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Degeneracy of Fermionic gas

  1. Dec 3, 2016 #1
    1. The problem statement, all variables and given/known data
    The question asked me to show that the entropy of a fermionic gas is
    ##S = -k_B \Sigma_i (1-f_i)\ln(1-f_i) +f_i\ln(f_i)##

    Using the Fermi-Dirac distribution so ##f_i = \frac{n_i}{g_i}##.

    2. Relevant equations


    3. The attempt at a solution
    The number of microstates ##\Omega## is
    ##\Omega =## Π##\frac{g_i!}{n_i!(g_i-n_i)!}## and using ##S = k_B ln(\Omega)## I've arrived at the expression:
    ##S = -k_B \Sigma_i g_i (1-f_i)\ln(1-f_i) +f_i\ln(f_i)##
    So there's an unwanted factor of ##g_i##. I'm told it is meant to be there, and I need to explain why it can be set to 1. Something to do with i being the state index and a quantum state can't be degenerate. Could someone explain that to me? Any help is much appreciated! :)
     
  2. jcsd
  3. Dec 6, 2016 #2
    Hi Kara386

    First observation here is that you need some big brackets in your equations - for example in your first equation, the [itex]k_B[/itex] should be multiplying everything, not just the first term.

    The business with the [itex]g_i[/itex] is just bookkeeping. In their formula they're using the index [itex]i[/itex] to label different states, and in your formula you're using the index [itex]i[/itex] to label different energies. Your formula includes a [itex]g_i[/itex] factor in it to allow for the possibility that there might be lots of states with the same energy.
     
  4. Dec 6, 2016 #3
    Thanks! I thought that might be it, because although originally the sum was over multiple states taken to be at the same energy, I think the introduction of the average ##f_i## allows the sum to be over individual states.

    Thanks for your help!
     
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