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Does torsion make parallel transport direction dependent?

  1. Nov 14, 2012 #1

    pervect

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    Torsion has propped up in a couple of recent threads, but none of my texts really cover it well.

    Does torsion make parallel transport direction dependent? I.e. if we parallel transport some vector v "forwards" along a curve, and then "backwards" along the very same curve to its starting point, do the non-commuting Christoffel symbols wind up making the vector v different than it was when it started?
     
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  3. Nov 14, 2012 #2

    bcrowell

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    No, I don't think so. Torsion might cause a gyroscope's axis to twist, say, like a right-handed screw as you transported it toward the constellation Sagittarius. Reversing the direction of transport would be like unscrewing the screw from the wood.

    It might be possible to get messed up here by two possible definitions of forward versus backward. You could intend it to mean (a) reversing the direction of the curve in spacetime, or (b) motion from spatial point 1 to spatial point 2 versus motion from 2 to 1. I don't think b is really interesting in general, because it's frame dependent. In case a, I think the answer to the question follows mathematically simply because the connection is a linear operator.
     
    Last edited: Nov 14, 2012
  4. Nov 14, 2012 #3

    PAllen

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    I think this can happen without torsion. If you parallel transport a 4-velocity along a null r-t path in a spherically symmetric field toward the central object, it 'rotates' toward the central object. If you then transport it away from the central object, it also rotates toward the central object. At least this is what I came up with in some notes of when I did this calculation long ago. I have not re-checked it.

    [Edit: even if the above is true, it may not be relevant to the question: the two paths are different. They reverse the radial change, but t moves forward for both. Thus, they are actually completely different paths (path of light from large r to small r; path of light from small r to large r. In spacetime, they are just different paths.]
     
    Last edited: Nov 14, 2012
  5. Nov 14, 2012 #4
    From what I understand, it is experimentally difficult to differentiate between GR and Torsional theories. If what you are suggesting were true, this would be a method for testing GR.
     
  6. Nov 14, 2012 #5

    bcrowell

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    My verbal description probably wasn't precise enough.

    One could say that the only possible result of any parallel transport, even in standard differential geometry with a torsion-free connection, is a rotation. This is true in the sense that parallel transport around an infinitesimal closed loop can never scale a vector. If you can't scale it, then the only other possible action is to "rotate" it in some sense.

    There are a couple of ways to see that there is a clear distinction between this type of "rotation" and the type you get with torsion. One way to characterize torsion is that it makes parallelograms fail to close, and this failure to close is proportional to L, where L is the length of the sides of the parallelogram. Another way to see the distinction is that torsion gives a handedness to the spacetime, whereas curvature doesn't. This is why people trying to make empirically testable models of torsion often make ones in which torsion field arises from the intrinsic spin of particles.

    The notion you're describing here is the one I was trying to describe in #2 as "b."
     
  7. Nov 15, 2012 #6
    Sorry for butting in on a really interesting conversation. I'm not sure this is quite true. One way of measuring curvature is to make a parallelogram out of geodesics. Their failure to close is a measure of geodesic deviation and can be used to compute curvature. See "Road to Reality", Penrose, Page 305 or so.

    Cheers
     
  8. Nov 15, 2012 #7

    George Jones

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    Actually, Penrose gives a rather nice description of the failure of the parallelogram to close in terms of both torsion and curvature. Another nice, but more technical desciption, is given in section 15.8 of "Differential Geometry and Lie Groups for Physicists" by Fecko.
     
  9. Nov 15, 2012 #8
    Following up on my own post I see that there is more to it than that. In the failure to close the parallelogram the curvature shows up in the third order parameter, while torsion shows in the second.
     
  10. Nov 15, 2012 #9

    pervect

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    That helps a lot. So if we consider Schild's ladder, it basically requires parallelograms to close (third order is good enough, second order isn't), which means that Schild's ladder defines torsion free notion of parallel transport.
     
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