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Boltzmann Entropy for micro state or macro state?

  1. Dec 21, 2012 #1
    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. jcsd
  3. Dec 21, 2012 #2


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    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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