# Arithmetic mean Fermi Dirac & Bose Einstein

1. Apr 29, 2015

### Frank Einstein

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. Apr 29, 2015

### ChrisVer

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).

3. Apr 29, 2015

### Staff: Mentor

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.

4. May 2, 2015

### Frank Einstein

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.

5. May 2, 2015

### 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.