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Magnetisation and susceptibility

  1. Nov 30, 2007 #1


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

    1. State a relationship between the free energy, F, and the magnetisation, M.

    2. State a partition function for the case of a system of N independent spin-1/2 paramagnets in a field, and derive an expression for its susceptibility.

    3. The attempt at a solution

    (1) Looking in my notes, I see that [tex]dF = - MdH - SdT[/tex] (*). It follows, then, that

    [tex]M = \left(-\frac{\partial F}{\partial H}\right)_{T}[/tex]

    That's the question answered, but I wonder how one arrives at this expression for F.

    I know that [tex]F = U - TS[/tex] and [tex]dW = - HdM[/tex], so it would seem that

    [tex]dF = TdS - dW - TdS - SdT = -dW - SdT = HdM - SdT[/tex]

    ie. not [tex]MdH[/tex]. Could someone show me how (*) is derived?

    (2) For the whole system, I find that

    [tex] Z = 2^{N}cosh^{N}(\beta\mu H) [/tex] ([tex]H[/tex] is my field, [tex]\mu[/tex] is the Bohr magneton )

    I thus find the free energy to be

    [tex]F = -kTN.log[2cosh(\beta\mu H)][/tex]

    And thus, by the relationship stated above,

    [tex]M = N\mu tanh(\frac{\mu H}{kT})[/tex]

    The susceptibility, I presume, is [tex]\frac{\partial M}{\partial H}\right)[/tex]. It comes out for me as

    [tex]\frac{N\mu^{2}}{kT} sech^{2}(\frac{\mu H}{kT})[/tex]

    (which, I find, tends to [tex]\frac{N\mu^{2}}{kT}[/tex] in the limit of low field or high temperature.

    Does that seem sensible?

    Last edited: Nov 30, 2007
  2. jcsd
  3. Nov 30, 2007 #2


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    Your post contains some confusion. First, your equation (*) for Helmholtz free energy should be written with dF rather than F.

    Here's a quick trick that may be helpful for treating the magnetization. How can we decide without derivation whether dW is HdM or MdH? Well, the differential energy dF is extensive (depends on volume), and only HdM depends differentially on a volume-dependent quantity M. Accordingly

    [tex]dF=HdM-SdT.[/tex] This should replace your (*).

    From this we can say that [tex]H=(\partial F/\partial M)_{T}[/tex].

    To get M, we need to define a new quantity, usually called Gibbs free energy,

    [tex]G=F-HM.[/tex] By Legendre transformation,



    [tex]M=-(\partial G/\partial H)_T[/tex].

    So I don't get your result for part 1...

    EDIT: oops, corrected signs
    Last edited: Nov 30, 2007
  4. Nov 30, 2007 #3


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    Sorry. That was a typo.

    Hmmm... I find in 'Statistical Mechanics, A Survival Guide', by Glazer and Wark, that the result [tex]dF = - MdB - SdT[/tex] (the one I have written down) is stated as a known result on pg. 34 (without proof: "from our knowledge of thermodynamics we know that for a magnetic system..."), followed by the expression

    [tex]M = \left(-\frac{\partial F}{\partial B}\right)_{T}[/tex]

    (I'm using H, of course, rather than B)

    But I'm not sure how they arrived at that expression for dF...
    Last edited: Nov 30, 2007
  5. Nov 30, 2007 #4


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    Last edited: Nov 30, 2007
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