# FLRW metric for flat universe

To derive the FLRW metric, we first make the assumption of isotropy and homogeneity of the universe. This enables us to write our general metric tensor in the following form:

$$ds^2 = c^2g_{00}(t)dt^2-dl^2$$

where $dl$ is the line element for a spacial surface. I am interested here in the case of a flat universe. As far as I understand, this means $dl^2 = dx^2+dy^2+dz^2$. Defining $t'$ such that $dt' = \sqrt{g_{00}}dt$, we get

$$ds^2 = c^2dt'^2-(dx^2+dy^2+dz^2)$$

which is clearly wrong since it is Minkowskian, and according to the internet the solution should be $ds^2 = c^2dt^2 - a^2(t)(dx^2+dy^2+dz^2)$ for the FLRW metric in the flat universe case. I have clearly a misunderstanding of either $dl^2$ or the validity of my change of variables; I suspect the latter. Where did I go wrong?

Everywhere I look for a detailed derivation of the FLRW metric, they do it for the case of a spherical universe, and then take some kind of limit for the radius $R$ which does not help me very much.

• Why don't you want to work in spherical polar coordinates? When you say $dl^2$ is the line element for spatial surface, does that mean for Euclidean space? – Rumplestillskin Jan 4 '17 at 1:28
• The most general homogeneous isotropic metric includes a factor of $a(t)^2$ in front of $dl^2$. – Javier Jan 4 '17 at 2:05
• It's also called a scale factor. FYI. – Rumplestillskin Jan 4 '17 at 2:24
• Homogeneous mean spatial sections look the same at any x, y, z. It does not say that they are the same at all times. You need the a(t). Your change of variables would otherwise be perfectly fine. Another way: the metric can have no x, y or z dependence (well, for flat anyway), but there is no time symmetry implied. – Bob Bee Jan 4 '17 at 2:57