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Deriving perfect fluid energy tensor from point particles

  1. May 31, 2016 #1
    1. The problem statement, all variables and given/known data

    For a system of discrete point particles the energy momentum takes the form

    [tex]T_{\mu \nu} = \sum_a \frac{p_\mu^{(a)}p_\nu^{(a)}}{p^{0(a)}} \delta^{(3)}(\vec{x}-\vec{x}^{(a)}),[/tex]

    where the index a labels the different particles. Show that, for a dense collection of particles with isotropically distributed velocities, we can smooth over the individual particle worldlines to obtain the perfect-fluid energy-momentum tensor.

    2. Relevant equations

    Energy-momentum tensor of a perfect fluid:

    [tex]
    T^{\mu \nu} = (\rho + p)U^\mu U^\nu + p \eta^{\mu \nu}
    [/tex]

    Here rho is the rest-frame energy density, p the isotropic rest-frame pressure, and U the four-velocity.


    3. The attempt at a solution

    I'm not really sure how to approach this problem. I would assume "smooth over" means average, so the only thing I can think of trying is

    [tex]
    \Delta s = \int \sqrt{\eta_{\mu \nu} \frac{dx^\mu}{d \lambda} \frac{dx^\nu}{d \lambda}} d \! \lambda,
    [/tex]

    [tex]
    T_{\mu \nu} = \frac{1}{\Delta s} \int \sum_a \frac{p_\mu^{(a)}p_\nu^{(a)}}{p^{0(a)}} \delta^{(3)}(\vec{x}-\vec{x}^{(a)}) \sqrt{\eta_{\mu \nu} \frac{dx^\mu}{d \lambda} \frac{dx^\nu}{d \lambda}} d \! \lambda,[/tex]

    but I'm not sure how to proceed with this integral, or if this is even the right approach. Can someone help me figure out how to approach this problem?
     
  2. jcsd
  3. Jun 5, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jun 7, 2016 #3
    I wanted to bump this one more time. Really don't know how to go about doing this.
     
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