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Arithmetic mean Fermi Dirac & Bose Einstein

  1. Apr 29, 2015 #1
    Hi everybody, I was doing one asignment form class, I was tasked to prove that in one system, the arimetic mean of FD and BE distributions is equal to MB's distribution for undishtingable particles.
    After doing the numbers I found out that it actually was, but I don't know why this happens, can someone explain this to me or point me to a place where I can learn that?
    Thanks.
     
  2. jcsd
  3. Apr 29, 2015 #2

    ChrisVer

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    Gold Member

    I think this is a demonstration that the quantum mechanical averaged values behave classically.
    Mind that the classical analogue of the Bose Einstein/Fermi Dirac distributions is the Maxwell Boltzmann distribution (give the number density as a function of the energy of the system).
     
  4. Apr 29, 2015 #3

    mfb

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    For all temperatures? That would surprise me. Fermi-Dirac includes particles at an energy way above the energies Maxwell-Boltzmann at low temperatures would give, and I don't see how MB would reproduce the step coming from Fermi-Dirac at the Fermi energy.
     
  5. May 2, 2015 #4
    Well, to be honest I have not advanced that much in theory lessons. All I know is that I have a system with two particles and three possible enrgies for them and I had to study the partiton function for each of the three distributions, I checked that the math were right and saw that it acturally worked so I asked the quastion.

    By the way thank you very much for your anwsers.
     
  6. May 2, 2015 #5

    mfb

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    Staff: Mentor

    Ah, for only two particles it might look different.
    I'm not convinced that it works out for all temperatures, however. In the limit of large temperature: sure.
     
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