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Magnetic susceptibility integral trouble

  • Thread starter maximus123
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  • #1
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Hello,

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

[itex] 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[/itex]
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

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

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


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


[itex]
\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}\int\mu_0\beta\vec{m_i^2}\,e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m\\
=\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\\

[/itex]
The last line simplification was achieved using the relation given at the top of the post (second equation down).
The second term then

[itex]
\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\\
=\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\\
=\frac{1}{Z}\frac{1}{Z}\frac{dZ}{d\vec{H_i}}\frac{dZ}{d\vec{H_i}}\frac{1}{\mu_0\beta}\\
=\frac{1}{Z}\frac{dZ}{d\vec{H_i}}\langle\vec{m_i}\rangle\\
=\langle\vec{m_i}\rangle^2\mu_0\beta
[/itex]
Giving a final result for the susceptibility as

[itex]
V\chi_0=\mu_0^2\beta^2\langle\vec{m_i}^2\rangle-\langle\vec{m_i}\rangle^2\mu_0\beta
[/itex]

So basically I have an extra factor of [itex]\mu_0\beta[/itex] 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.
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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$$=\mu_0\beta\frac{1}{Z}\int\vec{m_i^2}\,e^{\mu_0\vec{m}\cdot\vec{H}\beta}d^2m\\
=\mu_0^2\beta^2\langle\vec{m_i}^2\rangle\\
$$
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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