Register to reply

Derivative of metric tensor with respect to itself

by jdstokes
Tags: derivative, metric, respect, tensor
Share this thread:
jdstokes
#1
Aug16-08, 10:22 PM
P: 527
Is there an identity for [itex]\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}}[/itex]? Note raised and lowered indices.
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
jdstokes
#2
Aug16-08, 11:54 PM
P: 527
And it turns out there is!

[itex]\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[/itex]

[itex]\frac{\parital g^{\mu\nu}}{g_{\lambda\sigma}} = - g^{\lambda\mu}g^{\sigma\nu} [/itex].
masudr
#3
Aug18-08, 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.

Hurkyl
#4
Aug18-08, 12:38 AM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,091
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]
jdstokes
#5
Aug18-08, 01:12 AM
P: 527
Quote Quote by masudr View Post
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.
[itex]\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[/itex]
[itex]\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[/itex]

[itex]\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[/itex]

[itex]\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = -g^{\lambda\mu} g^{\sigma\nu}[/itex]
jdstokes
#6
Aug18-08, 01:18 AM
P: 527
I was just reading D'Inverno and happened to stumble across exercise 11.3 which is to show that

[itex]\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}} = - \frac{1}{2}(g^{\mu\lambda}g^{\nu\sigma}+g^{\sigma\mu}g^{\lambda\nu})[/itex]

which is actually equivalent to the result

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

because of symmetry of the metric tensor.

So it is correct after all.


Register to reply

Related Discussions
Derivative of the metric tensor Special & General Relativity 3
Covariant derivative of metric tensor Differential Geometry 3
Question about derivative of the metric tensor General Physics 0