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Homework Help: Magnetisation due to conduction electrons

  1. Dec 11, 2007 #1


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    1. The problem statement, all variables and given/known data

    Due to its spin, the electron possesses a magnetic moment μB.

    Treating the conduction electrons in a metal as a free electron gas, obtain an expression for the magnetization due to the magnetic moments of the conduction electrons, when placed in a magnetic field. Evaluate this expression at zero temperature.

    3. The attempt at a solution

    Ok. I have some hints for this Q, but they're confusing me.

    My vague guess would have been to try and calculate the number of electrons in the higher energy state (which, I think, would be the ones aligned with the field) and multiply by [tex]\frac{\mu_{\beta}}{V}[/tex] to obtain the overall magnetisation (the magnetic moment per unit volume).

    The hint I have here is that

    [tex]N_{\pm} = \frac{1}{2}\int_{0}^{\epsilon_{F}+\mu_{B}H} \rho(\epsilon) d\epsilon = \frac{4\pi V}{3h^3} (2m)^{3/2} (\epsilon_{F}\pm\mu_{B}H)^{3/2}[/tex]

    All this seems to be doing (to a beginner) is summing the possible states, not obtaining the actual number of either spin parallel/anti-parallel electrons (ie. occupants of states). And presumably the no. of states is vast. If we took the N+ version of that, I take it we'd be getting the no. of states up to the energy level of the electrons with the higher spin energy (and thus including the states of electrons with lower spin energy); if we took the N- version, just the lower energy states. [tex]N_{+} - N_{-}[/tex] would then, I presume, give us the number of states of the higher energy spins (most of which would be unoccupied).

    I am not sure why [tex]\rho(\epsilon)[/tex] has been written like this. By thinking about points in a positive octant (derivation not given here) one arrives at a no. of points

    [tex]G(\epsilon) = \frac{4\pi V}{3h^{3}}(2m\epsilon)^{3/2}[/tex]

    which can be expressed (I'll switch to his rho here)

    [tex]\rho(\epsilon) d\epsilon = \frac{dG(\epsilon)}{d\epsilon} = \frac{4\pi V}{h^{3}}(2m\epsilon)^{1/2} d\epsilon[/tex]

    [tex]\rho(\epsilon) d\epsilon[/tex], I take it, gives us the density of states. Surely we need some integral that multiplies that by a distribution function ([tex]e^{-\beta\epsilon}[/tex]?) in order to state the no. of states up to some energy [tex]\epsilon[/tex]

    Well, I'm a bit confused (isn't it obvious) and not really sure what I'm doing. If someone could shed light on this question and show me how to put this together, I think it would open up some of the other problems on the sheet. (I haven't seen any examples of this sort of thing, unfortunately).

    Last edited: Dec 11, 2007
  2. jcsd
  3. Dec 11, 2007 #2


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    There is a derivation of this (I think) in Kittel, Pauli magnetism.
  4. Dec 11, 2007 #3


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    Also, if you want an internet resource, try the online lecture notes 20 and 21 from http://spider.pas.rochester.edu:8080/phy418S05/lectures" [Broken] which is about Pauli paramagnetism of a free en gas.
    Last edited by a moderator: May 3, 2017
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