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

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?

• Sep 28, 2019 at 20:22

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.