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FermiDiracBoseEinsteinBoltzman derivationby Morgoth
Tags: derivation 
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#1
Jul112, 09:37 AM

P: 127

I am having hard times, trying to find out how the FermiDirac and BoseEinstein distributions give you at a limit the Boltzmann's one.
Let's see the FermiDirac one: <n_{i}>= 1/ { 1+ e^{[β(εiμ)] } } where β=1/kT, where T:Temperature and k the Boltzmann's constant. As we know the limits from quantum to classical physics for these are either at high Temperatures (so T→∞ So β→0) or low densities (n<<n_{Q}=(2πmkT/h^{2})^{3/2}). So I am trying to put on FermiDirac's distribution the limit β→0. I just want you to reconfirm my work: I multiplied on numerator and denominator with e^{[β(εiμ)] } getting: e^{[β(εiμ)] } / (e^{[β(εiμ)] }+1) Now again for β→0 I get e^{[β(εiμ)] } /2 which is half what I want to get.... 


#2
Jul112, 02:19 PM

Sci Advisor
Thanks
P: 4,160

Morgoth, If you only let T → ∞, then β → 0 and <n> → 1/2 = const, a uniform distribution which is what you got, and of course is not the classical limit. To get the classical limit you must take high temperature and low density. This is most easily done in terms of the fugacity, z = e^{βμ}.
<n_{i}> = z / (z + e^{βεi}). Let z approach 0, and then you get the MaxwellBoltzmann distribution, <n_{i}> = z e^{βεi} 


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