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center o bass
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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?
##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?