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The Arnowitt-Deser-Misner energy is the Hamiltonian boundary charge generating an asymptotic time translation in the ADM formalism. For a spatial metric tensor $h_{ij}$ in asymptotically Cartesian coordinates, use geometrized units $G=c=1$:Here $n^i$ is the outward Euclidean unit normal vector. Standard differentiable asymptotic flatness gives $h_{ij}={\delta }_{ij}+O(r^{-1})$ and ${\partial }_k{h}_{ij}=O(r^{-2})$; suitable additional decay and integrability conditions ensure that the limit exists. Together with ADM momentum, this charge forms the asymptotic energy-momentum four-vector.^[ref]

For a time-symmetric Schwarzschild spacetime slice in isotropic coordinates, $h_{ij}={\psi }^4{\delta }_{ij}$ with $\psi =1+M/(2r)$. Its surface integrand is $-2{\partial }_r{\psi }^4=4M/r^2+O(r^{-3})$. Integrating over $S_r$ therefore gives $E_{ADM}=M$. The corresponding slice of Minkowski spacetime has constant $h_{ij}={\delta }_{ij}$ and zero ADM energy. In a moving asymptotic frame, the energy component need not equal the invariant ADM mass.

A geometrized luminosity is dimensionless because in geometrized units an energy and a time have the same dimension. Starting from SI luminosity, multiply by $G/c^5$. The corresponding SI power unit is $c^5/G$. Radiating a fraction $f$ of a mass $M$ at constant SI luminosity $L$ takes $fM{c}^2/L$, or $f(GM/c^3)/L_{\mathrm{geom}}$ in seconds.

Geometrized units set $G=c=1$, allowing mass, length, and time to be expressed in common units. Restore $GM/c^2$ for a mass parameter in a radial metric tensor.

For two equal point masses at $\pm z$ falling from rest at $\pm z_0$ under Newtonian gravity, in geometrized units one has $\ddot{z}=-m/(4z^2)$ and $\dot{z}=-\sqrt{m(1/z-1/z_0)/2}$. The only nonzero component of the second mass moment tensor is $I_{zz}=2m{z}^2$, and its third derivative is $-m^2\dot{z}/z^2$. The trace-free mass quadrupole moment and quadrupole formula give the displayed power. From infinity, integrating down to $z=z_c$ gives $E=2\sqrt{2}{m}^{9/2}/(105z_c^{7/2})$. Relative to the total initial mass $2m$ the fraction is $\sqrt{2}(m/z_c{)}^{7/2}/105$. The weak-field slow-motion approximation must be distinguished from its extrapolation to a compact endpoint.

The Interior Schwarzschild metric describes a static spherical perfect fluid of constant energy density. Use geometrized units $G=c=1$, mass $M>0$, surface areal radius $R$ and $\rho =3M/(4\pi R^3)$. For $0\le r\le R$, defineThe metric and pressure areAt $r=R$, $p=0$ and the lapse equals $s$, matching the exterior Schwarzschild metric. The mass profile is $m(r)=M{r}^3/R^3$. The lapse can also be recovered from hydrostatic balance $p^{\prime }=-(\rho +p)N^{\prime }/N$: constant $\rho$ makes $(\rho +p)N$ constant, and the surface condition fixes $N=(3s-q)/2$.

At the centre,Finite nonnegative central pressure requires $s>1/3$, equivalently $R>9M/4$. At the equality limit the central pressure diverges even though $R>2M$. This makes the model a useful demonstration of relativistic pressure support and stellar compactness. Constant density is an idealization, rather than a realistic universal equation of state for dense matter.

A Kerr black hole is the rotating, stationary, axisymmetric vacuum spacetime described by the Kerr metric. In geometrized units $G=c=1$, let $M>0$ be its mass and $J=Ma$ its angular momentum. Its radial horizon polynomial isA Kerr black hole requires $|a|\le M$; equality is the extremal black hole limit. The Schwarzschild black hole is recovered at $a=0$. The outer Killing horizon rotates with angular velocity $a/(r_{+}^2+a^2)$. The exterior Kerr ergoregion is distinct from that horizon and permits rotational black-hole energy extraction.^[ref]

A Komar charge is an oriented surface integral constructed from a Killing vector field and the metric tensor. Let $k$ be the metric-dual differential one-form of a stationary Killing vector field. In four dimensions, with geometrized units $G=c=1$ and the standard future/outward orientation, its mass normalization iswhere $\star$ is the Hodge star operator and $S$ is a closed spacelike two-surface. The normalization of the Killing vector field fixes the time scale: for Schwarzschild spacetime, unit time translation at infinity gives the black-hole mass parameter.

The Killing equation relates the divergence of $dk$ to a contraction of the Ricci tensor. Consequently, homologous enclosing surfaces give the same mass in a vacuum region with zero cosmological constant. Matter or a nonzero cosmological constant introduces a volume contribution; an arbitrary finite-radius surface integral need not equal the asymptotic mass. Axial symmetry instead gives Komar angular momentum with a different normalization.

Use geometrized units $G=c=1$ and positive mass $M$. The Schwarzschild metric initially describes an exterior region, not the whole black hole:Here $r$ is the areal radius and $t$ is Schwarzschild time, normalized at infinity. We must distinguish a removable coordinate singularity from a genuine curvature singularity. The Kretschmann scalar isIt is finite at $r=2M$ and diverges at $r=0$. Thus $r=2M$ can be crossed in a better chart, whereas $r=0$ cannot be made a regular point by changing the coordinate chart.

First integrate the Schwarzschild tortoise coordinate:The radial metric becomes $f(-d{t}^2+d{r}_{\ast }^2)$. Set $v=t+r_{\ast }$; the ingoing Eddington-Finkelstein coordinates giveThe radial block has determinant minus one and smooth coefficients at $r=2M$, so it extends through the future Schwarzschild event horizon. Its ingoing radial null geodesics have constant $v$, while the outgoing radial null geodesics obey $dr/dv=f/2$. In the future interior both future-directed radial null families decrease $r$; $r=0$ is therefore an unavoidable future boundary for causal motion there. The outgoing chart $u=t-r_{\ast }$ similarly gives $d{s}^2=-fd{u}^2-2dudr+r^{2}d{\Omega }^2$, extending through the past horizon. A single one of these null charts does not display both horizons or the entire extension.

To regularize both horizons together, introduce Kruskal–Szekeres coordinates in the original exterior:ThenFor $r>0$ the right side defining $UV$ decreases monotonically from one to minus infinity. In fact its derivative is $-r{e}^{r/(2M)}/(4M^2)$. Thus it determines a unique smooth $r=r(UV)$ wherever $UV<1$, including $UV=0$. At $r=2M$ the radial metric coefficient is finite and nonzero, proving actual regularity of both horizons and their intersection, rather than merely relabelling the old coordinate divergence. Allowing all real $U,V$ with $UV<1$ produces the simply connected maximal analytic extension, the Kruskal spacetime.

There are four regions of the Kruskal spacetime. The original exterior is $U<0,V>0$; a second asymptotically flat exterior is $U>0,V<0$. The future black hole has $U>0,V>0$ and $0<r<2M$; the past white hole has $U<0,V<0$ and the same range of areal radius. The horizon branches are $U=0$ and $V=0$. Their intersection $U=V=0$ is a regular bifurcation surface, a two-sphere of radius $2M$, not the centre of the spacetime. Each point of the radial diagram represents a symmetry two-sphere.

Put $T_K=(V+U)/2$ and $X_K=(V-U)/2$. Radial null geodesics have slopes $d{T}_K=\pm d{X}_K$, and the curvature singularities haveThe upper branch is the future spacelike Schwarzschild singularity; the lower branch is the white hole past singularity. The Killing vector field extends as ${\partial }_t=(V{\partial }_V-U{\partial }_U)/(4M)$. It is future directed in the original exterior and past directed in the other exterior for the chosen global time orientation; inside, this stationary vector is spacelike. In the future interior, $r$ is a time coordinate decreasing toward the singularity. For example, a radial infaller of unit specific Killing energy obeys $dr/d\tau =-\sqrt{2M/r}$ and reaches $r=0$ from the horizon in finite proper time $4M/3$. The curvature divergence and this geodesic endpoint show why the extension stops there.

For the global causal picture, compactify the null coordinates using $p=\arctan U$, $q=\arctan V$ and let $\mathcal{T}=2(p+q)/\pi$, $\mathcal{X}=2(q-p)/\pi$. The conformal compactification rescales the metric to remove the coordinate factors from the radial metric while preserving its null directions. Since $UV=1$ gives $p+q=\pm \pi /2$ on the two singular branches, they become the horizontal lines $\mathcal{T}=\pm 1$. The resulting Penrose diagram displays the two exteriors, each with its own past null infinity, future null infinity and spatial infinity, plus the future and past interiors. The singularities are spacelike boundaries; the horizons are interior null surfaces. Future event horizons separate events that can escape to an exterior's future null infinity from events in the black-hole interior.

The time-symmetric spatial slice has a bridge between the exteriors, but it is not a traversable passage. Future-directed causal curves have nondecreasing $U$ and $V$. Moving from the right exterior to the left would require $V$ to decrease from positive to negative; the reverse passage would require $U$ to decrease. Both are impossible. This proves the non-traversability of the Schwarzschild bridge. The complete eternal vacuum extension has two exteriors, a black hole and a white hole; it has no causal route between its exteriors. A black hole made by stellar collapse is a different global spacetime: its matter-filled past replaces the white hole region, and it normally has only one exterior. The vacuum extension constructed here should not be mistaken for the collapse history of a single star.

With electric charge, spherical electrovacuum instead gives the Reissner-Nordstrom metric, in an electromagnetic normalization absorbing the charge conversion constants:For $0<|Q|<M$, there are two simple horizons. Outside $r_{+}$ the metric is static; between $r_{-}$ and $r_{+}$, $f<0$ and $r$ is timelike. Inside $r_{-}$, $f>0$ again, so $r$ is spacelike and the $r=0$ singularity is timelike. The outer horizon is the event horizon; the inner horizon is a Cauchy horizon. The same logarithmic tortoise-coordinate and exponential-null-coordinate method crosses each simple horizon locally, but one exponential scale does not regularize both simultaneously: their surface gravities have different magnitudes,Continuing the exact metric through successive inner horizons produces repeated blocks with further exteriors, and evolution beyond a Cauchy horizon is not uniquely fixed by data on the original Cauchy hypersurface. The timelike singularity can be avoided by some trajectories in this ideal extension, unlike the unavoidable spacelike singularity of the future Schwarzschild spacetime interior. However, the enormous inner-horizon blueshift makes this exact continuation unstable: perturbations can cause mass inflation. This is an important qualification on the global structure of a charged spherical black hole.

At $|Q|=M$ the two horizons coalesce, $f=(1-M/r{)}^2$ and surface gravity is zero. The horizon is a degenerate Killing horizon, the tortoise-coordinate divergence becomes a pole and the ordinary nondegenerate Kruskal spacetime construction must be replaced. The exterior develops an infinitely long spatial throat and the formal Hawking temperature vanishes. For $|Q|>M$, no real zero of $f$ remains: the electrovacuum solution has a naked singularity, not a black hole. Consequently arbitrarily allowing charge changes the causal structure, not just the horizon radius.

Use geometrized units $G=c=1$ until converting the clock readings. By the Birkhoff theorem, the spherical vacuum exterior is a portion of the Schwarzschild spacetime with fixed mass $M$:The exterior is not flat and is not radiating merely because the surface moves. When the surface crosses $r=2M$, this exterior must be represented in regular ingoing Eddington-Finkelstein coordinates, rather than treating the Schwarzschild-coordinate divergence as a physical barrier.

The Oppenheimer-Snyder model has a homogeneous pressureless matter interior. Since its initial expansion vanishes but its density is positive, the Friedmann equation requires positive spatial curvature. Choose a comoving radial angle $\chi$ and the boundary's proper time $\tau$, with the stopwatch set to zero at release. The interior FLRW metric and density areThe density law follows from the dust stress-energy tensor and its covariant conservation. The Friedmann equation and the initial rest condition giveIts contracting branch has the parameterizationIndeed $d\tau /d\eta =a$ and $\dot{a}=-\sin \eta /(1+\cos \eta )$ satisfy the displayed equation; $\eta =0$ is the turning point and $\eta =\pi$ is the final curvature singularity.

For homogeneous dust-ball matching, the boundary has areal radius $R=a\sin {\chi }_0$. Matching its mass to the exterior givesThis $M$ is the gravitational mass parameter, not the integral of rest density over the curved proper-volume element. The matching can also be checked directly. The outward angular extrinsic curvature inside is $K_{\theta \theta }=R\cos {\chi }_0$; outside it is $R\sqrt{1-2M/R+{\dot{R}}^2}$. The boundary motion hasso these curvatures agree. The induced metric is $-d{\tau }^2+R^{2}d{\Omega }^2$ on both sides and $K_{\tau \tau }=0$ because the pressure-free boundary is geodesic. No artificial surface stress layer is required.

The observer moving with this boundary meets the event horizon when $R=2M$. From $R=R_0(1+\cos \eta )/2$, the proper-time horizon crossing in homogeneous dust collapse occurs atThe singularity is reached at ${\tau }_s=\pi a_{\max }/2$, so the remaining clock time isThese are finite proper times, even though the external static time diverges at horizon crossing.

Convert the supplied rounded units consistently: $M=10^{38}$ Planck masses and $R_0=6.25\times 10^{39}$ Planck lengths. Thus $2M/R_0=0.032$, ${\chi }_0\simeq 0.17985$, and $a_{\max }\simeq 3.4939\times 10^{40}$ in Planck units. Multiplying geometric times by $5\times 10^{-44}$ seconds givesAt the accuracy justified by the rounded constants, these are about three milliseconds to horizon crossing and about seven microseconds more to the singularity. For $R_0\gg 2M$, expanding ${\eta }_H=\pi -2{\chi }_0$ gives $\Delta \tau \simeq (2/3)a_{\max }{\chi }_0^3\simeq 4M/3$, independently confirming the microsecond scale. The exact expression above retains the finite release-radius correction.

Strictly, this boundary clock reading is when the observer crosses the horizon, not when the horizon first exists anywhere. The event horizon inside an Oppenheimer-Snyder cloud is an outgoing radial null ray. Since $d\tau =ad\eta$, the interior metric is conformal to $-d{\eta }^2+d{\chi }^2+{\sin }^2\chi d{\Omega }^2$, and that ray has $d\chi /d\eta =1$. Tracing it back from $({\eta }_H,{\chi }_0)$ givesFor this cloud it starts at the centre after release, at comoving time ${\tau }_{\mathrm{birth}}\simeq 2.722\times 10^{-3}$ seconds. This answers the literal global-formation interpretation as well; it is not an event on the outer observer's worldline. The distinction matters because an event horizon is defined by the entire future causal escape problem, not by a locally detectable signal of formation.

For the observer at infinity, use the boundary's conserved Killing energy $E$ in the exterior:As $\tau \uparrow {\tau }_H$, $\dot{R}\to -E$ and $R-2M\simeq E({\tau }_H-\tau )$. Consequentlywhere ${\tau }_{\ast }>0$ fixes a harmless dimensionless logarithm. For actual outgoing light signals the relevant retarded time is $u=t-r_{\ast }(R)$. Since $r_{\ast }\simeq 2M\log (R-2M)+O(1)$,Therefore she never receives a signal showing the surface cross the horizon at any finite asymptotic time. Radiation emitted ever closer to crossing arrives ever later and is increasingly redshifted and diluted; photons emitted at or inside the event horizon cannot reach her. The formal crossing time at infinity is infinite, not the finite boundary stopwatch reading. Nor can the earlier global birth of the horizon be observed directly as a local flash from its central birth event.

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 $G=c=1$. For a stationary asymptotically flat spacetime, normalize the timelike Killing vector field $\xi$ to have squared norm $-1$ at infinity. The Komar mass isHere the oriented binormal is chosen so that an ordinary positive-mass static solution gives a positive integral. Equivalently, on a static slice with lapse $N=\sqrt{-{\xi }^2}$ and outward spatial unit normal $s^i$,This second expression fixes the orientation unambiguously and is convenient for the Majumdar–Papapetrou solution.

The spatial metric is ${\gamma }_{ij}=H^2{\delta }_{ij}$ and $N=H^{-1}$. On a large coordinate sphere, $s^i=H^{-1}{\hat{r}}^i$ and $dA=H^2{r}^{2}d\Omega$. ThusThe original PDF has the Euclidean norm in each source denominator, which is missing in the TeX aid. With that norm, the harmonic function has asymptoticsThereforeIf $M_i$ are retained as length parameters while $G$ is restored, the physical mass is ${\sum }_i{M}_i/G$. This is the charge at infinity. A Komar integral is not generally surface-independent here, since Einstein-Maxwell theory has the electromagnetic stress-energy tensor between the surfaces; in particular it should not be confused with a sum of vacuum horizon Komar charges.

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 $\kappa >0$.

Use geometrized units $G=c=1$ and metric signature $(-+++)$; $\tau$ is proper time. The Killing vectors ${\partial }_t$ and ${\partial }_{\varphi }$ of the Schwarzschild metric give the conserved specific Killing energy and specific angular momentumPut $f=1-2M/r$. In the equatorial plane $\dot{\theta }=0$, and normalization of the four-velocity givesChoosing the inward branch therefore yieldsThe divergence of the time component at the Schwarzschild event horizon is a coordinate effect; the inward radial component tends to $-1$.

Capture from infinity. To reach the event horizon from infinity, the radial square must remain nonnegative throughout $r>2M$. Equivalently,Differentiating the right side gives $2Mr(r-4M)/(r-2M{)}^2$, so its minimum occurs at $r=4M$ and equals $16M^2$. Thusis the necessary capture bound. At equality the radial numerator is $2M(r-4M{)}^2$. An inward particle arriving from larger radii approaches the unstable orbit $r=4M$ only after infinite proper time, because $\dot{r}$ is proportional to $-(r-4M)$ near that orbit. Actual plunges from infinity require $|h|<4M$. This is Schwarzschild marginally bound capture.