Register to reply

Gibbs distribution and Bose statistics

Share this thread:
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?
Phys.Org News Partner Physics news on
Technique simplifies the creation of high-tech crystals
Working group explores the 'frustration' of spin glasses
New analysis of oxide glass structures could guide the forecasting of melt formation in planetary interiors

Register to reply

Related Discussions
Fermi and Bose statistics... Quantum Physics 5
Canonical Bose-Einstein statistics Quantum Physics 3
Bose-Einstein Statistics Advanced Physics Homework 4
Is there a proof for Fermi and Bose statistics? Quantum Physics 14
Gravity and Bose-Einstienian statistics Quantum Physics 0