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The equilibrium isothermal equation of state is $p=c_s^2\rho$. Hydrostatic equilibrium gives $p_z=-\rho {\Omega }^{2}z$, henceHere $\Omega \ne 0$ is assumed. This is the same local harmonic-gravity balance as isothermal vertical structure in stellar gravity.

For $\gamma >1$, this equilibrium is stably stratified under adiabatic fluid perturbations. Its squared buoyancy frequency isA general perturbation can therefore support acoustic waves restored by compression and internal gravity waves restored by buoyancy, with nonoscillatory entropy or vortical disturbances also possible. The prescribed velocity is purely vertical and horizontally uniform. It describes vertical acoustic waves; it does not contain the horizontally varying circulation required for an oscillatory internal-gravity branch. For zero horizontal wavenumber, that branch has zero frequency. An isothermal equilibrium must not be confused with an isothermal perturbation: the latter would remove the buoyancy restoring force.

Introduce the fluid Lagrangian displacement $\xi (z)e^{i\sigma t}$, so $w=i\sigma \xi$. Use $\delta \rho ,\delta p$ for fluid Eulerian perturbations and $\Delta \rho ,\Delta p$ for fluid Lagrangian perturbations. The continuity equation and the adiabatic relation areThe fixed external gravitational potential is not perturbed. The vertical linearized Euler equations consequently giveSubstituting the preceding relations, and using $p_z=-\rho g$ and $p_{zz}=-{\rho }_{z}g-\rho g_z$, reduces its right-hand side toSince $g={\Omega }^{2}z$ and $g_z={\Omega }^2$, the same equation holds for $w$:The rigid fixed base supplies the essential condition $w(0)=0$. At infinity the condition is finite kinetic energy, not bounded velocity amplitude; a polynomial can grow while the Gaussian mass density makes its energy finite.

Put $x=z/H$ and $\lambda =({\sigma }^2/{\Omega }^2-1)/\gamma$. The equation becomes the Hermite differential equation in its probabilists' convention,For $w={\sum }_{j\ge 0}{a}_j{x}^j$, its recurrence isThe base condition makes $a_0=0$ and therefore every even coefficient zero. An odd series terminates at degree $2n+1$ when $\lambda =2n+1$. It remains to prove that a nonterminating odd solution cannot have finite energy; termination alone would not establish the requested exhaustion of modes.

For that purpose, set $y=e^{-x^2/4}w$. The finite-energy condition is $y\in L^2(0,\infty )$, and its equation isBecause the coefficients are even and the solution is odd at the base, extend $y$ oddly to the whole real line. Define lowering and raising operatorsIntegration by parts gives $\Vert ay{\Vert }^2=\lambda \Vert y{\Vert }^2$, so $\lambda \ge 0$. These identities apply to the square-integrable solution: multiplying its equation by a compactly cut off conjugate and taking larger cutoffs bounds $y^{\prime }$ and $xy$ in $L^2$, and removes the endpoint terms. The same argument applies to each lowered eigenfunction.

If $\lambda$ is not a nonnegative integer, repeatedly apply $a$. Each application lowers $\lambda$ by one and remains nonzero whenever its current value is positive, by the norm identity. After finitely many applications it would yield a nonzero square-integrable eigenfunction with negative $\lambda$, contradicting that identity. Thus $\lambda =N$ is a nonnegative integer. At $N=0$, $ay=0$ gives $y\propto e^{-x^2/4}$, which is even. Lowering reverses parity, so an initially odd eigenfunction can reach this even ground state only if $N$ is odd. Consequently $N=2n+1$. The recurrence then terminates and uniquely determines the odd solution from $a_1$.

The complete adiabatic vertical modes of a Gaussian atmosphere are thereforeHere ${\mathrm{He}}_j$ is the Probabilists' Hermite polynomial. For example the first shapes are $x$, $x^3-3x$, and $x^5-10x^3+15x$, with frequencies squared ${\Omega }^2(1+\gamma )$, ${\Omega }^2(1+3\gamma )$, and ${\Omega }^2(1+5\gamma )$. Either sign of each real frequency gives a time-harmonic solution. Gaussian moments make every displayed polynomial have finite kinetic energy. Without the fixed-base condition, finite energy alone would also admit even polynomials, which is why that condition is necessary.

In a first-order phase transition, two distinct global minima of the Landau free energy have equal values and exchange stability. The equilibrium order parameter jumps, and a first derivative of the equilibrium free energy is discontinuous. Depending on the direction through the phase diagram, that derivative can be magnetization, entropy or another conjugate quantity; a first-order transition need not have latent heat along every chosen path.

In a continuous phase transition, the equilibrium order parameter changes continuously while the curvature at the critical minimum vanishes. For $u>0$ and $h=0$, $M=0$ is stable for $r>0$, but for $r<0$ the stable minima have $M\simeq \pm \sqrt{-r/u}$. The order parameter therefore goes continuously to zero at $r=0$; the inverse magnetic susceptibility vanishes there.

For $u<0$, the sextic term stabilizes the Landau free energy. The nonzero minima can become favorable while $M=0$ still has positive curvature. At $r=3u^2/(16v)>0$ they coexist with $M=0$, so the order parameter jumps to magnitude $\sqrt{-3u/(4v)}$. Finding a stationary point alone does not locate the transition: equilibrium requires comparing the values of all competing minima. Spinodal boundaries describe loss of local stability, not coexistence.

Write $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant. For independent Ising spins, the one-site partition functions on the two sublattices are $z_B=2\cosh \beta (h+g)$ and $z_W=2\cosh \beta (h-g)$. Differentiating each logarithm with respect to its local field givesAt $h=0$, the two sublattices experience opposite fields, so spin reversal combined with sublattice exchange leaves the Hamiltonian invariant and gives $M_W=-M_B$.

For the antiferromagnetic Ising model, take $N$ sites, equal sublattice sizes and $q=2D$ neighbors per site. There are $Nq/2$ bonds counted once. The two-sublattice Ising mean-field approximation expands each bond asand drops only the product of fluctuations. Summing the remaining terms givesThe printed constant omits $N$; for a total Hamiltonian with unnormalized spin sums, this factor is necessary. Applying the independent-spin calculation now givesThe mean-field approximation is favored by large coordination number: averaging many neighbors reduces local relative fluctuations. The size of critical fluctuations still has to satisfy the Ginzburg criterion; high dimension is a reason for validity, not permission to discard every correlation in every regime.

Antisymmetric solutions and the transition. At $h=0$, putting $M_B=-M_W=M$ makes both equations reduce to $M=\tanh \beta (g+qJM)$. A zero-field continuous transition concerns $g=0$ as well: nonzero $g$ explicitly selects the staggered sign and rounds this critical bifurcation. Put $K=\beta qJ$. The graphical intersections of $M$ and $\tanh (KM)$ give only $M=0$ for $K\le 1$. For $K>1$ there are additionally two nonzero intersections $\pm M_{\ast }$: on $M>0$, $\tanh (KM)-M$ is strictly concave, initially increases, and is negative at $M=1$, so there is exactly one positive root. Spin reversal supplies the negative root. The origin becomes unstable in the staggered direction, whereas the nonzero roots have $K(1-M_{\ast }^2)<1$ and are stable. Expanding about the bifurcation givesThus $M_{-}=(M_B-M_W)/2$ turns on continuously, with $k_B{T}_c=qJ$ and order-parameter critical exponent $1/2$.

Uniform solutions. At $g=0$, imposing $M_B=M_W=M$ instead gives $M=\tanh \beta (h-qJM)$. The function $M-\tanh \beta (h-qJM)$ has derivative $1+\beta qJ{\mathrm{sech}}^2\beta (h-qJM)>0$, is negative at $M=-1$ and positive at $M=1$, and consequently has exactly one zero for every positive temperature. At $h=0$ this zero is $M=0$. Antiferromagnetic exchange opposes uniform polarization, so the uniform branch has no ferromagnetic ordering bifurcation. This does not remove the antisymmetric solutions below $T_c$, or exclude transitions of staggered order in the full problem at nonzero uniform field. In particular, uniqueness on the diagonal $M_B=M_W$ is not uniqueness of all solutions.

Free energy and stability. The appropriate off-shell two-sublattice Ising variational free energy is the energy minus temperature times the independent-spin entropy. DefineThere are $Nq/2$ bonds and $N/2$ spins of each kind, so the free-energy density per site isSince $s^{\prime }(m)=\mathrm{atanh}m$, stationarity givesand the analogous equation for $M_W$; these are exactly the two mean-field self-consistency equations. On a stationary solution, the identity $s(m)=m\mathrm{atanh}m-\ln (2\cosh (\mathrm{atanh}m))$ gives the equivalent on-shell expressionThe entropy form must be used to compare off-shell variations; treating this latter auxiliary expression as the variational potential would give spurious stability conclusions.

Using $s(m)=-\ln 2+m^2/2+m^4/12+O(m^6)$ and $M_B=M_{+}+M_{-}$, $M_W=M_{+}-M_{-}$ givesThis includes the requested quadratic expansion and first-order field terms. The positive quartic coefficient verifies a continuous phase transition in $M_{-}$ when the staggered quadratic coefficient changes sign at $k_{B}T=qJ$. The uniform quadratic coefficient remains positive.

Susceptibilities and a printed error. On a stable zero-field ordered branch let $M_B=-M_W=M$ and $a=\beta (1-M^2)$. Since the derivative of $\tanh x$ is $1-{\tanh }^{2}x$, differentiating the two equations with respect to the uniform field givesTheir sum and difference yieldThe difference equation is $(1-qJa){\chi }_{-}=0$; stability implies $qJa<1$ for $T<T_c$. Symmetry also explains the zero result: a uniform perturbation changes the two sublattice magnetizations equally to first order, and hence does not linearly change the staggered magnetization. This ${\chi }_{-}$ is a cross response to $h$, not the magnetic susceptibility to the staggered field $g$.

The displayed ${\chi }_{+}$ in the source instead contains $1+M^2$. It cannot follow from differentiating the stated equations. A decisive limiting check is $T\to 0$: $1-M^2$ vanishes exponentially and the actual uniform response tends to zero, whereas the printed expression tends to $1/(qJ)$. Above $T_c$, $M=0$ and the same linear system givesFor comparison, the actual staggered susceptibility is $\partial M_{-}/\partial g=a/(1-qJa)$, becoming $1/(k_{B}T-qJ)$ above $T_c$. Its divergence, rather than a divergence of the uniform response, detects antiferromagnetic criticality.

Use natural units, assume negligible neutrino chemical potentials, and approximate neutrino decoupling as occurring before electron-positron annihilation in cosmology. At high temperature, weak interactions maintain a common temperature for neutrinos, antineutrinos, electrons, Positrons and photons. Since ${\Gamma }_{\nu }/H\propto T^3$, the reaction rate eventually loses against expansion. Decoupled relativistic neutrinos then free stream; their momenta redshift as $a^{-1}$, so their distribution retains a temperature parameter satisfying $a{T}_{\nu }=\mathrm{constant}$.

The still-coupled electromagnetic bath receives the entropy of annihilating electrons and Positrons. Its effective number of relativistic entropy degrees of freedom is initiallywhere the two photon polarizations count as bosonic states and the four electron-positron spin states have the fermion entropy weight $7/8$. After the pairs disappear, $g_{\ast s}^{\mathrm{EM}}=2$. Cosmological entropy conservation in this coupled bath gives $g_{\ast s}^{\mathrm{EM}}(a{T}_{\gamma }{)}^3=\mathrm{constant}$. The three already decoupled neutrino species are excluded from this bath's entropy count. At decoupling $T_{\nu }=T_{\gamma }$, henceSubsequent expansion preserves this ratio in the idealized history. Residual interactions during annihilation give small corrections to the instantaneous-decoupling approximation. If a relic species later becomes massive, $T_{\nu }$ here remains the redshifting momentum-distribution parameter; it need not be its nonrelativistic kinetic temperature.

A single ordinary neutrino species includes one populated neutrino helicity and one antineutrino helicity. For negligible chemical potential, integration of the Fermi-Dirac distribution givesTherefore the number density of a relativistic neutrino relative to photons, with the antiparticle included, isThe factor $3/4$ is the fermionic number-density weight, distinct from the energy and entropy weight $7/8$. This relation continues to hold after this species becomes nonrelativistic if its comoving particle number is conserved.

Under the proposed matter-dominated, spatially flat model, the Friedmann equation requires the present relic-neutrino energy density to be approximately the critical density. Using the unreduced Planck mass $M_P=G^{-1/2}$ givesThus the critical-density mass of a thermal relic neutrino isTaking the paper's rounded values, $T_{\gamma 0}=2\times 10^{-13}\mathrm{GeV}$, $H_0=h/(4.5\times 10^{41}{\mathrm{GeV}}^{-1})$ and $M_P\sim 10^{19}\mathrm{GeV}$, givesFor $0.5<h<1$, this is of order $30$--$100\mathrm{eV}$ with these approximate constants. More precise constants change the coefficient, not the requested mass scale. This mass is much larger than the present momentum scale $T_{\nu 0}\simeq 1.4\times 10^{-4}\mathrm{eV}$, validating the nonrelativistic energy density approximation. The result assumes that one thermal relic species dominates the matter density; it is conditional on that model rather than a measurement of a neutrino mass.

Buoyancy and the convective criterion. Consider a small fluid parcel displaced outward by $\delta r$. Assume that sound waves keep its pressure equal to the ambient pressure, while heat exchange is slow enough that its displacement is an adiabatic process. Composition is uniform, so the parcel and its surroundings have the same mean molecular weight. For an ideal gas, $P\propto \rho T$, and an adiabatic parcel obeys $P\propto {\rho }^{\gamma }$. If $\delta \ln P<0$ is the pressure change on moving outward, thenThe density excess of the parcel at its new location is thereforeIf $\nabla >(\gamma -1)/\gamma$, this excess is negative: the outward-displaced parcel is lighter and accelerates farther outward. A downward displacement similarly continues downward. Conversely, a smaller gradient gives a restoring buoyancy force. The Schwarzschild criterion is thus stability for $\nabla <{\nabla }_{\mathrm{ad}}$ and instability for $\nabla >{\nabla }_{\mathrm{ad}}$, where the adiabatic temperature gradient is ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma$. For a monatomic gas,A composition gradient would require the additional density effect in the Ledoux criterion; it is absent in this calculation.

Opacity and pressure. The supplied opacity is characteristic of a cool hydrogen-rich stellar atmosphere dominated by negative hydrogen ion opacity. Mostly neutral hydrogen can attach electrons supplied by partially ionized metals; bound-free and free-free absorption involving the resulting $H^{-}$ ions produces a steep positive temperature dependence. The weak density dependence and strong temperature sensitivity can be represented by local power-law fits. At fixed composition the given pressure form is equivalent to $\kappa \propto {\rho }^{1/2}{T}^{17/2}$, because $P\propto \rho T$. This is an approximate fit over the relevant cool-atmosphere regime, not an exact opacity law at all temperatures; molecular opacity or hydrogen ionization eventually changes the regime.

For a geometrically thin atmosphere take the surface gravity of a star to be constant, $g_s=GM/R^2$. Combining $dP/dr=-\rho g_s$ with $d\tau /dr=-\kappa \rho$ gives hydrostatic equilibrium in optical depth,The grey atmosphere relation gives $T^8=T_e^8(3\tau +2{)}^2/16$. With negligible external pressure, $P(0)=0$, integration yieldsAn imposed nonzero external pressure would add $P(0{)}^{3/2}$; the boundary assumption is what fixes the integration constant here.

The factor in parentheses is $3\tau /[2(3\tau +2)]$. HenceThe grey-atmosphere convection onset with eighth-power temperature opacity occurs when this reaches $2/5$, soThe radiative continuation below this boundary would be unstable. Efficient convection instead keeps the deeper star approximately adiabatic.

Matching to the convective interior. Write $P=K{\rho }^{5/3}$. Eliminating density using the ideal gas law givesat fixed composition. Since ${\tau }_c$ is a fixed number, the matching temperature $T_b$ is proportional to $T_e$, while the matching pressure satisfies $P_b^{3/2}\propto g_s{T}_e^{-8}$. ConsequentlyThe polytropic mass-radius relation for index $3/2$ gives $K\propto R{M}^{1/3}$. Raising the matching relation to the ninth power and using $g_s\propto M/R^2$ therefore givesFinally the Stefan–Boltzmann law $L=4\pi R^2{\sigma }_{\mathrm{SB}}{T}_e^4$ yields the Hayashi relation with eighth-power temperature opacity,Composition and the opacity normalization are fixed in this comparison. The entropy coefficient $K$ is spatially constant in each convective model, but need not be the same for stars with different masses or radii.

Mechanical balance and equipotentials. Let $\Phi$ be the Newtonian gravitational potential and $\varpi$ the cylindrical distance from the rotation axis. For rigid angular velocity $\Omega$, a fluid element at rest in the rotating frame has no Coriolis acceleration. The centrifugal acceleration is ${\Omega }^2\varpi \hat{\varpi }$, so hydrostatic equilibrium becomesUsing the Poisson equation for gravity and ${\nabla }^2(x^2+y^2)=4$ gives

On any regular connected equipotential surface, every tangent displacement satisfies $d\varphi =0$ and therefore $dP=0$. Locally $P=P(\varphi )$, with $dP/d\varphi =-\rho$. This makes $\rho$ a function of $\varphi$ as well. Equivalently, taking the curl of the hydrostatic equation gives $\nabla \rho \times \nabla \varphi =0$. Thus $P,\rho$ and ${\nabla }^2\varphi$ are constant on each surface. The last conclusion follows from the modified Poisson equation; it does not constrain the magnitude of the first derivative $|\nabla \varphi |$ to be constant there.

For uniform composition, an equation of state $P=P(\rho ,T)$ determines $T$ uniquely at fixed $P,\rho$ in the usual stellar regime. For the gas-plus-radiation equation, the derivative with respect to $T$ is positive. Hence $T=T(\varphi )$, and the opacity and radiative conductivity are also functions of $\varphi$.

Surface flux and the requested sketch. Define $f(\varphi )=-\chi dT/d\varphi$, positive for the outward radiative flux. ThenOn an outer equipotential surface, $f$ is constant and the outward normal is parallel to $\nabla \varphi$. Its effective gravity is $g=|\nabla \varphi |$, so $F_n=fg$. The Stefan–Boltzmann law $F_n={\sigma }_{\mathrm{SB}}{T}_e^4$ gives the radiative gravity-darkening law,A rapidly rotating star is oblate. Centrifugal support reduces the effective gravity near the equator, which is therefore cooler and dimmer; the poles are hotter. The original sketch below uses a subcritical Roche model of a uniformly rotating star to show this geometry and the corresponding temperature variation. The material temperature on an interior equipotential is not the same quantity as the emergent effective temperature defined by surface flux.

Why purely local radiative balance fails. Differentiating the flux givesOn one equipotential, $f,f^{\prime },{\nabla }^2\varphi$ and the proposed source $\rho ϵ(\rho ,T)$ are all constant, but $|\nabla \varphi {|}^2$ generally varies with latitude. Unless $f^{\prime }=0$ or there is another special cancellation, no single value of $\rho ϵ$ can match this flux divergence at all points on the surface. This is the radiative-equilibrium obstruction in a rotating barotropic star. Slow stellar meridional circulation can transport heat to compensate; it is distinct from abandoning hydrostatic balance at leading order.

Integrated energy balance with circulation. The steady continuity equation is $\nabla \cdot (\rho \mathbf{v})=0$. Combining the entropy–enthalpy relation with hydrostatic balance givesIntegrate the energy equation over $V$ and apply the divergence theorem:Uniform composition makes $h=h(\rho ,T)$ constant on the enclosing equipotential, and $\varphi$ is constant there by definition. Pull $(h+\varphi )$ out of the last integral; the remaining net mass flux is zero by steady continuity. Thus the equipotential luminosity conservation with stellar circulation isThis does not require $\mathbf{v}\cdot \hat{n}=0$ pointwise on $S$: inward and outward circulation may cancel in the net mass flux.

Finally, $f$ is constant on $S$ and the divergence theorem applied to the effective potential givesTherefore $-\chi dT/d\varphi =L/(4\pi Gm-2{\Omega }^{2}V)$. Since $dP/d\varphi =-\rho$,The mass $m$ is the mass enclosed by this surface, not necessarily the total stellar mass. The denominator is positive for the regular hydrostatic surfaces used here; near loss of centrifugal confinement the assumed stellar model must be reconsidered. The derivation uses slow circulation, retaining hydrostatic/equipotential thermodynamics to leading order while allowing a nonzero advective heat flux.

Let $v_i={\dot{x}}_i$ and use a fixed arbitrary material subvolume $B_0$ in the reference configuration, with outward unit normal $N_I$. The reference mass density ${\rho }_0$ is time independent. The global momentum conservation law isThe divergence theorem, differentiation under the integral, and arbitrariness of $B_0$ give

The global conservation of energy law includes both internal energy and kinetic energy:Here $q^0$ is outgoing heat flux per reference area. Localizing givesSubtract the scalar product of the local momentum conservation law with $v$. Since $v_{i,I}={\dot{F}}_{iI}$, this proves the Lagrangian internal energy balance

For positive temperature $\theta$, the global entropy inequality isIts local excess is a nonnegative production density $\gamma$. Expanding the divergence of $q^0/\theta$ proves the Lagrangian entropy inequality in the requested equality form:No constitutive assumption on the heat flux was needed for these balance identities.

For the reference-configuration jump balances, make the orientation convention explicit. Write $[a]=a^{+}-a^{-}$ and let $n^0$ point from the minus side to the plus side. Let $c$ be the actual interface speed in direction $n^0$. A signed level function $\varphi$ with the plus side $\varphi >0$ obeys ${\varphi }_t=-c|{\nabla }_X\varphi |$ on the interface. For $a=a^{-}+[a]H(\varphi )$, the singular time derivative is $-c[a]{\delta }_S$; the singular reference divergence of $b$ is $n_I^0[b_I]{\delta }_S$. Thus a local conservation law $a_t=\mathrm{Div}b+$ bounded sources has the Rankine-Hugoniot conditionThis is also the shrinking moving-pillbox calculation of the global law: finite volume supplies have no surviving surface term.

To match the displayed positive-sign momentum jump in the paper, use its speed parameter as $V=-c$, so the interface travels in direction $-V{n}^0$. Assuming continuous ${\rho }_0$ and no surface force or energy supply, the jump laws are thenand, from ${\partial }_t({\rho }_0\eta )+\mathrm{Div}(q^0/\theta )\ge {\rho }_{0}r/\theta$,The last inequality is nonnegative entropy production concentrated on the interface. Temperature and flux in the entropy term are the traces on their respective sides; one cannot take a common temperature outside the jump unless it is continuous.

If $V$ is instead defined as the interface speed in direction $n^0$, all three $V$ terms above acquire a minus sign. In particular the printed positive-sign momentum formula requires the opposite speed convention; changing just the order of the jump does not remove this convention issue. If reference density is discontinuous, replace ${\rho }_0[a]$ by the jump $[{\rho }_{0}a]$ of the complete conserved density.

An equivalent internal-energy jump follows by subtracting the kinetic-energy jump using $[|v{|}^2/2]=\{v_i\}[v_i]$, where $\{a\}=(a^{+}+a^{-})/2$:The averaged traction appears because $[P_{Ii}{v}_i]=\{P_{Ii}\}[v_i]+[P_{Ii}]\{v_i\}$; using a one-sided traction would generally be incorrect.

For a reversible change of a simple compressible system with fixed composition, the first law isAn adiabatic change exchanges no heat. A reversible adiabatic change is isentropic, $dS=0$; adiabatic alone does not imply constant entropy for an irreversible process.

Define the Helmholtz free energy $F=E-TS$. Then $dF=-SdT-PdV$. Equality of its mixed partial derivatives gives the Maxwell relationHolding $T$ fixed in the first law and using this relation givesFor equilibrium radiation, $E=Ve(T)$ and $P=e(T)/3$, so this identity reads $e=T{e}^{\prime }/3-e/3$, or $T{e}^{\prime }=4e$. Integrating givesFinally in a reversible adiabatic expansion, $dE=-PdV$ becomes $4aV{T}^{3}dT+a{T}^{4}dV=-a{T}^{4}dV/3$. Dividing gives $3dT/T=-dV/V$, hence $V{T}^3$ is constant. This assumes equilibrium radiation throughout the expansion.

In the nondegenerate regime, $\mu$ lies well inside the gap, with $\mu \gg kT$ and $E_g-\mu \gg kT$. Then conduction occupations are approximately $e^{(\mu -E)/(kT)}$, while the hole probability in a valence state is approximately $e^{(E-\mu )/(kT)}$. Using the near-edge density of states and the supplied gamma integral givesThereforeThis assumes the edge approximations are valid over the thermally populated ranges. For an intrinsic semiconductor $n=p$, so each carrier density is proportional to $T^{3/2}{e}^{-E_g/(2kT)}$. Both mobile conduction electrons and holes contribute to conductivity, $\sigma =|e|(n{b}_e+p{b}_h)$ with mobilities $b_e,b_h$. The exponentially varying carrier densities explain strong temperature dependence even when the mobilities vary only algebraically.

Donors supply electrons and acceptors remove electrons to create holes. With complete ionization and negligible minority carriers, the majority densities are $n\simeq N_d$ on the n side and $p\simeq N_a$ on the p side. The corresponding isolated chemical potentials areCarrier diffusion at contact leaves positively charged donors and negatively charged acceptors in a depletion layer. At equilibrium the electrochemical potential is constant; the electron energy shift is $-|e|\varphi$. Hence the n-minus-p electric potential isAll $N$ quantities must have a consistent normalization: if $N_c,N_v$ are state densities per volume, $N_d,N_a$ are dopant densities rather than bare atom counts. The formula neglects incomplete ionization, degeneracy and changes in band structure.

At a common temperature the two equilibrium junction/contact effects do not drive a perpetual current. At different junction temperatures, electrons and holes transport entropy differently, giving different Seebeck coefficients on the two sides. The loop has a thermoelectric electromotive force, oriented as ${\int }_{T_1}^{T_2}(S_n-S_p)dT$, and therefore generally carries current when closed. This is the Seebeck effect: energy is supplied by heat flow from hot to cold. The temperature dependence of the junction barrier is useful intuition, but an exact terminal voltage is not obtained simply by subtracting two equilibrium built-in potentials; it also depends on bulk transport and contact conditions. The given equilibrium density-of-states data do not determine the numerical thermopowers.

The classical one-particle partition function, with phase-space cell $h^3$, isFor the homogeneous trap, rescale $r=V^{1/3}(kT{)}^{1/(2n)}u$. The spatial measure contributes $V(kT{)}^{3/(2n)}$, soHere $I_n=(2m\pi /h^2{)}^{3/2}{\int }_0^{\infty }4\pi u^2{e}^{-u^{2n}}du$, as specified; the last integral is $2\pi \Gamma (3/(2n))/n$.

Using the printed convention $Z_N=Z_1^N$, the free energy isThis convention treats the particles as labeled. For indistinguishable classical particles, $Z_N=Z_1^N/N!$ adds $kT\log N!$ to $F$. It changes the entropy by $-k\log N!$ but leaves the following force, energy and heat capacities unchanged. This normalization qualification is required to distinguish the printed expression from the usual Gibbs-corrected gas free energy.

The generalized force conjugate to the external parameter $V$ is $P=-(\partial F/\partial V{)}_T$, henceIt is the same equation of state as a hard-wall ideal gas, although $V$ now dilates a soft confining potential rather than being the occupied geometric volume.

Writing $L=\log V+a\log (kT)+\log I_n$, thermodynamic differentiation givesThe mean kinetic energy is $3NkT/2$ and the trap potential contributes $3NkT/(2n)$. At fixed $P$, $V=NkT/P$, so ${\partial }_{T}L{|}_P=(a+1)/T$. ConsequentlyThe same result follows from enthalpy $E+PV=(a+1)NkT$. These identities describe the classical ideal gas in a homogeneous soft trap.