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Magnetic susceptibility (Ising model)

  • Thread starter garyman
  • Start date
  • #1
19
0
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?
 

Answers and Replies

  • #2
olgranpappy
Homework Helper
1,271
3
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?
what is "[itex]M_s[/itex]"?

Also, you can use "\partial", i.e., "[itex]\partial[/itex]" for partial derivatives.
 
  • #3
Cyosis
Homework Helper
1,495
0
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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