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Consider $N$ atoms, each with magnetic moment $\mu$ in a zero field $H = 0$. Given the assumption that each moment is equally likely, what is the magnetization in the thermodynamic limit?


The idea of magnetization is fairly new to me; but, from reading online, I think that it is analogous to density. Furthermore, I know that the thermodynamic limit is when $N \rightarrow \infty$ and $V\rightarrow \infty$ with $N/V$ held constant.

I know that magnetization is defined as $dm/dV$, where $m$ is the elementary magnetic moment, and $V$ is the volume. When $V \rightarrow \infty$, the denominator of the $dm/dV$ term approaches $0$; so, is my answer just $0$?

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First of all, since the dipole distribution is uniform you can rewrite magnetization as $m/V$, where by $m$ I mean the total magnetization of $N$ atoms. Without losing any generality, let us assume that magnetic field $\vec H$ is applied along $z$ axis. Then, the part of the Hamiltonian that depends on magnetic field (energy of the dipole in field is a scalar product of the field and the dipole): \begin{equation} \mathcal{H_m}= -\mu H_z\sum\limits_{i=1}^N \cos{\theta_i}. \end{equation} Total magnetization is (see eq. (6.1.12) of Schwabl "Statistical mechanics" or eq. (52.1) Landau, Lifshitz "Statistical physics. Part 1. Vol 5." for discussion of Hamiltonian and magnetization). \begin{equation} m=-\langle{\partial \mathcal{H_m}}/{\partial H_z}\rangle=\mu \sum\limits_{i=1}^N \frac N\pi \frac \pi N \cos{\theta_i}=\frac{\mu N}\pi\int\limits_0^\pi d\theta \, \cos{\theta}=\frac{\mu N}\pi, \end{equation} where I've used $N>>1$ to justify rewriting the sum to integral in the second to last equality ($\pi/N$ becomes $d\theta$). Then desired magnetization is $\frac\mu \pi \frac N V$.

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  • $\begingroup$ Added references, but not sure about your probability distribution. $\endgroup$ – stinglikeabeer Sep 17 '18 at 1:20
  • $\begingroup$ Ok thanks. How about the equation for $U$ ? $\endgroup$ – user146017 Sep 17 '18 at 1:27
  • $\begingroup$ I've changed the notation from $U$ to $\mathcal{H_m}$ so it is more consistent with the aforementioned references. And added physical motivation of that contribution to energy (it is true for the electric field and electric dipole). $\endgroup$ – stinglikeabeer Sep 17 '18 at 2:05

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