# Metric tensor

1. Feb 11, 2013

Am i correct in thinking that in a simple $ds^2 =dt^2 - a^{2}d\underline{x}^{2}$ metric that:
$g_{\mu\nu}=diag(1,-a^2,-a^2,-a^2)$ and $g^{\mu\nu}=diag(1,-\frac{1}{a^2},-\frac{1}{a^2},-\frac{1}{a^2})$

thx

2. Feb 11, 2013

### Fredrik

Staff Emeritus
Yes.