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The FCC conventional unit cell has four atoms, so the cube containsatoms, neglecting surface corrections. A phonon is one excitation quantum of a vibrational normal mode; there can be many quanta per mode, and zero-point motion does not add half a phonon to the thermal population.

For the low-temperature thermal phonon number, integrate the Bose-Einstein distribution against the Debye model phonon density of states:At $10\mathrm{K}$, ${\Theta }_D/T=21.5$, so extending the upper limit to infinity has negligible effect at the precision required. The supplied integral yieldsThis is the thermally excited phonon number, not the number $3N$ of available modes.

The continuum approximation fails when the thermally occupied wavelengths become comparable with the cube size $L$. From the Debye temperature,A lowest nontrivial acoustic frequency is of order $v/L$; a standing-wave estimate ${\omega }_{\min }\sim \pi v/L$ corresponds to $\hslash {\omega }_{\min }/k_B\sim 0.25\mathrm{K}$. Periodic boundary conditions instead give $2\pi v/L$ and $0.50\mathrm{K}$. The precise spectrum depends on mechanical boundary conditions and branch speeds, neither of which is specified.

This finite-crystal phonon freeze-out requires summing discrete normal modes. For a finite positive lowest frequency, $N_{\mathrm{ph}}\sim g_{\min }{e}^{-\hslash {\omega }_{\min }/(k_{B}T)}$, rather than extrapolating the bulk $T^3$ law indefinitely. Rigid translations and rotations of an isolated crystallite are not internal thermal phonons. Surface acoustic modes can modify the intermediate crossover, but do not justify inventing a unique boundary spectrum.

The small parameter controlling the local Taylor expansion is the typical bond displacement divided by $\xi$. In the classical regime, the root-mean-square displacement is $x_{\mathrm{rms}}\simeq \sqrt{k_{B}T/\kappa }$. ThusFor example, the magnitude of the cubic term divided by the quadratic term is $|x|/\xi$, and subsequent terms carry further powers of this ratio. A physically safe bound-bond regime also has $k_{B}T\ll D$, or $T\ll 2.90\times 10^3\mathrm{K}$. These are the same parametric scale, differing only by a factor two. The inequalities are strong: they do not imply that a truncated local expansion is accurate close to several thousand kelvin.

At very low temperature, replace the classical fluctuation estimate by the zero-point motion as discussed in part ii. Its bond displacement is small compared with $\xi$ for the supplied parameters, so low temperature does not itself invalidate the local expansion. Actual melting, other neighbours and bond breaking are outside this one-dimensional bond model.

The large rises mark the opening of Ga and As K-shell photoionization channels at their respective thresholds. Above each edge, the outgoing photoelectron interferes with waves backscattered by neighboring atoms; that interference modulates the absorption transition probability. Near-edge resonances and multiple scattering require a different level of treatment from the simple high-energy single-scattering formula. Ga has four As nearest neighbors, while As has four Ga nearest neighbors, so the two edges sample different central atoms and scattering factors even though they probe the same nearest-neighbor bond lengths.

Write $k=\sqrt{2m_e(E-E_0)}/\hslash$ and $\chi =(\mu -{\mu }_0)/{\mu }_0$ relative to a smooth atomic background. The phase $2k{R}_j+{\varphi }_j(k)$ includes the outward-and-return propagation distance and atomic phase shifts. The factor $R_j^{-2}$ describes geometrical spreading; the backscattering amplitude depends on the neighbor species. The exponential $e^{-2R_j/\lambda (k)}$ accounts for loss of coherent photoelectron propagation. Averaging over Gaussian bond-length fluctuations gives $⟨e^{2ik\delta R}⟩=e^{-2k^2{\sigma }_j^2}$.

The printed polarization factor needs a notation correction: for K-shell dipole absorption it is proportional to $({\hat{\mathbf{R}}}_j\cdot \mathbf{ϵ}{)}^2={\cos }^2{\theta }_j$, not literally the cosine of a dimensional vector dot product. Four equivalent tetrahedral bonds have an isotropic sum of these projection weights. If the sum is over shells rather than individual neighbors, it also needs the coordination number. A common conventional amplitude contains $S_0^2{f}_j(k)/(k{R}_j^2)$; omitted normalization and $1/k$ factors must be absorbed in the printed $f_B$ convention rather than silently changing its dimensions.

These are the ingredients of extended X-ray absorption fine structure. A shell fit uses calculated or reference scattering amplitudes and phases, and refines bond distances, coordination and mean-square relative disorder together with an edge-energy correction. Fourier transformation separates approximate distance contributions, but the uncorrected peak position is shifted by the scattering phase and is not an exact bond length.^[ref]

Sample A has the larger Ga–As bond length. Its oscillations advance slightly faster with $k$: at higher $k$ its corresponding maxima lie to the left of B's. Reading several full cycles gives approximate periods $\Delta k_A\simeq 1.47{\mathring{\mathrm{A}}}^{-1}$ and $\Delta k_B\simeq 1.50{\mathring{\mathrm{A}}}^{-1}$. In the constant/common-phase-slope estimate, $R\propto 1/\Delta k$, soThis is a rough percent-level difference, not a precise fitted strain. Nonconstant atomic phase shifts affect the absolute distance inferred from a period and also limit this fractional estimate. The graph does not establish which alloy has which indium concentration or how much its bulk lattice parameter changed. Local Ga–As relaxation need not follow a uniform bulk strain exactly.

Low temperature reduces thermal relative motion, preserving high-$k$ oscillations and separating small phase changes more clearly. It does not remove zero-point motion or static alloy disorder. To obtain an illustrative disorder estimate, take the high-$k$ envelope of the plotted $k\chi (k)$ to fall from about $0.19$ at $k=7.5{\mathring{\mathrm{A}}}^{-1}$ to $0.12$ at $k=10.4{\mathring{\mathrm{A}}}^{-1}$. If other envelope factors vary slowly over this interval,The value is only an order-of-magnitude, model-dependent reading: the backscattering amplitude, mean free path and normalization have not been independently removed. In particular, ${\sigma }^2$ is the bond-projected mean-square relative displacement,including correlations and static disorder. It is not uniquely either atom's mean-square displacement. Equal uncorrelated projected displacements would each contribute ${\sigma }^2/2$, but that extra assumption is not supplied by the figure.

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.

At room temperature, electrons require quantum mechanics, including their occupied orbitals or bands and Fermi statistics. Heavy nuclei can often be treated as distinguishable particles near classical equilibrium lattice sites, because their exchange wavelengths are short. Their coupled displacements form phonons, however, and the occupation of a vibrational mode is classical only if $k_{B}T\gg \hslash \omega$ for that mode. Even at room temperature some high-frequency modes remain quantum mechanical.

At $1$ K, quantum lattice vibrations, zero-point motion and the scarcity of thermally excited phonons are essential. Many heavy nuclei still do not have overlapping exchange wavefunctions; quantum vibration and quantum exchange are distinct criteria. For the same spacing, the nuclear overlap threshold from the rough estimate is $T_N\simeq 211/A$ K, illustrating why the nuclear mass cannot be omitted from a blanket low-temperature claim.

Zero-point motion is the nonzero quantum fluctuation of position and momentum in a system's ground state. For a quantum harmonic oscillator, the ground state has energy $\hslash \omega /2$ and nonzero position and momentum variances even at zero temperature. It does not represent a thermal mixture and need not carry entropy. The large zero-point motion of light helium atoms helps explain why helium can remain a liquid at low pressure even as temperature approaches absolute zero.