Derivative of metric tensor with respect to itself

  1. Is there an identity for [itex]\frac{\partial g^{\mu\nu}}{\partial g_{\lambda\sigma}}[/itex]? Note raised and lowered indices.
     
  2. jcsd
  3. 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].
     
  4. 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.
     
    Last edited: Aug 18, 2008
  5. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    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]
     
    Last edited: Aug 18, 2008
  6. [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]
     
    Last edited: Aug 18, 2008
  7. 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.
     
    Last edited: Aug 18, 2008
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