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

Given this, my question is, how does one obtain the inverse metric $g^{\mu\nu}$? I read in some notes (e.g. here, top of page 2, and here, top of page 4) that it is given by $$g^{\mu\nu}=\eta^{\mu\nu}-\kappa h^{\mu\nu}+\kappa^{2}h^{\mu}_{\;\lambda}h^{\mu\lambda}+\cdots$$ Now I know how to get the expression to first-order by writing $g^{\mu\nu}=\eta^{\mu\nu}+\delta g^{\mu\nu}$, and then using that $$\delta g^{\mu\nu}=-g^{\mu\lambda}\delta g_{\lambda\sigma}g^{\sigma\nu}=-\kappa g^{\mu\lambda}h_{\lambda\sigma}g^{\sigma\nu}=-\kappa \eta^{\mu\lambda}\eta^{\sigma\nu}h_{\lambda\sigma}+\mathcal{O}(\kappa^{2})$$ However, I'm unsure how to obtain the higher order terms. 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?

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  • $\begingroup$ If you're going above first order, to be consistent you'll have to raise and lower the indices with the full perturbed metric and then throw away terms than come out higher order in h than that you are considering. $\endgroup$ – R. Rankin Jul 3 '17 at 11:32
  • $\begingroup$ @R.Rankin That's what I thought, but then how do the authors in the links I gave get the expression for the inverse metric that I've put in my post? $\endgroup$ – user35305 Jul 3 '17 at 11:45
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One particularly effective and quick way to write this is to write the metric as $g=\eta+\kappa h$, so that

$$g^{-1}=(\eta+\kappa h)^{-1}=\eta^{-1}(\textbf{1}+\kappa h\eta^{-1})^{-1}$$

Then we just use the expansion

$$(\textbf{1}+\epsilon\textbf{A})^{-1}=\textbf{1}-\epsilon\textbf{A}+\epsilon^2\textbf{A}^2+\cdots,$$

which holds for matrices just as it does for numbers. The desired result is found immediately, as well as higher order terms.

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  • $\begingroup$ Thanks for your answer. I didn't realise that the Taylor expansion for $(1-x)^{-1}$ carried over straightforwardly for matrices. How does one prove this? $\endgroup$ – user35305 Jul 4 '17 at 9:13
  • $\begingroup$ @user35305 If you assume the inverse of $1+ \epsilon \mathbf{A}$ is some linear combination of $\mathbf{A}^n$ and write that $(\mathbf{1} + \epsilon \mathbf{A})^{-1} (\mathbf{1} +\epsilon \mathbf{A}) = (\mathbf{1} +\epsilon \mathbf{A})(a_0 \mathbf{1} + a_1 \mathbf{A} + \ldots ) = \mathbf{1}$ and equate coefficients you obtain the expression. $\endgroup$ – user110503 Jul 5 '17 at 15:22
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For your Minkowski background metric:

$$g_{\mu\nu}=\eta_{\mu\nu}+\kappa h_{\mu\nu}$$

We have that the perturbation can be written as:

$$\delta g_{\mu\nu}=g_{\mu\nu}-\eta_{\mu\nu}=\kappa h_{\mu\nu}$$

We also know that, at first order:

$$g^{\mu\nu}=\eta^{\mu\nu}-\kappa h^{\mu\nu}$$

Now we want to find it's covariant form, which goes like:

$$\delta g^{\mu\nu}=-g^{\mu\lambda}\delta g_{\lambda\rho}g^{\rho\nu}$$

Now simply substitute into this equation from our other equations:

$$=-\left(\eta^{\mu\lambda}-\kappa h^{\mu\lambda}\right)\left(\kappa h_{\lambda\rho}\right)\left(\eta^{\rho\nu}-\kappa h^{\rho\nu}\right)$$

Throwing away the third order term we obtain:

$$=-\kappa h^{\mu\nu}+\eta^{\mu\lambda}\kappa h_{\lambda\rho}\kappa h^{\rho\nu}+\kappa h^{\mu\lambda}\kappa h_{\lambda\rho}\eta^{\rho\nu}$$

$$=-\kappa h^{\mu\nu}+\kappa h_{\rho}^{\mu}\kappa h^{\rho\nu}+\kappa h^{\mu\lambda}\kappa h_{\lambda}^{\nu}\eta$$ Since the metric must be symmetric, so must the perturbation be also hence we can write:

$$\delta g^{\mu\nu}=-\kappa h^{\mu\nu}+2\kappa h_{\rho}^{\mu}\kappa h^{\rho\nu}$$

Now I got a factor of 2 different from your reference, Which I think can be eliminated by applying requirement for the total metric:

$$g^{\mu\nu}g_{\mu\nu}=\delta_{\mu}^{\mu}$$ But I think you get the Idea, it's a process that just grows outrageously in tediousness with each higher order. Cheers!! (:

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  • $\begingroup$ Thanks for the answer. The problem I find with this though is that one derives $\delta g^{\mu\nu}=-h^{\mu\nu}$ using $\delta g^{\mu\nu}=-g^{\mu\lambda}\delta g_{\lambda\sigma}g^{\sigma\nu}$ and truncating at first-order, so how can one then plug this back into the same relation again? Also, one can only eliminate the factor of 2 by simply disregarding it which seems a bit dodgy. $\endgroup$ – user35305 Jul 3 '17 at 13:29
  • $\begingroup$ @R. Rankin You missed a minus sign in the 4th equation. $\endgroup$ – Avantgarde Jul 3 '17 at 14:33
  • $\begingroup$ @Avantgarde thanks, I'll be expanding this answer after work as per OPs request $\endgroup$ – R. Rankin Jul 3 '17 at 23:08
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This is a relatively old question which lacks a formally complete answer. On finding myself in the need of the inverse of a metric and on not being able to find a proper treatment elsewhere (on casual browsing), I have decided to put a proper formal treatment here.

Following the treatment given here, one can (super-)easily derive the inverse metric to all order of perturbation theory without using ad-hoc relations. I've arranged the following in three steps.

Step - 1: Correct Statement of Problem

The metric whose inverse we intend to determine must be written in a more formal fashion:

$$ g_{\mu\nu} = \eta_{\mu\nu} + \epsilon \ ^{(1)}h_{\mu\nu} + \frac{\epsilon^2}{2!} \ ^{(2)}h_{\mu\nu} + \cdots $$ For later convenience, we move all perturbations into $H_{\mu\nu}$: $$g_{\mu\nu} = \eta_{\mu\nu} + H_{\mu\nu} $$

This way of stating the problem is essentially different than stated by OP in the question. I hope the notation does not need any explanations.

Step-2: And the inverse is

Let us write the inverse as: b

$$ g^{\mu\nu} = (g_{\mu\nu})^{-1}$$ $$ = \eta^{\mu \alpha} \ (\delta{^\alpha_\nu} + \eta^{\alpha\beta}H_{\beta\nu})^{-1}$$

We first note, that we can contract the background metric inside the brackets: $ H{^\alpha_\nu} = \eta^{\alpha\beta}H_{\beta\nu}$. Further, to deal with the brackets, as suggested by Bob in another response, we use the binomial expansion: $$ (1+x)^{-1} = 1 - x + x^2 -x^3 +\cdots $$

And, after a few steps of index gymnastics we reach:

$$ g^{\mu\nu} = \eta^{\mu\nu} - H^{\mu\nu} + H^{\mu\rho}H{_\rho^\nu} - H^{\mu\rho}H{_\rho^\beta}H{_\beta^\nu} + \cdots $$

Are we done?

Step-3: The expansion parameter

The beauty of this arrangement lies in the following realization: $$ H^{\mu\nu} \xrightarrow{\text{can only give rise to terms with}} \epsilon^1, \epsilon^2, \epsilon^3 \cdots$$ $$ H^{\mu\rho}H{_\rho^\nu} \xrightarrow{\text{can only give rise to terms with}} \epsilon^2, \epsilon^3, \epsilon^4 \cdots $$ $$H^{\mu\rho}H{_\rho^\beta}H{_\beta^\nu}\xrightarrow{\text{can only give rise to terms with}} \epsilon^3, \epsilon^4, \epsilon^5 \cdots $$

Hence to get to a useful expression of the inverse, we must arrange the inverse in powers of $\epsilon$.

Doing a bit of work, we get following terms at order $\epsilon^n$:

(note overall sign comes from the last equation in step-2)

  1. $n=0$ $$\frac{1}{0!}(\eta^{\mu \nu}$$
  2. $n=1$ $$ \frac{1}{1!}(- h^{1\mu \nu}) $$
  3. $n=2$ $$ \frac{1}{2!}(2 h^{1}{}_{a}{}^{\nu} h^{1\mu a} - h^{2\mu \nu}) $$
  4. $n=3$ $$\frac{1}{3!}( -6 h^{1}{}_{a}{}^{b} h^{1}{}_{b}{}^{\nu} h^{1\mu a} + 3 h^{1\mu a} h^{2}{}_{a}{}^{\nu} + 3 h^{1}{}_{a}{}^{\nu} h^{2\mu a} - h^{3\mu \nu})$$

As should be obvious on carefully following the above treatment, the final answer neatly looks like:

$$ g^{\mu\nu} = \eta^{\mu \nu} - \epsilon h^{1\mu \nu} + \tfrac{1}{2} \epsilon^2 (2 h^{1}{}_{a}{}^{\nu} h^{1\mu a} - h^{2\mu \nu}) + \tfrac{1}{6} \epsilon^3 (-6 h^{1}{}_{a}{}^{b} h^{1}{}_{b}{}^{\nu} h^{1\mu a} + 3 h^{1\mu c} h^{2}{}_{c}{}^{\nu} + 3 h^{1}{}_{d}{}^{\nu} h^{2\mu d} - h^{3\mu \nu})$$

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For example, in gravitational wave theory to construct the Pseudo-energy momentum tensor a la Issacson tensor you actually need a perturbed generic background to second order. So let be $g_{ab}(\lambda)$ one.parameter family in the way that $$g_{ab}(\lambda)=\tilde{g}_{ab}+\lambda h^{1}_{ab}+\lambda ^2 h^2_{ab}$$ then clearly the inverse is gonna be given by $$g^{ab}\equiv(g_{ab}(\lambda))^{-1}$$ so at first order in $\lambda$ the we need to perform the first derivative respect to the parameter $$\frac{d}{d\lambda}(g_{ab}(\lambda))^{-1}\lvert_{\lambda=0}=-\tilde{g}^{ac}\tilde{g}^{bd}h^1_{cd}=-h^{ab}$$in the same way for second order in $\lambda$ you need the second derivative $$\frac{d^2}{d\lambda^2}(g_{ab}(\lambda))^{-1}\lvert_{\lambda=0}=-\left(\frac{d}{d\lambda}(g_{ab}(\lambda))^{-2}\frac{d}{d\lambda}g_{ab}(\lambda)+(g_{ab})^{-2}\frac{d^2}{d\lambda^2}g_{ab}(\lambda)\right)\lvert_{\lambda=0}$$ $$=-\left(-2\tilde{g}^{af}\tilde{g}^{bg}\tilde{g}^{cd}h^1_{fc}h^1_{dg}+\tilde{g}^{ac}\tilde{g}^{bd}h^{2}_{cd}\right)$$ $$=2h^{1ac}h^{1b}_c-h^{2ab}$$

so to construct the complete inverse metric up to the second order you need this generic form

$$g^{ab}(\lambda)=g^{ab}(0)+\frac{d}{d\lambda}(g^{ab}(\lambda))\lvert_{\lambda=0}+\frac{1}{2}\frac{d^2}{d\lambda^2}(g^{ab}(\lambda))\lvert_{\lambda=0}$$

plugging the quantities that we have already computed you obtain $$g^{ab}(\lambda)=\tilde{g}^{ab}-\lambda h^{1ab}+\lambda^2(h^{1ac}h^{1b}_c-\frac{1}{2}h^{2ab})$$ a check that you should do to keep all in order is for example check the common delta relation with the total metric

$$g^{ac}(\lambda)g_{cb}(\lambda)=\delta^a_b$$

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