- #1

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For non-relativistic, the usual value is E=3/2 kT for three degrees of freedom.

For ultra-relativistic, it is E=3kT.

Does anyone have a mathematical explanation for that, or a link to a derivation?

Thanks.

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- Thread starter Piano man
- Start date

- #1

- 75

- 0

For non-relativistic, the usual value is E=3/2 kT for three degrees of freedom.

For ultra-relativistic, it is E=3kT.

Does anyone have a mathematical explanation for that, or a link to a derivation?

Thanks.

- #2

- 988

- 178

[itex]<X> = \frac{\int X e^{-E/kT} d^3 p}{\int e^{-E/kT} d^3 p}[/itex]

Energy E, momentum p, etc.

You can do this integral by going into spherical coordinates for the momentum, integrating out the angles, and using the appropriate expression for the total energy. In the general case, it's rather complicated, but in the nonrelativistic and ultrarelativistic limits, it's easier.

With the help of page 376 of Abramowitz and Stegun - Page index, I was able to find the appropriate integrals in the general case. For that, one substitutes

p = m*sinh(u)

E = m*cosh(u)

and then integrates with (9.6.24), giving

[itex]<E> = m\frac{K_4(x) - K_0(x)}{2(K_3(x) - K_1(x))}[/itex]

where x = m/kT and the K's are modified Bessel functions of the second kind. A&S also gives some formulas for the K's for the large and small limits of x.

- #3

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- 178

[itex]P_{ij} = n <v_i p_j> = n<\frac{p_i p_j}{E}>[/itex]

for number density n, which becomes

[itex]P_{ij} = P \delta_{ij} ,\ P = n<\frac{p^2}{3E}>[/itex]

One can solve this case by integration by parts:

[itex]\int e^{-E/kT} \frac{p^2}{3E} p^2 dp = \frac13 \int e^{-E/kT} p^3 dE = kT \int e^{-E/kT} p^2 dp[/itex]

yielding P = nkT as the general solution.

I've been working in the classical limit, of course.

You can find more details at Partition function (statistical mechanics) (Wikipedia)

- #4

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Thank you very much. That gives me something to work from.

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