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Parallel transport 
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#1
Apr411, 04:17 AM

P: 894

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? 


#2
Apr411, 06:58 AM

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P: 17,340

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 


#3
Apr411, 02:23 PM

P: 894

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] 


#4
Apr411, 03:16 PM

Sci Advisor
P: 1,682

Parallel transport
The Riemann tensor provides this information.



#5
Apr411, 03:28 PM

PF Gold
P: 4,087



#6
Apr411, 03:36 PM

P: 894

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? 


#7
Apr411, 03:39 PM

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P: 1,682




#8
Apr411, 03:43 PM

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P: 17,340




#9
Apr411, 08:21 PM

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P: 1,682

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]. 


#10
Apr411, 08:27 PM

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PF Gold
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#11
Apr411, 08:41 PM

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P: 1,682

[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. 


#12
Apr411, 09:30 PM

P: 894

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 AharonovBohm case with a solenoid. So what geometric object talks about the curvature in this case? Do we need something else? 


#13
Apr411, 09:37 PM

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PF Gold
P: 5,060

[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. 


#14
Apr711, 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 singlevalued 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. 


#15
Apr811, 01:47 AM

Sci Advisor
P: 1,594

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 ndimensional 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 1form 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 1form. 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 nonabelian for n > 2, so you need to be careful in what order you multiply all the matrices. This can be represented schematically as a pathordered 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. It is simpler to think of a 2dimensional 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 changeofbasis 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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