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We can write a the covariant form of a perturbed Minkowski background to second order as

$$ g_{\mu \nu} = \eta_{\mu \nu} + \kappa h^{(1)}_{\mu \nu} + \kappa^2h^{(2)}_{\mu \nu}$$

where $\kappa$ is just used to track the order of terms.

Now, I understand that the contravariant form at second order is,

$$ g^{\mu \nu} = \eta^{\mu \nu} - \kappa h_{(1)}^{\mu \nu} + \kappa^2(2 {{h^{(1)}}_{\alpha}}^{\nu} h_{(1)}^{\mu \alpha}-h_{(2)}^{\mu \nu})$$

My question is if I have some algebraic expressions for the covariant terms, how do I compute the contravariant terms? e.g. how would I obtain $h_{(2)}^{\mu \nu}$ given $h^{(1)}_{\mu \nu}, h^{(2)}_{\mu \nu}$? I understand that at first order we can simply lower/raise indices via the unperturbed background metric, but surely this does not apply as we go to higher orders?

My question is somewhat related to the questions here and here, but I believe it is slightly different, since I am not concerned with deriving an tensor expression for $g^{\mu \nu}$ but instead with how to explicitly determine the algebraic value of the terms

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  • $\begingroup$ Your equations seem to have some issues. (Repeated + sign in the first, indices of the second order term of the second equation are inconsistent) Could you fix them? $\endgroup$
    – TimRias
    Sep 2, 2020 at 7:59
  • $\begingroup$ Apologies, thanks for the catch. Fixed. $\endgroup$ Sep 2, 2020 at 8:02

2 Answers 2

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The first and second order perturbations are defined to be tensors on the background spacetime. Therefore you can raise and lower the indices with the background metric. This is why the coefficients of the inverse metric take a funny form.

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    $\begingroup$ So $h_{(2)}^{\mu \nu} = \eta^{\mu \alpha} \eta^{\nu \beta} h_{\alpha \beta}$? $\endgroup$ Sep 2, 2020 at 8:04
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    $\begingroup$ If the second h also has a (2), then yes. $\endgroup$
    – TimRias
    Sep 2, 2020 at 8:12
  • $\begingroup$ My typing is terrible today! Thanks for your help $\endgroup$ Sep 2, 2020 at 8:17
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The defining property of the inverse metric is $$g_{\mu\nu}g^{\nu\alpha} = \delta_{\mu}^{\;\,\alpha}\tag{1}\label{eq:inverse}$$ We want to make an expansion $$g_{\mu\nu} = \eta_{\mu\nu} + \kappa h_{\mu\nu}^{(1)} + \kappa^2 h_{\mu\nu}^{(2)}\tag{2}\label{eq:linearExp}$$ Now you can plug eq.\eqref{eq:linearExp} into the defining property \eqref{eq:inverse}, to obtain $$\left( \eta_{\mu\nu} + \kappa h_{\mu\nu}^{(1)} + k^2 h_{\mu\nu}^{(2)}\right) g^{\nu\sigma} = \delta_{\mu}^{\;\,\sigma}$$ You can contract with the flat metric $\eta^{\alpha\mu}$ and get: $$\begin{align} \left( \delta^{\alpha}_{\;\,\nu} + \kappa {h^{\alpha}_{\;\,\nu}}^{(1)} + \kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)}\right) g^{\nu\sigma} &= \eta^{\alpha\sigma}\\ g^{\alpha\sigma} + \left(\kappa {h^{\alpha}_{\;\,\nu}}^{(1)} + \kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)}\right) g^{\nu\sigma} &= \eta^{\alpha\sigma} \end{align}$$ where we have defined the perturbations with upper indices to be corresponding contractions with the flat metric. We can extract from the last line the expression: $$g^{\alpha\sigma} = \eta^{\alpha\sigma} - \kappa {h^{\alpha}_{\;\,\nu}}^{(1)}g^{\nu\sigma} - \kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)}g^{\nu\sigma}\tag{3}\label{eq:selfConsistent}$$ One can then insert this equation in itself (in the right hand side) to achieve an expansion in $\kappa$ self-consistently. Let us expand up to order $\kappa^2$ like this: $$\begin{align} g^{\alpha\beta} &= \eta^{\alpha\beta} - \kappa {h^{\alpha}_{\;\,\nu}}^{(1)}g^{\nu\beta} - \kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)}g^{\nu\beta}\\[6pt] &=\eta^{\alpha\beta} -\kappa \color{blue}{(\eta^{\nu\beta} - \kappa {h^{\nu}_{\;\,\lambda}}^{(1)}g^{\lambda\beta} - \kappa^2 {h^{\nu}_{\;\,\lambda}}^{(2)}g^{\lambda\beta})} {h^{\alpha}_{\;\,\nu}}^{(1)} \\[6pt] &\quad -\kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)} \color{blue}{(\eta^{\nu\beta} - \kappa {h^{\nu}_{\;\,\lambda}}^{(1)}g^{\lambda\beta} - \kappa^2 {h^{\nu}_{\;\,\lambda}}^{(2)}g^{\lambda\beta})} \end{align} $$ where the blue terms are insertions of \eqref{eq:selfConsistent}. We can in principle repeat the process up to an arbitrary order, but we are interested on getting the expression up to order $\kappa^2$ $$\begin{align} g^{\alpha\beta} &= \eta^{\alpha\beta} -\kappa \eta^{\nu\beta} {h^{\alpha}_{\;\,\nu}}^{(1)} + \kappa^2 {h^{\nu}_{\;\,\lambda}}^{(1)}g^{\lambda\beta}{h^{\alpha}_{\;\,\nu}}^{(1)} -\kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)} \eta^{\nu\beta} + \mathcal{O}(\kappa^3)\\[6pt] &= \eta^{\alpha\beta} -\kappa {h^{\alpha\beta}}^{(1)} + \kappa^2 {h^{\nu}_{\;\,\lambda}}^{(1)}{h^{\alpha}_{\;\,\nu}}^{(1)}\color{blue}{(\eta^{\lambda\beta} - \kappa {h^{\lambda}_{\;\,\mu}}^{(1)}g^{\mu\beta} - \kappa^2 {h^{\lambda}_{\;\,\mu}}^{(2)}g^{\mu\beta})} -\kappa^2 {h^{\alpha\beta}}^{(2)} + \mathcal{O}(\kappa^3)\\[6pt] &= \eta^{\alpha\beta} -\kappa {h^{\alpha\beta}}^{(1)} + \kappa^2 {h^{\nu}_{\;\,\lambda}}^{(1)}{h^{\alpha}_{\;\,\nu}}^{(1)}\eta^{\lambda\beta} -\kappa^2 {h^{\alpha\beta}}^{(2)} + \mathcal{O}(\kappa^3)\\[6pt] &= \eta^{\alpha\beta} -\kappa {h^{\alpha\beta}}^{(1)} + \kappa^2 {h^{\nu\beta}}^{(1)}{h^{\alpha}_{\;\,\nu}}^{(1)} - \kappa^2 {h^{\alpha\beta}}^{(2)} + \mathcal{O}(\kappa^3)\\[6pt] &= \eta^{\alpha\beta} - \kappa {h^{\alpha\beta}}^{(1)} + \kappa^2\left({h^{\nu\beta}}^{(1)}{h^{\alpha}_{\;\,\nu}}^{(1)} - {h^{\alpha\beta}}^{(2)}\right) + \mathcal{O}(\kappa^3) \end{align} $$

I hope the procedure is clear, however as you can see I don't get the factor 2 in front of the $h^{(1)}h^{(1)}$ term. Let me know if you spot mistakes.

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