What happens to $g^{\alpha\beta}_{,\sigma}=-g^{\alpha\mu}g^{\beta\nu}g_{\mu\nu,\sigma}$ when $g_{\mu\nu}\rightarrow \eta_{\mu\nu}$ (weak field limit)?

Question

The equation $$g^{\alpha\mu}_{\,\,\,\, ,\sigma}\,g_{\mu\nu} + g^{\alpha\mu}\,g_{\mu\nu,\sigma} = (g^{\alpha\mu}g_{\mu\nu})_{,\sigma} = \delta^\alpha_{\nu,\sigma} = 0 $$ gives the useful relation $$g^{\alpha\beta}_{\,\,\,\, ,\sigma} = -g^{\alpha\mu} g^{\beta\nu}g_{\mu\nu,\sigma} \quad\quad(*)$$ between the derivative of the metric and its inverse.

Here's what puzzles me. In any linearized treatment of gravity, where we assume $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ where $\eta$ is the Minkowski metric and $|h_{\mu\nu}| \ll 1$, we approximate $g_{\mu\nu} \approx \eta_{\mu\nu}$ whenever it is a multiplicative factor; thus, we raise and lower indices with $\eta_{\mu\nu}$, thus:

$$h_\mu^\nu = g^{\nu\kappa}h_{\mu\kappa}=\eta^{\nu\kappa}h_{\mu\kappa}.$$

Moreover, in flat spacetime, the Minkowski metric commutes with differentiation (just as the normal metric commutes with covariant differentiation):

$$h_{\mu,\sigma}^\nu = \eta^{\nu\kappa}h_{\mu\kappa,\sigma}.$$

Now look at (*). In the weak field approximation, the right hand side is: $$-g^{\alpha\mu} g^{\beta\nu}g_{\mu\nu,\sigma} \approx -\eta^{\alpha\mu} \eta^{\beta\nu}g_{\mu\nu,\sigma} = -g^{\alpha\beta}_{\,\,\,\, ,\sigma}$$

which contradicts $(*)$! In other words, the minus sign, which is clearly correct in the general case, seems to be wrong in the weak field approximation where $g_{\mu\nu} \rightarrow \eta_{\mu\nu}$.

What am I overlooking here?

Answer 1

First of all, we do not generally use $g_{\mu\nu}=\eta_{\mu\nu}$ when it is a multiplicative factor, but we only discard the terms of the order $O(h^2)$. Second, we have the following relation for the inverse metric in linearized gravity: $$g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}.$$ You can check the derivation of this here. With this we have

$$-g^{\alpha\mu}g^{\beta\nu}g_{\mu\nu,\sigma}=-(\eta^{\alpha\mu}-h^{\alpha\mu})(\eta^{\beta\nu}-h^{\beta\nu})h_{\mu\nu,\sigma}$$ $$=-\eta^{\alpha\mu}\eta^{\beta\nu}h_{\mu\nu,\sigma}+O(h^2)=-h^{\alpha\beta}_{\hspace{.2cm},\sigma}=g^{\alpha\beta}_{\hspace{.2cm},\sigma}$$

in accordance with your equation (*). Note that we used $\eta_{\mu\nu,\sigma}=\eta^{\mu\nu}_{\hspace{.2cm},\sigma}=0.$