# Raising and Lowering Indices of a Perturbed Metric

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).

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.

• Could you please show me how second order terms will arise if we raise index by g rather than the base metric Dec 24, 2020 at 13:49

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 $$h^2, h^3$$.

• Is it for similar reasons that the Christoffel symbol for the perturbed metric is given by $\Gamma_{ab}^c = \frac{1}{2} g^{(B)ab} (\partial_a h_{bd} + \partial_b h_{ad} - \partial_d h_{ab})$?
– Tom
Jun 25, 2020 at 18:21
• @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 $\kappa$. Jun 25, 2020 at 18:26
• 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?
– Tom
Jun 25, 2020 at 18:39
• Well, this special case it conisedered sometimes, however, I don't think that it is most common situation Jun 25, 2020 at 20:27
• @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 Jun 27, 2020 at 18:27