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Pressure of an electron gas at 0K

  1. Apr 19, 2008 #1
    What is the pressure exerted by a Fermi electron gas at 0K? I know that from Pauli's exclusion principle there must be a non zero pressure even at 0K, but what is the quantitative relation between this pressure and the electron concentration? How do I go about to derive this?
    Last edited: Apr 19, 2008
  2. jcsd
  3. Apr 19, 2008 #2


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    You mean you want to derive the Electron Degeneracy Pressure?
  4. Apr 19, 2008 #3


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    By electron gas at 0K, does one infer the electrons are not moving, and there positive charges to neutralize, or are the electons packed (which doesn't seem like much of a gas)?

    I would expect a fair amount of coulomb repulsion on whatever unit cell.

    Pressure is also equivalent to energy density.
  5. Apr 19, 2008 #4

    Ken G

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    You are talking about electron degeneracy pressure. What you do is, use the density to determine the volume of the "box" each electron occupies (1/density). Now do a quantum mechanical "particle in a box" calculation to find the energy of that electron. Now take the derivative with respect to volume of the energy-- that's the pressure.

    Physically, what is happening here is that even at 0K, the electron wave function has a second derivative so that it can stay inside the box of volume allowed to each electron (to keep them from infringing on the states of the other Fermions). That second derivative (a la Schroedinger) implies the presence of kinetic energy, even at 0K. The presence of kinetic energy means there is momentum flux, and where there's momentum flux, there's pressure.
  6. Apr 19, 2008 #5
    Thanks Ken G.. I have been able to find the electron gas pressure as a function of the electron concentration using the approach tht u mentioned.
  7. Apr 19, 2008 #6
    I'm assuming this isn't homework =)
    Have a look at my derivation here: http://www.physics.thetangentbundle.net/wiki/Statistical_mechanics/Fermi_gas [Broken]
    and let me know if anything needs clarification.
    Last edited by a moderator: May 3, 2017
  8. Apr 19, 2008 #7

    Ken G

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    Yes, this is the exact approach, involving a lengthy yet suitably elegant calculation that even includes arbitrary (half-integer) spins. To be clear, my description above is heuristic, intended to approximate the physics simply and with the right scalings, moreso than getting an accurate answer. The approximate way can almost be done in your head if you know the exact particle-in-the-box solution for a 1/density sized box, and you get that the pressure will come out roughly 2/3 of the density times the Fermi energy, whereas lbrits' exact approach comes out 2/5 of the density times the Fermi energy. So I'm about 5/3 high, expressly because my way does not involve ever calculating a Fermi energy. You get what you pay for.

    Physically, what this all means is you get about a 40% reduction in the ground state energy if you let all the electrons spread over the whole volume (say the whole white dwarf star), but stack them to higher and higher energies as the lower states fill, rather than just have them all be excited to identical energies in the ground state of their own little boxes. That comparison makes for a cute problem for anyone bored with the standard degeneracy pressure fare!
    Last edited by a moderator: May 3, 2017
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