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Homework Help: Fermi gas

  1. Mar 1, 2008 #1
    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?
     
  2. jcsd
  3. Mar 2, 2008 #2

    olgranpappy

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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
    [tex]
    \mu \approx E_{\rm Fermi}(1 - O((T/E_{\rm Fermi})^2)
    [/tex]
     
  4. Mar 2, 2008 #3
    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)
     
    Last edited: Mar 2, 2008
  5. Mar 2, 2008 #4

    olgranpappy

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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.
     
  6. Mar 2, 2008 #5
    what happens to the chemical potential as T increases to T>E_F?
     
  7. Mar 3, 2008 #6

    olgranpappy

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    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
    [tex]
    e^{-(E-\mu)/T}
    [/tex]
    which can result from the fermi (bose) distribution
    [tex]
    \frac{1}{e^{(E-\mu)/T}\pm 1}
    [/tex]
    if \mu is negative and large in magnitude. I.e., [itex]e^{|\mu|/T}>>1[/itex].
     
  8. Mar 4, 2008 #7
    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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