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Lightlike radial null geodesic - how do we know it has constant theta and phi?

  1. Jul 28, 2012 #1

    andrewkirk

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    Consider a light ray emanating from the origin of a FLRW coordinate system in a homogeneous, isotropic universe. The initial velocity of that ray will have only x0 (t) and x1 (r) components. In papers I have seen it is assumed that its velocity will continue to have zero circumferential components: x2 (θ) and x3 ([itex]\phi[/itex]), in other words that θ and [itex]\phi[/itex] are constant.

    A loose argument for this is that, for the geodesic to develop any circumferential components would identify a preferred direction in space, thereby contradicting the isotropy assumption. I find this unconvincing, as the 'direction' is a coordinate-dependent artifact, and hence does not necessarily have any physical significance. The isotropy assumption is a coordinate-independent statement about the nature of the spacetime, not about a particular coordinate system (well, perhaps it does contain information about the time coordinate, as it seems to state that the constant-time hypersurfaces are isotropic, but those hypersurfaces can be parameterised in an infinity of different ways, so there's nothing significant about a particular spherical choice of coordinates as in the FLRW system.).

    Is there a more rigorous argument as to why the null geodesic cannot have any circumferential components?
     
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  3. Jul 28, 2012 #2

    fzero

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    The isotropy argument is perfectly physical, but this also follows from the geodesic equation, since

    $$\Gamma^\theta_{rr}=\Gamma^\theta_{rt}=\Gamma^\theta_{tt}=0$$

    and similarly for ##\phi##. I think your concerns about other coordinate systems is just explained by noting that the spherical symmetry would look different in arbitrary coordinates
     
  4. Aug 9, 2012 #3

    andrewkirk

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    Geodesic Equation:
    [itex]\frac{d^2x^\theta}{d\lambda^2}\ +\ \Gamma^\theta_{\alpha\beta}\ \frac{dx^\alpha}{d\lambda} \frac{dx^\beta}{d\lambda}[/itex]

    Thank you for your reply fzero. I can see that those zero values of the Christoffel symbols are correct and at first I thought that was enough to guarantee the result, via the geodesic equation, but on reflection, I think that alone is not sufficient to give the result.

    Although the zero values of [itex]\Gamma^\theta_{rr}\text{, }\Gamma^\theta_{rt}\text{, }\Gamma^\theta_{tt}[/itex] ensure that those terms of the geodesic equation are zero everywhere, there are plenty of other Christoffel symbols in that equation that are nonzero, such as [itex]\Gamma^\theta_{\theta r}[/itex].

    At first I thought we could disregard terms like that because [itex]\frac{dx^\theta}{d\lambda}(0)=\frac{d^2x^\theta}{d\lambda^2}(0)=0[/itex]. I now see that that is insufficient argument. There are plenty of functions for which the first and second derivatives at a point are zero but which subsequently become nonzero, for example [itex]y=x^3\text{ at }x=0[/itex].

    What arguments can be employed together with the above observation about the Christoffel symbols to reach a conclusion that [itex]\frac{dx^\theta}{d\lambda}=0[/itex] everywhere along the radial geodesic? Perhaps using some additional info from the metric?
     
    Last edited: Aug 9, 2012
  5. Aug 11, 2012 #4

    andrewkirk

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    I have managed to complete the proof. The attached file is a TeX formatted version of the proof, together with a corollary that a vector with no circumferential components that is parallel transported along the radial curve does not gain any circumferential components.

    The proof relied heavily on the observations by fzero that [itex]\Gamma^\theta_{tt}= \Gamma^\theta_{rt} = \Gamma^\theta_{rr}=0[/itex], but also had to use some nonzero values of other Christoffel symbols, which turned out to lead to key cancellations.
     

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