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Null geodesics of the FRW metric

  1. Dec 10, 2014 #1
    When working with light-propagation in the FRW metric
    $$ds^2 = - dt^2 + a^2 ( d\chi^2 + S_k(\chi) d\Omega^2)$$
    most texts just set $$ds^2 = 0$$ and obtain the equation
    $$\frac{d\chi}{dt} = - \frac{1}{a}$$
    for a light-ray moving from the emitter to the observer.

    Question1: Do we not strictly speaking also have to check that the above equation actually specifies a geodesic?

    Setting ##ds^2 = 0## does not automatically guarantee that the obtained relation specifies a geodesic, right?

    Question2: Is there a quick way to verify that the above curve indeed is a null-geodesic?
     
  2. jcsd
  3. Dec 10, 2014 #2

    PeterDonis

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    Staff: Mentor

    Yes.

    I don't know of any quicker way than finding an affine parametrization of the curve and plugging in to the geodesic equation, but someone else might.
     
  4. Dec 10, 2014 #3

    PAllen

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    Well, the given metric displays spherical symmetry. Then, for a radial null path, there is one solution. Then, if it is not a geodesic, what could choose a direction?

    I would thus say, it is an interesting consistency check (which I have done for Kruskal coordinates) to verify satisfaction of the geodesic equation. However, for purposes of doing the least work for a valid conclusion, it is superfluous (in this particular case).
     
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