Null geodesics of the FRW metric

  • #1

Main Question or Discussion Point

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

  • #2
PeterDonis
Mentor
Insights Author
2019 Award
28,301
8,051
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.
 
  • #3
PAllen
Science Advisor
2019 Award
7,967
1,260
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).
 

Related Threads on Null geodesics of the FRW metric

  • Last Post
2
Replies
26
Views
5K
Replies
14
Views
3K
  • Last Post
Replies
3
Views
945
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
897
  • Last Post
Replies
7
Views
973
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
14
Views
1K
Top