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For the nonrotating BTZ black hole family with $M<0$ and angular period $2\pi$, the circumference of a small circle divided by its proper radius tends to $2\pi \sqrt{-M}$. A smooth axis requires this ratio to be $2\pi$, so only $M=-1$ gives regular Anti-de Sitter spacetime. Other values have a conical singularity, with deficit angle $\delta$ or an angular excess when $\delta <0$. The lapse $r^2/{\ell }^2-M$ is strictly positive, so outward radial null geodesics escape from arbitrarily near the axis: the defect is a naked singularity, despite finite local curvature away from it.

A radial infalling geodesic in Schwarzschild spacetime has $dt/ds=E/f$ and $dr/ds=-\sqrt{E^2-f}$, where $f=1-2M/r$. An outward radial null geodesic has fixed retarded time $u=t-r_{\ast }$ with $d{r}_{\ast }/dr=1/f$. Consequently $du/ds=(E+\sqrt{E^2-f})/f$. This factor stretches a short emitted pulse's duration, and its reciprocal multiplies the received frequency at infinity. Near horizon crossing $s_H$, it becomes $du/ds\sim 4M/(s_H-s)$, including both gravitational redshift and the emitter's Doppler shift. The local short-pulse approximation requires its emitted duration to be much smaller than $s_H-s$.

// Target: general-relativity.bigb

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.

Write $M=-a^2$ with $a>0$. On a constant-time slice, the proper distance from the axis to radius $r$ and the circumference at that radius areHence $C/s\to 2\pi a$. A smooth axis in a Riemannian metric requires the Euclidean limit $C/s\to 2\pi$, so regularity forces $a=1$, or $M=-1$. For $a\ne 1$ the conical singularity of negative-mass BTZ geometry has deficit angle $2\pi (1-a)$; for $a>1$ this is an angular excess. Even integer $a>1$ is singular with the stipulated angular period $2\pi$: a multiple covering does not make the tip a smooth point.

At $M=-1$, set $r=\ell \sinh \chi$. The metric tensor becomeswhich is regular at $\chi =0$ and describes Anti-de Sitter spacetime with nonperiodic time. For every $M<0$, $f=a^2+r^2/{\ell }^2$ stays positive. An outward radial null geodesic obeys $dt/dr=1/f$, and reaches the timelike conformal boundary in finite coordinate time, since ${\int }_0^{\infty }dr/f=\pi \ell /(2a)$. Thus no event horizon hides the conical axis. The geometry is nakedly singular except at $M=-1$. Its local curvature away from the axis is that of Anti-de Sitter spacetime; the argument is a global angular-identification test, not curvature blowup.

A future trapped surface is a smooth compact spacelike two-surface without boundary whose two future-directed orthogonal null expansions are strictly negative. In four-dimensional general relativity, the Penrose singularity theorem states: a time-oriented globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a trapped surface, and the null convergence condition $R_{ab}{k}^a{k}^b\ge 0$ for every null vector is future null-geodesically incomplete. With the Einstein field equations, the null energy condition implies this null convergence condition; a cosmological constant drops out of the null contraction.

The Kruskal spacetime is an example. In its black hole interior, use future null normals in ingoing Eddington-Finkelstein coordinates:For a round sphere of areal radius $r$, its area is $4\pi r^2$ and its null expansions areBoth are negative for $0<r<2M$. The vacuum Einstein field equations give $R_{ab}=0$, and a two-ended Kruskal Cauchy hypersurface is noncompact. The future radial null geodesics reaching $r=0$ in finite affine parameter provide precisely the incompleteness predicted by the Penrose singularity theorem. A trapped surface at $r<2M$ is strictly trapped; the horizon sphere has one zero null expansion instead.

A radial null geodesic in a spherically symmetric spacetime has zero angular momentum and fixed angular coordinates. Its projection is a null curve of the two-dimensional radial Lorentzian metric.

In the exterior of Schwarzschild spacetime, $d{r}_{\ast }/dr=(1-2M/r{)}^{-1}$. Outgoing radial null geodesics therefore have constant $u=t-r_{\ast }$. At fixed receiver radius $R$, their arrival Schwarzschild time is $u+r_{\ast }(R)$, so differences in $u$ give arrival-time differences. A static receiver's proper time includes the factor $\sqrt{1-2M/R}$.

// Target: optics.bigb