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Derivative of the metric tensor

  1. Mar 10, 2008 #1
    Could anybody help to spot the inconsistency in the following reasoning?

    When calculating the normal derivative of the metric tensor I get:

    [tex] \partial_\mu g^{\rho \sigma} = g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} + 2 \partial_\mu g^{\rho \sigma}, [/tex] (1)

    which means that:

    [tex] g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} = -\partial_\mu g^{\rho \sigma}. [/tex] (2)

    And I don't see how this could be.

    That's how I get this result:

    \partial_\mu g^{\rho \sigma} =
    \partial_\mu (g^{\rho \lambda} g^{\sigma \gamma} g_{\lambda \gamma}) =
    g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} + g^{\rho \lambda} g_{\lambda \gamma} \partial_\mu g^{\sigma \gamma} + g_{\lambda \gamma} g^{\sigma \gamma} \partial_\mu g^{\rho \lambda} =
    g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} + \delta^\rho_\gamma \partial_\mu g^{\sigma \gamma} + \delta^\sigma_\lambda \partial_\mu g^{\rho \lambda} =
    = g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} + 2 \partial_\mu g^{\rho \sigma}.
    [/tex] (3)

    Could anybody show how to get directly the right hand side of equation (2) from the left hand side, or show where the mistake in the equation (3) is?

    Thanks a lot.
  2. jcsd
  3. Mar 10, 2008 #2
    Unless I've misunderstood something terribly obvious, it's trivial to get what you want. We know that

    [tex]\partial_cg^{ab} = g^{ad}g^{be}\partial_cg_{de} + 2\partial_cg^{ab}[/tex]

    Agreed? Now subtract [itex]2\partial_cg^{ab}[/itex] from both sides and you get

    [tex]g^{ad}g^{be}\partial_cg_{de} = \partial_cg^{ab} - 2\partial_cg^{ab} = -\partial_cg^{ab}[/tex]

    which is what you're looking for.
  4. Mar 10, 2008 #3


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    Last edited: Mar 10, 2008
  5. Mar 12, 2008 #4
    Thank you shoehorn and samalkhaiat for your replies, it helped a lot. Especially samalkhaiat. That's what I needed.
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