# Magnetic susceptibility (Ising model)

Hi,

I'm slightly confused about how to prove that:

$$\chi$$=$$\vartheta$$<M>/$$\vartheta$$H

is equal to...

$$\chi$$=(<M2>-<M>2 )/ T

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

and know that E=-J$$\sum$$SiSj-H$$\sum$$Si

But seem to get lost in the differentiation. I am going aobut this the correct way?

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olgranpappy
Homework Helper
Hi,

I'm slightly confused about how to prove that:

$$\chi$$=$$\vartheta$$<M>/$$\vartheta$$H

is equal to...

$$\chi$$=(<M2>-<M>2 )/ T

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

and know that E=-J$$\sum$$SiSj-H$$\sum$$Si

But seem to get lost in the differentiation. I am going aobut this the correct way?
what is "$M_s$"?

Also, you can use "\partial", i.e., "$\partial$" for partial derivatives.

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