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The black-hole area theorem is a statement about classical future event horizons, with both a local focusing condition and global horizon regularity assumptions. Assume the null energy condition, an appropriately predictable exterior and no future termination of horizon generators before the focusing argument can be applied. Future-complete generators provide a particularly direct sufficient version of the last assumption. Strong asymptotic predictability is the usual global condition used to exclude the pathological alternatives; the energy condition by itself is not the whole theorem.

On a smooth part of the horizon, choose a future-directed affinely parametrized generator ${\ell }^a$ and affine parameter $\lambda$. Its expansion $\theta$ is the fractional rate of change of an infinitesimal transverse area element:The horizon is a null hypersurface, so its generator congruence is hypersurface-orthogonal and has zero twist. The transverse screen metric is positive definite, hence ${\sigma }_{ab}{\sigma }^{ab}\ge 0$. In four dimensions the Null Raychaudhuri equation isThe Einstein field equations and the null energy condition imply the null convergence condition $R_{ab}{\ell }^a{\ell }^b\ge 0$, so ${\theta }^{\prime }\le -{\theta }^2/2$.

Suppose ${\theta }_0<0$ at ${\lambda }_0$. As long as the congruence remains smooth, integrating $d(1/\theta )/d\lambda \ge 1/2$ shows focusing within affine distance at most $2/|{\theta }_0|$. Equivalently, comparison with the equality solution givesIts denominator tends to zero at finite future affine parameter. The corresponding transverse area collapses, giving a conjugate point to a spacelike surface. Beyond a focal point a generator cannot remain on an achronal boundary, since nearby points can then be joined by a timelike curve. But a regular future event horizon is such a boundary, and its generators cannot leave it to the future. With the stated future-extension assumption this is a contradiction. Therefore $\theta \ge 0$ on the smooth horizon.

Following generators from one horizon cross-section to a later one now gives a noncontracting area map. Generators entering at past endpoints or merger crease sets can add area, whereas the global hypotheses exclude generators disappearing from the future horizon. ConsequentlyThis is the future-complete horizon focusing proof of the area theorem. In $D$ spacetime dimensions, the coefficient $1/2$ is replaced by $1/(D-2)$ and the same proof gives a focusing distance at most $(D-2)/|{\theta }_0|$. It therefore also applies to the three-dimensional horizon cross-sections of the preceding five-dimensional example.

A cosmological constant does not invalidate the local focusing step. Contract the modified Einstein field equations with the null generator. Since $g_{ab}{\ell }^a{\ell }^b=0$, both the scalar-curvature term and the $\Lambda$ term disappear:Thus a cosmological constant preserves null convergence, whatever its sign. Under suitable global assumptions for the relevant asymptotic region, the black-hole area argument still works. A nonzero $\Lambda$ changes the asymptotic geometry and can introduce cosmological horizons, so the statement concerns the appropriate future black-hole horizon; it is not an assertion that every horizon-like surface automatically obeys the same area ordering.

A classical real canonical massless scalar field also preserves the area theorem's energy condition. Its stress tensor hasThis proof is independent of whether the field gradient is spacelike, timelike or null: the metric term vanishes and the remaining real square is nonnegative. It is the null energy condition for a canonical scalar field. Adding this field to other matter satisfying the null energy condition retains null convergence. With the same global regularity assumptions, area still cannot decrease; when the derivative along a generator is nonzero, it supplies an additional nonnegative focusing source.

The conclusion uses the given minimally coupled classical stress tensor. A wrong-sign kinetic term or a nonminimal curvature coupling would require another calculation. Likewise, a renormalized quantum stress-energy tensor need not obey the classical pointwise condition; the quantum radiation considered next is not covered by this classical proof.

Use a free real scalar field obeying a linear Klein-Gordon equation, for example $(\square -{\mu }^2-\xi R)\varphi =0$. This linearity is the assumption that permits mode evolution by a Bogoliubov transformation. Work with a complete set of normalized modes, or normalizable wave packets when the spectrum is continuous. The stationary asymptotic regions supply time translations relative to which positive-frequency solutions and their complex conjugates can be distinguished.

For solutions define the conserved Klein-Gordon inner productHere $\Sigma$ is a Cauchy surface with future-directed normal. Its conservation follows by applying the wave equation to the divergence of its current, provided boundary flux is included or vanishes. Let $u_j$ be a positive-frequency basis in the past and $v_s$ one in the future, each propagated as a solution through the intermediate geometry. Normalize them by $(u_i,u_j)={\delta }_{ij}$, $(u_i^{\ast },u_j^{\ast })=-{\delta }_{ij}$ and $(u_i,u_j^{\ast })=0$, with the corresponding identities for the future basis. The two field expansions areThe canonical commutation relations give $[a_i,a_j^{†}]={\delta }_{ij}$ and $[b_s,b_t^{†}]={\delta }_{st}$. The in-vacuum satisfies $a_j|0_{-}⟩=0$, while the out-vacuum satisfies $b_s|0_{+}⟩=0$. They are generally different vacua.

Fix the coefficient convention by expanding future modes in past modes:The minus sign is due to the negative norm of the conjugate modes. Taking the Klein-Gordon inner product with the field, which is antilinear in its first entry, givesBoth frequency sectors are needed: if $\beta \ne 0$, a future annihilation operator contains a past creation operator. This is the mechanism of particle number from Bogoliubov coefficients, not a classical nonzero mean field.

Computing $[b_s,b_t^{†}]$ and $[b_s,b_t]$ explicitly gives the canonical identities for a bosonic Bogoliubov transformation:Completeness also gives the inverse $a_j={\sum }_s({\alpha }_{sj}{b}_s+{\beta }_{sj}^{\ast }{b}_s^{†})$. Thus the evolved past vacuum, viewed in the future Fock space, obeysFor a finite-mode treatment with invertible $\alpha$, solve this condition by a multimode squeezed vacuum:The canonical identities make $Z$ symmetric. Commuting $b_s$ through the exponential gives $b_s|\Psi ⟩={\sum }_t{Z}_{st}{b}_t^{†}|\Psi ⟩$, which directly verifies the vacuum condition. Its normalization is $|C|=\det (I-Z{Z}^{†}{)}^{1/4}$. A general initial state is obtained by applying its initial creation-operator polynomial and expressing those operators in the future basis. In infinitely many modes, a common unitary Fock representation requires the creation-mixing coefficients to be Hilbert-Schmidt; finite wave-packet counts avoid treating delta-normalized modes as individual normalizable states. These are the bosonic mode mixing implementability conditions, rather than an assumption of arbitrary finite total creation over an infinite time interval.

For the requested number operator of one future mode, set $N_s=b_s^{†}{b}_s$. Expectations below are in the evolved past vacuum $|\Psi ⟩$, equivalently in $|0_{-}⟩$ using the Heisenberg-mode operators above. Acting on the past vacuum gives $b_s|0_{-}⟩=-{\sum }_j{\beta }_{sj}^{\ast }{a}_j^{†}|0_{-}⟩$, whose squared norm isThe mean particle number vanishes exactly when that mode has no negative-frequency mixing.

The second moment needs the anomalous contractionThe canonical commutation relations give $N_s^2=b_s^{†2}{b}_s^2+N_s$. Moreover,Its vacuum and two-particle pieces are orthogonal. Applying the commutation relations twice showsThe two pairings each contribute $n_s^2$ to the squared norm. Hence the full answer isThus $\mathrm{Var}(N_s)=n_s(n_s+1)+|c_s{|}^2$. These are the out-mode number fluctuations in the in-vacuum. Omitting $c_s$ would assume a property not given in the general question. A thermal single mode has $c_s=0$, giving $⟨N_s^2⟩=2n_s^2+n_s$. A single-mode squeezed vacuum state instead has $|c_s{|}^2=n_s(n_s+1)$, giving $⟨N_s^2⟩=3n_s^2+2n_s$.

If the counted observable is the total number over a finite set of future modes, its moments follow by retaining cross-mode correlations as well. Define $C_{st}=⟨b_s^{†}{b}_t⟩$ and $M_{st}=⟨b_s{b}_t⟩$. The same vacuum pairings give $⟨N_s{N}_t⟩=C_{ss}{C}_{tt}+|C_{st}{|}^2+|M_{st}{|}^2+{\delta }_{st}{C}_{ss}$. Thus $⟨N_{\mathrm{tot}}⟩=\mathrm{tr}C$ and $⟨N_{\mathrm{tot}}^2⟩=(\mathrm{tr}C{)}^2+\mathrm{tr}(C^2)+\mathrm{tr}(M^{†}M)+\mathrm{tr}C$, with $C_{st}={\sum }_j{\beta }_{sj}{\beta }_{tj}^{\ast }$ and $M_{st}=-{\sum }_j{\alpha }_{sj}^{\ast }{\beta }_{tj}^{\ast }$. Independent thermal counts cannot be assumed for a general correlated squeezed state.

For Hawking radiation, choose the incoming vacuum at past null infinity of a collapse spacetime and propagate late outgoing modes backwards. Near a nonextremal forming event horizon, the logarithmic tortoise-coordinate divergence leads to the Hawking exponential ray mapwhere $u$ is late retarded time, $U$ is an early regular affine null coordinate and $\kappa >0$ is the final surface gravity. A late positive-frequency mode $e^{-i\omega u}$ therefore becomes proportional to $(U_H-U{)}^{i\omega /\kappa }$ on rays escaping just before horizon formation. This is not positive-frequency with respect to $U$: it has both frequency signs and hence nonzero $\beta$.

The thermal factor can be seen directly. With $x=U_H-U>0$, $a=\omega /\kappa$ and a regulator $\epsilon >0$, the relevant Fourier integrals areAs $\epsilon \downarrow 0$, the arguments of the two complex factors approach $\pm \pi /2$. Their squared-modulus ratio for the suppressed and enhanced frequency branches is $e^{-2\pi a}$. Thus the thermal ratio of Hawking Bogoliubov coefficients, combined with the canonical normalization, yields the bosonic occupationin units $\hslash =c=k_B=1$. Scattering through the exterior potential multiplies the asymptotic occupation by the appropriate greybody factor.

A complete future mode description includes modes crossing the horizon as well as modes reaching future null infinity. The global incoming vacuum develops correlations between these sectors; restricting to the exterior gives approximately thermal outgoing occupation rather than a claim that the complete pure state became a thermal mixed state. This is the particle-production picture of a collapsing black hole, not an incoming thermal bath on an eternal geometry. Including backreaction turns the emitted positive energy into black-hole mass loss. The renormalized quantum stress tensor is not subject to the pointwise classical positivity used in the black-hole area theorem, so this does not contradict its classical proof.

An orthonormal coframe is a collection of independent differential forms $E^{\hat{a}}=e^{\hat{a}}{}_{\mu }d{x}^{\mu }$ satisfying $g_{\mu \nu }={\eta }_{\hat{a}\hat{b}}{e}^{\hat{a}}{}_{\mu }{e}^{\hat{b}}{}_{\nu }$, with $\eta =\mathrm{diag}(-1,1,1,1)$. Locally one can diagonalize the metric tensor and rescale the resulting covectors, or use Gram-Schmidt orthogonalization for a symmetric bilinear form, choosing a timelike direction first. The coframe is not unique: a local Lorentz transformation preserves the metric. For the Levi-Civita connection, solve Cartan's first structure equation and metric compatibility together:These determine the connection 1-forms. The antisymmetry applies to the lowered pair of indices; a mixed time-space pair instead has ${\omega }^0{}_i={\omega }^i{}_0$. Next Cartan's second structure equation gives the curvature 2-forms, and reading off their coefficients determines the Riemann curvature tensor:This also states the curvature-sign convention. Contracting gives the Ricci tensor, and its coordinate components are $R_{\mu \nu }=e^{\hat{a}}{}_{\mu }{e}^{\hat{b}}{}_{\nu }{R}_{\hat{a}\hat{b}}$.

On a static patch with $V>0$, chooseWrite $h=\sqrt{V}$, $a=h^{\prime }$, $b=h/r$, and $c=\cot \theta /r$. The exterior derivative givesThe nonzero independent connection 1-forms are consequentlywith ${\omega }^1{}_0={\omega }^0{}_1$, spatial mixed pairs antisymmetric, and the remaining forms zero. For example,Similarly ${\Omega }^0{}_2={\omega }^0{}_1\wedge {\omega }^1{}_2=-ab{E}^0\wedge E^2$, and the angular derivatives in $d{\omega }^1{}_3$ cancel against ${\omega }^1{}_2\wedge {\omega }^2{}_3$. The resulting Cartan curvature forms for a static spherical metric areFor the last entry, $d(-c{E}^3)=E^2\wedge E^3/r^2$ and ${\omega }^2{}_1\wedge {\omega }^1{}_3=-V{E}^2\wedge E^3/r^2$. The remaining curvature 2-forms follow from ${\Omega }_{ab}=-{\Omega }_{ba}$.

Put $A=V^{\prime \prime }/2$, $B=V^{\prime }/(2r)$, and $C=(1-V)/r^2$. Reading the Riemann curvature tensor from the forms and contracting gives the diagonal orthonormal Ricci tensorFor example $R_{00}=R^1{}_{010}+R^2{}_{020}+R^3{}_{030}=A+2B$, while $R_{22}=-B-B+C$. All off-diagonal entries vanish because the curvature 2-forms only involve their own index planes. In the coordinate basis the nonzero components areThe angular Einstein field equations with the negative cosmological constant give $(rV{)}^{\prime }=1+3r^2/l^2$. Integrating introduces one real constant, conventionally written $-2m$:Substitution gives $V^{\prime \prime }/2+V^{\prime }/r=3/l^2$, so the time and radial Einstein field equations hold as well. The solution is the Schwarzschild--anti-de Sitter black hole family, including its zero and negative mass limits.

For $m>0$, $V$ increases strictly from $-\infty$ at $r=0$ to $+\infty$ at infinity. Its unique positive simple zero $r_h$ satisfies $r_h^3+l^2{r}_h=2m{l}^2$. This zero is a coordinate singularity, not a curvature singularity: with the ingoing coordinate $v=t+\int dr/V$, the radial metric is $-Vd{v}^2+2dvdr$ and is regular there. Inside the future black-hole region, $\nabla r={\partial }_v+V{\partial }_r$ is future timelike, so every future-directed causal trajectory decreases $r$ and cannot return to the asymptotic anti-de Sitter boundary. Conversely, every exterior radius admits an outward radial null geodesic reaching that boundary, since ${\int }_r^{\infty }dr/V$ converges. Thus $m>0$ gives an event horizon at the unique positive zero of $V$. Here the event horizon is defined relative to the asymptotic timelike boundary of Anti-de Sitter spacetime. For $m=0$, $V=1+r^2/l^2>0$ and the spacetime is pure Anti-de Sitter spacetime, with no black-hole event horizon. For $m<0$, $V=1+2|m|/r+r^2/l^2>0$, so there is also no event horizon; the mass term produces a naked singularity at $r=0$. The field equation itself does not impose a positive mass integration constant.

Use geometrized units with $\hslash =k_B=1$ initially, and write $f(r)=1-2M/r$, $h(r)=1-Q^2/(Mr)$. The given nonextremality assumption makes $h_H=h(2M)=1-Q^2/(2M^2)>0$. Thus the static Killing vector field has a simple zero of its norm atThe exterior $r>2M$ is static. For radial null geodesics, $dt/dr=\pm 1/f$, independently of $h$. With $v=t+\int dr/f$, the metric iswhich is regular at $r_H$ since $h_H>0$. The outgoing null equation is $dr/dv=f/2$. Inside the future horizon these rays move to smaller radius, while outside they can escape to infinity. Hence the zero is the event horizon. The inner zero of $h$ is not another regular Killing horizon: the Killing norm there diverges rather than vanishes. The strict charge bound matters when making the regular extension.

For the Euclidean black-hole regularity condition, continue $t=i{t}_E$, consistent with the sign of the given quantum time-translation operator. Set $x=r-2M$. To leading order $f=x/(2M)$ and $h=h_H$, so the radial–time part of the Euclidean metric isIntroduce the radial proper length coordinate $\rho$ by $x=h_H{\rho }^2/(8M)$. ThenThe first two terms are a polar plane with angle $t_E/(4M)$. Smoothness at its origin requiresA different period would make a conical singularity. The time coordinate is normalized to proper time at infinity, so no further asymptotic redshift factor is needed. The cancellation of $h_H$ proves the charge independence of the Euclidean horizon period under a regular radial conformal factor; this argument would fail at $h_H=0$.

To justify the thermal interpretation, analytic continuation of the time-evolution operator $U(t)=e^{iHt}$ gives $U(i{t}_E)=e^{-H{t}_E}$. Its matrix element is an imaginary-time propagation kernel. Identifying the initial and final configurations and integrating over them takes its trace, giving the Euclidean path integral $Z=\mathrm{Tr}{e}^{-\beta H}$. In an energy eigenbasis each state has weight $e^{-\beta E}$, exactly the Boltzmann factor of a canonical ensemble. Thus the geometrically required period is the inverse Hawking temperature. With dimensions restored, an imaginary time period $\Delta t_E$ gives $\beta =\Delta t_E/\hslash =1/(k_{B}T)$. ConsequentlyHere $\mathcal{M}$ is mass in ordinary units; for a neutral hole $M=G\mathcal{M}/c^2$ and the physical imaginary-time period is $8\pi G\mathcal{M}/c^3$. Bosonic thermal fields are periodic and fermionic fields antiperiodic over that same Euclidean period. The relation to the canonical density operator, not just dimensional reasoning, identifies the temperature.

For an evaporation estimate take a neutral, isolated Schwarzschild black hole and approximate its radiated power with the Stefan–Boltzmann law. Its area is $A=16\pi G^2{\mathcal{M}}^2/c^4$, and the photon Stefan-Boltzmann constant is ${\sigma }_{SB}={\pi }^2{k}_B^4/(60{\hslash }^3{c}^2)$. Substituting the temperature givesThe black-hole evaporation speeds up as the mass decreases. Integrating, rather than dividing the initial rest energy by the initial power, givesThe corresponding initial Hawking temperature is about $6\times 10^{-8}$K. This is the Stefan-Boltzmann estimate of Schwarzschild evaporation time at the level of accuracy requested. Greybody factors and other emitted species alter the numerical coefficient, but not the leading cubic scaling of a large isolated hole. The estimate assumes negligible accretion or ambient radiation absorption; a stellar-mass hole in the present cosmic radiation bath is not currently losing mass at that isolated-hole rate. Finally the semiclassical formula does not describe the Planck-scale endpoint: extrapolating it to zero mass is an estimate of the macroscopic lifetime, not a proof of what the final quantum object is.

Use geometrized units, and denote the beacon's proper time by $s$ to distinguish it from the short emitted duration $\tau$. Take $s=0$ at release. The period $M$, the duration $\tau$, and the wavelength $\lambda$ are measured by the beacon's own clock. Its radial timelike geodesic has conserved Killing energy per unit rest mass $E=fdt/ds$, where $f=1-2M/r$. Initially $dr/ds=0$ at $r=8M$, so $E=\sqrt{f(8M)}=\sqrt{3}/2$. Normalization of the four-velocity givesFor the Schwarzschild radial fall from rest at a finite radius, introduceIndeed $ds/d\eta =8M(1+\cos \eta )$ and $dr/d\eta =-4M\sin \eta$ reproduce the inward radial equation. At the Schwarzschild event horizon, $r=2M$ and $\cos {\eta }_H=-1/2$. HenceThis is the finite falling-clock time to the event horizon. By contrast, Schwarzschild time obeys $dt/ds=E/f$ and diverges at crossing. Near the horizon $r-2M\simeq E(s_H-s)$, so $dt/ds\simeq 2M/(s_H-s)$ and $t\simeq -2M\log ((s_H-s)/M)+\text{constant}$.

Every outward optical pulse emitted while $r>2M$ can reach a distant observer, and no pulse emitted after entry into the future black-hole region can do so. The infinite coordinate crossing time does not produce infinite emissions, because the pulse clock runs in proper time. If there is a pulse at release, the escape times are $s_n=nM$ for $n=0,\dots ,23$. Thus 24 pulses are received, counting the pulse at release; 23 are emitted subsequently during the fall. More generally the PDF does not specify the phase of the pulse clock. For a first emission at $s=\delta$ with $0\le \delta <M$, the finite pulse count for a falling Schwarzschild beacon isThe ceiling handles a pulse scheduled exactly at crossing correctly: it cannot escape. This count concerns pulses associated with the fall, not any pulses emitted before release, and assumes an ideal observer able to detect arbitrarily redshifted signals.

To find the stretching of an optical pulse, use the Schwarzschild tortoise coordinate $r_{\ast }=r+2M\log (r/(2M)-1)$. An outward radial null geodesic has constant Schwarzschild retarded null coordinate $u=t-r_{\ast }$. At an observer's fixed large radius $R$, its arrival time is $t_R=u+r_{\ast }(R)$; therefore only the varying emission value of $u$ contributes to the interval between arriving wavefronts. Differentiating along the falling source givesThe outgoing pulse redshift from a radial Schwarzschild infaller includes both gravitational redshift and the emitter's inward Doppler shift. A short pulse emitted at radius $r_e$ has, to leading order in its duration,For a finite receiver radius, replace $\mathcal{D}$ by $\sqrt{f(R)}\mathcal{D}$ to measure the receiver's proper time and frequency; this correction is negligible for $R\gg M$. The inverse scaling of frequency follows because the number of carrier cycles is unchanged when the pulse is stretched. Here $\nu =1/\lambda$ is ordinary frequency, not angular frequency, in units with light speed one.

Under the release-phase convention, the last emission is at $s_e=23M$. Solve $8({\eta }_e+\sin {\eta }_e)=23$ to obtain ${\eta }_e\simeq 1.94374184$, $r_e\simeq 2.54255983M$, and $\mathcal{D}(r_e)\simeq 7.49122492$. Because $\tau \ll M$ and this pulse begins $0.683364M$ before crossing, the local short-pulse approximation applies. Thus its approximate observed duration and frequency areWithout fixing that phase, these numerical coefficients are not unique: put $s_e=\delta +(N(\delta )-1)M$, find $r_e$ from the parameterization, and use the boxed general formulas. For a last pulse close to crossing, with $\Delta =s_H-s_e\ll M$ but $\tau \ll \Delta$, the horizon redshift of a Schwarzschild infaller gives the useful estimatesIf a phase instead places the pulse within one pulse-duration of crossing, $\Delta \lesssim \tau$, a single constant stretching factor is inappropriate. The received extent of a complete escaping pulse is exactly $u(s_e+\tau )-u(s_e)$; a pulse straddling crossing has an escaping tail extending to arbitrarily late arrival times, whose instantaneous frequency tends to zero. The finite count remains valid, while its last-pulse duration then has no finite constant multiple of $\tau$. This identifies precisely what the unspecified pulse phase can change.

Use geometrized units and assume $M>0$, $|Q|\le M$, so the Reissner-Nordstrom spacetime contains a black hole. Its outer event horizon is $r_{+}=M+\sqrt{M^2-Q^2}$. The classical absorption cross-section is obtained in geometrical optics, by counting incident null geodesics captured from infinity. Spherical symmetry lets us use the equatorial plane. Write $f(r)=1-2M/r+Q^2/r^2$, and let a dot denote differentiation with respect to an affine parameter. The conserved Killing energy and angular momentum are $E=f\dot{t}$ and $L=r^2\dot{\varphi }$. Null normalization giveswhere the impact parameter is $b=L/E$ and $W=f/r^2$. An incoming ray turns around if the angular-momentum barrier reaches $E^2$ before the ray reaches the event horizon. The limiting captured ray approaches the exterior unstable circular null geodesic, at the maximum of $W$.

The stationary-point equation isThe outer root is the photon sphere radiusIt lies outside $r_{+}$: for $|Q|\le M$ it ranges from $3M$ down to $2M$, whereas $r_{+}$ ranges from $2M$ down to $M$. More precisely $r_{\mathrm{ph}}>2M$ except in the extremal case, and $r_{+}\le 2M$; even at extremality $r_{\mathrm{ph}}=2M>r_{+}=M$. At this root,so the circular null geodesic is unstable. The smaller stationary root lies at or inside the outer horizon and does not set capture from infinity. For $b<b_c$, the radial equation has no exterior turning point and the ray enters the event horizon; for $b>b_c$, it turns around. The critical impact parameter is thereforeThe stationary-point equation gives $Q^2=(3M{r}_{\mathrm{ph}}-r_{\mathrm{ph}}^2)/2$, hence $f(r_{\mathrm{ph}})=(r_{\mathrm{ph}}-M)/(2r_{\mathrm{ph}})$. Since a parallel incident beam is captured over a disk of radius $b_c$, the photon capture cross-section of a Reissner-Nordstrom black hole isThe equality case approaches the unstable orbit indefinitely, but it is a measure-zero boundary of the captured disk and does not change the cross-section.

For the suggested parameterization, choose $\psi$ with $Q=3M\sin \psi /\sqrt{8}$ and $\cos \psi \ge 0$. Then $\sqrt{9M^2-8Q^2}=3M\cos \psi$, so the same cross-section isThe black-hole range has $|\sin \psi |\le \sqrt{8}/3$, hence $\cos \psi \ge 1/3$. For $Q=0$, $r_{\mathrm{ph}}=3M$, $f(r_{\mathrm{ph}})=1/3$, and $b_c=3\sqrt{3}M$, givingthe classical Schwarzschild black hole light-capture result. As an additional limit check, $|Q|=M$ gives $r_{\mathrm{ph}}=2M$, $b_c=4M$, and ${\sigma }_{\mathrm{abs}}=16\pi M^2$. These are geometric capture areas, larger than the corresponding horizon areas; the relevant barrier is the photon sphere, not the horizon's apparent disk. The result applies to light in the short-wavelength limit, while a finite-frequency wave absorption cross-section depends on the greybody factors.

Use geometrized units $G=c=1$. The Physical-process first law of black-hole mechanics concerns a small perturbation of a rotating black hole in an initially stationary spacetime by infalling matter, followed by relaxation to another stationary spacetime. For an uncharged black hole, or neutral accretion with charge held fixed, its first-order statement isHere $\kappa >0$ is the background surface gravity, ${\Omega }_H$ is the background horizon angular velocity, and $\delta A$ is the change in event horizon area. The energy $\delta E$ and angular momentum $\delta J$ are the fluxes carried into the event horizon; when these account for the changes of the black hole, they equal $\delta M$ and its change of angular momentum. The result is a linearized law, not an exact identity for a finite violent accretion event. The perturbation must be small enough that focusing does not produce new caustics or invalidate the expansion about the existing generators.

Let $k^a$ be the stationary Killing vector field normalized at infinity, $m^a$ the axial Killing vector field with $2\pi$-periodic orbits, andOn the background Killing horizon, ${\chi }^a$ is its future-directed null generator and satisfies ${\chi }^b{\nabla }_b{\chi }^a=\kappa {\chi }^a$. Choose a parameter $v$ with $\chi ={\partial }_v$. The null expansion in this parameter is $\theta ={\partial }_v\log dA$. The generator is not affinely parametrized; if $\ell ={\partial }_{\lambda }$ is affine, then $\chi =\kappa \lambda \ell$ after an appropriate choice of affine origin, with $\lambda \propto e^{\kappa v}$. Rescaling the Null Raychaudhuri equation givesThe null twist vanishes because the generators are normal to the event horizon. By the Einstein field equations, $R_{ab}{\chi }^a{\chi }^b=8\pi T_{ab}{\chi }^a{\chi }^b$. Background null expansion and null shear vanish on a Killing horizon of the background stationary spacetime. Thus $\theta$ and $\hat{\sigma }$ are first order, their squares are second order, and the linearized equation isThe final stationary spacetime supplies the boundary condition $\theta \to 0$ in the future. Explicitly,This future boundary condition reflects the global definition of the event horizon. For localized infall, $\theta$ also tends to zero in the remote past. Integrating the linearized equation along the generators, and then over a background cross-section with area element $d{A}_0$, givesUsing the background area element introduces only second-order corrections.

With orientations chosen so that future infalling positive energy has positive flux, the conserved Killing energy and axial angular momentum giveConsequently $\delta E-{\Omega }_H\delta J={\int }_{\mathcal{H}}\delta T_{ab}{\chi }^a{\chi }^{b}dvd{A}_0$, proving the displayed Physical-process first law of black-hole mechanics. Restoring $G$ replaces $8\pi$ by $8\pi G$. The null energy condition makes the integral nonnegative and therefore implies area increase in this regime; that condition is needed for the sign conclusion, not for the linearized flux identity itself. Charged infall would require the additional electrostatic work term, and an extremal background needs separate treatment because the proof uses $\kappa >0$.

Let $p^a$ be the future-directed causal four-momentum of the fragment captured by the Kerr black hole. In signature $(-+++)$ its conserved Killing energy and axial angular momentum areThe future-directed null generator of the Killing horizon is ${\chi }^a=k^a+{\Omega }_H{m}^a$, where ${\Omega }_H$ is the Kerr horizon angular velocity. At a crossing of the future event horizon, the scalar product of a future-directed causal vector with a future-directed null vector is nonpositive. For example, in a local orthonormal frame, $\chi =\alpha (1,\mathbf{n})$ with $|\mathbf{n}|=1$ and $\alpha >0$, while $p^0\ge |\mathbf{p}|$; hence $p\cdot \chi =\alpha (-p^0+\mathbf{p}\cdot \mathbf{n})\le 0$. ThereforeThis is a local causal restriction, so it applies also to an extremal Kerr black hole. A massive fragment crossing a regular future event horizon has a strict inequality; equality is the reversible limiting case, with causal four-momentum approaching the generator direction.