Question:
How does the mean kinetic energy of molecules in a relativistic fluid depend on temperature?
remember.kelly
2008-05-20 20:32:52 UTC
I answered the non relativistic case on this page:
http://answers.yahoo.com/question/index?qid=20080520111803AANm37R&r=w#TZ9qWjD8AFr6I28GKNYJ

Before plowing ahead with the present problem, make sure my previous solution is correct, which gave the same solution as known for the ideal gas. The present question concerns an interacting gas also, but you may want to consider a ideal gas if the other turns out to be too difficult.
Three answers:
?
2008-05-21 14:38:19 UTC
The case of an interacting relativistic gas is too difficult to discuss here. An ideal gas with a relativistic temperature is considered in the same way as in the non-relativistic case.



The distribution function is given by the standard expression (isotropy is assumed),



f(p) dp = C p^2 exp(-E/T) dp.



Here, E is the energy, C is constant, and dp is a scalar. The difference with the non-relativistic case is the relation between the energy E and the momentum p:



E = √ (p^2 c^2 + m^2 c^4).



One can rewrite f in terms of the Lorentz gamma-factor, γ = 1/√(1-v^2), p = mc v γ (velocity is in units of c):



f(γ) d γ = C γ^2 v exp(-γ/θ) d γ,



where



θ=T/mc^2.



The distribution function is normalized by the condition ∫ f d γ=1 (to avoid nonessential constants, I take a unit particle density). To calculate the integral, make the substitution



γ = cosh(x), v= √(γ^2-1)/γ = tanh(x).



Then



γ^2 v d γ = cosh(x) sinh^2(x) =

cosh^3(x) - cosh(x) = (1/4) (cosh(3x)-cosh(x)).



Use the integral formula for the Bessel function K_n(z):



K_n(z) = exp( -z cosh(x) ) cosh(n x) dx,



the integration limits are from 0 to infinity, and apply the recursion relation for the Bessel function



(2n/z) K_n (z)= K_{n+1} (z) - K_{n-1}



with n=2. This gives



C = 1/( θ K_2(1/θ) ).



The kinetic energy is equal to W = mc^2 (γ-1),

and the average kinetic energy is



= mc^2 C ∫ (γ-1) γ^2 v exp(-γ/θ) d γ



The first integral contains the factor



γ^3 v d γ = cosh^2(x) sinh^2(x) =

(1/4) sinh^2(2x) = (1/8) (cosh (4x) - 1).



Using above formulas, after some algebra we get



= mc^2 ( (3/4) K_3 - K_2 + (1/4) K_1 ) / K_2.



Arguments of Bessel functions K_n are (1/θ).



Check the low temperature limit, θ << 1. Asymptotics of Bessel functions at z>>1 are



K_n (z) -> √(π/2z) exp(-z) [ 1 + (4n^2-1)/(8z) ].



In this limit, -> mc^2 (θ/8) [ (3/4)*35 - 15 +(1/4)*3 ] =

(T/8) [ 27 - 15] = 12 T/8 = 3 T/2 ,

which coincides with the non-relativistic case.
Charles M
2008-05-21 11:44:33 UTC
Any discussion of relativistic fluid dynamics must include the stress-energy-momentum tensor Tμν. It is related to the Einstein Tensor thus:

Tμν = 8πGμν

Without a detailed discussion of tensor mechanics over Riemannian manifolds, you can at best merely brush the subject of relativistic fluid mechanics. I'm not sure what level you are at, but this requires post graduate level physics gymnastics to be done right.



Here is an excellent source on relativistic fluids:

Amsden, A.A., Harlow, F.H., and Nix, J.R., “Relativistic Nuclear Fluid Dynamics”, Phys. Rev. C, 15, 2059–2071, (1977).
Frst Grade Rocks! Ω
2008-05-21 05:07:39 UTC
I will star this and hopefully it will attract the attention of either Zo Maar or PTG. They and a few other irregulars have the the knowledge and the skill to handle it. I ain't gotta prayer.



I bet that someone working either with plasma in the interior of a star or in laser fusion, or looking at the interior of a neutron star may have solved it.



It would also be interesting to think about the boltzman distribution and how relativistic speeds would effect it.



Best regards


This content was originally posted on Y! Answers, a Q&A website that shut down in 2021.
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