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For a monovalent three-dimensional free electron gas, the spin-summed molar density of states is $g(E_F)=3N_A/(2E_F)$. Combining the shell estimate $C_{el}=3k_B^{2}g(E_F)T$ with $C_{la}=(12{\pi }^4/5)R(T/{\theta }_D{)}^3$ gives the displayed nonzero crossover. The controlled Sommerfeld expansion replaces $3$ by ${\pi }^2/3$ and multiplies $T_{\ast }^2$ by ${\pi }^2/9$. A Fermi energy alone does not determine the DOS in an unspecified band structure.

For a three-dimensional parabolic band and fixed positive $\mu -E_c$, the Sommerfeld expansion gives $n(T)=n(0)[1+{\pi }^2(k_{B}T/(\mu -E_c){)}^2/8+O(T^4)]$. This increase differs from the usual fixed-density calculation, in which the chemical potential decreases with temperature. The exact density is proportional to $T^{3/2}{F}_{1/2}[(\mu -E_c)/(k_{B}T)]$.

For a smooth single-spin density of states $g$ near a positive Fermi energy and fixed total particle number, the Sommerfeld expansion gives $\mu ={ϵ}_F-({\pi }^2/6)(g_F^{\prime }/g_F)(k_{B}T{)}^2+O(T^4)$. The spin contribution to the total-moment magnetic susceptibility, defined by differentiating with respect to $B$, is $2{\mu }_B^2\int g(E)[-n^{\prime }(E)]dE$. Expanding about $\mu$ and including the shift of chemical potential gives the displayed formula. The $g_F^{\prime \prime }$ term is thermal broadening; the $-g_F^{\prime 2}/g_F$ term enforces fixed particle number. A band edge or singular density of states within $k_{B}T$ of ${ϵ}_F$ requires a separate analysis. Orbital magnetic response is excluded.

A normal Fermi liquid is an interacting fermion phase continuously connected to a Fermi gas without broken symmetry. Its Landau quasiparticles are dressed particle/hole excitations near the Fermi surface, with the same conserved quantum numbers, finite quasiparticle residue and a decay width small compared with their excitation energy. Their reference dispersion already includes zero-temperature interaction renormalization; “bare quasiparticle” here does not mean a free atom.^[ref]

The rate assumes $k_{B}T,|\xi |\ll E_F$, $\xi =ϵ-\mu$, regular scattering amplitudes and density of states, equilibrium Fermi-Dirac distributions, and two-body collisions conserving energy and momentum. Pauli exclusion principle restricts the available phase space. It describes a regular three-dimensional normal liquid, not a one-dimensional interacting electron liquid. On the surface, ${\tau }^{-1}\propto T^2$; at $T=0$ and $\xi >0$, it is proportional to ${\xi }^2$. Hence excitations become increasingly sharp. The quoted asymmetric occupation factor is for particle out-scattering: a hole decay or full spectral width also needs the complementary in-scattering process; it does not imply exponentially stable deep holes.

Symmetry $f_{ab}=f_{ba}$ cancels the quadratic functional's factor one-half, giving the thermal Landau quasiparticle energy shiftHere $a=(\mathbf{k},\lambda )$, and the derivative is with respect to an occupation, not particle density. Replacing the homogeneous momentum sum by an energy integral defines a smooth angularly averaged kernel $H_a(ϵ)$. With $\delta n=n_F(ϵ;\mu ,T)-\Theta ({\mu }_0-ϵ)$,Fixed density gives $\delta \mu =-({\pi }^2/6)(k_{B}T{)}^2{D}^{\prime }({\mu }_0)/D({\mu }_0)+O(T^4)$; fixed chemical potential instead has $\delta \mu =0$. Thus ${\widetilde{ϵ}}_a={ϵ}_a+O(T^2)$ in either case, provided the kernels are smooth and no instability intervenes. This is the Sommerfeld expansion; singular spectra require separate treatment.^[ref]

The leading entropy is therefore that of quasiparticles with their zero-temperature effective mass:$D^{\ast }$ here is the total, spin-summed density of states. The plotted low-temperature slopes require the enhanced masses $5.4m$ at molar volume $26.2\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $2.7m$ at $36.8\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, rather than the bare-mass predictions. At fixed molar particle number, ${\gamma }_{\mathrm{mol}}\propto m^{\ast }{V}_{\mathrm{mol}}^{2/3}$. Agreement is restricted to the low-temperature limit: the measured curves depart markedly from a fixed-mass ideal-gas extrapolation at higher temperature. This supports long-lived quasiparticles with smooth low-energy parameters, but heat capacity alone does not measure their lifetime or prove adiabatic continuity.

For $T\ll T_F$, only electrons within an energy window of order $kT$ around the Fermi energy can change occupation appreciably, because of the Pauli exclusion principle. The active fraction is of order $kT/E_F=T/T_F$. Giving each of these electrons energy of order $kT$ estimates an excitation energy $U-U_0\sim N_{e}k{T}^2/T_F$, and thereforePer mole of conduction electrons, $N_{e}k=R$, which is the quoted $2RT/T_F$. Per mole of metal atoms with $z$ model conduction electrons each, the conduction-electron count in the molar electronic heat capacity instead gives $2zRT/T_F$. The numerical factor is only approximate: the Sommerfeld expansion gives $C_V=({\pi }^2/2)N_{e}kT/T_F$ for a three-dimensional free-electron gas.

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.

Pauli exclusion principle prevents most filled electronic states from changing occupation at $T\ll T_F$. Only a shell of width of order $k_{B}T$ around the Fermi energy participates. Since the density of states near that energy is of order $n/E_F$ per volume and energy, the fraction of active electrons is of order $k_{B}T/E_F=T/T_F$. Their typical change of energy is of order $k_{B}T$, so the thermal energy increment per volume scales asDifferentiating with respect to temperature gives the electronic heat capacityThe simple shell argument fixes the linear law, not its coefficient. A Sommerfeld expansion using the spin-summed free-electron density of states $g(E_F)=3n/(2E_F)$ gives $C_e=({\pi }^2/3)g(E_F)k_B^{2}T$, hence $A={\pi }^2/2$, consistent with the supplied value. The result excludes strong band-structure or interaction corrections and requires a degenerate electron population.

The density of states $g(E)$ counts one-electron states per energy interval, including both electron spin orientations. It may be a total or a per-volume quantity; $U$ must have the matching normalization. The Fermi-Dirac distributionis the occupation probability of such a state. In the fixed-volume free electron model, the integration variable $E$ and the density of states do not depend on $T$. The occupation depends explicitly on $T$ and implicitly through the chemical potential $\mu (T)$ needed to preserve electron number. Thermal expansion or a changing electronic band structure would require additional effects outside this model.

At $T\ll T_F$, only states within order $k_{B}T$ of the Fermi energy change their occupation. There are order $g(E_F)k_{B}T$ such states, each carrying an excitation energy of order $k_{B}T$. Thus $U(T)-U(0)\propto g(E_F)(k_{B}T{)}^2$ and the electronic heat capacity is proportional to $g(E_F)k_B^{2}T$. Using the supplied coefficient,The shell argument fixes the scaling, not the numerical coefficient; the controlled Sommerfeld expansion fixes that coefficient at fixed electron number. Filled states far below $E_F$ cannot accept arbitrary thermal excitations because of the Pauli exclusion principle.

For the degenerate electron gas, let $D_g=Am/(\pi {\hslash }^2)$ include both spin states. The Sommerfeld expansion givesFor the gapless symmetric intrinsic semiconductor, part (d) gave the same coefficient with $D_s=A{m}^{\ast }/(\pi {\hslash }^2)$. Thus for equal sample areas and equal spin and valley degeneracies,The equality often intended in this comparison requires the same density of states. The source sets $m_e^{\ast }=m_h^{\ast }$ but does not explicitly equate them to the electron gas mass $m$. Without that additional assumption the ratio is $m^{\ast }/m$, or more generally $D_s/D_g$ if sample areas or degeneracy factors differ. For example, $m^{\ast }=2m$ with equal areas satisfies the stated equal-effective mass condition in the semiconductor and gives ratio two. The filled Fermi sea does not add a factor $N_e/N_c$: only the thermal shell near its Fermi energy contributes to its heat capacity.

The Wiedemann-Franz law relates the electronic part of thermal conductivity to electrical conductivity:For degenerate electrons with elastic scattering and a transport function smooth over $k_{B}T$, a Sommerfeld expansion makes the same Fermi-level transport factor appear in charge conduction and in heat conduction. Their ratio cancels that factor and leaves the displayed Lorenz number. Heat conductivity is defined at zero electric current, so thermoelectric subtraction is part of the precise transport expression.

The law does not relate total solid heat conductivity to charge conductivity when phonons contribute substantially. Strong inelastic scattering, sharp energy dependence, nondegenerate carriers or nonstandard electronic regimes can change the ratio. Low-temperature residual elastic-scattering transport is a common setting where the standard value is recovered. The unit W·ohm/K² follows because ${\kappa }_e/(\sigma T)$ contains one extra inverse temperature beyond thermal conductivity itself.

The occupation probability of an individual one-electron state is the Fermi-Dirac distribution,At $T=0$, it is a step: one below $E_F$ and zero above it. Heating rounds the step over an energy range of order a few $k_{B}T$, creating empty states below and occupied states above the chemical potential. At every positive temperature, $f(\mu ,T)=1/2$.

For fixed $N,V$ in the three-dimensional free electron model, the chemical potential is determined by $N=\int g(ϵ)f(ϵ,T)dϵ$; it is not exactly constant. In the degenerate regime $T\ll T_F$, the Sommerfeld expansion givesIt is therefore reasonable in a simple low-temperature sketch to set $\mu \simeq E_F$, provided this approximation is stated. The figure includes the leading small shift. At much higher temperature the fixed-$E_F$ approximation fails and the distribution approaches the classical dilute limit. The plotted occupation is $f$, rather than the energy distribution $gf$.

Only a shell of energy width of order $k_{B}T$ near the Fermi energy can change occupation: much lower states are blocked by the Pauli exclusion principle. The thermal-shell estimate of electronic heat capacity countsTo recover the rough coefficient requested in the paper, assign each of these active particles an excitation energy of order $3k_{B}T/2$. ThenThe differentiation matters: the active shell itself widens as temperature rises. This argument explains the small linear-in-$T$ electronic heat capacity; its shell width and average excitation energy are heuristic, so it does not establish a precise numerical prefactor.

A controlled calculation from the Fermi-Dirac distribution gives a nearby but different result. At fixed particle number, the Sommerfeld expansion yieldsThus $9/2$ is a rough estimate of the exact free-electron coefficient ${\pi }^2/2\simeq 4.935$, not an exact consequence of the stated density of states. Both require $T\ll T_F$.

Assume sodium supplies one conduction electron per atom. The supplied Fermi energy givesThe gas is strongly degenerate at room temperature. For one mole, $N=N_A$ and $N_A{k}_B=R=8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. Using the requested estimate,The supplied mass density determines the answer per unit volume: $N/V=(970/0.023)N_A=2.54\times 10^{28}{\mathrm{m}}^{-3}$. ThereforeThe density is not needed for the molar heat capacity once $E_F$ is supplied. Using the precise Sommerfeld expansion coefficient instead gives $0.327\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

Either electronic estimate is only about $1\%$ of the stated measured total. Most of the total heat capacity comes from lattice phonons, whose high-temperature Dulong-Petit law limit is $3R=24.94\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The remaining difference from $28\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ need not be electronic: anharmonic lattice effects and the difference between heat capacity at constant pressure and heat capacity at constant volume also contribute. The paper does not specify the measurement constraint, so the two totals should not be equated exactly.

Keep the electron number fixed, soand $\mu (0)=E_F$. Heating smooths the Fermi-Dirac distribution across the Fermi energy, creating holes below $\mu$ and particles above it. The sign of the low-temperature adjustment of chemical potential follows from the local slope of the density of states.

In one dimension, $g\propto {ϵ}^{-1/2}$ decreases with energy. At fixed $\mu$, smoothing removes more electrons below it than it adds above it, so $\mu$ initially rises. In three dimensions, $g\propto {ϵ}^{1/2}$ increases, so $\mu$ falls. In two dimensions the constant density of states makes the leading changes cancel, so $\mu$ stays extremely close to $E_F$ in the degenerate regime.

For quantitative checks, a Sommerfeld expansion gives, for $k_{B}T\ll E_F$,For a constant two-dimensional density of states above an unbounded parabolic band bottom, the number integral can be done exactly:soIt actually decreases, but its low-temperature correction is exponentially small rather than quadratic. “Constant in two dimensions” is therefore a low-temperature approximation, not an exact statement for every $T$.

The one-dimensional rise is also only an initial trend. In the high-temperature Maxwell-Boltzmann distribution limit, the ideal $d$-dimensional electron gas has $\mu \simeq k_{B}T\ln [n_d{\lambda }_T^d/2]$, where $n_d=N_e/L^d$ and ${\lambda }_T=\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$ is the thermal de Broglie wavelength. It eventually becomes negative and decreases in every dimension. The one-dimensional chemical potential thus passes through a maximum. This high-temperature statement refers to the stipulated unbounded free-electron dispersion relation, not a claim that a real finite-width crystal band remains parabolic indefinitely.

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.

For a degenerate Fermi gas, the Pauli exclusion principle restricts thermally changed occupations to an energy shell of width $k_{B}T$ around the Fermi energy. The number of active electrons is of order $g(E_F)k_{B}T\sim N{k}_{B}T/E_F$; each carries an excitation energy of order $k_{B}T$. Thus the thermal energy grows as $N(k_{B}T{)}^2/E_F$ and the electronic heat capacity is proportional to $T$.

More precisely, the fixed-particle-number Sommerfeld expansion givesHere $g$ is the total spin-summed density of states for the sample. For a three-dimensional parabolic band, $g(E_F)=3N/(2E_F)$, yieldingThe linear law concerns the electronic contribution; the low-temperature lattice heat capacity usually adds a $T^3$ term.

At $T\ll T_F$, Pauli exclusion restricts thermally changed occupations to an energy interval of order $k_{B}T$ about the Fermi energy. The affected fraction is of order $T/T_F$; their typical excitation energy is of order $k_{B}T$, so the thermal energy correction is proportional to $T^2$ and the heat capacity is linear in $T$.

More precisely, with $g(E_F)$ the total density of one-particle states per unit energy, including spin,For a three-dimensional ideal gas with quadratic dispersion, $g(E_F)=3N/(2E_F)$, henceThe fixed-particle-number Sommerfeld expansion gives $U=3N{E}_F/5+({\pi }^2/4)N(k_{B}T{)}^2/E_F+\cdots$. The temperature dependence of the chemical potential must be included when obtaining this expression; holding it artificially fixed changes the intermediate energy correction.

The Fermi-Dirac distribution gives the mean occupancy of a single fermion state:Its maximum occupancy of one expresses the Pauli exclusion principle. A spatial orbital with two spin states can hold two electrons because they are different single-particle states. At zero temperature, the occupation is a step at the Fermi energy $E_F$; at low nonzero temperature, occupation changes appreciably only in an energy shell of width of order $k_{B}T$ about the chemical potential.

Consequently, most conduction electrons cannot absorb small thermal energies into already occupied states. Only a fraction of order $T/T_F$ participates, where $T_F=E_F/k_B$. For a normal metal with a smooth density of states $g(E_F)$, including both spin states and using the total-system normalization, the Sommerfeld expansion at fixed particle number givesand thereforeFor a three-dimensional free electron gas, $g(E_F)=3N/(2E_F)$, so $C_{\mathrm{el}}=({\pi }^2/2)N{k}_B(T/T_F)$. This is much smaller than a classical equipartition estimate at $T\ll T_F$. The chemical potential must adjust with temperature to maintain $N$; treating it as fixed while differentiating an energy integral gives the wrong fixed-$N$ calculation in general.

The lattice adds a Debye contribution $C_{\mathrm{ph}}={\beta }_D{T}^3$ at temperatures well below the Debye temperature. HenceA plot of measured $C/T$ against $T^2$ can separate the electronic intercept from the phonon slope. calorimetry must account for the heat capacity of the sample holder, heat leaks, thermal equilibration and thermometer calibration. A magnetic impurity contribution or a superconducting transition can invalidate this two-term fit; the prediction is for the normal metallic state in an appropriate low-temperature interval. Differences from a free-electron coefficient can reveal quasiparticle effective-mass enhancement or nontrivial electronic structure rather than a failure of Fermi statistics.

For a complete single-particle state, the Fermi-Dirac distribution isHere ${\mu }_{\mathrm{ch}}$ denotes the chemical potential, distinct from the magnetic-moment symbol in (b). At zero temperature it approaches the Fermi energy $E_F$ and the occupation has a sharp edge. At $T\ll T_F=E_F/k_B$, thermal changes occur within an energy interval of order $k_{B}T$ about that edge.

The Fermi-edge heat-capacity counting argument uses the fact that only order $g(E_F)k_{B}T$ electrons can be rearranged because states well below the edge are blocked by the Pauli exclusion principle. Each active excitation carries energy of order $k_{B}T$, so $U(T)-U(0)\sim g(E_F)(k_{B}T{)}^2$ and $C_{V,e}\propto T$. For a smooth density of states, the Sommerfeld expansion givesHere $g$ is the total single-particle density of states, including spin. In a three-dimensional free-electron gas, $g(E_F)=3N/(2E_F)$, giving $C_{V,e}=({\pi }^2/2)N{k}_B(T/T_F)$. This is the low-temperature normal-state electronic contribution, not the lattice heat capacity or a description across a superconducting transition.^[ref]

Fermion occupation and ordinary metallic heat capacity. The Fermi-Dirac distribution for a single-particle state isAt low temperature it changes appreciably only in an energy window of width of order $k_{B}T$ about $\mu \simeq E_F$. If the full density of states $D(E)$ is smooth and nonzero there, only about $D(E_F)k_{B}T$ states change population. Each excited particle or hole carries energy of order $k_{B}T$ relative to the Fermi surface, so $U(T)-U(0)$ is of order $D(E_F)(k_{B}T{)}^2$. Differentiation therefore gives $C_{\mathrm{el}}\propto T$.

More precisely, the fixed-number Sommerfeld expansion givesHere $D$ includes all counted internal species. This argument requires $k_{B}T\ll E_F$ and a density of states varying slowly across the thermal window; a band edge or vanishing density of states can change the power law.

Two-dimensional linear dispersion at zero chemical potential. Let $g$ count identical positive-energy branches, keeping the state-counting convention explicit. The number of wavevector states in a ring of radius $k$ is $gA(2\pi kdk)/(2\pi {)}^2$. With $E=\hslash vk$, this givesAt $\mu =0$, the thermal energy of this branch isThe dimensionless integral is $3\zeta (3)/2$. Differentiating givesThis is the heat capacity for a two-dimensional linear fermion dispersion. A physical neutral particle–hole-symmetric Dirac cone also has electron holes in its filled negative-energy band. Their excitation energy gives an equal contribution, doubling $C_{+}$ for the same cone count. The $T^2$ scaling is unchanged. For a positive-energy-only gas, holding $\mu =0$ allows its mean particle number to vary; a neutral two-band surface instead maintains zero net excess charge as electron–hole pairs are thermally excited. These conventions should not be conflated when specifying an exact coefficient.

Nonzero chemical potential. The ordinary metallic regime isThe thermal window then lies far from the cone tip. Since $D(E)\propto E$, its fractional change across that window is only of order $k_{B}T/\mu$. Treating it as locally constant restores the Sommerfeld result $C_{\mathrm{el}}\propto D(\mu )T\propto \mu T$. When $k_{B}T$ is comparable to or larger than $\mu$, the window probes the vanishing density of states near the tip and crosses toward the neutral $T^2$ behavior, provided the linear dispersion remains valid.

Bulk phonons and the surface-to-bulk comparison. For the topological insulator model, assume bulk electronic excitations are frozen out while the conducting surface remains active. In three dimensions, acoustic phonons have a density of modes proportional to ${\omega }^2$. Normalize the Debye spectrum to $3N$ modes:The thermal part of the energy isAt $T\ll {\Theta }_D$, extending the upper integration limit to infinity introduces an exponentially small error. With $x=\hslash \omega /(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$,Temperature-independent zero-point energy makes no contribution to this heat capacity. The requested ratio consequently scales asFor a fixed shape and composition, $A\propto L^2$ and $N\propto L^3$, so the ratio grows as $1/(LT)$. Smaller samples and lower temperatures favor the surface contribution. It dominates if the dimensionless right-hand side, including its state-counting prefactor, exceeds unity. This is the surface-electron and bulk-phonon heat-capacity crossover.

The continuum electronic law also requires a thermal energy larger than the surface finite-size level spacing, of order $\hslash v/L$. The phonon continuum has its own finite-size cutoff. Therefore dominance obtained by formally taking $T\to 0$ cannot always be inferred from the two continuum power laws alone.

There is a numerical-normalization qualification in the paper. Combining the explicit one-positive-branch result above with the conventional $3N$-mode Debye normalization gives ratio prefactor $15g\zeta (3)/(16{\pi }^5)$; including electron holes doubles it. For $g=1$, this is approximately $0.00368$ before that doubling, rather than the supplied $5/(12{\pi }^2)\simeq 0.0422$. The problem specifies neither a cone/spin multiplicity that equates these nor another mode normalization. Its supplied prefactor is therefore used as the stipulated approximation for the requested numerical estimate, rather than claimed as an exact consequence of the single-branch count.

Numerical estimate. A cubic sample of volume $1\mathrm{m}{\mathrm{m}}^3$ has side $L=10^{-3}\mathrm{m}$. Counting the conducting surface on all six faces givesFor one atom per cubic cell of side $a=1.5\times 10^{-9}\mathrm{m}$,Use $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ and $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. Inserting the stated $v=5\times 10^5\mathrm{m}{\mathrm{s}}^{-1}$, ${\Theta }_D=500\mathrm{K}$ and $T=1\mathrm{K}$ into the paper's supplied prefactor givesThus bulk phonons dominate at this temperature in this estimate. If only one face contributes, the ratio is $1.22\times 10^{-6}$; if there are $\nu$ chemical atoms per unit cell, divide the corresponding estimate by $\nu$. These geometrical and crystallographic factors are not supplied explicitly.

With the same prefactor and geometry, extrapolating $1/T$ would suggest equality near $7.3\mu \mathrm{K}$. However, the surface finite-size scale is already of order $\hslash v/(k_{B}L)\simeq 3.8\mathrm{mK}$ for this specimen. That extrapolated crossover lies outside the continuum electronic regime, so it is not a reliable prediction of the sample's ultimate lowest-temperature heat capacity.