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Gibbs distribution and Bose statistics

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Aug20-08, 02:21 PM
P: 23
For photon gas chemical potential is zero. It is because derivate of free energy with respect to number (T, V fixed) of particles is zero in equilibrium. (free energy has minimum).
I was wondering why cannot I apply this reasoning to conclude (wrong) that chemical potential is zero for any Bose gas?


When deriving Bose statistics one applies Gibbs distribution for variable number of particles (grand canonical) to set of particles in the gas which are in a given quantum state. Then apply formula for grand potential (equal to -PV), mean number of particles in given quantum state is then obtained as derivate of grand potential with respect to chemical potential. So one gets Bose distribution. But if I say now that chemical potential is equal to derivate of free energy with respect to number of particles and latter is equal to zero (when gas is in equilibrium) I get wrong conclusion. Why?
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