Register to reply 
Gibbs distribution and Bose statistics 
Share this thread: 
#1
Aug2008, 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? 


Register to reply 
Related Discussions  
Fermi and Bose statistics...  Quantum Physics  5  
Canonical BoseEinstein statistics  Quantum Physics  3  
BoseEinstein Statistics  Advanced Physics Homework  4  
Is there a proof for Fermi and Bose statistics?  Quantum Physics  14  
Gravity and BoseEinstienian statistics  Quantum Physics  0 