# Null geodesics of the FRW metric

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?

## Answers and Replies

PeterDonis
Mentor
2020 Award
Do we not strictly speaking also have to check that the above equation actually specifies a geodesic?

Yes.

Is there a quick way to verify that the above curve indeed is a null-geodesic?

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.

PAllen
Science Advisor
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).