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

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 ($\phi$), in other words that θ and $\phi$ 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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 Recognitions: Gold Member Homework Help Science Advisor 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

 Quote by fzero this also follows from the geodesic equation, since $$\Gamma^\theta_{rr}=\ \Gamma^\theta_{rt}=\ \Gamma^\theta_{tt}=0$$ and similarly for ##\phi##.
Geodesic Equation:
$\frac{d^2x^\theta}{d\lambda^2}\ +\ \Gamma^\theta_{\alpha\beta}\ \frac{dx^\alpha}{d\lambda} \frac{dx^\beta}{d\lambda}$

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 $\Gamma^\theta_{rr}\text{, }\Gamma^\theta_{rt}\text{, }\Gamma^\theta_{tt}$ 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 $\Gamma^\theta_{\theta r}$.

At first I thought we could disregard terms like that because $\frac{dx^\theta}{d\lambda}(0)=\frac{d^2x^\theta}{d\lambda^2}(0)=0$. 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 $y=x^3\text{ at }x=0$.

What arguments can be employed together with the above observation about the Christoffel symbols to reach a conclusion that $\frac{dx^\theta}{d\lambda}=0$ everywhere along the radial geodesic? Perhaps using some additional info from the metric?

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

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 $\Gamma^\theta_{tt}= \Gamma^\theta_{rt} = \Gamma^\theta_{rr}=0$, but also had to use some nonzero values of other Christoffel symbols, which turned out to lead to key cancellations.
Attached Files
 radial geodesic.pdf (59.7 KB, 3 views)