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How to reduce fermi-dirac to maxwell-boltzmann in a solid?

  1. Mar 21, 2010 #1
    For indistinguishable particles we use fermi-dirac(FD) or bose-einstein(BE), and for distinguishable we use maxwell-boltzmann(MB).For the distinguishable case our prof gave us the example of atoms in solid, because the positions of the atoms are fixed, so they are distinguishable, thus satisfy MB statistics.
    But the so called "fixed" i think is just extremely narrow wave packets with very small overlaps, so in the strict sense atoms in solid are still indistinguishable, then how can we reduce FD or BE to MB, from the condition "wave packets are narrow with small overlaps".
    Like in the treatment of dilute gas, when we assume the occupation number is much smaller than the number of degeneracies for each energy level, from the math FD and BE reduce to MB nicely. So it puzzled me whether for the atoms in solid we can find a nice way to reduce to MB.
  2. jcsd
  3. Apr 22, 2010 #2
    Well, no reply for a long time...Maybe I did not state the question clear enough, let me try again:
    For atoms in solid, the overlaps between wavefunctions are so small that we can treat them as distinguishable, hence the MB statistics.
    But let's take a alternative point of view: atoms in solid are actually indistinguishable, so it's always correct to use FD statistics (let say they're fermions), but we said we could use MB because overlaps are small, which means that [tex]{\Omega _{MB}} \approx {\Omega _{FD}}[/tex] when overlaps are small. So how do we get "[tex]{\Omega _{MB}} \approx {\Omega _{FD}}[/tex]" from "overlaps are small"
    For reference:
    MB:[tex]{\Omega _{MB}} = \prod\limits_{j = 1}^n {\frac{{g_j^{{N_j}}}}{{{N_j}!}}} [/tex]
    FD:[tex]{\Omega _{FD}} = \prod\limits_{j = 1}^n {\frac{{{g_j}!}}{{{N_j}!\left( {{g_j} - {N_j}} \right)!}}} [/tex]
    Where gj is the degeneracy for different energies and Nj is the occupation number
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