A law about spectra, not about heat

The third law arrives in most treatments as a statement about cooling, and the first rung of this ladder treats it that way: absolute zero is unreachable, because every stage of cooling removes a fixed fraction of what is left rather than a fixed amount.

That is the operational form, it is correct, and it says nothing about why. The statistical form does, and it is a statement about a spectrum rather than about heat:

$$S(T \to 0) = k\ln g,$$

with $g$ the number of states the system’s ground state has. It vanishes when the ground state is unique and not otherwise.

What the restatement changes #

Three things become clearer at once, and each was awkward in the operational form.

Residual entropy stops being an exception. Ice keeps 3.41 joules per kelvin per mole because its ground state has $(3/2)^N$ arrangements, all of the same energy, all satisfying the ice rules — a count of arrangements and nothing else. That is a degenerate ground state, $k\ln g$ is not zero, and the law is being obeyed rather than excused.

“Approaching zero temperature” acquires a scale. The operational statement says nothing about how cold is cold, and the statistical one says exactly: cold enough that the system is in its ground state, which means cold compared with the gap to the first excited state. Different degrees of freedom in the same lump of matter have wildly different gaps, so “at zero temperature” is a different temperature for each of them.

And the unattainability follows rather than being assumed. If two entropy curves at different values of some external parameter must meet at $T = 0$ — which they must, if both approach the same $k\ln g$ set by a spectrum neither parameter changes — then the staircase of the first rung has steps that shrink to nothing, and no finite number of them reaches the floor. Nernst’s unattainability statement is a consequence of the degeneracy statement, in that direction and not the other.

What the counts are #

The carbon monoxide case is the one that shows the two kinds of residual entropy apart, and it is worth separating carefully because they are usually presented together.

A CO molecule in a crystal can sit either way round, and the two orientations differ in energy by very little, so a count of two per molecule gives $R\ln 2 = 5.76$ joules per kelvin per mole. The measured residual entropy is 4.6.

The shortfall is not an error. The two orientations are not exactly degenerate — the molecule has a small dipole moment, so neighbouring molecules prefer to align — and on the way down some ordering happens before the rotation freezes. So the true ground state is unique or nearly so, and the residual entropy is a glass’s rather than an ice’s: a property of the cooling rather than of the spectrum.

Ice is different. Its degeneracy is protected by geometry — the ice rules are a constraint that an enormous number of configurations satisfy exactly, with no energy separating them — so the count is the answer and no amount of patience changes it. Pauling’s estimate of $(3/2)^N$ is within one per cent of the measurement.

That distinction is the useful content of the rung. Both look like residual entropy on a calorimeter, and one is a statement about a spectrum and the other about an experiment.

The gaps, and why they are so different #

The figure of ordering temperatures spans ten orders of magnitude, and where each degree of freedom sits on it is a question about an energy scale, which is worth working through because it is the same arithmetic every time.

Lattice vibrations. A phonon’s energy is $\hbar\omega$ with $\omega$ set by the interatomic force constants and the atomic mass, and for an ordinary solid the Debye temperature is a few hundred kelvin. So a solid at room temperature is already partly frozen out vibrationally, which is why heat capacities fall below the classical value long before anything else does.

Electron spins in a paramagnetic salt. The spins interact through their magnetic dipole fields, and a Bohr magneton at a nanometre gives an energy of order $10^{-25}$ joules, which is a few tens of millikelvin. Below that they order and their entropy is gone.

Nuclear spins. The nuclear magneton is smaller than the Bohr magneton by the mass ratio, about 1,836, so the dipolar energy is smaller by its square — a factor of $3.4 \times 10^6$. Millikelvin becomes nanokelvin, and that single ratio is why nuclear demagnetisation reaches six orders of magnitude further than electronic.

Every one of those numbers is a coupling energy divided by Boltzmann’s constant. Nothing else enters, and the enormous range on the figure is the range of coupling energies in matter — from an electronvolt for a chemical bond down to $10^{-32}$ joules for two nuclei a lattice spacing apart.

The entropy nobody has taken out #

The most striking consequence is what a sample at a very low temperature is actually like.

Copper’s nuclei have spin 3/2, so each has four states and each mole of copper carries $R\ln 4 = 11.5$ joules per kelvin per mole of nuclear-spin entropy. Nothing removes it until something orders the spins, and the only thing that can is their own mutual interaction — which for copper is dipolar and correspondingly feeble.

The ordering temperature is about 58 nanokelvin. Above that the spins are as disordered as they were at room temperature.

So a copper sample cooled to a microkelvin — which is a considerable technical achievement — has a lattice entropy that is entirely negligible and a nuclear spin entropy that is unchanged from its room-temperature value. Its total entropy is dominated by a degree of freedom that has not begun to cool in the relevant sense, and there is no violation anywhere: the spins are simply not in their ground state, because a microkelvin is twenty times their ordering temperature.

That is also what makes nuclear demagnetisation the coldest technique there is. The staircase of the first rung works by removing entropy from a spin system magnetically and then letting it take entropy from whatever it is attached to, and the amount available is $R\ln(2I+1)$ per mole. Electron spins run out at a millikelvin, where their own interactions order them. Nuclear spins have six orders of magnitude further to go, and the lowest temperatures ever reached — a few hundred picokelvin in a rhodium nuclear spin system — are the bottom of that staircase.

Where the entropy actually goes #

There is a question the figures invite and do not answer: when a spin system orders, where does its entropy go?

Into the lattice, as heat, and that is the whole mechanism of magnetic cooling run in reverse. The ordering releases $R\ln g$ of entropy times the temperature, which at a nanokelvin is a minute amount of energy — but the heat capacity of everything else at a nanokelvin is minute too, so the temperature rise is not negligible.

That balance is what limits how long a nuclear-demagnetisation experiment lasts. The cold stage is warmed by every stray heat leak — vibration, radioactivity in the apparatus, cosmic rays passing through — and the time it stays cold is the entropy available divided by the leak. A well-built stage holds a few tens of microkelvin for days and a few hundred picokelvin for hours.

It also explains why the lowest temperatures ever reached are spin temperatures rather than temperatures of everything. In a rhodium sample at 280 picokelvin, the nuclear spin system is at that temperature and the lattice is at a few microkelvin, because the two are coupled only weakly and the spins were cooled directly. Whether that counts as a temperature of the sample is a question about how strongly two subsystems have to be coupled before their separate temperatures stop being separate — the same question a spin system pushed past infinite temperature raises from the other end — and at these temperatures they do not.

The measurement that made the law usable #

The third law’s practical content is that entropies have no arbitrary constant, and that is what made chemical thermodynamics possible — so it is worth saying how the constant is fixed in practice, because the method is this rung’s statement used backwards.

Take a substance, cool it as low as the apparatus reaches, and integrate $C_p/T$ upward, adding a latent heat at each transition. That gives the entropy at any temperature relative to whatever was left at the bottom — and the third law says that what was left is $k\ln g$, which for a good crystal is zero. So the integral is an absolute entropy.

Absolute entropies are what equilibrium constants are computed from. The position of a chemical equilibrium is fixed by a free-energy difference, the free energy contains $-TS$, and without an absolute $S$ the whole calculation would carry an unknown constant per species. Every tabulated equilibrium constant that was computed rather than measured rests on the law of this rung.

The check that it works is that the calorimetric entropies agree with the ones computed from spectroscopy — from measured vibrational frequencies and rotational constants, using the counting that defines entropy directly. They agree for most substances, and where they disagree the difference is a residual entropy with an identifiable count behind it. That agreement, across hundreds of substances, is the real evidence for the law: not a single decisive experiment but a systematic consistency between two entirely different routes to the same number.

What would count as a violation #

It is worth being precise about what the law forbids, since the two figures above are full of non-zero entropies at low temperature.

A violation would be a system demonstrably in its ground state, with that ground state demonstrably unique, and a measured entropy that did not go to zero. Nothing of the kind has been found, and the law is regarded as secure.

What is not secure is the converse. There is no general proof that every system has a non-degenerate ground state, and there are models — spin ices, certain frustrated antiferromagnets, some quantum spin liquids — whose ground-state degeneracy grows with the size of the system, giving a residual entropy per particle that survives to zero temperature in the thermodynamic limit.

Spin ice is the best-studied case, and it is the same combinatorics as water ice transplanted into a magnetic system: the moments on a pyrochlore lattice obey a two-in-two-out rule, the count is Pauling’s, and the measured residual entropy of dysprosium titanate matches $R\ln(3/2)$ to a few per cent. Whether that degeneracy survives to absolute zero or is lifted by some weaker interaction at a temperature nobody has reached is exactly the open question, and it has the same shape as the glass question of the rung below: an extrapolation into a region the experiment cannot enter.

The staircase, seen from this rung #

The first rung of this ladder derives unattainability from a picture of two entropy curves that must meet at the axis, and it is worth saying where that requirement comes from now that the statistical form is available.

The two curves are the entropy against temperature at two field strengths, and the cooling cycle steps between them. What forces them to meet at $T = 0$ is that both approach $k \ln g$ with the same $g$: the applied field changes the energies of the spin states but not how many there are, so the ground-state degeneracy is a property the field cannot alter. Two curves approaching the same limit must converge, so each step of the staircase is shorter than the last, and no finite number of them reaches the floor.

That argument fails exactly where the degeneracy is field-dependent, and there is such a case. A field large enough to make the ground state unique where it was degenerate — by splitting a Kramers doublet, say — changes $g$ from two to one, and the two curves then approach different limits. The staircase argument does not apply, and neither does unattainability in that form.

What happens instead is that the system’s own internal field takes over: no laboratory field can be reduced below the field the spins make for each other, so the last step is limited by the material rather than by the apparatus. That is why the first rung’s figures carry a residual internal field as an option, and it is what actually stops a demagnetisation stage.

Where this stops being right #

The law is about equilibrium and low temperatures are where equilibrium is hardest. Every measurement at a nanokelvin involves timescales of hours against relaxation times that can be longer, so whether a system is in its ground state or merely stuck in a nearby one is an experimental question that has to be settled sample by sample.

“Degeneracy” is exact and real spectra are not. Any two states of a finite system differ in energy by something, so every degeneracy is approximate and the question is whether the splitting is small compared with $kT$. That makes the third law a statement whose meaning depends on the temperature reached, which is uncomfortable and is the honest position — an approximation announcing where it stops rather than concealing it.

The nuclear entropies quoted are per mole of a specific nuclide. Natural copper is two isotopes with different spins and different moments, so the numbers are averages, and a stage built from an enriched isotope behaves measurably differently.

And nothing here is about the interior of a black hole. Horizon entropy is proportional to area and does not go to zero as anything cools, and whether it counts as a degeneracy of a ground state is one of the central open questions of the subject rather than an exception to be noted.

What the pictures cannot show #

The bars in the second figure represent entropies per mole and the entropy of a mole of anything is a number with $10^{23}$ in it. What is drawn is $R\ln g$ — a logarithm of a count so large it has no name — and the drawing makes it look like a modest quantity, which for the purposes of a heat balance it is and for the purposes of understanding what is being counted it is not.

Nor can the first figure show the coupling. Each degree of freedom is drawn as a separate bar at a separate temperature, and what makes them a single sample is that they exchange energy — slowly, at these temperatures, and not at all in the limit. A picture of four independent bars is a picture of the limit in which the sample is four samples.

Where the ladder stands #

Five rungs stand on third-law. The first found the staircase that never reaches the floor. The second found ice’s residual entropy and where the number comes from. The third found a bounded spectrum letting a temperature pass through infinity. The fourth found a residual entropy belonging to an experiment rather than to a substance. This one restates the law as a condition on a spectrum, and the previous four become cases of it.

The habit worth carrying away is about which of two statements of a law to keep. When a law has an operational form and a structural one, the structural form usually explains the exceptions and the operational form usually generates them. Unattainability is a fact about apparatus and follows from the degeneracy statement; the degeneracy statement is a fact about spectra and follows from nothing here. Preferring the second is what turns residual entropy from an anomaly into a prediction.

What is left on this ladder is the question the spin-ice case raises and this rung cannot settle: whether a ground-state degeneracy that grows with system size is a real thermodynamic property or an artefact of an interaction nobody has reached yet. That is a question about the very bottom of a spectrum, it is being pursued in frustrated magnets and in quantum spin liquids, and it is one of the few places where the third law is doing work rather than being obeyed.