Parallel transport to explain motion of light near black hole?

  • #26
PeterDonis
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I don't think either statement implies the other in general.
Doesn't #1 imply #2? #1 says that the change in the vector must be orthogonal to the vector. (More precisely, the argument that establishes #1 establishes this--at least I think it does*.) Any such change can only change the direction of the vector, not its magnitude.

I agree that #2 does not imply #1 for a null vector, so the two statements are not equivalent.

* - The argument that establishes #1 is that ##R_{aacd} = 0##. If we use orthogonal coordinates, this statement is the same as saying that the change in the vector must be orthogonal to the vector. If we use non-orthogonal coordinates, that is not necessarily the case; however, since we can always find orthogonal coordinates, I think the argument still establishes that the change in the vector must be orthogonal to the vector.
 
  • #27
PAllen
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Doesn't #1 imply #2? #1 says that the change in the vector must be orthogonal to the vector. (More precisely, the argument that establishes #1 establishes this--at least I think it does*.) Any such change can only change the direction of the vector, not its magnitude.

I agree that #2 does not imply #1 for a null vector, so the two statements are not equivalent.

* - The argument that establishes #1 is that ##R_{aacd} = 0##. If we use orthogonal coordinates, this statement is the same as saying that the change in the vector must be orthogonal to the vector. If we use non-orthogonal coordinates, that is not necessarily the case; however, since we can always find orthogonal coordinates, I think the argument still establishes that the change in the vector must be orthogonal to the vector.
Well, a null vector is orthogonal to itself ... by definition of norm=0. So where does that leave you?
 
  • #28
PeterDonis
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a null vector is orthogonal to itself
Hm, good point; for a null vector, if the argument for #1 implied that the change had to be orthogonal, but could not be a rescaling, it would be impossible to change a null vector at all by parallel transporting around a closed curve. So that implication is too strong.
 
  • #29
PAllen
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To clarify my doubt on this:

1) We know that parallel transport can never scale a non-null vector at all, so what what happens on closed curve is moot (closed or open curve can change direction but not magnitude).

2) We know that for a null vector parallel transport can rescale it as well change direction (while keeping null norm, always).

I am not yet convinced of any argument (given (2)), that transport around a closed curve cannot result in a re-scaling only.
 
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  • #30
martinbn
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If you parallel transport a vector around a loop you can end up with a vector with a diffrent orientation. Then why can't you end up with a vector at 180 degrees angle with the original i.e. a scalor multiple, where the scalor is -1? It seems quit possible. Start at the north pole facing south, walk along a meridian till you reach the equator, move along the equator for half of the circumference, then go back north to the pole. You end up facing the other way compare to you start.
 
  • #31
bcrowell
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If you parallel transport a vector around a loop you can end up with a vector with a diffrent orientation. Then why can't you end up with a vector at 180 degrees angle with the original i.e. a scalor multiple, where the scalor is -1? It seems quit possible. Start at the north pole facing south, walk along a meridian till you reach the equator, move along the equator for half of the circumference, then go back north to the pole. You end up facing the other way compare to you start.
You're right, that's a clear counterexample. I think the problem is my assertion that the result generalizes from infinitesimal parallelograms to larger paths. Now that I think more carefully about it, it's clearly wrong. If you add a bunch of vectors that are not parallel to a given line, you can still get a sum that's parallel to the line.
 
  • #32
bcrowell
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After posting a bunch of wrong stuff about the rescaling of null vectors, I think I finally understand it better. The antisymmetry of the Riemann tensor on its first two indices is exactly the condition needed so that parallel transport around an infinitesimal closed loop doesn't change inner products. For such a loop lying in a fixed plane, the set of possible actions by Riemann tensors are therefore the six-dimensional set of linear transformations that preserve inner products. In other words, they are the set of possible Lorentz transformations. Such a transformation can certainly take a null vector and rescale it.
 

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