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

Tags: gas energy, gravitational field, ideal gas, statistical physics
 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?
 Sci Advisor Thanks P: 2,094 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 $$H=\frac{\vec{p}^2}{2m}+m g z.$$ The classical probility distribution is thus given by $$P(\vec{x},\vec{p})=\frac{1}{Z} \exp[-\beta H(\vec{x},\vec{p})].$$ The mean kinetic energy for particles in a cubic box of length $L$ thus is $$\langle E_{\text{kin}} \rangle=\frac{3}{2} T$$ and the mean potential energy $$\langle m g z \rangle=m g L \left (1+\frac{1}{\exp(m g L/T)-1} \right )-T.$$

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