Parallel transport


by JustinLevy
Tags: parallel, transport
JustinLevy
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#1
Apr4-11, 04:17 AM
P: 886
If I know the metric everywhere, and I specify a closed path, how can I calculate whether a vector parallel transported around the path will return to being the same vector or not?

I assume there is some simple integral to describe this, but I'm not sure how to write it down. Unfortunately, wikipedia isn't helping much at the moment either. If someone could just explain to me how to do this, it would be quite helpful.

Also if after going around the path the vector is rotated, is the angle this makes with the original vector somehow related to Berry's phase?
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DaleSpam
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#2
Apr4-11, 06:58 AM
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Hi Justin,

There is a thread where Anamitra and I worked through several examples of parallel transport on a sphere and in the Schwarzschild metric.

http://www.physicsforums.com/showthread.php?t=423334
See in particular:
http://www.physicsforums.com/showpos...4&postcount=69
http://www.physicsforums.com/showpos...&postcount=114
JustinLevy
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#3
Apr4-11, 02:23 PM
P: 886
Thanks for the links. gah, everything is in pdf's.

I guess the issue is everything I've read give the condition defining parallel transport. Using the notation from the link you gave:
[tex]\frac{dA^\mu}{d\tau} + \Gamma^{\mu}{}_{\nu\lambda}A^\nu \frac{dx^\lambda}{d\tau} = 0[/tex]

What I'd like instead, is some geometric object that tells how a vector evolves along the transport. Like
[tex] A^\mu(\tau\!=\!1) = A^\mu(\tau\!=\!0) + \int_0^1 \mathrm{'object'} \ d\tau[/tex]

I assume this is related to the Christoffel symbol. Since everything depends on the path, I'm not sure how to invert the first equation to get such an object.

In case there is an easier way to do this, what I ultimately want is a formula for how a vector is changed along a closed path. For example, if the closed path is specified by [itex]0\le\tau\le 1[/itex], then look at
[tex]g_{\mu\nu}A^\mu(0)A^\nu(1)[/tex]

bapowell
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#4
Apr4-11, 03:16 PM
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Parallel transport


The Riemann tensor provides this information.
Mentz114
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#5
Apr4-11, 03:28 PM
PF Gold
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This reminds me of the Wilson Loop

http://en.wikipedia.org/wiki/Wilson_loop

An extract :

Actually, if A is viewed as a connection over a principal G-bundle, the equation above really ought to be "read" as the parallel transport of the identity around the loop which would give an element of the Lie group G.

Note that a path-ordered exponential is a convenient shorthand notation common in physics which conceals a fair number of mathematical operations. A mathematician would refer to the path-ordered exponential of the connection as "the holonomy of the connection" and characterize it by the parallel-transport differential equation that it satisfies.
JustinLevy
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#6
Apr4-11, 03:36 PM
P: 886
Quote Quote by bapowell View Post
The Riemann tensor provides this information.
Can you please expand upon this?
For example, in the integral equation how do I specify what 'object' is in terms of the Riemann tensor and the vector and the path?
bapowell
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#7
Apr4-11, 03:39 PM
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Quote Quote by Mentz114 View Post
This reminds me of the Wilson Loop
Yes, the change in transported vector about a closed loop on an n-dimensional Riemannian manifold can be generated by a rotation. The holonomy group is therefore [itex]O(n)[/itex].
DaleSpam
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#8
Apr4-11, 03:43 PM
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Quote Quote by JustinLevy View Post
I guess the issue is everything I've read give the condition defining parallel transport. Using the notation from the link you gave:
[tex]\frac{dA^\mu}{d\tau} + \Gamma^{\mu}{}_{\nu\lambda}A^\nu \frac{dx^\lambda}{d\tau} = 0[/tex]

What I'd like instead, is some geometric object that tells how a vector evolves along the transport. Like
[tex] A^\mu(\tau\!=\!1) = A^\mu(\tau\!=\!0) + \int_0^1 \mathrm{'object'} \ d\tau[/tex]
But the condition defining parallel transport gives you a differential equation. Just solve that differential equation to get A at any point.
bapowell
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#9
Apr4-11, 08:21 PM
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From Nakahara pg. 217:

In your notation, a vector [itex]A^\mu(\tau = 0)[/itex] parallel transported an infinitesimal distance [itex]\epsilon^\mu[/itex] along a curve C to the point [itex]\tau = 1[/itex] becomes

[tex]A^\mu(\tau=1) = A^\mu(\tau = 0) - A^\kappa(\tau = 0)\Gamma^\mu_{\nu \kappa}(\tau = 0)\epsilon^\nu[/tex].
PAllen
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#10
Apr4-11, 08:27 PM
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Quote Quote by bapowell View Post
From Nakahara pg. 217:

In your notation, a vector [itex]A^\mu(\tau = 0)[/itex] parallel transported an infinitesimal distance [itex]\epsilon^\mu[/itex] along a curve C to the point [itex]\tau = 1[/itex] becomes

[tex]A^\mu(\tau=1) = A^\mu(\tau = 0) - A^\kappa(\tau = 0)\Gamma^\mu_{\nu \kappa}(\tau = 0)\epsilon^\nu[/tex].
I think the OP meant a substantial finite distnce, not an infitesimal distance. Dalespam expressed what I was thinking: you've got a differential equation. You solve it (in principle), and you have a value of the vector at every point along the path. If the path is closed, and the limit from one direciton is differnt from the other, you have intrinsic curvature, otherwise you don't.
bapowell
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#11
Apr4-11, 08:41 PM
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Quote Quote by PAllen View Post
I think the OP meant a substantial finite distnce, not an infitesimal distance. Dalespam expressed what I was thinking: you've got a differential equation. You solve it (in principle), and you have a value of the vector at every point along the path. If the path is closed, and the limit from one direciton is differnt from the other, you have intrinsic curvature, otherwise you don't.
My post should apply to finite distances as well. The invariance of a vector parallel transported around a closed path implies zero intrinsic curvature for infinitesimally small loops as well (in fact, this is typically the way the Riemann tensor is motivated -- it 'measures' the intrinsic geometry of a surface by sampling the local (differential) neighborhood of a point):

[tex]\delta A^\mu = A^\kappa(0)R^\mu_{\kappa \lambda \nu} \epsilon^\lambda \delta^\nu[/tex]

where [itex]\epsilon[/itex] and [itex]\delta[/itex] are displacements along the two sides of the parallelogram making up the closed loop.

EDIT: OK, I understand your point PAllen now that I reread your post. Indeed, my expressions necessarily only apply to infinitesimals. I'm unsure of exactly how the integration would be done for a finite displacement, however, since I think you will have a different connection coefficient between each differentially separated coordinate patch along the finite curve.
JustinLevy
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#12
Apr4-11, 09:30 PM
P: 886
Quote Quote by bapowell View Post
From Nakahara pg. 217:

In your notation, a vector [itex]A^\mu(\tau = 0)[/itex] parallel transported an infinitesimal distance [itex]\epsilon^\mu[/itex] along a curve C to the point [itex]\tau = 1[/itex] becomes

[tex]A^\mu(\tau=1) = A^\mu(\tau = 0) - A^\kappa(\tau = 0)\Gamma^\mu_{\nu \kappa}(\tau = 0)\epsilon^\nu[/tex].
Consider the cosmic string solutions which are locally flat spacetime everywhere outside of the cosmic string, but there is an angular deficit if you travel all the way around. So by comparing lengths of different paths, it is clearly possible to tell that there is some kind of curvature. Yet since the spacetime is locally flat everywhere, shouldn't the Riemann tensor and Christoffel symbols be zero?

I somehow thought I'd be able to see this "topological" curvature, using the parallel transport around the cosmic string. Is the answer no? Your answer seems to say that is the case. I kind of hoped this would be like a Berry's Phase in the Aharonov-Bohm case with a solenoid.

So what geometric object talks about the curvature in this case? Do we need something else?
PAllen
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#13
Apr4-11, 09:37 PM
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Quote Quote by bapowell View Post
My post should apply to finite distances as well. The invariance of a vector parallel transported around a closed path implies zero intrinsic curvature for infinitesimally small loops as well (in fact, this is typically the way the Riemann tensor is motivated -- it 'measures' the intrinsic geometry of a surface by sampling the local (differential) neighborhood of a point):

[tex]\delta A^\mu = A^\kappa(0)R^\mu_{\kappa \lambda \nu} \epsilon^\lambda \delta^\nu[/tex]

where [itex]\epsilon[/itex] and [itex]\delta[/itex] are displacements along the two sides of the parallelogram making up the closed loop.

EDIT: OK, I understand your point PAllen now that I reread your post. Indeed, my expressions necessarily only apply to infinitesimals. I'm unsure of exactly how the integration would be done for a finite displacement, however, since I think you will have a different connection coefficient between each differentially separated coordinate patch along the finite curve.

[tex]
\frac{dA^\mu}{d\tau} + \Gamma^{\mu}{}_{\nu\lambda}A^\nu \frac{dx^\lambda}{d\tau} = 0
[/tex]

Now I'm confused. We have a differential equation along some specified path (with tau as the path parameter), see post #8. Solve it, in principle, and you have the vector uniquely determined at each point on the path. This is what I understood Dalespam was saying.
Sam Gralla
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#14
Apr7-11, 11:36 PM
P: 95
Hi Justin,

The differential equation you're talking about is hard to solve, but comes up often enough that people have given its solutions a name, the "parallel propagator". More specifically they'll define a bitensor (tensor depending on two spacetime points x,x') that tells you the new vector at x given the old at x'. Synge's book is the classic reference, but I'd recommend the clear treatment in Eric Poisson's review on the motion of point paticles in curved spacetime. You won't find explicit expressions except in very simple spacetimes. However, there is a series expansion near a point in a general spacetime.

-Sam

[EDIT] Two seconds later I realized/remembered you're asking about closed loops. The parallel propagator is usually only defined for geodesics and definitely wouldn't be for closed loops. It'd be hard to get a single-valued bitensor out that way :). Anyway maybe the information helps you in some way. Most of the time people will define some kind of "propagator" for differential equations they can't solve explicitly.
Ben Niehoff
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#15
Apr8-11, 01:47 AM
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Justin,

As someone else mentioned, what you are trying to do is in fact a Wilson line, where the "gauge group" is the structure group of the manifold; i.e., SO(n) for an n-dimensional manifold (or SO(p,q) if the manifold has indefinite signature). I say 'SO' rather than 'O' because parallel transport is continuous (however, we could potentially have O(n) as the structure group if the manifold is not orientable).

The thing you need to integrate is the connection form [itex]\omega^a{}_b[/itex], or "spin connection", which is a 1-form defined as

[tex]\omega_\mu{}^a{}_b \; dx^\mu = e_\lambda{}^a \; e^\nu{}_b \; \Gamma_\mu{}^\lambda{}_\nu \; dx^\mu[/tex]

where [itex]e_\lambda{}^a[/itex] are the frames. [itex]\omega[/itex] is then an [itex]\mathfrac{so}(n)[/itex]-valued 1-form. It has a direct geometric interpretation: when you contract it with a vector X, it spits back an infinitesimal rotation matrix that tells you how the tangent space is rotated when parallel transporting in the direction of X.

So, to get some finite parallel transport, you need to integrate this object along the path. The catch is, SO(n) is non-abelian for n > 2, so you need to be careful in what order you multiply all the matrices. This can be represented schematically as a path-ordered integral:

[tex]R(\gamma) = \mathcal{P} \exp \int_\gamma \omega[/tex]

And then R will be the total rotation matrix that tells you the result of parallel transport along the path [itex]\gamma[/itex]. However, this expression is not very useful for calculating anything; it's usually more efficient to just solve the differential equation you wrote down earlier.



Quote Quote by JustinLevy View Post
Consider the cosmic string solutions which are locally flat spacetime everywhere outside of the cosmic string, but there is an angular deficit if you travel all the way around. So by comparing lengths of different paths, it is clearly possible to tell that there is some kind of curvature. Yet since the spacetime is locally flat everywhere, shouldn't the Riemann tensor and Christoffel symbols be zero?

I somehow thought I'd be able to see this "topological" curvature, using the parallel transport around the cosmic string. Is the answer no? Your answer seems to say that is the case. I kind of hoped this would be like a Berry's Phase in the Aharonov-Bohm case with a solenoid.

So what geometric object talks about the curvature in this case? Do we need something else?
The catch here is that the connection coefficients do not vanish everywhere!

It is simpler to think of a 2-dimensional cone, which exhibits exactly the same effect: it has a deficit angle concentrated at a single point. You can unroll the cone into a plane with a wedge cut out. What you will notice in this case is that the cone cannot be covered by a single Cartesian coordinate chart: there is a discontinuity when jumping across the wedge. Therefore, even though the connection coefficients vanish throughout the chart, they must have a delta function when jumping between two such charts. Since your path of integration crosses from one chart to the next, you must include this delta function. You can compute this explicitly using the change-of-basis formula for the Christoffel symbols if you are careful.

Note, of course, that you CAN cover the whole cone using polar coordinates. But in these coordinates, the connection coefficients do not vanish!

Furthermore, in either coordinate system, if you are careful you should be able to compute the curvature, and find that it is a delta function concentrated at the apex of the cone.


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