Chemical Potential in a Degenerate Fermi Gas

[indigojoker]

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?

 

[olgranpappy]

the chemical potential is very close to the fermi energy since the temperature is very much less than the fermi energy
$$\mu \approx E_{\rm Fermi}(1 - O((T/E_{\rm Fermi})^2)$$

 

[indigojoker]

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)

 

[olgranpappy]

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.

 

[indigojoker]

what happens to the chemical potential as T increases to T>E_F?

 

[olgranpappy]

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
$$e^{-(E-\mu)/T}$$
which can result from the fermi (bose) distribution
$$\frac{1}{e^{(E-\mu)/T}\pm 1}$$
if \mu is negative and large in magnitude. I.e., $e^{|\mu|/T}>>1$.

 

[genneth]

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.