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Consider a stipulated dilute collection of independent two-level systems, each with longitudinal magnetic moment $s\mu$, $s=\pm 1$, and energy $E_s=-s\mu B_0$. The prescribed vacuum/applied magnetic flux density is $B_0={\mu }_0{H}_{\mathrm{app}}$. Ignore interactions, orbital magnetic response, and the small sample-field correction. This is the magnetic version of the existing independent spin-one-half two-level system, with alignment probabilities rather than an interacting Curie–Weiss model.

For temperature $T>0$, the canonical ensemble gives the one-moment partition function $Z_1=2\cosh x$, with $x=\mu B_0/(k_{B}T)$. Taking the expected value over its two Boltzmann weights givesHere $n$ is the number density, $k_B$ the Boltzmann constant, and $\tanh$ the hyperbolic tangent. For $|x|\ll 1$, its small-argument expansion of the hyperbolic tangent gives $M\simeq {\mu }_{0}n{\mu }^2{H}_{\mathrm{app}}/(k_{B}T)$, a Curie's law response in the applied-$H$ convention. For large $|x|$, $M$ approaches the saturation magnetization $n\mu$ with sign set by the field, and its differential magnetic susceptibility falls towards zero.

Both plots use stipulated $n=10^{26}{\mathrm{m}}^{-3}$, $\mu =9.2740101\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and ${\mu }_0=1.25663706\times 10^{-6}\mathrm{T}\mathrm{m}{\mathrm{A}}^{-1}$. The moment is approximately one Bohr magneton; the chosen vacuum permeability is a rounded model input. At $T=2\mathrm{K}$, $M_s=927.401\mathrm{A}{\mathrm{m}}^{-1}$ and ${\chi }_{\mathrm{model}}(0)=3.91410\times 10^{-4}$. The linear approximation overestimates $M$ by less than $3\%$ for $|x|\le 0.3$, corresponding to $|{\mu }_0{H}_{\mathrm{app}}|\le 0.89324\mathrm{T}$ or $|H_{\mathrm{app}}|\le 0.71082\mathrm{MA}{\mathrm{m}}^{-1}$. These parameters keep the modeled response weak; neglected magnetic feedback is of relative order ${\chi }_{\mathrm{model}}(0)$, rather than an explanation of a material's ordered phase.

At $B_0=0$, cooling increases the zero-field slope as $1/T$. At fixed $B_0\ne 0$, cooling instead gives ${\chi }_{\mathrm{model}}\sim 4{\mu }_{0}n{\mu }^2{e}^{-2\mu |B_0|/(k_{B}T)}/(k_{B}T)$: a saturated moment can hardly change further. The order of the zero-field and zero-temperature limits matters. This independent model has no finite-temperature collective phase transition; the diverging zero-field slope alone does not establish one.

The condition $L\ll na$ implies $|\tau a|\ll 1$. The small-argument expansion of the hyperbolic tangent gives $\tanh x=x+O(x^3)$, soUsing $f=k_{B}T\tau$ yields the entropic Hooke law for a one-dimensional chainThus the effective spring constant is $k_{B}T/(n{a}^2)$.

At fixed tension, the exact extension isIncreasing the temperature decreases the positive argument of the hyperbolic tangent, so the chain contracts. In the small-extension regime this becomeswhich is inversely proportional to $T$.