# Inversion of a metric

I am currently reading a paper by Bredberg $$et.al$$ arXiv:1101.2451 titled "From Navier-Stokes to Einstein". In this paper, the authors have considered a metric of the form $$\begin{eqnarray}ds^2_{p+2} = -r d\tau^2+2 d\tau dr +dx^idx_i\\-2(1-\frac{r}{r_c})v_idx^id\tau-2\frac{v_i}{r_c}dx^idr\\+(1-\frac{r}{r_c})\Big[(v^2+2p)d\tau^2+\frac{v_iv_j}{r_c}dx^idx^j\Big]+\big(\frac{v^2}{r_c}+\frac{2P}{r_c}\Big)d\tau dr\\ +\mathcal{O(\epsilon^3)} \end{eqnarray}$$

The first line conains terms of the order $$\mathcal{O(\epsilon^0)}$$, second line of the order $$\mathcal{O}(\epsilon)$$ and the third line of the order $$\mathcal{O}(\epsilon^2)$$. Here $$v_i \sim \mathcal{O}(\epsilon)$$, $$P \sim \mathcal{O}(\epsilon^2)$$. The authors have considered timelike hypersurfaces $$r = r_c$$. I am trying to find out the unit normal to these hypersurfaces $$n^a$$. The Unit normal to these hypersurfaces are given by $$n^{\mu}\partial_{\mu} = \frac{1}{\sqrt{r_c}}\partial_{\tau}+\sqrt{r_c}(1-\frac{P}{r_c})\partial_{r}+\frac{v^i}{\sqrt{r_{c}}}\partial_i+\mathcal{O}(\epsilon^3)$$. To achieve this form of the contravariant unit normal $$n^a$$, i need to invert the metric and find $$g^{\mu \nu}$$ upto order $$\mathcal{O(\epsilon^2)}.$$ I am currently stuck in this. Any help regarding the inversion of this metric either via hand or by CAS would be highly regarded. PS: Notation:- the spacetime dimension is $$p+2$$ with $$\tau$$ being the time coordinate ,$$r$$ being the radial coordinate and $$i = 1,2,....p$$ being the angular coordinates.

$$g_{\mu \nu} = \bar{g}_{\mu \nu} + h_{\mu \nu}$$
so that every other quantity is expressed as an infinite series in $$h_{\mu \nu}$$. Then, the inverse metric up to $$\mathcal{O}(h^2)$$ is given by
$$g^{\mu \nu } = \bar{g}^{\mu \nu } - h^{\mu \nu } + h_{\alpha }{}^{\nu } h^{\mu \alpha }$$
which you can easily compute once you know $$\bar{g}_{\mu \nu }$$ and $$h_{\mu \nu }$$. Note that all contractions, traces, etc. are taken with respect to the background metric $$\bar{g}_{\mu \nu}$$.