Bar for Math Input
Switch the input display between 2-D Math and Maple Input notation.
This page describes how the context bar can be used to switch the input display between 2-D Math and Maple Input notation at a Maple prompt.
• When Math is selected, the input display mode is 2-D math notation:
• When Text is selected, the input display mode is Maple Input notation (sometimes called 1-D math):
At a Maple prompt, you can toggle the input mode by selecting Math or Text from the context bar.
Note: Set the default input display mode from the Display tab of the Options dialog. See Options>Display.
You can also use the F5 shortcut key, which at a Maple prompt toggles between 2-D math input and Maple Input.
In document mode, the context bar lets you switch between entering plain text and math. For more information on switching between text and math input, see Select Entry Mode. | {"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": 0.9989991784095764, "perplexity": 4192.7471106158555}, "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-2022-27/segments/1656104668059.88/warc/CC-MAIN-20220706060502-20220706090502-00312.warc.gz"} |
http://tex.stackexchange.com/questions/5170/mapsto-vs-rightarrow/5171 | # Mapsto vs. rightarrow
This is perhaps half a math question and half a typesetting one. In what circumstances do I use \mapsto and when \rightarrow? I feel like the standard seems to be: when you're specifying the operator you use \rightarrow, but if you're just saying that one domain maps to another you use \mapsto. Is that correct?
e.g. I would say Q\times\Gamma\mapsto Q but \delta:Q\times\Gamma\rightarrow Q.
-
You use \mapsto to denote the actual function mapping. For example, consider the function $f:\mathbb N\to\mathbb N$ given by $f(n)=5n$. You could write that second part as $n\mapsto5n$.
The \mapsto notation is useful for talking about a function without inventing a name for it, e.g. "the map $\mathbb{R}\to \mathbb{R}$ given by $x\mapsto x^2$ is not injective". Note that \to is a nice shorthand for \rightarrow, which at the same time is more semantic. – Villemoes Nov 10 '10 at 10:36
I'd prefer using \colon instead of :; in fact I'd even more prefer $f\from\mathbb N\to\mathbb N$, with \newcommand*\from{\colon}. – Hendrik Vogt Nov 10 '10 at 14:41
@Hendrik: I like the idea of using \from. I think I've seen equal numbers of : and \colon in books (or the equivalent spacing) so I think it's just up to personal preference (or editor preference, really). I have no strong feelings on that one. – TH. Nov 11 '10 at 5:43 | {"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": 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": 0.9675387740135193, "perplexity": 994.2407728472002}, "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-2015-18/segments/1429246655589.82/warc/CC-MAIN-20150417045735-00222-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/expressing-polar-equation-as-a-cartesian-equation.638002/ | # Expressing polar equation as a Cartesian equation
1. Sep 22, 2012
### Bipolarity
1. The problem statement, all variables and given/known data
Express the following equation in Cartesian form
$$r = 1 - cos(θ)$$
2. Relevant equations
$$x = r*cos(θ)$$
$$y = r*sin(θ)$$
$$r^{2} = x^{2} + y^{2}$$
$$tan(θ) = \frac{y}{x}$$
3. The attempt at a solution
I have no idea... a hint would be nice thanks!
BiP
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Sep 22, 2012
### gabbagabbahey
If $x=r\cos\theta$, what is $\cos\theta$? If $x^2+y^2=r^2$, and $r\geq 0$, what is $r$?
3. Sep 22, 2012
### Bipolarity
Thanks!
BiP
Similar Discussions: Expressing polar equation as a Cartesian equation | {"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": 0.9849138259887695, "perplexity": 1918.1056204193326}, "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-2017-51/segments/1512948572676.65/warc/CC-MAIN-20171215133912-20171215155912-00091.warc.gz"} |
https://mafiadoc.com/homomorphisms-between-mapping-class-groups_5bbd9ca1097c47e4188b45e6.html | Homomorphisms between mapping class groups
Nov 8, 2010 - Benson Farb, Chris Leininger and especially Johanna Mangahas for many very interesting conversations on the topic of this paper. The.
HOMOMORPHISMS BETWEEN MAPPING CLASS GROUPS
arXiv:1011.1855v1 [math.GT] 8 Nov 2010
JAVIER ARAMAYONA & JUAN SOUTO Abstract. Suppose that X and Y are surfaces of finite topological type, where X has genus g ≥ 6 and Y has genus at most 2g − 1; in addition, suppose that Y is not closed if it has genus 2g − 1. Our main result asserts that every non-trivial homomorphism Map(X) → Map(Y ) is induced by an embedding, i.e. a combination of forgetting punctures, deleting boundary components and subsurface embeddings. In particular, if X has no boundary then every non-trivial endomorphism Map(X) → Map(X) is in fact an isomorphism. As an application of our main theorem we obtain that, under the same hypotheses on genus, if X and Y have finite analytic type then every non-constant holomorphic map M(X) → M(Y ) between the corresponding moduli spaces is a forgetful map. In particular, there are no such holomorphic maps unless X and Y have the same genus and Y has at most as many marked points as X.
1. Introduction Throughout this article we will restrict our attention to connected orientable surfaces of finite topological type, meaning of finite genus and with finitely many boundary components and/or cusps. We will feel free to think about cusps as marked points, punctures or topological ends. If a surface has empty boundary and no cusps, it is said to be closed. The mapping class group Map(X) of a surface X of finite topological type is the group of isotopy classes of orientation preserving homeomorphisms fixing pointwise the union of the boundary and the set of punctures. We denote by T (X) and by M(X) = T (X)/ Map(X) the Teichm¨ uller space and moduli space of X, respectively. The first author has been partially supported by M.E.C. grant MTM2006/14688. The second author has been partially supported by NSF grant DMS-0706878, NSF Career award 0952106, and the Alfred P. Sloan Foundation. 1
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1.1. The conjecture. The triad formed by the mapping class group Map(X), Teichm¨ uller space T (X) and moduli space M(X) is often compared with the one formed by SLn Z, the symmetric space SOn \ SLn R, and the locally symmetric space SOn \ SLn R/ SLn Z; here SLn Z stands as the paradigm of an arithmetic lattice in a higher rank semisimple algebraic group. This analogy has motivated many, possibly most, advances in the understanding of the mapping class group Map(X). For example, Grossman [15] proved that Map(X) is residually finite; Birman, Lubotzky and McCarthy [7] proved that the Tits alternative holds for subgroups of Map(X); the Thurston classification of elements in Map(X) mimics the classification of elements in an algebraic group [44]; Harvey [18] introduced the curve complex in analogy with the rational Tits’ building; Harer’s [17] computation of the virtual cohomological dimension of Map(X) follows the outline of Borel and Serre’s argument for arithmetic groups [8], etc... On the other hand, the comparison between Map(X) and SLn Z has strong limitations; for instance the mapping class group has finite index in its abstract commensurator [23], does not have property (T) [1] and has infinite dimensional second bounded cohomology [5]. In addition, it is not known if the mapping class group contains finite index subgroups Γ with H 1 (Γ; R) 6= 0. With the dictionary between Map(X) and SLn Z in mind, it is natural to ask to what extent there is an analog of Margulis’ superrigidity in the context of mapping class groups. This question, in various guises, has been addressed by a number of authors in recent times. For instance, Farb-Masur [14] proved that every homomorphism from an irreducible lattice in a higher-rank Lie group to a mapping class group has finite image. Notice that, on the other hand, mapping class groups admit non-trivial homomorphisms into higher-rank lattices [30]. With the same motivation, one may try to understand homomorphisms between mapping class groups; steps in this direction include the results of [2, 3, 19, 22, 25, 37]. In the light of this discussion we propose the following general conjecture, which states that, except in some low-genus cases (discussed in Example 1), some version of Margulis’ superrigidity holds for homomorphisms between mapping class groups: Superrigidity conjecture. Margulis’ superrigidity holds for homomorphisms φ : Map(X) → Map(Y ) between mapping class groups as long as X has at least genus three.
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The statement of the superrigidity conjecture is kept intentionally vague for a good reason: different formulations of Margulis’ superrigidity theorem suggest different forms of the conjecture. For instance, recall that the geometric version of superrigidity asserts that any homomorphism Γ → Γ0 between two lattices in simple algebraic groups of higher rank is induced by a totally geodesic immersion MΓ → MΓ0 between the locally symmetric spaces associated to Γ and Γ0 . One possible way of interpreting the superrigidity conjecture for mapping class groups is to ask whether every homomorphism between mapping class groups of, say, surfaces of finite analytic type, induces a holomorphic map between the corresponding moduli spaces. Remark. There are examples [2] of injective homomorphisms Map(X) → Map(Y ) which map some pseudo-Anosovs to multi-twists and hence are not induced by any isometric embedding M(X) → M(Y ) for any reasonable choice of metric on M(X) and M(Y ). This is the reason why we prefer not to ask, as done by Farb and Margalit (see Question 2 of [3]), whether homomorphisms between mapping class groups are geometric. The Lie theoretic version of superrigidity essentially asserts that every homomorphism Γ → Γ0 between two irreducible lattices in higher rank Lie groups either has finite image or extends to a homomorphism between the ambient groups. The mapping class group Map(X) is a quotient of the group of homeomorphisms (resp. diffeomorphisms) of X but not a subgroup [39, 36], and thus there is no ambient group as such. A natural interpretation of this flavor of superrigidity would be to ask whether every homomorphism Map(X) → Map(Y ) is induced by a homomorphism between the corresponding groups of homeomorphisms (resp. diffeomorphisms). Finally, one has the folkloric version of superrigidity: every homomorphism between two irreducible higher rank lattices is one of the “obvious” ones. The word “obvious” is rather vacuous; to give it a little bit of content we adopt Maryam Mirzakhani’s version of the conjecture above: every homomorphism between mapping class groups has either finite image or is induced by some manipulation of surfaces. The statement manipulation of surfaces is again vague, but it conveys the desired meaning. 1.2. The theorem. Besides the lack of counterexamples, the evidence supporting the superrigidity conjecture is limited to the results in [3, 19, 22, 25, 37]. The goal of this paper is to prove the conjecture,
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with respect to any of its possible interpretations, under suitable genus bounds. Before stating our main result we need a definition: Definition 1. Let X and Y be surfaces of finite topological type, and consider their cusps to be marked points. Denote by |X| and |Y | the compact surfaces obtained from X and Y by forgetting all their marked points. By an embedding ι:X→Y we will understand a continuous injective map ι : |X| → |Y | with the property that whenever y ∈ ι(|X|) ⊂ |Y | is a marked point of Y in the image of ι, then ι−1 (y) is also a marked point of X. Note that forgetting a puncture, deleting a boundary component, and embedding X as a subsurface of Y are examples of embeddings. Conversely, every embedding is a combination of these three building blocks; compare with Proposition 3.1 below. It is easy to see that every embedding ι : X → Y induces a homomorphism Map(X) → Map(Y ). Our main result is that, as long as the genus of Y is less than twice that of X, the converse is also true: Theorem 1.1. Suppose that X and Y are surfaces of finite topological type, of genus g ≥ 6 and g 0 ≤ 2g − 1 respectively; if Y has genus 2g − 1, suppose also that it is not closed. Then every nontrivial homomorphism φ : Map(X) → Map(Y ) is induced by an embedding X → Y . Remark. As we will prove below, the conclusion of Theorem 1.1 also applies to homomorphisms φ : Map(X) → Map(Y ) when both X and Y have the same genus g ∈ {4, 5}. We now give some examples that highlight the necessity for the genus bounds in Theorem 1.1. Example 1. Let X be a surface of genus g ≤ 1; if g = 0 then assume that X has at least four marked points or boundary components. The mapping class group Map(X) surjects onto PSL2 Z ' (Z/2Z) ∗ (Z/3Z). In particular, any two elements α, β ∈ Map(Y ) with orders two and three, respectively, determine a homomorphism Map(X) → Map(Y ); notice that such elements exist if Y is closed, for example. Choosing α and β appropriately, one can in fact obtain infinitely many conjugacy classes of homomorphisms Map(X) → Map(Y ) with infinite image and with the property that every element in the image is either pseudoAnosov or has finite order.
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Example 1 shows that some lower bound on the genus of X is necessary in the statement of Theorem 1.1. Furthermore, since Map(X) has non-trivial abelianization if X has genus 2, there exist homomorphisms from Map(X) into mapping class groups of arbitrary closed surfaces Y that are not induced by embeddings. On the other hand, we expect Theorem 1.1 to be true for surfaces of genus g ∈ {3, 4, 5}. Remark. Recall that the mapping class group of a punctured disk is a finite index subgroup of the appropriate braid group. In particular, Example 1 should be compared with the rigidity results for homomorphisms between braid groups, and from braid groups into mapping class groups, due to Bell-Margalit [3] and Castel [11]. Next, observe that an upper bound on the genus of the target surface is also necessary in the statement of Theorem 1.1 since, for instance, the mapping class group of every closed surface injects into the mapping class group of some non-trivial connected cover [2]. Moreover, the following example shows that the bound in Theorem 1.1 is in fact optimal: Example 2. Suppose that X has non-empty connected boundary and let Y be the double of X. Let X1 , X2 be the two copies of X inside Y , and for x ∈ X denote by xi the corresponding point in Xi . Given a homeomorphism f : X → X fixing pointwise the boundary and the cusps define fˆ : Y → Y, fˆ(xi ) = (f (x))i ∀xi ∈ Xi The map f → fˆ induces a homomorphism Map(X) → Map(Y ) which is not induced by any embedding. 1.3. Applications. After having established that in Theorem 1.1 a lower bound for the genus of X is necessary and that the upper bound for the genus of Y is optimal, we discuss some consequences of our main result. First, we will observe that, in the absence of boundary, every embedding of a surface into itself is in fact a homeomorphism; in light of this, Theorem 1.1 implies the following: Theorem 1.2. Let X be a surface of finite topological type, of genus g ≥ 4 and with empty boundary. Then any non-trivial endomorphism φ : Map(X) → Map(X) is induced by a homeomorphism X → X; in particular φ is an isomorphism.
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Remark. The analogous statement of Theorem 1.2 for injective endomorphisms was known to be true by the work of Ivanov and McCarthy [25, 24, 37]. Theorem 1.2, as well as other related results discussed in Section 11, are essentially specializations of Theorem 1.1 to particular situations. Returning to the superrigidity conjecture, recall that in order to prove superrigidity for (cocompact) lattices one may associate, to every homomorphism between two lattices, a harmonic map between the associated symmetric spaces, and then use differential geometric arguments to show that this map is a totally geodesic immersion. This is not the approach we follow in this paper, and neither Teichm¨ uller space nor moduli space will play any role in the proof of Theorem 1.1. As a matter of fact, reversing the logic behind the proof of superrigidity, Theorem 1.1 will actually provide information about maps between moduli spaces, as we describe next. Suppose that X and Y are Riemann surfaces of finite analytical type. Endow the associated Teichm¨ uller spaces T (X) and T (Y ) with the standard complex structure. The latter is invariant under the action of the corresponding mapping class group and hence we can consider the moduli spaces M(X) = T (X)/ Map(X), M(Y ) = T (Y )/ Map(Y ) as complex orbifolds. Suppose now that X and Y have the same genus and that Y has at most as many marked points as X. Choosing an identification between the set of marked points of Y and a subset of the set of marked points of X, we obtain a holomorphic map M(X) → M(Y ) obtained by forgetting all marked points of X which do not correspond to a marked point of Y . Different identifications give rise to different maps; we will refer to these maps as forgetful maps. In Section 12 we will prove the following result: Theorem 1.3. Suppose that X and Y are Riemann surfaces of finite analytic type and assume that X has genus g ≥ 6 and Y genus g 0 ≤ 2g − 1; in the equality case g 0 = 2g − 1 assume that Y is not closed. Then, every non-constant holomorphic map f : M(X) → M(Y ) is a forgetful map. As a direct consequence of Theorem 1.3 we obtain:
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Corollary 1.4. Suppose that X and Y are Riemann surfaces of finite analytic type and assume that X has genus g ≥ 6 and Y genus g 0 ≤ 2g − 1; in the equality case g 0 = 2g − 1 assume that Y is not closed. If there is a non-constant holomorphic map f : M(X) → M(Y ), then X and Y have the same genus and X has at least as many marked points as Y . In order to prove Theorem 1.3 we will deduce from Theorem 1.1 that the map f is homotopic to a forgetful map F . The following result, proved in Section 12, will immediately yield the equality between f and F : Proposition 1.5. Let X and Y be Riemann surfaces of finite analytical type and let f1 , f2 : M(X) → M(Y ) be homotopic holomorphic maps. If f1 is not constant, then f1 = f2 . Recall that the Weil-Peterson metric on moduli space is K¨ahler and has negative curvature. In particular, if the moduli spaces M(X) and M(Y ) were closed, then Proposition 1.5 would follow directly from the work of Eells-Sampson [12]. In order to prove Proposition 1.5 we simply ensure that their arguments go through in our context. 1.4. Strategy of the proof of Theorem 1.1. We now give a brief idea of the proof of Theorem 1.1. The main technical result of this paper is the following theorem: Proposition 1.6. Suppose that X and Y are surfaces of finite topological type of genera g ≥ 6 and g 0 ≤ 2g − 1 respectively; if Y has genus 2g − 1, suppose also that it is not closed. Every nontrivial homomorphism φ : Map(X) → Map(Y ) maps (right) Dehn twists along non-separating curves to (possibly left) Dehn twist along non-separating curves. Given a non-separating curve γ ⊂ X denote by δγ the Dehn twist associated to γ. By Proposition 1.6, φ(δγ ) is a Dehn twist along some non-separating curve φ∗ (γ) ⊂ Y . We will observe that the map φ∗ preserves disjointness and intersection number 1. In particular, φ∗ maps chains in X to chains in Y . In the closed case, it follows easily that there is a unique embedding X → Y which induces the same map on curves as φ∗ ; this is the embedding provided by Theorem 1.1. In the presence of boundary and/or cusps the argument is rather involved, essentially because one needs to determine which cusps and boundary components are to be filled in.
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Hoping that the reader is now convinced that Theorem 1.1 follows after a moderate amount of work from Proposition 1.6, we sketch the proof of the latter. The starting point is a result of Bridson [9] which asserts that, as long as X has genus at least 3, any homomorphism φ : Map(X) → Map(Y ) maps Dehn twists to roots of multitwists. The first problem that we face when proving Proposition 1.6 is that Bridson’s result does not rule out that φ maps Dehn twists to finite order elements. In this direction, one may ask the following question: Question 1. Suppose that φ : Map(X) → Map(Y ) is a homomorphism between mapping class groups of surfaces of genus at least 3, with the property that the image of every Dehn twist along a non-separating curve has finite order. Is the image of φ finite? The answer to this question is trivially positive if ∂Y 6= ∅, for in this case Map(Y ) is torsion-free. We will also give an affirmative answer if Y has punctures: Theorem 1.7. Suppose that X and Y are surfaces of finite topological type, that X has genus at least 3, and that Y is not closed. Then any homomorphism φ : Map(X) → Map(Y ) which maps a Dehn twist along a non-separating curve to a finite order element is trivial. For closed surfaces Y we only give a partial answer to the above question; more concretely, in Proposition 5.1 we will prove that, as long as the genus of Y is in a suitable range determined by the genus of X, then the statement of Theorem 1.7 remains true. Remark. At the end of section 5 we will observe that a positive answer to question 1 would imply that the abelianization of finite index subgroups in Map(X) is finite. This is conjectured to be the case. We continue with the sketch of the proof of Proposition 1.6. At this point we know that for every non-separating curve γ ⊂ X, the element φ(δγ ) is a root of a multitwist and has infinite order. We may thus associate to γ the multicurve φ∗ (γ) supporting the multitwist powers of φ(δγ ). In principle, and also in practice if Y has sufficiently large genus, φ(δγ ) could permute the components of φ∗ (γ). However, under the genus bounds in Theorem 1.1, we deduce from a result of Paris [41] that this is not the case. Once we know that φ(δγ ) fixes each component of φ∗ (γ), a simple counting argument yields that φ∗ (γ) is actually a single curve. This implies that φ(δγ ) is a root of some power of the Dehn twist along φ∗ (γ). In the last step, which we now describe, we will obtain that φ(δγ ) is in fact a Dehn twist.
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Denote by Xγ the surface obtained by deleting from X an open regular neighborhood of γ, and let Yφ0∗ (γ) be the surface obtained by deleting φ∗ (γ) from Y . The centralizer of the Dehn twist δγ in Map(X) is closely related to Map(Xγ ) and the same is true for the centralizer of φ(δγ ) and Map(Yφ0∗ (γ) ). More concretely, the homomorphism φ induces a homomorphism Map(Xγ ) → Map(Yφ0∗ (γ) ) Moreover, it follows from the construction that φ(δγ ) is in fact a power of the Dehn twist along φ∗ (γ) if the image of this homomorphism is not centralized by any finite order element in Map(Yφ0∗ (γ) ). This will follow from an easy computation using the Riemann-Hurwitz formula together with the next result: Proposition 1.8. Suppose that X and Y are surfaces of finite topological type. If the genus of X is at least 3 and larger than that of Y , then there is no nontrivial homomorphism φ : Map(X) → Map(Y ). Remark. If X is closed, Proposition 1.8 is due to Harvey and Korkmaz [19]. Their argument makes heavy use of torsion in Map(X); therefore, it cannot be used for general surfaces. Once we know that φ(δγ ) is a power of a Dehn twist, it follows from the braid relation that this power has to be ±1, as we needed to prove. This finishes the sketch of the proof of Proposition 1.6 and hence of Theorem 1.1. Before concluding the introduction, we would like to mention a related result due to Bridson and Vogtmann [10] on homomorphisms between outer automorphism groups of free groups, namely: Theorem (Bridson-Vogtmann). Suppose n > 8. If n is even and n < m ≤ 2n, or if n is odd and n < m ≤ 2n − 2, then every homomorphism Out(Fn ) → Out(Fm ) factors through a homomorphism Out(Fn ) → Z/2Z. We remark that in their proof, Bridson and Vogtmann make very heavy use of the presence of rather large torsion subgroups in Out(Fn ). Rather on the contrary, in the present paper torsion is an annoyance. In particular, the proof of the Bridson-Vogtmann theorem and that of Theorem 1.1 are completely different. In spite of that, we would like to mention that the Bridson-Vogtmann theorem played a huge role in this paper: it suggested the possibility of understanding homomorphisms between mapping class groups Map(X) → Map(Y ) under the assumption that the genus of Y is not much larger than that of X.
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Acknowledgements. The authors wish to thank Martin Bridson, Benson Farb, Chris Leininger and especially Johanna Mangahas for many very interesting conversations on the topic of this paper. The authors are also grateful to Michel Boileau and Luis Paris for letting them know about the work of Fabrice Castel. The first author wishes to express his gratitude to Ser Peow Tan and the Institute for Mathematical Sciences of Singapore, where parts of this work were completed. 2. Generalities In this section we discuss a few well-known facts on mapping class groups. See [13, 24] for details. Throughout this article, all surfaces under consideration are orientable and have finite topological type, meaning that they have finite genus, finitely many boundary components and finitely many punctures. We will feel free to consider cusps as marked points, punctures, or ends homeomorphic to S1 × R. For instance, if X is a surface with, say, 10 boundary components and no cusps, by deleting every boundary component we obtain a surface X 0 with 10 cusps and no boundary components. A simple closed curve on a surface is said to be essential if it does not bound a disk containing at most one puncture; we stress that we consider boundary-parallel curves to be essential. From now on, by a curve we will mean an essential simple closed curve. Also, we will often abuse terminology and not distinguish between curves and their isotopy classes. We now introduce some notation that will be used throughout the paper. Let X be a surface and let γ be an essential curve not parallel to the boundary of X. We will denote by Xγ the complement in X of the interior of a closed regular neighborhood of γ; we will refer to the two boundary components of Xγ which appear in the boundary of the regular neighborhood of γ as the new boundary components of Xγ . We will denote by Xγ0 the surface obtained from Xγ by deleting the new boundary components of Xγ ; equivalently, Xγ0 = X \ γ. A multicurve is the union of a, necessarily finite, collection of pairwise disjoint, non-parallel curves. Given two multicurves γ, γ 0 we denote their geometric intersection number by i(γ, γ 0 ). A cut system is a multicurve whose complement is a connected surface of genus 0. Two cut systems are said to be related by an elementary move if they share all curves but one, and the remaining two curves intersect exactly once. The cut system complex of a surface X is the simplicial graph whose vertices are cut systems on X and where two
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cut systems are adjacent if the corresponding cut systems are related by an elementary move. 2.1. Mapping class group. The mapping class group Map(X) of a surface X is the group of isotopy classes of orientation preserving homeomorphisms X → X which fix the boundary pointwise and map every cusp to itself; here, we also require that the isotopies fix the boundary pointwise. We will also denote by Map∗ (X) the group of isotopy classes of all orientation preserving homeomorphisms of X. Observe that Map(X) is a subgroup of Map∗ (X) only in the absence of boundary; in this case Map(X) has finite index in Map∗ (X). While every element of the mapping class group is an isotopy class of homeomorphisms, it is well-known that the mapping class group cannot be realized by a group of diffeomorphisms [39], or even homeomorphisms [36]. However, for our purposes the difference between actual homeomorphisms and their isotopy classes is of no importance. In this direction, and in order to keep notation under control, we will usually make no distinction between mapping classes and their representatives. 2.2. Dehn twists. Given a curve γ on X, we denote by δγ the (right) Dehn twist along γ. It is important to remember that δγ is solely determined by the curve γ and the orientation of X. In other words, it is independent of any chosen orientation of γ. Perhaps the main reason why Dehn twists appear so prominently in this paper is because they generate the mapping class group: Theorem 2.1 (Dehn-Lickorish). If X has genus at least 2, then Map(X) is generated by Dehn twists along non-separating curves. There are quite a few known concrete sets of Dehn twists which generate the mapping class group. We will consider the so-called Humphries generators [21]; see Figure 1 for a picture of the involved curves. a1
b1
a2
b2
a3
c
b3
bg
rk
r1
r2
Figure 1. The Humphries generators: Dehn twists along the curves ai , bi , c and ri generate Map(X).
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Algebraic relations among Dehn twists are often given by particular configurations of curves. We now discuss several of these relations; see [13, 16, 33] and the references therein for proofs and details: Conjugate Dehn twists. For any curve γ ⊂ X and any f ∈ Map(X) we have δf (γ) = f δγ f −1 Hence, Dehn twists along any two non-separating curves are conjugate in Map(X). Conversely, if the Dehn twist along γ is conjugate in Map(X) to a Dehn twist along a non-separating curve, then γ is nonseparating. Observe that Theorem 2.1 and the fact that Dehn twists along any two non-separating curves are conjugate immediately imply the following very useful fact: Lemma 2.2. Let X be a surface of genus at least 3 and let φ : Map(X) → G be a homomorphism. If δγ ∈ Ker(φ) for some γ ⊂ X non-separating, then φ is trivial. Disjoint curves. Suppose γ, γ 0 are disjoint curves, meaning i(γ, γ 0 ) = 0. Then δγ and δγ 0 commute. Curves intersecting once. Suppose that two curves γ and γ 0 intersect once, meaning i(γ, γ 0 ) = 1. Then δγ δγ 0 δγ = δγ 0 δγ δγ 0 This is the so-called braid relation; we say that δγ and δγ 0 braid. It is known [16] that if γ and γ 0 are two curves in X and k ∈ Z is such that |k · i(γ, γ 0 )| ≥ 2, then δγk and δγk0 generate a free group F2 of rank 2. In particular we have: Lemma 2.3. Suppose that k ∈ Z \ {0} and that γ and γ 0 are curves such that δγk and δγk0 satisfy the braid relation, then either γ = γ 0 or k = ±1 and i(γ, γ 0 ) = 1. Chains. Recall that a chain in X is a finite sequence of curves γ1 , . . . , γk such that i(γi , γj ) = 1 if |i − j| = 1 and i(γi , γj ) = 0 otherwise. Let γ1 , . . . , γk be a chain in X and suppose first that k is even. Then the boundary ∂Z of a regular neighborhood of ∪γi is connected and we have (δγ1 δγ2 . . . δγk )2k+2 = δ∂Z If k is odd then ∂Z consists of two components ∂1 Z and ∂2 Z and the appropriate relation is (δγ1 δγ2 . . . δγk )k+1 = δ∂Z1 δ∂Z2 = δ∂Z2 δ∂Z1 These two relations are said to be the chain relations.
13
Lanterns. A lantern is a configuration in of seven curves a, b, c, d, x, y and z in X as represented in figure 2.
a
x b
y
d
c z
Figure 2. A lantern If seven curves a, b, c, d, x, y and z in X form a lantern then the corresponding Dehn twists satisfy the so-called lantern relation: δa δb δc δd = δx δy δz Conversely, it is due to Hamidi-Tehrani [16] and Margalit [33] that, under mild hypotheses, any seven curves whose associated Dehn twists satisfy the lantern relation form a lantern. More concretely: Proposition 2.4 (Hamidi-Tehrani, Margalit). Let a, b, c, d, x, y, z be essential curves whose associated Dehn twists satisfy the lantern relation δa δb δc δd = δx δy δz If the curves a, b, c, d, x are paiwise distinct and pairwise disjoint, then a, b, c, d, x, y, z is a lantern. In the course of this paper we will continuously discriminate against separating curves. By a non-separating lantern we understand a lantern with the property that all the involved curves are non-separating. It is well-known, and otherwise easy to see, that X contains a nonseparating lantern if X has genus at least 3. In particular we deduce that, as long as X has genus g ≥ 3, every non-separating curve belongs to a non-separating lantern. 2.3. Centralizers of Dehn twists. Observe that the relation f δγ f −1 = δf (γ) , for f ∈ Map(X) and γ ⊂ X a curve, implies that Z(δγ ) = {f ∈ Map(X) | f (γ) = γ},
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JAVIER ARAMAYONA & JUAN SOUTO
where Z(δγ ) denotes the centralizer of δγ in Map(X). Notice that Z(δγ ) is also equal to the normalizer N (hδγ i) of the subgroup of Map(X) generated by δγ . An element in Map(X) which fixes γ may either switch the sides of γ or may preserve them. We denote by Z0 (δγ ) the group of those elements which preserve sides; observe that Z0 (δγ ) has index at most 2 in Z(δγ ). The group Z0 (δγ ) is closely related to two different mapping class groups. First, let Xγ be the surface obtained by removing the interior of a closed regular neighborhood γ × [0, 1] of γ from X. Every homeomorphism of Xγ fixing pointwise the boundary and the punctures extends to a homeomorphism X → X which is the identity on X \ Xγ . This induces a homomorphism Map(Xγ ) → Map(X); more concretely we have the following exact sequence: (2.1)
0 → Z → Map(Xγ ) → Z0 (δγ ) → 1
Here, the group Z is generated by the difference δη1 δη−1 of the Dehn 2 twists along η1 and η2 , the new boundary curves of Xγ . Instead of deleting a regular neighborhood of γ we could also delete γ from X. Equivalently, let Xγ0 be the surface obtained from Xγ by deleting the new boundary curves of Xγ . Every homeomorphism of X fixing γ induces a homeomorphism of Xγ0 . This yields a second exact sequence (2.2)
0 → hδγ i → Z0 (δγ ) → Map(Xγ0 ) → 1
2.4. Multitwists. To a multicurve η ⊂ X we associate the group Tη = h{δγ , γ ⊂ η}i ⊂ Map(X) generated by the Dehn twists along the components of η. We refer to the elements in Tη as multitwists along η. Observe that Tη is abelian; more concretely, Tη is isomorphic to the free abelian group with rank equal to the number of components of η. Let η ⊂ X be a multicurve. An element f ∈ Tη which does not belong to any Tη0 , for some η 0 properly contained in η, is said to be a generic multitwist along η. Conversely, if f ∈ Map(X) is a multitwist, then the support of f is the smallest multicurve η such that f is a generic multitwist along η. Much of what we just said about Dehn twists extends easily to multitwists. For instance, if η ⊂ X is a multicurve, then we have Tf (η) = f Tη f −1
15
for all f ∈ Map(X). In particular, the normalizer N (Tη ) of Tη in Map(X) is equal to N (Tη ) = {f ∈ Map(X)|f (η) = η} On the other hand, the centralizer Z(Tη ) of Tη is the intersection of the centralizers of its generators; hence Z(Tη ) = {f ∈ Map(X)|f (γ) = γ for every component γ ⊂ η} Notice that N (Tη )/Z(Tη ) acts by permutations on the set of components of η. For further use we observe that if the multicurve η happens to be a cut system, then N (Tη )/Z(Tη ) is in fact isomorphic to the group of permutations of the components of η. Denote by Z0 (Tη ) the subgroup of Z(Tη ) fixing not only the components but also the sides of each component. Notice that Z(Tη )/Z0 (Tη ) is a subgroup of (Z/2Z)|η| and hence is abelian. Observe that it follows from the definition of the mapping class group and from the relation δf (γ) = f δγ f −1 that every Dehn twist along a boundary component of X is central in Map(X). In fact, as long as X has at least genus 3, such Dehn twists generate the center of Map(X): Theorem 2.5. If X has genus at least 3 then the group T∂X generated by Dehn twists along the boundary components of X is the center of Map(X). Moreover, we have 1 → T∂X → Map(X) → Map(X 0 ) → 1 where X 0 is the surface obtained from X by deleting the boundary. Notice that if X is a surface of genus g ∈ {1, 2}, with empty boundary and no marked points, then the center of Map(X) is generated by the hyperelliptic involution. 2.5. Roots. It is a rather surprising, and annoying, fact that such simple elements in Map(X) as Dehn twists have non-trivial roots [34]. Recall that a root of f ∈ Map(X) is an element g ∈ Map(X) for which there is k ∈ Z with f = g k . Being forced to live with roots, we state here a few simple but important observations: Lemma 2.6. Suppose that f ∈ Tη and f 0 ∈ Tη0 are generic multitwists along multicurves η, η 0 ⊂ X. If f and f 0 have a common root, then η = η0. Lemma 2.7. Suppose that f ∈ Tη is a generic multitwist along a multicurve η and f 0 ∈ Map(X) is a root of f . Then f 0 (η) = η and hence f 0 ∈ N (Tη ).
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If η is an essential curve then Tη = hδη i; hence N (Tη ) = Z(δη ) is the subgroup of Map(X) preserving η. Recall that Z0 (δη ) is the subgroup of Z(δη ) which preserves sides of η. Lemma 2.8. Suppose that δη ∈ Map(X) is a Dehn twist along an essential curve η. For f ∈ Z0 (δη ) the following are equivalent: • f is a root of a power of δη , and • the image of f in Map(Xη0 ) under the right arrow in (2.2) has finite order. Moreover, f is itself a power of δη if and only if the image of f in Map(Xη0 ) is trivial. 2.6. Torsion. In the light of Lemma 2.8, it is clear that the existence of roots is closely related to the presence of torsion in mapping class groups. While it is known that every mapping class group always contains a finite index torsion-free subgroup, this is not going to be of much use here. The fact that the mapping class group of a surface with boundary is torsion-free is going to be of more importance. Theorem 2.9. If X is a surface with nonempty boundary, then Map(X) is torsion-free. Similarly, if X has marked points, then finite subgroups of Map(X) are cyclic. The key to understand torsion in mapping class groups is the resolution by Kerckhoff [26] of the Nielsen realization problem: the study of finite subgroups of the mapping class group reduces to the study of groups of automorphism of Riemann surfaces. For instance, it follows from the classical Hurewitz theorem that the order of such a group is bounded from above solely in terms of the genus of the underlying surface. Below we will need the following bound, due to Maclachlan [31] and Nakajima [40], for the order of finite abelian subgroups of Map(X). Theorem 2.10. Suppose that X has genus g ≥ 2. Then Map(X) does not contain finite abelian groups with more than 4g + 4 elements. We remark that if g ≤ 5 all finite subgroups, abelian or not, of Map(X) have been listed [28, 29]. In the sequel we will make use of this list in the case that g = 3, 4. Finally, we observe that a finite order diffeomorphism which is isotopic to the identity is in fact the identity. This implies, for instance, ¯ is obtained from X by filling in punctures, and τ : X → X that if X is a finite order diffeomorphism representing a non-trivial element in ¯ is non-trivial as well. Map(X), then the induced mapping class of X
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2.7. Centralizers of finite order elements. By (2.1) and (2.2), centralizers of Dehn twists are closely related to other mapping class groups. Essentially the same is true for centralizers of other mapping classes. We now discuss the case of torsion elements. The following result follows directly from the work of Birman-Hilden [6]: Theorem 2.11 (Birman-Hilden). Suppose that [τ ] ∈ Map(X) is an element of finite order and let τ : X → X be a finite order diffeomorphism representing [τ ]. Consider the orbifold O = X/hτ i and let O∗ be the surface obtained from O by removing the singular points. Then we have a sequence 1 → h[τ ]i → Z([τ ]) → Map∗ (O∗ ) where Map∗ (O∗ ) is the group of isotopy classes of all homeomorphisms O∗ → O∗ . Hidden in Theorem 2.11 we have the following useful fact: two finite order diffeomorphisms τ, τ 0 : X → X which are isotopic are actually conjugate as diffeomorphisms (see the remark in [4, p.10]). Hence, it follows that the surface O∗ in Theorem 2.11 depends only on the mapping class [τ ]. Abusing notation, in the sequel we will speak about the fixed-point set of a finite order element in Map(X). 3. Homomorphisms induced by embeddings In this section we define what is meant by an embedding ι : X → Y between surfaces. As we will observe, any embedding induces a homomorphism between the corresponding mapping class groups. We will discuss several standard examples of such homomorphisms, notably the so-called Birman exact sequences. We will conclude the section with a few observations that will be needed later on. Besides the possible differences of terminology, all the facts that we will state are either well known or simple observations in 2-dimensional topology. A reader who is reasonably acquainted with [13, 24] will have no difficulty filling the details. 3.1. Embeddings. Let X and Y be surfaces of finite topological type, and consider their cusps to be marked points. Denote by |X| and |Y | the compact surfaces obtained from X and Y , respectively, by forgetting all the marked points, and let PX ⊂ |X| and PY ⊂ |Y | be the sets of marked points of X and Y . Definition. An embedding ι : X → Y is a continuous injective map ι : |X| → |Y | such that ι−1 (PY ) ⊂ PX . An embedding is said to be a homeomorphism if it has an inverse which is also an embedding.
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We will say that two embeddings ι, ι0 : X → Y are equivalent or isotopic if there if a continuous injective map f : |Y | → |Y | which is isotopic (but not necessarily ambient isotopic) to the identity relative to the set PY of marked points of Y and which satisfies f ◦ ι = ι0 . Abusing terminology, we will often say that equivalent embeddings are the same. Given an embedding ι : X → Y and a homeomorphism f : X → X which pointwise fixes the boundary and the marked points of X, we consider the homeomorphism ι(f ) : Y → Y given by ι(f )(x) = (ι◦f ◦ι−1 )(x) if x ∈ ι(X) and ι(f )(x) = x otherwise. Clearly, ι(f ) is a homeomorphism which pointwise fixes the boundary and the marked points of Y . In particular ι(f ) represents an element ι# (f ) in Map(Y ). It is easy to check that we obtain a well-defined group homomorphism ι# : Map(X) → Map(Y ) characterized by the following property: for any curve γ ⊂ X we have ι# (δγ ) = δι(γ) . Notice that this characterization immediately implies that if ι and ι0 are isotopic, then ι# = ι0# . 3.2. Birman exact sequences. As we mentioned above, notable examples of homomorphisms induced by embeddings are the so-called Birman exact sequences, which we now describe. Let X and Y be surfaces of finite topological type. We will say that Y is obtained from X by filling in a puncture if there is an embedding ι : X → Y and a marked point p ∈ PX , such that the underlying map ι : |X| → |Y | is a homeomorphism, and ι−1 (PY ) = PX \ {p}. If Y is obtained from X by filling in a puncture we have the following exact sequence: (3.1)
1
/
π1 (|Y | \ PY , ι(p))
/
Map(X)
ι#
/
Map(Y )
/
1
The left arrow in (3.1) can be described concretely. For instance, if γ is a simple loop in |Y | based ι(p) and avoiding all other marked points of Y , then the image of the element [γ] ∈ π1 (|Y | \ PY , ι(p)) in Map(X) is the difference of the two Dehn twists along the curves forming the boundary of a regular neighborhood of ι−1 (γ). Similarly, we will say that Y is obtained from X by filling in a boundary component if there is an embedding ι : X → Y , with ι−1 (PY ) = PX , and such that the complement in |Y | of the image of the underlying map |X| → |Y | is a disk which does not contain any marked point of
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Y . If Y is obtained from X by filling in a boundary component then we have the following exact sequence: (3.2)
1 → π1 (T 1 (|Y | \ PY )) → Map(X) → Map(Y ) → 1
Here T 1 (|Y | \ PY ) is the unit-tangent bundle of the surface |Y | \ PY . We refer to the sequences (3.1) and (3.2) as the Birman exact sequences. 3.3. Other building blocks. As we will see in the next subsection, any embedding is a composition of four basic building blocks. Filling punctures and filling boundary components are two of them; next, we describe the other two types. Continuing with the same notation as above, we will say that Y is obtained from X by deleting a boundary component if there is an embedding ι : X → Y with ι(PX ) ⊂ PY and such that the complement of the image of the underlying map |X| → |Y | is disk containing exactly one point in PY . If Y is obtained from X by deleting a boundary component then we have (3.3)
1 → Z → Map(X) → Map(Y ) → 1
where, Z is the group generated by the Dehn twist along the forgotten boundary component. Finally, we will say that ι : X → Y is a subsurface embedding if ι(PX ) ⊂ PY and if no component of the complement of the image of the underlying map |X| → |Y | is a disk containing at most one marked point. Notice that if ι : X → Y is a subsurface embedding, then the homomorphism ι# : Map(X) → Map(Y ) is injective if and only if ι is anannular, i.e. if no component of the complement of the image of the underlying map |X| → |Y | is an annulus without marked points; compare with (2.1) above. 3.4. General embeddings. Clearly, the composition of two embeddings is an embedding. For instance, observe that filling in a boundary component is the same as first forgetting it and then filling in a puncture. The following proposition, whose proof we leave to the reader, asserts that every embedding is isotopic to a suitable composition of the elementary building blocks we have just discussed: Proposition 3.1. Every embedding ι : X → Y is isotopic to a composition of the following three types of embedding: filling punctures, deleting boundary components, and subsurface embeddings. In particular, the homomorphism ι# : Map(X) → Map(Y ) is injective if and only if ι is an anannular subsurface embedding.
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Notation. In order to avoid notation as convoluted as T 1 (|Y | \ PY ), most of the time we will drop any reference to the underlying surface |Y | or to the set of marked point PY ; notice that this is consistent with taking the freedom to consider punctures as marked points or as ends. For instance, the Birman exact sequences now read 1 → π1 (Y ) → Map(X) → Map(Y ) → 1, if Y is obtained from X by filling in a puncture, and 1 → π1 (T 1 Y ) → Map(X) → Map(Y ) → 1, if it is obtained filling in a boundary component. We hope that this does not cause any confussion. 3.5. Two observations. We conclude this section with two observations that will be needed below. First, suppose that ι : X → Y is an embedding and let η ⊂ X be a multicurve. The image ι(η) of η in Y is an embedded 1-manifold, but it does not need to be a multicurve. For instance, some component of ι(η) may not be essential in Y ; also two components of ι(η) may be parallel in Y . If this is not the case, that is, if ι(η) is a multicurve in Y , then it is easy to see that ι# maps the subgroup Tη of multitwists supported on η isomorphically onto Tι(η) . We record this observation in the following lemma: Lemma 3.2. Let ι : X → Y be an embedding and let η ⊂ X be a multicurve such that • every component of ι(η) is essential in Y , and • no two components of ι(η) are parallel in Y , Then ι(η) is also a multicurve in Y and the homomorphism ι# maps Tη ⊂ Map(X) isomorphically to Tι(η) ⊂ Map(Y ). Moreover, the image of a generic multitwist in Tη is generic in Tι(η) . Finally, we recall the well-known fact that the Birman exact sequence (3.1) does not split: Lemma 3.3. Suppose that X is a surface of genus g ≥ 3 with empty boundary and a single puncture. If Y is the closed surface obtained from X by filling in the puncture, then the exact sequence (3.1) 1 → π1 (Y ) → Map(X) → Map(Y ) → 1 does not split. Proof. The mapping class group Map(Y ) of the closed surface Y contains a non-cyclic finite subgroup; namely one isomorphic to Z/2Z ×
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Z/2Z. By Theorem 2.9 such a subgroup does not exist in Map(X), which proves that there is no splitting of (3.1). In Lemma 3.3 we proved that in a very particular situation one of the two Birman exact sequences does not split. Notice however that it follows from Theorem 1.1 that they never do; this also follows from the work of Ivanov-McCarthy [25]. 4. Triviality theorems In this section we remind the reader of two triviality theorems for homomorphisms from mapping class groups to abelian groups and permutation groups; these results are widely used throughout this paper. The first of these results is a direct consequence of Powell’s theorem [42] on the vanishing of the integer homology of the mapping class group of surfaces of genus at least 3: Theorem 4.1 (Powell). If X is a surface of genus g ≥ 3 and A is an abelian group, then every homomorphism Map(X) → A is trivial. We refer the reader to Korkmaz [27] for a discussion of Powell’s theorem and other homological properties of mapping class groups. As a first consequence of Theorem 4.1 we derive the following useful observation: Lemma 4.2. Let X, Y and Y¯ be surfaces of finite topological type, and let ι : Y → Y¯ be an embedding. Suppose that X has genus at least 3 and that φ : Map(X) → Map(Y ) is a homomorphism such that the composition φ¯ = ι# ◦ φ : Map(X) → Map(Y¯ ) is trivial. Then φ is trivial as well. Proof. By Proposition 3.1 the embedding ι : Y → Y¯ is a suitable composition of filling in punctures and boundary components, deleting boundary components and subsurface embeddings. In particular, we may argue by induction and assume that ι is of one of these four types. For the sake of concreteness suppose ι : Y¯ → Y is the embedding associated to filling in a boundary component; the other cases are actually a bit easier and are left to the reader. We have the following diagram: Map(X)
1
/
π1 (T 1 Y¯ )
LLL LLLφ¯ φ LLL L& ι# / Map(Y / Map(Y ) ¯) /
1
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JAVIER ARAMAYONA & JUAN SOUTO
The assumption that φ¯ is trivial amounts to supposing that the image of φ is contained in π1 (T 1 Y¯ ). The homomorphism π1 (T 1 Y¯ ) → π1 (Y¯ ) yields another diagram: Map(X)
1
/
Z
/
π1
JJ JJ φ0 JJ φ JJ J% 1¯ /π (T Y )
¯)
1 (Y
/
1
Since every nontrivial subgroup of the surface group π1 (Y¯ ) has nontrivial homology, we deduce from Theorem 4.1 that φ0 is trivial. Hence, the image of φ is contained in Z. Applying again Theorem 4.1 above we deduce that φ is trivial, as it was to be shown. Before stating another consequence of Theorem 4.1 we need a definition: Definition 2. A homomorphism φ : Map(X) → Map(Y ) is said to be irreducible if its image does not fix any essential curve in Y ; otherwise we say it is reducible. Remark. Recall that we consider boundary parallel curves to be essential. In particular, every homomorphism Map(X) → Map(Y ) is reducible if Y has non-empty boundary. Let φ : Map(X) → Map(Y ) be a reducible homomorphism, where X has genus at least 3, and let η ⊂ Y be a multicurve which is componentwise invariant under φ(Map(X)); in other words, φ(Map(X)) ⊂ Z(Tη ). Moreover, notice that Theorem 4.1 implies that φ(Map(X)) ⊂ Z0 (Tη ), where Z0 (Tη ) is the subgroup of Z(Tη ) consisting of those elements that fix the sides of each component of η. Now let Yγ0 = Y \ η be the surface obtained by deleting η from Y . Composing (2.2) as often as necessary, we obtain an exact sequence as follows: (4.1)
1 → Tη → Z0 (Tη ) → Map(Yη0 ) → 1
The same argument of the proof of Lemma 4.2 shows that φ is trivial if the composition of φ and the right homomorphism (4.1) is trivial. Hence we have: Lemma 4.3. Let X, Y be surfaces of finite topological type. Suppose that X has genus at least 3 and that φ : Map(X) → Map(Y ) is a nontrivial reducible homomorphism fixing the multicurve η ⊂ Y . Then φ(Map(X)) ⊂ Z0 (Tη ) and the composition of φ with the homomorphism (4.1) is not trivial.
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The second trivially theorem, due to Paris [41], asserts that the mapping class group of a surface of genus g ≥ 3 does not have subgroups of index less than or equal to 4g + 4; equivalently, any homomorphism from the mapping class group into a symmetric group Sk is trivial if k ≤ 4g + 4: Theorem 4.4 (Paris). If X has genus g ≥ 3 and k ≤ 4g + 4, then there is no nontrivial homomorphism Map(X) → Sk where the latter group is the group of permutations of the set with k elements. Before going any further we should mention that in [41], Theorem 4.4 is only stated for closed surfaces. However, the proof works as it is also for surfaces with boundary and or punctures. We leave it to the reader to check that this is the case. As a first consequence of Theorem 4.1 and Theorem 4.4 we obtain the following special case of Proposition 1.8: Proposition 4.5. If X has genus at least 3 and Y at most genus 2, then every homomorphism φ : Map(X) → Map(Y ) is trivial. Proof. Assume for concreteness that Y has genus 2; the cases of genus 0 and genus 1 are in fact easier and are left to the reader. Notice that by Lemma 4.2, we may assume without lossing generality that Y has empty boundary and no marked points. Recall that Map(Y ) has a central element τ of order 2, namely the hyperelliptic involution. As we discussed above, we identify the finite order mapping class τ with one of its finite order representatives, which we again denote by τ . The surface underlying the orbifold Y /hτ i is the 6-punctured sphere S0,6 . By Theorem 2.11 we have the following exact sequence: 1 → hτ i → Map(Y ) → Map∗ (S0,6 ) → 1 where Map∗ (S0,6 ) is, as always, the group of isotopy classes of all orientation preserving homeomorphisms of S0,6 . Therefore, any homomorphism φ : Map(X) → Map(Y ) induces a homomorphism φ0 : Map(X) → Map∗ (S0,6 ) By Paris’ theorem, the homomorphism obtained by composing φ0 with the obvious homomorphism Map∗ (S0,6 ) → S6 , the group of permutations of the punctures, is trivial. In other words, φ0 takes values in Map(S0,6 ). Since the mapping class group of the standard sphere S2 is trivial, Lemma 4.2 implies that φ0 is trivial. Therefore, the image of φ is contained in the abelian subgroup hτ i ⊂ Map(Y ). Finally, Theorem 4.1 implies that φ is trivial, as we had to show.
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5. Getting rid of the torsion We begin this section reminding the reader of a question posed in the introduction: Question 1. Suppose that φ : Map(X) → Map(Y ) is a homomorphism between mapping class groups of surfaces of genus at least 3, with the property that the image of every Dehn twist along a non-separating curve has finite order. Is the image of φ finite? In this section we will give a positive answer to the question above if the genus of Y is exponentially bounded by the genus of X. Namely: Proposition 5.1. Suppose that X and Y are surfaces of finite topological type with genera g and g 0 respectively. Suppose that g ≥ 4 and that either g 0 < 2g−2 − 1 or g 0 = 3, 4. Any homomorphism φ : Map(X) → Map(Y ) which maps a Dehn twist along a non-separating curve to a finite order element is trivial. Under the assumption that Y is not closed, we obtain in fact a complete answer to the question above: Theorem 1.7. Suppose that X and Y are surfaces of finite topological type, that X has genus at least 3, and that Y is not closed. Then any homomorphism φ : Map(X) → Map(Y ) which maps a Dehn twist along a non-separating curve to a finite order element is trivial. Recall that by Theorem 2.9, the mapping class group of a surface with non-empty boundary is torsion-free. Hence we deduce from Lemma 2.2 that it suffices to consider the case that ∂Y = ∅. From now on, we assume that we are in this situation. The proofs of Proposition 5.1 and Theorem 1.7 are based on Theorem 4.1, the connectivity of the cut system complex, and the following algebraic observation: Lemma 5.2. For n ∈ N, n ≥ 2, consider Zn endowed with the standard action of the symmetric group Sn by permutations of the basis elements e1 , . . . , en . If V is a finite abelian group equipped with an Sn action, then for any Sn -equivariant epimorphism φ : Zn → V one of the following two is true: (1) Either the restriction of φ to Zn−1 × {0} is surjective, or (2) V has order at least 2n and cannot be generated by fewer than n elements. Moreover, if (1) does not hold and V 6= (Z/2Z)n then V has at least 2n+1 elements.
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Proof. Let d be the order of φ(e1 ) in V and observe that, by Sn equivariance, all the elements φ(ei ) also have order d. It follows that (dZ)n ⊂ Ker(φ) and hence that φ descends to an epimorphism φ0 : (Z/dZ)n → V Our first goal is to restrict to the case that d is a power Q of a prime. a In order to do this, consider the prime decomposition d = j pj j of d, where pi 6= pj and ai ∈ N. By the Chinese remainder theorem we have Y a Z/dZ = Z/pj j Z j
Hence, there is a Sn -equivariant isomorphism Y a (Z/dZ)n = (Z/pj j Z)n j a
Consider the projection πj : Zn → (Z/pj j Z)n and observe that if the a restriction to Zn−1 × {0} of φ0 ◦ πj surjects onto φ0 ((Z/pj j Z)n ) for all j, then φ(Zn−1 × {0}) = V . Supposing that this were not the case, replace φ by φ0 ◦ πj and V a by φ((Z/pj j Z)n ). In more concrete terms, we can assume from now on that d = pa is a power of a prime. At this point we will argue by induction. The key claim is the following surely well-known observation: Claim. Suppose that p is prime. The only Sn -invariant subgroups W of (Z/pZ)n are the following: • The trivial subgroup {0}, • (Z/pZ)n itself, • E = {(a, a, . . . , a) ∈ (Z/pZ)n |a = 0, . . . , p − 1}, and • F = {(a1 , . . . , an ) ∈ (Z/pZ)n |a1 + · · · + an = 0}. Proof of the claim. Suppose that W ⊂ (Z/pZ)n is not trivial and take v = (vi ) ∈ W nontrivial. If v cannot be chosen to have distinct entries then W = E. So suppose that this is not case and choose v with two distinct entries, say v1 and v2 . Consider the image v 0 of v under the transposition (1, 2). By the Sn -invariance of W we have v 0 ∈ W and hence v − v 0 ∈ W . By construction, v − v 0 has all entries but the two first ones equal to 0. Moreover, each of the first two entries is the negative of the other one. Taking a suitable power we find that (1, −1, 0, . . . , 0) ∈ W . By the Sn -invariance of W we obtain that every element with one 1, one −1 and 0 otherwise belongs to W . These elements span F .
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We have proved that either W is trivial, or W = E or F ⊂ W . Since the only subgroups containing F are F itself and the total space, the claim follows. Returning to the proof of Lemma 5.2, suppose first that a = 1, i.e. d = p is prime. The kernel of the epimorphism φ0 : (Z/pZ)n → V is a Sn -invariant subspace. Either φ0 is injective, and thus V contains pn ≥ 2n elements, or its kernel is one of the spaces E or F provided by the claim. Since the union of either one of them with (Z/pZ)n−1 × {0} spans (Z/pZ)n , it follows that the restriction of φ to Zn−1 ×{0} surjects onto V . This concludes the proof if a = 1. Suppose that we have proved the result for a − 1. We can then consider the diagram: /
0
0
/
(Z/pa−1 Z)n 0
a−1
φ ((Z/p
/
/
(Z/pa Z)n
(Z/pZ)n
φ0 n
Z) )
/
V
/
0
V /φ ((Z/pa−1 Z)n )
/
/
0
0
Observe that if one of the groups to the left and right of V on the bottom row has at least 2n elements, then so does V . So, if this is not the case we may assume by induction that the restriction of the left and right vertical arrows to (Z/pa−1 Z)n−1 × {0} and (Z/pZ)n−1 × {0} are epimorphisms. This shows that the restriction of φ0 to (Z/pa Z)n−1 ×{0} is also an epimorphism. It follows that either V has at least 2n elements or the restriction of φ to Zn−1 × {0} is surjective, as claimed. Both the equality case and the claim on the minimal number of elements needed to generate V are left to the reader. We are now ready to prove: Lemma 5.3. Given n ≥ 4, suppose that g > 0 is such that 2n−2 −1 > g or g ∈ {3, 4}. If Y is surface of genus g ≥ 3, V ⊂ Map(Y ) is a finite abelian group endowed with an action of Sn , and φ : Zn → V is a Sn -equivariant epimomorphism, then the restriction of φ to Zn−1 × {0} is surjective. Proof. Suppose, for contradiction, that the restriction of φ to Zn−1 ×{0} is not surjective. Recall that by the resolution of the Nielsen realization problem [26] there is a conformal structure on Y such that V can be represented by a group of automorphisms. Suppose first that 2n−2 − 1 > g. Since we are assuming that the restriction of φ to Zn−1 × {0} is not surjective, Lemma 5.2 implies that
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V has at least 2n elements. Then: 2n = 4(2n−2 − 1) + 4 > 4g + 4, which is impossible since Theorem 2.10 asserts that Map(Y ) does not contain finite abelian groups with more than 4g + 4 elements. Suppose now that g = 4. If n ≥ 5 we obtain a contradiction using the same argument as above. Thus assume that n = 4. Since 24+1 = 32 > 20 = 4 · 4 + 4, it follows from the equality statement in Lemma 5.2 that V is isomorphic to (Z/2Z)4 . Luckily for us, Kuribayashi-Kuribayashi [28] have classified all groups of automorphisms of Riemann surfaces of genus 3 and 4. From their list, more concretely Proposition 2.2 (c), we obtain that (Z/2Z)4 cannot be realized as a subgroup of the group of automorphisms of a surface of genus 4, and thus we obtain the desired contradiction. Finally, suppose that g = 3. As before, this case boils down to ruling out the possibility of having (Z/2Z)4 acting by automorphisms on a Riemann surface of genus 3. This is established in Proposition 1.2 (c) of [28]. This concludes the case g = 3 and thus the proof of the lemma. Remark. One could wonder if in Lemma 5.3 the condition n ≥ 4 is necessary. Indeed it is, because the mapping class group of a surface of genus 3 contains a subgroup isomorphic to (Z/2Z)3 , namely the group H(8, 8) in the list in [28]. After all this immensely boring work, we are finally ready to prove Proposition 5.1. Proof of Proposition 5.1. Recall that a cut system in X is a maximal multicurve whose complement in X is connected; observe that every cut system consists of g curves and that every non-separating curve is contained in some cut system. Given a cut system η consider the group Tη generated by the Dehn twists along the components of η, noting that Tη ' Zg . Any permutation of the components of η can be realized by a homeomorphism of X. Consider the normalizer N (Tη ) and centralizer Z(Tη ) of Tη in Map(X). As mentioned in Section 2.4, we have the following exact sequence: 1 → Z(Tη ) → N (Tη ) → Sg → 1, where Sg denotes the symmetric group of permutations of the components of η. Observe that the action by conjugation of N (Tη ) onto Tη induces an action Sg = N (Tη )/Z(Tη ) y Tη which is conjugate to the standard action of Sg y Zg . Clearly, this action descends to an action Sg y φ(Tη ).
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Seeking a contradiction, suppose that the image under φ of a Dehn twist δγ along a non-separating curve has finite order. Since all the Dehn twists along the components of η are conjugate to δγ we deduce that all their images have finite order; hence φ(Tη ) is generated by finite order elements. On the other hand, φ(Tη ) is abelian because it is the image of an abelian group. Being abelian and generated by finite order elements, φ(Tη ) is finite. It thus follows from Lemma 5.3 that the subgroup of Tη generated by Dehn twists along g − 1 components of η surjects under φ onto φ(Tη ). This implies that φ(Tη ) = φ(Tη0 ) 0 whenever η and η are cut systems which differ by exactly one component. Now, since the cut system complex is connected [20], we deduce that φ(δα ) ∈ φ(Tη ) for every non-separating curve α. Since Map(X) is generated by Dehn twists along non-separating curves, we deduce that the image of Map(X) is the abelian group φ(Tη ). By Theorem 4.1, any homomorphism Map(X) → Map(Y ) with abelian image is trivial, and thus we obtain the desired contradiction. Before moving on we discuss briefly the proof of Theorem 1.7. Suppose that Y is not closed. Then, every finite subgroup of Map(Y ) is cyclic by Theorem 2.9. In particular, the bound on the number of generators in Lemma 5.2 implies that if V ⊂ Map(Y ) is a finite abelian group endowed with an action of Sn and φ : Zn → V is a Sn -equivariant epimomorphism then the restriction of φ to Zn−1 × {0} is surjective. Once this has been established, Theorem 1.7 follows with the same proof, word for word, as Proposition 5.1. Remark. Let X and Y be surfaces, where Y has a single boundary component and no cusps. Let G be a finite index subgroup of Map(X) and let φ : G → Map(Y ) be a homomorphism. A simple modification of a construction due to Breuillard-Mangahas [32] yields a closed surface Y 0 containing Y and a homomorphism φ0 : Map(X) → Map(Y 0 ) such that for all g ∈ G we have, up to isotopy, φ0 (g)(Y ) = Y and φ0 (g)|Y = φ(g). Suppose now that G could be chosen so that there is an epimorphism G → Z. Assume further that φ : G → Map(Y ) factors through this epimorphism and that the image of φ is purely pseudo-Anosov. Then, every element in the image of the extension φ0 : Map(X) → Map(Y ) either has finite order or is a partial pseudo-Anosov. A result of Bridson [9], stated as Theorem 6.1 below, implies that every Dehn twist in
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Map(X) is mapped to a finite order element in Map(Y ). Hence, the extension homomorphism φ0 produces a negative answer to Question 1. We have hence proved that a positive answer to Question 1 implies that every finite index subgroup of Map(X) has finite abelianization. 6. The map φ∗ In addition to the triviality results given in Theorems 4.1 and 4.4, the third key ingredient in the proof of Theorem 1.1 is the following result due to Bridson [9]: Theorem 6.1 (Bridson). Suppose that X, Y are surfaces of finite type and that X has genus at least 3. Any homomorphism φ : Map(X) → Map(Y ) maps roots of multitwists to roots of multitwists. A remark on the proof of Theorem 6.1. In [9], Theorem 6.1 is proved for surfaces without boundary only. However, Bridson’s argument remains valid if we allow X to have boundary. That the result can also be extended to the case that Y has non-empty boundary needs a minimal argument, which we now give. Denote by Y 0 the surface obtained from Y by deleting all boundary components and consider the homomorphism π : Map(Y ) → Map(Y 0 ) provided by Theorem 2.5. By Bridson’s theorem, the image under π ◦ φ of a Dehn twist δγ is a root of a multitwist. Since the kernel of π is the group of multitwists along the boundary of Y , it follows that φ(δγ ) is also a root of a multitwist, as claimed. A significant part of the sequel is devoted to proving that under suitable assumptions the image of a Dehn twist is in fact a Dehn twist. We highlight the apparent difficulties in the following example: Example 3. Suppose that X has a single boundary component and at least two punctures. By [15], the mapping class group Map(X) is residually finite. Fix a finite group G and an epimorphism π : Map(X) → G. It is easy to construct a connected surface Y on which G acts and which contains |G| disjoint copies Xg (g ∈ G) of X with gXh = Xgh for all g, h ∈ G. Given x ∈ X, denote the corresponding element in Xg by xg . If f : X → X is a homeomorphism fixing pointwise the boundary and punctures, we define fˆ : Y → Y with fˆ(xg ) = (f (x))π([f ])g for xg ∈ Xg and fˆ(y) = π([f ])(y) for y ∈ / ∪g∈G Xg ; here [f ] is the element in Map(X) represented by f .
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Notice that fˆ does not fix the marked points of Y ; in order to by-pass this difficulty, consider Y¯ the surface obtained from Y by forgetting all marked points, and consider fˆ to be a self-homeomorphism of Y¯ . It is easy to see that the map f 7→ fˆ induces a homomorphism φ : Map(X) → Map(Y¯ ) with some curious properties, namely: • If γ ⊂ X is a simple closed curve which bounds a disk with at least two punctures then the image φ(δγ ) of the Dehn twist δγ along γ has finite order. Moreover, δγ ∈ Ker(φ) if and only if δγ ∈ Ker(π). • If γ ⊂ X is a non-separating simple closed curve then φ(δγ ) has infinite order. Moreover, φ(δγ ) is a multitwist if δγ ∈ Ker(π); otherwise, φ(δγ ) is a non-trivial root of a multitwist. Observe that in the latter case, φ(δγ ) induces a non-trivial permutation of the components of the multicurve supporting any of its multitwist powers. This concludes the discussion of Example 3. While a finite order element is by definition a root of a multitwist, Proposition 5.1 ensures that, under suitable bounds on the genera of the surfaces involved, any non-trivial homomorphism Map(X) → Map(Y ) maps Dehn twists to infinite order elements. From now on we assume that we are in the following situation: (*) X and Y are orientable surfaces of finite topological type, of genus g and g 0 respectively, and such that one of the following holds: • Either g ≥ 4 and g 0 ≤ g, or • g ≥ 6 and g 0 ≤ 2g − 1. Remark. It is worth noticing that the reason for the genus bound g ≥ 6 in Theorem 1.1 is that 2g−2 − 1 < 2g − 1 if g < 6. Assuming (*), it follows from Proposition 5.1 that any non-trivial homomorphism φ : Map(X) → Map(Y ) maps Dehn twists δγ along non-separating curves γ to infinite order elements in Map(Y ). Furthermore, it follows from Theorem 6.1 that there is N such that φ(δγN ) is a non-trivial multitwist. We denote by φ∗ (γ) the multicurve in Y supporting φ(δγN ), which is independent of the choice of N by Lemma 2.6. Notice that two multitwists commute if and only if their supports do not intersect; hence, φ∗ preserves the property of having zero intersection number. Moreover, the uniqueness of φ∗ (γ) implies that for
31
any f ∈ Map(X) we have φ∗ (f (γ)) = φ(f )(φ∗ (γ)). Summing up we have: Corollary 6.2. Suppose that X and Y are as in (*) and let φ : Map(X) → Map(Y ) be a non-trivial homomorphism. For every non-separating curve γ ⊂ X, there is a uniquely determined multicurve φ∗ (γ) ⊂ Y with the property that φ(δγ ) is a root of a generic multitwist in Tφ∗ (γ) . Moreover the following holds: • i(φ∗ (γ), φ∗ (γ 0 )) = 0 for any two disjoint non-separating curves γ and γ 0 , and • φ∗ (f (γ)) = φ(f )(φ∗ (γ)) for all f ∈ Map(X). In particular, the multicurve φ∗ (γ) is invariant under φ(Z(δγ )). The remainder of this section is devoted to give a proof of the following result: Proposition 6.3. Suppose that X and Y are as in (*); further, assume that Y is not closed if it has genus 2g − 1. Let φ : Map(X) → Map(Y ) be an irreducible homomorphism. Then, for every non-separating curve γ ⊂ X the multicurve φ∗ (γ) is a non-separating curve. Recall that a homomorphism φ : Map(X) → Map(Y ) is irreducible if its image does not fix any curve in Y , and that if φ is irreducible then ∂Y = ∅; see Definition 2 and the remark following. Before launching the proof of Proposition 6.3 we will establish a few useful facts. Lemma 6.4. Suppose X and Y satisfy (*) and that φ : Map(X) → Map(Y ) is an irreducible homomorphism. Let Y¯ be obtained from Y by filling in some, possibly all, punctures of Y , and let φ¯ = ι# ◦ φ : Map(X) → Map(Y¯ ) be the composition of φ with the homomorphism ι# induced by the embedding ι : Y → Y¯ . For every non-separating curve γ ⊂ X we have: • ι(φ∗ (γ)) is a multicurve, and • φ¯∗ (γ) = ι(φ∗ (γ)). In particular, ι yields a bijection between the components of φ∗ (γ) and φ¯∗ (γ). Proof. First observe that, arguing by induction, we may assume that Y¯ is obtained from Y by filling in a single cusp. We suppose from now on that this is the case and observe that it follows from Lemma 4.2 that φ¯ is not trivial. Notice also that since Y and Y¯ have the same genus, φ¯∗ (γ) is well-defined by Corollary 6.2.
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By definition of φ∗ and φ¯∗ , we can choose N ∈ N such that φ(δγN ) ¯ N ) are generic multitwists in Tφ (γ) and Tφ¯ (γ) . In particular, and φ(δ ∗ γ ∗ it follows from Lemma 3.2 that in order to prove Lemma 6.4 it suffices to show that ι(φ∗ (γ)) does not contain (1) inessential components, or (2) parallel components. Claim 1. ι(φ∗ (γ)) does not contain inessential components. Proof of Claim 1. Seeking a contradiction, suppose that a component η of φ∗ (γ) is inessential in Y¯ . Since Y¯ is obtained from Y by filling in a single cusp, it follows that η bounds a disk in Y with exactly two punctures. Observe that this implies that for any element F ∈ Map(Y ) we have either F (η) = η or i(F (η), η) > 0. On the other hand, if f ∈ Map(X) is such that i(f (γ), γ) = 0 then we have i(φ(f )(η), η) ≤ i(φ(f )(φ∗ (γ)), φ∗ (γ)) = i(φ∗ (f (γ)), φ∗ (γ)) = 0 We deduce that η = φ(f )(η) ⊂ φ∗ (f (γ)) for any such f . Since any two non-separating curves in X are related by an element of Map(X) we obtain: (?) If γ 0 is a non-separating curve in X with i(γ, γ 0 ) = 0 then η = φ(δγ 0 )(η) and η ⊂ φ∗ (γ 0 ). Choose γ 0 ⊂ X so that X \(γ ∪γ 0 ) is connected. It follows from (?) that if γ 00 is any other non-separating curve which either does not intersect γ or γ 0 we have φ(δγ 00 )(η) = η. Since the mapping class group is generated by such curves, we deduce that every element in φ(Map(X)) fixes η, contradicting the assumption that φ is irreducible. This concludes the proof of Claim 1. We use a similar argument to prove that ι(φ∗ (γ)) does not contain parallel components. Claim 2. ι(φ∗ (γ)) does not contain parallel components. Proof of Claim 2. Seeking again a contradiction suppose that there are η 6= η 0 ⊂ φ∗ (γ) whose images in Y¯ are parallel. Hence, η ∪ η 0 bounds an annulus which contains a single cusp. As above, it follows that for any element f ∈ Map(Y ) we have either f (η ∪ η 0 ) = η ∪ η 0 or i(f (η), η) > 0. By the same argument as before, we obtain that φ(Map(X)) preserves η ∪ η 0 . Now, it follows from either Theorem 4.1 or Theorem 4.4 that φ(Map(X)) cannot permute η and η 0 . Hence φ(Map(X)) fixes η, contradicting the assumption that φ is irreducible. As we mentioned above, Lemma 6.4 follows from Claim 1, Claim 2 and Lemma 3.2.
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Continuing with the preliminary considerations to prove Proposition 6.3, recall that the final claim in Corollary 6.2 implies that φ(δγ ) preserves the multicurve φ∗ (γ). Our next goal is to show that, as long as φ is irreducible, the element φ(δγ ) preserves every component of φ∗ (γ). Lemma 6.5. Suppose that X and Y are as in (*) and let φ : Map(X) → Map(Y ) be an irreducible homomorphism. If γ ⊂ X is a non-separating simple closed curve, then φ(Z0 (δγ )) fixes every component of φ∗ (γ). Hence, φ(Z0 (δγ )) ⊂ Z0 (Tφ∗ (γ) ). Recall that Z0 (δγ ) is the subgroup of Map(X) fixing not only γ but also the two sides of γ and that it has at most index 2 in the centralizer Z(δγ ) of the Dehn twist δγ . Proof. We first prove Lemma 6.5 in the case that Y is closed. As in Section 2, we denote by Xγ the surface obtained by deleting the interior of a closed regular neighborhood of γ from X. Recall that by (2.1) there is a surjective homomorphism Map(Xγ ) → Z0 (δγ ) Consider the composition of this homomorphism with φ and, abusing notation, denote its image by φ(Map(Xγ )) = φ(Z0 (δγ )). By Corollary 6.2, the subgroup φ(Map(Xγ )) of Map(Y ) acts on the set of components of φ∗ (γ) and hence on Y \ φ∗ (γ). Since Y is assumed to be closed and of at most genus 2g − 1 we deduce that Y \ φ∗ (γ) has at most |χ(Y )| = 2g 0 − 2 ≤ 4g − 4 components. Since the surface Xγ has genus g − 1 ≥ 3, we deduce from Theorem 4.4 that φ(Map(Xγ )) fixes each component of Y \ φ∗ (γ). Suppose now that Z is a component of Y \ φ∗ (γ) and let η be the set of components of φ∗ (γ) contained in the closure of Z. Noticing that 4 − 4g ≤ χ(Y ) ≤ χ(Z) ≤ −|η| + 2 we obtain that η consists of at most 4g−2 components. Since φ(Map(Xγ )) fixes Z, it acts on the set of components of η. Again by Theorem 4.4, it follows that this action is trivial, meaning that every component of φ∗ (γ) contained in the closure of Z is preserved. Since Z was arbitrary, we deduce that φ(Map(Xγ ) preserves every component of φ∗ (γ) as claimed. Lemma 4.3 now implies that φ(Z0 (δγ )) = φ(Map(Xγ )) ⊂ Z0 (Tφ∗ (γ) ). This concludes the proof of Lemma 6.5 in the case that Y is closed. We now turn our attention to the general case. Recall that the assumption that φ is irreducible implies that ∂Y = ∅. Let Y¯ be the surface obtained from Y by closing up all the cusps and denote by
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φ¯ : Map(X) → Map(Y¯ ) the composition of φ with the homomorphism ι# : Map(Y ) → Map(Y¯ ) induced by the embedding ι : Y → Y¯ . By the ¯ On the other hand, Lemma 6.4 above, Lemma 6.5 holds true for φ. shows that for any γ ⊂ X non-separating there is a bijection between φ∗ (γ) and φ¯∗ (γ). Thus the claim follows. Note that Lemma 6.5 yields the following sufficient condition for a homomorphism between mapping class groups to be reducible: Corollary 6.6. Suppose that X and Y are as in (*) and let φ : Map(X) → Map(Y ) be a non-trivial homomorphism. Let γ and γ 0 be distinct, disjoint curves on X such that X \ (γ ∪ γ 0 ) is connected. If the multicurves φ∗ (γ) and φ∗ (γ 0 ) share a component, then the homomorphism φ : Map(X) → Map(Y ) is reducible. Proof. Suppose that φ is irreducible and observe that Map(X) is generated by Dehn twists along curves α which are disjoint from γ or γ 0 . For any such α we have δα ∈ Z0 (δγ ) ∪ Z0 (δγ 0 ). In particular, it follows from Proposition 6.5 that φ(Map(X)) fixes every component of φ∗ (γ) ∩ φ∗ (γ 0 ). The assumption that φ was irreducible implies that φ∗ (γ) ∩ φ∗ (γ 0 ) = ∅. We are now ready to prove Proposition 6.3: Proof of Proposition 6.3. Let γ be a non-separating curve on X. Extend γ to a multicurve η ⊂ X with 3g − 3 components γ1 , . . . , γ3g−3 , and such that the surface X \ (γi ∪ γj ) is connected for all i, j. Since δγi and δγj are conjugate in Map(X) we deduce that φ∗ (γi ) and φ∗ (γj ) have the same number K of components for all i, j. Since φ is irreducible, Corollary 6.6 implies that φ∗ (γi ) and φ∗ (γj ) do not share any components for all i 6= j. This shows that ∪i φ∗ (γi ) is the union of (3g − 3)K distinct curves. Furthermore, since δγi and δγj commute, we deduce that ∪i φ∗ (γi ) is a multicurve in Y . Suppose first that Y has genus g 0 ≤ 2g−2. In the light of Lemma 6.4, it suffices to consider the case that Y is closed. Now, the multicurve ∪i φ∗ (γi ) has at most 3g 0 − 3 ≤ 3(2g − 2) − 3 < 6g − 6 components. Hence: 6g − 6 K< ≤ 2, 3g − 3 and thus the multicurve φ∗ (γ) consists of K = 1 components; in other words, it is a curve. It is non-separating because otherwise the multicurve ∪i φ∗ (γi ) would consist of 3g − 3 separating curves, and a closed surface of genus g 0 ≤ 2g − 2 contains at most 2g − 3 disjoint separating curves. This concludes the proof of the proposition in the case that Y has genus at most 2g − 2.
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Suppose now that Y has genus g 0 = 2g − 1 and that Y is not closed. Again by Lemma 6.4, we can assume that Y has a single puncture, which we consider as a marked point. In this case, the multicurve ∪i φ∗ (γi ) consists of at most 3g 0 − 2 = 6g − 5 curves. Since we know that ∪i φ∗ (γi ) is the union of (3g − 3)K distinct curves, we deduce K ≤ 2. In the case that ∪i φ∗ (γi ) has fewer than 6g − 6 components, we proceed as before. Therefore, it remains to rule out the possibility of having exactly 6g − 6 components. Suppose, for contradiction, that ∪i φ∗ (γi ) has 6g − 6 components. Since Y has genus 2g − 1 and exactly one marked point, the complement of ∪i φ∗ (γi ) in Y is a disjoint union of pairs of pants, where exactly one of them, call it P , contains the marked point of Y . Now, the boundary components of P are contained in the image under φ∗ of curves a1 , a2 , a3 ∈ {γ1 , . . . , γ3g−3 }. Assume, for the sake of concreteness, that ai 6= aj whenever i 6= j; the remaining case is dealt with using minor modifications of the argument we give here. Suppose first that the multicurve α = a1 ∪ a2 ∪ a3 does not disconnect X and let α0 6= α be another multicurve with three components satisfying: (1) X \ α0 is connected, (2) i(α, α0 ) = 0, and (3) X \ (γ ∪ γ 0 ) is connected for all γ, γ 0 ∈ α ∪ α0 . Notice that since X \ α and X \ α0 are homeomorphic, there is f ∈ Map(X) with f (α) = α0 . Now, P 0 = φ(f )(P ) is a pair of pants which contains the marked point of Y . Taking into account that ∂P ⊂ φ∗ (α) and ∂P 0 ⊂ φ∗ (α0 ) we deduce from (2) that i(∂P, ∂P 0 ) = ∅ and hence that P = P 0 . Since α0 6= α we may assume, up to renaming, that a1 6⊂ α0 . Since φ(f )(∂P ) = ∂P 0 and ∂P ∩ φ∗ (a1 ) 6= ∅, we deduce that is i such that φ∗ (ai ) ∩ φ∗ (f (a1 )) contains a boundary curve of P . In the light of (3), it follows from Corollary 6.6 that φ is reducible; this contradiction shows that X \ α cannot be connected. If X \α is not connected, then it has two components, as X \(a1 ∪a2 ) is connected. Suppose first that neither of the two components Z1 , Z2 of X \ α is a (possibly punctured) pair of pants and notice that this implies that Z1 and Z2 both have positive genus. Let P1 ⊂ Z1 be an unpunctured pair of pants with boundary ∂P1 = a1 ∪ a2 ∪ a03 and let P2 ⊂ Z2 be second unpunctured a pair of pants with Z2 \ P2 connected and with boundary ∂P2 = a3 ∪a01 ∪a02 where a01 and a02 are not boundary parallel in Z2 ; compare with Figure 3. Notice that Z10 = (Z1 ∪ P2 ) \ P1 is homeomorphic to Z1 . Similarly, Z20 = (Z2 ∪ P1 ) \ P2 is homeomorphic to Z2 . Finally notice also that Zi0 contains the same punctures as Zi
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Figure 3. for i = 1, 2. It follows from the classification theorem of surfaces that there is f ∈ Map(X) with f (Z1 ) = Z10 and f (Z2 ) = Z20 . In particular, f (α) = α0 where α0 = a01 ∪ a02 ∪ a03 . We highlight a few facts: (1) There is f ∈ Map(X) with f (α) = α0 , (2) i(α, α0 ) = 0, and (3) X \ (γ ∪ γ 0 ) is connected for all γ, γ 0 ∈ {a1 , a2 , a01 , a02 }. As above, we deduce that φ(f )(∂P ) = ∂P 0 and that for all i = 1, 2, 3 there is j such that φ∗ (ai ) ∩ φ∗ (f (aj )) contains a boundary curve of P . In the light of (3), it follows again from Corollary 6.6 that φ is reducible. We have reduced to the case that one of the components of X \ α, say Z1 , is a (possibly punctured) pair of pants. We now explain how to reduce to the case that Z1 is a pair of pants without punctures. Let a03 ⊂ Z1 be a curve which, together with a3 , bounds an annulus A ⊂ Z1 such that Z1 \ A does not contain any marked points. Notice that we may assume without loss of generality that the multicurve γ1 ∪· · ·∪γ3g−3 above does not intersect a03 . It follows that i(φ∗ (a03 ), ∪φ∗ (γi )) = 0. Next, observe that a pants decomposition of Y consists of 3(2g − 1) − 3 + 1 = 6g − 5 curves. Since φ∗ (a03 ) has two components and ∪φ∗ (γi ) has 6g − 6 components, we deduce that there exists i such that φ∗ (a03 ) and φ∗ (γi ) share a component. If i 6= 3, property (3) and Corollary 6.6 imply that φ is reducible, since a03 ∪ γi does not separate X. It thus follows that φ∗ (a03 ) and φ∗ (a3 ) share a component, and so ∂P ⊂ φ∗ (a1 ∪ a2 ∪ a03 ). Summing up, it remains to rule out the possibility that Z1 is a pair of pants without punctures. Choose α0 ⊂ X satisfying: (1) α0 bounds a pair of pants in X, (2) i(α, α0 ) = 0, and (3) X \ (γ ∪ γ 0 ) is connected for all γ, γ 0 ∈ α ∪ α0 . Now there is f ∈ Map(X) with f (α) = α0 and we can repeat word by word the argument given in the case that X \ α was connected. After having ruled out all possibilities, we deduce that ∪i φ∗ (γi ) cannot have 6g − 6 components. This concludes the proof of Proposition 6.3.
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7. Proof of Proposition 1.8 In this section we show that every homomorphism Map(X) → Map(Y ) is trivial if the genus of X is larger than that of Y . As a consequence we obtain that, under suitable genus bounds, the centralizer of the image of a non-trivial homomorphism between mapping class groups is torsion-free. Proposition 1.8. Suppose that X and Y are orientable surfaces of finite topological type. If the genus of X is at least 3 and larger than that of Y , then every homomorphism φ : Map(X) → Map(Y ) is trivial. Recall that Proposition 1.8 is due to Harvey-Korkmaz [19] in the case that both surfaces X and Y are closed. Proof. We will proceed by induction on the genus of X. Notice that Proposition 4.5 establishes the base case of the induction and observe that by Lemma 4.2 we may assume that Y is has empty boundary and no cusps. Suppose now that X has genus g ≥ 4 and that we have proved Proposition 1.8 for surfaces of genus g − 1. Our first step is to prove the following: Claim. Under the hypotheses above, every homomorphism Map(X) → Map(Y ) is reducible. Proof of the claim. Seeking a contradiction, suppose that there is an irreducible homomorphism φ : Map(X) → Map(Y ), where Y has smaller genus than X. Let γ ⊂ X be a non-separating curve. Observing that X and Y satisfy (*), we deduce that φ∗ (γ) is a non-separating curve by Proposition 6.3 and that φ(Z0 (δγ )) ⊂ Z0 (δφ∗ (γ) ) by Lemma 6.5. By (2.2), Z0 (δφ∗ (γ) ) dominates Map(Yφ0∗ (γ) ) where Yφ0∗ (γ) = Y \ φ∗ (γ). On the other hand, we have by (2.1) that Z0 (δγ ) is dominated by the group Map(Xγ ) where Xγ is obtained from X by deleting the interior of a closed regular neighborhood of γ. Since φ∗ (γ) is non-separating, the genus of Yφ0∗ (γ) and Xγ is one less than that of Y and X, respectively. The induction assumption implies that the induced homomorphism Map(Xγ ) → Map(Yφ0∗ (γ) ) is trivial. The last claim in Lemma 4.3 proves that the homomorphism Map(Xγ ) → Z0 (δφ∗ (γ) ) ⊂ Map(Y ) is also trivial. We have proved that Z0 (δγ ) ⊂ Ker(φ). Since Z0 (δγ ) contains a Dehn twist along a non-separating curve, we deduce that φ
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is trivial from Lemma 2.2. This contradiction concludes the proof of the claim. Continuing with the proof of the induction step in Proposition 1.8, suppose there exists a non-trivial homomorphism φ : Map(X) → Map(Y ). By the above claim, φ is reducible. Let η ⊂ Y be a maximal multicurve in Y which is componentwise fixed by φ(Map(X)), and notice that φ(Map(X)) ⊂ Z0 (Tη ) by Lemma 4.3. Consider φ0 : Map(X) → Map(Yη0 ), the composition of φ with the homomorphism (4.1). The maximality of the multicurve η implies that φ0 is irreducible. Since the genus of Yη0 is at most equal to that of Y , we deduce from the claim above that φ0 is trivial. Lemma 4.3 implies hence that φ is trivial as well. This establishes Proposition 1.8 As we mentioned before, a consequence of Proposition 1.8 is that, under suitable assumptions, the centralizer of the image of a homomorphism between mapping class groups is torsion-free. Namely, we have: Lemma 7.1. Let X and Y be surfaces of finite topological type, where X has genus g ≥ 3 and Y has genus g 0 ≤ 2g. Suppose moreover that Y has at least one (resp. three) marked points if g 0 = 2g − 1 (resp. g 0 = 2g). If φ : Map(X) → Map(Y ) is a non-trivial homomorphism, then the centralizer of φ(Map(X)) in Map(Y ) is torsion-free. The proof of Lemma 7.1 relies on Proposition 1.8 and the following consequence of the Riemann-Hurwitz formula: Lemma 7.2. Let Y be a surface of genus g 0 ≥ 0 and let τ : Y → Y be a nontrivial diffeomorphism of prime order, representing an element in Map(Y ). Then τ has F ≤ 2g 0 + 2 fixed-points and the underlying 0 surface of the orbifold Y /hτ i has genus at most g¯ = 2g +2−F . 4 Proof. Consider the orbifold Y /hτ i and let F be the number of its singular points; observe that F is also equal to the number of fixed points of τ since τ has prime order p. Denote by |Y /hτ i| the underlying surface of the orbifold Y /hτ i. The Riemann-Hurwitz formula shows that (7.1)
2 − 2g 0 = χ(Y ) = p · χ(|Y /hτ i|) − (p − 1) · F
After some manipulations, (7.1) shows that F =
2g 0 − 2 + p · (2 − 2¯ g) p−1
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where g¯ is the genus of |Y /hτ i|. Clearly, the quantity on the right is maximal if g¯ = 0 and p = 2. This implies that F ≤ 2g 0 + 2, as claimed. Rearranging (7.1), we obtain g¯ =
2g 0 + (2 − F )(p − 1) 2p
Again this is maximal if p is as small as possible, i.e. p = 2. Hence 0 . g¯ ≤ 2g +2−F 4 We are now ready to prove Lemma 7.1. Proof of Lemma 7.1. First, if Y has non-empty boundary there is nothing to prove, for in this case Map(Y ) is torsion-free. Therefore, assume that ∂Y = ∅. Suppose, for contradiction, that there exists [τ ] ∈ Map(Y ) non-trivial, of finite order, and such that φ(Map(X)) ⊂ Z([τ ]). Let τ : Y → Y be a finite order diffeomorphism representing [τ ]. Passing to a suitable power, we may assume that the order of τ is prime. Consider the orbifold Y /hτ i as a surface with the singular points marked, and recall that by Theorem 2.11 we have the following exact sequence: 1
/
h[τ ]i
/
Z([τ ])
β
/
Map∗ (Y /hτ i)
On the other hand, we have by definition 1 → Map(Y /hτ i) → Map∗ (Y /hτ i) → SF → 1 where F is the number of punctures of Y /hτ i. Again, F is equal to the number of fixed points of τ since τ has prime order. Observe that Lemma 7.2 gives that F ≤ 2g 0 + 2 ≤ 4g + 2; hence, it follows from Theorem 4.4 that the composition of the homomorphism Map(X)
φ
/
Z([τ ])
β
/
Map∗ (Y /hτ i)
/
SF
is trivial; in other words, (β ◦ φ)(Map(X)) ⊂ Map(Y /hτ i). Our assumptions on the genus and the marked points of Y imply, by the genus bound in Lemma 7.2, that Y /hτ i has genus less than g. Hence, the homomorphism β ◦ φ : Map(X) → Map(Y /hτ i) is trivial by Proposition 1.8. This implies that the image of φ is contained in the abelian group h[τ ]i. Theorem 4.1 shows hence that φ is trivial, contradicting our assumption. This concludes the proof of Lemma 7.1 The following example shows that Lemma 7.1 is no longer true if Y is allowed to have genus 2g and fewer than 3 punctures.
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Example 4. Let X be a surface with no punctures and such that ∂X = S1 . Let Z be a surface of the same genus as X, with ∂Z = ∅ but with two punctures. Regard X as a subsurface of Z and consider the two-to-one cover Y → Z corresponding to an arc in Z \ X joining the two punctures of Z. Every homomorphism X → X fixing poinwise the boundary extends to a homeomorphism of Z fixing the punctures and which lifts to a unique homeomorphism Y → Y which fixes the two components of the preimage of X under the covering Y → Z. The image of the induced homomorphism Map(X) → Map(Y ) is centralized by the involution τ associated to the two-to-one cover Y → Z. Moreover, if X has genus g then Y has genus 2g and 2 punctures. 8. Proof of Proposition 1.6 We are now ready to prove that under suitable genus bounds, homomorphisms between mapping class groups map Dehn twists to Dehn twists. Namely: Proposition 1.6. Suppose that X and Y are surfaces of finite topological type, of genus g ≥ 6 and g 0 ≤ 2g − 1 respectively; if Y has genus 2g − 1, suppose also that it is not closed. Every nontrivial homomorphism φ : Map(X) → Map(Y ) maps (right) Dehn twists along non-separating curves to (possibly left) Dehn twists along non-separating curves. Remark. The reader should notice that the proof of Proposition 1.6 applies, word for word, to homomorphisms between mapping class groups of surfaces of the same genus g ∈ {4, 5}. We will first prove Proposition 1.6 under the assumption that φ is irreducible and then we will deduce the general case from here. Proof of Proposition 1.6 for irreducible φ. Suppose that φ is irreducible and recall that this implies that ∂Y = ∅. Let γ ⊂ X be a nonseparating curve. Thus φ∗ (γ) is also a non-separating curve, by Proposition 6.3. We first show that φ(δγ ) is a power of δφ∗ (γ) . Let Xγ be the complement in X of the interior of a closed regular neighborhood of γ and Yφ0∗ (γ) = Y \ φ∗ (γ) the connected surface obtained from Y by removing φ∗ (γ). We have that: (?) Xγ and Yφ0∗ (γ) have genus g − 1 ≥ 3 and g 0 − 1 ≤ 2g − 2 respectively. Moreover, observe that Yφ0∗ (γ) has two more punctures than Y ; in particular, Yγ0 has at least 3 punctures if it has genus 2g − 2.
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By (2.1) and (2.2) we have epimorphisms Map(Xγ ) → Z0 (φ(δγ ))
and
Z0 (δφ∗ (γ) ) → Map(Yφ0∗ (γ) ).
In addition, we know that φ(Z0 (δγ )) ⊂ Z0 (δφ∗ (γ) ) by Lemma 6.5. Composing all these homomorphisms we get a homomorphism φ0 : Map(Xγ ) → Map(Yφ0∗ (γ) ) It follows from Lemma 2.2 that the restriction of φ to Z0 (δγ ) is not trivial because the latter contains a Dehn twist along a non-separating curve; Lemma 4.3 implies that φ0 is not trivial either. Since δγ centralizes Z0 (δγ ), it follows that φ0 (δγ ) ∈ Map(Yφ0∗ (γ) ) centralizes the image of φ0 . Now, the definition of φ∗ (γ) implies that some power of φ(δγ ) is a power of the Dehn twist δφ∗ (γ) . Hence, the first claim of Lemma 2.8 yields that φ0 (δγ ) has finite order, and thus φ0 (δγ ) ∈ Map(Yφ0∗ (γ) ) is a finite order element centralizing φ(Map(Xγ )). By (?), Lemma 7.1 applies and shows that φ0 (δγ ) is in fact trivial. The final claim of Lemma 2.8 now shows that φ(δγ ) is a power of δφ∗ (γ) ; in other words, there exists N ∈ Z \ {0} such that φ(δγ ) = δφN∗ (γ) . It remains to prove that N = ±1. Notice that N does not depend on the particular non-separating curve γ since any two Dehn twists along non-separating curves are conjugate. Consider a collection γ1 , . . . , γn of non-separating curves on X, with γ = γ1 , such that the Dehn twists δγi generate Map(X) and i(γi , γj ) ≤ 1 for all i, j (compare with Figure 1). Observe that the N -th powers of the Dehn twists along the curves {φ∗ (γi )} generate φ(Map(X)). It follows hence from the assumption that φ is irreducible that the curves {φ∗ (γi )} fill Y (compare with the proof of Lemma 10.1 below). Thus, since φ∗ preserves disjointness by Corollary 6.2, there exists γ 0 ∈ {γ1 , . . . , γn } such that i(γ, γ 0 ) = 1 and i(φ∗ (γ), φ∗ (γ 0 )) ≥ 1. Since i(γ, γ 0 ) = 1, the Dehn twists δγ and δγ 0 braid. Thus, the N -th powers δφN∗ (γ) = φ(δγ ) and δφN∗ (γ 0 ) = φ(δγ 0 ) of the Dehn twists along φ∗ (γ) and φ∗ (γ 0 ) also braid. Since i(φ∗ (γ), φ∗ (γ 0 )) ≥ 1, Lemma 2.3 shows that i(φ∗ (γ), φ∗ (γ 0 )) = 1 and N = ±1, as desired. Before moving on, we remark that in final argument of the proof of the irreducible case of Theorem 1.6 we have proved the first claim of the following lemma: Lemma 8.1. Suppose that X, Y are as in the statement of Proposition 1.6, and let and φ : Map(X) → Map(Y ) be an irreducible homomorphism. Then the following holds: • i(φ∗ (γ), φ∗ (γ 0 )) = 1 for all curves γ, γ 0 ⊂ X with i(γ, γ 0 ) = 1.
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• If a, b, c, d, x, y and z is a lantern with the property that no two curves chosen among a, b, c, d and x separate X, then φ∗ (a), φ∗ (b), φ∗ (c), φ∗ (d), φ∗ (x), φ∗ (y) and φ∗ (z) is a lantern in Y . We prove the second claim. By the irreducible case of Proposition 1.6 we know that if γ is any component of the lantern in question, then φ∗ (γ) is a single curve and φ(δγ ) = δφ∗ (γ) . In particular notice that the Dehn-twists along φ∗ (a), φ∗ (b), φ∗ (c), φ∗ (d), φ∗ (x), φ∗ (y) and φ∗ (z) satisfy the lantern relation. Since a, b, c, d, x are pairwise disjoint, Corollary 6.2 yields that the curves φ∗ (a), φ∗ (b), φ∗ (c), φ∗ (d), φ∗ (x) are also pairwise disjoint. Moreover, the irreducibility of φ, the assumption that that no two curves chosen among a, b, c, d and x separate X, and Corollary 6.6 imply that the curves φ∗ (a), φ∗ (b), φ∗ (c), φ∗ (d) and φ∗ (x) are pairwise distinct. Thus, the claim follows from Proposition 2.4. We are now ready to treat the reducible case of Proposition 1.6. Proof of Proposition 1.6 for reducible φ. Let φ : Map(X) → Map(Y ) be a non-trivial reducible homomorphism, and let η be the maximal multicurve in Y which is componentwise fixed by φ(Map(X)). Recall the exact sequence (4.1): 1 → Tη → Z0 (Tη ) → Map(Yη0 ) → 0 Lemma 4.3 shows that φ(Map(X)) ⊂ Z0 (Tη ) and that the composition φ0 : Map(X) → Map(Yη0 ) of φ and the homomorphism Z0 (Tη ) → Map(Yη0 ) is not trivial. Observe that φ0 is irreducible because η was chosen to be maximal. The surface Yη0 may well be disconnected; if this is the case, Map(Yη0 ) is by definition the direct product of the mapping class groups of the connected components of Yη0 . Noticing that the sum of the genera of the components of Yη0 is bounded above by the genus of Y , it follows from the bound g 0 ≤ 2g − 1 and from Proposition 1.8 that Yη0 contains at a single component Yη00 on which φ(Map(X)) acts nontrivially. Hence, we can apply the irreducible case of Proposition 1.6 and deduce that φ0 : Map(X) → Map(Yη00 ) maps Dehn twists to possibly left Dehn twists. Conjugating φ by an outer automorphism of Map(X) we may assume without loss of generality that φ0 maps Dehn twists to Dehn twists. Suppose now that a, b, c, d, x, y and z form a lantern in X as in Lemma 8.1; such a lantern exists because X has genus at least 3. By Lemma 8.1 we obtain that the images of these curves under φ0∗ also form a lantern. In other words, if S ⊂ X is the four-holed sphere with
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boundary a ∪ b ∪ c ∪ d then there is an embedding ι : S → Yη00 ⊂ Yη0 such that for any γ ∈ {a, . . . , z} we have φ0 (δγ ) = δι(γ) Identifying Yη00 with a connected component of Yη0 = Y \ η we obtain an embedding ˆι : S → Y . We claim that for any γ in the lantern a, b, c, d, x, y, z we have φ(δγ ) = δˆι(γ) . A priori we only have that, for any such γ, both φ(δγ ) and δˆι(γ) project to the same element δι(γ) under the homomorphism Z0 (Tη ) → Map(Yη0 ). In other words, there is τγ ∈ Tη with φ(δγ ) = δˆι(γ) τγ . Observe that since any two curves γ, γ 0 in the lantern a, b, c, d, x, y, z are non-separating, the Dehn twists δγ and δγ 0 are conjugate in Map(X). Therefore, their images under φ are also conjugate in φ(Map(X)) ⊂ Z0 (Tη ). Since Tη is central in Z0 (Tη ), it follows that in fact τγ = τγ 0 for any two curves γ and γ 0 in the lantern. Denote by τ the element of Tη so obtained. On the other hand, both δa , . . . , δz and δˆι(a) , . . . , δˆι(z) satisfy the lantern relation and, moreover, τ commutes with everything. Hence 1 = φ(δa )φ(δb )φ(δc )φ(δd )φ(δz )−1 φ(δy )−1 φ(δx )−1 = −1 −1 −1 −1 = δˆι(a) τ δˆι(b) τ δˆι(c) τ δˆι(d) τ τ −1 δˆι−1 δˆι(y) τ δˆι(x) = (z) τ −1 −1 = δˆι(a) δˆι(b) δˆι(c) δˆι(d) δˆι−1 (z) δˆ ι(y) δˆ ι(x) τ = τ
Hence, we have proved that φ(δa ) = δˆι(a) τ = δˆι(a) In other words, the image under φ of the Dehn twist along some, and hence every, non-separating curve is a Dehn twist. 9. Reducing to the irreducible In this section we explain how to reduce the proof of Theorem 1.1 to the case of irreducible homomorphisms between mapping class groups of surfaces without boundary. 9.1. Weak embeddings. Observe there are no embeddings X → Y if X has no boundary but Y does (compare with Corollary 11.1 below). We are going to relax the definition of embedding to allow for this possibility. For this purpose, it is convenient to regard X and Y as possibly non-compact surfaces without marked points; recall that we declared ourselves to be free to switch between cusps, marked points and ends.
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Definition. Let X and Y be possibly non-compact surfaces of finite topological type without marked points. A weak embedding ι : X → Y is a topological embedding of X into Y . Given two surfaces X and Y without marked points there are two, ˆ and Yˆ with sets P ˆ and P ˆ essentially unique, compact surfaces X X Y ˆ \ P ˆ and Y = Yˆ \ P ˆ . We will of marked points and with X = X X Y say that a weak embedding ι : X → Y is induced by an embedding ˆ P ˆ ) → (Yˆ , P ˆ ) if there is a homeomorphism f : Y → Y which ˆι : (X, X Y is isotopic to the identity relative to PYˆ , and ˆι|X = f ◦ ι. It is easy to describe which weak embeddings are induced by embeddings: A weak embedding ι : X → Y is induced by an embedding if and only if the image ι(γ) of every curve γ ⊂ X which bounds a ˆ containing at most one marked point bounds a disk in Yˆ disk in X which again contains at most one marked point. Since ι(γ) bounds a disk without punctures if γ does, we can reformulate this equivalence in terms of mapping classes: Lemma 9.1. A weak embedding ι : X → Y is induced by an embedding if and only if δι(γ) is trivial in Map(Y ) for every, a fortiori non-essential, curve γ ⊂ X which bounds a disk with a puncture. Notice that in general a weak embedding X → Y does not induce a homomorphism Map(X) → Map(Y ). On the other hand, the following proposition asserts that if a homomorphism Map(X) → Map(Y ) is, as far as it goes, induced by a weak embedding, then it is induced by an actual embedding. Proposition 9.2. Let X and Y be surfaces of finite type and genus at least 3. Suppose that φ : Map(X) → Map(Y ) is a homomorphism such that there is a weak embedding ι : X → Y with the property that for every non-separating curve γ ⊂ X we have φ(δγ ) = δι(γ) . Then φ is induced by an embedding X → Y . Proof. Suppose that a ⊂ X bounds a disk with one puncture and consider the lantern in X given in Figure 4. We denote the boldprinted curves by a, b, c, d and the dotted lines by x, y, z; observe that a is the only non-essential curve in the lantern. By the lantern relation and because a is not essential we have (9.1)
1 = δa = δx δy δz δb−1 δc−1 δd−1
The images under ι of the curves a, b, c, d, x, y, z also form a lantern in Y and hence we obtain (9.2)
−1 −1 −1 δι(a) = δι(x) δι(y) δι(z) δι(b) δι(c) δι(d)
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Figure 4. A lantern in X where one of the curves is non-essential and all the others are non-separating. The assumption in the Proposition implies that the image under the homomorphism φ of the right side of (9.1) is equal to the right side of (9.2). This implies that δι(a) is trivial. Lemma 9.1 shows now that the weak embedding ι is induced by an embedding, which we again denote by ι. Let ι# the homomorphism induced by ι. Since, by assumption, φ(δγ ) = δι(γ) = ι# (δγ ) for all non-separating curves γ, and since the Dehn twists along these curves generate the mapping class group, we deduce that φ = ι# . In particular, φ is induced by an embedding, as we needed to prove. 9.2. Down to the irreducible case. Armed with Proposition 9.2, we now prove that it suffices to establish Theorem 1.1 for irreducible homomorphisms. Namely, we have: Lemma 9.3. Suppose that Theorem 1.1 holds for irreducible homomorphisms. Then it also holds for reducible ones. Proof. Let X and Y be surfaces as in the statement of Theorem 1.1 and suppose that φ : Map(X) → Map(Y ) is a non-trivial reducible homomorphism. Let η be a maximal multicurve in Y whose every component of η is invariant under φ(Map(X)); by Lemma 4.3, φ(Map(X)) ⊂ Z0 (Tη ). Consider, as in the proof of Proposition 1.6, the composition φ0 : Map(X) → Map(Yη0 ) of φ and the homomorphism in (4.1). Lemma 4.2 shows that φ0 is non-trivial; moreover, it is irreducible by the maximality of η. Now, Proposition 1.6 implies that for any γ non-separating both φ(δγ ) = δφ∗ (γ) and φ0 (δγ ) = δφ0∗ (γ) are Dehn twists. As in the proof of the
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reducible case of Proposition 1.6 we can consider Yη0 = Y \ η as a subsurface of Y . Clearly, φ∗ (γ) = φ0∗ (γ) after this identification. Assume that Theorem 1.1 holds for irreducible homomorphisms. Since φ0 is irreducible, we obtain an embedding ι : X → Yη0 inducing φ0 . Consider the embedding ι : X → Yη0 as a weak embedding ˆι : X → Y . By the above, φ(δγ ) = δˆι(γ) , for every γ ⊂ X nonseparating. Finally, Proposition 9.2 implies that φ is induced by an embedding. 9.3. No factors. Let φ : Map(X) → Map(Y ) be a homomorphism as in the statement of Theorem 1.1. We will say that φ factors if there ¯ an embedding ¯ι : X → X, ¯ and a homomorphism are a surface X, ¯ → Map(Y ) such that the following diagram commutes: φ¯ : Map(X) (9.3)
Map(X)
LLL LLLφ LLL L& / ¯ Map(Y ) Map(X) ¯ ι#
φ¯
Since the composition of two embeddings is an embedding, we deduce that φ is induced by an embedding if φ¯ is. Since a homomorphism Map(X) → Map(Y ) may factor only finitely many times, we obtain: Lemma 9.4. If Proposition 1.6 holds for homomorphisms φ : Map(X) → Map(Y ) which do not factor, then it holds in full generality. Our next step is to prove that any irreducible homomorphism φ : Map(X) → Map(Y ) factors if X has boundary. We need to establish the following result first: Lemma 9.5. Suppose that X and Y are as in the statement of Theorem 1.1 and let φ : Map(X) → Map(Y ) be an irreducible homomorphism. Then the centralizer of φ(Map(X)) in Map(Y ) is trivial. Proof. Suppose, for contradiction, that there is a non-trivial element f in Z(φ(Map(X))); we will show that φ is reducible. Noticing that the genus bounds in Theorem 1.1 are more generous than those in Lemma 7.1, we deduce from the latter that f has infinite order. Let γ ⊂ X be a non-separating curve and recall that φ(δγ ) is a Dehn twist by Proposition 1.6. Since f commutes with φ(δγ ) it follows that f is not pseudo-Anosov. In particular, f must be reducible; let η be the canonical reducing multicurve associated to f [7]. Since φ(Map(X)) commutes with f we deduce that φ(Map(X)) preserves η. We will
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prove that φ(Map(X)) fixes some component of η, obtaining hence a contradiction to the assumption that φ is irreducible. The arguments are very similar to the arguments in the proof of Lemma 6.4 and Lemma 6.5. First, notice that the same arguments as the ones used to prove Lemma 6.4 imply that some component of η is fixed if some component of Y \η is a disk or an annulus. Suppose that this is not the case. Then Y \ η has at most 2g 0 − 2 ≤ 4g − 4 components. Hence Theorem 4.4 implies that φ(Map(X)) fixes every component of Y \ η. Using again that no component of Y \η is a disk or an annulus we deduce that every such component C has at most 2g 0 + 2 ≤ 4g − 2 boundary components. Hence Theorem 4.4 implies that φ(Map(X)) fixes every component of ∂C ⊂ η. We have proved that some component of η is fixed by φ(Map(X)) and hence that φ is reducible, as desired. We can now prove: Corollary 9.6. Suppose that X and Y are as in Theorem 1.1 and that ∂X 6= ∅. Then every irreducible homomorphism φ : Map(X) → Map(Y ) factors. Proof. Let X 0 = X \ ∂X be the surface obtained from X by deleting the boundary and consider the associated embedding ι : X → X 0 . By Theorem 2.5, the homomorphism ι# : Map(X) → Map(X 0 ) fits in the exact sequence 1 → T∂X → Map(X) → Map(X 0 ) → 1 where T∂X is the center of Map(X). It follows from Lemma 9.5 that if φ is irreducible, then T∂X ⊂ Ker(φ). We have proved that φ descends to φ0 : Map(X 0 ) → Map(Y ) and hence that φ factors as we needed to show. Combining Lemma 9.3, Lemma 9.4 and Corollary 9.6 we deduce: Proposition 9.7. Suppose that Theorem 1.1 holds if • X and Y have no boundary, and • φ : Map(X) → Map(Y ) is irreducible and does not factor. Then, Theorem 1.1 holds in full generality.
10. Proof of Theorem 1.1 In this section we prove the main result of this paper, whose statement we now recall:
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Theorem 1.1. Suppose that X and Y are surfaces of finite topological type, of genus g ≥ 6 and g 0 ≤ 2g − 1 respectively; if Y has genus 2g − 1, suppose also that it is not closed. Then every nontrivial homomorphism φ : Map(X) → Map(Y ) is induced by an embedding X → Y . Remark. As mentioned in the introduction, the same conclusion as in Theorem 1.1 applies for homomorphisms φ : Map(X) → Map(Y ) if both X and Y have the same genus g ∈ {4, 5}. This will be shown in the course of the proof. By Proposition 9.7 we may assume that X and Y have no boundary, that φ is irreducible and that it does not factor. Moreover, by Proposition 1.6, the image of a Dehn twist δγ along a non-separating curve is either the right or the left Dehn twist along the non-separating curve φ∗ (γ). Notice that, up to composing φ with an outer automorphism of Map(Y ) induced by an orientation reversing homeomorphism of Y , we may actually assume that φ(δγ ) is actually a right Dehn twist for some, and hence every, non-separating curve γ ⊂ X. In light of this, from now on we will assume without further notice that we are in the following situation: Standing assumptions: X and Y have no boundary; φ is irreducible and does not factor; φ(δγ ) = δφ∗ (γ) for all γ ⊂ X non-separating. Under these assumptions, we now prove φ is induced by a homeomorphism. We will make extensive use of the concrete set of generators of Map(X) given in Figure 1, which we include here as Figure 5 for convenience. The reader should have Figure 5 constantly in mind through the rest of this section. a1
b1
a2
b2
a3
c
b3
bg
rk
r1
r2
Figure 5. Dehn twists along the curves ai , bi , c and ri generate Map(X). Notice that the sequence a1 , b1 , a2 , b2 , . . . , ag , bg in Figure 5 forms a chain; we will refer to it as the ai bi -chain. We refer to the multicurve
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r1 ∪ · · · ∪ rk as the ri -fan. The curve c would be worthy of a name such as el pendejo or el pinganillo but we prefer to call it simply the curve c. Observe that all the curves in Figure 5 are non-separating. Hence, it follows from Proposition 6.3 that φ∗ (γ) is a non-separating curve for any curve γ in the collection ai , bi , ri , c. We claim that the images under φ∗ of all these curves in fill Y . Lemma 10.1. The image under φ∗ of the ai bi -chain, the ri -fan and the c-curve fill Y . Proof. Since the Dehn twists along the curves ai , bi , ri , c generate Map(X), any curve in Y which is fixed by the images under φ of all these Dehn twists is fixed by φ(Map(X)). Noticing that the image under φ of a Dehn twist along any of ai , bj , rl , c is a Dehn twist along the φ∗ -image of the curve, the claim follows since φ is assumed to be irreducible. Suppose that γ, γ 0 are two distinct elements of the collection ai , bi , ri , c. We now summarize several of the already established facts about the relative positions of the curves φ∗ (γ), φ∗ (γ 0 ): (1) If i(γ, γ 0 ) = 0 then i(φ∗ (γ), φ∗ (γ 0 )) = 0 by Corollary 6.2. (2) If γ, γ 0 are distinct and disjoint, and X \ (γ ∪ γ 0 ) is connected, then φ∗ (γ) 6= φ∗ (γ 0 ) by Corollary 6.6. (3) If i(γ, γ 0 ) = 1 then i(φ∗ (γ), φ∗ (γ 0 ) = 1 by Lemma 8.1. Notice that these properties do not ensure that φ∗ (ri ) 6= φ∗ (rj ) if i 6= j. We denote by R ⊂ Y the maximal multicurve with the property that each one of its components is homotopic to one of the curves φ∗ (ri ). Notice that R = ∅ if and only if X has at most a puncture and that in any case R has at most as many components as curves has the ri -fan. The following lemma follows easily from (1), (2) and (3) above: Lemma 10.2. With the notation of Figure 5 the following holds: • The image under φ∗ of the ai bi -chain is a chain of the same length in Y . • Every component of the multicurve R intersects the φ∗ -image of the ai bi -chain exactly in the component φ∗ (bg ). • The curve φ∗ (c) is disjoint from every curve in R, intersects φ∗ (b2 ) exactly once, and is disjoint from the images of the other curves in the ai bi -chain. At first glance, Lemma 10.2 yields the desired embedding without any further work, but this is far from true. A first problem is that R may have too few curves. Also, the curves in the ri -fan come equipped with a natural ordering and that we do not know yet that φ∗ preserves
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this ordering. Before tackling these problems, we clarify the position of φ∗ (c): Lemma 10.3. Let Z be a regular neighborhood of the ai bi -chain, noting that c ⊂ Z. Then there is an orientation preserving embedding F : Z → Y such that φ∗ (γ) = F (γ) for γ = ai , bi , c and i = 1, . . . , g. Proof. The image under φ∗ of the ai bi -chain is a chain of the same length, by Lemma 10.2. Let Z 0 be a regular neighborhood of the φ∗ image of the ai bi -chain. Since regular neighborhoods of any two chains of the same length are homeomorphic in an orientation preserving manner, there is an orientation preserving embedding F : Z → Z0 with F (ai ) = φ∗ (ai ) and F (bi ) = φ∗ (bi ) for all i. It remains to prove that F can be chosen so that F (c) = φ∗ (c). Let Z0 ⊂ Z be the subsurface of X filled by a1 , b1 , a2 , b2 and observe that, up to isotopy, c ⊂ Z0 . The boundary of Z0 is connected and, by the chain relation (see Section 2) we can write the Dehn twist along ∂Z0 as: δ∂Z0 = (δa1 δb2 δa2 δb2 )10 Hence we have φ(δ∂Z0 ) = (φ(δa1 )φ(δb2 )φ(δa2 )φ(δb2 ))10 = (δφ∗ (a1 ) δφ∗ (b2 ) δφ∗ (a2 ) δφ∗ (b2 ) )10 = (δF (a1 ) δF (b2 ) δF (a2 ) δF (b2 ) )10 = δF (∂Z0 ) where the last equality follows again from the chain relation. Since c is disjoint from ∂Z0 we have that δc and δ∂Z0 commute. Hence φ∗ (c) does not intersect F (∂Z0 ). On the other hand, since φ∗ (c) intersects φ∗ (b2 ) ⊂ F (Z0 ), we deduce from Lemma 10.2 that φ∗ (c) ⊂ F (Z0 ). Observe now that F (Z) \ (∪φ∗ (ai ) ∪ φ∗ (bi )) ' Z \ (∪ai ∪ bi ) is homeomorphic to an annulus A. It follows from Lemma 10.2 that the intersection of φ∗ (c) with A is an embedded arc whose endpoints are in the subsegments of ∂A corresponding to φ∗ (b2 ). There are two choices for such an arc. However, there is an involution τ : F (Z) → F (Z) with τ (φ∗ (ai )) = φ∗ (ai ) and τ (φ∗ (bi )) = φ∗ (bi ) and which interchanges these two arcs. It follows that, up to possibly replacing F by τ ◦ F , we have F (c) = φ∗ (c), as we needed to prove. At this point we are ready to prove the first cases of Theorem 1.1.
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Proof of Theorem 1.1 (X is closed or has one puncture). Let Z be as in Lemma 10.3 and note that, from the assumptions on X, the homomorphism σ# : Map(Z) → Map(X) induced by the embedding σ : Z → X is surjective. Let φˆ : Map(Z) → Map(Y ) be the composition of σ# and φ. By Lemma 10.3, φˆ is induced by an embedding of F :Z →Y. Suppose first that X has one puncture. In particular, there is a weak embedding X → Z ⊂ X which is homotopic to the identity X → X. The composition of this weak embedding X → Z and of the embedding F : Z → Y is a weak embedding which, in the sense of Proposition 9.2, induces φ. It follows from Proposition 9.2 that φ is induced by an embedding, as we needed to prove. The case that X is closed is slightly more complicated. The assumption that φ is irreducible and that a collection of curves in F (Z) fills Y , implies that Y \ F (Z) is a disk containing at most one puncture. If Y \ F (Z) is a disk without punctures, then the we can clearly extend the map F to a continuous injective map X → Y . Since any continuous injective map between closed connected surfaces is a homeomorphism, it is a fortiori an embedding and we are done in this case. It remains to rule out the possibility that X is closed and Y has one puncture. Suppose that this is the case and let Y¯ be the surface obtained from Y by filling in its unique puncture. We can now apply the above argument to the induced homomorphism φ¯ : Map(X) → Map(Y¯ ), to deduce that φ¯ is induced by an embedding X → Y¯ . Since any embedding from a closed surface is a homeomorphism we deduce that φ¯ is an isomorphism. Consider φ ◦ φ¯−1 : Map(Y¯ ) → Map(Y ) and notice that composing φ ◦ φ¯−1 with the filling-in homomorphism Map(Y ) → Map(Y¯ ) we obtain the identity. Hence, φ ◦ φ¯−1 is a splitting of the Birman exact sequence 1 → π1 (Y¯ ) → Map(Y ) → Map(Y¯ ) → 1, which does not exist by Lemma 3.3. It follows that Y cannot have a puncture, as we needed to prove. We continue with the preparatory work needed to prove Theorem 1.1 in the general case. From now on we assume that X has at least 2 punctures. Standing assumption: X has at least 2 punctures. Continuing with the proof of Theorem 1.1, we prove next Y has the same genus as X.
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Lemma 10.4. Both surfaces X and Y have the same genus g. Proof. With the same notation as in Lemma 10.3 we need to prove that S = Y \ F (Z) is a surface of genus 0. By Lemma 10.1 the arcs ρi = φ∗ (ri ) ∩ S fill S. Denote by S¯ the surface obtained by attaching a disk along F (∂Z) ⊂ ∂S and notice that the arcs ρi can be extended ¯ Moreover, every curve in S¯ to a collection of disjoint curves ρ¯i in S. either agrees or intersects one of the curves ρ¯i more than once, which is impossible if S has genus at least 1; this proves Lemma 10.4. Notice that if η ⊂ X is separating, all we know about φ(δη ) is that it is a root of a multitwist by Bridson’s Theorem 6.1; in particular, φ(δη ) may be trivial or have finite order. If this is not the case, we denote by φ∗ (η) the multicurve supporting any multitwist power of φ(δη ). Observe that if η bounds a disk with punctures then, up to replacing the ai bi -chain by another such chain, we may assume that i(η, ai ) = i(η, bi ) = 0 for all i. In particular, φ∗ (η) does not intersect any of the curves φ∗ (ai ) and φ∗ (bi ). It follows that no component of φ∗ (η) is non-separating. We record our conclusions: Lemma 10.5. Suppose that η ⊂ X bounds a disk with punctures and that φ(δη ) has infinite order. Then every component of the multicurve φ∗ (η) separates Y . Our next goal is to bound the number of cusps of Y : S Lemma 10.6. Every connected component of Y \ ( i φ∗ (ai ) ∪ R) contains at most single puncture. In particular Y has at most as many punctures as X. Recall that R ⊂ Y is the maximal multicurve with the property that each one of its components is homotopic to one of the curves φ∗ (ri ). Proof. Observe that Lemma 10.1 and Lemma 10.3 imply that the union of R and the image under φ∗ of the ai bi -chain fill Y . In particular, every component of the complement in Y of the union of R and all the curves φ∗ (ai ) and φ∗ (bi ) contains at most one puncture of Y . It follows from Lemma 10.2 that the multicurve ∪φ∗ (bi ) does not separate any of the components of the complement of (∪φ∗ (ai )) ∪ R in Y . We have proved the first claim. It follows again from Lemma 10.2 that the multicurve ∪φ∗ (ai ) ∪ R separates Y into at most k components where k ≥ 2 is the number of punctures of X. Thus, Y has at most as many punctures as X. So far, we do not know much about the relative positions of the curves in R; this will change once we have established the next three lemmas.
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Lemma 10.7. Suppose that a, b ⊂ X are non-separating curves that bound an annulus A. Then φ∗ (a) and φ∗ (b) bound an annulus A0 in Y ; moreover, if A contains exactly one puncture and φ∗ (a) 6= φ∗ (b), then A0 also contains exactly one puncture. Proof. Notice that A is disjoint from a chain of length 2g − 1. Since φ∗ maps chains to chains (Lemma 10.2), since it preserves disjointness (Corollary 6.2) and since Y has the same genus as X (Lemma 10.4), we deduce that φ∗ (∂A) consists of non-separating curves which are contained in an annulus in Y . The first claim follows. Suppose that φ∗ (a) 6= φ∗ (b); up to translating by a mapping class, we may assume that a = a1 and that b is a curve disjoint from (∪ai )∪(∪ri ) and with i(b, b1 ) = 1 and i(b, bi ) = 0 for i = 2, . . . , g (compare with the dashed curve in Figure 6). Since φ∗ preserves disjointness and a1
b1
a2
b
b2
a3
b3
c
bg
rk
r1
r2
Figure 6. intersection number one (Lemma 8.1), it follows that the annulus A0 bounded by φ∗ (a) = φ∗ (a1 ) and contained in one of the two S φ∗ (b) is S connected components of Y \ ( i φ∗ (ai ) ∪ i φ∗ (ri )) adjacent to φ∗ (a1 ). By Lemma 10.6, each one of these components contains at most a puncture, and thus the claim follows. Lemma 10.8. Let γ, γ 0 ⊂ X be non-separating curves bounding an annulus with one puncture. Then φ∗ (γ) 6= φ∗ (γ 0 ). Proof. We will prove that if φ∗ (γ) = φ∗ (γ 0 ), then φ factors in the sense of (9.3); notice that this contradicts our standing assumption. Suppose φ∗ (γ) = φ∗ (γ 0 ), noting that Proposition 1.6 implies that φ(δγ ) = φ(δγ 0 ). Let p be the puncture in the annulus bounded by ¯ surface obtained from X by filling in the γ and γ 0 . Consider the X puncture p and the Birman exact sequence (3.1): ¯ p) → Map(X) → Map(X) ¯ →1 1 → π1 (X, ¯ Let α ∈ π1 (X, ¯ p) be the unique associated to the embedding X → X. essential simply loop contained in the annulus bounded by γ ∪ γ 0 . The
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image of α under the left arrow of the Birman exact sequence is δγ δγ−1 0 . ¯ Hence, α belongs to the kernel of φ. Since π1 (X, p) has a set of generators consisting of translates of α by Map(X) we deduce that that ¯ p) ⊂ Ker(φ). This shows that φ : Map(X) → Map(Y ) factors π1 (X, ¯ and concludes the proof of Lemma 10.8. through Map(X) Lemma 10.9. Let a, b ⊂ X be non-separating curves which bound an annulus with exactly two punctures. Then φ∗ (a) and φ∗ (b) bound an annulus A0 ⊂ Y with exactly two punctures. Moreover, if x ⊂ A is any non-separating curve in X separating the two punctures of A, then φ∗ (x) ⊂ A0 and separates the two punctures of A0 . Proof. Let x ⊂ A be a curve as in the statement. Suppose first that φ∗ (a) 6= φ∗ (b). Consider the annuli A0 , A01 and A02 in Y with boundaries ∂A0 = φ∗ (a) ∪ φ∗ (b), ∂A01 = φ∗ (a) ∪ φ∗ (x), ∂A02 = φ∗ (x) ∪ φ∗ (b). By Lemmas 10.7 and 10.8, the annuli A01 and A02 contain exactly one puncture. Finally, since φ∗ (x) does not intersect φ∗ (a)∪φ∗ (b), it follows that A0 = A01 ∪ A02 and the claim follows. It remains to rule out the possibility that φ∗ (a) = φ∗ (b). Seeking a contradiction, suppose that this is the case. Consider curves d, c, y, z as in Figure 7 and notice that a, b, c, d, x, y, z is a lantern, and that y
x
d a
b
z c
Figure 7. The back dots represent cusps c, d are not essential. In particular, the lantern relation reduces to δa δb = δx δy δz . Applying φ we obtain δφ2∗ (a) = δφ∗ (x) δφ∗ (y) φ(δz ) By Bridson’s theorem, φ(δz ) is a root of a multitwist. Since δa and δz commute, we have that δφ2∗ (a) φ(δz )−1 is also a root of a multitwist. By the same arguments as above, there are annuli A01 , A02 in Y , each containing at most one puncture, with boundaries ∂A01 = φ∗ (a) ∪ φ∗ (x), ∂A02 = φ∗ (a) ∪ φ∗ (y)
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Observe that i(φ∗ (x), φ∗ (y)) is even. If i(φ∗ (x), φ∗ (y)) > 2, then by [16, Theorem 3.10] the element δφ∗ (x) δφ∗ (y) is relatively pseudo-Anosov, and hence not a multitwist. If i(φ∗ (x), φ∗ (y)) = 2, then we are in the situation of Figure 8, meaning that we can extend φ∗ (a), φ∗ (x), φ∗ (y)
Figure 8. The solid lines are φ∗ (a), φ∗ (x) and φ∗ (y) and the black dots are cusps. ˆ φ∗ (x), φ∗ (y), zˆ with cˆ, dˆ not-essential and ˆb to a lantern φ∗ (a), ˆb, cˆ, d, non-separating. From the lantern relation we obtain: δφ−1 δ −1 δ = δz−1 ˆ δˆb ∗ (y) φ∗ (x) φ∗ (a) This implies that φ(δz ) = δz−1 ˆ δˆb δφ∗ (a) is a multitwist whose support contains non-separating components. This contradicts Lemma 10.5 and so we deduce that, if φ∗ (a) = φ∗ (b) then φ∗ (x) and φ∗ (y) cannot have positive intersection number. Finally, we treat the case i(φ∗ (x), φ∗ (y)) = 0. Since x and y are disjoint from a, it follows that the left side of δφ−1 δ −1 δ 2 = φ(δz ) ∗ (x) φ∗ (y) φ∗ (a) is a multitwist supported on a multicurve contained in φ∗ (a) ∪ φ∗ (x) ∪ φ∗ (y). Since these three curves are non-separating, it follows from Lemma 10.5 that φ(δz ) = Id. This shows that φ∗ (a) = φ∗ (x) = φ∗ (y). Since a and x bound an annulus which exactly one puncture, we get a contradiction to Lemma 10.8. Thus, we have proved that φ∗ (a) 6= φ∗ (b); this concludes the proof of Lemma 10.9 We are now ready to finish the proof of Theorem 1.1. Proof of Theorem 1.1. Continuing with the same notation and standing assumptions, we now introduce orderings on the ri -fan and the multicurve R ⊂ Y . In order to do so, observe that the union of the multicurve ∪ai and any of the curves in the ri -fan separates X. Similarly, notice that by Lemma 10.2 the union of the multicurve ∪φ∗ (ai ) and any of the components of R is a multicurve consisting of g + 1 nonseparating curves. Since Y has genus g, by Lemma 10.4, we deduce
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that the union of the multicurve ∪φ∗ (ai ) and any of the components of R separates Y . We now define our orderings: • Given two curves ri , rj in the ri -fan we say that ri ≤ rj if ri and c are in the same connected component of X \(a1 ∪· · ·∪ag ∪rj ). Notice that the labeling in Figure 5 is such that ri ≤ rj for i ≤ j. • Similarly, given two curves r, r0 ∈ R we say that r ≤ r0 if r and φ∗ (c) are in the same connected component of X \ (φ∗ (a1 ) ∪ · · · ∪ φ∗ (ag ) ∪ r0 ). The minimal element of the ri -fan, the curve r1 in Figure 5, is called the initial curve in the ri -fan; we define the initial curve of the multicurve R in an analogous way. We claim that its image under φ∗ is the initial curve of R: Claim. φ∗ (r1 ) is the initial curve in R. Proof of the claim. Suppose, for contradiction, that φ∗ (r1 ) is not the initial curve in R. Consider, besides the curves in Figure 5, a curve c0 as in Figure 9. In words, c and c0 bound an annulus with exactly two punctures and (10.1)
i(c0 , ri ) = 0 ∀i ≥ 2 and i(c0 , ai ) = 0 ∀i
Notice that by Lemma 10.9, φ∗ (c) and φ∗ (c0 ) bound an annulus A which contains exactly two punctures. a1
b1
a2
b2
a3
c
b3
c'
bg
rk
r1
r2
Figure 9. The dotted curve c0 and c bound an annulus with two punctures. Since φ(r1 ) is not the initial curve, then i(φ∗ (c0 ), ∪φ∗ (rj )) = 0 for all j, as i(c0 , rj ) = 0 for all j > 1. Also, by disjointness i(φ∗ (c0 ), φ∗ (ai )) = 0 for all i. Since the boundary ∂A = φ∗ (c) ∪ φ∗ (c0 ) of the annulus A is disjoint of ∪φ∗ (ai ) ∪ φ∗ (ri ) it is contained in one of the connected components of X \ (∪φ∗ (ai ) ∪ φ∗ (ri )). However, each one of these components contains at most one puncture, by Lemma 10.6. This contradicts Lemma 10.9, and thus we have established the claim.
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We are now ready to prove that φ∗ induces an order preserving bijection between the ri -fan and the multicurve R. Denote the curves in R by ri0 , labeled in such a way that ri0 ≤ rj0 if i ≤ j. By the previous claim, φ∗ (r1 ) = r10 . Next, consider the curve r2 , and observe that r1 and r2 bound an annulus with exactly one puncture. Hence, Lemma 10.8 yields that φ∗ (r1 ) = r10 and φ∗ (r2 ) also bound an annulus with exactly one puncture. In particular, φ∗ (r2 ) cannot be separated from r10 by any component of R. This proves that φ∗ (r2 ) = r20 . We now consider the curve r3 . The argument just used for r2 implies that either φ∗ (r3 ) = r30 or φ∗ (r3 ) = r10 . The latter is impossible, as the curves r1 and r3 bound an annulus with exactly two punctures and hence so do φ∗ (r1 ) = r10 and φ∗ (r3 ), by Lemma 10.9. Thus φ∗ (r3 ) = r30 . Repeating this argument as often as necessary we obtain that the map φ∗ induces an injective, order preserving map from the ri -fan to R. Since by definition R has at most as many components as the ri -fan, we have proved that this map is in fact an order preserving bijection. Let Z ⊂ X be a regular neighborhood of the ai bi -chain, and recall c ⊂ Z. By Lemma 10.3 there is an orientation preserving embedding F : Z → Y such that φ∗ (γ) = F (γ) for γ = ai , bi , c (i = 1, . . . , g). We choose Z so that it intersects every curve in the ri -fan in a segment. Observe that Lemma 10.2 implies that F can be isotoped so that F (Z ∩ (∪ri )) = F (Z) ∩ R The orderings of the ri -fan and of R induce orderings of Z ∩ (∪ri ) and Z ∩ R. Since the map φ∗ preserves both orderings we deduce that F preserves the induced orderings of Z ∩ (∪ri ) and F (Z) ∩ R.
Z ∩ r2 Z ∩ r1 ∂Z
∂Z
Figure 10. Attaching the first (left) and second (right) annuli along ∂Z. Let k be the number of curves in the ri -fan, and thus in R. We successively attach k annuli along the boundary ∂Z of Z, as indicated in Figure 10. In this way we get a surface Z1 naturally homeomorphic to X. We perform the analogous operation on ∂F (Z), thus obtaining
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a surface Z2 which is naturally homeomorphic to Y . Since the map φ∗ is preserves the orderings of the ri -fan and of R, we get that the homeomorphism F : Z → F (Z) extends to a homeomorphism F¯ : X → Y such that F¯ (γ) = φ∗ (γ) for every curve γ in the collection ai , bi , c, ri . It follows that the homomorphisms φ and F¯# both map the Dehn twist along γ to the Dehn twist along φ∗ (γ) and, in particular, to the same element in Map(Y ). Since the Dehn twists along the curves ai , bi , c, ri generate Map(X), we deduce φ = F# . This finishes the proof of Theorem 1.1. 11. Some consequences of Theorem 1.1 We now present several consequences of Theorem 1.1; each of them results from imposing extra conditions on the surfaces involved and then reinterpreting the word “embedding” in that specific situation. Namely, observe that Proposition 3.1 immediately implies the following: Corollary 11.1. Suppose that X and Y are surfaces of finite topological type and that ι : X → Y is an embedding. (1) If X is closed, i.e. if X has neither boundary nor marked points, then ι is a homeomorphism. (2) If X has no boundary, then ι is obtained by forgetting a (possibly empty) collection of punctures of X. In particular, ι# : Map(X) → Map(Y ) is surjective, and it is injective if and only if ι is a homeomorphism. (3) If X = Y and X has no boundary, then ι is a homeomorphism. (4) If X = Y and ι is a subsurface embedding, then ι is (isotopic to) a homeomorphism. Combining the first part of Corollary 11.1 and Theorem 1.1, we obtain: Corollary 11.2. Suppose that X and Y are surfaces of finite topological type, of genus g ≥ 6 and g 0 ≤ 2g − 1 respectively; if Y has genus 2g − 1, suppose also that it is not closed. If X is closed then every nontrivial homomorphism φ : Map(X) → Map(Y ) is induced by a homeomorphism X → Y ; in particular φ is an isomorphism.
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As mentioned in the introduction, if there are no restrictions on the genus of Y then Corollary 11.2 is far from true. Indeed, Theorem 1 of [2] shows that for every closed surface X there exist a closed surface Y and an injective homomorphism Map(X) → Map(Y ). Moving away from the closed case, if X is allowed to have marked points and/or boundary then there are numerous non-trivial embeddings of X into other surfaces. That said, the second part of Corollary 11.1 tells us that if X has no boundary, then every subsurface embedding of X into another surface is necessarily a homeomorphism. Hence we have: Corollary 11.3. Suppose that X and Y are surfaces of finite topological type, of genus g ≥ 6 and g 0 ≤ 2g − 1 respectively; if Y has genus 2g − 1, suppose also that it is not closed. If X has empty boundary then any injective homomorphism φ : Map(X) → Map(Y ) is induced by a homeomorphism X → Y ; in particular φ is an isomorphism. Again, if there are no restrictions on the genus of Y then Corollary 11.3 is simply not true; see Section 2 of [25] and Theorem 2 of [2]. By Corollary 11.1 (3), if X has no boundary then any embedding ι : X → X is a homeomorphism. Observing that Theorem 1.1 applies for homomorphisms between surfaces of the same genus g ≥ 4, (see the remark following the statement of the theorem) we deduce: Theorem 1.2. Let X be a surface of finite topological type, of genus g ≥ 4 and with empty boundary. Then any non-trivial endomorphism φ : Map(X) → Map(X) is induced by a homeomorphism X → X; in particular φ is an isomorphism. Note that Theorem 1.2 fails if X has boundary. However, by Corollary 11.1 (4), any subsurface embedding X → X is isotopic to a homeomorphism. Therefore, we recover the following result due to IvanovMcCarthy [25] (see [22] and [37] for related earlier results): Corollary 11.4 (Ivanov-McCarthy). Let X be a surface of finite topological type, of genus g ≥ 4. Then any injective homomorphism φ : Map(X) → Map(X) is induced by a homeomorphism X → X; in particular φ is an isomorphism. 12. Proof of Theorem 1.3 Given a Riemann surface X of finite analytic type, endow the associated Teichm¨ uller space T (X) with the standard complex structure.
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Recall that Map(X) acts discretely on T (X) by biholomorphic automorphisms. In particular, we can consider the moduli space M(X) = T (X)/ Map(X) as a complex orbifold; by construction it is a good orbifold, meaning that its universal cover is a manifold. Suppose now that Y is another Riemann surface of finite analytic type. We will consider maps f : M(X) → M(Y ) in the category of orbifolds. Since M(X) and M(Y ) are both good obifolds we can associate to every such map f a homomorphism f∗ : Map(X) → Map(Y ) and a holomorphic map f˜ : T (X) → T (Y ) which is f∗ -equivariant, that is, ˜ ˜ f (γx) = f∗ (γ) f (x) ∀γ ∈ Map(X) and x ∈ T (X) and such that the following diagram commutes: T (X)
M(X)
f
/
/
T (Y )
M(Y )
Here both vertical arrows are the standard projections. Remark. In the remainder of this section we will treat M(X) and M(Y ) as if they were manifolds. This is justified by two observations. First, every statement we make holds indistinguishable for manifolds and for orbifolds. And second, in all our geometric arguments we could pass to a manifold finite cover and work there. We hope that this does not cause any confusion. Suppose now that X and Y have the same genus and that Y has at most as many marked points as X. Choosing an identification between the set of marked points of Y and a subset of the set of marked points of X, we obtain a holomorphic map M(X) → M(Y ) obtained by forgetting all marked points of X which do not correspond to a marked point of Y . Different identifications give rise to different maps; we will refer to these maps as forgetful maps.
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In the rest of the section we will prove Theorem 1.3, whose statement we now recall: Theorem 1.3. Suppose that X and Y are Riemann surfaces of finite analytic type and assume that X has genus g ≥ 6 and Y genus g 0 ≤ 2g − 1; if g 0 = 2g − 1 assume further that Y is not closed. Then, every non-constant holomorphic map f : M(X) → M(Y ) is a forgetful map. In order to prove Theorem 1.3, we will first deduce from Theorem 1.1 that there is a forgetful map F : M(X) → M(Y ) homotopic to f ; then we will modify an argument due to Eells-Sampson to conclude that that f = F . In fact we will prove, without any assumptions on the genus, that any two homotopic holomorphic maps between moduli spaces are equal: Proposition 1.5. Let X and Y be Riemann surfaces of finite analytic type and let f1 , f2 : M(X) → M(Y ) be homotopic holomorphic maps. If f1 is not constant, then f1 = f2 . Armed with Proposition 1.5, we now conclude the proof of Theorem 1.3. Proof of Theorem 1.3 from Proposition 1.5. Suppose that f : M(X) → M(Y ) is holomorphic and not constant. It follows from the latter assumption and from Proposition 1.5 that f is not homotopic to a constant map. In particular, the induced homomorphism f∗ : Map(X) → Map(Y ) is non-trivial. Thus, it follows from Theorem 1.1 that f∗ is induced by an embedding. Now, since X has empty boundary, Corollary 11.1 (2) tells us that every embedding X → Y is obtained by filling in a collection of punctures of X. It follows that there is a forgetful map F : M(X) → M(Y ) with F∗ = f∗ . We deduce that F and f are homotopic to each other because the universal cover T (Y ) of M(Y ) is contractible. Hence, Proposition 1.5 shows that f = F , as we needed to prove. The remainder of this section is devoted to prove Proposition 1.5. Recall at this point that T (X) admits many important Map(X)-invariant metrics. In particular, we will endow: • T (X) with McMullen’s K¨ahler hyperbolic metric [38], and • T (Y ) with the Weil-Petersson metric.
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The reason why we do not endow T (X) with the Weil-Petersson metric is encapsulated in the following observation: Lemma 12.1. Every holomorphic map f : T (X) → T (Y ) is Lipschitz. Proof. Denote by T (X)T and T (Y )T the Teichm¨ uller spaces of X and Y , respectively, both equipped with the Teichm¨ uller metric. Consider f as a composition of maps T (X)
Id
/
T (X)T
f
/
T (Y )T
Id
/
T (Y )
By [38, Theorem 1.1] the first arrow is bi-lipschitz. By Royden’s theorem [43], the middle map is 1-Lipschitz. Finally, the last arrow is also Lipschitz because the Teichm¨ uller metric dominates the Weil-Petersson metric up to a constant factor [38, Proposition 2.4]. We will also need the following fact from Teichm¨ uller theory: Lemma 12.2. There exists a collection {Ki }n∈N of subsets of M(X) with the following properties: S • M(X) = n∈N Ki , • Kn ⊂ Kn+1 for all n, • there is L > 0 such that Kn+1 is contained within distance L of Kn for all n, and • the co-dimension one volume of ∂Kn decreases exponentially when n → ∞. Recall that T (X), and hence M(X), has been endowed with McMullen’s K¨ahler hyperbolic metric. Proof. Let M(X) be the Deligne-Mumford compactification of the moduli space M(X); recall that points Z ∈ M(X) \ M(X) are surfaces with k nodes (k ≥ 1). Wolpert [45] proved that every point in M(X) \ M(X) has a small neighborhood UZ in M(X) whose intersection UZ = UZ ∩ M(X) with M(X) is bi-holomorphic to a neighborhood of (0, . . . , 0) in ((D∗ )k × (D)dimC (T (X))−k )/G where G is a finite group; here D∗ and D are the punctured and unpunctured open unit disks in C. We may assume, without loss of generality, that UZ ' ((D∗ )k × (D)dimC (T (X))−k )/G where D ⊂ D is the disk of Euclidean radius 21 centered at 0.
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Endow now D∗ and D with the hyperbolic metric. For n ∈ N ∪ {0} let Dn ⊂ D be the disk such that the hyperbolic distance between ∂D∗ and ∂Dn∗ is equal to n in D∗ . We set: UZ (n) = ((Dn∗ )k × (Dn )dimC (T (X))−k )/G Observe that, with respect to the hyperbolic metric, the volume of ∂UZ (n) decreases exponentially with n. Since M(X) \ M(X) is compact, we can pick finitely many sets UZ1 , . . . , UZr such that M(X) \ ∪i UZi is compact. For n ∈ N we set Kn = M(X) \ ∪i UZi (n) By construction, M(X) = ∪n Kn and Kn ⊂ Kn+1 ∀n We claim that the sets Kn satisfy the rest of the desired properties. First, recall that Theorem 1.1 of [38] implies that the Teichm¨ uller and K¨ahler hyperbolic metrics on M(X) are bi-Lipschitz equivalent to each other. In particular, it suffices to prove the claim if we consider M(X) equipped with the Teichm¨ uller metric. It is due to Royden [43] that the Teichm¨ uller metric agrees with the Kobayashi metric. It follows that the inclusion UZi ⊂ M(X) is 1-Lipschitz when we endow UZi with the product of hyperbolic metrics and the M(X) with the Teichm¨ uller metric. p By the choice of Dn , every point in UZi (n + 1) is within distance dimC (T (X)) of UZi (n) with respect to the hyperbolic metric. Hence, the same is true with respect to the Teichm¨ uller metric. Therefore, Kn+1 is contained within a fixed Teichm¨ uller distance of Kn . Noticing that ∂Kn ⊂ ∪i=1,...,r ∂UZi (n), that the volume of ∂UZi (n) decreases exponentially with respect to the hyperbolic metric, and that 1-Lipschitz maps contract volume, we deduce that the volume of ∂Kn also decreases exponentially with respect to the Teichm¨ uller metric. This concludes the proof of Lemma 12.2. We are almost ready to prove Proposition 1.5. We first remind the reader of a few facts and definitions on the energy of maps. Suppose that N and M are Riemannian manifolds and that f : N → M is a smooth map. The energy density of f at x ∈ N is defined to be: Ex (f ) =
dim RN X i=1
kdfx vi k2M
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where v1 , . . . , vdimR N is an arbitrary orthonormal basis of Tx M . The energy of f is then the integral of the energy density: Z Ex (f )d volN (x) E(f ) = N
Here d volN is the Riemannian volume form of N . Suppose now that N and M are K¨ahler and let ωN and ωM be the respective K¨ahler forms. Recall that dimC N ωN = ωX ∧ · · · ∧ ωX
is a volume form on N ; more concretely, it is a constant multiple of the Riemannian volume form d volX , where the constant depends only on dimC N . Pulling back the K¨ahler form ωM via f : N → M , we also have the dimC N −1 top-dimensional form (f ∗ ωM )ωN on N . A local computation due to Eells and Sampson [12] shows that for all x ∈ N we have (12.1)
dimC N −1 Ex (f )d volN ≥ c · (f ∗ ωM )ωN
where c > 0 is a constant which again only depends on the dimension dimC N . Moreover, equality holds in (12.1) if and only if f is holomorphic at x. Remark. We stress that the proof of (12.1) is infinitesimal. In particular, it is indifferent to any global geometric property of the involved manifolds such as completeness. We finally have all the ingredients needed to prove Proposition 1.5: Proof of Proposition 1.5. Suppose that f0 , f1 : M(X) → M(Y ) are holomorphic maps and recall that we have endowed M(Y ) with the Weil-Petersson metric and M(X) with McMullen’s K¨ahler hyperbolic metric. We remind the reader that both metrics are K¨ahler and have finite volume. Suppose that f0 and f1 are homotopic and let Fˆ : [0, 1] × M(X) → M(Y ) be a homotopy (as orbifold maps). The Weil-Petersson metric is negatively curved but not complete. However, it is geodesically convex. This allows to consider also the straight homotopy F : [0, 1] × M(X) → M(Y ), F (t, x) = ft (x) determined by the fact that t 7→ ft (x) is the geodesic segment joining f0 (x) and f1 (x) in the homotopy class of Fˆ ([0, 1] × {x}). Clearly, ft is smooth for all t.
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Given a vector V ∈ Tx M(X) the vector field t 7→ d(ft )x V is a Jacobi field along t 7→ ft (x). Since the Weil-Peterson metric is negatively curved, the length of Jacobi fields is a convex function. Now, Lemma 12.1 gives that f0 and f1 are Lipschitz, and therefore the length of d(ft )x V is bounded by the length of V times a constant which depends neither on t nor on x. It follows that the maps ft : M(X) → M(Y ) are uniformly Lipschitz. In particular, they have finite energy E(ft ) < ∞. In fact, the same convexity property of Jacobi fields shows that, for any x, the function t 7→ Ex (ft ) is convex. This implies that the energy function t 7→ E(ft ) is also convex. Moreover, it is strictly convex unless both holomorphic maps f0 and f1 either agree or are constant. Since the last possibility is ruled out by our assumptions we have: Fact. Either f0 = f1 or the function t 7→ E(ft ) is strictly convex.
Supposing that f0 6= f1 we may assume that E(f0 ) ≥ E(f1 ). By the fact above, for all t ∈ (0, 1): we have (12.2)
E(ft ) < E(f0 )
We are going to derive a contradiction to this assertion. In order to do so, let Kn ⊂ M(X) be one of the sets provided by Lemma 12.2 and denote by ωX and ωY the K¨ahler forms of M(X) and M(Y ) respectively. Since the K¨ahler forms are closed, we deduce from Stokes theorem that Z dim (T (X))−1 d (F ∗ ωY )ωX C 0= [0,t]×Kn Z dim (T (X))−1 = (F ∗ ωY )ωX C ∂([0,t]×Kn ) Z Z dimC (T (X))−1 dim (T (X))−1 ∗ = (F ωY )ωX − (F ∗ ωY )ωX C {t}×Kn {0}×Kn Z dim (T (X))−1 + (F ∗ ωY )ωX C [0,t]×∂Kn Z Z dimC (T (X))−1 dim (T (X))−1 ∗ = (ft ωY )ωX − (f0∗ ωY )ωX C Kn Kn Z dim (T (X))−1 + (F ∗ ωY )ωX C [0,t]×∂Kn
Below we will prove: R dim (T (X))−1 Claim. limn→∞ [0,t]×∂Kn (F ∗ ωY )ωX C = 0.
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Assuming the claim we conclude with the proposition. From the claim and the computation above we obtain: Z Z dimC (T (X))−1 dimC (T (X))−1 ∗ ∗ =0 − (f0 ωY )ωX lim (ft ωY )ωX n→∞
Kn
Kn
Taking into account that both maps ft and f0 are Lipschitz and that M(X) has finite volume, we deduce that Z Z dimC (T (X))−1 dim (T (X))−1 ∗ (ft ωY )ωX = (f0∗ ωY )ωX C M(X)
M(X)
We obtain now from (12.1) Z dim (T (X))−1 E(ft ) ≥ c (ft∗ ωY )ωX C M(X) Z dim (T (X))−1 =c (f0∗ ωY )ωX C = E(f0 ) M(X)
where the last equality holds because f0 is holomorphic. This contradicts (12.2). It remains to prove the claim. Proof of the claim. Let d = dimR T (X) and fix (t, x) ∈ [0, 1] × ∂Kn . Let E1 , . . . , Ed be an orthonormal basis of T(t,x) ([0, 1] × ∂Kn ). We have ∗ (F ωY )(E1 , E2 ) · ωX (E3 , E4 ) · . . . · ωX (Ed−1 , Ed ) = hdF(t,x) E1 , i · dF(t,x) E2 iY hE3 , iE4 iX . . . hEd−1 , Ed iX = kdF(t,x) k2 where h·, ·iX and h·, ·iY are the Riemannian metrics on M(X) and M(Y ) and where kdF(t,x) k is the operator norm of dF(t,x) . From this computation we deduce that there is a constant c > 0 depending only on the dimension such that Z Z dim (T (X))−1 ∗ C ≤c (F ωY )ωX kdF(t,x) k2 d vol[0,t]×∂Kn (t, x) [0,t]×∂Kn
[0,t]×∂Kn
Recall that F : [0, 1] × M(X) → M(Y ) is the straight homotopy between the holomorphic (and hence Lipschitz) maps f0 and f1 . Let L be a Lipschitz constant for these maps and fix, once and for all, a point x0 ∈ K0 ⊂ M(X). As we mentioned above, the restriction of F to {t} × M(X) is L-Lipschitz for all t. Hence, the only direction that ∂ dF(t,x) can really increase is the ∂t direction. Since F is the straight
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homotopy, we have ∂ kdF(t,x) kY = dM(Y ) (f0 (x), f1 (x)) ∂t ≤ 2LdM(X) (x, x0 ) + dM(Y ) (f0 (x0 ), f1 (x0 )) It follows that there are constants A, B depending only on L and the base point x0 such that for all (t, x) we have kdF(t,x) k ≤ A · dM(X) (x0 , x)2 + B Summing up we have Z dim (T (X))−1 ∗ C ≤ (12.3) (F ωY )ωX [0,t]×∂Kn 2 ≤ A · max dM(X) (x, x0 ) + B vol(∂Kn ) x∈∂Kn
By construction, maxx∈∂Kn dM(X) (x, x0 ) is bounded from above by a linear function of n. On the other hand, vol(∂Kn ) decreases exponentially. This implies that the right side of (12.3) tends to 0 with n → ∞. This proves the claim. Having proved the claim, we have concluded the proof of Proposition 1.5. References [1] J. Andersen, Mapping Class Groups do not have Kazhdan’s Property (T), math.QA/0706.2184v1. [2] J. Aramayona, C. Leininger and J. Souto, Injections of mapping class groups, Geom. Topol. 13 (2009), no. 5. [3] R. Bell and D. Margalit, Braid groups and the co-Hopfian property, J. Algebra 303 (2006). [4] M. Bestvina, T. Church and J. Souto, Some groups of mapping classes not realized by diffeomorphisms, to appear in Commentarii Mathematici Helvetici. [5] M. Bestvina and K. Fujiwara, Bounded cohomology of subgroups of mapping class groups, Geom. Topol. 6 (2002). [6] J. Birman and H. Hilden, On isotopies of homeomorphisms of Riemann surfaces, Annals of Math., 97 (1973). [7] J. Birman, A. Lubotzky and J. McCarthy, Abelian and solvable subgroups of the mapping class groups, Duke Math. J. 50 (1983). [8] A. Borel and J.-P. Serre, Corners and arithmetic groups, Comment. Math. Helv. 48 (1973). [9] M. Bridson, Semisimple actions of mapping class groups on CAT(0) spaces, preprint, arXiv:0908.0685. [10] M. Bridson and K. Vogtmann, Abelian covers of graphs and maps between outer automorphism groups of free groups, preprint , arXiv:1007.2598. [11] F. Castel, Repr´esentations g´eom´etriques des groupes de tresses, preprint.
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[12] J. Eells, J. H. Sampson, Harmonic mappings of Riemannian manifolds, Amer. J. Math. 86 (1964). [13] B. Farb and D. Margalit, A primer on mapping class groups, to appear in Princeton University Press. [14] B. Farb, H. Masur, Superrigidity and mapping class groups, Topology 37, No.6 (1998). [15] E. Grossman, On the residual finiteness of certain mapping class groups, J. London Math. Soc. (2) 9 (1974/75) [16] H. Hamidi-Tehrani, Groups generated by positive multitwists and the fake lantern problem, Algebr. Geom. Topol. 2 (2002) [17] J. Harer, The virtual cohomological dimension of the mapping class group of an orientable surface, Invent. Math. 84 (1986). [18] W. Harvey, Boundary structure of the modular group, in Riemann surfaces and related topics: Proceedings of the 1978 Stony Brook Conference Ann. of Math. Stud., 97, Princeton Univ. Press, 1981. [19] W. Harvey and M. Korkmaz, Homomorphisms from mapping class groups, Bull. London Math. Soc. 37 (2005), no. 2. [20] A. Hatcher and W. Thurston, A presentation for the mapping class group of a closed orientable surface, Topology 19 (1980). [21] S. P. Humphries, Generators for the mapping class group, in Topology of lowdimensional manifolds (Proc. Second Sussex Conf., Chelwood Gate, 1977), pp. 44–47, Lecture Notes in Math., 722, Springer, Berlin, 1979. [22] N. Ivanov, Automorphisms of Teichm¨ uller modular groups, in Topology and geometry—Rohlin Seminar, 199–270, Lecture Notes in Math., 1346, Springer, Berlin, 1988. [23] N. Ivanov, Automorphism of complexes of curves and of Teichm¨ uller spaces, Internat. Math. Res. Notices 1997. [24] N. Ivanov, Mapping class groups, in Handbook of Geometric Topology, NorthHolland, 2002. [25] N. Ivanov and J. McCarthy, On injective homomorphisms between Teichm¨ uller modular groups. I., Invent. Math. 135 (1999), no. 2. [26] S. Kerckhoff, The Nielsen realization problem, Ann. of Math. (2) 117 (1983). [27] M. Korkmaz, Low-dimensional homology groups of mapping class groups: a survey, Turkish J. Math. 26 (2002), no. 1. [28] A. Kuribayashi and I. Kuribayashi, Automorphism groups of compact Riemann surfaces of genera three and four, J. Pure Appl. Algebra 65 (1990). [29] A. Kuribayashi and I. Kuribayashi, On automorphism groups of compact Riemann surfaces of genus 5, Proc. Japan Acad. Ser. A Math. Sci. 63 (1987). [30] E. Looijenga, Prym representations of mapping class groups, Geometria Dedicata 64, (1997). [31] C. Maclachlan, Abelian groups of automorphisms of compact Riemann surfaces, Proc. London Math. Soc. (3) 15 (1965). [32] J. Mangahas, Uniform uniform exponential growth of subgroups of the mapping class group, Geom. Funct. Anal. 19 (2010), no. 5. [33] D. Margalit, A lantern lemma, Algebr. Geom. Topol. 2 (2002). [34] D. Margalit and S. Schleimer, Dehn twists have roots, Geometry and Topology 13 (2009).
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[35] G. A. Margulis, Arithmeticity of the irreducible lattices in the semisimple groups of rank greater than 1, Invent. Math. 76 (1984), no. 1. [36] V. Markovic, Realization of the mapping class group by homeomorphisms, Invent. Math. 168 (2007), no. 3. [37] J. McCarthy, Automorphisms of surface mapping class groups. A recent theorem of N. Ivanov, Invent. Math. 84 (1986), no. 1. [38] C. T. McMullen, The moduli space of Riemann surfaces is K¨ ahler hyperbolic, Ann. of Math. (2) 151 (2000), no. 1. [39] S. Morita, Characteristic classes of surface bundles, Invent. Math. 90 (1987), no. 3. [40] S. Nakajima, On abelian automorphism groups of algebraic curves, J. London Math. Soc. (2) 36 (1987) [41] L. Paris, Small index subgroups of the mapping class group, to appear in Journal of Group Theory. [42] J. Powell, Two theorems on the mapping class group of a surface, Proc. Amer. Math. Soc. 68 (1978). [43] H. L. Royden, Automorphisms and isometries of Teichm¨ uller space, in Advances in the Theory of Riemann Surfaces (Proc. Conf., Stony Brook, N.Y., 1969) pp. 369–383 Ann. of Math. Studies, No. 66. Princeton Univ. Press, Princeton, N.J. [44] W. P. Thurston, On the geometry and dynamics of diffeomorphisms of surfaces, Bull. Amer. Math. Soc. 19 (1988). [45] S. Wolpert, Riemann surfaces, moduli and hyperbolic geometry, in Lectures on Riemann Surfaces, World Scientific, 1989.
Department of Mathematics, National University of Ireland, Galway. [email protected] Department of Mathematics, University of Michigan, Ann Arbor. [email protected] | {"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": 0.9743284583091736, "perplexity": 949.4375284617907}, "config": {"markdown_headings": false, "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-2018-51/segments/1544376829997.74/warc/CC-MAIN-20181218225003-20181219011003-00016.warc.gz"} |
https://www.physicsforums.com/threads/extrinsic-intrinsic-curvature.30000/ | # Extrinsic/Intrinsic curvature
1. Jun 9, 2004
### Gza
Could someone give me an intuitive example of extrinsic and intrinsic curvature. That would be much appreciated, thanks in advance.
2. Jun 9, 2004
Last edited by a moderator: Apr 20, 2017
3. Jun 10, 2004
### Gza
Yes, that was very helpful. I had to dig up a softer book on a treatment of tensors, but it still served its purpose, thanks again.
4. Jun 11, 2004
### pmb_phy
The cylinder is an excellant example of zero curvature. It is also an excellant example of a manifold for which there are infinitely many geodesics between any two points on the surface.
Pete
5. Jun 12, 2004
### Gza
Maybe i'm referring to the wrong concept, but I thought a circle had a curvature inverse of its radius, so wouldn't the curved part of the cylinder have curvature?
6. Jun 12, 2004
### pmb_phy
You're refering to a different kind of curvature. In the case of the cylinder - when someone says that the surface has zero curvature they mean that there is no "intrinsic" curvature. However it does have an "extrinsic" curvature.
Pete
7. Sep 9, 2004
### mathwonk
As I recall, again from reading Spivak some 35 years ago, Gaussian curvature of a surface at a point p, may be defined as the product of the curvature of the two (perpendicular) curves through p having respectively maximum and minimum curvature as curves. So for a cylinder, you are right that the curve of maximal curvature through the point is a circle of positive curvature, but the curve through the point with minimum curvature is a line with curvature zero, so the product, the curvature of the surface, is zero. Intuitively this is true because the cylinder can be flattened out without tearing it, so really it is not curved as a surface.
I do not know what intrinsic and extrinsic curvature mean but i can guess. Curvatiure is determined by a way of emasuring lengths i.e. a "metric". If a surface like a doughnut for instance is embedded in three space then there are many ways to define a length on it. There is the "extrinsic length" which is just the restriction to the doughnut of the notion of euclidean length. The associated curvatuire would be the extrinsic curvature. E.g. it was extrinsic curvature we were discussing above for the cylinder.
But it seems intuitively clear to me that we could define length differently, in a such a way that the length on (the surface of) a doughnut agreed with the extrinsic length on a cylinder and then the curvature of a doughnut surface would be zero.
So really all curvature is intrinsic, since it is determined by the metric, but you may call the metric and the associated curvature extrinsic if ity happens to agree with that of the embedding space. This is just a plausible guess, but not an uninformed one.
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https://www.groundai.com/project/higher-order-terms-in-the-inflaton-potential-and-the-lower-bound-on-the-tensor-to-scalar-ratio-r/ | Higher order terms in the inflaton potential and the lower bound on the tensor to scalar ratio r
# Higher order terms in the inflaton potential and the lower bound on the tensor to scalar ratio r
C. Destri Dipartimento di Fisica G. Occhialini, Università Milano-Bicocca and INFN, sezione di Milano-Bicocca, Piazza della Scienza 3, 20126 Milano, Italia. H. J. de Vega LPTHE, Université Pierre et Marie Curie (Paris VI) et Denis Diderot (Paris VII), Laboratoire Associé au CNRS UMR 7589, Tour 24, 5ème. étage, Boite 126, 4, Place Jussieu, 75252 Paris, Cedex 05, France Observatoire de Paris, LERMA. Laboratoire Associé au CNRS UMR 8112.
61, Avenue de l’Observatoire, 75014 Paris, France.
N. G. Sanchez Observatoire de Paris, LERMA. Laboratoire Associé au CNRS UMR 8112.
61, Avenue de l’Observatoire, 75014 Paris, France.
July 18, 2019
###### Abstract
The MCMC analysis of the CMB+LSS data in the context of the Ginsburg-Landau approach to inflation indicated that the fourth degree double–well inflaton potential in new inflation gives an excellent fit of the present CMB and LSS data. This provided a lower bound for the ratio of the tensor to scalar fluctuations and as most probable value , within reach of the forthcoming CMB observations. In this paper we systematically analyze the effects of arbitrarily higher order terms in the inflaton potential on the CMB observables: spectral index and ratio . Furthermore, we compute in close form the inflaton potential dynamically generated when the inflaton field is a fermion condensate in the inflationary universe. This inflaton potential turns out to belong to the Ginsburg-Landau class too. The theoretical values in the plane for all double well inflaton potentials in the Ginsburg-Landau approach (including the potential generated by fermions) fall inside a universal banana-shaped region . The upper border of the banana-shaped region is given by the fourth order double–well potential and provides an upper bound for the ratio . The lower border of is defined by the quadratic plus an infinite barrier inflaton potential and provides a lower bound for the ratio . For example, the current best value of the spectral index , implies is in the interval: . Interestingly enough, this range is within reach of forthcoming CMB observations.
###### pacs:
98.80.Cq,05.10.Cc,11.10.-z
## I Introduction
The current WMAP data are validating the single field slow-roll scenario WMAP5 (). Single field slow-roll models provide an appealing, simple and fairly generic description of inflation libros (); revius (). This inflationary scenario can be implemented using a scalar field, the inflaton with a Lagrangian density
L=a3(t)[˙φ22−(∇φ)22a2(t)−V(φ)], (1)
where is the inflaton potential. Since the universe expands exponentially fast during inflation, gradient terms are exponentially suppressed and can be neglected. At the same time, the exponential stretching of spatial lengths classicalize the physics and permits a classical treatment. One can therefore consider an homogeneous and classical inflaton field which obeys the evolution equation
¨φ+3H(t)˙φ+V′(φ)=0, (2)
in the isotropic and homogeneous Friedmann-Robertson-Walker (FRW) metric
ds2=dt2−a2(t)d→x2, (3)
which is sourced by the inflaton. Here stands for the Hubble parameter. The energy density and the pressure for a spatially homogeneous inflaton are given by
ρ=˙φ22+V(φ),p=˙φ22−V(φ). (4)
The scale factor obeys the Friedmann equation,
H2(t)=13M2Pl[12˙φ2+V(φ)]. (5)
In order to have a finite number of inflation efolds, the inflaton potential must vanish at its absolute minimum
V′(φmin)=V(φmin)=0. (6)
These two conditions guarantee that inflation is not eternal. Since the inflaton field is space-independent inflation is followed by a matter dominated era (see for example ref. bibl ()).
Inflation as known today should be considered as an effective theory, that is, it is not a fundamental theory but a theory of a condensate (the inflaton field) which follows from a more fundamental one. In order to describe the cosmological evolution it is enough to consider the effective dynamics of such condensates. The inflaton field may not correspond to any real particle (even unstable) but is just an effective description while the microscopic description should come from a Grand Unification theory (GUT) model.
At present, there is no derivation of the inflaton model from a microscopic GUT theory. However, the relation between the effective field theory of inflation and the microscopic fundamental theory is akin to the relation between the effective Ginsburg-Landau theory of superconductivity gl () and the microscopic BCS theory, or like the relation of the sigma model, an effective low energy theory of pions, photons and nucleons (as skyrmions), with the corresponding microscopic theory: quantum chromodynamics (QCD).
In the absence of a microscopic theory of inflation, we find that the Ginsburg-Landau approach is a powerful effective theory description. Such effective approach has been fully successful in several branches of physics when the microscopic theory is not available or when it is very complicated to solve in the regime considered. This is the case in statistical physics, particle physics and condensed matter physics. Such GL effective theory approach permits to analyse the physics in a quantitative way without committing to a specific model gl ().
The Ginsburg-Landau framework is not just a class of physically well motivated inflaton potentials, among them the double and single well potentials. The Ginsburg-Landau approach provides the effective theory for inflation, with powerful gain in the physical insight and analysis of the data. As explained in this paper and shown in the refs. mcmc ()-bibl (), the analysis of the present set of CMB+LSS data with the effective theory of inflation, favor the double well potential. Of course, just analyzing the present data without this powerful physical theory insight, does not allow to discriminate between classes of models, and so, very superficially and incompletely, it would seem that almost all the potentials are still at the same footing, waiting for the new data to discriminate them.
In the Ginsburg-Landau spirit the potential is a polynomial in the field starting by a constant term gl (). Linear terms can always be eliminated by a constant shift of the inflaton field. The quadratic term can have a positive or a negative sign associated to unbroken symmetry (chaotic inflation) or to broken symmetry (new inflation), respectively.
As shown in refs. mcmc (); bibl () a negative quadratic term and a negligible cubic term in new inflation provides a very good fit to the CMB+LSS data, (the inflaton starts at or very close to the false vacuum ). The analysis in refs.mcmc (); bibl () showed that chaotic inflation is clearly disfavoured compared with new inflation. Namely, inflaton potentials with are favoured with the inflaton starting to evolve at .
We can therefore ignore the linear and cubic terms in . I f we restrict ourselves for the moment to fourth order polynomial potentials, eq.(6) and imply that the inflaton potential is a double well (broken symmetric) with the following form:
V(φ)=−12m2φ2+14λφ4+m44λ=14λ(φ2−m2λ)2. (7)
The mass term and the coupling are naturally expressed in terms of the two energy scales which are relevant in this context: the energy scale of inflation and the Planck mass GeV,
m=M2MPl,λ=y8N(MMPl)4. (8)
Here is the quartic coupling.
The MCMC analysis of the CMB+LSS data combined with the theoretical input above yields the value for the coupling mcmc (); bibl (). turns out to be order one consistent with the Ginsburg-Landau formulation of the theory of inflation bibl ().
According to the current CMB+LSS data, this fourth order double–well potential of new inflation yields as most probable values: mcmc (); bibl (). This value for is within reach of forthcoming CMB observations apjnos (). For the best fit value , the inflaton field exits the horizon in the negative concavity region intrinsic to new inflation. We find for the best fit mcmc (); bibl (),
M=0.543×1016GeV for the scale of inflation andm=1.21×1013GeV for the inflaton mass. (9)
It must be stressed that in our approach the amplitude of scalar fluctuations allows us to completely determine the energy scale of inflation which turns out to coincide with the Grand Unification energy scale (well below the Planck energy scale). Namely, we succeed to derive the energy scale of inflation without the knowledge of the value of from observations. guarantees the validity of the effective theory approach to inflation. The fact that the inflaton mass is implies the appearance of infrared phenomenon as the quasi-scale invariance of the primordial power.
Since the inflaton potential must be bounded from below , the highest degree term must be even and with a positive coefficient. Hence, we consider polynomial potentials of degree where .
The request of renormalizability restricts the degree of the inflaton potential to four. However, since the theory of inflation is an effective theory, potentials of degrees higher than four are in principle acceptable.
A given Ginsburg-Landau potential will be reliable provided it is stable under the addition to the potential of terms of higher order. Namely, adding to the th order potential further terms of order and should only produce small changes in the observables. Otherwise, the description obtained could not be trusted. Since, the highest degree term must be even and positive, this implies that all even terms of order higher or equal than four should be positive.
Moreover, when expressed in terms of the appropriate dimensionless variables, a relevant dimensionless coupling constant can be defined by rescaling the inflaton field. This coupling turns out to be of order where is the number of efolds since the cosmologically relevant modes exit the horizon till the end of inflation showing that the slow-roll approximation is in fact an expansion in 1sN (). It is then natural to introduce as coupling constant . This is consistent with the stability of the results in the above sense. Generally speaking, the Ginsburg-Landau approach makes sense for small or moderate coupling.
Odd terms in the inflaton field are allowed in in the effective theory of inflation. Choosing an even function of implies that is a symmetry of the inflaton potential. At the moment, as stated in ciri (); bibl (), we do not see reasons based on fundamental physics to choose a zero or a nonzero cubic term, which is the first non-trivial odd term. Only the phenomenology, that is the fit to the CMB+LSS data, decides on the value of the cubic and the higher order odd terms. The MCMC analysis of the WMAP plus LSS data shows that the cubic term is negligible and therefore can be ignored for new inflation mcmc (); bibl (). CMB data have also been analyzed at the light of slow-roll inflation in refs. otros ().
In the present paper we systematically study the effects produced by higher order terms () in the inflationary potential on the observables and .
We show in this paper that all curves for a large class of double–well potentials of arbitrary high order in new inflation fall inside the universal banana region depicted in fig. 10. Moreover, we find that the curves for even double–well potentials with arbitrarily positive higher order terms lie inside the universal banana region [fig. 10]. This is true for arbitrarily large values of the coefficients in the potential.
Furthermore, the inflaton field may be a condensate of fermion-antifermion pairs in a grand unified theory (GUT) in the inflationary background. In this paper we explicitly write down in closed form the inflaton potential dynamically generated as the effective potential of fermions in the inflationary universe. This inflaton potential turns out to belong to the Ginsburg-Landau class of potentials considered in this paper. We find that the corresponding curves lie inside the universal banana region provided the one-loop part of the inflaton potential is at most of the same order as the tree level piece. Therefore, a lower bound for the ratio tensor/scalar fluctuations is present for all potentials above mentioned. For the current best value of the spectral index bibl (); WMAP5 () the lower bound turns out to be .
Namely, the shape of the banana region fig. 10 combined with the value for the spectral index yields the lower bound . If one consider low enough values for (in disagreement with observations) can be arbitrarily small within the GL class of inflaton potentials.
The upper border of the universal region tells us that for . Therefore, we have inside the region within the large class of potentials considered here
0.021
Interestingly enough is within reach, although borderline for the Planck satellite apjnos ().
Among the simplest potentials in the Ginsburg-Landau class, the one that best reproduces the present CMB+LSS data, is the fourth order double–well potential eq.(7), yielding as most probable values: . Our work here shows that adding higher order terms to the inflaton potential does not really improve the data description in spite of the addition of new free parameters. Therefore, the fourth order double–well potential gives a robust and stable description of the present CMB/LSS data and provides clear predictions to be contrasted with the forthcoming CMB observations apjnos ().
There is an abundant literature on slow-roll inflationary potentials and the cosmological parameters and including new inflation and in particular hilltop inflation lily (); otros2 (); otros3 (); otros4 ().
The question on whether a lower bound for is found or not depends on whether the Ginsburg-Landau (G-L) effective field approach to inflation is used or not. Namely, within the G-L approach, the new inflation double well potential determines a banana shaped relationship which for the observed value determines a lower bound on . The analysis of the CMB+LSS data within the G-L approach which we performed in refs. mcmc (); bibl () shows that new inflation is preferred by the data with respect to chaotic inflation for fourth degree potentials, and that the lower bound on is then present. Without using the powerful physical G-L framework such discrimination between the two classes of inflation models is not possible and the lower bound for does not emerge. Other references in the field (i. e. lily (); kinney08 (); PeirisEasther ()) do not work within the Ginsburg-Landau framework, do not find lower bounds for and cannot exclude arbitrarily small values for , much smaller than our lower bound .
This paper is organized as follows: in section II we present in general inflaton potentials of arbitrary high degree, specializing then to fourth and sixth–order polynomial potentials and displaying their corresponding curves. Sec. III contains the th order double–well polynomial inflaton potentials with arbitrary random coefficients and their curves. Sec. IV presents the limits of these polynomial potentials and we present in sec. V the exponential potential and its infinite coupling limit. In sec. V we compute the inflaton potential from dynamically generated fermion condensates in a de Sitter space-time displaying their curves. Finally, we present and discuss the universal banana region in sec. VII together with our conclusions.
## Ii Physical parametrization for inflaton potentials
We start by writing the inflaton potential in dimensionless variables as ciri ()
V(φ)=M4v(φMPl), (10)
where is the energy scale of inflation and is a dimensionless function of the dimensionless field argument . Without loss of generality we can set . Moreover, provided we can choose without loss of generality .
In the slow-roll regime, higher time derivatives in the equations of motion can be neglected with the final well known result for the number of efolds
N=−∫ϕendϕexitdϕv(ϕ)v′(ϕ), (11)
where is the inflaton field at horizon exit. To leading order in we can take to be the value at which attains its absolute minimum , which must be zero since inflation must stop after a finite number of efolds bibl ().
Then, in chaotic inflation we have , with for , while in new inflation we have with for . We consider potentials that can be expanded in Taylor series around , with a non-vanishing quadratic (mass) term.
It is convenient to rescale the inflaton field in order to conveniently parametrize the higher order potential. We define a coupling parameter by rescaling the inflaton and its potential keeping invariant the quadratic term, that is
v(ϕ)=1gv1(ϕ√g) (12)
For a potential expanded in power series around we write:
v1(u)=c0∓12u2+∑k≥3ckkuk (13)
Then, replacing
u=ϕ√g, (14)
we find
v(ϕ)=c0g∓12ϕ2+∑k≥3gk/2−1kckϕk. (15)
The positive sign in the quadratic term corresponds to chaotic inflation (in which case ), while the negative sign corresponds to new inflation (in which case is chosen such that vanishes at its absolute minimum).
Clearly plus the set of coefficients provide an overcomplete parametrization of the inflaton potential which we will now reduce. In the case of chaotic inflation a convenient choice is , so that
v(ϕ)=12ϕ2+√gc33ϕ3+g4ϕ4+∑k≥5gk/2−1kckϕk[chaotic inflation] (16)
which represents a generic higher order perturbation of the trinomial chaotic inflation studied in refs. mcmc ().
In the case of new inflation, where , it is more convenient to set without loss of generality that . In order to have appropriate inflation, must be the absolute minimum of and the closest one to the origin on the positive semi–axis. That is,
v′1(1)=−1+∑k≥3ck=0 (17)
and then fixes from eq.(12) the constant term in the potential
c0=12−∑k≥3ckk (18)
We thus get for the inflaton potential
v1(u)=12(1−u2)+∑k≥3ckk(uk−1)[new inflation], (19)
corresponding to
v(ϕ)=12(1g−ϕ2)+∑k≥3ckk(gk/2−1ϕk−1g)[new inflation] (20)
For the coupling and the field using eq.(14),
g=1ϕ2min=M2Plφ2min,u=ϕϕmin=φφmin. (21)
From eq.(11) it now follows that the parameter can be expressed as the integral
y(u)=8∫uumindxv1(x)v′1(x),u≡√gϕexit, (22)
where,
g=y(u)8N, (23)
with for chaotic inflation and for new inflation. Eq.(22) can be regarded as a parametrization of and in terms of the rescaled exit field . Clearly, as a function of , is uniformly of order . is numerically of order as long as is of order one. As we shall see, typically both at horizon exit and are of order one. We have for new inflation and for chaotic inflation.
In what follows we therefore use instead of as a coupling constant and make contact with eq.(10) by setting
φ=MPl√8Nyu,V(φ)=8NM4yv1(√y8NφMPl). (24)
We can easily read from this equation the order of magnitude of and since is given by eq.(9) and and are of order one. Hence, and .
As we will see below, the coupling (or ) is the most relevant coupling since it is related to the inflaton rescaling: the tensor–scalar ratio and the spectral index vary in a more relevant manner with than with the rest of the parameters in the potential eq.(15).
By construction the function has the following properties
• ;
• for in chaotic inflation;
• for in new inflation;
• as ;
• as in chaotic inflation;
• as in new inflation.
In terms of this parametrization and to leading order in , the tensor to scalar ratio and the spectral index read:
r=y(u)N[v′1(u)v1(u)]2,ns−1=−38r+y(u)4Nv′′1(u)v1(u) (25)
Notice that both and are of order for generic inflation potentials in this Ginsburg-Landau framework as we see from eq.(25). Moreover, the running of the scalar spectral index from eq.(24) and its slow-roll expression turn out to be of order
dnsdlnk=−y2(u)32N2{v′1(u)v′′′1(u)v21(u)+3[v′1(u)]4v41(u)−4[v′1(u)]2v′′1(u)v31(u)}.
and therefore can be neglected bibl (). Such small estimate for is in agreement with the present data WMAP5 () and makes the running unobservable for a foreseeable future.
Since can be inverted for any , these two relations can also be regarded as parametrizations and in terms of the coupling constant .
We are interested in the region of the plane obtained from eq.(25) by varying (or ) and the other parameters in the inflaton potential. We call this region.
From now on, we will restrict to new inflation.
For a generic [with the required global properties described above] we can determine the asymptotic of , since they follow from the weak coupling limit and from the strong coupling limit . When , then and from the property above,
r=8N+O(u−1)=0.13333…+O(u−1) (26)
and
ns=1−2N+O(u−1)=0.9666…+O(u−1). (27)
When we have in new inflation and then,
ns≃1+2Nlogu⟶−∞,r≃−8Nu2loguv1(0)⟶0+. (28)
We see that in the strong coupling regime becomes very small and becomes well below unity. However, the slow-roll approximation is valid for and in any case, the WMAP+LSS results exclude WMAP5 (). Therefore, the strong coupling limit is ruled out.
Eq.(25) for can be rewritten using eq.(22) in the suggestive form,
r=64Ny(u)[dlny(u)du]−2 (29)
Since may be small only in case is large (the logarithmic derivative of has a milder effect for large .) Therefore, we only find in a strong coupling regime. Notice that is much smaller than in the strong coupling regime [eq.(21)].
Let us now study large classes of physically meaningful inflaton potentials in order to provide generic bounds on the region of the plane within an interval of surely compatible with the WMAP+LSS data for , namely . To gain insight into the problem, we consider first the cases amenable to an analytic treatment, leaving the generic cases to a numerical investigation. As we will see below, the boundaries of the region turn out to be described parametrically by the analytic formulas (32) and (50).
### ii.1 The fourth degree double–well inflaton potential
The case when the is the standard double–well quartic polynomial
has been studied in refs. mcmc (); bibl (). In the general framework outlined above we have for this case,
v1(u)=14(u2−1)2=14−12u2+14u4,λ=y8N(MMPl)4,m=M2MPl. (30)
By explicitly evaluating the integral in eq. (22) one obtains
y(u)=u2−1−logu2, (31)
and then, from eq. (25)
ns=1−1N3u2+1(1−u2)2(u2−1−logu2),r=1N16u2(1−u2)2(u2−1−logu2) (32)
where . As required by the general arguments above, is a monotonically decreasing function of , ranging from till when increases from till . In particular, when vanishes quadratically as,
y(u)u→1−=12(1−u2)2.
The concavity of the potential eq.(30) for the inflaton field at horizon crossing takes the value
v′′1(u)=3u2−1.
We see that vanishes at , that is at . (This is usually called the spinodal point revi ()). Therefore,
v′′1(u)>0fory<0.431946…andv′′1(u)<0fory>0.431946…. (33)
Our MCMC analysis of the CMB+LSS data combined with the theoretical model eq.(30) yields mcmc (); bibl () deep in the negative concavity region .
The negative concavity case for is specific to new inflation eq.(30). can be expressed as a linear combination of the observables and as
ns−1+38r=y(u)4Nv′′1(u)v1(u)
As expected in the general framework presented above, the limit implies weak coupling , that is, the potential is quadratic around the absolute minimum and we find,
nsy→0=1−2N,ry→0=8N,uy→0=1, (34)
which coincide with and for the monomial quadratic potential in chaotic inflation.
In the limit which implies (strong coupling), we have
uy→+∞=e−(y+1)/2→0+
and
nsy≫1=1−yN,ry≫1=16yNe−y−1. (35)
Notice that the slow-roll approximation is no longer valid when the coefficient of becomes much larger than unity. Hence, the results in eq.(35) are valid for . We see that in this strong coupling regime (see fig. 1), becomes very small and becomes well below unity. However, the WMAP+LSS results exclude WMAP5 (). Therefore, this strong coupling limit is ruled out.
For the fourth order double–well inflaton potential, the relation defined by eq.(32) is a single curve depicted with dotted lines in fig. 1. It represents the upper border of the banana shaped region in fig. 1.
Notice that there is here a maximum value for , namely with bibl (). The curve has here two branches: the lower branch in which increases with increasing and the upper branch in which decreases with increasing .
### ii.2 The sixth–order double–well inflaton potential
We consider here new inflation described by a six degree even polynomial potential with broken symmetry. According to eq. (10) and eq. (12) we then have
V(φ)=M4gv1(√gφMPl),v1(u)=c0−12u2+c44u4+c66u6. (36)
where for stability we assume . Moreover, if we regard this case as a higher order correction to the quartic double–well potential, then is positive.
The inflaton potential eq.(36) is a particular case of eq.(13). The conditions eqs. (17) and (18) that the absolute minimum of be at yields
c4+c6=1,c0=12−14c4−16c6 (37)
It is convenient to use as free parameter so that and . Thus,
v1(u)=12(1−u2)−1−b4(1−u4)−b6(1−u6)=112(1−u2)2(3+b+2bu2) (38)
where in order to ensure that .
The integral in eq. (22) can be explicitly evaluated with the result
y(u)=8∫u1dxv1(x)v′1(x)=23(u2−1)−13(3+b)logu2+(1+b)23blog1+bu21+b (39)
According to the general arguments presented above [see the lines below eq. (22)] one can verify that is a monotonically decreasing function of for , where .
The scalar index and the tensor–scalar ratio are evaluated from eq. (25) as
r=yN[12u(1+bu2)(1−u2)(3+b+2bu2)]2,ns=1−38r+3y(u)N5bu4+3(1−b)u2−1(1−u2)2(3+b+2bu2) (40)
Various curves are plotted in fig. 1 for several values of in the interval sweeping the region . We see that for increasing [namely, for increasing sextic coupling and decreasing quartic coupling, see eq.(38)] the curves move down and right, sweeping the banana-shape region depicted on fig. 1.
Clearly, is a variable more relevant than . Changing moves and in the whole available range of values, while changing only amounts to displacements transverse to the banana region in the plane. In particular, for a given becomes smaller for increasing .
We see in fig. 1 two important limiting curves: the and the curves. When the function reduces to the fourth order double-well potential eq.(30) and we recover its characteristic curve . When the potential has no quartic term and reduces to the quadratic plus sixth order potential:
v1(u)b→1=16(1−u2)2(2+u2)=13−12u2+16u6. (41)
In summary, the quadratic plus quartic broken–symmetry potential describes the upper/left border of the banana–shaped region of fig. 1, while the quadratic plus sextic broken–symmetry potential describes its lower/right border.
## Iii Higher–order even polynomial double-well inflaton potentials
The generalization of the sixth order inflaton potential with broken symmetry to arbitrarily higher orders is now straightforward:
V(φ)=M4gv1(√gφMPl),v1(u)=12(1−u2)+n∑k=2c2k2k(u2k−1), (42)
with the constraint eq.(17)
n∑k=2c2k=1 (43)
which guarantees that is an extreme of .
We consider here the case when all higher coefficients are positive or zero :
c2k≥0,k=2,…,n
such that is the unique positive minimum.
We determine the shape of the region for arbitrary positive or zero values of the coefficients [subject to the constraint (43)], performing a large number of simulations with different setups. After producing coefficients we numerically computed the function following eq.(22)
y(u)=4∫1udxx1−x2+∑nk=2c2kk(x2k−1)1−∑nk=2c2kx2k−2
and obtain the curves from eq. (25) by plotting directly vs. .
Uniform distributions of coefficients are obtained by setting
c2k=(n∑j=1logξj)−1logξk,k=1,2,…,n (44)
where the numbers are independently and uniformly distributed in the unit interval. We used the parametrization eq.(44) also when the are chosen according to other rules.
For example, in figs. 2-3 we plot the results when , that is for the ten degree polynomial. In this case we let and take independently the values or , for a total of 78 distinct configurations of coefficients. For better clarity, in figs. 2-3 we also include the two border cases and .
For higher values of we extracted the numbers at random within the unit interval. In particular, for the highest case considered, , we used three distributions: in the first, the were all extracted independently and uniformly over the unit interval; in the second we set and extracted the independently and uniformly; in the third we picked at random four freely varying and fixed to 1 the remaining 45 ones (that is we picked at random four possibly non–zero , setting the rest to zero); the values of the four free were chosen at random in the same set of values of the case. The results of these simulations are shown in fig. 5.
As evident from fig. 3, where the curves are split in upper/lower branches with growing/decreasing and especially from fig. 5, the case of the quadratic plus th order polynomial provides a bound to the banana region from below. That is, for any fixed value of , the quadratic plus th order polynomial provides the lowest value for .
One sees from fig. 5 that some blue curves go beyond the slashed red curve for the quadratic plus potential on the right upper border of the banana region . Namely, the right upper border of the region is not given by the quadratic plus potential while this potential provides the lower border of the region.
We performed many other tests with intermediate values of and several other distributions, including other dependent distributions, with characteristic values for growing linearly with or decreasing in a power–like or exponential way. In all cases, the results were consistent with those given above.
It is also important to observe that the class of potentials considered, that is arbitrary even polynomials with positive or zero couplings, is a class of weakly coupled models. This is evident from fig. 4, were is plotted vs. the coupling , which remains of order one when decreases well below the current experimental limits. This weak coupling is the reason why the addition of higher even monomials to these potentials causes only minor quantitative changes to the shape of the curves.
The inflaton potential eq.(42) in the original inflaton field takes therefore the form
V(φ)=4NM4y(u)⎧⎨⎩1−y(u)8Nφ2M2Pl+n∑k=2c2kk(y(u)8N)kφ2kM2kPl⎫⎬⎭,
Therefore, since the coupling is we have the -th term in the potential suppressed by the -th power of as well as by the factor .
In particular, the quartic term
y(u)c432N(MMPl)4φ4
possesses a very small quartic coupling since . Notice that these suppression factors are natural in the GL approach and come from the ratio of the two relevant energy scales here: the Planck mass and the inflation scale . When the GL approach is not used these suppression factors do not follow in general.
The validity of the GL approach relies on the wide separation between the scale of inflation and the higher energy scale (corresponding to the underlying unknown microscopic theory as discussed in ref. 1sN ().) It is not necessary to require in the GL approach but to impose 1sN ()
V(φ)≪M4Pland hencev1(ϕ√g)≪1012g.
This last condition gives an upper bound for the inflaton field depending on the large argument behavior of . We get for example:
φ≪106MPlforv1(u)u→∞∼u2,φ≪2600MPlforv1(u)u→∞∼u4.
The validity of the effective GL theory relies on that separation of scales and the GL approach allows to determine the scale of inflation as GeV (at the GUT scale) and well below the Planck scale using the amplitude of the scalar fluctuations from the CMB data mcmc (); bibl ().
Inflaton potentials containing terms of arbitrary high order in the inflaton are considered in ref. lily (), sec. 25.3.2 without using the GL approach and within the small field hypothesis . Smallness conditions on the expansion coefficients are required in ref. lily (). This is actually not needed in the GL approach, whose validity relies only on the wide separation of scales between and , at least in the case of even polynomials with positive coefficients.
## Iv The quadratic plus the 2nth order double-well inflaton potential
In order to find the observationally interesting right and down border of the banana we consider the quadratic plus the th order potential for new inflation ho (),
v1(u)=12(1−u2)+12n(u2n−1). (45)
As in the general case eq.(19), we choose the absolute minimum at . The customary relation eq.(22) takes here the form ho (),
y(u)=4n∫1udxxn(1−x2)+x2n−11−x2n−2where0
This integral can be expressed as a sum of terms including logarithms and arctangents PBM ().
In the weak coupling limit and take the values of the quadratic monomial potential eqs.(26)-(27) ho (); bibl ():
ns−1y→0=−2N=−0.0333…,ry→0=8N=0.1333…, (47)
while in the strong coupling limit at fixed and take the values
ns≃1+2Nlogu⟶−∞,r≃−16Nnn−1u2logu⟶0+,
in accordance with the general formula eq.(28). In fig. 6 we plot vs. for the potential eq.(45) and the exponents and . We see that for vs. tends towards a limiting curve. For we reach the upper end of the curve [the monomial quadratic potential eq.(47)] while for large the left and lower end of the curve is reached. However, the current CMB–LSS data rule out this strong coupling part of the curve for .
### iv.1 The n→∞ limit at fixed u.
Let us first compute eq.(46) for at fixed . Since in the integrand of eq.(46),
limn→∞x2n=0.
and eq.(46) reduces to
y(u)n→∞=4n∫1udxx[n(1−x2)−1]=2[u2−1−lnu2+O(1n)].
Hence, eq.(46) becomes
y(u)n→∞=2(−lnu2−1+u2)where0
which is just twice the result found in the quartic double–well potential, eq. (31). Notice that eq.(45) in the limit becomes
limn→∞v1(u)={12(1−u2)foru<1+∞foru>1.. (49)
From eqs.(25), (45) and (48) we find for and in the limit
ns−1n→∞=−1N2u2+1(1−u2)2(−lnu2−1+u2), (50) (51) rn→∞=8Nu2(1−u2)2(−lnu2−1+u2). (52)
Now, in the limiting cases and (at ), that is, the strong coupling limit and the the weak coupling limit , respectively, we obtain from eqs.(50)
limu→ | {"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": 0.9293637275695801, "perplexity": 883.7981031785399}, "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-2019-51/segments/1575540490743.16/warc/CC-MAIN-20191206173152-20191206201152-00311.warc.gz"} |
https://infoscience.epfl.ch/record/129945 | Infoscience
Journal article
# Second-order hyperbolic S.P.D.E.'s driven by homogeneous Gaussian noise on a hyperplane
We study a class of hyperbolic stochastic partial di®erential equations in Euclidean space, that includes the wave equation and the telegraph equation, driven by Gaussian noise concentrated on a hyperplane. The noise is assumed to be white in time but spatially homogeneous within the hyperplane. Two natural notions of solutions are function-valued solutions and random field solutions. For the linear form of the equations, we identify the necessary and sufficient condition on the spectral measure of the spatial covariance for existence of each type of solution, and it turns out that the conditions di®er. In spatial dimensions 2 and 3, under the condition for existence of a random field solution to the linear form of the equation, we prove existence and uniqueness of a random field solution to non-linear forms of the equation. | {"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": 0.9388861060142517, "perplexity": 257.0576907377049}, "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/1500549424610.13/warc/CC-MAIN-20170723202459-20170723222459-00635.warc.gz"} |
https://www.physicsforums.com/threads/magnetic-vector-potential-of-infinite-wire.119419/ | # Magnetic vector potential of infinite wire
1. May 1, 2006
### siddharth
I don't know if this thread belongs in Introductory Physics or here, so please feel free to move it if you wish.
The question is
Find the vector potential a distance $r$ from an infinite straight wire carrying a current $I$
I know that the vector potential can be given by
$$\frac{\mu_0}{4 \pi} \int \frac{I \vec{dl}}{r}$$
The problem is that in this case, the wire goes to infinity. So that that doesn't work.
I tried it in another way. Since $$B = \nabla \times A$$ and the direction of A is generally the direction of the current, it reduces to
$$\left(\frac{d A_r}{dz} - \frac{ d A_z}{dr}\right) \hat{e_{\phi}} = \frac{\mu_0 I}{2 \pi r} \hat{e_{\phi}}$$
ie,
$$\frac{d A_z}{dr} = -\frac{\mu_0 I}{2 \pi r}$$
and
$$A = -\frac{\mu_0 I}{2 \pi} ln\left(\frac{r}{r_0}\right) \hat{e_z}$$
Is that right? Also, how can we generally say that A is along the direction of the current?
Last edited: May 1, 2006
2. May 1, 2006
### Tom Mattson
Staff Emeritus
No, it's fine here.
Yes, it does. The improper integral you would obtain does in fact converge.
I don't think so, but it looks like it's close. What's $r_0$?
You needn't say that at all. Remember that the potentials are related to the physical fields via differential operators. That relationship affords you some gauge freedom. You can replace $\vec{A}$ with $\vec{A}+\nabla\chi$ provided that $\chi$ is continuously differentiable. You will get the same magnetic field, even if $\nabla\chi$ isn't in the direction of the current.
3. May 1, 2006
### siddharth
Does it?
Doing it that way, I get
$$A = \left(\frac{\mu_0 I}{4 \pi}\right) ln(\sec \theta + \tan \theta)$$
where $\theta$ is the angle made between the lines joining the point where I'm finding the potential to the center of the wire and the ends.
So as $\theta$ goes to pi/2 or -pi/2, doesn't it diverge?
$r_0$ is some point where I have defined A to be 0.
Yeah, I understand. But the thing is, in most cases A in fact does point along the direction of the current.
4. May 1, 2006
### Physics Monkey
Hi siddharth,
The integral defining A does indeed diverge, however it is possible to extract a finite part which produces the correct physics. Extracting the finite parts from such integrals is a fairly common technique for dealing with these infinite geometries.
In your case the integral you want to do (after a change of variables) is $$\frac{\mu_0 I}{4 \pi} \int^{\Lambda}_{-\Lambda} dz \frac{1}{\sqrt{\rho^2 + z^2}},$$ which you can write as $$\frac{\mu_0 I}{2 \pi} \int^\Lambda_0 dz \frac{1}{\sqrt{\rho^2 + z^2}} = \frac{\mu_0 I}{2 \pi} \log{\left(\frac{\Lambda}{\rho} + \sqrt{1 + \frac{\Lambda^2}{\rho^2}} \right)}$$. For large cutoff you can approximate the argument of the log as $$\frac{\mu_0 I}{2 \pi} \log{\left(\frac{2 \Lambda}{\rho} \right)} = - \frac{\mu_0 I}{2 \pi} \log{\left(\frac{\rho}{\rho_0}\right)} + \frac{\mu_0 I}{2 \pi} \log{\left(\frac{2 \Lambda}{\rho_0}\right)}$$. Now look what has happened! Yes, the vector potential is formally infinite if you let the cutoff go to infinity, but all the physical results (magnetic field) are independent of the cutoff. You are able to extract the relevant physics from a formally divergent integral. Needless to say, this is a very powerful methodology that you will meet again and again as you progress through physics.
Also, the vector potential often points along the current because in the Coulomb gauge one has the formula $$\vec{A} = \frac{\mu_0}{4 \pi} \int d^3 x' \frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|}$$. It's clear then that as long as the current points in one direction, A will point in that direction in the Coulomb gauge.
Hope this helps.
Last edited: May 1, 2006
5. May 2, 2006
### siddharth
That's fantastic!
I'm curious, why does it work? How come the relevant physics comes out so nicely?
Thanks a ton for your help.
Last edited: May 2, 2006
6. May 2, 2006
### Tom Mattson
Staff Emeritus
I was wrong about that, as PM pointed out. You have to do some mathematical trickery that I had forgotten about (it's been a while).
As I said, that's a choice. You can choose the gauge that's convenient for you.
7. May 2, 2006
### Physics Monkey
Isn't it cool! Physics is so awesome.
What's going on is essentially a separation of different scales. As far as charged particles close to a long wire are concerned, it doesn't matter how long the wire is as long as it's very long. You could never figure out what the length of the wire is by performing local experiments. Since the length of the wire is unobservable, you might suspect that it enters the vector potential only as a pure gauge term. This suspicion is confirmed when you carry out the calculation, and you can easily check for yourself that is indeed possible to eliminate $$\Lambda$$ by choosing a different gauge. It is quite instructive to carry out calculations of this type for a variety of infinite geometries. You will find that the infinite geometry gives you some divergence, but the divergence is always unobservable. Such calculations give you glimpse of the very subtle and deep nature of field theory. It's all very beautiful stuff really.
Similar Discussions: Magnetic vector potential of infinite wire | {"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": 0.9387573003768921, "perplexity": 376.18608753080287}, "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/1492917125719.13/warc/CC-MAIN-20170423031205-00060-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://forum.mysensors.org/topic/11421/read-the-digoo-dg-r8h-temp-humidity-sensor/3 | # Read the Digoo DG-R8H temp/humidity sensor
• Hi there all,
I know that usually all the node is maked by DIY... and there is some that are really greate.
But before knowing MySensore I have bought 8 pieces of Digoo DG-R8H sensors.
This one are really nice at less than 4 Euro. But have not display
I make it work using OpenMqttGateway but I want to try the MySensors also if possible.
So is there anyone that have use this sensors please ?
Thanks all
Denis
• @DenisJ these sensors are not compatible with mysensors. You can however setup a separate mysensors network and run these at the same time.
• Yes in the mean time I understand that MySensor have all the personalized ask/answer structure.
But I still study cause I like much this type of home network.
Anyway maybe there can be a personalized gateway part only for the incompatible sensors.
I'll try to study it when I'll be more "old" in this
Thanks a lot for the answer
Denis
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21 | {"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": 0.8203423023223877, "perplexity": 3747.3235947266653}, "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-2020-50/segments/1606141176864.5/warc/CC-MAIN-20201124140942-20201124170942-00475.warc.gz"} |
http://www.gnu.org/software/gsl/manual/html_node/Vector-operations.html | Next: , Previous: Exchanging elements, Up: Vectors [Index]
#### 8.3.8 Vector operations
Function: int gsl_vector_add (gsl_vector * a, const gsl_vector * b)
This function adds the elements of vector b to the elements of vector a. The result a_i \leftarrow a_i + b_i is stored in a and b remains unchanged. The two vectors must have the same length.
Function: int gsl_vector_sub (gsl_vector * a, const gsl_vector * b)
This function subtracts the elements of vector b from the elements of vector a. The result a_i \leftarrow a_i - b_i is stored in a and b remains unchanged. The two vectors must have the same length.
Function: int gsl_vector_mul (gsl_vector * a, const gsl_vector * b)
This function multiplies the elements of vector a by the elements of vector b. The result a_i \leftarrow a_i * b_i is stored in a and b remains unchanged. The two vectors must have the same length.
Function: int gsl_vector_div (gsl_vector * a, const gsl_vector * b)
This function divides the elements of vector a by the elements of vector b. The result a_i \leftarrow a_i / b_i is stored in a and b remains unchanged. The two vectors must have the same length.
Function: int gsl_vector_scale (gsl_vector * a, const double x)
This function multiplies the elements of vector a by the constant factor x. The result a_i \leftarrow x a_i is stored in a.
Function: int gsl_vector_add_constant (gsl_vector * a, const double x)
This function adds the constant value x to the elements of the vector a. The result a_i \leftarrow a_i + x is stored in a. | {"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": 0.8274130821228027, "perplexity": 2730.1585246524096}, "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/1427131297281.13/warc/CC-MAIN-20150323172137-00093-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://infoscience.epfl.ch/record/228889 | ## Optimal control for power generating kites
Numerical solutions for optimal power generation in a pumping power cycle for Airborne Wind Energy is presented. The problem is essentially an infinite horizon optimal control problem which is then attempted to solve by three methods, finite time horizon approximation, search for the optimal periodic limit cycle and time transformation of the infinite horizon to a finite half-open space. We find the optimal periodic limit cycle formulation to be the most computationally efficient (least time consuming) of the three formulations. While the finite time approximation yields a solution close to the optimal limit cycle, it suffers due to nonlinearities in the system failing to give a consistent limit cycle and also requires large computational time. The time transformation method fails to solve the problem at all as it requires problems to have solutions, exponentially converging to a steady state which is not the case for our system in the time domain.
Year:
2014
Laboratories: | {"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": 0.9003719091415405, "perplexity": 374.9136587033071}, "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/1518891813818.15/warc/CC-MAIN-20180221222354-20180222002354-00202.warc.gz"} |
http://mathhelpforum.com/calculus/186803-proving-root-cubic-polynomial-print.html | # Proving the root of a cubic polynomial
• August 27th 2011, 11:15 AM
terrorsquid
Proving the root of a cubic polynomial
How, would I go about proving the root exists for:
$x^3+5x^2-4x-1$ in the interval [0,1]
I'm a little lost as to how I should start.
Thanks.
• August 27th 2011, 11:18 AM
Ackbeet
Re: Proving the root of a cubic polynomial
Try using the Intermediate Value Theorem. Does that give you any ideas?
• August 27th 2011, 08:04 PM
terrorsquid
Re: Proving the root of a cubic polynomial
It seems like I'm not doing anything though :D f(0) = -1 and f(1) = 1 and all you say is therefore by the intermediate value there must exist an x such that f(x) = 0 because f(0) is negative and f(1) is positive?
Am I missing any steps in the proof? Just lean on the theorem and do nothing?
• August 27th 2011, 08:49 PM
Prove It
Re: Proving the root of a cubic polynomial
Quote:
Originally Posted by terrorsquid
It seems like I'm not doing anything though :D f(0) = -1 and f(1) = 1 and all you say is therefore by the intermediate value there must exist an x such that f(x) = 0 because f(0) is negative and f(1) is positive?
Am I missing any steps in the proof? Just lean on the theorem and do nothing?
If the function goes from negative to positive, or vice versa, then it must cross the x-axis somewhere in between. That's all you need to do :)
• August 27th 2011, 09:05 PM
CaptainBlack
Re: Proving the root of a cubic polynomial
Quote:
Originally Posted by Prove It
If the function goes from negative to positive, or vice versa, then it must cross the x-axis somewhere in between. That's all you need to do :)
Continuous function
CB
• August 28th 2011, 10:12 AM
Prove It
Re: Proving the root of a cubic polynomial
Quote:
Originally Posted by CaptainBlack
Continuous function
CB
That is very true, and I should have pointed it out, but since polynomials are continuous and the function given by the OP is a polynomial, I thought it went without speaking. But thanks :) | {"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": 1, "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": 0.8745241165161133, "perplexity": 575.4378366532532}, "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-2014-23/segments/1406510268734.38/warc/CC-MAIN-20140728011748-00221-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/167280-number-homomorphisms-between-two-groups.html | # Thread: Number of homomorphisms between two groups
1. ## Number of homomorphisms between two groups
Hi,
The exact question in the book is "Determine the number of homomorphisms from the additive group Z15 to the additive group Z10" (Zn is cyclic group of integers mod n under addition)
Now if the question asks to find the number of homomorphisms from Z15 onto Z10, then by the First Isomorphism Theorem I can prove that none exit. But if homomorphisms from Z15 into Z10 are allowed to be counted, how do I do it?
2. You've got a pretty good start - the idea is to use the First Isomorphism Theorem. You more or less identified that $\displaystyle \mathbb{Z}_{15}$ needs to map onto some kind of subgroup of $\displaystyle \mathbb{Z}_{10}$. It can't be $\displaystyle \mathbb{Z}_{10}$, because $\displaystyle 15/K=10$ has no integer solution. So there are three other possibilities: $\displaystyle \mathbb{Z}_{15}$ maps into (the subgroup isomorphic to) $\displaystyle \mathbb{Z}_{5}$, $\displaystyle \mathbb{Z}_{2}$, or $\displaystyle \{e\}$.
Try applying a similar argument.
3. Originally Posted by sashikanth
Hi,
The exact question in the book is "Determine the number of homomorphisms from the additive group Z15 to the additive group Z10" (Zn is cyclic group of integers mod n under addition)
Now if the question asks to find the number of homomorphisms from Z15 onto Z10, then by the First Isomorphism Theorem I can prove that none exit. But if homomorphisms from Z15 into Z10 are allowed to be counted, how do I do it?
Let f be a homomorphism from Z15 to Z10. The image of f has to be a subgroup of Z10. Thus, the possible cases are |im f|=1, 2, 5, 10 by Lagrange's theorem. The kernel of f also has to be a subgroup of Z15. Thus, by the first isomophism theorem, |im f| has to be either 1 or 5.
Case 1: |im f|=1. There is only one such homomorphism, a trivial one.
Case 2: |im f|=5. There are four such homomorphisms, i.e., \eulerphi(5) = 4.
For case 2, f(1) has to generate the group, im f. It follows that the number of choices for f(1) is \eulerphi(5) = 4.
Thus, the total number is 5.
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# find the number of homomorphusm
Click on a term to search for related topics. | {"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": 0.814357340335846, "perplexity": 459.2958547812648}, "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-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00533.warc.gz"} |
https://www.physicsforums.com/threads/definition-of-a-rotating-frame-in-gr.380136/ | # Definition of a rotating frame in GR?
1. Feb 20, 2010
### bcrowell
Staff Emeritus
What is a definition in GR that correctly captures the concept that a frame is rotating? Is it enough to say that it's stationary but not static?
2. Feb 20, 2010
### atyy
Last edited: Feb 21, 2010
3. Feb 21, 2010
### edpell
One observes no fictitious centrifugal force. ?
4. Feb 21, 2010
### bcrowell
Staff Emeritus
Thanks for the link, atyy! Hmm...I think what he's saying with the timelike congruences is essentially equivalent to the idea that a particular observer can check whether the Sagnac effect exists. For instance, say you have a rotating disk. You can make a timelike congruence consisting of world-lines at rest relative to the axis, or a congruence consisting of world-lines at rest relative to the disk. In the latter case, you get a Sagnac effect at every point in space.
I guess my question was awfully vague, but this may help to point me in the right direction to make it more well defined. It seems straightforward to define the right notion for a local observer: do you get a Sagnac effect? I had in mind more the question of whether there was any way to say anything globally.
I don't think this works, because by the equivalence principle a centrifugal force is equivalent to a gravitational force.
5. Feb 21, 2010
### atyy
So I googled a bit and came across Ashtekar and Magnon, 1975 about the Sagnac effect in GR. They discuss two definitions of rotation which are absolute. One is the rotation of a timelike vector field, the other is the rotation of a Fermi transported tetrad. And somehow the Sagnac effect links both of them, and they also say rotation is only a "local" concept in GR. I haven't read the paper beyond that.
6. Feb 21, 2010
### bcrowell
Staff Emeritus
Cool, thanks! That makes sense to me. The Sagnac effect is something you can measure locally, and the absence of a Sagnac effect (locally) is equivalent to staticity (locally). So I think the answer to my original question is probably that there is no way to say in general whether a frame is globally rotating, but you can do it locally, and my proposed definition works.
Similar Discussions: Definition of a rotating frame in GR? | {"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": 0.8819981813430786, "perplexity": 566.6008140088722}, "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/1502886102663.36/warc/CC-MAIN-20170816212248-20170816232248-00318.warc.gz"} |
https://www.lessonplanet.com/teachers/the-antarctic-ecosystem-where-would-it-be-without-krill | # The Antarctic Ecosystem: Where Would It Be Without Krill?
##### This The Antarctic Ecosystem: Where Would It Be Without Krill? lesson plan also includes:
Students investigate the importance of krill to the Antarctic ecosystem by researching the animals that depend on it and drawing a food web. They conclude by writing paragraphs explaining the potential consequences of a decline in krill populations.
Concepts
Resource Details | {"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": 0.8020596504211426, "perplexity": 4753.368249068252}, "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-2018-43/segments/1539583511216.45/warc/CC-MAIN-20181017195553-20181017221053-00203.warc.gz"} |
https://www.physicsforums.com/threads/why-does-the-gravitational-force-decrease-below-the-earths-surface.908462/ | # Why does the gravitational force decrease below the Earth's surface?
1. Mar 21, 2017
### Steven Hansel
1. The problem statement, all variables and given/known data
Why does gravity force decreases below earth's surface?
2. Relevant equations
F = G (m1 m2)/r^2
F = force of gravity
G = Universal gravity constant
m1 and m2 = mass of the objects
r = distance between two objects
g = GM/r^2
g = gravity of the object
G = universal gravity constant
M = mass of the object
r = distance/radius between two objects
a = g
g1 = g0 (1-h/r)
g1 = gravity below earth surface
g0 = gravity in the earth surface
h = height of the object
r = distance between two objects
3. The attempt at a solution
I'm really confused on why does gravity force decreases below earth's surface. Gravity becomes stronger when the radius/distance become shorter between two objects. Now, an object is getting shorter radius/distance to the planet why the gravitational force decreases? Isn't it supposed to be stronger? I looked up at the internet and it said "The body will be attracted by the mass of the Earth which is enclosed in a sphere of radius (R - h)" Does this mean that gravity decreases because there are mass in that enclosed sphere?
So, will the gravitational force increases if i create a really deep hole below earth surface and put an object into the deep hole?
Last edited: Mar 21, 2017
2. Mar 21, 2017
### Bandersnatch
Can you write down the gravitational force equation and list all the variables in it?
3. Mar 21, 2017
### Steven Hansel
Done!
4. Mar 21, 2017
### Bandersnatch
Ok, consider an uniform solid sphere (a ball) composed of some material of density ρ. The sphere has mass M and radius R. Can you express the mass of the sphere in terms of its radius and density?
5. Mar 21, 2017
### Steven Hansel
Msphere = 4/3π.r^3.p
i think replace M with volume of a sphere and density of the sphere? into the equation of g = 4π.r^3.p/3.(r-h)^2
Correct me if i'm wrong thanks!
6. Mar 21, 2017
### Bandersnatch
so far so good.
Forget h. We're looking at gravitational acceleration at the surface of the sphere.
You forgot G and the factor of 1/3 from the volume equation. The equation should read:
$g=G \frac{4 \pi R^3 \rho}{3 r^2}$
Where R is the radius of the sphere and r is the distance to the centre of the gravitational field. When you're above the surface of a sphere, these two are different. But they're equal when standing on the surface of a sphere, which lets you write:
$g=CR$ where C are all the constants grouped together.
Is everything clear so far? Can you see what happens if you dig down?
7. Mar 21, 2017
### Steven Hansel
Wow, thanks! that really give me an insight, i was confused at the sphere part but now i understand. thanks again!
8. Mar 21, 2017
### PeroK
If you put an object in a very deep hole, then the Earth's mass above the object will be trying to pull it out of the hole, and no longer be pulling it towards the centre.
9. Mar 21, 2017
### Steven Hansel
That really helped me more, because of the radius decreasing as we go to the core of the earth, the mass of earth/radius into the core also decreases. thanks a lot!
10. Mar 21, 2017
### Staff: Mentor
That's not a good description as it contradicts the shell theorem. The object should feel no net force from any of the planet's mass above it. It only feels force from the mass below it, and said mass is less the deeper the object is.
11. Mar 21, 2017
### BvU
For the benefit of Steven:
It looks as if PerOk and Gneill disagree. They don't. They just have a different idea of 'above'. For P it's in a direction away from the object and away from the center of the earth towards the surface and for GNeill it's everything that's further away from the center of the earth than the object. PeroK: dome, Neill: shell.
Makes a difference.
The nice thing is that indeed the 'above' from PerOk and his 'below' (but outside the sphere with a radius equal to the distance from the object to the center of the earth) exactly compensate each other: no net force as Gneill states.
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https://www.physicsforums.com/threads/angular-momentum-along-a-sloping-line.823882/ | Homework Help: Angular momentum along a sloping line
1. Jul 18, 2015
rpthomps
1. The problem statement, all variables and given/known data
A 1.0 kg particle is moving at a constant 3.5 m/s along the line y=0.62x +1.4, where x and y are in meters and where the motion is toward the positive x and y directions. Find its angular momentum about the origin
2. Attempt at a solution
$L=Iw\\\\L=myv\\\\L=(1)(0.62x+1.4)(3.5)$
Not sure what to do with x though. If I set x=0, this just evaluates the momentum at a point not over the line. The line is infinite, so I would have thought the momentum evaluates to infinity as well but the answer is 4.2
2. Jul 18, 2015
Suraj M
Firstly, how did you get this formula? By the definition?
Rethink your substitution for y. What is $y$ by definition?
3. Jul 19, 2015
rpthomps
You're right. There is a problem with my relationship.
The trig part doesn't seem to simplify to nicely though...
4. Jul 19, 2015
haruspex
What is this point P you have chosen? Just consider the point where the trajectory crosses the y axis.
5. Jul 19, 2015
rpthomps
Can I use that position because angular momentum will be conserved for the whole trip and thus will be the same along the path of the mass and the position you suggested is the simplest to calculate?
6. Jul 19, 2015
haruspex
Yes.
7. Jul 19, 2015
rpthomps
Then thank you sir for your help! Really appreciated.
8. Jul 21, 2015
Suraj M
OP, since you've got the answer, it might help you in the future to know the formula for the perpendicular distance of a point from a line, which would simplify the calculation as there would be no angle involved in the calculation.
Do you happen to have a formula like that? if you did you'll get your d and hence answer would just be mvd.
9. Jul 26, 2015
rpthomps
Thank you.
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http://cms.math.ca/cjm/msc/32W20?fromjnl=cjm&jnl=CJM | location: Publications → journals
Search results
Search: MSC category 32W20 ( Complex Monge-Ampere operators )
Expand all Collapse all Results 1 - 3 of 3
1. CJM Online first
Zheng, Tao
The Chern-Ricci flow on Oeljeklaus-Toma manifolds We study the Chern-Ricci flow, an evolution equation of Hermitian metrics, on a family of Oeljeklaus-Toma (OT-) manifolds which are non-Kähler compact complex manifolds with negative Kodaira dimension. We prove that, after an initial conformal change, the flow converges, in the Gromov-Hausdorff sense, to a torus with a flat Riemannian metric determined by the OT-manifolds themselves. Keywords:Chern-Ricci flow, Oeljeklaus-Toma manifold, Calabi-type estimate, Gromov-Hausdorff convergenceCategories:53C44, 53C55, 32W20, 32J18, 32M17
2. CJM 2013 (vol 66 pp. 1413)
Zhang, Xi; Zhang, Xiangwen
Generalized Kähler--Einstein Metrics and Energy Functionals In this paper, we consider a generalized Kähler-Einstein equation on Kähler manifold $M$. Using the twisted $\mathcal K$-energy introduced by Song and Tian, we show that the existence of generalized Kähler-Einstein metrics with semi-positive twisting $(1, 1)$-form $\theta$ is also closely related to the properness of the twisted $\mathcal K$-energy functional. Under the condition that the twisting form $\theta$ is strictly positive at a point or $M$ admits no nontrivial Hamiltonian holomorphic vector field, we prove that the existence of generalized Kähler-Einstein metric implies a Moser-Trudinger type inequality. Keywords:complex Monge--Ampère equation, energy functional, generalized Kähler--Einstein metric, Moser--Trudinger type inequalityCategories:53C55, 32W20
3. CJM 2009 (vol 62 pp. 218)
Xing, Yang
The General Definition of the Complex Monge--Ampère Operator on Compact Kähler Manifolds We introduce a wide subclass ${\mathcal F}(X,\omega)$ of quasi-plurisubharmonic functions in a compact Kähler manifold, on which the complex Monge-Ampère operator is well defined and the convergence theorem is valid. We also prove that ${\mathcal F}(X,\omega)$ is a convex cone and includes all quasi-plurisubharmonic functions that are in the Cegrell class. Keywords:complex Monge--Ampère operator, compact Kähler manifoldCategories:32W20, 32Q15
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https://en.wikiversity.org/wiki/Electromagnetic_wave | # Electromagnetic wave
Subject classification: this is a physics resource.
Search for Electromagnetic wave on Wikipedia.
## Electromagnetic wave
Electromagnetic waves, which are synchronized oscillations of electric and magnetic fields that propagate at the speed of light through a vacuum. The oscillations of the two fields are perpendicular to each other and perpendicular to the direction of energy and wave propagation, forming a transverse wave.
Electromagnetic waves are produced whenever charged particles are accelerated, and these waves can subsequently interact with other charged particles.
Vector equation of Electromagnetic wave
${\displaystyle \nabla \cdot E=0}$
${\displaystyle \nabla \times E={\frac {1}{T}}E}$
${\displaystyle \nabla \cdot B=0}$
${\displaystyle \nabla \times B={\frac {1}{T}}B}$
Electromagnetic wave equation
${\displaystyle \nabla ^{2}E=-\omega E}$
${\displaystyle \nabla ^{2}B=-\omega B}$
Electromagnetic wave function
${\displaystyle E=ASin\omega t}$
${\displaystyle B=ASin\omega t}$
${\displaystyle \omega ={\sqrt {\frac {1}{T}}}=C=\lambda f}$
${\displaystyle T=\mu \epsilon }$
Electromagnetic radiation is associated with those EM waves that are free to propagate themselves ("radiate") without the continuing influence of the moving charges that produced them, because they have achieved sufficient distance from those charges. Thus, EMR is sometimes referred to as the far field. In this language, the near field refers to EM fields near the charges and current that directly produced them, specifically, electromagnetic induction and electrostatic induction phenomena.
EM waves carry energy, momentum and angular momentum away from their source particle and can impart those quantities to matter with which they interact. Quanta of EM waves are called photons, whose rest mass is zero, but whose energy, or equivalent total (relativistic) mass, is not zero so they are still affected by gravity.
Electromagnetic radiation travels as a moment at speed
${\displaystyle v=\omega ={\sqrt {\frac {1}{T}}}=C=\lambda f}$
Carry energy level
${\displaystyle E=pv=pC=p\lambda f=hf}$
Where
${\displaystyle h=p\lambda }$
From above
${\displaystyle p={\frac {h}{\lambda }}}$
${\displaystyle \lambda ={\frac {h}{p}}={\frac {C}{f}}}$
Electromagnetic radiation is in the form of a Quanta , h , whose rest mass is zero . EM travels as Electromagnetic wave at speed of light carries an energy level of a Photon , hf
Photon (Quanta's energy) . ${\displaystyle E_{h}=hf=h({\frac {\omega }{2\pi }})=\hbar \omega }$
Quanta (Massless particle) . ${\displaystyle h=p\lambda }$
Moment. ${\displaystyle p={\frac {h}{\lambda }}=h{\frac {k}{2\pi }}=\hbar k}$
### Electromagnetic spectrum
The wavefront of electromagnetic waves emitted from a point source (such as a lightbulb) is a sphere. The position of an electromagnetic wave within the electromagnetic spectrum could be characterized by either its frequency of oscillation or its wavelength
The Electromagnetic spectrum includes, in order of increasing frequency and decreasing wavelength:
Electromagnetic radiation carries an energy level
${\displaystyle E=pv=pC=p\lambda f=hf}$
This energy is quantized by a quantity called quanta
${\displaystyle h=p\lambda }$
Electromagnetic radiation carries an energy level
${\displaystyle E=hf}$
This energy is quantized by a quantity called quanta
${\displaystyle h=p\lambda }$
Which displays a duality of Wave-Particle like
Wave like . ${\displaystyle \lambda ={\frac {h}{p}}}$
Particle like . ${\displaystyle p={\frac {h}{\lambda }}}$
There are two states that photon are found Radiant Photon and Electric Photon
Radiant Photon is found at threshold frequency , fo
${\displaystyle f=f_{o}={\frac {C}{\lambda _{o}}}}$
${\displaystyle E=hf_{o}}$
${\displaystyle h=p\lambda _{o}}$
${\displaystyle p={\frac {h}{\lambda _{o}}}}$
${\displaystyle \lambda _{o}={\frac {h}{p}}={\frac {C}{f_{o}}}}$
Electric Photon is found at frequency greater than the threshold frequency , f > fo
${\displaystyle f>f_{o}>{\frac {C}{\lambda _{o}}}}$
${\displaystyle E=hf}$
${\displaystyle h=p\lambda }$
${\displaystyle p={\frac {h}{\lambda }}}$
${\displaystyle \lambda _{o}={\frac {h}{p}}={\frac {C}{f}}}$
Photon cannot exist in 2 states at the same time
${\displaystyle \Delta p\Delta \lambda >{\frac {1}{2}}{\frac {h}{2\pi }}={\frac {h}{4\pi }}={\frac {\hbar }{2}}}$
#### Penertration
Radiant photon (Beta photon) does not penertrate into matter . Eletric photon (Gamma photon) peneratrates into matter to create heat transfer in matter and can free electron from matter's substances' atom accoding to Photoelectric effect
#### Deflection
Photon enters a magnetic field will be deflected
Alpha photon will be deflected upward
Beta photon will be not be deflected and travel straight
Gamma photon will be deflected downward | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 35, "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": 0.9595023989677429, "perplexity": 1228.378244529225}, "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-2020-29/segments/1593655900335.76/warc/CC-MAIN-20200709131554-20200709161554-00311.warc.gz"} |
https://www.physicsforums.com/threads/relative-velocity-and-momentum.276249/ | # Relative velocity and momentum
1. Dec 1, 2008
### Maiia
1. The problem statement, all variables and given/known data
Tｏｎｙ （ｏｆ ｍａｓｓ ５０ｋｇ）ｃｏａｓｔｓ ｏｎ ｈｉｓ ｂｉｃｙｃｌｅ （ｏｆ ｍａｓｓ ５ ｋｇ）ａｔ ａ ｃｏｎｓｔａｎｔ ｓｐｅｅｄ ｏｆ ６ｍ／ｓ， ｃａｒｒｙｉｎｇ ａ １２ ｋｇ ｐａｃｋ． Ｔｏｎｙ ｔｈｒｏｗｓ ｈｉｓ ｐａｃｋ ｆｏｒｗａｒｄ， ｉｎ ｔｈｅ ｄｉｒｅｃｔｉｏｎ ｏｆ ｈｉｓ ｍｏｔｉｏｎ， ａｔ ２ｍ／ｓ ｒｅｌａｔｉｖｅ ｔｏ ｔｈｅ ｓｐｅｅｄ ｏｆ ｔｈｅ ｂｉｃｙｃｌｅ ｊｕｓｔ ｂｅｆｏｒｅ ｔｈｅ ｔｈｒｏｗ． Ｗｈａｔ ｉｓ ｔｈｅ ｉｎｉｔｉａｌ ｍｏｍｅｎｔｕｍ ｏｆ ｔｈｅ ｓｙｓｔｅｍ （Ｔｏｎｙ， ｔｈｅ ｂｉｃｙｃｌｅ， ａｎｄ ｔｈｅ ｐａｃｋ）？
Ａｎｓｗｅｒ ｉｎ ｕｎｉｔｓ ｏｆ ｋｇｍ／ｓ．
Ｄｏｅｓ ｒｅｌａｔｉｖｅ ｖｅｌｏｃｉｔｙ ｈｅｒｅ ｍｅａｎ ｔｈａｔ ｔｈｅ ｐａｃｋ ｉｓ ａｃｔｕａｌｌｙ ｍｏｖｉｎｇ ａｔ ８ｍ／ｓ？Ｉ＇ｍ ａ ｌｉｔｔｌｅ ｃｏｎｆｕｓｅｄ ａｓ ｔｏ ｗｈａｔ ｒｅｌａｔｉｖｅ ｖｅｌｏｃｉｔｙ ｉｓ． To get initial momentum, I ｗｏｕｌｄ ｊｕｓｔ ａｄｄ ａｌｌ ｔｈｅｉｒ ｍｏｍｅｎｔｕｍｓ ｔｏｇｅｔｈｅｒ， ｒｉｇｈｔ？Ｂｅｃａｕｓｅ ｔｈｅｙ ｈａｖｅ ｄｉｆｆｅｒｅｎｔ ｓｐｅｅｄｓ？ Ｂｅｃａｕｓｅ Ｉ don't think Ｉ would be able to add their masses together and multiply by the velocity.
2. Dec 1, 2008
### turin
Yes, relative to the speed of the bike just means that you should imagine that the bike is at rest and the earth is moving 6 m/s backward. So, yes, the pack is moving at 8 m/s relative to the earth. However, are you sure this is relevant to the question that the problem asks?
3. Dec 1, 2008
### Maiia
hmm i guess not b/c the pack should be moving at the same speed as the boy and the bike..but it would be if i were asked to find the momentum of the system after the pack was thrown i think..
4. Dec 1, 2008
Careful ... | {"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": 0.9430956244468689, "perplexity": 53.362885996887734}, "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/1476988717963.49/warc/CC-MAIN-20161020183837-00209-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://space.stackexchange.com/questions/59175/how-to-find-the-hyperbolic-angle-given-the-mean-anomaly | How to find the hyperbolic angle given the mean anomaly?
I'm modelling a hyperbolic gravity assist trajectory around Jupiter and trying to calculate the coordinates for each hour interval before/after passing periapsis. I've calculated $$M_h = 0.0176$$ is the mean anomaly 1 hour from periapsis, but how can I determine the corresponding hyperbolic angle, i.e solve this equation for H, given e = 1.3893:
$$M_h = 0.0176 = e~\rm sinh\it(H) - H$$
Using iterative calculations I know the answer is approximately $$H = 0.04$$, but I'm hoping to solve the equation above "precisely".
Like the corresponding eccentric anomaly for elliptical orbits, there is no closed-form formula for going from mean anomaly to hyperbolic anomaly.
You're going to have to use some sort of numerical method to go in that direction. Newton-Raphson tends to converge quickly enough.
$${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}}$$
With Mean anomaly at the chosen time as $$M$$, we'll look for zeroes on the function: $$f(H_n) = e \sinh H_n - H_n -M$$ And we'll need its first derivative: $$f'(H_n) = e \cosh H_n - 1$$
And we'll iterate with:
$$H_{n+1} = H_n - \frac{e \sinh H_n - H_n -M}{e \cosh H_n -1}$$
In almost every case I've used it, it's been useful to set the initial guess of the hyperbolic anomaly $$H_0$$ equal to the Mean Anomaly $$M$$. Given your chosen parameters of eccentricity $$e = 1.3893$$ and mean anomaly $$M = 0.0176$$, these are the values pulled up from a quick Google Sheets Spreadsheet:
Iteration Hyperbolic Anomaly
$$H_0$$ $$\underline{0.0}176$$
$$H_1$$ $$\underline{0.0451}9085695$$
$$H_2$$ $$\underline{0.045154584}33$$
$$H_3$$ $$\underline{0.04515458422}$$
... ...
Newton-Raphson's convergence is typically quadratic, resulting in roughly doubling the number of correct digits each iteration. We're at the three significant figures of your mean anomaly value by $$H_1$$, and by $$H_3$$ the iterated value doesn't change under the floating-point precision Google Sheets can handle.
One more thing: The convergence of using the Newton-Raphson method above with Kepler's equations gets slower as orbital eccentricity approaches $$e=1$$. If Orbital eccentricity was $$e=1.01$$, it would take until $$H_7$$ to get three significant figures stable from the specified Mean Anomaly. At $$e=1.001, H_{21}$$, and $$e=1.0001, H_{74}$$, and so forth. If your hyperbolas are extremely near-parabolic, you may need to look into an alternate method to calculate position as a function of time. | {"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": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9760357141494751, "perplexity": 448.36011637058647}, "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-2022-21/segments/1652662631064.64/warc/CC-MAIN-20220527015812-20220527045812-00759.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=337175 | # Fractions, Exponents, and x :(
by tunaizgood
Tags: exponents, fractions
P: 1 I simply just don't get fractions and exponents i was given this problem: [6x^(4/5)-3x(2/3)]/3x^(1/3) i haven't gotten anywhere and i'm stuck help thanks in advance
Sci Advisor HW Helper PF Gold P: 12,016 What problem? What are you supposed to do? All you have done so far is to present an expression. What are you supposed to do with it? Paint the expression on a subway station, perhaps? (That would be easy, but technically illegal)
Math
Emeritus
Thanks
PF Gold
P: 38,898
Quote by tunaizgood I simply just don't get fractions and exponents i was given this problem: [6x^(4/5)-3x(2/3)]/3x^(1/3)
Is there supposed to be a "^" before the 2/3?
i haven't gotten anywhere and i'm stuck help thanks in advance
So this is
$$\frac{6x^{\frac{4}{5}}- 3x^{\frac{2}{3}}}{3x^{\frac{1}{3}}}$$
and you want to simplify it? I would start by factoring out a "3"
$$\frac{3(2x^{\frac{4}{5}}- x^{\frac{2}{3}})}{3x^{\frac{1}{3}}}$$
$$=\frac{2x^{\frac{4}{5}}- x^{\frac{2}{3}}}{x^{\frac{1}{3}}}$$
Now write the $x^{1/3}$ in the denominator as $x^{-1/3}$ in the numerator and use the "laws of exponents": $x^ax^b= x^{a+b}$
Related Discussions Precalculus Mathematics Homework 13 Calculus & Beyond Homework 3 Precalculus Mathematics Homework 4 General Math 9 General Math 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": 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": 0.9303523302078247, "perplexity": 1003.2612490424796}, "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-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00644-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://physics.hmc.edu/course/3/ | ## Physics 23 — Special Relativity
Einstein’s special theory of relativity is developed from the premises that the laws of physics are the same in all inertial frames and that the speed of light is a constant. The relationship between mass and energy is explored and relativistic collisions analyzed. The families of elementary particles are described.
Course page
https://sakai.claremont.edu:8443/portal | {"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": 0.9658136367797852, "perplexity": 212.7848995280849}, "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-2019-09/segments/1550247500089.84/warc/CC-MAIN-20190221051342-20190221073342-00143.warc.gz"} |
https://www.physicsforums.com/threads/period-of-oscillation-for-vertical-spring.722354/ | # Period of Oscillation for vertical spring
1. Nov 12, 2013
### conniebear14
1. The problem statement, all variables and given/known data
A mass m=.25 kg is suspended from an ideal Hooke's law spring which has a spring constant k=10 N/m. If the mass moves up and down in the earth's gravitational field near earth's surface find period of oscillation.
2. Relevant equations
T=1/f period equals one over frequency
T= 2pi/w two pi/angular velocity
f=w/2pi
w= (k/m)^1/2
T=2pi/sqrt(k/m)
3. The attempt at a solution
Using these equations I found periods for springs that were horizontally gliding, my question is can I use these same formulas for a vertical spring? Does gravity have to be taken into account?
2. Nov 12, 2013
### haruspex
Yes. Since it partly offsets the tension in the spring, it could affect the period. But I'm not asserting that it does. Think about where the mid point of the oscillation will be in terms of spring extension.
3. Nov 12, 2013
### conniebear14
Okay, for this problem lets not take gravity into account. Are my equations correct? Can I use the same approach that I used for a horizontal spring?
4. Nov 12, 2013
### haruspex
I don't understand. I thought I just advised you to take gravity into account. Just write down the equation for ƩF=ma.
Draft saved Draft deleted
Similar Discussions: Period of Oscillation for vertical spring | {"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": 0.9780760407447815, "perplexity": 1475.9137836040793}, "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/1502886105961.34/warc/CC-MAIN-20170820015021-20170820035021-00682.warc.gz"} |
https://byjus.com/question-answer/which-of-the-following-is-called-the-pacemaker-of-the-heart-sinoatrial-nodeatrioventricular-nodepurkinje-fibresbundle/ | Question
# Which of the following is called the pacemaker of the heart?Sinoatrial nodeAtrioventricular nodePurkinje fibresBundle of His
Solution
## The correct option is A Sinoatrial nodeThe sinoatrial node is called the pacemaker of the heart as it sets the pace at which the heart beats.. The electrical impulse generated by the sinoatrial node sets the pace at which the heart beats. It is located in the right atrium. The sinoatrial node is located in the right atrium. It is also called the pacemaker The electrochemical impulse generated by the sinoatrial node stimulates the atrioventricular node and the impulse is further carried to the two ventricles through the bundle of His and purkinje fibres from the atrioventricular node. The impulse causes the contraction of the auricles and ventricles. The atrioventricular node is situated between the right atrium and the right ventricle. Bundle of His is situated in the interventricular septa.
Suggest corrections | {"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": 0.9009369015693665, "perplexity": 2313.969409046455}, "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-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00618.warc.gz"} |
https://arxiv.org/abs/1709.02709 | math-ph
(what is this?)
# Title: Large Strebel graphs and $(3,2)$ Liouville CFT
Abstract: 2D quantum gravity is the idea that a set of discretized surfaces (called map, a graph on a surface), equipped with a graph measure, converges in the large size limit (large number of faces) to a conformal field theory (CFT), and in the simplest case to the simplest CFT known as pure gravity, also known as the gravity dressed (3,2) minimal model. Here we consider the set of planar Strebel graphs (planar trivalent metric graphs) with fixed perimeter faces, with the measure product of Lebesgue measure of all edge lengths, submitted to the perimeter constraints. We prove that expectation values of a large class of observables indeed converge towards the CFT amplitudes of the (3,2) minimal model.
Comments: 35 pages, 6 figures, misprints corrected, presentation of appendix A modified Subjects: Mathematical Physics (math-ph); High Energy Physics - Theory (hep-th) MSC classes: 05C10, 33C10, 57R20 (Primary) 81T40, 05C80, 30F30 (Secondary) Report number: IPHT-T17/139 Cite as: arXiv:1709.02709 [math-ph] (or arXiv:1709.02709v2 [math-ph] for this version)
## Submission history
From: Séverin Charbonnier [view email]
[v1] Fri, 8 Sep 2017 14:05:28 GMT (262kb,D)
[v2] Wed, 13 Sep 2017 13:35:05 GMT (262kb,D) | {"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": 0.838894784450531, "perplexity": 3756.501463547004}, "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-2017-43/segments/1508187825147.83/warc/CC-MAIN-20171022060353-20171022080353-00269.warc.gz"} |
http://mathhelpforum.com/differential-equations/176880-extended-linearity-principle.html | # Math Help - Extended Linearity Principle
1. ## Extended Linearity Principle
How can I solve these differential equations by using the Extended Linearity Principle.
a) Find the general solution of the differential equation
dy/dt = -4 y+ 3e^(-t)
b) Solve the initial-value problem
dy/dt +3y = cos 2t, y(0) = -1
2. Well, the Extended Linearity Principle says that you can solve the DE by adding together the homogeneous solution and any particular solution. So what's the homogeneous solution? | {"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": 0.9829554557800293, "perplexity": 1810.0258862921805}, "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/1466783403508.34/warc/CC-MAIN-20160624155003-00171-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/114290-matrices-cramer-s-rule-help-needed.html | # Thread: MATRICES (cramer's rule) help needed
1. ## MATRICES (cramer's rule) help needed
MATRICES
Solve the following . ( Cramer's Rule )
Q) x+2y+3z=7
3x+4y+2z=9
2x+5y+4z=9
2. Originally Posted by raza9990
MATRICES
Solve the following . ( Cramer's Rule )
Q) x+2y+3z=7
3x+4y+2z=9
2x+5y+4z=9
What help do you want? You cite Cramer's rule. Do you know what that is?
Cramer's rule says that the solutions of a system of equations
$a_{11}x+ a_{12}y+ a_{13}z= b_1$
$a_{21}x+ a_{22}y+ a_{23}z= b_2$
$a_{31}x+ a_{32}y+ a_{33}z= b_3$
are given by
$x=\frac{\left|\begin{array}{ccc}b_1 & a_{12} & a_{13} \\ b_2 & a_{22} & a_{23} \\ b_3 & a_{32} & a_{33}\end{array}\right|}{\left|\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|}$
$y=\frac{\left|\begin{array}{ccc}a_{11} & b_2 & a_{13}\\ a_{21} & b_2 & a_{23}\\ a_{31} & b_3 & a_{33}\end{array}\right|}{\left|\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|}$
$z=\frac{\left|\begin{array}{ccc}a_{11} & a_{12} & b_1 \\ a_{21} & a_{22} & b_2 \\ a_{31} & a_{32} & b_3\end{array}\right|}{\left|\begin{array}{ccc}a_{ 11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|}$
Put in the numbers for your equations and calculate the determinants. Have you done that yet?
3. Originally Posted by HallsofIvy
What help do you want? You cite Cramer's rule. Do you know what that is?
Cramer's rule says that the solutions of a system of equations
$a_{11}x+ a_{12}y+ a_{13}z= b_1$
$a_{21}x+ a_{22}y+ a_{23}z= b_2$
$a_{31}x+ a_{32}y+ a_{33}z= b_3$
are given by
$x=\frac{\left|\begin{array}{ccc}b_1 & a_{12} & a_{13} \\ b_2 & a_{22} & a_{23} \\ b_3 & a_{32} & a_{33}\end{array}\right|}{\left|\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|}$
$y=\frac{\left|\begin{array}{ccc}a_{11} & b_2 & a_{13}\\ a_{21} & b_2 & a_{23}\\ a_{31} & b_3 & a_{33}\end{array}\right|}{\left|\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|}$
$z=\frac{\left|\begin{array}{ccc}a_{11} & a_{12} & b_1 \\ a_{21} & a_{22} & b_2 \\ a_{31} & a_{32} & b_3\end{array}\right|}{\left|\begin{array}{ccc}a_{ 11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|}$
Put in the numbers for your equations and calculate the determinants. Have you done that yet?
plz can u solve it for me.
i cant figure it out
4. You've been given the complete set-up for applying Cramer's Rule. All that remains is the plug-in-chug, computing the numerical values of the determinants (perhaps in your calculator or other technology), and simplifying the resulting fractions.
Where are you stuck? Please be complete. Thank you! | {"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": 12, "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": 0.9486604928970337, "perplexity": 1673.8240982429902}, "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/1510934805242.68/warc/CC-MAIN-20171119004302-20171119024302-00789.warc.gz"} |
https://infoscience.epfl.ch/record/177041 | Infoscience
Journal article
# Soliton Instabilities and Vortex Street Formation in a Polariton Quantum Fluid
Exciton polaritons have been shown to be an optimal system in order to investigate the properties of bosonic quantum fluids. We report here on the observation of dark solitons in the wake of engineered circular obstacles and their decay into streets of quantized vortices. Our experiments provide a time-resolved access to the polariton phase and density, which allows for a quantitative study of instabilities of freely evolving polaritons. The decay of solitons is quantified and identified as an effect of disorder-induced transverse perturbations in the dissipative polariton gas. | {"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": 0.8310610055923462, "perplexity": 1590.8302490144833}, "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-2017-09/segments/1487501170651.78/warc/CC-MAIN-20170219104610-00165-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/35082-proving-print.html | # Proving...
• Apr 19th 2008, 02:52 AM
Simplicity
Proving...
Q: Using the definitions of $\mathrm{cosh} x$ and $\mathrm{sinh} x$ in terms of $e^x$ and $e^{-x}$, prove that $\mathrm{cosh} 2x = 2 \mathrm{cosh}2x - 1$.
• Apr 19th 2008, 03:06 AM
mr fantastic
Quote:
Originally Posted by Air
Q: Using the definitions of $\mathrm{cosh} x$ and $\mathrm{sinh} x$ in terms of $e^x$ and $e^{-x}$, prove that $\mathrm{cosh} 2x = 2 \mathrm{cosh}2x - 1$.
Prove: $\mathrm{cosh} 2x = 2 \mathrm{cosh}^2 x - 1$.
$\cosh (2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{(e^x + e^{-x})^2 - 2}{2} = \frac{(e^x + e^{-x})^2}{2} - \frac{2}{2} = 2 \left(\frac{e^x + e^{-x}}{2}\right)^2 - 1$ .......
• Apr 19th 2008, 03:24 AM
Simplicity
Quote:
Originally Posted by mr fantastic
Prove: $\mathrm{cosh} 2x = 2 \mathrm{cosh}^2 x - 1$.
$\cosh (2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{(e^x + e^{-x})^2 \mathbf{- 2}}{2} = \frac{(e^x + e^{-x})^2}{2} - \frac{2}{2} = 2 \left(\frac{e^x + e^{-x}}{2}\right)^2 - 1$ .......
How did you get $-2$?
• Apr 19th 2008, 03:34 AM
mr fantastic
Quote:
Originally Posted by Air
How did you get $-2$?
$(e^{x} + e^{-x})^2 = e^{2x} + 2 (e^x)(e^{-x}) + e^{-2x} = e^{2x} + 2 + e^{-2x}$.
Therefore $e^{2x} + e^{-2x} = (e^{x} + e^{-x})^2 - 2$. | {"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": 0.9928101301193237, "perplexity": 3669.8752041128396}, "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/1476988720356.2/warc/CC-MAIN-20161020183840-00415-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://quant.stackexchange.com/questions/11518/simulating-state-space-model-with-ar1-dynamics | Simulating state space model with AR(1) dynamics
I asked a question similar to this previously:
https://dsp.stackexchange.com/questions/16341/simulating-a-state-space-model
However I think I have a better handle on it now and want to re-ask it:
I simply want to simulate data from a state space model where the state variables follow an AR(1) process (see the code in the first link above).
Given the burn in issues (see link below) I assume it's better to determine empirically how many observations are required until the system reaches its theoretical unconditional variance then ensure that I simulate at least 2 times that amount before using the x(t) generated by the AR(1) process in my state space model.
http://www.mathworks.co.uk/help/econ/simulate-stationary-arma-processes.html
Q1.) If I want to simulate data from my state space model is it necessary to ensure that the AR(1) process is in it's equilibrium state first?
Q2.) From the simulated data I will be able to estimate the observation and state error variances as well as the AR(1) parameters and the unconditional mean and variance of the process (averaged over many sample runs). Assuming these empirical values all match their corresponding theoretical ones, can I then be fully satisfied that the state space model which is based on the AR(1) process has been implemented correctly?
Q3.) How can I estimate what the likely error bounds should be on the parameters that I propose to estimate in Q2?
Baz
-
1 Answer
Q1- for AR(1) only one 1 lag, ie burn in, should be sufficient. However, you could do 50 to feel comfortable.
Q2- Matching the theoretical one is not a possibility
Q3. (update) AIC/BIC tests on the simulated series can help select the best one. You can get the logL values from KF or estimate functions in Matlab.
-
Thanks!! 1 lag I will try it. Seems a little weird though> I can simulate the process 50 times (as shown in the first link) and then take the variance of the AR(1) process as different time steps for processes with a high auto-correlation value say 0.99 it cab take a few hundred steps before they settle to their steady state theoretical variance. It would seem that prior to this any estimation process is trying to hit a moving target? – Bazman Jun 3 '14 at 9:55
Q2. The whole point of simulated the AR(1) process is so I can generate simluated data (i.e.) with known parameter values. Then fit the model shown at the bottom of p34 onwards here: pure.au.dk/portal-asb-student/files/48326397/…. As noted above I can check that the variance and means are correct across samples but any one individual sample will be subject to stochastic variation (and unfortunately in practice I can only use one sample). The idea is to estimate the mean and covariance of the state variable X using a Kalman filter, then to – Bazman Jun 3 '14 at 9:59
Q2 continued.) then compute the log-likelihood across all parameter values. As a final step the loglikelihood is used as the objective function for an optimization process to find the original variables. Right now I am simulating the AR(1) process and trying to apply the Kalman Filter/optimisation process but the results are not really that good even when i give the optimizer good initial guesses. Should this work? If not why not and how can I test the Kalman Filter/optimization scheme for this model set up? – Bazman Jun 3 '14 at 10:05
q3.) How do I calculate the p-value? – Bazman Jun 3 '14 at 10:06
To find the best AR simulated series that best fits the parameters, check the LogL, or AIC/BIC (aicbic Matlab function) from estimate function in Matlab. KF will give you the best estimate implied in the simulated series. p-values wont help you here, you would get them from regression. – user12348 Jun 3 '14 at 10:57 | {"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": 0.8401542901992798, "perplexity": 649.531930041814}, "config": {"markdown_headings": false, "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-2015-22/segments/1432207932596.84/warc/CC-MAIN-20150521113212-00305-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://talkstats.com/tags/exact-tests/ | # exact tests
1. ### Interpretation of Monte Carlo Significance for Chi-Square Analysis
I am currently utilizing a 4x7 chi-square and cannot collapse any columns or rows, resulting in expected cell counts of less than 5 in 14 cells (50%), minimum expected value is 1.36. I included a monte carlo test of significance to account for this, however, I am unclear on how to interpret... | {"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": 0.9774612784385681, "perplexity": 952.2862554725215}, "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-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00473.warc.gz"} |
https://pirsa.org/22100002 | PIRSA:22100002
# Review Talk: A primer on the covariant phase space formalism
### APA
Fiorucci, A. (2022). Review Talk: A primer on the covariant phase space formalism. Perimeter Institute. https://pirsa.org/22100002
### MLA
Fiorucci, Adrien. Review Talk: A primer on the covariant phase space formalism. Perimeter Institute, Oct. 03, 2022, https://pirsa.org/22100002
### BibTex
``` @misc{ pirsa_22100002,
doi = {10.48660/22100002},
url = {https://pirsa.org/22100002},
author = {Fiorucci, Adrien},
keywords = {Quantum Gravity},
language = {en},
title = {Review Talk: A primer on the covariant phase space formalism},
publisher = {Perimeter Institute},
year = {2022},
month = {oct},
note = {PIRSA:22100002 see, \url{https://pirsa.org}}
}
```
Adrien Fiorucci Technische Universität Wien
Collection
Talk Type
Subject
## Abstract
This lecture aims at introducing the notion of asymptotic symmetries in gravity and the derivation of the related surface charges by means of covariant phase space techniques. First, after a short historical introduction, I will rigorously define what is meant by “asymptotic symmetry” within the so-called gauge-fixing approach. The problem of fixing consistent boundary conditions and the formulation of the variational principle will be briefly discussed. In the second part of the lecture, I will introduce the covariant phase space formalism, as conceived by Wald and coworkers thirty years ago, which adapts the Hamiltonian formulation of classical mechanics to Lagrangian covariant field theories. With the help of this fantastic tool, I will elaborate on the construction of canonical surface charges associated with asymptotic symmetries and address the crucial questions of their conservation and integrability on the phase space. In the third and last part, I will conclude with an analysis of the algebraic properties of the surface charges, describing in which sense they represent the asymptotic symmetry algebra in full generality, without assuming conservation or integrability. For pedagogical purposes, the theoretical concepts will be illustrated throughout in the crucial and well-known case of radiative asymptotically flat spacetimes in four dimensions, as described by Einstein’s theory of General Relativity, and where many spectacular and unexpected features appear even in the simplest case of historical asymptotically Minkowskian boundary conditions. In particular, I will show that the surface charge algebra contains the physical information on the flux of energy and angular momentum at null infinity in the presence of gravitational radiation. | {"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": 0.8450592160224915, "perplexity": 441.0208819912669}, "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-2023-14/segments/1679296948900.50/warc/CC-MAIN-20230328232645-20230329022645-00611.warc.gz"} |
https://math.stackexchange.com/questions/2483800/interpretation-of-curvature-formula-for-a-parametric-curve | # Interpretation of Curvature formula for a parametric curve
Given $S(t) = (x(t), y(t))$, the curvature at any point on S is given by below formula:
$$K = \dfrac{S'(t) \times S''(t)}{|S'(t)|^{3/2}}$$
Where $S'(t)$ is the first order derivative of $S(t)$ and $S''(t)$ is the second order derivative of $S(t).$
1. I know that the first order derivative gives the tangent vector function to the curve, but How do i interpret the second order derivative of a parametric curve?
2. In the above formula for curvature, how does more "perpendicular-ness" between $S'(t)$ and $S''(t)$ increases the curvature of the curve?
## 1 Answer
Here's an intuitive explanation. The second derivative can be interpreted as acceleration. If you decompose acceleration in a tangential and orthogonal part with respect to the curve then the tangential part does not contribute to a change in direction (it only changes speed along the curve). So the curvature –how much you turn the steering wheel– depends only on the orthogonal part and your speed.
• [+1] Good explanation... and this orthogonal part will be at its maximum when S' and S'' are orthogonal – Jean Marie Oct 22 '17 at 6:20 | {"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": 0.9200148582458496, "perplexity": 297.41410691290343}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2019-26/segments/1560628001014.85/warc/CC-MAIN-20190627075525-20190627101525-00053.warc.gz"} |
https://en.m.wikibooks.org/wiki/Numerical_Methods_Qualification_Exam_Problems_and_Solutions_(University_of_Maryland)/August_2002 | # Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/August 2002
## Problem 2
Suppose there is a quadrature formula ${\displaystyle \int _{a}^{b}f(x)dx\approx w_{a}f(a)+w_{b}f(b)+\sum _{j=1}^{n}w_{j}f(x_{j})\!\,}$ which produces the exact integral whenever ${\displaystyle f\!\,}$ is a polynomial of degree ${\displaystyle 2n+1\!\,}$ . Here the nodes ${\displaystyle \{x_{j}\}_{j=1}^{n}\!\,}$ are all distinct. Prove that the nodes lies in the open interval ${\displaystyle (a,b)\!\,}$ and the weights ${\displaystyle w_{a},w_{b}\!\,}$ and ${\displaystyle \{w_{j}\}_{j=1}^{n}\!\,}$ are positive.
## Solution 2
### All nodes lies in (a,b)
Let ${\displaystyle \{x_{i}\}_{i=1}^{l}\!\,}$ be the nodes that lie in the interval ${\displaystyle (a,b)\!\,}$ .
Let ${\displaystyle q_{l}(x)=\prod _{i=1}^{l}(x-x_{i})\!\,}$ which is a polynomial of degree ${\displaystyle l\!\,}$ .
Let ${\displaystyle p_{n}(x)=\prod _{i=1}^{n}(x-x_{i})=q_{l}(x)\prod _{i=1}^{n-l}(x-x_{i})\!\,}$ which is a polynomial of degree ${\displaystyle n>l\!\,}$ .
Then
${\displaystyle \langle p_{n},q_{l}\rangle =\int _{a}^{b}q_{l}^{2}(x)\underbrace {\prod _{i=1}^{n-l}(x-x_{i})} _{r(x)}\neq 0\!\,}$
since ${\displaystyle r(x)\!\,}$ is of one sign in the interval ${\displaystyle (a,b)\!\,}$ since for ${\displaystyle i=1,2,\ldots n-l\!\,}$ , ${\displaystyle x_{i}\not \in (a,b).\!\,}$
This implies ${\displaystyle q_{l}\!\,}$ is of degree ${\displaystyle n\!\,}$ since otherwise
${\displaystyle \langle p_{n},q_{l}\rangle =0\!\,}$
from the orthogonality of ${\displaystyle p_{n}\!\,}$ . | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 22, "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": 0.969703197479248, "perplexity": 1336.2210552514878}, "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-2021-25/segments/1623488559139.95/warc/CC-MAIN-20210624202437-20210624232437-00375.warc.gz"} |
http://groupprops.subwiki.org/wiki/Cyclic_group | # Cyclic group
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This article is about a basic definition in group theory. The article text may, however, contain advanced material.
VIEW: Definitions built on this | Facts about this: (facts closely related to Cyclic group, all facts related to Cyclic group) |Survey articles about this |
VIEW RELATED: Analogues of this | [SHOW MORE]
This article defines a group property that is pivotal (i.e., important) among existing group properties
View a list of pivotal group properties | View a complete list of group properties [SHOW MORE]
This is a family of groups parametrized by the natural numbers, viz, for each natural number, there is a unique group (upto isomorphism) in the family corresponding to the natural number. The natural number is termed the parameter for the group family
## Definition
No. Shorthand A group is termed cyclic (sometimes, monogenic or monogenous) if ... A group is termed cyclic if ...
1 modular arithmetic definition it is either isomorphic to the group of integers or to the group of integers modulo n for some positive integer . Note that the case gives the trivial group. or for some positive integer . Note that the case gives the trivial group.
2 generating set of size one it has a generating set of size 1. there exists a such that .
3 quotient of group of integers it is isomorphic to a quotient of the group of integers it is isomorphic to a quotient group of the group of integers , i.e., there exists a surjective homomorphism from to .
### Equivalence of definitions
Further information: Equivalence of definitions of cyclic group
The second and third definition are equivalent because the subgroup generated by an element is precisely the set of its powers. The first definition is equivalent to the other two, because:
• The image of under a surjective homomorphism from to must generate
• Conversely, if an element generates , we get a surjective homomorphism by
## Particular cases
VIEW: groups satisfying this property | groups dissatisfying this property
VIEW: |
Cyclic group of order
1 Trivial group
2 Cyclic group:Z2
3 Cyclic group:Z3
4 Cyclic group:Z4
5 Cyclic group:Z5
6 Cyclic group:Z6
7 Cyclic group:Z7
8 Cyclic group:Z8
9 Cyclic group:Z9
## Metaproperties
Metaproperty name Satisfied? Proof Statement with symbols
subgroup-closed group property Yes cyclicity is subgroup-closed If is a cyclic group and is a subgroup of , is also a cyclic group.
quotient-closed group property Yes cyclicity is quotient-closed If is a cyclic group and is a normal subgroup of , the quotient group is also a cyclic group.
finite direct product-closed group property No cyclicity is not finite direct product-closed It is possible to have cyclic groups and such that the external direct product is not a cyclic group. In fact, if both and are nontrivial finite cyclic groups and their orders are not relatively prime to each other, or if one of them is infinite, the direct product will not be cyclic.
## Relation with other properties
This property is a pivotal (important) member of its property space. Its variations, opposites, and other properties related to it and defined using it are often studied
### Stronger properties
Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions
finite cyclic group both cyclic and a finite group |
group of prime order Finite cyclic group|FULL LIST, MORE INFO
odd-order cyclic group |
### Weaker properties
Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions
abelian group any two elements commute cyclic implies abelian abelian not implies cyclic Epabelian group, Locally cyclic group, Residually cyclic group|FULL LIST, MORE INFO
metacyclic group has a cyclic normal subgroup with a cyclic quotient group (obvious) metacyclic not implies cyclic Characteristically metacyclic group, Group with metacyclic derived series|FULL LIST, MORE INFO
polycyclic group has a subnormal series where all the successive quotient groups are cyclic groups Characteristically metacyclic group, Characteristically polycyclic group, Finitely generated abelian group, Metacyclic group|FULL LIST, MORE INFO
locally cyclic group every finitely generated subgroup is cyclic |
group whose automorphism group is abelian cyclic implies abelian automorphism group abelian automorphism group not implies abelian Locally cyclic group|FULL LIST, MORE INFO
group of nilpotency class two the inner automorphism group is abelian (via abelian) (via abelian) Group whose automorphism group is abelian, Group whose inner automorphism group is central in automorphism group|FULL LIST, MORE INFO
nilpotent group (via abelian) (via abelian) Abelian group, Epinilpotent group, Group of nilpotency class two, Group whose automorphism group is nilpotent|FULL LIST, MORE INFO
finitely generated group has a finite generating set cyclic means abelian with a generating set of size one any finite non-cyclic group such as the Klein four-group Polycyclic group|FULL LIST, MORE INFO
finitely generated abelian group finitely generated and abelian follows from separate implications for finitely generated and abelian Klein four-group is a counterexample. |
finitely generated nilpotent group finitely generated and nilpotent (via finitely generated abelian) (via finitely generated abelian) Finitely generated abelian group|FULL LIST, MORE INFO
supersolvable group (via finitely generated abelian) (via finitely generated abelian) Characteristically metacyclic group, Characteristically polycyclic group, Finitely generated abelian group, Metacyclic group|FULL LIST, MORE INFO
solvable group Abelian group, Metabelian group, Metacyclic group, Nilpotent group, Polycyclic group|FULL LIST, MORE INFO
## Facts
• There is exactly one cyclic group (upto isomorphism of groups) of every positive integer order : namely, the group of integers modulo . There is a unique infinite cyclic group, namely
• For any group and any element in it, we can consider the subgroup generated by that element. That subgroup is, by definition, a cyclic group. Thus, every group is a union of cyclic subgroups. Further information: Every group is a union of cyclic subgroups
## References
### Textbook references
Book Page number Chapter and section Contextual information View
Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347More info 54 formal definition
Groups and representations by Jonathan Lazare Alperin and Rowen B. Bell, ISBN 0387945261More info 3 definition introduced in paragraph
Topics in Algebra by I. N. HersteinMore info 39 Example 2.4.3 definition introduced in example
A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613More info 9
An Introduction to Abstract Algebra by Derek J. S. Robinson, ISBN 3110175444More info 47
Finite Group Theory (Cambridge Studies in Advanced Mathematics) by Michael Aschbacher, ISBN 0521786754More info 2
Algebra by Serge Lang, ISBN 038795385XMore info 9
Algebra (Graduate Texts in Mathematics) by Thomas W. Hungerford, ISBN 0387905189More info 33 defined as cyclic subgroup
Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632More info 46 Page 46: leading to point (2.7), Page 47, Point (2.9) | {"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": 38, "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": 0.8218908905982971, "perplexity": 1769.7196740788238}, "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-2014-23/segments/1405997877881.80/warc/CC-MAIN-20140722025757-00231-ip-10-33-131-23.ec2.internal.warc.gz"} |
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