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The Landau criterion asks whether creating an excitation can lower a moving superfluid's energy. For a weakly coupled, heavy obstacle moving through a uniform liquid initially at zero temperature, let $ϵ(p)>0$ be the excitation energy in the fluid rest frame, with momentum magnitude $p$. Transferring momentum $\mathbf{p}$ from an obstacle moving at velocity $\mathbf{v}$ supplies, to leading order in its negligible recoil, energy $\mathbf{v}\cdot \mathbf{p}$. Emission requiresFor an isotropic dispersion relation, the easiest direction is parallel to $\mathbf{v}$, giving the necessary kinematic conditionAn infimum includes cases in which the limiting slope occurs only as $p\to 0$. A finite-mass obstacle of mass $M$ instead loses $\mathbf{v}\cdot \mathbf{p}-p^2/(2M)$, so the corresponding emission inequality includes the extra positive recoil term $p^2/(2M)$ on the left. This is an energy and momentum balance, not a calculation of an emission rate.

The Landau criterion excludes spontaneous emission into the specified spectrum below its threshold. Above it, kinematic permission alone does not guarantee strong coupling or rapid decay. It is not a universal observed flow threshold: walls, channel geometry and quantum vortex nucleation can enable other decay routes, and finite temperature permits scattering from existing excitations. Using only bulk phonons and rotons leaves those processes outside the model.

Landau's 1941 energy and momentum argument used a roton approximation different from the finite-momentum minimum illustrated below.^[ref]

Landau received his Nobel Prize in Physics in 1962 while hospitalized in Moscow after a car accident.^[ref]

For a deliberately simple Bose-Einstein condensation comparison, stipulate an ideal free-particle excitation law $ϵ=p^2/(2m)$. Then $ϵ/p=p/(2m)$ has infimum zero. Macroscopic one-particle occupation therefore supplies no positive emission threshold by this argument alone. By contrast a linear phonon law $ϵ=cp$ yields $v_L=c$. The dispersion relation and the collective response contain information that the occupation criterion alone does not supply; Bose-Einstein condensation and thermodynamic superfluidity must be tested separately.

To check the picture algebraically, on the stipulated roton branch,Thus tangency occurs at $q_L=\sqrt{4.3}$, slightly above the roton energy minimum. At an energy minimum $e^{\prime }=0$, but a tangent through the origin needs $e^{\prime }=e/q>0$. Minimizing energy and minimizing the energy-to-momentum ratio are different operations. This geometric construction makes the criterion useful even when the full dispersion relation is supplied numerically rather than by a simple formula.

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.

A roton is an excitation near a local minimum of a superfluid's dispersion relation at nonzero momentum $p_0$. The displayed local approximation defines a positive energy gap $\Delta$ and positive curvature effective mass $m_r$; it is not a free helium atom or a macroscopic quantum vortex.^[ref] In the stipulated plot, these correspond to $p_0=2p_{\ast }$, $\Delta =0.6E_{\ast }$ and $m_r=p_{\ast }^2/(4E_{\ast })$. The plot illustrates how a low energy at appreciable momentum can lower the Landau criterion below the speed of sound, without assigning any measured parameter to real helium.

Landau's 1947 revision joined the low-momentum linear part to a spectrum with a finite-momentum minimum, retaining the macroscopic superfluid theory while changing its excitation model.^[ref]