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From what I've read, in the framework of linearised gravity, one perturbs the metric around a Minkowski background, $\eta_{\mu\nu}$, such that $$g_{\mu\nu}(x)=\eta_{\mu\nu}+h_{\mu\nu}(x)\tag{1}$$ where $h_{\mu\nu}(x)$ is a small perturbation, i.e. $$\big\lvert h_{\mu\nu}\big\rvert<<1.\tag{2}$$

The inverse metric is then found by assuming the following ansatz: $$g^{\mu\nu}(x)=\eta^{\mu\nu}+\tilde{h}^{\mu\nu}\tag{3}$$ where $\tilde{h}^{\mu\nu}$ is also small (i.e. $\big\lvert \tilde{h}_{\mu\nu}\big\rvert<<1$).

Using this, it is easy to find that $$g^{\mu\nu}(x)=\eta^{\mu\nu}-h^{\mu\nu} \tag{4}$$ to first order.

My question is, what is the justification for this ansatz? Is it simply that one expects the inverse metric to have a similar form to the metric in order to satisfy $$g^{\mu\alpha}g_{\alpha\nu}=\delta^{\mu}_{\;\nu}~?\tag{5}$$

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Hint: Ansatz (3) is not necessary. To derive eq. (4) from eqs. (1), (2) & (5), use instead that for an infinitesimal variation $$ gg^{-1}~=~{\bf 1}\qquad\Rightarrow\qquad \delta (g^{-1})~=~- g^{-1}(\delta g)g^{-1},\tag{A}$$ where $g$ is an arbitrary invertible matrix (with lower indices), and $g^{-1}$ is the inverse matrix (with upper indices). Can you see how eq. (A) is derived?

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    $\begingroup$ Ah ok. Is eq.(A) found by noting that $\delta(g^{-1}g)=\delta(g^{-1})g+g^{-1}\delta g=0$, and so $\delta(g^{-1})=-g^{-1}(\delta g)g^{-1}$?! So given this, does one simply use that an infinitesimal perturbation of the inverse metric around a Minkowski background should give: $$g^{\mu\nu}=\overline{g}^{\mu\nu}+\delta g^{\mu\nu}=\eta^{\mu\nu}+\delta g^{\mu\nu}$$ where $\overline{g}^{\mu\nu}=\eta^{\mu\nu}$ is the background metric one is perturbing around... $\endgroup$
    – user35305
    Commented Nov 26, 2016 at 19:14
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    $\begingroup$ ... and since $\delta g_{\mu\nu}= h_{\mu\nu}$, we have from eq.(A) that $$\delta g^{\mu\nu}=-\left(\eta^{\mu\alpha}+\delta g^{\mu\alpha}\right)h_{\alpha\beta}\left(\eta^{\beta\nu}+\delta g^{\beta\nu}\right)=-\eta^{\mu\alpha}h_{\alpha\beta}\eta^{\beta\nu}=-h^{\mu\nu}$$ to first order. Hence, $$g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}$$ Would this be correct? $\endgroup$
    – user35305
    Commented Nov 26, 2016 at 19:15
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic ♦
    Commented Nov 26, 2016 at 19:15
  • $\begingroup$ Ok cool. Is the reason why both the metric and its inverse have the form $g_{\mu\nu}=\eta_{\mu\nu}+\delta g_{\mu\nu}$, and $g^{\mu\nu}=\eta^{\mu\nu}+\delta g^{\mu\nu}$, respectively, simply because one is perturbing both $g_{\mu\nu}$ and $g^{\mu\nu}$ around a Minkowski background (obviously it wouldn't make any sense to expand a metric and its inverse around different backgrounds)? In general, would one have $g_{\mu\nu}=\bar{g}_{\mu\nu}+\delta g_{\mu\nu}$ and $g_{\mu\nu}=\bar{g}_{\mu\nu}+\delta g_{\mu\nu}$, respectively, where $\bar{g}_{\mu\nu}$ is some known background metric?! $\endgroup$
    – user35305
    Commented Nov 26, 2016 at 19:26
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic ♦
    Commented Nov 26, 2016 at 19:56
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I am late, but I'll answer anyway. In linearized gravity, $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$. We wish to find the expression for $g^{\mu \nu}$ up to linear order, which in general, must be some linear combination of $\eta^{\mu \nu}$ and $h^{\mu \nu}$. To that end, we write:

$g^{\mu \nu} = a\eta^{\mu \nu} + b h^{\mu \nu}$ where $a,b$ are coefficients to be determined

$g_{\mu \nu} g^{\nu \lambda} = (\eta_{\mu \nu} + h_{\mu \nu})(a\eta^{\nu \lambda} + b h^{\nu \lambda}) = \delta_\mu^\lambda$

$\Rightarrow a \delta_\mu^\lambda + (a+b)h_\mu^{\ \lambda} = \delta_\mu^\lambda \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ [$O(h^2)$ term neglected in the previous step]

$\Rightarrow a=1, b=-1$

So the inverse metric tensor in linearized gravity turns out to be $g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu}$.

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    $\begingroup$ as stressed by the other answer, one need not assume any ansatz for the inverse metric. $\endgroup$
    – AccidentalFourierTransform
    Commented May 1, 2017 at 18:47

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