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Consider $N$ noninteracting bosons in one internal quantum state, at thermal equilibrium in the three-dimensional harmonic oscillator potentialHere $m$ is the mass, the positive ${\omega }_i$ are angular frequencies, and $\bar{\omega }$ is their geometric mean. Measure the one-particle energy $ϵ$ and chemical potential $\mu$ from the ground-state energy. The quantum harmonic oscillator levels then have $ϵ=\hslash ({\omega }_x{n}_x+{\omega }_y{n}_y+{\omega }_z{n}_z)$ with nonnegative integer $n_i$ and reduced Planck constant $\hslash =h/(2\pi )$.

When $ϵ\gg \hslash {\omega }_i$, counting lattice points can be approximated by the volume of a tetrahedron in the positive octant of $(n_x,n_y,n_z)$:This is a density of states, with units of inverse energy. The factor $1/6$ is the unit tetrahedron volume; differentiating produces the factor $1/2$. Treat the single ground state separately: its macroscopic population would be lost by replacing every level with a smooth density of states.

For $k_{B}T\gg \hslash {\omega }_i$, integrate the Bose-Einstein distribution over excited energy eigenstates. As $\mu$ approaches zero from below, their maximum population isHere $k_B$ is the Boltzmann constant and $\zeta$ the Riemann zeta function. A short proof uses $1/(e^x-1)={\sum }_{\ell \ge 1}{e}^{-\ell x}$ and ${\int }_0^{\infty }{x}^2{e}^{-\ell x}dx=2/{\ell }^3$. The nonnegative terms permit exchanging summation and integration by Tonelli's theorem, giving ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$. Near zero the integrand is proportional to $x$, so its integral converges. That finite excited-state capacity is the mechanism of Bose-Einstein condensation.

Set $N_{\mathrm{ex}}^{\mathrm{sat}}(T_c)=N$. Particle conservation gives, to this continuum accuracy,Above $T_c$, the macroscopic condensate fraction is zero and the chemical potential adjusts below zero. This cubic power belongs to a three-dimensional harmonic trap; it is not the homogeneous three-dimensional ideal gas exponent. For $N=10^5$, $\bar{\omega }/(2\pi )=100\mathrm{Hz}$ and $\zeta (3)\simeq 1.202056903$, the model gives $T_c\simeq 209.5\mathrm{nK}$. At $T=0.8T_c\simeq 167.6\mathrm{nK}$, $N_0/N=0.488$, or approximately $48800$ condensed bosons. These are chosen parameters and calculated values.

Finite $N$ gives a smooth crossover rather than a sharp phase transition. The exact ideal gas sum over excited quantum harmonic oscillator levels differs from the continuum integral, especially when $k_{B}T$ is not large relative to every $\hslash {\omega }_i$. A sharp trapped-gas limit takes $N\to \infty$, $\bar{\omega }\to 0$ with $N{\bar{\omega }}^3$ fixed and fixed frequency ratios. Interactions also alter the critical temperature, condensate fraction and release profile; a dilute gas can have appreciable interaction energy compared with the trap level spacing. The cubic law is a useful baseline, not a fit guaranteed for either 1995 experiment.

For a homogeneous ideal gas in $d$ dimensions with $A,\alpha >0$, the density of states scales as ${ϵ}^{d/\alpha -1}$. At limiting chemical potential zero, the Bose-Einstein distribution is proportional to $1/ϵ$ near zero energy. The excited-particle integral is finite exactly when $d/\alpha >1$. Thus positive-temperature condensation at fixed nonzero density is possible precisely for $0<\alpha <d$. A nonconserved particle number, as for ordinary equilibrium photons, removes the fixed-number saturation mechanism even when the integral is finite.

Bose-Einstein statistics describes identical noninteracting bosons in equilibrium. Any occupation number is allowed in a one-particle state; the resulting Bose-Einstein distribution is $n(E)=[e^{\beta (E-\mu )}-1{]}^{-1}$.

Bosonic exchange symmetry is the symmetric exchange statistics of identical bosons. Commuting creation operators build the symmetric bosonic Fock space; a normalized one-particle mode has occupation states $(a^{†}{)}^n|0⟩/\sqrt{n!}$ for every nonnegative integer $n$. At thermal equilibrium this symmetry leads to Bose-Einstein statistics and the Bose-Einstein distribution. Exchange symmetry itself does not require thermal equilibrium.

Commuting creation operators make multiparticle states invariant under exchange of their labels. With $[a,a^{†}]=1$, the normalized number states $(a^{†}{)}^r|0⟩/\sqrt{r!}$ exist for all $r\ge 0$, giving unrestricted bosonic occupation numbers. The single-mode thermal sum is $Z={\sum }_{r\ge 0}{e}^{-\beta Er}=(1-e^{-\beta E}{)}^{-1}$ and the mean occupation is $(e^{\beta E}-1{)}^{-1}$. Thus the exchange symmetry and the Bose-Einstein distribution are two expressions of the same bosonic algebra.

Choose a positive constant $\Sigma$ and the illustrative interface lawHere $T_h$ and $T_c$ are the hot and cold interface-side temperatures, not distant thermometer readings. This is a low-temperature phonon model under restrictive assumptions, not a universal fit to Kapitza's measured exponent or a material-property table. Its coefficient $\Sigma$ is not the Stefan-Boltzmann constant: the carriers in this model are phonons, rather than photons crossing a vacuum.

The fourth power has a simple counting explanation. In the three-dimensional low-temperature Debye model, a linear phonon dispersion relation gives a density of states proportional to ${\omega }^2$. A phonon has energy $\hslash \omega$ and Bose-Einstein distribution occupation $[\exp (\hslash \omega /k_{B}T)-1{]}^{-1}$. If propagation speeds and the angle-averaged transmission introduce no additional frequency scale over the occupied modes, the outgoing energy flux has the formExtending the upper integration limit assumes occupied modes lie well below the cutoff. Balancing the two outgoing fluxes, with coefficients satisfying zero net heat flux at equal temperatures, gives the chosen difference of fourth powers. Frequency-dependent transmission, different excitations or additional transport channels can change this conclusion.

The finite-drop area-specific thermal resistance follows by factoring a difference of powers:Near thermal equilibrium, write $T_h=T+\delta T$ and $T_c=T$. Expansion gives $q=4\Sigma T^3\delta T+O((\delta T{)}^2)$, so the differential area-specific Kapitza resistance isAt a finite bias, varying only the hot side gives $(\partial q/\partial T_h{)}^{-1}=1/(4\Sigma T_h^3)$; this is generally different from $r_{K,\sec }$. Stating what is held fixed makes a differential resistance reproducible.

For an original normalized example, choose reference scales $T_0>0$ and $A_0>0$, and setFor contact area $A=2A_0$, the total heat transfer rate is $\dot{Q}=2.1472A_0\Sigma T_0^4$. The finite-drop resistance satisfies $\Sigma T_0^3{r}_{K,\sec }=0.2/1.0736=125/671\approx 0.18629$. Using the equilibrium tangent at $T_0$ instead would predict $j=4(0.2)=0.8$, illustrating why a substantial temperature rise requires the nonlinear law. Doubling the common equilibrium temperature reduces $R_K$ by eight in this chosen model.

A series check can be done without guessing any material properties. Stipulate a hot-side slab with dimensionless areal bulk resistance $b_s=\Sigma T_0^3{L}_s/k_s=0.10$, and neglect the cold-side bulk drop. At the same $j=1.0736$, the hot reservoir must have $T_H/T_0=1.2+b_{s}j=1.30736$. The imposed reservoir drop splits into $0.10736$ in the bulk and $0.20$ at the interface. Applying the fourth-power law directly to $T_H$ would assign the bulk drop to the interface and overestimate the heat flux.

Finally, transfer heat at rate $Aq$ between maintained interface-side temperatures $T_h>T_c>0$. The hot side loses entropy at rate $Aq/T_h$, and the cold side gains it at rate $Aq/T_c$. With no interface entropy accumulation, the entropy production per unit area isFor the normalized example it is ${\dot{S}}_{\mathrm{prod}}/(A\Sigma T_0^3)=1.0736(1-1/1.2)=0.178933\dots$. The same energy crosses both sides in a steady state, while total entropy increases, consistently with the Second law of thermodynamics. A finite temperature jump during heat transfer therefore expresses dissipation, not a failure of energy conservation.

For transverse coupling with equal mode amplitudes $\lambda$, Fermi's golden rule and the oscillator creation operator and annihilation operator givewhere the first index is the final spin and $g$ is the density of states per unit angular frequency. Decay creates a quantum and carries the factor $n_{\Omega }+1$; excitation annihilates one and carries $n_{\Omega }$. For nonuniform resonant occupations, $n_{\Omega }$ is their density-weighted mean.

With a thermal Bose-Einstein distribution, $n_{\Omega }/(n_{\Omega }+1)=e^{-\beta \hslash \Omega }$. The two-state master equation therefore has the Boltzmann distribution as its unique stationary populations when ${\Gamma }_0>0$. The population deviation relaxes at rate ${\Gamma }_0(2n_{\Omega }+1)$. If there is no resonant coupling, both rates vanish and stationarity does not determine the populations.

For three-dimensional Debye model mode density $D(\omega )=9N{\omega }^2/{\omega }_D^3$, the Bose-Einstein distribution becomes $n_{\omega }\simeq k_{B}T/(\hslash \omega )$ when $T\gg {\Theta }_D$. Integrating the occupations gives $N_{\mathrm{ph}}\simeq (9/2)NT/{\Theta }_D$. The number of phonon quanta is not the number $3N$ of vibrational modes and need not be bounded by it. If a stipulated electron-phonon scattering model makes resistivity proportional to this phonon number with a temperature-independent coefficient, the high-temperature electrical resistivity is proportional to $T$.

Integrating the Bose-Einstein distribution against the three-dimensional Debye model phonon density of states gives $N_{\mathrm{ph}}=9N(T/{\Theta }_D{)}^3{\int }_0^{{\Theta }_D/T}{x}^2/(e^x-1)dx$. For $T\ll {\Theta }_D$, expand $(e^x-1{)}^{-1}={\sum }_{j\ge 1}{e}^{-jx}$ and integrate termwise: the limiting integral is $2{\sum }_{j\ge 1}{j}^{-3}=2\zeta (3)$. This counts thermal quanta, not available modes or half quanta of zero-point energy.

For a dilute nonrelativistic gas with $x=m/T\gg 1$ and zero chemical potential, the Bose-Einstein distribution approaches the Maxwell-Boltzmann distribution. Expanding $E=m+p^2/(2m)+\cdots$ in the number density integral gives $n\sim g(mT/(2\pi ){)}^{3/2}{e}^{-x}$. Since $\rho =mn+3nT/2+\cdots$ and $P=nT$, the entropy density at zero chemical potential gives $s=n(x+5/2+\cdots )$ and hence the displayed leading expression. This includes the internal multiplicity $g$; a single real scalar field has $g=1$.

Define $\hslash {\omega }_0=E_2-E_1$ and use radiation energy density $u(\omega )$ per unit angular frequency, excluding the zero-point contribution. The Angular-frequency Einstein coefficients give rates per atom: $B_{12}u({\omega }_0)$ for absorption, $B_{21}u({\omega }_0)$ for stimulated emission, and $A_{21}$ for spontaneous emission. Stimulated emission adds a photon to the stimulating mode; spontaneous emission persists with no incident photons.

In thermal equilibrium the Boltzmann distribution gives $N_2/N_1=e^{-\hslash {\omega }_0/(k_{\mathrm{B}}T)}$. Balancing up and down rates yieldsMatching the Planck law for every temperature fixesNo degeneracy factor is needed for these nondegenerate levels.^[ref]

For plates of large area $A$ and spacing $d$, the surviving low-frequency transverse electromagnetic mode has one polarization, $\mathbf{E}\parallel \hat{\mathbf{z}}$, and in-plane wavevector. Periodic in-plane counting givesMultiplying this single-polarization two-dimensional photon density of states by $\hslash \omega$ and the Bose-Einstein distribution produces the thermal radiation between parallel conducting plates:The same rate balance therefore givesThese $B$ coefficients refer to the radiation polarization in their respective environments: isotropic free-space $B$ must not automatically be equated to the coefficient for a specified $z$-polarized field.

For hydrogen, take the quantization axis along $z$, neglect spin-orbit coupling, and assume perfect plates, a wide in-plane region and an unmodified atomic wavefunction: $a_{\mathrm{B}}\ll d\ll c/{\omega }_0$, with negligible surface-induced mixing or shifts. The electric-dipole selection rules for hydrogen require $\Delta l=\pm 1$; the allowed $z$ field has $\Delta m_{\ell }=0$. Hence the polarization-selective spontaneous emission between conducting plates givesThe thin-cavity enhancement of a normal electric dipole applies: the $2p_0\to 1s$ dipole points along $z$ and couples to the enhanced mode density. For $m_{\ell }=\pm 1$, the transition dipoles lie in the $x$–$y$ plane and couple to none of these low-frequency modes: the electric-dipole decay rate is zero in the ideal model, so this channel's lifetime is infinite. Real finite, imperfect plates and additional decay or mixing channels limit the lifetime. The order-one orientation factor for the allowed transition does not change the requested lifetime scaling.

The thermal oscillator occupation is the Bose-Einstein distribution. With the rate notation ${\Gamma }_{\eta \sigma }$ for a transition from $\sigma$ to $\eta$, the master equation for the up population isAt thermal equilibrium, assuming ${\Gamma }_0>0$, vanishing time derivative is detailed balance between the two spin states. HenceorTogether with normalization, this gives $p_{\uparrow }=1/(1+e^{\beta \hslash \Omega })$ and $p_{\downarrow }=e^{\beta \hslash \Omega }/(1+e^{\beta \hslash \Omega })$, the Boltzmann distribution for the energy eigenvalues $\pm \hslash \Omega /2$. At zero temperature, absorption vanishes and emission relaxes the spin into its down ground state.

The nonzero-rate condition is necessary: if $\lambda =0$ or $g(\Omega )=0$, both rates vanish, so every initial pair of populations is stationary within this approximation. In that case stationarity alone cannot imply a thermal distribution. The equilibrium conclusion therefore presumes coupling to resonant oscillator modes.

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 Debye model treats the crystal as an isotropic elastic continuum with independent harmonic vibrational modes. There are three acoustic polarizations, approximated by linear dispersions and a common effective speed. A cutoff is chosen to retain $3N$ modes for $N$ atoms. The model neglects anharmonic interactions and the detailed high-wavevector dispersion. In a multi-atom primitive cell, optical branches require additional treatment; copper has one atom per primitive cell even though its conventional cubic cell contains four atoms.

A frequency interval contains, per volume,Each phonon mode has Bose-Einstein distribution occupation $(e^{\hslash \omega /(k_{B}T)}-1{)}^{-1}$. The fixed zero-point energy contributes no heat capacity. Its thermal energy density isFor $T\ll {\theta }_D$, extending the upper limit to infinity gives $u_{\mathrm{ph}}=(3{\pi }^4/5)n{k}_B{T}^4/{\theta }_D^3$. ThereforeThe cubic law reflects the three-dimensional acoustic mode density proportional to ${\omega }^2$. It is a low-temperature limit, not a valid extrapolation through the classical high-temperature regime.

At fixed chain length, each harmonic mode contributes a Bose–Einstein thermal occupation and a temperature-independent zero-point energy. The latter contributes no heat capacity. With $x=\hslash \omega /(k_{B}T)$,At low temperature only the approximately linear part of the acoustic dispersion is excited, so $g(\omega )\simeq 1/(\pi v_s)$. Using ${\int }_0^{\infty }{x}^2{e}^x/(e^x-1{)}^{2}dx={\pi }^2/3$ givesIt is linear in temperature, rather than the three-dimensional Debye $T^3$ law. At high temperature every scalar oscillator tends to $k_B$, henceThe cutoff temperature is ${\Theta }_m=\hslash {\omega }_m/k_B=305.5$ K. The curve rises smoothly from zero toward this plateau; ${\Theta }_m$ sets a crossover scale, not a sharp transition. In normalized variables its initial slope is $C_V/(N{k}_B)\simeq (2\pi /3)T/{\Theta }_m$. The low-temperature continuum limit also requires temperature above the finite-size level spacing; uniform center-of-mass translation is treated separately. Below the lowest nonzero mode energy a finite chain does not retain a continuum linear law down to arbitrary temperature.

For thermal photons, the available thermal length is ${\ell }_T=\hslash c/(k_{B}T)$. The number of occupied modes per volume scales as ${\ell }_T^{-3}$, while their characteristic energy is of order $k_{B}T=\hslash c/{\ell }_T$. Thus $n_{\gamma }\propto {\ell }_T^{-3}$ and $u_{\gamma }={\rho }_{\gamma }{c}^2\propto \hslash c{\ell }_T^{-4}$. More explicitly, the two photon polarizations and the Bose-Einstein distribution giveconsistent with the rounded coefficients supplied in the source.

Use its definition $s=n_{\gamma }/n_B$, with $n_B={\rho }_B/m_p={\Omega }_B{\rho }_c/m_p$ and ${\rho }_c$ the mass-equivalent cosmological critical density. Writing $Q=k_{B}T/(\hslash c)$, the photon energy formula gives $Q^4={\Omega }_{\gamma }{\rho }_{c}c/(0.66\hslash )$. ThereforeThis photon-to-baryon ratio from density parameters has leading coefficient $0.328\simeq 0.33$, as required. For an illustrative estimate, take the stated ${\rho }_c=10^{-26}\mathrm{kg}{\mathrm{m}}^{-3}$, ${\Omega }_{\gamma }=5\times 10^{-5}$ and ${\Omega }_B=0.05$. Then $\boxed{s\sim 1.4\times 10^9}$, i.e. of order a billion photons per baryon. The source alone supplies no numerical density parameters, so an estimate must state such additional inputs.

Strictly, this $s$ is the photon-to-baryon ratio, rather than the thermodynamic entropy per baryon itself. For a thermal photon gas, $S_{\gamma }/(k_B{N}_B)=[4u_{\gamma }/(3k_{B}T{n}_{\gamma })]s\simeq 3.6s$ with the exact blackbody coefficients. The ratio used in the source is proportional to the radiation entropy per baryon under adiabatic evolution.

Work in natural units $k_B=\hslash =c=1$, initially neglecting helium and excited hydrogen states. The reaction $p+e^{-}\leftrightarrow H+\gamma$ is in chemical equilibrium when ${\mu }_p+{\mu }_e={\mu }_H$, since equilibrium photons have zero chemical potential. Substitution of the nonrelativistic Maxwell-Boltzmann distribution for the three densities givesFor ground-state hydrogen use $g_e=g_p=2$, $g_H=4$, and $m_H\simeq m_p$ in the translational prefactor. Writing its Hydrogen binding energy as $I$, this becomes $(m_{e}T/(2\pi ){)}^{3/2}{e}^{-I/T}$. Charge neutrality gives $n_p=n_e=X_e{n}_B$, while baryon conservation gives $n_H=(1-X_e)n_B$. Hence the hydrogen-only Saha equation isThe photon Bose-Einstein distribution supplies $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, where $\zeta (3)$ is the usual Riemann zeta function value in the thermal number integral. With baryon-to-photon ratio ${\eta }_{\gamma }=n_B/n_{\gamma }$, substitution yieldsThis reproduces the requested numerical coefficient, with the printed $\xi (3)$ interpreted as the thermal $\zeta (3)$ value. There is a normalization ambiguity in the PDF: it writes $\eta =n_B/s$ without defining $s$. To reproduce this coefficient, $s$ must mean photon number density. If $s$ has its usual entropy density meaning, the Saha baryon-abundance normalization instead givesThe two forms agree through ${\eta }_{\gamma }=[{\pi }^4{g}_{\ast s}/(45\zeta (3))]{\eta }_s$, with the entropy sector specified. A baryon-to-entropy ratio cannot simply be inserted into the photon-normalized expression.

As the universe cools, the Boltzmann factor favors neutral atoms and cosmological recombination sharply lowers $X_e$. For an order-one ionized fraction the photon-normalized equation gives the balanceup to an order-one ionization-fraction factor. The small baryon abundance means that there are very many photons per baryon: even a small high-energy photon tail can ionize most atoms. Thus the recombination temperature is far below the microscopic binding scale $I\simeq 13.6\mathrm{eV}$, typically a few tenths of an eV for the standard abundance, rather than of order $I$.

Photon decoupling is a related but distinct kinetic event. The Thomson scattering rate $n_e{\sigma }_T$ falls relative to $H$ as free electrons disappear, and photons begin free streaming. The precise last-scattering epoch depends on integrated optical depth, not only the local equality of rates. At still lower density the effective recombination rate per electron also drops below the expansion rate. Residual electron freeze-out leaves a small nonzero ionized fraction, larger than the vanishingly small equilibrium Saha prediction; radiative atomic bottlenecks mean the entire history cannot be obtained from equilibrium alone.

The temperature comparison in the last request needs an interpretation. If “reionisation temperature” means the microscopic thermal ionization scale, the large photon-to-baryon ratio explains why cosmological recombination requires $T\ll I$. If it means later cosmic reionization, there is no universal thermal threshold: stars or other sources supply ionizing photons and can reheat gas. The cosmological photon background is actually colder at that later epoch, because $T_{\gamma }\propto a^{-1}$. Its temperature therefore cannot literally be higher than the background temperature at recombination merely because hydrogen is reionized.

In a large volume, one one-particle state occupies momentum-space volume $h^3/V$. A spherical shell of radius $p$ has volume $4\pi p^{2}dp$, so for spin degeneracy $g$ the density of states is $4\pi gV{p}^{2}dp/h^3$. With ultra-relativistic dispersion $E=cp$ and zero chemical potential, the Bose-Einstein distribution givesThe convergent dimensionless integral is ${\pi }^4/15$, obtained by expansion into exponentials and summing $\sum n^{-4}$. Therefore $ϵ\propto T^4$ and $\alpha =4$. Photons are massless bosons with two polarizations; equilibrium emission and absorption do not conserve photon number and force ${\mu }_{\gamma }=0$.

For recombination, chemical equilibrium imposes ${\mu }_p+{\mu }_e={\mu }_H+{\mu }_{\gamma }={\mu }_H$, including rest energies in the chemical potentials. Divide the three nonrelativistic number-density expressions to obtainHere $m_p{c}^2+m_e{c}^2-m_H{c}^2=I$. Neglect the relative proton/hydrogen mass difference and take the ground-state degeneracy ratio $g_p{g}_e/g_H=1$ (electron and proton spins give $2\cdot 2$ and hydrogen gives $4$). Charge neutrality gives $n_p=n_e$. This yields the stated Saha equation

With $n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$, the ratio asked for isEquivalently the prefactor is $16\pi \zeta (3)\eta (2\pi {)}^{-3/2}[kT/(m_e{c}^2){]}^{3/2}$. Because photons vastly outnumber baryons, even a small high-energy tail can keep hydrogen ionized. Neutralization therefore requires exponential suppression of that tail far below $kT=I$, explaining the much lower recombination temperature. The equilibrium formula is an estimate; the actual expanding-universe recombination kinetics eventually depart from Saha equilibrium.

Hawking radiation is a prediction of quantum field theory in curved spacetime, not of classical collapse alone. Consider a free massless Klein-Gordon field in the collapse vacuum for Hawking radiation: initially no incoming particles from past null infinity, with the short-distance state regular for freely falling observers through the forming event horizon. At late times use the approximately fixed exterior Schwarzschild metric; neglect backreaction during the observation interval. First set $G=c=\hslash =k_B=1$.

The exterior metric has $f(r)=1-2M/r$. Introduce the tortoise coordinate and outgoing and ingoing null coordinates,The surface gravity is $\kappa =1/(4M)$. The logarithm in $r_{\ast }$ makes a late escaping ray spend a long exterior coordinate time close to the event horizon. One of the regular outgoing Kruskal–Szekeres coordinates is proportional to $-e^{-\kappa u}$. Smooth propagation backwards through the collapsing body relates this regular coordinate to the incoming coordinate on past null infinity. If $v_H$ labels the last ray able to escape, the late-time Hawking exponential ray map is consequentlyHere the symbol $u(v)$ is the outgoing retarded time evaluated on the ray; the first equation is the leading asymptotic relation, with subleading corrections that vanish at late times. The constant $C$ depends on collapse history, whereas the exponential rate is determined by the final surface gravity. The relation follows from a nondegenerate Killing horizon; a linear ray map would not have the same effect.

An outgoing positive-frequency mode at future null infinity behaves as $p_{\omega }\sim e^{-i\omega u}/\sqrt{4\pi \omega }$, where $\omega >0$ is measured with respect to the asymptotic stationary time. In the near-horizon geometric optics approximation its backward-propagated form on past null infinity isThis is not purely positive frequency in $v$. Its decomposition into incoming positive-frequency modes $f_{{\omega }^{\prime }}\sim e^{-i{\omega }^{\prime }v}/\sqrt{4\pi {\omega }^{\prime }}$ and their complex conjugates is a Bogoliubov transformation:The coefficients follow from the Klein-Gordon inner product, or equivalently from the Fourier transform with the indicated flux normalization. Writing $y=v_H-v$ and $a=\omega /\kappa$, the two Fourier integrals, after a convergence regulator, areThe equality is the defining integral of the Gamma function, continued to these arguments from the right half-plane. Up to phases of unit modulus, $\alpha$ is $\sqrt{{\omega }^{\prime }/\omega }{I}_{+}/(2\pi )$ and $\beta$ is $\sqrt{{\omega }^{\prime }/\omega }{I}_{-}/(2\pi )$. Taking $ϵ\downarrow 0$ on the principal branches, the arguments of $ϵ\pm i{\omega }^{\prime }$ tend to $\pm \pi /2$. Thus $|I_{+}|\propto e^{\pi a/2}$ and $|I_{-}|\propto e^{-\pi a/2}$, giving the thermal ratio of Hawking Bogoliubov coefficientsIn particular, using $|\Gamma (1+ia){|}^2=\pi a/\sinh (\pi a)$,The collapse-dependent phases and $C$ disappear from the ratio. The canonical identities for a bosonic Bogoliubov transformation state that the positive-frequency norm minus the negative-frequency norm is one for a normalized outgoing mode. Combining that identity with the thermal ratio yields the late-time mean occupationThese are the Bose-Einstein distribution and the Hawking temperature. Strictly, monochromatic modes have delta-function normalization, and $\int d{\omega }^{\prime }|{\beta }_{\omega {\omega }^{\prime }}{|}^2$ contains the infinite-duration divergence $\int d{\omega }^{\prime }/{\omega }^{\prime }$. Normalized wave packets with a finite frequency band and a finite retarded-time window remove this artifact. For sufficiently late packets the occupation is the frequency average of the displayed thermal factor, tending to it for a narrow band. Fermionic fields instead give the corresponding denominator $e^{2\pi \omega /\kappa }+1$ because their canonical normalization uses anticommutators.

Four-dimensional propagation adds an important qualification to the word thermal. Separation of the massless Klein-Gordon equation into spherical harmonics gives an exterior radial scattering potentialOnly a fraction ${\Gamma }_{\ell }(\omega )$ of the outgoing mode is transmitted to future null infinity; this is the greybody factor. The scalar particle flux there isThe factor $2\ell +1$ counts the angular modes, and multiplication by $\omega$ gives the spectral energy flux. An observer at infinity therefore receives an outgoing thermal population filtered by curvature scattering, rather than an exact unfiltered blackbody spectrum. In physical units, for mass $M_{\mathrm{phys}}$,There is no incoming thermal bath in the collapse vacuum for Hawking radiation. The outgoing flux has positive Killing energy; its counterpart inside the event horizon can carry negative Killing energy. Including the expectation value of the stress-energy tensor in the gravitational evolution allows the mass to decrease. The fixed-background derivation describes the slowly evolving semiclassical regime and does not determine the endpoint of evaporation. Quantum expectation values need not satisfy the classical null energy condition, so this loss of area is not a contradiction of the classical second law of black-hole mechanics. The essential late-time universality is the combination of a regular initial quantum state and the exponential redshift at the newly formed event horizon.