 Solution

Take $\mu >0$ to be the magnitude of either spin projection's magnetic moment, so the independent magnetic levels are $E_{\pm }=\mp \mu B$. Their partition function and magnetic energy arewhere $\beta =1/(k_{B}T)$ and $x=\beta \mu B$. The canonical entropy is consequentlyThis independent-spin magnetic entropy counts the two orientation states of each independent moment; translational entropy and interactions are excluded.

Along the magnetic-entropy demagnetization adiabat, constant magnetic entropy fixes $x$, hence $B/T$ is constant. Lowering the field therefore gives the ideal adiabatic demagnetization relationA useful cycle first magnetizes isothermally while releasing heat to a reservoir, then thermally isolates the sample and lowers the field. Cooling the entire sample requires its other heat capacities and spin-lattice equilibration to be accounted for. Magnetic interactions, residual fields and heat leaks prevent extrapolating the independent-moment formula to an attainable zero temperature.^[ref]