# Derivative of the metric with respect to inverse metric

This might be a simple question but what is $$\frac{\partial g_{\mu\nu}}{\partial g^{\sigma\rho}}$$, with proof please if possible.

Hint $$\frac{\partial}{\partial g^{\sigma\rho}}(g_{\mu\nu}g^{\nu\lambda})=\frac{\partial}{\partial g^{\sigma\rho}}(\ \delta_\mu^\lambda)=0$$ You can use the product rule on the first term to get $$\frac{\partial g_{\mu\nu}}{\partial g^{\sigma\rho}} g^{\nu\lambda}+ g_{\mu\nu} \frac{\partial g^{\nu\lambda}}{\partial g^{\sigma\rho}}=0$$ For a general tensor we have $$\frac{\partial g^{\nu\lambda}}{\partial g^{\sigma\rho}}=\delta^\nu_\sigma\delta^\lambda_\rho$$ but since the metric tensor is symmetric the individual components are not independent.For example $$\frac{\partial g^{31}}{\partial g^{13}}=\frac{\partial g^{13}}{\partial g^{13}}=1$$ Accounting for this symmetry we get $$\frac{\partial g^{\nu\lambda}}{\partial g^{\sigma\rho}}=\tfrac 1 2(\delta^\nu_\sigma\delta^\lambda_\rho+\delta^\nu_\rho\delta^\lambda_\sigma).$$ To see where this factor one half comes from go to this answer.

Finally we get

$$\frac{\partial g_{\mu\nu}}{\partial g^{\sigma\rho}}=-g_{\mu\rho}g_{\sigma\nu}$$ for general tensors and $$\frac{\partial g_{\mu\nu}}{\partial g^{\sigma\rho}}=-\tfrac 1 2(g_{\mu\rho}g_{\sigma\nu}+g_{\mu\sigma}g_{\rho\nu})$$ for the metric tensor

• $- \frac{1}{2} ( g_{\mu\sigma} g_{\nu\rho} + g_{\nu\sigma} g_{\mu\rho})$ ?
– ODE
Commented Apr 12, 2020 at 9:58
• I added more to the hint Commented Apr 12, 2020 at 19:05
• $-g_{\mu\rho}g_{\nu\sigma}$ ? can you please give me the answer ?
– ODE
Commented Apr 13, 2020 at 10:46
• That's correct (using what I said). I forgot that the metric is symmetric so I added more. I also added the solution explicitly for later reference. Commented Apr 13, 2020 at 13:50
• As pointed out by @ODE, your expression for the derivative is wrong, since for example it leads to $\partial g^{00}/\partial g^{00}=2$. Commented Apr 13, 2020 at 13:52

For fun, here's how to do it in a coordinate-free manner :

The inverse metric is the matrix inverse of the metric, so that, if we vary our metric, we can use the binomial identity :

$$$$(g + h)^{-1} = g^{-1} - g^{-1} (I + hg^{-1})^{-1} h g^{-1}$$$$

In our case :

$$\begin{eqnarray} \frac{\delta_h g^{-1}}{\delta g} &=& \lim_{\varepsilon \to 0} \frac{(g + \varepsilon h)^{-1} - g^{-1}}{\varepsilon} \\ &=& - \lim_{\varepsilon \to 0} \frac{g^{-1} (I + \varepsilon hg^{-1})^{-1} \varepsilon h g^{-1}}{\varepsilon} \\ &=& - \lim_{\varepsilon \to 0} \frac{\varepsilon g^{-1} h g^{-1} + \mathcal{O}(\varepsilon^2)}{\varepsilon} \\ &=& -g^{-1} h g^{-1} \end{eqnarray}$$

Therefore, the variation in the direction of a perturbation $$h$$ is $$g^{-1} h g^{-1}$$ (in more physicky notation, $$\delta g^{-1} = g^{-1} \delta g g^{-1}$$). If we pick a specific direction such as one of the component, this means

$$$$\delta g^{ab} = -g^{ac} \delta g_{cd} g^{db}$$$$

This is pretty much what is expected. As the inverse of the inverse metric is the metric, the relation holds the other way around, as well.

• Maybe you should say "indices-free" rather than "coordinate-free". A matrix representation for $g$ requires a choice of coordinates. Commented May 26, 2020 at 9:45
• Uh and you missed two minuses in the last equation and in the last row of the second equation. Commented May 26, 2020 at 9:47

For a metric tensor the answer should be

$$\frac{\partial g_{\mu\nu}}{\partial g^{\sigma\rho}}=-\frac{n(\sigma\rho)}{2}\left(g_{\mu\sigma}g_{\nu\rho}+g_{\nu\sigma}g_{\mu\rho}\right),$$

where $$n(\sigma\rho)=2-\delta_{\sigma\rho}$$ and $$\delta_{\sigma\rho}$$ is the Kronecker delta, i.e. the function is 1 if $$\sigma=\rho$$ and 2 if $$\sigma\neq\rho$$.