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Basic thing with general relativity

  1. May 23, 2008 #1
    (This could be a dumb question, but in any case I'll be happy if the answer is simple)

    I understood that the path of a freely moving object in space satisfies the geodesic equation, but I'm not fully sure how precisely to deal with the four momentum of the object. Have I understood this correctly, if I think that the four momentum is a tangent vector to the path, and that the time evolution of the four momentum is obtained by parallel transport along the path?
  2. jcsd
  3. May 23, 2008 #2


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    For the 4-momentum to be constant requires ( D is covariant differentation )

    [tex]\frac{Dp^{\mu}}{D\tau} = m\frac{D^2x^\mu}{D\tau^2} = 0[/tex]

    which is the geodesic equation.

    I think your conclusion is correct.
  4. May 23, 2008 #3

    George Jones

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    when the path is parameterized by proper time divided by rest mass. If the path is parameterized by proper time, then the tangent vector is the 4-velocity of the object. If some other parameter is used (e.g., the square of the proper time for positive proper time), then the tangent vector to the path is not necessarily easily physically identified.

    Yes, subject to the above.
  5. May 24, 2008 #4
    I see. With massless objects the path cannot be parametrized by proper time, so there should be something else. Suppose I have some arbitrary parametrization of a path of some object whose mass could be zero or non-zero. If I somehow know correctly the four momentum at some point, will the parallel transport always give the four momentum correctly elsewhere along the path too?
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