5
$\begingroup$

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

$\endgroup$

3 Answers 3

6
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ $ - \frac{1}{2} ( g_{\mu\sigma} g_{\nu\rho} + g_{\nu\sigma} g_{\mu\rho})$ ? $\endgroup$
    – ODE
    Commented Apr 12, 2020 at 9:58
  • $\begingroup$ I added more to the hint $\endgroup$ Commented Apr 12, 2020 at 19:05
  • $\begingroup$ $-g_{\mu\rho}g_{\nu\sigma}$ ? can you please give me the answer ? $\endgroup$
    – ODE
    Commented Apr 13, 2020 at 10:46
  • $\begingroup$ 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. $\endgroup$ Commented Apr 13, 2020 at 13:50
  • 1
    $\begingroup$ As pointed out by @ODE, your expression for the derivative is wrong, since for example it leads to $\partial g^{00}/\partial g^{00}=2$. $\endgroup$ Commented Apr 13, 2020 at 13:52
2
$\begingroup$

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 :

\begin{equation} (g + h)^{-1} = g^{-1} - g^{-1} (I + hg^{-1})^{-1} h g^{-1} \end{equation}

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

\begin{equation} \delta g^{ab} = -g^{ac} \delta g_{cd} g^{db} \end{equation}

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.

$\endgroup$
2
  • $\begingroup$ Maybe you should say "indices-free" rather than "coordinate-free". A matrix representation for $g$ requires a choice of coordinates. $\endgroup$ Commented May 26, 2020 at 9:45
  • $\begingroup$ Uh and you missed two minuses in the last equation and in the last row of the second equation. $\endgroup$ Commented May 26, 2020 at 9:47
0
$\begingroup$

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 $\sigma\neq\rho$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.