Deriving perfect fluid energy tensor from point particles

Click For Summary
SUMMARY

The discussion focuses on deriving the perfect fluid energy-momentum tensor from a system of discrete point particles. The energy-momentum tensor is expressed as T_{\mu \nu} = \sum_a \frac{p_\mu^{(a)}p_\nu^{(a)}}{p^{0(a)}} \delta^{(3)}(\vec{x}-\vec{x}^{(a)}) and is related to the perfect fluid tensor T^{\mu \nu} = (\rho + p)U^\mu U^\nu + p \eta^{\mu \nu}. Participants explore the method of averaging particle contributions to achieve a smooth representation of the energy-momentum tensor, with references to previous solutions and discussions on the approach to the integral involved.

PREREQUISITES
  • Understanding of energy-momentum tensors in general relativity.
  • Familiarity with four-velocity and its role in relativistic physics.
  • Knowledge of delta functions and their application in physics.
  • Basic calculus, particularly integration techniques in the context of physics.
NEXT STEPS
  • Study the derivation of the perfect fluid energy-momentum tensor in detail.
  • Investigate the role of isotropic velocity distributions in fluid dynamics.
  • Explore the mathematical properties of delta functions in physics.
  • Review previous solutions on similar topics, such as those found on Physics Stack Exchange.
USEFUL FOR

Physicists, graduate students in theoretical physics, and researchers interested in the foundations of fluid dynamics in the context of general relativity.

mjordan2nd
Messages
173
Reaction score
1

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


[/B]
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



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?
 
Physics news on Phys.org
Go ahead and post your solution and we can go from there.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 38 ·
2
Replies
38
Views
1K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
689
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K