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Boltzmann Entropy for micro state or macro state? 
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#1
Dec2112, 08:36 AM

P: 9

From theory, we know that Boltzmann entropy for a given distribution, defined through a set of occupancy numbers {ni}, of the macrostate M, is given by:
S=k log(Ω{ni}) where omega is the number of microstates for the previously given set of occupancy number, {ni} . Assuming that the system is in equilibrium, we get omega to be predominantly the number of microstates which fill up the entire 6ND gamma space. Using counting principles in the 6 dimensional mu space we get omega to be equal to product(1/[factorial(ni)]). My question is would interchanging particle labels (such that {ni} does not change) result in a new microstate? If yes, it means that particles are no longer indistinguishable. If no, then entropy becomes zero since there is just 1 microstate for the given {ni} in gamma space. I would appreciate if anyone clears this doubt. 


#2
Dec2112, 09:08 AM

P: 789

I think energy does not completely define a microstate. Momentum does. So each energy state has a number of momentum states and these particles are distinguishable by their momentum (but again, not by any "identity"). You have to take this degeneracy into account.



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