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A little geodesic help

  1. May 1, 2008 #1
    I understand what geodesics are and how to calculate them from Christoffel symbols and all that. But I've just come across a question I have no idea about. I've been given the dust filled Friedmann solution:

    ds^2 = -dt^2 + a(t)^2 (dX^2 + X^2 dO^2) (O=omega)

    And been told to show that radial geodesics obey:

    a^2 (dX/dT) = k where k constant and T is proper time.

    And hence (dt/dT)^2 = 1 + k/a^2

    How do I introduce proper time? And where to I get an expression for dX/dT from? I'm completely stumped...and help would be much appreciated!
     
  2. jcsd
  3. May 1, 2008 #2

    George Jones

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    On a timelike worldline, dT^2 = -dt^2.

    What does "radial" give you?
     
  4. May 1, 2008 #3
    It should mean that dO^2 = 0 so that there's no angular component to the worldline. So this gives:

    ds^2 = -dT^2 + a^2 dX^2

    I've also realised that for the second part I can use g(ab)t(a)t(b) = -1 for timelike geodesics, which means that (dt/dT)^2 = 1 + a^2(dX/dT)^2 which gives the second part assuming that (dX/dT) = k/a^2, but I still can't think how to show this first part...
     
  5. May 1, 2008 #4

    George Jones

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    I think you meant

    ds^2 = -dt^2 + a^2 dX^2.

    Using my substitution gives,

    -dT^2 = - dt^2 + a^2 dX^2,

    which is the same as

    Even though it's for a different situation, this post might help.
     
  6. May 1, 2008 #5
    Wonderful. I managed to obtain the required relation by extremising the action under X. Thanks so much for the fantastic help you give!

    Just one more quick question about that link in your last post: can you always equate the conserved quantity relating to (dt/dT) to E/m or is that only for the Schwarzschild metric?
     
  7. May 1, 2008 #6

    George Jones

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    For large r, Scwharzschild spacetime is like Minkowski spacetime, and in Minkowski spacetime, dt/dT = E/m. This led to the choice of label. I think it is common to use this label in any spacetime for which this is true. Not all spacetimes, however, have this property.
     
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