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I Perturbative expansion of the metric and its inverse

  1. Jun 29, 2017 #1
    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. jcsd
  3. Jun 30, 2017 #2
    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.
     
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