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Differential geometry question

  1. Apr 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Can some one please explain to me how to show that
    [tex]J^{\alpha}{ }_{;\alpha}={1\over{\sqrt{-g}}}\partial_\alpha(\sqrt{-g}J^\alpha)[/tex]

    2. Relevant equations
    [tex]\Gamma^\gamma{}_{\alpha\beta}={1\over 2}g^{\gamma\delta}(g_{\delta\alpha,\beta}+g_{\delta\beta,\alpha}-g_{\alpha\beta,\delta})[/tex]
    [tex]\partial_\alpha\sqrt{-g}=-{1\over 2}\sqrt{-g}g^{\mu\nu}g_{\mu\nu,\alpha}[/tex] (I think this is correct).

    3. The attempt at a solution
    Here's what I've tried
    [tex]J^{\alpha}{ }_{;\alpha}=g^{\alpha\beta}J_{\beta;\alpha}[/tex]
    [tex]=g^{\alpha\beta}(J_{\beta,\alpha}-{1\over 2}J_\gamma g^{\gamma\delta}(g_{\delta\alpha,\beta}+g_{\delta\beta,\alpha}-g_{\alpha\beta,\gamma})).[/tex]
    Now turning to the other side
    [tex]={1\over{\sqrt{-g}}}(J^\alpha\partial_\alpha\sqrt{-g}+\sqrt{-g}\partial_\alpha J^\alpha)[/tex]
    [tex]={1\over{\sqrt{-g}}}(-{1\over 2}\sqrt{-g}g^{\mu\nu}g_{\mu\nu,\alpha}J^\alpha+\sqrt{-g}J^\alpha{}_{,\alpha})[/tex]
    Then cancel the sqrt(-g). But here I'm stuck.
    Last edited: Apr 19, 2008
  2. jcsd
  3. Apr 19, 2008 #2
    Never mind. I got it. Just had to turn the crank a little. Cheers.
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