Boltzmann Entropy for micro state or macro state?

  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. 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.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook