Register to reply 
Parallel transport 
Share this thread: 
#1
Feb2912, 10:45 PM

Sci Advisor
P: 2,802

Assuming that we are working with an infinitessimally small region of a manifold so that we can consider only first order effects, does parallel transport in the absence of torsion necessarily "close the quadrilateral"? What I mean is, if I have two vectors (very small vectors) V and U, and I parallel transported V along U, and vice versa, will the resulting 4 vectors (2 original, 2 parallel transported) form a closed quadrilateral?
I'm trying to get an intuition for torsion, and it seems to me that torsion would be the quantity which prevented the closing of the quadrilateral. I just wanted to confirm this, it seems right to me, but I can't be sure. 


#2
Mar112, 12:38 AM

Sci Advisor
P: 1,593

What you have is correct as long as the vector fields U and V commute. For noncommuting vector fields on flat space, you would have a pentagon, where the last piece is given by the commutator of the vector fields. So in curved space, torsion measures the failure of this pentagon to close. Or thought of another way, it measures the failure of quadrilaterals to close, after taking into account the fact that vector fields might not commute.
There are many ways to write the torsion tensor in various notation systems, but I think the easiest to interpret is [tex]T(X,Y) = \nabla_X Y  \nabla_Y X  [X,Y][/tex] 


#3
Mar112, 02:15 AM

Sci Advisor
P: 2,802

I did try to take account of the lie bracket.
Since the Lie bracket operation is valid with no mention of the connection, I would think that it was always the "closer of quadrilaterals". The 5 sides of the pentagon are the vector fields U, V, evaluated at the base point, U, V evaluated at the "tips" of each other, and [U,V] closing the quadrilateral. Is this not always valid? I haven't mentioned anything about parallel transport at this point, so how could a definition of parallel transport affect this picture? When you have parallel transport with no torsion, is it not required that U, V, and the parallel transported U and V (as distinct from the U and V which naturally reside "at each others' tips") close the quadrilateral? This discussion would be much easier if I could draw this out haha. 


#4
Mar112, 02:39 AM

Sci Advisor
P: 2,802

Parallel transport
I made a picture to show what I was talking about.
This is my conception of what things look like (infinitessimally of course) in the absence of torsion. I used deltas instead of nablas to denote the covariant derivative because I can't find the nabla symbol in word. I have also suppressed the subscripts. The outer quadrilateral doesn't close, and requires [U,V] to close it, but the inner one does. One can easily see from that picture then that [itex]\nabla_v u\nabla_u v=[u,v][/itex] which implies no torsion. I seem to have derived a "proof", so to speak, from this picture, that it is always the case that the inside quadrilateral will always close in the absence of torsion. In this "proof", though, I just wantonly add and subtract vectors like I was in Euclidean space, so I wasn't sure that it's valid even in the infinitesimal case since I haven't checked how errors scale. My conception of torsion, then, is that it will open up that inner quadrilateral, leaving the outer pentagon untouched. This will obviously make the above equality that I just wrote not true. Is this a valid picture? 


#5
Mar112, 02:47 PM

Sci Advisor
P: 1,593

I think that makes sense.
There is another way to think of torsion that makes it clearer why it is called "torsion": First off, any two neighboring tangent spaces at points [itex]x[/itex] and [itex]x + dx[/itex] are related by some matrix in [itex]GL(n)[/itex]. Therefore, a connection is really just an object that tells you, for each possible displacement [itex]dx[/itex], what matrix to use in order to "glue" these tangent spaces together. Since the neighboring points are infinitesimally close together, the matrix in [itex]GL(n)[/itex] must be infinitesimally close to the identity; hence the connection is a 1form that takes values in the Lie algebra [itex]\mathfrak{gl}(n)[/itex]. Then we say that [itex]GL(n)[/itex] is the "structure group" of the manifold. There are some various properties you might ask of a manifold that correspond to "reduction of the structure group" to some subgroup of [itex]GL(n)[/itex]. For example, if the manifold is orientable, then its structure group must be in [itex]SL(n)[/itex]. We could also ask that the connection is "metric compatible", meaning that it preserves the lengths of vectors. In this case, the structure group will be [itex]O(n)[/itex], or [itex]SO(n)[/itex] if the manifold is also orientable. In any case, the Lie algebras of [itex]O(n)[/itex] and [itex]SO(n)[/itex] are the same. Suppose the connection is metriccompatible, such that all parallel transports correspond to some [itex]SO(n)[/itex] rotation. Now let us travel along a geodesic with tangent vector X. Hence [tex]\nabla_X X = 0[/tex] As you may notice, the geodesic equation doesn't care about the torsion. Since X appears twice, only the symmetric part of the connection participates, and the torsion is purely the antisymmetric part. Hence if two connections [itex]\nabla[/itex] and [itex]\nabla'[/itex] differ only by torsion, then they have the same geodesics. But what happens to the rest of the tangent space as it is parallel transported along a geodesic? The parallel transport is in [itex]SO(n)[/itex], and the geodesic equation tells us that along this particular path, one vector (namely X) must stay fixed. Hence the rest of the tangent space must rotate under [itex]SO(n1)[/itex], which is the subgroup of [itex]SO(n)[/itex] that leaves X fixed. So, the rest of the tangent space "twists around" the geodesic. So then this is what torsion means. Connections differing only by torsion must have the same geodesics, and if we travel along a geodesic, the only degrees of freedom left are those that twist around the geodesic. Hence torsion corresponds to the amount of twisting around a geodesic, as we parallel transport along the geodesic. If I have time later, I'll see if I can put this into symbols, so you can see exactly how torsion mathematically relates to twisting around geodesics. Or maybe you can give it a shot. 


#6
Mar112, 04:38 PM

Sci Advisor
P: 2,802

By "twist around", you mean like rotation in the plane(s) perpendicular to the plane that includes the X vector and a small piece of the curve that is at the point of interest (this plane is sometimes called the osculating plane right)?



#7
Mar112, 06:00 PM

Sci Advisor
P: 1,593

No. X is the only thing that has to remain fixed, by the geodesic equation. No vectors orthogonal to X need to remain fixed.
For example, in 5 dimensions, X can be fixed, and ALL vectors perpendicular to X can be rotating under SO(4). 


#8
Mar212, 03:24 PM

Sci Advisor
P: 2,802

Ah, so if we reduce the SO(n) possible rotations to SO(n1), we actually reduce the possible rotations by several planes right. For example, SO(4) has 6 possible planes of rotation, and SO(3) only has 3. I'm just trying to figure out which planes of rotations are "forbidden".
Let's work in 3D for now because that's easy to visualize. If I don't travel along geodesics, my tangent vectors can rotate in any of the 3 planes. If I do travel along a geodesic in the xdirection, i.e. imagine my geodesic coincides with the xcoordinate axis for some stretch, which plane can I still rotate in? SO(2) implies I can only rotate in 1 plane right. By symmetry this would have to be the yz plane right? This is the socalled binormal plane right (binormal to the osculating plane). Tryin to visualize this hehe. 


#9
Mar212, 03:45 PM

Sci Advisor
P: 2,802

Ah visualizing this in the shower made me get it. I can rotate in any plane perpendicular to the tangent vector to the geodesic (which is what you said lol, but now I got it). This seems so obvious now that I "saw" the picture in my head. Thanks! In my picture above, Torsion would correspond to my parallel translated vectors possibly twisting (rotating) into or out of the page right. =D
It may be a little bit before I can put it into mathematical language though...hehe. Follow up question. MTW states that torsion violates the equivalence principle, how so? If torsion keeps all geodesics the same, I don't see how it would violate the equivalence principle. 


#10
Mar212, 03:45 PM

Sci Advisor
P: 1,593

The twist rotations will always be in the subspace orthogonal to the geodesic. So whatever collection of planes make up that subspace.
The planes that are forbidden are ANY planes that contain the tangent vector to the geodesic. 


#11
Mar212, 03:47 PM

Sci Advisor
P: 2,802

Yes, I got that now, thanks for clearing that up for me. =]



#12
Mar212, 03:49 PM

Sci Advisor
P: 1,593




#13
Mar212, 03:52 PM

Sci Advisor
P: 2,802

Could I use this to then see which situation I'm in: free falling in gravitational field, or free free floating in empty space?
Would spinning particles precess in two different ways? Does this mean that EinsteinCartan theory violates the equivalence principle? (At least the Einstein or Strong versions, perhaps not the weak version)? I believe it is only the weak equivalence principle that has been tested to like 10^10 precision right. 


#14
Mar212, 04:08 PM

Sci Advisor
P: 1,593

You'll have to tell me what definitions you're using. People use the words "equivalence principle" to mean a lot of different things.



#15
Mar312, 01:19 AM

Sci Advisor
P: 2,802

Basically, for the weak equivalence principle, I just mean m_i=m_g, i.e. objects with same mass fall at the same rate.
For the strong equivalence principle, I mean, that there are not experiments which can locally tell if a frame is falling freely in a gravitational field, or floating freely in space. 


Register to reply 
Related Discussions  
Parallel transport  Special & General Relativity  14  
Parallel Transport  Advanced Physics Homework  3  
Parallel Transport  Special & General Relativity  20  
Parallel transport  Introductory Physics Homework  0  
Lie transport vs. parallel transport  Differential Geometry  10 