Magnetic susceptibility integral trouble

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SUMMARY

The discussion focuses on deriving the magnetic susceptibility integral, specifically the equation for susceptibility, Vχ₀. The user presents their calculations involving the partition function Z, magnetic moment m, and the external magnetic field H, ultimately aiming to express the susceptibility as Vχ₀ = μ₀²β²⟨mᵢ²⟩ - ⟨mᵢ⟩²μ₀β. They encounter an issue with an extra factor of μ₀β in their calculations, which prevents them from achieving the expected result. A correction is suggested regarding the interpretation of the integral involving ⟨mᵢ²⟩.

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maximus123
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Hello,

This is really more of an algebra question. Here is the main integral I am using for the susceptibility

V\chi_0=\frac{d\langle \vec{m_i}\rangle}{d\vec{H_i}}=\frac{1}{Z}\int\vec{m_i}\frac{d}{d\vec{H_i}}e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m-\frac{1}{Z^2}\frac{dZ}{d\vec{H_i}}\int\vec{m_i}e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m
Z is the partition function, m is the magnetic moment, H is the field, beta is just a constant in these calculations.

I have it that

\langle\vec{m_i}\rangle=\frac{1}{\mu_0\beta}\frac{1}{Z}\frac{dZ}{d\vec{H_i}}

And ultimately I need to show that the susceptibility equation can be expressed as


V\chi_0=\mu_0\beta(\langle\vec{m_i}^2\rangle-\langle\vec{m_i}\rangle^2)
I can get pretty close. If we look at the first term in the susceptibility equation at the top of the post


<br /> \frac{1}{Z}\int\vec{m_i}\frac{d}{d\vec{H_i}}e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m\\<br /> =\frac{1}{Z}\int\mu_0\beta\vec{m_i^2}\,e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m\\<br /> =\mu_0\beta\frac{1}{Z}\int\vec{m_i^2}\,e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m\\<br /> =\mu_0^2\beta^2\langle\vec{m_i}^2\rangle\\<br /> <br />
The last line simplification was achieved using the relation given at the top of the post (second equation down).
The second term then

<br /> \frac{1}{Z^2}\frac{dZ}{d\vec{H_i}}\int\vec{m_i}e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m\\<br /> =\frac{1}{Z}\frac{1}{Z}\frac{dZ}{d\vec{H_i}}\int\vec{m_i}e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m\\<br /> =\frac{1}{Z}\frac{1}{Z}\frac{dZ}{d\vec{H_i}}\frac{dZ}{d\vec{H_i}}\frac{1}{\mu_0\beta}\\<br /> =\frac{1}{Z}\frac{dZ}{d\vec{H_i}}\langle\vec{m_i}\rangle\\<br /> =\langle\vec{m_i}\rangle^2\mu_0\beta<br />
Giving a final result for the susceptibility as

<br /> V\chi_0=\mu_0^2\beta^2\langle\vec{m_i}^2\rangle-\langle\vec{m_i}\rangle^2\mu_0\beta<br />

So basically I have an extra factor of \mu_0\beta in the first term so I can't factorize it out to achieve the result I am supposed to. My apologies for the long winded nature of the post but if you could point out where I've made a mistake it would be greatly appreciated.
 
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maximus123 said:
$$=\mu_0\beta\frac{1}{Z}\int\vec{m_i^2}\,e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m\\
=\mu_0^2\beta^2\langle\vec{m_i}^2\rangle\\
$$
The last line simplification was achieved using the relation given at the top of the post (second equation down).​

The last line line is incorrect. You should be able to interpret the expression ##\frac{1}{Z} \int\vec{m_i^2}\,e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m##.
 

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