Why is the gas energy in a g-field 5/2KNkT ?

by LeakedPaladin
Tags: gas energy, gravitational field, ideal gas, statistical physics
LeakedPaladin is offline
Mar19-13, 04:46 PM
P: 1
I am messing around with classical gasses using a demon algorithm, trying to reproduce the behavior of ideal gasses in various situations. Exitingly, this is working very well. Here are my findings so far:

Classical gas, 3D : E=1.497 NkT - the usual 3/2 NkT result

Classical gas, 2D : E=0.989 NkT - 2 DoF, so 2/2 NkT

Relativistic gas, 3D : E = 2.992 NkT - as expected, can be derived from equipartition

Classical gas, uniform gravitational field, 3D : E = 2.486 NkT

Now the last one quite surprises me. Why is that he case? Intuitevely I would expect 3 DoF on |p| and 3 DoF on x, so 6*1/2 NkT = 3NkT. Is there a hidden constraint?
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vanhees71 is offline
Mar20-13, 04:45 AM
Sci Advisor
P: 2,150
The equipartition theorem holds true only for phase-space degrees of freedom that enter quadratically in the Hamiltonian. For an ideal gas in a homogeneous gravitational field you have for the single-particle Hamiltonian
[tex]H=\frac{\vec{p}^2}{2m}+m g z.[/tex]
The classical probility distribution is thus given by
[tex]P(\vec{x},\vec{p})=\frac{1}{Z} \exp[-\beta H(\vec{x},\vec{p})].[/tex]
The mean kinetic energy for particles in a cubic box of length [itex]L[/itex] thus is
[tex]\langle E_{\text{kin}} \rangle=\frac{3}{2} T[/tex]
and the mean potential energy
[tex]\langle m g z \rangle=m g L \left (1+\frac{1}{\exp(m g L/T)-1} \right )-T.[/tex]

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