# Derivative of metric tensor with respect to itself

by jdstokes
Tags: derivative, metric, respect, tensor
 P: 527 Is there an identity for $\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}}$? Note raised and lowered indices.
 P: 527 And it turns out there is! $\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = \frac{\partial}{\partial g_{\lambda\sigma}} (g_{\alpha\beta}g^{\alpha\mu}g^{\beta\nu}) = \delta^\lambda_\alpha\delta^\sigma_\beta g^{\alpha\mu} g^{\beta\nu} + g_{\alpha\beta}\frac{g^{\alpha\mu}}{g_{\lambda\sigma}}g^{\beta\nu} + g_{\alpha\beta}g^{\alpha\mu}\frac{\partial g^{\beta\nu}}{g_{\lambda\sigma}}\implies$ $\frac{\parital g^{\mu\nu}}{g_{\lambda\sigma}} = - g^{\lambda\mu}g^{\sigma\nu}$.
 P: 932 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.
Emeritus
PF Gold
P: 16,101

## Derivative of metric tensor with respect to itself

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

$$\frac{\partial g_{ab}}{\partial g_{cd}} = \frac{\partial (g_{ae} g_{bf} g^{ef})}{\partial g_{cd}}$$
P: 527
 Quote by 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.
$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = \frac{\partial}{\partial g_{\lambda\sigma}} (g_{\alpha\beta}g^{\alpha\mu}g^{\beta\nu}) = \delta^\lambda_\alpha\delta^\sigma_\beta g^{\alpha\mu} g^{\beta\nu} + g_{\alpha\beta}\frac{g^{\alpha\mu}}{g_{\lambda\sig ma}}g^{\beta\nu} + g_{\alpha\beta}g^{\alpha\mu}\frac{\partial g^{\beta\nu}}{g_{\lambda\sigma}}\implies$
$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = g^{\lambda\mu} g^{\sigma\nu} + \delta^{\nu}_\alpha \frac{g^{\alpha\mu}}{g_{\lambda\sig ma}} + \delta^\mu_\beta\frac{\partial g^{\beta\nu}}{\partial g_{\lambda\sigma}}\implies$

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

$\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = -g^{\lambda\mu} g^{\sigma\nu}$
 P: 527 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.

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