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Magnetic susceptibility (Ising model)

  1. Apr 29, 2009 #1
    Hi,

    I'm slightly confused about how to prove that:

    [tex]\chi[/tex]=[tex]\vartheta[/tex]<M>/[tex]\vartheta[/tex]H

    is equal to...

    [tex]\chi[/tex]=(<M2>-<M>2 )/ T

    I've expressed <M> as [tex]\sum[/tex]Msexp(-E/kBT) /
    [tex]\sum[/tex]exp(-E/kBT)

    and know that E=-J[tex]\sum[/tex]SiSj-H[tex]\sum[/tex]Si

    But seem to get lost in the differentiation. I am going aobut this the correct way?
     
  2. jcsd
  3. May 1, 2009 #2

    olgranpappy

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    Homework Helper

    what is "[itex]M_s[/itex]"?

    Also, you can use "\partial", i.e., "[itex]\partial[/itex]" for partial derivatives.
     
  4. May 2, 2009 #3

    Cyosis

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    Homework Helper

    Firstly [tex]M=\sum_i S_i \neq M_s[/tex]? I can't really envision what [itex]M_s[/itex] would be at all. Secondly I personally find it easiest to solve this problem by first noticing that [tex]\langle M \rangle= k_b T \frac{\partial \log Z}{\partial h}[/tex] then compute [tex]\frac{\partial \langle M \rangle}{\partial h}[/tex].
     
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