Deriving perfect fluid energy tensor from point particles

  • Thread starter mjordan2nd
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  • #1
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Homework Statement


[
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.

Homework Equations


[/B]
Energy-momentum tensor of a perfect fluid:

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

Here [itex]\rho[/itex] is the rest-frame energy density, p the isotropic rest-frame pressure, and [itex]U[/itex] the four-velocity.

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?
 

Answers and Replies

  • #3
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Go ahead and post your solution and we can go from there.
 

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