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A cooling method that reduces an applied magnetic field while suppressing heat transfer to or from the material. For a suitable material exhibiting paramagnetism, an approximately reversible field reduction can lower its temperature at approximately constant entropy.

The preparation matters as much as the field reduction. First establish thermal equilibrium with a cold heat bath, then raise the magnetic field while maintaining the bath temperature: this is magnetization in an isothermal process. The bath accepts heat and the material loses magnetic entropy. Next open a heat switch, thermally isolating the working material, and lower the magnetic field. If the sweep maintains internal thermal equilibrium with negligible entropy production, this second leg is an isentropic process. An adiabatic process alone does not guarantee constant entropy.

The attraction is a controllable energy scale: changing the magnetic field changes the separation of magnetic energy levels, even across a thermally insulating gap. This lets an already cold solid become colder without needing a colder contacting heat bath. A practical refrigerator also needs a load, a period of absorbing its heat, and a reset that rejects accumulated entropy to the bath. A single field sweep has finite cooling capacity.

Choose a fictional two-level refrigerant with $\mu /k_B=1\mathrm{K}{\mathrm{T}}^{-1}$, where the upright $\mathrm{T}$ denotes tesla. This chosen moment is a convenient model parameter, not a fitted property of the historical salt. At bath temperature $T_b=1\mathrm{K}$, raise $B$ from $0.20\mathrm{T}$ to $1.00\mathrm{T}$. Then $x$ increases from $0.20$ to $1.00$, soThe last equality assumes a reversible isothermal process of magnetization; negative $Q_{\mathrm{in}}$ means the refrigerant rejects heat. The bath gains $0.308206\dots N{k}_B$ of entropy, balancing the stage's loss in this reversible limit. An appreciable temperature difference during real heat transfer adds entropy production.

Now open the heat switch and reduce $B$ from $1.00\mathrm{T}$ to $0.10\mathrm{T}$. Constant $x=1$ gives $T_f=0.10\mathrm{K}$. Throughout this ideal cooling leg, $p_{-}=0.880797\dots$, $p_{+}=0.119203\dots$, and $S_{\mathrm{mag}}=0.365334\dots N{k}_B$. Neither state becomes more probable. Field reduction shrinks the energy gap by ten while preserving the Boltzmann distribution's dimensionless ratio of gap to temperature.

There is a subtle energy check. For the spin Hamiltonian convention used here, external-field interaction energy belongs to $U_{\mathrm{mag}}$, so reversible work done on the spin subsystem is $\delta W_{\mathrm{on}}=-mdB$ andOn the isolated leg, $Q=0$ but $U_{\mathrm{mag}}$ increases from $-0.761594\dots N{k}_B\mathrm{K}$ to $-0.0761594\dots N{k}_B\mathrm{K}$, becoming less negative as the gap shrinks. Integrating $-mdB$ at constant $x=1$ supplies exactly $+0.685435\dots N{k}_B\mathrm{K}$. Cooling therefore need not decrease this chosen spin internal energy. An account of work exchanged with a magnet circuit must specify where field interaction energy is assigned before comparing signs. The occupation ratio and temperature prediction do not depend on that bookkeeping choice.

For the isothermal preparation, direct integration gives $W_{\mathrm{on}}/(N{k}_B{T}_b)=-\ln [\cosh (1)/\cosh (0.20)]=-0.413913\dots$. The endpoint internal energy difference is $-0.722119\dots N{k}_B{T}_b$, equal to $Q_{\mathrm{in}}+W_{\mathrm{on}}$. This independently checks both the rejected heat and the magnetic-work sign in the chosen convention.

Three distinct effects spoil indefinite $T\propto B$ cooling: extra entropy outside the spin subsystem, additional magnetic energy scales, and irreversible heat or work input. They can be distinguished with a thermodynamic derivative. For a smooth equilibrium Helmholtz free energy obeying $dF=-SdT-mdB$, a Maxwell relation givesHere $C_B$ is the total heat capacity participating in the stage's thermal equilibrium. The ideal model has $(\partial m/\partial T{)}_B=-C_{B,\mathrm{mag}}/B$. If its spins share temperature with a field-independent lattice and load of combined heat capacity $C_{\mathrm{other}}>0$, thenThe positive slope still means cooling when $dB<0$, but its smaller magnitude gives less cooling per fractional field reduction. For example, a low-temperature Debye model gives a lattice entropy proportional to $T^3$. As that lattice entropy falls, the spins take up the released entropy; $x$ can no longer remain constant. A weak enough coupling to neglect its equilibrium interaction energy can still be strong enough to exchange heat on the sweep timescale.

Interactions between magnetic moments, including exchange interaction and the fields of neighboring magnetic dipole moments, introduce energy scales that do not vanish with the applied $B$. Once they compete with $k_{B}T$ or $\mu B$, the independent-site partition function fails. Correlations and possible magnetic ordering alter the available magnetic entropy and the cooling curve. Simply setting the external magnetic field to zero does not eliminate those interactions. Different refrigerants therefore need their own measured equilibrium properties; there is no universal final temperature from the applied-field ratio alone.

Finally, a real open heat switch has residual thermal conductance; supports, wires and radiation carry incoming heat. Field changes may also dissipate energy. Locally near thermal equilibrium, with incoming heat rate ${\dot{Q}}_{\mathrm{leak}}$ and internal entropy production rate ${\dot{S}}_{\mathrm{prod}}$, the stage balance becomesThis uses a common stage temperature; a non-equilibrated load requires separate balances. The sweep must allow spin-lattice relaxation while remaining fast enough to limit accumulated parasitic heat. At fixed $B$, the same balance explains warming and finite hold time. Useful refrigeration is the controlled absorption of a load's heat during that finite interval, followed by a renewed isothermal process of magnetization to export its entropy.

Magnetic entropy counts the uncertainty in the occupied magnetic microstates. For independent localized magnetic moments with one-site probabilities $p_j$,where $N$ counts sites and $k_B$ is the Boltzmann constant. This is the Gibbs entropy for the magnetic degrees of freedom. Equal occupation of two states gives $N{k}_B\ln 2$; certain occupation of one state gives zero. Calling this disorder is useful only if the relevant states and their probabilities are specified. In an interacting material, correlations generally prevent replacing the many-body Gibbs entropy by this sum of independent one-site contributions.

Increasing the magnetic field at fixed positive temperature favors lower magnetic energy levels, reducing magnetic entropy. During ideal isolated demagnetization, both the level separation and temperature fall together, leaving the occupation probabilities unchanged. Thus the independent-spin idealization cools without increasing spin disorder along the isolated leg. When a real crystal's vibrating lattice also cools, its lost entropy can instead be taken up by the magnetic degrees of freedom. The conserved quantity is the total entropy of the thermally isolated stage, including any attached load in thermal equilibrium.

Use the independent spin-one-half two-level system with fixed $N$ distinguishable sites and Hamiltonian levels $E_{\pm }=\pm \mu B$, where $\mu >0$ is the magnitude of the field-axis magnetic moment and $B>0$ is a uniform externally controlled magnetic field. Assume positive temperature, no interactions between sites, and no appreciable lattice or load entropy. The two levels differ by $2\mu B$. With inverse temperature $\beta$, the canonical ensemble gives the one-site and total partition functionsHere the minus subscript denotes the lower energy $-\mu B$. Factorization of the partition function requires additive energy and independent sites; there is no particle-exchange factor $N!$ for these fixed distinguishable sites.

Differentiate the partition function, holding $B$ fixed when differentiating with respect to $\beta$. The internal energy, Helmholtz free energy, total field-axis magnetic moment $m$, and magnetic entropy areThe extensive moment $m$ is distinct from magnetization, the moment per unit volume. Direct substitution of $p_{\pm }$ into the Gibbs entropy gives the same $s(x)$ independently of the free-energy derivative.

At fixed $B$, the magnetic heat capacity isStrict decrease makes $s$ invertible on $x>0$. Constant magnetic entropy therefore fixes $x$, not $B$ or $T$ separately:The partition function may be used to label these equilibrium states even though the isolated cooling leg has no external heat bath. Actual thermalization of the lattice and load additionally requires a sufficiently fast spin-lattice relaxation; perfectly uncoupled spins by themselves cannot refrigerate an attached object.

The limiting cases are instructive. For $x\ll 1$, $s(x)=\ln 2-x^2/2+O(x^4)$, $m\simeq N{\mu }^{2}B/(k_{B}T)$, and $C_{B,\mathrm{mag}}\simeq N{k}_B{x}^2$. The moment response reproduces Curie's law. For $x\gg 1$, $s(x)\simeq (1+2x)e^{-2x}$ and the upper-state occupation is exponentially small. Taking $B\to 0$ at fixed $T>0$ gives $s\to \ln 2$; taking $T\to 0$ at fixed $B>0$ gives $s\to 0$. The origin has a path-dependent limit. Extrapolating a line of constant $x$ to $B=T=0$ does not construct a usable zero-temperature state or prove that a real material retains independent degenerate spins there.