How does adding heat cause helium-3 to freeze? [duplicate]

Question

According to here at 3MPa helium-3 actually goes from liquid to solid at 0.1 kelvin, this is the complete opposite of how normal materials behave, how is this possible?

Answer 1

I quote from Richardson's article "The Pomeranchuk Effect":

"In 1950, I. Pomeranchuk, a well-known particle theorist, suggested that melting ³He could be cooled by squeezing it. At the time of his suggestion ³He was quite rare and had not yet even been liquefied. He observed that at low enough temperatures the thermal phenomena in condensed ³He would be dominated by spin properties instead of phonon properties. The liquid of ³He would obey Fermi statistics with an entropy proportional to the temperature, much like the free electrons in a good metal. On the other hand, the entropy of solid ³He would be that of the disordered collection of weakly interacting spin-1/2 nuclei. At temperatures greater than the (then) expected nuclear-magnetic ordering temperatures less than 1 mK, the entropy per mole of solid ³He would be S=R ln 2, independent of temperature until the high-temperature phonon modes of the solid become important. (The Debye temperature of solid ³He is approximately 30 K.)... The entropy of solid ³He exceeds that of liquid ³He at temperatures less than 0.3 K." [emph. added]

The Clausius–Clapeyron relation tells us that the pressure–temperature slope for a phase transition is ΔS/ΔV, where ΔS and ΔV are the change in entropy and volume upon transition. With melting/evaporation/sublimation, we typically see an entropy increase coupled with a volume increase, but this is not required; perhaps the most common counterexample is water, which shrinks with melting. This gives the water phase diagram a reentrant melting slope, i.e., a negative slope, up to around 2000 atm:

That same reentrant behavior, in conjunction with the switching of sides such that solid lies on the higher-temperature side, is the aspect of interest in the helium-3 phase diagram:

In that case, it arises because—unusually—the solid state can have a higher entropy that the liquid state. The higher-entropy state is always the higher-temperature equilibrium state. (The reason is that the Gibbs free energy, which Nature tends to minimize, contains a negative coefficient in front of the entropy term. This ultimately arises from Nature's tendency to maximize entropy.) The entropy change ΔS for melting is thus negative under these particular conditions for this particular material.

In Low-Temperature Physics, Enss and Hunklinger write

"...the melting curve of ³He...exhibits a pronounced minimum at a temperature of about T = 320 mK. ...we can conclude that the entropy of solid ³He is larger than that of liquid ³He at temperatures below the minimum. ...The essential contribution to the entropy comes from nuclear spins. In the liquid, the entropy varies at low temperatures proportional to T as expected for a Fermi gas... In solid ³He the atoms are strongly localized and the Fermi-gas model is not applicable. At high temperatures the orientation of the localized spins is statistical and their contribution to the entropy is S= = R ln 2, where R is the universal gas constant... With decreasing temperatures a transition to an antiparallel arrangement of the spins is found. The transition temperature to the antiferromagnetic state is 0.9 mK. At T = 0.32 K, the temperature at which the minimum occurs, the entropies of liquid and solid ³He are equal."

See also Farcher's great description of what's happening with the nuclear spins from a previous question on this phenomenon.