# Explanation for Robertson-Walker being Conformally Flat

1. Oct 8, 2012

### Airsteve0

I understand that in order for a space to be conformally flat that the vanishing of the Weyl tensor is a necessary condition. I was wondering if the following arguments are also (if not the same) in arguing that Robertson-Walker spacetime is conformally flat.

Starting with the Robertson-Walker metric:

$ds^{2}$=$a^{2}(t)\left(\frac{dr^{2}}{1-Kr^{2}}+r^{2}(d\theta^{2}+sin(\theta)^{2}d\phi^{2})\right)-dt^{2}$

Could it be argued that there exists coordinates $x^{\gamma}$ such that:

$ds^{2}$=$f(x^{\gamma})\eta_{\alpha\beta}dx^{\alpha}dx^{β}$

where $\eta_{\alpha\beta}$ is flat. As such, the 4 dimensional space would possess signature of (+ or -) 2 and be describable as locally Minkowskian.

2. Oct 10, 2012