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Komar mass of Schwarzschild

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data
    The Komar mass of a Schwarzschild geometry can be written as [itex]\frac{1}{4\pi}\int_{S}n^{\alpha}\sigma_{\beta} \nabla_{\alpha} \xi^{\beta}dA[/itex], where [itex]n^{\alpha}[/itex] and [itex]\sigma_{\beta}[/itex] are timelike and spacelike normal vectors respectively. How does one actually go about evaluating this integral?


    2. Relevant equations



    3. The attempt at a solution
    I've simplified it down to [itex]\frac{1}{4\pi}\int_{S}n^{\alpha}\sigma_{\beta} \Gamma^{\beta}_{\alpha t} dA[/itex] but I have no idea how to continue from there. Wouldn't [itex]n^{\alpha}\sigma_{\beta}[/itex] be zero since they are orthogonal? Also, what is dA? I know it includes factors from the metric but which ones?
     
  2. jcsd
  3. May 8, 2012 #2
    It's the surface element of the 2-surface you defined, [itex] dA = \sqrt{g^{(2)}} d^2x [/itex] where [itex]g^{(2)}[/itex] is the determinant of induced metric on this surface.

    So in Schwarzschild, I can choose my vectors n and σ in the obvious way, normalizing them to n2=-1, σ2=+1,
    [tex] n = (-\sqrt{1-\frac{2M}{r}}, 0, 0, 0) [/tex]
    [tex] \sigma = (0, \sqrt{1-\frac{2M}{r}}^{-1}, 0, 0) [/tex]
    Then the 2-surface is just a sphere, with [itex] \sqrt{-g^{(2)}} = r^2 \sin(\theta) [/itex], [itex] dA = r^2 d\Omega [/itex], so I get
    [tex] I = \frac{1}{4\pi}\int_{S}n^{\alpha}\sigma_{\beta} \Gamma^{\beta}_{\alpha t} dA = \frac{1}{4\pi}\int_{S}n^{t} g_{rr} \sigma^{r} \Gamma^{r}_{t t} dA = \frac{1}{4\pi}\int_{S} -\sqrt{1-\frac{2M}{r}}(1-\frac{2M}{r})^{-1}(1-\frac{2M}{r})^{-1/2} \frac{M}{r^2} (1-\frac{2M}{r}) dA
    [/tex]
    [tex]
    = -\frac{1}{4\pi}\int_{S} \frac{M}{r^2} r^2 d\Omega = -\frac{M}{4\pi}\int_S d\Omega = -M
    [/tex]
    So maybe there's a sign wrong somewhere, or something, but otherwise it goes like that :)
     
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