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

  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. jcsd
  3. vanhees71

    vanhees71 3,329
    Science Advisor
    2014 Award

    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]
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

Draft saved Draft deleted