Adiabatic Demagnetization and Magnetic Cooling

Ordinary refrigeration runs out of coolants below about one kelvin: pumped liquid helium-4 stalls near 1 K, helium-3 near 0.3 K, and there is no colder liquid to pump. To go lower you stop cooling a fluid and start cooling a spin system. Adiabatic demagnetization uses a paramagnetic salt — a lattice of nearly independent magnetic moments — as the working substance. First you magnetize it isothermally, lining the moments up and squeezing entropy out into a helium bath; then you thermally isolate it and slowly turn the field off. The moments re-randomize, and because they can only draw the required entropy from the lattice vibrations, the whole crystal cools. Proposed independently by Debye in 1926 and Giauque in 1927 and first realized by Giauque and MacDougall in 1933, the method opened the millikelvin range and, in its nuclear version, the microkelvin range. The physics is governed by one fact: for an ideal paramagnet the entropy depends on field and temperature only through the ratio $B/T$ . Along a reversible adiabat the entropy is fixed, so $B/T$ is fixed, and the sample cools in exact proportion to the field: $T_f/T_i = B_f/B_i$ . This article derives that isentrope arithmetic from the two-level magnetic entropy, shows why a residual internal field puts a floor under the final temperature, and connects the shrinking cooling steps to the Third Law — the statistical reason absolute zero cannot be reached in a finite number of stages. The detailed magnetic equation of state (the full Brillouin function for general spin $J$ ) is developed in the companion two-level paramagnet article and only recalled here.

Key takeaways

Below ~1 K no fluid refrigerant works; you cool a paramagnetic salt instead, using its magnetic entropy as the reservoir.
Cycle: isothermally magnetize (entropy $\to$ bath), then adiabatically demagnetize under thermal isolation (temperature drops).
For an ideal paramagnet the entropy is a function of $B/T$ alone, so a reversible (isentropic) demagnetization keeps $B/T$ constant: $T_f/T_i = B_f/B_i$ .
A residual internal field $b$ replaces $B$ by $\sqrt{B^2+b^2}$ and floors the final temperature — you cannot demagnetize to true zero field, so you cannot reach $T=0$ .
Each stage removes a fixed fraction of the temperature and the entropy gap between the field-on and field-off curves shrinks to zero at $T=0$ : the Third-Law unattainability of absolute zero.
Key idea: turning the field off forces the moments to re-absorb entropy, and with the lattice thermally isolated that entropy can only come from cooling the crystal — refrigeration by disordering spins.

The core insight

The magnetic entropy of an ideal paramagnet depends on $B$ and $T$ only through $x = \mu B/k_BT$ . Holding entropy fixed while lowering the field therefore holds $x$ fixed, and the temperature falls in exact proportion to the field, $T_f = T_i\,(B_f/B_i)$ — a spin refrigerator whose cooling ratio you dial with a magnet.

Where this lives today

Adiabatic demagnetization refrigerators (ADRs) are the workhorse for reaching tens of millikelvin without the moving parts or gravity dependence of a dilution refrigerator, which makes them the cooling method of choice for space-borne detectors — the X-ray microcalorimeters on satellite observatories are cooled this way, because an ADR needs only a magnet and a heat switch and works in free fall. Nuclear demagnetization stages, sitting below a dilution refrigerator, take specialized laboratories into the microkelvin regime for studies of superfluid helium-3 and nuclear magnetism. The same magnetocaloric physics — entropy shuttled by an applied field — is being engineered near room temperature with giant-magnetocaloric alloys for solid-state refrigeration research. The magnetic equation of state underlying all of this, the Brillouin function for a general spin $J$ and the Curie-law susceptibility, is developed in the two-level paramagnet article; here we used only the entropy it implies.

The physical picture

Imagine a crystal of a dilute paramagnetic salt — a chrome alum or a cerium magnesium nitrate, where magnetic ions sit far apart and barely feel one another. Each ion carries a magnetic moment that, in a field, prefers to point along $\mathbf{B}$ . At high temperature and low field the moments are completely disordered: every orientation is equally likely and the spin entropy is at its maximum, $k_B\ln(2J{+}1)$ per ion. Turn on a strong field while the salt is in good thermal contact with a liquid-helium bath, and the moments line up; their entropy drops, and the heat of ordering flows out to the bath. The salt is now cold and ordered, holding much less spin entropy than it did. Now cut the thermal link so the salt is isolated, and slowly reduce the field. The moments want to re-randomize — disorder is entropically favored once the aligning field weakens — but re-randomizing costs entropy, and the only place that entropy can come from is the lattice vibrations. So the spins pull energy out of the lattice to pay for their own disordering, and the crystal's temperature drops. The magnetic degrees of freedom act as a sponge that was wrung out at high field and now soaks entropy back up as the field falls, chilling everything it touches.

Figure

Entropy-temperature diagram: the magnetize-then-demagnetize cooling path

There is a single decisive statement hiding in that diagram. Because the field-off and field-on entropy curves both start at zero at $T=0$ and separate only at finite temperature, the horizontal (constant-entropy) demagnetization step always lands at a lower temperature — but by an amount that shrinks as you get colder, because the two curves are squeezing together. No single stage reaches $T=0$ , and no finite stack of stages does either. That geometric squeeze is the Third Law made visible, and it is the punchline of the second figure.

Figure

Third-Law squeeze: the two entropy curves merge at T=0, so the cooling steps shrink

The core equations

A single ion of moment $\mu$ (spin- $\tfrac12$ for concreteness) has two orientations in a field, with energies $\mp\mu B$ ; everything follows from its two-level partition function.

Z_1 = e^{+\mu B/k_BT} + e^{-\mu B/k_BT} = 2\cosh\!\left(\frac{\mu B}{k_BT}\right) \qquad (1)

where: $\mu$ is the ionic magnetic moment, $B$ the applied field, $T$ the temperature, and $k_B$ Boltzmann's constant. The lower-energy state ( $-\mu B$ ) is the one aligned with the field; the dimensionless group $x \equiv \mu B/k_BT$ measures alignment energy against thermal energy. The magnetic entropy per ion follows from $s=-\partial f/\partial T$ with $f=-k_BT\ln Z_1$ , and collapses to a function of the single variable $x$ .

\frac{s}{k_B} = \ln\!\big(2\cosh x\big) - x\tanh x, \qquad x=\frac{\mu B}{k_BT} \qquad (2)

where: $s$ is the entropy per ion. At $x=0$ (no field or infinite temperature) $s=k_B\ln 2$ , the fully disordered value; as $x\to\infty$ (strong field or low temperature) $s\to 0$ , the fully aligned value. For general spin $J$ the same structure holds with $\ln 2\to\ln(2J{+}1)$ . Because $s$ depends on $B$ and $T$ only through $x=\mu B/k_BT$ , entropy is a function of the ratio $B/T$ alone.

s = s\!\left(\frac{B}{T}\right) \qquad (3)

where: this is the defining property of an ideal (non-interacting) paramagnet. It is exactly what makes the refrigerator predictable: a curve of $s$ versus $T$ at one field is the same curve rescaled at another field. A reversible demagnetization is isentropic, so $s$ — and therefore $x=\mu B/k_BT$ — is held constant, giving the central result.

\frac{T_f}{T_i} = \frac{B_f}{B_i} \qquad (4)

where: $T_i,B_i$ are the temperature and field before demagnetization and $T_f,B_f$ after. The cooling ratio is set entirely by the field ratio; lowering the field by a factor of 100 lowers the temperature by a factor of 100. Real salts are not perfectly ideal: residual dipole and exchange interactions act like a small internal field $b$ , so $B$ is replaced by an effective $\sqrt{B^2+b^2}$ .

\frac{T_f}{T_i} = \frac{\sqrt{B_f^2+b^2}}{\sqrt{B_i^2+b^2}} \qquad (5)

where: $b$ is the residual internal field (of order $10^{-2}$ – $10^{-1}$ T for common salts). Even at $B_f=0$ the effective field cannot fall below $b$ , so $T_f$ cannot fall below $T_i\,b/\sqrt{B_i^2+b^2}$ — the floor that keeps the salt above absolute zero.

Derivation AThe magnetic entropy and why it depends only on B/T

The goal is to derive equation (2) for the spin- $\tfrac12$ magnetic entropy from the two-level partition function, and to make explicit the property that turns this into a refrigerator: that the entropy is a function of $B/T$ alone. The strategy is the standard canonical route — build the free energy from $Z_1$ , differentiate to get the entropy — but with care about the one subtlety that makes magnetic cooling work: every appearance of $B$ and $T$ enters bundled into the single ratio $x=\mu B/k_BT$ , so the entropy inherits that dependence and nothing else. We recall the two-level canonical machinery from the Boltzmann-factor and canonical-partition-function articles rather than rebuilding it.

Step 1. Start from the single-ion free energy $f=-k_BT\ln Z_1 = -k_BT\ln(2\cosh x)$ with $x=\mu B/k_BT$ . Note that $x$ carries all the field and temperature dependence; $\mu$ and $k_B$ are constants.

Step 2. Differentiate to get the entropy, $s=-(\partial f/\partial T)_B$ . Writing $f=-k_BT\ln(2\cosh x)$ and using $\partial x/\partial T = -x/T$ , the product rule gives $s = k_B\ln(2\cosh x) + k_BT\,\dfrac{2\sinh x}{2\cosh x}\cdot\dfrac{x}{T}$ , and the second term is $k_B\,x\tanh x$ . Careful with the sign: $-\partial f/\partial T$ produces $+k_B\ln(2\cosh x)$ from the explicit $T$ and $-k_Bx\tanh x$ from the implicit $T$ inside $x$ .

Step 3. Collect: $s/k_B = \ln(2\cosh x) - x\tanh x$ , equation (2). Check the anchors: at $x=0$ , $\cosh 0=1$ and $\tanh 0=0$ , so $s=k_B\ln 2$ — maximal disorder. As $x\to\infty$ , $\ln(2\cosh x)\to x$ and $x\tanh x\to x$ , so the two large terms cancel and $s\to 0$ — full alignment, no entropy.

Step 4. Read off the structural point. Every $T$ and $B$ in equation (2) sits inside $x=\mu B/k_BT$ . Therefore $s=s(B/T)$ : the entropy is constant along any ray $B\propto T$ in the $(T,B)$ plane. This is equation (3), and it is exactly the property that will let a fixed-entropy process trade field for temperature. Verified: the companion verifier evaluates $s(x)$ at $x=0.1,0.5,1,2,3$ (obtaining $0.688,0.582,0.365,0.090,0.017$ in units of $k_B$ ), confirms $s(0)=\ln2$ and $s\to0$ at large $x$ using a numerically stable form, and checks $\mathrm{d}s/\mathrm{d}x=-x\,\mathrm{sech}^2x\le 0$ symbolically.

Go deeper · Derivation B · 2 minThe isentrope arithmetic and the Third-Law floor

Derivation BThe isentrope arithmetic and the Third-Law floor

Here we prove the cooling law (4), $T_f/T_i=B_f/B_i$ , and then show why the same entropy function forbids reaching absolute zero. The strategy is to impose the isentropic condition on the result of the previous derivation — constant $s$ means constant $x$ — and read off the temperature change; then to reintroduce a residual field and the Third Law to see the floor. The subtlety worth flagging is that the clean ratio law is exact only for an ideal paramagnet; the residual internal field is not a small correction but the very thing that sets the lowest attainable temperature.

Step 1. A demagnetization performed slowly under thermal isolation is reversible and adiabatic, hence isentropic: $s$ is held fixed. By equation (3), fixing $s$ fixes $x=\mu B/k_BT$ .

Step 2. Constant $x$ between the initial and final states means $\dfrac{\mu B_i}{k_BT_i}=\dfrac{\mu B_f}{k_BT_f}$ . Cancel $\mu/k_B$ and rearrange: $\dfrac{T_f}{T_i}=\dfrac{B_f}{B_i}$ , equation (4). Lowering the field cools the salt in exact proportion.

Step 3. Insert the residual field. Real interactions mean the effective field never vanishes; replacing $B\to\sqrt{B^2+b^2}$ in $x$ gives equation (5). At $B_f=0$ this leaves $T_f=T_i\,b/\sqrt{B_i^2+b^2}$ , a strictly positive floor. You cannot demagnetize below the internal field, so you cannot cool below this temperature in one stroke.

Step 4. Connect to the Third Law. As $T\to 0$ the entropy of the salt tends to the same limit (zero for a simple paramagnet) independently of the field, so the field-on and field-off entropy curves merge at $T=0$ . The vertical entropy gap $\Delta s(T)$ that each isothermal magnetization exploits therefore shrinks to zero as $T\to0$ ; each demagnetization step gains a smaller temperature drop than the last, and an infinite number of finite steps would be required to reach $T=0$ . That is the unattainability statement of the Third Law, recalled by name from the Third-Law article, expressed here as a property of the magnetic entropy. Verified: the verifier reproduces the ideal isentrope $T_f=T_i(B_f/B_i)=0.01$ K for the worked example (both by the ratio law and by holding $x$ fixed numerically), confirms the residual-field floor gives $T_f>0$ at $B_f=0$ , and checks that the entropy gap between the two field curves $\to 0$ as $T\to0$ while remaining sizeable near $1$ K.

Limiting cases

Each limit of equation (2) and the cooling law is a physical check on the construction.

Case	Result	Meaning
Strong field / low $T$ ( $x\gg1$ )	$s\to0$	Moments fully aligned; the spin reservoir is "empty" and can absorb no more entropy.
Zero field / high $T$ ( $x\to0$ )	$s\to k_B\ln 2$	Moments fully disordered; maximum spin entropy available to soak up on demagnetization.
Ideal paramagnet ( $b=0$ )	$T_f/T_i=B_f/B_i$	Cooling ratio equals field ratio exactly; no floor — but no real salt is ideal.
$T\to0$ , any field	$s\to0$	Field-on and field-off curves merge; steps shrink — the Third-Law unattainability of $T=0$ .

Common trapThinking adiabatic demagnetization can reach absolute zero

It cannot, and the reason is structural, not technical. Two effects put a floor under the final temperature. First, no real salt is an ideal paramagnet: residual dipolar and exchange interactions act like an internal field $b$ , so demagnetizing to $B_f=0$ still leaves the effective field at $b$ and the temperature at $T_i\,b/\sqrt{B_i^2+b^2}$ , not zero. Second, and more fundamentally, the Third Law forces the field-on and field-off entropy curves to meet at $T=0$ ; the entropy gap each stage relies on vanishes there, so the temperature drop per stage shrinks toward zero and no finite number of stages reaches absolute zero. Chaining stages (each starting where the last ended, and switching to a working substance with smaller $b$ — electronic salts, then nuclear spins) pushes ever lower, but the destination recedes as you approach it. Absolute zero is a limit, not a reachable state.

Worked examples

Example 1 — A single demagnetization from 1 K to 10 mK. A paramagnetic salt sits in a field $B_i=1$ T at $T_i=1$ K after isothermal magnetization against a helium bath. The thermal link is cut and the field is lowered reversibly to $B_f=0.01$ T. Find the final temperature, treating the salt as ideal.

Step 1. The demagnetization is isentropic, so $x=\mu B/k_BT$ is constant. Whatever value $\mu/k_B$ takes, only the ratio matters: $\mu B_i/(k_BT_i)=\mu B_f/(k_BT_f)$ .

Step 2. Apply equation (4): $T_f = T_i\,\dfrac{B_f}{B_i} = 1\ \mathrm{K}\times\dfrac{0.01\ \mathrm{T}}{1\ \mathrm{T}} = \mathbf{0.01\ K = 10\ mK}$ . A hundredfold field reduction gives a hundredfold temperature reduction.

Step 3. Reality check with a residual field $b=0.005$ T. Now $T_f = T_i\sqrt{B_f^2+b^2}/\sqrt{B_i^2+b^2} = 1\times\sqrt{0.01^2+0.005^2}/\sqrt{1^2+0.005^2} \approx \mathbf{0.0112\ K}$ — slightly above the ideal 10 mK, and if you demagnetized fully to $B_f=0$ you would still be stuck at $T_i\,b/\sqrt{B_i^2+b^2}\approx 5\ \mathrm{mK}$ , never zero. The takeaway: the cooling ratio equals the field ratio for an ideal salt, and the residual internal field sets the lowest temperature a given salt can reach.

Example 2 — Staging down and the Giauque milestone. Giauque and MacDougall's 1933 experiment started near $T_i=1.5$ K in a strong field and reached about $T_f=0.25$ K after demagnetization. Interpret this and estimate how many decades of cooling a chained refrigerator can span.

Step 1. The single-stage cooling factor was $T_i/T_f = 1.5/0.25 = \mathbf{6}$ — a demagnetization from high field to a low residual field of roughly one-sixth the effective initial field.

Step 2. Chaining stages multiplies the factors. If each stage removes a fixed fraction so that $T_{n}=T_0\,r^{\,n}$ with $r=0.1$ , then five stages take $1$ K to $T_5=1\times(0.1)^5=\mathbf{10^{-5}\ K=10\ \mu K}$ . In practice electronic salts stall in the millikelvin range because their residual field $b$ is large; switching the working substance to nuclear moments, whose interactions are thousands of times weaker, resets $b$ and lets nuclear demagnetization reach microkelvin and below.

Step 3. The floor is never zero. Each stage's temperature drop is limited by the merging entropy curves, so the sequence of temperatures is bounded below by a positive number — the arithmetic of stages approaches $T=0$ geometrically but never arrives. The takeaway: magnetic cooling proceeds in stages, each dialing temperature down by a field ratio, with the working substance's residual field — electronic, then nuclear — setting how low each generation of stage can go.

Origins · 1 minabsolute zero and how this was discovered

absolute zero

the T→0 limit of thermodynamics · The unreachable floor of the temperature scale, approached but never attained

Absolute zero, $T=0$ on the thermodynamic scale, is the temperature at which a system's entropy reaches its minimum; the Third Law of Thermodynamics forbids reaching it in any finite sequence of processes. Adiabatic demagnetization is the technique that took experiments closest to it: proposed independently by Peter Debye (1926) and William Francis Giauque (1927), it was first realized by Giauque and MacDougall in 1933, reaching about a quarter of a kelvin. Giauque received the 1949 Nobel Prize in Chemistry for this low-temperature work and its bearing on the Third Law. The nuclear variant, pioneered by Kurti and Simon in the 1950s, has since reached the microkelvin and even nanokelvin range — ever closer to a limit that, by the Third Law, always recedes.

History

1905

— Langevin builds the classical statistical theory of paramagnetism and the Langevin function

1926

— Debye proposes cooling by adiabatic demagnetization of a paramagnetic salt

1927

— Giauque independently proposes the same method for reaching sub-kelvin temperatures

1933

— Giauque and MacDougall demagnetize gadolinium sulfate and reach about 0.25 K

1935

— de Haas and Wiersma reach a few millikelvin with dilute salts

1949

— Giauque awarded the Nobel Prize in Chemistry for low-temperature research

1956

— Kurti and Simon demonstrate nuclear adiabatic demagnetization into the microkelvin range

Going further

References. Giauque & MacDougall, "Attainment of Temperatures Below 1° Absolute by Demagnetization of Gd₂(SO₄)₃·8H₂O," Phys. Rev. 43, 768 (1933) · Debye, Ann. Phys. 81, 1154 (1926) · Pobell, Matter and Methods at Low Temperatures, 3rd ed., Ch. 9–10 · Reif, Fundamentals of Statistical and Thermal Physics, §5.10 · Kittel & Kroemer, Thermal Physics, Ch. 12 (paramagnetic cooling and the Third Law).

Provenance. A verifier should reproduce: (1) the spin- $\tfrac12$ entropy $s/k_B=\ln(2\cosh x)-x\tanh x$ at $x=0.1,0.5,1,2,3$ and its $x\to0$ ( $\ln2$ ) and $x\to\infty$ ( $0$ ) limits with a numerically stable evaluation; (2) $\mathrm{d}s/\mathrm{d}x=-x\,\mathrm{sech}^2x$ symbolically; (3) the ideal isentrope $T_f=T_i(B_f/B_i)=0.01$ K both by the ratio law and by holding $x$ fixed; (4) the residual-field floor $T_f>0$ at $B_f=0$ ; (5) the entropy gap between field curves vanishing as $T\to0$ ; (6) the geometric staging $T_n=T_0 r^n$ and the Giauque factor $1.5/0.25=6$ .

Builds on: Third Law of Thermodynamics · Temperature, Pressure, and Chemical Potential as Entropy Derivatives · Pauli Paramagnetism. Leads to: The Two-Level Paramagnet and the Brillouin Function · Nuclear demagnetization to microkelvin temperatures · Unattainability of absolute zero.

Why can't I just keep pumping on liquid helium to go below 1 K?

Pumping lowers the vapor pressure and hence the boiling temperature of a liquid, but the vapor pressure falls exponentially as the liquid cools, so the cooling power collapses. Pumped helium-4 stalls near 1 K and helium-3 near 0.3 K, and there is no colder liquid to pump. Below that you must cool a system whose entropy you can remove by a non-thermal handle — a magnetic field acting on spins — rather than by evaporating atoms. That is exactly what adiabatic demagnetization provides: the field, not a pump, does the entropy bookkeeping.

What is the difference between the isothermal and adiabatic steps?

They are the two strokes of the cycle and they do opposite jobs. In the isothermal magnetization step the salt stays in contact with a bath at fixed temperature while you raise the field; the moments order, their entropy drops, and the released heat flows out to the bath — you are loading the spin reservoir by wringing entropy out of it. In the adiabatic demagnetization step you cut the thermal link and lower the field; entropy is now conserved, the moments re-disorder, and they pull the required entropy out of the lattice, cooling the whole sample. One step exports entropy at constant temperature; the other trades field for temperature at constant entropy.

Does the field do work that cools the sample, like a gas expanding?

Not in the way an expanding gas cools by doing $p\,\mathrm{d}V$ work on its surroundings. The cooling here is an entropy effect: at constant entropy, lowering the field forces the spin subsystem to increase its entropy (more disorder is favored as the aligning field weakens), and because the total entropy is fixed the lattice entropy must fall — which means the lattice cools. The magnetic work bookkeeping is real, but the clean statement is the isentropic one: $s(B/T)$ constant forces $T\propto B$ . The analogy to adiabatic gas expansion ( $T\propto p^{(\gamma-1)/\gamma}$ at constant entropy) is exact in spirit — both are isentropic expansions of a working substance.

Why does a residual internal field matter so much?

Because it sets the lowest temperature the salt can reach. The ideal law $T_f/T_i=B_f/B_i$ would let you cool without limit by demagnetizing to $B_f=0$ . But dipolar and exchange interactions between the moments never switch off; they act like an internal field $b$ , so the effective field is $\sqrt{B^2+b^2}$ and can never drop below $b$ . The final temperature therefore bottoms out at roughly $T_i\,b/B_i$ . Electronic salts have $b\sim10^{-1}$ – $10^{-2}$ T and stall in the millikelvin range; nuclear moments interact thousands of times more weakly, so their $b$ is far smaller and nuclear demagnetization reaches microkelvin. Choosing a working substance is choosing a value of $b$ .

Is this the same as how the Third Law is usually stated?

Yes — adiabatic demagnetization is the classic illustration of the Third Law's unattainability form. The Nernst statement that entropy tends to a field-independent constant as $T\to0$ means the entropy-versus-temperature curves for different fields converge at absolute zero. Each demagnetization step exploits the gap between two such curves, and since the gap closes at $T=0$ , the temperature reduction per step shrinks to nothing there. No finite number of steps can bridge a gap that has vanished, which is precisely "absolute zero cannot be reached in a finite number of operations." The full statement and its consequences are developed in the Third Law article.

How is nuclear demagnetization different from the electronic version?

It is the same physics with a different working substance. Electronic paramagnetic salts use the magnetic moments of unpaired electrons; their mutual interactions (the residual field $b$ ) stall them around a millikelvin. Nuclear moments are about a thousand times smaller, so their interactions — and hence $b$ — are roughly a million times weaker in energy, letting the ratio law carry the temperature far lower before the floor is hit. A nuclear demagnetization stage is therefore mounted below a conventional refrigerator that pre-cools it into the millikelvin range and magnetizes the nuclei; demagnetizing them then reaches microkelvin. The cost is that nuclear moments are so weakly coupled that they take a long time to come into equilibrium with the electrons and lattice.