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Compare uniform three-dimensional ideal gases at the same number density n, with fermion mass $m_F$, spin degeneracy $g_F$, boson mass $m_B$ and a single boson spin component. The Fermi temperature has $k_B{T}_F={\hslash }^2(6{\pi }^{2}n/g_F{)}^{2/3}/(2m_F)$, while Bose saturation gives $k_B{T}_B=2\pi {\hslash }^2[n/\zeta (3/2){]}^{2/3}/m_B$. Eliminating density yieldsThe degeneracy is essential; using a spinless Fermi sea changes the numerical estimate. This is an ideal-gas comparison and does not remove interaction effects in a dense liquid.

The Fermi temperature expresses a Fermi energy as a temperature scale, $T_F=E_F/k_B$, where $k_B$ is the Boltzmann constant. In the convention $E_F>0$ measured from the relevant band minimum, $T\ll T_F$ is the regime of a strongly degenerate Fermi gas. For an interacting Fermi liquid, an effective Fermi scale can be introduced through its quasiparticle dispersion; its convention should be specified.

Metallic atomic densities place very many electrons in a small volume. The Pauli exclusion principle prevents all these fermions occupying the lowest spatial state, so filling the available states produces a Fermi energy of order electronvolts. At room temperature, $k_{\mathrm{B}}T\simeq 0.026\mathrm{eV}$, much smaller than that scale. Equivalently, ordinary metals are at $T\ll T_F=E_F/k_{\mathrm{B}}$, where $T_F$ is the Fermi temperature. Their electrons are quantum degenerate even at ordinary laboratory temperatures.

The Fermi temperature isInserting it into the question's quoted estimate givesAs explained in (d), that count is per mole of conduction electrons. Using the stipulated six electrons per tungsten atom gives $1.8025\times 10^{-3}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ per mole of tungsten in the same rough model. The accurate free-electron Sommerfeld expansion coefficient would instead give $4.4474\times 10^{-3}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ per mole of tungsten. These are different approximations and counting conventions, not three interchangeable exact answers.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

In the free electron model, conduction electrons occupy independent plane-wave states with energy ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ in a uniform mean background. Crystal-periodic band structure and explicit electron–electron correlations are omitted. Electrons still obey the Pauli exclusion principle and Fermi–Dirac statistics; free does not mean distinguishable classical particles. A scattering time can be added phenomenologically to describe transport.

The Fermi energy is the highest occupied single-particle energy at zero temperature, measured here from the bottom of the parabolic band. The Fermi temperature is the corresponding scale $T_F=E_F/k_B$, not the actual thermal temperature of the sample. Metallic electron densities are large, so filling all distinct states up to the required momentum gives Fermi energies of several electronvolts, equivalent to tens of thousands of kelvin or more. At ordinary temperatures, only a narrow region around the Fermi surface has appreciably changing occupations.

A room-temperature metal and its lattice are nevertheless at room temperature, not at $T_F$. High occupied kinetic energies are present even in the zero-temperature ground state and cannot all be liberated as heat without violating the allowed occupancy structure. A metal below skin temperature often feels cold because it transfers heat rapidly out of the skin; its thermal conductivity and thermal effusivity matter, not the value of its Fermi temperature. At the same actual temperature, a poorly conducting material can therefore feel warmer while neither material is thermodynamically hotter.

For a nondegenerate two-dimensional electron gas, the two quadratic momentum terms supply $k_{B}T/2$ each by the equipartition theorem. There is no lattice restoring potential energy for a free electron. ThusOn cooling below the Fermi temperature, the Pauli exclusion principle restricts the thermally active electrons to within approximately $k_{B}T$ of the Fermi energy. The linear low-temperature heat capacity of a two-dimensional Fermi gas isThe heat capacity therefore decreases from its classical value as quantum degeneracy becomes important. The particle number of electrons and sample area are held fixed when differentiating the internal energy.

The nuclear Fermi gas model treats protons and neutrons as two species of spin one-half fermions moving approximately independently in a confining nuclear mean field. At a temperature small compared with their Fermi temperature, the Pauli exclusion principle fills the lowest available momentum states. It does not allow all nucleons to sit at zero momentum.

For species $i=p,n$, two spin states occupy each momentum-space cell $(2\pi \hslash {)}^3/V$. Counting a filled sphere giveswhere $n_i=N_i/V$. The Fermi energy is measured from the mean-field well bottom; it need not equal the energy needed to remove a nucleon into vacuum. Occupied-state averaging givesEven at zero temperature there is kinetic energy and degeneracy pressure. As an explicitly illustrative input, total density $n=0.16{\mathrm{fm}}^{-3}$ and $N=Z$ give $p_{F}c\simeq 263\mathrm{MeV}$ and $E_F\simeq 37\mathrm{MeV}$. These numbers illustrate the formula; they are not additional data supplied by the paper.

Separate proton and neutron seas explain a kinetic contribution to nuclear symmetry energy. At fixed $n$, put $\delta =(N-Z)/A$ and define $E_F$ at $n_p=n_n=n/2$. ThenIncreasing one species' population raises its Fermi surface and costs energy, favoring similar populations before electrostatic and interaction terms are included.

pauli blocking suppresses collisions deep in a filled sea because scattering needs unoccupied final states. An energetic external probe can knock a nucleon into an empty state and reveal its initial Fermi motion, making the model useful for broad inclusive scattering distributions. A finite surface can be approximated using a position-dependent nuclear mean field.

A free gas cannot produce nuclear binding or saturation: those require an attractive interaction and its density dependence. The smooth gas picture also omits discrete nuclear shells, nuclear spin-orbit splitting, nuclear pairing, short-distance correlations and deformation. It cannot assign all ground-state spins from a momentum sphere. Its useful prediction is the scale and occupancy of nucleon motion, with a surrounding mean field supplied separately.

For the electronic and lattice heat-capacity crossover in a monovalent metal, to obtain a numerical estimate from $E_F$ alone, assume monovalent potassium is a three-dimensional free electron gas with a parabolic band. Including both spins, the molar density of states is $g(E_F)=3N_A/(2E_F)$. Using the supplied approximate coefficient $3$, rather than silently replacing it by ${\pi }^2/3$, givesEquating it to the stated low-temperature Debye heat capacity yields the nonzero crossoverWe used $k_B=8.61733\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$ and $E_F=2\mathrm{eV}$. This is self-consistently far below ${\theta }_D=100\mathrm{K}$ and the Fermi temperature. Using the exact Sommerfeld coefficient instead gives $0.954\mathrm{K}$; that is a change of approximation, not a discrepancy in the algebra. Without a band/DOS assumption, the supplied Fermi energy does not by itself determine $g(E_F)$.

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.

One conduction electron per atom gives number density $n=\rho N_A/M\simeq 8.50\times 10^{28}{\mathrm{m}}^{-3}$. Filling two spin states per momentum state in the three-dimensional Fermi gas gives $k_F=(3{\pi }^{2}n{)}^{1/3}$ andThe Fermi temperature greatly exceeds ordinary temperatures, explaining why these conduction electrons are degenerate rather than a classical Maxwell gas. The numerical result uses the stipulated free-electron band and bare electron mass, rather than a detailed copper band structure.

A spin-half gas has two spin states. Filling the Fermi sphere at zero temperature gives $n=k_F^3/(3{\pi }^2)$, soWith $m_3\simeq 5.008\times 10^{-27}\mathrm{kg}$, the number density is $1.62\times 10^{28}{\mathrm{m}}^{-3}$ and $k_F\simeq 7.82\times 10^9{\mathrm{m}}^{-1}$. ThusThe associated Fermi temperature isFermionic degeneracy matters on cooling to a few kelvin or below in this model, and is pronounced for $T\ll T_F$. Liquid helium-3 is interacting; its quasiparticle effective mass and thermodynamics need not equal the bare-mass free-gas predictions. The requested calculation is the ideal Fermi-gas scale.

Only states within an energy window of order $k_{B}T$ around the Fermi energy can change occupation in a degenerate metal. A Sommerfeld expansion gives electronic energy above its zero-temperature value proportional to $g(E_F)(k_{B}T{)}^2$, hence $C_e=\gamma T$. Three-dimensional linear-dispersion acoustic phonons have density of states proportional to ${\omega }^2$; integrating their Bose occupations gives thermal energy proportional to $T^4$ and the Debye heat capacity $C_{mph}=\beta T^3$. Thus $C/T=\gamma +\beta T^2$ at sufficiently low temperature in the normal state.

Reading the straight low-temperature fit gives approximately $\gamma \simeq 33\mathrm{mJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-2}$ and $\beta \simeq 3.8\mathrm{mJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-4}$. These estimates agree with the reported fit to this material in the original experiment.^[ref] Treating the graph values as approximate, the molar density of states isThe stipulated two electrons per formula unit give $n=2N_A/V_m\simeq 2.62\times 10^{27}{\mathrm{m}}^{-3}$. For a single spin-degenerate parabolic band, the molar density of states is $g(E_F)=3(2N_A)/(2E_F)=3N_A/E_F$. HenceThese are model-dependent effective parameters inferred from the measured heat capacity and assumed carrier count, rather than an independent determination of a unique band structure.

Using the Debye prefactor stated in the question givesThis prefactor counts three effective modes per formula unit. Consistently choosing the same count for the Debye wavevector gives $k_D=(6{\pi }^2{N}_A/V_m{)}^{1/3}\simeq 4.3\times 10^9{\mathrm{m}}^{-1}$. The corresponding average sound speed isThere is a normalization choice here: a formula unit has 31 atoms. With the conventional full atomic-mode count, the molar Debye coefficient contains an additional factor 31, and both ${\Theta }_D$ and $k_D$ increase by $31^{1/3}$. That convention gives ${\Theta }_D\simeq 251\mathrm{K}$ and $k_D\simeq 1.34\times 10^{10}{\mathrm{m}}^{-1}$, while leaving the inferred sound speed unchanged. The mode count in the Debye model must be used consistently; mixing the printed temperature with an atomic-count wavevector gives an erroneous sound speed.

The upward curvature above about $3\mathrm{K}$ indicates low-energy vibrations missing from a pure acoustic $T^3$ model. In this cage compound, the original analysis fits an additional localized mode with $\hslash \Omega /k_B\simeq 32\mathrm{K}$, assigned to a rattling mode of a barium guest atom.^[ref] At temperatures high compared with all relevant phonon energies, the full lattice approaches the Dulong-Petit law, $C_{mlattice}\to 3(31)R=93R\simeq 773\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The reduced three-mode model alone approaches $3R$ and omits the other branches. An approximately linear electronic term can still be added while the carriers remain degenerate, but cannot be extrapolated beyond their Fermi temperature.

In a conductor, mobile electrons redistribute in response to an electric potential, producing an induced charge density that opposes its source. Finite electronic compressibility means a finite density change requires a change of local kinetic Fermi energy. Screening is therefore not perfect at every microscopic distance. On scales comparable to atomic structure or the Fermi wavelength, a continuum local-density model itself fails.

Thomas-Fermi screening treats electrons as a locally uniform degenerate free electron gas, replaces the ionic background by a smooth neutralizing charge, and neglects nonlocal response, exchange and correlation corrections. The electrochemical chemical potential is uniform in equilibrium. Because an electron has charge $-e$, its electrostatic energy is $-eV$, givingAt zero potential $\mu =E_{F0}$. Linearizing at fixed $\mu$, with $|eV|\ll E_{F0}$, givesInsert this in Poisson's equation to obtain $({\nabla }^2-{\xi }^{-2})V=-{\rho }_{\mathrm{free}}/{ϵ}_0$, where the Thomas-Fermi screening length isAway from a point source, direct radial differentiation shows ${\nabla }^2(e^{-r/\xi }/r)={\xi }^{-2}{e}^{-r/\xi }/r$. The decaying solution is therefore $V=C{e}^{-r/\xi }/r$. Integrating Poisson's equation over a small sphere containing charge $Q$ fixes $C=Q/(4\pi {ϵ}_0)$. For a sufficiently large source, the linearized formula is only a far-field approximation, not a justified description of its singular core.

For monovalent sodium, the electron number density is its atomic number density,This gives $E_{F0}=3.15\mathrm{eV}$ andRoom temperature is far below the corresponding Fermi temperature, so degeneracy is well justified, and sodium's simple conduction band makes a free-electron estimate useful. However $k_F\xi =0.615$ and the mean atomic spacing is $n_0^{-1/3}=0.340\mathrm{nm}$. The potential varies on a scale shorter than $k_F^{-1}$ and the atomic spacing, so the slowly varying local-density and continuum assumptions are not quantitatively controlled at the screening length itself. A nonlocal treatment includes Friedel oscillations, absent from the pure exponential.

Nearly unchanged interlayer force constants in two-layer and twenty-layer films are consistent with very local screening and bonding forces: reducing the distant bulk need not strongly change the force between adjacent layers. They also indicate that metallic electronic response can survive in the thin film. The observation does not by itself validate the numerical Thomas-Fermi approximation, establish an identical full phonon spectrum, or rule out surface-induced electronic changes; it directly constrains the measured local force constant.

Set $\lambda \simeq d$ with the convention of part b. ThenThus ordinary solid-state temperatures are far below this rough overlap scale. The electrons need quantum statistics and quantum states throughout that regime. The number is an order-of-magnitude wavelength criterion, not an exact Fermi temperature: electron density, electronic band structure, effective mass and the numerical convention all matter.

Use thermal de Broglie wavelength ${\lambda }_T=h/\sqrt{2\pi m{k}_{B}T}$ and spacing $a=n^{-1/3}$. Degeneracy onset from thermal wavelength occurs atThe numerical factor depends on the convention for a typical momentum, but the scale ${\hslash }^2{n}^{2/3}/(m{k}_B)$ does not. For fermions the Fermi temperature $T_F={\hslash }^2(3{\pi }^{2}n{)}^{2/3}/(2m{k}_B)$ gives a comparable scale for spin degeneracy two.

Representative estimates, with the assumed densities stated explicitly, are:

Here $u$ is the atomic mass constant. Thus the characteristic orders are $10^5\mathrm{K}$ for electrons, sub-microkelvin for a dilute heavy atomic gas, and $10^{12}\mathrm{K}$ for neutrons. There is no density-independent quantum-degeneracy temperature for atomic gases: for example, taking $n=2.5\times 10^{25}{\mathrm{m}}^{-3}$ and mass $4u$ instead gives about $0.065\mathrm{K}$. These are illustrative input estimates, not values supplied by the paper. Denser neutron-star cores require relativistic and interaction corrections; the estimate identifies the degeneracy scale rather than a precise stellar temperature.

For a single-particle state of energy $ϵ$, put $z=e^{\mu /(k_{B}T)}$. The equilibrium mean occupancies areThe Maxwell-Boltzmann distribution is the dilute classical limit. The Fermi-Dirac distribution applies to identical fermions with at most one particle in each complete single-particle state by the Pauli exclusion principle. The Bose-Einstein distribution applies to identical bosons with no such upper occupation bound. The chemical potential fixes total particle number; these are mean state occupancies, not three separately normalized single-particle probability densities. For a number-conserving ideal Bose gas, $\mu$ cannot exceed the lowest single-particle energy.

Take $\mathbf{k}$ to be a wavevector and count periodic-box states with cell volume $(2\pi {)}^3/V$ in momentum space, parameterized by $\mathbf{k}=\mathbf{p}/\hslash$. Write $g_s$ for internal degeneracy. The number of states in a spherical shell is $g_{s}V4\pi k^{2}dk/(2\pi {)}^3$. Since $ϵ=\alpha k$,Thus the one-particle canonical partition function isHere $\beta =1/(k_{B}T)$, $g_s=1$ for one nondegenerate branch, and $g_s=2$ for the spin-half case below. The integral uses ${\int }_0^{\infty }{u}^2{e}^{-u}du=2$. For classical indistinguishable particles, $Z_N=Z_1^N/N!$, givingThe factor three in the energy arises from the linear dispersion relation and three-dimensional state counting; the usual nonrelativistic quadratic-dispersion value $3N{k}_{B}T/2$ does not apply.

For spin-half fermions at $T=0$, all states up to the Fermi energy are occupied. With $g_s=2$,Integrating the occupied energy states givesThese are the corresponding ultrarelativistic ideal Fermi gas formulae with general slope $\alpha$. The nonzero ground-state internal energy and degeneracy pressure reflect the Pauli exclusion principle: particles cannot all occupy the lowest-momentum state. Formally extrapolating the classical equations gives $U,p\to 0$, but Maxwell–Boltzmann statistics is not valid at fixed positive $n$ as $T\to 0$.

At large $T$, the number equation gives $z\simeq N/Z_1\ll 1$, so expanding the Fermi-Dirac distribution recovers the Maxwell-Boltzmann distribution. A convenient crossover scale is the Fermi temperature,For $T\gg T_F$, both gases have $p\simeq n{k}_{B}T$. At fixed $n$, the Fermi curve begins at $p_0=n{ϵ}_F/4$, has zero initial slope, lies above the classical line, and approaches it at high temperature. For example its low-temperature expansion is $p/(n{ϵ}_F)=1/4+({\pi }^2/6)(T/T_F{)}^2+\cdots$. The plot evaluates the finite-temperature occupation integrals at fixed number density.

If the source's symbol $\mathbf{k}$ is interpreted literally as momentum rather than a wavevector, the state-space cell is $h^3/V$. In that convention replace $\alpha$ in the wavevector-based coefficients and energy scales above by $\hslash \alpha$. The dimensionless thermodynamics and the relation $p=U/(3V)$ are unchanged.

Thermal effusivity measures a material's response to transient conductive contact:where $k$ is thermal conductivity, $\rho$ is mass density and $c_p$ is specific heat capacity at constant pressure in the thermal model. Its units are $\mathrm{W}{\mathrm{s}}^{1/2}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. It combines the ability to conduct heat with the ability to store thermal energy.

Consider two homogeneous semi-infinite bodies initially at uniform temperatures $T_1>T_2$. Put them in perfect planar contact, neglecting thermal resistance at the interface and using constant material properties. The heat equation similarity solution gives the common interface temperatureTo derive it, the outgoing heat flux from body 1 is $e_1(T_1-T_i)/\sqrt{\pi t}$, while the incoming heat flux to body 2 is $e_2(T_i-T_2)/\sqrt{\pi t}$. Equating them gives both the weighted temperature andEqual effusivities give the arithmetic mean; a much larger $e_2$ holds the interface closer to the initial temperature of body 2.

This explains why objects at equal actual temperature can feel differently cold on contact: the skin's heat loss depends on the transient exchange. It does not measure an electronic Fermi temperature. ^[ref] The ideal $t^{-1/2}$ initial flux is limited in practice by imperfect contact and finite response times; finite thickness matters once the thermal diffusion length reaches an outer boundary.