# Why does light travel radially in the FRW universe?

1. Oct 6, 2014

### center o bass

When deriving different properties from the FRW-metric $$ds^2 = -dt^2 + a^2 (d\chi^2 + S_k^2(\chi) d\Omega^2)$$ -- considering the propagation of light such that $ds^2 = 0$ -- one always assumes $d\Omega = 0$. But how do we know that $d\Omega$ always vanish for propagating light?

2. Oct 6, 2014

### center o bass

This is obvious for k=0; but why is it also true for the other geometries?

3. Oct 6, 2014

### Chalnoth

In general, light rays will have non-zero angular motion. But as we are orienting the coordinate system such that we are at the center, the only light rays we see are the ones that are pointed directly towards the center. If they had non-zero $d\Omega$, then they wouldn't be pointed directly at the center and we wouldn't see them.

Curvature doesn't change this argument at all.

4. Oct 7, 2014

### Chronos

Sounds like somebody is trying to backdoor a personal theory here.

5. Oct 8, 2014

### center o bass

I do not see why the argument that "the only light rays we see are the ones that are pointed directly towards the center" remains true when curvature is added to the picture. This is because curvature bends light; which is indeed the reason why we can see stars that are really behind the sun (the first successful test of GR). The light that travel from these stars certainly must have dΩ≠0 for otherwise they would be absorbed by it (the sun).

6. Oct 9, 2014

### Orodruin

Staff Emeritus
What do you think the $\chi$ coordinate describes?

You could very well have light rays travelling along geodesics with $d\Omega\neq 0$ just as you can have straight lines in $\mathbb R^2$ that do not cross the origin. However, given a straight line you can always choose your coordinate system such that it does cross the origin.