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I have seen in GR that if a metric is a perturbation of some base metric $g^{(B)}_{\mu \nu}$ such that

$g_{\mu \nu} = g^{(B)}_{\mu \nu} + h_{\mu \nu},$

then

$g^{\mu \nu} = g^{(B) \mu \nu} - h^{\mu \nu}.$

Does this mean that $g^{(B) \mu \nu}$ is the inverse metric such that $ g^{(B) \alpha \beta} g^{(B)}_{\beta \gamma} = \delta^{\alpha}_{\gamma}$ and that $h^{\mu \nu}$ is obtained by raising two indices of $h_{\mu \nu}$ with $ g^{(B) \alpha \beta} $? (I haven't matched the indices on the last one but hopefully get my meaning, you raise one index with the inverse of the base metric, then raise another one).

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You raise and lower the indices with the original metric $g$ , but to fist order it is the same as doing it with the $g^{(B)}$ metric.

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  • $\begingroup$ Could you please show me how second order terms will arise if we raise index by g rather than the base metric $\endgroup$
    – Shashaank
    Dec 24, 2020 at 13:49
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Yes, here you are working in linear approximation, so all terms of highers orders in $h$ are negligible, so $h^{\mu \nu} = g^{(B) \mu \alpha} g^{(B) \nu \beta} h_{\alpha \beta}$, if you replace $g^{(B)}$ by $g$, the difference will be only in terms of order $h^2, h^3$.

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  • $\begingroup$ Is it for similar reasons that the Christoffel symbol for the perturbed metric is given by $\Gamma_{ab}^c = \frac{1}{2} g^{(B)ab} (\partial_a h_{bd} + \partial_b h_{ad} - \partial_d h_{ab})$? $\endgroup$
    – Tom
    Jun 25, 2020 at 18:21
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    $\begingroup$ @Tom yes, for the same reason, maybe a more obvious way to see this, is to write $h_{\mu \nu} = \kappa \tilde{h}_{\mu \nu}$, where $\kappa \ll 1$, and consider it as an expansion in $\kappa$. $\endgroup$
    – spiridon_the_sun_rotator
    Jun 25, 2020 at 18:26
  • $\begingroup$ Also in GR, is it common to consider a perturbation which is time-dependent but uniform in space, so not depending on the spatial coordinates? $\endgroup$
    – Tom
    Jun 25, 2020 at 18:39
  • $\begingroup$ Well, this special case it conisedered sometimes, however, I don't think that it is most common situation $\endgroup$
    – spiridon_the_sun_rotator
    Jun 25, 2020 at 20:27
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    $\begingroup$ @Tom restricting the perturbation to the one, which will preserve the conformal structure is not property of a metric, but a perturbation itself. For example $ f(r, \theta) (dr^2 + r^2 d \theta^2)$ will in fact also be conformally equivalent to flat metric. For some special reason, you may be interested in perturbations such, that preserve the structure $f(x) d x^2$, i.e of form $h = h(x) dx^2$. However, in general case the perturbation will push the metric from confomally flat form $\endgroup$
    – spiridon_the_sun_rotator
    Jun 27, 2020 at 18:27

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