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Second derivative of a metric and the Riemann curvature tensor

  1. Apr 30, 2013 #1
    I can't see how to get the following result. Help would be appreciated!

    This question has to do with the Riemann curvature tensor in inertial coordinates.

    Such that, if I'm not wrong, (in inertial coordinates) [tex]R_{abcd}=\frac{1}{2} (g_{ad,bc}+g_{bc,ad}-g_{bd,ac}-g_{ac,bd})[/tex]
    where [itex]",_i"[/itex] denotes [itex]\partial \over \partial x^i[/itex].

    How does [tex]g_{ab,cd}=-\frac{1}{3}(R_{acbd}+R_{adbc})[/tex]?
    ______
    So
    [tex]-\frac{1}{3}(R_{acbd}+R_{adbc})=-\frac{1}{6}(g_{bc,ad}+g_{ad,bc}-g_{ba,cd}-g_{cd,ab}+g_{bd,ac}+g_{ac,bd}-g_{ab,cd}-g_{cd,ab})[/tex]
    [tex]=-\frac{1}{6}(g_{bc,ad}+g_{ad,bc}+g_{bd,ac}+g_{ac,bd})+\frac{1}{3}g_{cd,ab}+\frac{1}{3}g_{ab,cd}[/tex]
    But how does [tex]\frac{1}{6}(g_{bc,ad}+g_{ad,bc}+g_{bd,ac}+g_{ac,bd})+\frac{1}{3}g_{cd,ab}=\frac{2}{3}g_{ab,cd}[/tex]
    ____
    ...Have I made a mistake?
     
  2. jcsd
  3. Apr 30, 2013 #2

    Bill_K

    User Avatar
    Science Advisor

    Clearly it doesn't! I'd be curious to know where you saw the relationship.

    The RHS is known as the symmetrized Riemann tensor, Sabcd ≡ −1/3(Racbd + Radbc). It has 20 independent components, just like the Riemann tensor, with symmetries Sabcd = Sbacd = Sabdc = Scdab and Sabcd + Sacdb + Sadbc = 0. Conversely the Riemann tensor can be given in terms of the symmetrized tensor as Rabcd = Sadcb - Sacdb.
     
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