# Derivative of metric tensor with respect to itself

1. Aug 16, 2008

### jdstokes

Is there an identity for $\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}}$? Note raised and lowered indices.

2. Aug 16, 2008

### jdstokes

And it turns out there is!

$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = \frac{\partial}{\partial g_{\lambda\sigma}} (g_{\alpha\beta}g^{\alpha\mu}g^{\beta\nu}) = \delta^\lambda_\alpha\delta^\sigma_\beta g^{\alpha\mu} g^{\beta\nu} + g_{\alpha\beta}\frac{g^{\alpha\mu}}{g_{\lambda\sigma}}g^{\beta\nu} + g_{\alpha\beta}g^{\alpha\mu}\frac{\partial g^{\beta\nu}}{g_{\lambda\sigma}}\implies$

$\frac{\parital g^{\mu\nu}}{g_{\lambda\sigma}} = - g^{\lambda\mu}g^{\sigma\nu}$.

3. Aug 18, 2008

### masudr

Neat! I'm going to transcribe your LaTeX with the partial deriv. signs put in ('cos my brain can't read it otherwise) and ask you about the final step:

$$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = \frac{\partial}{\partial g_{\lambda\sigma}} (g_{\alpha\beta}g^{\alpha\mu}g^{\beta\nu}) = \delta^\lambda_\alpha\delta^\sigma_\beta g^{\alpha\mu} g^{\beta\nu} + g_{\alpha\beta}\frac{\partial g^{\alpha\mu}}{\partial g_{\lambda\sigma}}g^{\beta\nu} + g_{\alpha\beta}g^{\alpha\mu}\frac{\partial g^{\beta\nu}}{\partial g_{\lambda\sigma}}\implies \frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = - g^{\lambda\mu}g^{\sigma\nu}$$

So, how did you get to the final step? I ran into (something probably stupid) a problem trying to bring to factor out the derivative term.

Last edited: Aug 18, 2008
4. Aug 18, 2008

### Hurkyl

Staff Emeritus
I'm somewhat suspicious of this for two reasons.

1. The components of the metric tensor are not independent variables
2. I don't think it makes sense to ask for derivatives w.r.t. components of the metric tensor

$$\frac{\partial g_{ab}}{\partial g_{cd}} = \frac{\partial (g_{ae} g_{bf} g^{ef})}{\partial g_{cd}}$$

Last edited: Aug 18, 2008
5. Aug 18, 2008

### jdstokes

$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = \frac{\partial}{\partial g_{\lambda\sigma}} (g_{\alpha\beta}g^{\alpha\mu}g^{\beta\nu}) = \delta^\lambda_\alpha\delta^\sigma_\beta g^{\alpha\mu} g^{\beta\nu} + g_{\alpha\beta}\frac{g^{\alpha\mu}}{g_{\lambda\sig ma}}g^{\beta\nu} + g_{\alpha\beta}g^{\alpha\mu}\frac{\partial g^{\beta\nu}}{g_{\lambda\sigma}}\implies$
$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = g^{\lambda\mu} g^{\sigma\nu} + \delta^{\nu}_\alpha \frac{g^{\alpha\mu}}{g_{\lambda\sig ma}} + \delta^\mu_\beta\frac{\partial g^{\beta\nu}}{\partial g_{\lambda\sigma}}\implies$

$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = g^{\lambda\mu} g^{\sigma\nu} + \frac{\partial g^{\nu\mu}}{\partial g_{\lambda\sigma}} + \frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}}\implies$

$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = -g^{\lambda\mu} g^{\sigma\nu}$

Last edited: Aug 18, 2008
6. Aug 18, 2008

### jdstokes

I was just reading D'Inverno and happened to stumble across exercise 11.3 which is to show that

$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = - \frac{1}{2}(g^{\mu\lambda}g^{\nu\sigma}+g^{\sigma\mu}g^{\lambda\nu})$

which is actually equivalent to the result

$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = -g^{\lambda\mu} g^{\sigma\nu}$

because of symmetry of the metric tensor.

So it is correct after all.

Last edited: Aug 18, 2008