Inverse metric in Newtonian limit of GR

Question

I am reading Carroll's book. So looking at the Newtonian limit we write $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ where $h_{\mu\nu}$ is some small perturbation. He says that because $g^{\mu\nu}g_{\nu\sigma} = \delta^{\mu}_{\sigma}$ then, to first order in $h$, we have $g^{\mu\nu} = \eta^{\mu\nu} - h^{\mu\nu}$ where $h^{\mu\nu} = \eta^{\mu \rho} \eta^{\nu \sigma}h_{\rho \sigma}$. I don't see how this is the form of $g^{\mu \nu}$ to first order in $h$ though. Suppose $g$ does indeed have that form, then \begin{align*} g^{\mu \nu}g_{\nu \sigma} &= (\eta^{\mu \nu} - h^{\mu \nu})(\eta_{\nu \sigma} + h_{\nu \sigma}) \\ &= \eta^{\mu \nu} \eta_{\nu \sigma} + \eta^{\mu \nu} h_{\nu \sigma} - h^{\mu \nu} \eta_{\nu \sigma} - h^{\mu \nu}h_{\nu \sigma} \end{align*}

We want this be equal to $\eta^{\mu \nu} \eta_{\nu \sigma}$, meaning all those other terms have to be quadratic in $h$, but none of them appear to be?

Answer 1

Perhaps this is an easy way to see it. To first order in $h$, any tensor $T^{\mu \nu}$ can be written $T^{\mu \nu} = t^{\mu \nu} + \lambda h^{\mu \nu}$ for some $\lambda$ and tensor $\boldsymbol{t}$.

Now, for the metric tensor, we must have that $g^{\mu \nu} = \eta^{\mu \nu} + \lambda h^{\mu \nu}$ by matching with the $\mathcal{O}(0)$ term, i.e. $g^{\mu \nu} g_{\nu \sigma} = \delta^{\mu}_{\sigma} + \mathcal{O}(h)$.

Now it is just a simple expansion up to second order: $g^{\mu \nu} g_{\nu \sigma} = \delta^{\mu}_{\sigma} + \lambda h^{\mu}_{\sigma} + h^{\mu}_{\sigma} + \mathcal{O}(h^2)$, and $\lambda = -1$ follows.