= Solution
For two localized <energy> levels $0,\delta$ with degeneracies $d_0,d_1$, let $x=\delta/(k_BT)$. Canonical equilibrium gives
$$
z=d_0+d_1e^{-x},\qquad p_1=\frac{d_1e^{-x}}z,\qquad
u=\delta p_1,\qquad s=k_B(\ln z+xp_1),\qquad
c=k_Bx^2p_1(1-p_1).
$$
Here $p_1$ is the total upper-level occupation <probability> and $u,s,c$ are per unit. For nondegenerate levels the <entropy> rises from zero to $k_B\ln2$, while the <heat capacity> vanishes at both low and high <temperatures> and has a Schottky maximum near $k_BT=0.417\delta$. A <spin one-half> in a field, or a localized defect with two configurations, supplies an example. Higher levels must be thermally inaccessible; unequal degeneracies change the populations and peak. Distinguishable localized units have total <partition function> $z^N$, unlike a gas of indistinguishable particles sharing just two orbitals.
For the magnetic atom, choose level labels with $E_m=-mg\mu_BB$, $m=-J,\ldots,J$. Put $x=T_B/T=g\mu_BB/(k_BT)$. Its <partition function> and exact <entropy of a finite magnetic multiplet> are
$$
\boxed{Z_J(x)=\sum_{m=-J}^{J}e^{mx}
=\frac{\sinh[(2J+1)x/2]}{\sinh(x/2)},\qquad
\frac{s_J}{k_B}=\ln Z_J-x\frac{d\ln Z_J}{dx}.}
$$
The ratio form has removable limit $Z_J(0)=2J+1$. Since $s_J$ is a function of $x$ only, it depends on field and <temperature> solely through $B/T$. For fixed $J$ at small $x$, <symmetry> removes the odd <powers> and the supplied sum of $m^2$ gives
$$
Z_J=(2J+1)\left[1+\frac{J(J+1)}6x^2+O(x^4)\right],
\qquad
\boxed{s_J=k_B\left[\ln(2J+1)-\frac{J(J+1)}6\frac{T_B^2}{T^2}+O(x^4)\right].}
$$
For large $J$, the expansion requires $J|x|\ll1$, not merely $|x|\ll1$.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2012/paper-1-tsp2012-magnetic-entropy.png]
{title=Two-level heat and magnetic entropy}
{description=Original Schottky heat-capacity curve and exact finite-spin entropies show the entropy deficit removed by isothermal magnetization. Quadratic high-temperature expansions are shown only over a limited small-field interval.}
For <demagnetization with a linear-heat-capacity load>, integrate its specified $C_L/T$:
$$
S_L(T)=S_L(0)+\frac{Mk_B}{\theta}T.
$$
Isothermal <magnetization> at $T_i$ removes spin <entropy>; the bath receives the corresponding heat. During the following reversible, thermally isolated demagnetization, it is the combined <entropy> of spins and load that stays constant. At the minimum-field endpoint $B_f=0$, the ideal independent spins reach $s_{\max}=k_B\ln(2J+1)$. Therefore
$$
Ns_J(T_i,B_i)+S_L(T_i)=Nk_B\ln(2J+1)+S_L(T_{\min}),
$$
$$
\boxed{N\left[k_B\ln(2J+1)-s_J(T_i,B_i)\right]
=\frac{Mk_B}{\theta}(T_i-T_{\min}).}
$$
The bracket is the initial <magnetic entropy deficit>. The PDF's intermediate equality uses the initial <entropy> itself instead; with its preceding definition of $s$, that equality omits the maximum-entropy term. The corrected balance recovers its final <temperature> formula:
$$
\boxed{T_{\min}=T_i-\frac{N\theta J(J+1)g^2\mu_B^2B_i^2}{6Mk_B^2T_i^2}.}
$$
Without the small-field expansion, the exact independent-spin prediction is $T_i-(N\theta/M)[\ln(2J+1)-s_J(T_i,B_i)/k_B]$. As a check, zero initial field gives zero <entropy> deficit and no cooling. The incorrectly printed intermediate balance would fail this check.
The result requires the load law throughout the <temperature> interval, hence $T_i\ll\theta$; a small initial <Zeeman splitting>, quantitatively $JT_{B,i}/T_i\ll1$; positive final <temperature>; negligible additional <heat capacities> and heat leaks; thermal equilibration between subsystems; and a reversible sweep slow enough to maintain equilibrium. A negative extrapolated <temperature> is a breakdown of this approximation, not a cooling prediction. Spins alone would keep $B/T$ fixed, but the attached load changes that adiabat by transferring <entropy> to them as it cools.
Interactions invalidate the product of independent one-site <partition functions>. The <entropy> must be computed from the many-body <energy> spectrum or a justified interacting approximation. <Correlations> and possible magnetic ordering introduce an interaction scale and can remove the independent-spin <entropy> available for cooling near zero field. Internal fields and ordering commonly limit practical demagnetization. The detailed correction and any residual <entropy> depend on the interaction Hamiltonian, dimensionality and frustration; neither a universal ordering <temperature> nor a numerical final-temperature correction is specified here.
Back to article page
