Past exam of the general relativity and cosmology and condensed matter physics paper of Part B of the physics course of the University of Oxford / 2011 / Section VI / 8

The Brillouin function is odd, increases monotonically through zero, and approaches $\pm 1$ for $x\to \pm \infty$. Its small-$x$ slope is $(J+1)/(3J)$. For $x\gg 1$, the magnetic ions overwhelmingly occupy the largest field-aligned moment projection, so the magnetization reaches magnetic saturation:The sketch shows several $J$ values; all have the same limiting values, while their slopes differ.

The $1/x$ terms in the Brillouin function cancel. Keeping the linear terms givesConsequently,Use the dimensionless SI magnetic susceptibility $\chi =\partial M/\partial H_{\mathrm{applied}}$ and $B_{\mathrm{applied}}={\mu }_0{H}_{\mathrm{applied}}$, as in the source's molecular-field convention. The Curie law isIf susceptibility is instead defined as $\partial M/\partial B$, its coefficient is $C/{\mu }_0$ and is dimensionful. The SI convention above makes the subsequent Curie–Weiss law and $\lambda$ consistent.

For positive $\lambda$, a fluctuation of magnetization produces a Weiss molecular field in the same direction, reinforcing alignment. The mean-field self-consistency equation isLinearizing in the paramagnetic phase givesAt zero applied field, the right-hand side's slope at $M=0$ is $T_C/T$. Below $T_C$ this exceeds one, and the saturation of the Brillouin function allows stable nonzero solutions of opposite sign: spontaneous ferromagnetism. Above $T_C$, only the zero-field solution $M=0$ remains stable. The stated linear susceptibility is the paramagnetic branch for $T>T_C$; it must not be continued as a negative susceptibility below the transition.

Using the supplied iron parameters givesAn internal Weiss molecular field depends on temperature through $M(T)$. Its low-temperature saturation estimate uses $M_s=ng{\mu }_{B}J\simeq 1.58\times 10^6\mathrm{A}{\mathrm{m}}^{-1}$:At and above $T_C$ the zero-applied-field magnetization vanishes in this model, so this is a saturation estimate, not the field at $T_C$.

Use the paper's stipulated face-centered cubic lattice, which has twelve nearest neighbours. The saturated one-ion alignment scale isDividing the effective field among twelve equivalent neighbours gives the per-bond exchange interaction scaleTo specify the convention, write the Heisenberg model as $\mathcal{H}=-\mathcal{J}{\sum }_{⟨ij⟩}{\mathbf{J}}_i\cdot {\mathbf{J}}_j$, with dimensionless angular-momentum operators and each bond counted once. Then $g{\mu }_B{B}_e=12\mathcal{J}⟨J_z⟩$, so $\mathcal{J}=3k_B{T}_C/[12J(J+1)]\simeq 11.2\mathrm{meV}$ here. A convention with $-2{\mathcal{J}}^{\prime }$ instead has ${\mathcal{J}}^{\prime }=5.62\mathrm{meV}$. The total saturated interaction energy per ion is $-12\mathcal{J}{J}^2/2\simeq -67.4\mathrm{meV}$; the factor of one half avoids counting each bond twice.