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Considering the FRW metric with perturbations; how can I calculate the Einstein tensor (without a very very disgusting expression which comes from the variation of the difference between the Ricci tensor minus the metric tensor times the curvature)?

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  • $\begingroup$ Have you read anything on the cosmological perturbation theory? What "disgusting expression" you don't like? $\endgroup$ – OON Nov 1 '17 at 15:24
  • $\begingroup$ The disgusting expression is the straight forward try to solution- $\delta (R_{\alpha \beta}-g_{\mu \nu} R)$ $\endgroup$ – user Nov 1 '17 at 15:44
  • $\begingroup$ Sorry, but that doesn't tell me anything about your standards of beauty=) neither about how exactly you tried to perform those perturbations and how far have you simplified that expression. Usually in the cosmological perturbation theory you separate different transverse and longitudinal components. So again, have you read anything on that? There are textbooks and lectures available... $\endgroup$ – OON Nov 1 '17 at 15:55
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The problem concerns standard gravitational perturbation theory. Instead of expanding around say Minkowski space, one expands around the FLRW background.

Using the gauge symmetries and scalar-vector-tensor decomposition, one can reduce the form of the perturbed metric to an incredibly simple form, namely,

$$g_{\mu\nu} = g^{\mathrm{FLRW}}_{\mu\nu} + h_{\mu\nu}= a(t)^2 \begin{pmatrix} 1+2\Psi & 0 & 0 & 0\\ 0& 2\Phi -1 & 0 & 0\\ 0& 0 & 2\Phi -1 & 0\\ 0& 0& 0 & 2\Phi -1 \end{pmatrix}$$

One approach to finding the Einstein equations is noting to first order $\delta R_{\mu\nu} = -\frac12 \Delta_L h_{\mu\nu}$ where $\Delta_L$ is the Lichnerowicz operator. Alternative, one can in this case simply plug in the perturbed metric.

Defining the Hubble parameter, $\mathcal H = \dot a a^{-1}$, we have $G_{00} = 3\mathcal H^2 + 2\nabla^2 \Phi - 6\mathcal H \Phi'$. The spatial part mixed with the time component is,

$$G_{0i} = 2\partial_i(\Phi' + \mathcal H \Psi)$$

and finally the spatial part - the messiest - is,

$$G_{ij} = -(2\mathcal H' +\mathcal H^2)\delta_{ij} + \partial_i \partial_j (\Phi - \Psi)$$ $$ + \left[\nabla^2 (\Psi - \Phi) + 2\Phi'' + 2(2\mathcal H' + \mathcal H^2)(\Phi + \Psi) + 2\mathcal H \Psi' + 4\mathcal H \Phi' \right]\delta_{ij}.$$

Compared to the perturbation equations in full generality, even with gauge fixing, this is a relatively manageable expression.

It can be further simplified depending on the scenario. Ignoring anisotropic stress, $\Phi = \Psi$ which greatly reduces the equations and in some instances gives us only a Laplace equation to solve.

The situation doesn't get any better than this. In fact, having done perturbation theory of solutions to string theory, I can say the situation can be a lot worse. General relativity is horrendously non-linear, there's no avoiding that.

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  • $\begingroup$ Well, as you mention scalar-vector-tensor decomposition would be good to als onote that here you consider only scalar sector and omit gravitational waves $\endgroup$ – OON Nov 1 '17 at 16:00
  • $\begingroup$ Just in case, I know, they decouple at linear level. Simply saying the tensor perturbations exist too=) $\endgroup$ – OON Nov 1 '17 at 16:02
  • $\begingroup$ @OON Yes, didn't want to scare the OP any more :) $\endgroup$ – JamalS Nov 1 '17 at 16:06
  • $\begingroup$ And $\delta G$ is the difference between the perturbed G and the unperturbed one? $\endgroup$ – user Nov 1 '17 at 17:30

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