Why is the gas energy in a gfield 5/2KNkT ?by LeakedPaladin Tags: gas energy, gravitational field, ideal gas, statistical physics 

#1
Mar1913, 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? 



#2
Mar2013, 04:45 AM

Sci Advisor
Thanks
P: 2,150

The equipartition theorem holds true only for phasespace degrees of freedom that enter quadratically in the Hamiltonian. For an ideal gas in a homogeneous gravitational field you have for the singleparticle 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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