# Deriving perfect fluid energy tensor from point particles

## Homework Statement

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

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

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

Energy-momentum tensor of a perfect fluid:

$$T^{\mu \nu} = (\rho + p)U^\mu U^\nu + p \eta^{\mu \nu}$$

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

## The Attempt at a Solution

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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

$$\Delta s = \int \sqrt{\eta_{\mu \nu} \frac{dx^\mu}{d \lambda} \frac{dx^\nu}{d \lambda}} d \! \lambda,$$

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

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

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