# FRW metric spacelike slice

In Steven Winberg's Cosmology book (p. 6) he says that the spatial part (co-moving part) of the FRW metric can be written as $$\tilde g_{ij} = \delta_{ij}+K \frac{x^i x^j}{1- K \mathbf x ^2}$$ These qoordinates are "quasi-cartesian". I can get back $$\tilde g _{11} = \frac{1}{1-K \mathbf x^2}$$ by using $$x^1=r$$. But, for the other coordinates I can't find the right components using this relations. If I use $$x^2=\theta$$ then I can't get anything. Any help will be highly appreciated.

## 1 Answer

Your equation gives the components of the spatial metric in cartesian coordinates: i.e. $$x^1 \equiv x$$, $$x^2 \equiv y$$, and $$x^3 \equiv z$$. You cannot use $$x^1=r$$.

If you want to get the metric in a different coordinate system (like spherical coordinates) you just need to be able to write $$x$$, $$y$$, and $$z$$ in terms of the new coordinates (i.e. $$r$$, $$\theta$$, and $$\phi$$) and use the normal way tensor components transform:

$$g'_{ij} = g_{kl} \frac{\partial x^k}{\partial x'^i}\frac{\partial x^l}{\partial x'^j},$$ where the primed coordinates are your new coordinates and there as an implicit sum over indices $$k$$ and $$l$$.

In your case it's probably easier (but totally equivalent) to write out the proper length $$ds^2$$ in terms of $$dx$$, $$dy$$, and $$dz$$ and then write the $$dx$$, $$dy$$, and $$dz$$ in terms of $$dr$$, $$d\theta$$, and $$d\phi$$. Then you'll end up with $$ds^2$$ in terms of $$dr$$, $$d\theta$$, and $$d\phi$$ and can read off $$\tilde{g}_{rr}$$, $$\tilde{g}_{\theta\theta}$$, $$\tilde{g}_{\phi\phi}$$, $$\tilde{g}_{r\theta}$$, etc.