- #1

maximus123

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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]

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]

[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 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]

[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.