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Gibbs distribution and Bose statistics

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ala
#1
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?

Explanation:

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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