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Derivative of metric tensor with respect to itself

  1. Aug 16, 2008 #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. Aug 16, 2008 #2
    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. Aug 18, 2008 #3
    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. Aug 18, 2008 #4

    Hurkyl

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    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. Aug 18, 2008 #5
    [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. Aug 18, 2008 #6
    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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