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Christofle symbol and determinant of metric tensor

  1. Dec 30, 2008 #1
    Hi, every one I'm newbie here. I have a few problem with my study about GR.
    Here's a problem


    Could I prove these relation by change index (in 1st term ) from a -> d and also d -> a?

    and let's defined [tex]{g}=det{\\g_{ab}\\}[/tex]

    [tex]{g^{ab}}\partial_c(\\g_{ab})=\frac{1}{g} \partial_c(\\g)[/tex]

    How I prove these equation ? Any one got an idea?
  2. jcsd
  3. Dec 31, 2008 #2


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    Yes. (You also have to use that the metric is a symmetric tensor: [itex]g_{ab}=g_{ba}[/itex]).

    I haven't given much thought to this specific identity, but the only time I proved an identity involving the determinant of the metric, I had to use this crap.
  4. Dec 31, 2008 #3
    By definition [itex]\sqrt{-g}[/itex] is a tensor density of weight one. As a consequence of this and the fact that the covariant derivative of the metric is zero, one has the result

    [tex]\nabla_a\sqrt{-g} = \partial_a\sqrt{-g} - \Gamma^b_{\phantom{b}ab}\sqrt{-g} = 0[/tex]

    It's trivial to go from here to the result you want to prove.
  5. Jan 2, 2009 #4


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    You can check Ray d'Inverno's book on GR; in chapter 7 I believe he derives these things quite extensively :)
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