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Energy-momentum conservation equation

  1. Feb 28, 2007 #1

    cristo

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    We have the metric [tex]ds^2=-e^{2\Phi}dt^2+e^{2\Lambda}dr^2+r^2d\Omega^2[/tex], and the energy momentum tensor takes the form [tex]t^{ab}=(\rho+p)u^au^b+pg^{ab}[/tex] where the 4-velocity is [tex]u=e^{-\Phi}\partial_t[/tex], and [itex]\Phi[/itex] and [itex]\Lambda[/itex] are functions of r only.

    I'm asked to show that the ebergy-momentum conservation equation [tex]\nabla_aT^{ab}=0[/tex] is equivalent to [tex](\rho+p)\frac{d\Phi}{dr}=-\frac{dp}{dr}[/tex].

    Now, I'm after some advice on how to calculate this. I know all the connection coefficients (they were given in the question). I tried setting up the four equations, one for each value of b, and then attempting to solve these, but each of these equations has 12 terms in, so I figured there must be an easier way to do this!

    Does anyone know whether there is another way, or if there are any simplifications I can make; or do I need to plough through the 4 equations?

    Thanks
     
  2. jcsd
  3. Feb 28, 2007 #2

    cristo

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    Ok, I've got the answer. Just decided to compute the r equation, and it gave the result.
     
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