# Isotric an homogenous implied form of FRW metric

1. Feb 20, 2015

### binbagsss

So in deriving the metric, the space-time can be foliated by homogenous and isotropic spacelike slices.

And the metric must take the form:

$ds^{2}=-dt^{2}+a^{2}(t)\gamma_{ij}(u)du^{i}du^{j}$,

where $\gamma_{ij}$ is the metric of a spacelike slice at a constant t

QUESTION:
So I've read that:
1) Homogenity would be broken if the a(t) was taken outside the metric
2) By isotropicity there can be no cross-terms dtdx, dtdy, dtdz.

What I know:
homogenous means the same throughout - translationally invariant.
isotropic means the same in every direction - rotationally invariant.

But I'm struggling to see how 1) and 2) follow from this. As stupid as it sounds, I dont really see where time comes in when these properties are only on the spacelike slices.

Cheers.

2. Feb 20, 2015

### Staff: Mentor

No; "must" is incorrect. The correct statement is "one can always choose coordinates in which the metric takes the form..." See further comments below.

Reference please? A general note: statements like "I've read" or "I read somewhere" are red flags that you should be giving a specific reference, not a general vague statement. We can't tell whether what you've read is reliable if we can't read it ourselves.

What does "taken outside the metric" mean? Does it mean it multiplies the entire RHS, instead of just the spatial part? I'm going to assume it does in the rest of this post.

More precisely, the fact that the spacetime is isotropic means we can always choose coordinates in which there are no cross terms. See further comments below.

Yes.

Yes.

Is it both 1) and 2) that you don't understand, or just 1)? Your next statement seems to relate to 1), not 2).

The spacelike slices depend on the coordinates you choose. Statements 1) and 2) are really statements about how you can choose coordinates. I pointed that out above in my response to 2), about what "isotropic" means, but it goes for "homogeneous" too. "Homogeneous" does not mean you can't have $a(t)$ multiply the time part of the metric as well as the spatial part; it means that in a homogeneous spacetime, you can choose coordinates such that the time part of the metric is just $- dt^2$, a constant, independent of the coordinates. If you choose other coordinates for a homogeneous spacetime, then the time part of the metric might have a function $a(t)$ multiplying it (or a function of any of the coordinates). That doesn't mean the spacetime isn't homogeneous; it means you chose coordinates differently.