- #1

- 560

- 2

$$ds^2 = - dt^2 + a^2 ( d\chi^2 + S_k(\chi) d\Omega^2)$$

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?