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Metricity equation

  1. Mar 21, 2014 #1
    ∇[itex]_{\mu}[/itex]g[itex]_{\alpha\beta}[/itex]=0 is the metricity equation (∇ the covariant derivative).
    However, I wondered whether also ∇[itex]_{\mu}[/itex]g[itex]^{\alpha\beta}[/itex]=0 holds?
    When I checked it, it did not. But I really want it to hold, so I was wondering whether I made a mistake.
    Thank you.
     
  2. jcsd
  3. Mar 21, 2014 #2

    WannabeNewton

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    Hi kuecken. In fact ##\nabla_{\mu}g_{\alpha\beta} = 0 \Leftrightarrow \nabla_{\mu}g^{\alpha\beta} = 0##.

    This is actually very easy to see:

    ##0 = \nabla_{\mu}\delta^{\alpha}{}{}_{\beta} = \nabla_{\mu}(g^{\alpha \gamma}g_{\gamma \beta}) = g^{\alpha\gamma}\nabla_{\mu}g_{\gamma \beta} + g_{\gamma\beta}\nabla_{\mu}g^{\alpha \gamma}## holds identically hence if ##\nabla_{\mu}g_{\alpha\beta} = 0## then ##\nabla_{\mu}g^{\alpha\beta} = 0## and similarly for the converse.
     
  4. Mar 21, 2014 #3
    This is the expression I arrived at. However, can I argue that the g[itex]_{\gamma\beta}[/itex] in front of ∇ is arbitrary? Or how do I do that last step
     
  5. Mar 21, 2014 #4

    WannabeNewton

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    So we have ##g_{\gamma\beta}\nabla_{\mu}g^{\alpha\gamma} = 0##. Contract both sides with ##g^{\nu\beta}## to get ##\delta^{\nu}{}{}_{\gamma}\nabla_{\mu}g^{\alpha\gamma} = \nabla_{\mu}g^{\alpha\nu} = 0##.
     
  6. Mar 21, 2014 #5

    dextercioby

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    Because of the metricity condition, one can define the so-called contravariant derivatives:

    [tex] \nabla^{\mu}T_{\alpha} = g^{\mu\nu} \nabla_{\nu}T_{\alpha}[/tex]

    [tex] = \nabla_{\nu}\left(g^{\mu\nu}T_{\alpha}\right)[/tex]

    that is freely take out and put back the metric tensor under the covariant/contravariant differentiation sign.
     
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