Finding the inverse metric of a metric close to Minkowski metric [closed]

Question

Let $g_{uv}=\eta_{uv}+h_{uv}$ be a metric with $\mid h_{uv} \mid$ very small so that the metric is close to the Minkowski metric.

Then we can write the inverse metric $g^{uv}=\eta^{uv}+k^{uv}$ with $\mid k^{uv} \mid$ being very small as well.

Now, how can I express $k^{uv}$ in terms of $h_{uv}$? I contracted the metric with the inverse metric to obtain $k^{wv}(\eta_{vu}+h_{uv})=-\eta^{wv}h_{vu}$. However, I cannot proceed from this. I cannot deal with the summed indices...Could anyone please help me?

Answer 1

Impose $(\eta_{\mu\nu} + h_{\mu\nu}) (\eta^{\nu\rho} + k^{\nu\rho}) = \delta^\rho_\mu$, then expand the left hand side at first order in the perturbation of the metric, that is discard the term $hk$:

$$ \delta^\rho_\mu= \delta^\rho_\mu + \eta_{\mu\nu}k^{\nu\rho} + h_{\mu\nu}\eta^{\nu\rho} $$

From this you conclude

$$ \eta_{\mu\nu}k^{\nu\rho}= -h_{\mu\nu}\eta^{\nu\rho} $$

And then

$$ {k^\rho}_\mu = - {h^\rho}_\mu \rightarrow k^{\rho\mu} = - h^{\rho\mu} $$

The inverse metric is then $g^{\mu\nu} = \eta^{\mu\nu} - h^{\mu\nu}$

If you want to do second order perturbation theory you should keep the $hk$ term. Try to do it as an exercise

Note: in perturbation theory you raise and lower indeces with the unperturbed metric, which is the Minkowskian metric in this case but it can be any metric in general.