Derivative of metric tensor with respect to itselfby jdstokes Tags: derivative, metric, respect, tensor 

#1
Aug1608, 10:22 PM

P: 527

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




#2
Aug1608, 11:54 PM

P: 527

And it turns out there is!
[latex]\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[/latex] [latex]\frac{\parital g^{\mu\nu}}{g_{\lambda\sigma}} =  g^{\lambda\mu}g^{\sigma\nu} [/latex]. 



#3
Aug1808, 12:15 AM

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:
[tex] \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} [/tex] 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. 



#4
Aug1808, 12:38 AM

Emeritus
Sci Advisor
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 Anyways, its probably better to start with [tex] \frac{\partial g_{ab}}{\partial g_{cd}} = \frac{\partial (g_{ae} g_{bf} g^{ef})}{\partial g_{cd}} [/tex] 



#5
Aug1808, 01:12 AM

P: 527

[latex]\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[/latex] [latex]\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[/latex] [latex]\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = g^{\lambda\mu} g^{\sigma\nu}[/latex] 



#6
Aug1808, 01:18 AM

P: 527

I was just reading D'Inverno and happened to stumble across exercise 11.3 which is to show that
[latex]\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} =  \frac{1}{2}(g^{\mu\lambda}g^{\nu\sigma}+g^{\sigma\mu}g^{\lambda\nu})[/latex] which is actually equivalent to the result [latex]\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = g^{\lambda\mu} g^{\sigma\nu}[/latex] because of symmetry of the metric tensor. So it is correct after all. 


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