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Pyotr Leonidovich Kapitza's 1938 test compared helium flow through optically flat glass disks. Suspension set the tube-to-bath level difference; a cathetometer read levels. The gap was estimated as $0.5\mu \mathrm{m}$. Levels barely moved over minutes above the transition, but equalized in seconds below. A laminar flow viscosity calculation gave only an upper bound.^[ref]

The essential control is to compare the same restriction and level readout at different temperatures. A changed pressure head is not itself evidence for a changed material response: the driving head must be included when interpreting the flow. For an idealized comparison at equal vapor pressure, negligible surface tension corrections and nearly constant mass density $\rho$, hydrostatic equilibrium givesHere $\Delta h$ is the difference in free-surface heights and $g$ is gravitational acceleration. This explains why following a liquid level provides both a driving pressure and a transported volume; it does not turn a finite-resolution observation into proof of zero viscosity.

Entropy at absolute zero. The third law of thermodynamics concerns the low-temperature limit of equilibrium states. In the zero-entropy formulation, the entropy of a perfect crystal with a nondegenerate ground state tends to zero as $T\to 0$. The statistical interpretation is $S=k_B\ln \Omega$: a unique accessible ground state has $\Omega =1$. The entropy-difference formulation of the third law of thermodynamics states that entropy differences associated with regular changes of mechanical parameters tend to zero as absolute zero is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate ground state can carry residual entropy.

For a phase of fixed composition and pressure,A finite limiting entropy constrains this integral. Ordinary equilibrium heat capacities therefore vanish as temperature approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic phonons and $C\propto T$ for a normal degenerate Fermi liquid. A constant positive heat capacity extrapolated to zero would give a logarithmically divergent entropy difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.

Gibbs free energy. For fixed particle numbers,Thus, when $S\to 0$, the Gibbs free energy approaches its zero-temperature value with a horizontal tangent:The equality $G(0,p)=H(0,p)$ uses the enthalpy $H=U+pV$; neither limiting value needs to be zero. The low-temperature Gibbs free-energy expansion makes the approach quantitative. For $C_p=a{T}^r$ with $r>0$ and $S(0)=0$,The second derivative is ${\partial }_T^{2}G=-C_p/T\le 0$, although it need not tend to zero. For a reaction between such equilibrium phases, ${\partial }_T\Delta G=-\Delta S\to 0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine residual entropy $S_0$ is retained instead, $G=G(0,p)-S_{0}T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.

Why helium remains a liquid. Helium atoms attract each other weakly and are exceptionally light. Their zero-point energy is large compared with the binding that would localize them in a solid. Both helium-4 and helium-3 can therefore remain liquid down to the lowest temperatures at low pressure; substantial applied pressure is needed to solidify them. This is not a violation of the third law. A quantum liquid can have a unique low-entropy ground state, and zero-point motion does not itself imply thermal entropy. The third law does not require every substance to crystallize at atmospheric pressure.

A helium-4 atom has integer spin zero and is a boson. Below the lambda point near $2.17\mathrm{K}$ at saturated vapor pressure, its liquid is superfluid. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the condensate fraction need not equal the superfluid fraction. The two-fluid model of superfluidity separates an entropy-carrying normal component of a superfluid from the superfluid component, whose entropy is zero. At sufficiently low temperature the dominant thermal excitations are acoustic phonons: their linear dispersion in three dimensions gives $U-U_0\propto V{T}^4$, $C_V\propto V{T}^3$ and $S\propto V{T}^3$. Higher-energy rotons are exponentially suppressed. Thus the normal component of a superfluid and entropy disappear as $T\to 0$, while superfluid flow and quantum motion can persist.

A helium-3 atom has spin $1/2$ and is a fermion. Well below its Fermi temperature, but above its superfluid transition, liquid helium-3 is a degenerate interacting Fermi liquid. Only quasiparticles close to the Fermi surface are thermally excited, giving $C_V\propto T$ and $S\propto T$. At temperatures of order a millikelvin, helium-3 also becomes superfluid through formation of Cooper pairs of fermions. This is not Bose-Einstein condensation of individual helium-3 atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic quasiparticle contribution to heat capacity is exponentially suppressed at temperatures well below the pairing gap. Gapless collective modes can still give thermal terms following a power law. The entropy nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.^[ref]

Helium-3 also illustrates the importance of nuclear spin entropy. Over an intermediate low-temperature range, localized spins in the solid can retain approximately $k_B\ln 2$ of entropy per atom, while the degenerate liquid has much less. Reversible conversion of liquid to the higher-entropy solid then absorbs heat; adiabatic compression along the melting curve can cool the sample. This is Pomeranchuk cooling. At still lower temperature, interactions between the nuclear spins remove this nearly free-spin entropy, so a constant $k_B\ln 2$ cannot be extrapolated to absolute zero.^[ref]

Liquefaction and cooling methods. A gas must be below its liquid-gas critical point to form a distinct liquid phase by compression. Compression above the critical temperature alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression heat, cools the compressed gas in a heat exchanger and uses expansion to lower its temperature further.

A throttling valve operates approximately at constant specific enthalpy when kinetic-energy changes and heat leakage are negligible. The Joule-Thomson coefficient isusing consistent molar quantities, or consistent total quantities for a fixed amount. A pressure drop cools the gas if ${\mu }_{\mathrm{JT}}>0$. An ideal gas has zero Joule-Thomson coefficient; real intermolecular interactions are essential. In particular, room-temperature helium warms on throttling, so it must first be brought into a region of positive Joule-Thomson coefficient. Expansion through an engine or turbine can instead remove work and produce approximately isentropic cooling.

A regenerative liquefier sends the cold returning gas through a counterflow heat exchanger to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces liquid. Modern helium liquefaction commonly combines work-producing expansion with a final throttle. ^[ref]

Once liquid exists, pumping away vapour lowers its saturation pressure and boiling temperature. The escaping vapour carries latent heat away, cooling the remaining liquid. At still lower temperatures, a dilution refrigerator exploits the different entropy carried by helium-3 in concentrated and dilute helium mixtures; adiabatic demagnetization and Pomeranchuk cooling provide other routes.

Approaching zero. The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach absolute zero in finitely many stages. As the entropy available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero entropy cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. Quantum liquids and refrigeration methods can approach $0\mathrm{K}$ while remaining consistent with vanishing equilibrium entropy; they do not provide a finite-operation route to absolute zero.

The two-fluid model of superfluidity separates mass current into a superfluid fraction and a normal component of a superfluid:The normal component carries entropy and responds to shear viscosity; an ideal superfluid flow is entropy-free. These are interpenetrating hydrodynamic responses, not two permanently distinguishable sets of atoms. Thermal elementary excitations carry momentum relative to the condensate and generate the normal response.

The Andronikashvili experiment gives an operational picture. A closely spaced stack of disks oscillates torsionally in He-II. Viscous coupling entrains the normal response between disks, changing the oscillator's inertia, while vortex-free superfluid need not follow the rotation. The temperature-dependent inertia measures the normal component under the apparatus's coupling conditions. Counterflow also permits second sound, primarily an entropy/temperature wave rather than the ordinary density wave of first sound.

Condensate depletion is a different quantity. The condensate fraction counts occupation of a single-particle mode, while ${\rho }_s$ measures the response to phase gradients or motion. In a weak three-dimensional homogeneous Bose gas at zero temperature,so interactions produce quantum depletion even without thermal excitations. In a translation-invariant Galilean system at zero temperature, the superfluid mass density can nevertheless equal the total density. He-II has strong interaction-induced depletion but a nearly complete zero-temperature superfluid response. Lattices, disorder and finite temperature require their own response calculation; condensate and superfluid fractions should not be equated in general.

The Landau criterion follows by considering an excitation of momentum $\mathbf{p}$ in a moving fluid. Its laboratory-frame creation cost can be $ϵ(\mathbf{p})-\mathbf{v}\cdot \mathbf{p}$. A necessary energetic condition against this decay channel isFor the low-momentum Bogoliubov spectrum of a weakly interacting Bose gas, $ϵ\simeq c_{s}p$, giving the sound-speed threshold for that channel. The phonon–roton spectrum of He-II can instead have its minimum ratio near the roton region. Vortex nucleation, boundaries and a moving obstacle can lower a measured critical velocity; the ideal excitation criterion is not a guarantee of dissipation-free flow at every smaller speed in every apparatus. An ideal noninteracting Bose gas has a condensate but quadratic excitations with $v_L=0$, another distinction between condensation and robust superfluidity.

A cold-gas test moved a blue-detuned laser beam through a trapped condensate and measured heating as its speed changed. The repulsive optical obstacle supplies a controlled perturbation; strong heating appeared above a threshold. Its interpretation must account for the inhomogeneous cloud and excitation/nucleation mechanism, rather than identify every observed threshold with the central sound speed.^[ref]

A superleak is a narrow or porous connection that strongly suppresses the viscous normal component of a superfluid while allowing appreciable superfluid flow. Treating ${\mathbf{v}}_n\simeq 0$ is a regime assumption, not a perfect material filter for particular atoms. The thermomechanical effect uses that distinction to maintain a temperature difference during chemical equilibration. The idealization fails when normal leakage or dissipative quantum vortex motion becomes appreciable.

The two-fluid model of superfluidity gives the normal component of a superfluid and the superfluid independent velocities.

An elementary uniform-flow example makes the distinction operational. Write the mass flux per unit area and the ideal entropy flux asHere $s$ is specific entropy, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and ${\rho }_s,{\rho }_n$ are response mass densities. The mass density ${\rho }_s$ used here gives the three-dimensional phase stiffness coefficient ${\rho }_s(\hslash /m{)}^2$ when ${\mathbf{v}}_s=(\hslash /m)\nabla \theta$ is substituted into the kinetic energy density ${\rho }_s{v}_s^2/2$. For a stipulated ${\rho }_s=3\rho /4$, ${\rho }_n=\rho /4$ and $v_n=3u$ along a channel, zero net mass flux requires $v_s=-u$. Yet $j_S=3\rho su$ is positive. Opposing velocities can therefore transport entropy while transporting no net mass.

This example explains why excellent apparent heat transport need not be ordinary diffusion of heat through a stationary liquid. In the ideal zero-net-mass-flux limit, the heat flux associated with this entropy transport is $T{j}_S$. Viscous losses, quantum vortices and external heat exchange require additional terms. The model also supports second sound.