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Energy-Momentum Tensor of Perfect Fluid

  1. May 23, 2015 #1
    1. The problem statement, all variables and given/known data

    I am given this metric: ##ds^2 = - c^2dt^2 + a(t)^2 \left( dx^2 + dy^2 + dz^2 \right)##. The non-vanishing christoffel symbols are ##\Gamma^t_{xx} = \Gamma^t_{yy} = \Gamma^t_{zz} = \frac{a a'}{c^2}## and ##\Gamma^x_{xt} = \Gamma^x_{tx} = \Gamma^y_{yt} = \Gamma^y_{ty} = \Gamma^z_{zt} = \Gamma^z_{tz} = \frac{a'}{a}##.

    The energy momentum tensor can be written as ##T^{\alpha \beta} = \left( \rho + \frac{P}{c^2} \right)u^{\alpha}u^\beta + Pg^{\alpha \beta}##. Show that ##\frac{d(\rho a^3)}{da} + \frac{3Pa^2}{c^2} = 0##.


    2. Relevant equations


    3. The attempt at a solution

    I shall let c=1 and u = (1,0,0,0) for simplicity.

    I know that the conservation requirement gives
    [tex]\nabla_\alpha T^{\alpha \beta} = 0 [/tex]
    Letting c=1, we have the relativistic continuity equation as
    [tex]u^\alpha \left(\nabla_\alpha \rho \right) + (\rho + P)\left( \nabla_\alpha u^\alpha \right) = 0 [/tex]

    For ##\nabla_\alpha \rho = \partial_\alpha \rho## and ##\nabla_\alpha u^\alpha = \partial_\alpha + \Gamma^\alpha_{\alpha \mu} u^\mu =\partial_\alpha u^\alpha + u^t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) = 3c \left( \frac{a'}{a} \right)##.

    Thus we have
    [tex]u^\alpha \partial_\alpha \rho + 3(\rho + P)\left( \frac{a'}{a} \right) = 0 [/tex]
    [tex]\partial_t \rho + 3(\rho + P)\left( \frac{a'}{a} \right) = 0 [/tex]

    How do I proceed?
     
  2. jcsd
  3. May 23, 2015 #2
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