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.
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
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)
<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 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 />
<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.