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Apparently it is to be shown that for electrons under an external magnetic field, in the limit as $B\to 0 $ $$ \chi = \frac{dM}{dB} \approx \frac{n\,\mu^{*^2}}{k\,T}\,\frac{f_{1/2}(z)}{f_{3/2}(z)} $$

So this is the Pauli paramagnetism problem and I understand the effect and the physics, but cannot seem to get the requested ratio of $f_{\nu}(z)$ functions.

Here is how I have been doing it:

With the grand canonical ensemble formalism I have $$ \langle N_{\pm} \rangle = \int_0^{\infty} d\epsilon\, g(\epsilon)\,\frac{1}{\exp^{\beta(\epsilon-\mu\pm \mu_B\,B)}} $$ where $$ g(\epsilon) = \frac{2}{\sqrt{\pi}}\,\left(\frac{2\,\pi\,m}{h^2}\right)^{3/2}\;V\,\sqrt{\epsilon} $$ and is the density of states from translational motion. I then say that $$ \langle M \rangle = \mu_B\,(\langle N_- \rangle - \langle N_+ \rangle ) = \cdots = Const*V*[f_{1/2}(\beta\,\mu+\mu_B\,B)-f_{1/2}(\beta\,\mu-\mu_B\,B)] $$

This is fine I think except that when I calculate $\chi$ I will get fermi-dirac functions with -1/2 and not the ratio I need.

$$ \chi = \lim_{B\to 0} \frac{\partial\,M}{\partial\,B} = \cdots ? $$

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Anybody? Or is this too....? –  nate Apr 9 '13 at 4:33

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