Chemical Potential in a Degenerate Fermi Gas

indigojoker
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in a Fermi gas, we know that when the temperature is much less than the Fermi energy, it becomes a degenerate gas. does this mean the chemical potential of the system be very large?
 
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the chemical potential is very close to the fermi energy since the temperature is very much less than the fermi energy
<br /> \mu \approx E_{\rm Fermi}(1 - O((T/E_{\rm Fermi})^2)<br />
 
where did you get the formula for the chemical potential?

So the chemical potential becomes a large negative number as temperature increases? I am trying to show that at high temperatures, the chemical potential is the same as an ideal gas.

(i am considering the 2-d case)
 
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that formula is for T << E_F
which is always the case for a metal (since all metals melt well before T=E_F)...

it is derived, for example, in Ashcroft and Mermin "Solid State Physics" chapter 2. See, Eq. 2.77.
For the 2d case see A+M chapter 2 problem number 1. for the classical limit see A+M chapter 2 problem 3.
 
what happens to the chemical potential as T increases to T>E_F?
 
for high temperatures the system will be a gas. if the temperature is high enough it will be a classical gas for which the Boltzmann distribution will hold--i.e., for either fermions or bosons the mean occupation number is very low and proportional to
<br /> e^{-(E-\mu)/T}<br />
which can result from the fermi (bose) distribution
<br /> \frac{1}{e^{(E-\mu)/T}\pm 1}<br />
if \mu is negative and large in magnitude. I.e., e^{|\mu|/T}&gt;&gt;1.
 
Indigojoker: as far as I know, there is no analytic expression for the chemical potential of a non-interacting fermi gas. I remember doing this derivation at some point, and I think I went via the canonical partition function: F=kT ln Z, and \mu=dF/dN. It's not possible to evaluate the expression directly, but you should be able to show that in the high T limit it would tend to the same form as the ideal gas.
 
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