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Non equilibrium boson distribution function

  1. Oct 22, 2014 #1
    In statistical mechanics the boson distribution function has the well known form
    ##f = \frac{1}{e^{E/T} - 1},##
    (in the special case of zero chemical potential). As one considers the non-equilibrium variant this generalize to
    ##f = \frac{1}{e^{\frac{E}{T(1+ \Theta)}} - 1},##
    for some function ##\Theta##. Now, is there any intuitive (or rigorous) explanation of why this is the correct form for the non-equilibrium distribution?
  2. jcsd
  3. Oct 25, 2014 #2
    Given that I haven't studied NE thermodynamics, this is my attempt:

    If the expression is correct (and I haven't found a reference), it is likely correct as an approximation for small deviations from a local equilibrium. Then we can apply Onsager's relations. Onsager looked at the the entropy density s, so

    du = T ds

    when neglecting chemical potential. [ http://en.wikipedia.org/wiki/Onsager_reciprocal_relations ]

    The classical derivation of the Bose-Einstein partition statistics uses S = k lnW to identify β = 1/kT from dE = T dS (again neglecting C.P.). [ http://en.wikipedia.org/wiki/Maxwell–Boltzmann_statistics#Derivation_from_canonical_ensemble ] You have normalized k =1.

    Small perturbations in entropy density would show up as perturbations in the energy density du = T (1 + Θ) ds under the approximation.* We get E/T(1+Θ) where it was previously E/T.

    * The convenient functional form, since T = constant is a possible constraint.
    Last edited: Oct 25, 2014
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