# I Perturbative expansion of the metric and its inverse

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1. Jun 29, 2017

### Frank Castle

As I understand it, in the context of cosmological perturbation theory, one expands the metric tensor around a background metric (in this case Minkowski spacetime) as $$g_{\mu\nu}=\eta_{\mu\nu}+\kappa h_{\mu\nu}$$ where $h_{\mu\nu}$ is a metric tensor and $\kappa <<1$.

My question is, how does one determine the inverse metric $g^{\mu\nu}$? I've read some notes that state the result: $$g^{\mu\nu}=\eta^{\mu\nu}-\kappa h^{\mu\nu}+\kappa^{2}h^{\mu}_{\,\lambda}h^{\lambda\nu}+\mathcal{O}(\kappa^{3})$$
I know how to get the expression to first order by writing $g^{\mu\nu}=\eta^{\mu\nu}+\delta g^{\mu\nu}$ and using that $$\delta g^{\mu\nu}=g^{\mu\lambda}\delta g_{\lambda\sigma}g^{\sigma\nu}=-g^{\mu\lambda} h_{\lambda\sigma}g^{\sigma\nu}=-\eta^{\mu\lambda}\eta^{\sigma\nu}h_{\lambda\sigma}+\mathcal{O}(\kappa^{2})$$ I am unsure how to include higher order contributions. Furthermore, how can one justify raising and lowering the indices of $h_{\mu\nu}$ with $\eta_{\mu\nu}$ if one includes such higher order terms?

2. Jun 30, 2017

### Frank Castle

Sorry, meant to put symmetric tensor here.

Also, it should be $\delta g^{\mu\nu}=-g^{\mu\lambda}\delta g_{\lambda\sigma}g^{\sigma\nu}$.

Apologies for the errors.