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Parallel transport and geodecics

  1. Jun 9, 2012 #1
    So from what I understand if you pass a vector (using parallel transport) through a closed curve where there is curvature in the interior, the vector will come back not to it's original vector but with a changed sense. However if the vector is on a geodesic it will not change its sense after it gets parallel transported but isn't this a contradiction (since it's a closed curve with curvature in the interior) or is this just the definition of a geodesic?
     
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  3. Jun 9, 2012 #2
    I don't think it is possible to have a closed curve geodesic, since a geodesic is a world line for a particle in free fall. You can form a closed curve comprised of segments of geodesics, and if you parallel transport around one of these, the vector will change its sense.

    Chet
     
  4. Jun 9, 2012 #3
    I think kidphysics is thinking of the often quoted example of transporting a vector along a great circle around a sphere which is an analogy to a geodesic.
     
  5. Jun 9, 2012 #4

    WannabeNewton

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    A geodesic parallel transports its own tangent vector, [itex]\triangledown _{\mathbf{U}}\mathbf{U} = 0[/itex]; nothing is being said about arbitrary vectors being transported along the curve.
     
  6. Jun 9, 2012 #5

    Dale

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    Yes, but since parallel transport also preserves the dot product I think that you could probably generalize it to arbitrary vectors.
     
  7. Jun 9, 2012 #6

    WannabeNewton

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    Yes indeed it does but generalize what sorry?
     
  8. Jun 9, 2012 #7
    yes this is what I was referencing sorry for not including this. So this was during a susskind lecture and he was saying that if you parallel transport a vector along the geodesic then it returns full circle and is not displaced by any angle. I assume this is just a property of geodesics they are the only closed paths which has curvature in the interior such that when a vector is parallel transported along that closed path it is invariant to changing orientation. I was just looking for an affirmative.
     
  9. Jun 9, 2012 #8

    Dale

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    If a closed geodesic parallel transports its own tangent vector then it must parallel transport all other vectors also since it preserves the dot product also, right?
     
  10. Jun 9, 2012 #9

    Dale

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    It is true for geodesics on a sphere, but I am not sure that it is true in other manifolds.
     
  11. Jun 9, 2012 #10

    A.T.

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    I don't think so. Try a closed geodesic path on a cone that encloses the apex.
     
  12. Jun 9, 2012 #11

    Dale

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    Is there such a geodesic on a cone? I didn't think that a cone had any closed geodesics, but I must admit that I am having trouble visualizing it for sure. I am definitely less confident about this than most of my posts.
     
  13. Jun 9, 2012 #12

    WannabeNewton

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    Well the geodesic will maintain [itex]\triangledown _{\mathbf{U}}g(U,U) = 0[/itex], with [itex]\mathbf{U}[/itex] being the tangent to the geodesic, but I'm not seeing why this or the definition of the geodesic would imply [itex]\triangledown _{\mathbf{U}}\mathbf{V} = 0[/itex] for some arbitrary vector [itex]\mathbf{V}[/itex]. I'm probably missing something so I apologize in advance.
     
  14. Jun 9, 2012 #13

    Dale

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    No need to apologize, I am not confident that I am right either.
     
  15. Jun 9, 2012 #14

    PeterDonis

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    A *timelike* geodesic is, yes. And a null geodesic is a world line for a light ray in "free fall" (i.e., no waveguides or other stuff present). But there are also spacelike geodesics. Those can be closed if the manifold is curved. For example, yuiop gave the example of a great circle on a sphere.

    (There are spacetimes, such as the Godel universe, where there are closed timelike curves; but I'm not sure whether those curves can be geodesics. I think they can, but I'm not positive.)

    The apex is a singularity of the manifold; I believe there is a theorem that says parallel transport around a closed geodesic brings all vectors back to the same vectors, but only if the closed geodesic doesn't enclose a singularity.
     
  16. Jun 9, 2012 #15

    WannabeNewton

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    Sorry, I was thinking of segments of geodesics being connected together to form a closed loop. But yeah for a single closed geodesic I agree with what you said before.
     
  17. Jun 9, 2012 #16

    A.T.

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    If the opening angle is less than 60° it has closed geodesics around the apex.
    Replace the pointy tip with a small spherical dome. Or simply look at the geodesics on an ellipsoid of revolution.
     
  18. Jun 9, 2012 #17

    Dale

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    Cool, I didn't know that.
     
  19. Jun 9, 2012 #18

    A.T.

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    Me neither. I figured it out after you questioned it. So better check it. It seems that the direction change after a loop around the apex is:

    2*pi*sin(opening_angle / 2)
     
  20. Jun 9, 2012 #19

    Bill_K

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    The surface of a cone is metrically equivalent to flat space. A cone is flat space with a sector removed and the two "edges" glued together. The geodesics on it are straight lines. It will be possible to have a geodesic that hits both "edges" if and only if the sector you removed is greater than 180o, and if so, it will go all the way around.
     
  21. Jun 9, 2012 #20

    A.T.

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    That was my idea too. And if you remove more than 270° it will go all the way around twice, right?
     
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