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Fermi energy of non relativistic electrons

  1. Feb 18, 2006 #1
    we have non relativistic electrons at absolute zero. we need to show that the total energy is 3N*Ef/5 where Ef is the Fermi energy. this calculation I can do. we then need to show that the Pressure is equal to 2U/3V where U is the total energy. I did this by noting the dU=TdS - PdV and at absolute zero, S=0 therefore dS=0, so P = -dU/dV. using the explicit expression for U in terms of V you can get the required answer easily. I originally tried deriving this by finding the partition function for the fermions explicitly, then finding the helmholtz free energy, and then finding the pressure from this, using e.g
    however this method did not appear to work (it gave me very messy mathmatical expressions that did not look right). I'm not too sure why you can't get the answer this way. Is it because the equations for U,S,F in terms of the partition function only apply to Boltzmann statistics, and not quantum stats?

  2. jcsd
  3. Feb 18, 2006 #2

    Physics Monkey

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    You definitely can get the answer by directly using the partition function. In fact, it is possible to prove for any T that [tex] P = 2U/3V [/tex] for a free Fermi gas. I don't know what exactly you're referring to in terms of your messy math expressions, but at [tex] T=0 [/tex] the partition function is a fairly simple object for free fermions. Perhaps you could be more specific about your trouble with the mathematics?
  4. Feb 18, 2006 #3


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    Just to add a little side note here :

    Sachi, the partition function for non-interacting electrons is the same as that for the ideal gas. In 3-dimensions (for a non-interacting gas), the partition function goes like the volume times the cube of a Gaussian integral (in the limit of large Fermi energy). Perhaps, if you can't find where your error is, you could look back into your derivations for the ideal gas equation of state and specific heat.
  5. Feb 21, 2006 #4
    the last post threw up some very interesting issues. for fermions at absolute zero we note that n(E)dE is equal to g(E)dE multiplied by two (to take into account the two different spins), where g(E)dE is the density of states. the problem with my initial method was that the integral goes from E=0 to Ef, where Ef is the Fermi energy, and this limit is itself a function of the Volume, so we have to explicitly evaluate the integral by doing integration by parts and stick in our expression for Ef at every stage which gets messy. the idea about assuming that Ef= infinity is interesting, because it would make the maths much easier, but it's not clear to me why this assumption is valid. there is also the problem that once the partition function has been derived (and is presumably similar to the one for a perfect gas), how do we convert the partition function into a pressure, as I'm pretty sure that the expressions for U,F,S e.g F=-NKbTln(Zn) only apply to classical and not Quantum stats.
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