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The advanced time $v=t+r_{\ast }$ or retarded time $u=t-r_{\ast }$ combines Schwarzschild time with the Schwarzschild tortoise coordinate to remove one horizon-coordinate singularity. The advanced choice gives ingoing Eddington-Finkelstein coordinates; the retarded choice is outgoing.

A Finkelstein diagram plots areal radius $r$ against $T=v-r$ in ingoing Eddington-Finkelstein coordinates. Its radial null directions show explicitly that both future light-ray families decrease $r$ inside a Schwarzschild black hole. It is not a compactified Penrose diagram.

With $f=1-2\mu /r$, the time coordinate $t^{\prime }=t+(2\mu /c)\log |r/(2\mu )-1|$ gives $d{s}^2=f{c}^{2}d{t}^{\prime 2}-(4\mu c/r)d{t}^{\prime }dr-(1+2\mu /r)d{r}^2-r^{2}d{\Omega }^2$. Its radial metric tensor block has determinant $-c^2$, so it extends regularly across the future Schwarzschild event horizon. The advanced null time of ingoing Eddington-Finkelstein coordinates is $v=t^{\prime }+r/c$, not $t^{\prime }$ itself. Radial null directions satisfy $dr/(cd{t}^{\prime })=-1$ or $(1-2\mu /r)/(1+2\mu /r)$.

In the ingoing Eddington-Finkelstein coordinates metric $d{s}^2=-f(r)d{v}^2+2dvdr+r^{2}d{\Omega }^2$, where $f=1-2M/r$, the tangent $U=-E{\partial }_r$ is affine when $E>0$ is constant along each ray. The displayed radial partner obeys $N^2=0$ and $U\cdot N=-1$. Since ${\Gamma }^v{}_{rv}={\Gamma }^a{}_{rr}=0$ and ${\Gamma }^r{}_{rv}=-f^{\prime }/2$, its derivative ${\partial }_r{N}^r=f^{\prime }/(2E)$ cancels the connection term, giving ${\nabla }_{U}N=0$. Its screen is the angular two-plane, with optical tensor ${\hat{B}}_{AB}=-Er{\gamma }_{AB}$, null expansion $-2E/r$, and zero null shear and null twist.

A horizon separates regions according to a specified causal, dynamical or geometrical criterion. The criterion must be stated: an event horizon, an apparent horizon, a Killing horizon and a Cauchy horizon answer different questions. Some horizons describe a black hole, others an observer's acceleration or the universe's expansion. A vanishing coordinate metric component alone does not establish any of these definitions.

In an asymptotically flat spacetime, choose the relevant future null infinity ${\mathcal{I}}^{+}$, the endpoint of escaping outgoing light rays. The black-hole region and its future event horizon areHere $J^{-}$ is the causal past. Events in $\mathcal{B}$ cannot send a causal signal to the chosen infinity. This is a global definition: identifying the horizon requires the full future development, so a local measurement need not reveal whether an event lies on it. Smooth portions of an event horizon are null hypersurfaces generated by null geodesics. Generators may join at past caustics or crease sets, so the horizon need not be smooth everywhere. A past event horizon is defined by the time-reversed escape criterion; in the maximally extended Schwarzschild solution it bounds a white-hole region.

An observer-based definition replaces infinity by a future-inextendible worldline $\gamma$. Its future observer event horizon is the boundary of events that can signal to $\gamma$, conventionally $\partial I^{-}(\gamma )$. A past observer horizon similarly bounds events reachable from its past worldline. Such horizons depend on the observer and need not imply a black hole.

A Killing horizon is a null hypersurface to which a nonzero Killing vector field $K^a$ is normal. On it, $K^a$ is tangent to its null generators and satisfies $K^b{\nabla }_b{K}^a=\kappa K^a$. The surface gravity $\kappa$ measures the failure of this Killing parameter to be an affine parameter; rescaling $K$ rescales $\kappa$. A degenerate Killing horizon has $\kappa =0$. Stationary black-hole horizons often are Killing horizons, whereas a general evolving event horizon need not have any Killing symmetry. The condition $K^a{K}_a=0$ alone is insufficient: the null vector must also be normal to a null hypersurface.

The Schwarzschild spacetime illustrates several of these notions. In units $G=c=1$, let $f=1-2M/r$. Its metric is $d{s}^2=fd{t}^2-f^{-1}d{r}^2-r^{2}d{\Omega }^2$, and $r=2M$ is the future event horizon of the black-hole region in the appropriate extension. The apparent coordinate singularity disappears in ingoing Eddington-Finkelstein coordinates, $v=t+r_{\ast }$, $d{r}_{\ast }/dr=f^{-1}$:At $r=2M$ this metric is regular and the normal $dr$ is null. Future outgoing radial rays satisfy $dr/dv=f/2$; inside the horizon their areal radius decreases even on the outgoing family. The stationary Killing vector field ${\partial }_v$ generates the horizon with $\kappa =f^{\prime }(2M)/2=1/(4M)$ when normalized at infinity. The Kretschmann scalar is $48M^2/r^6$, finite at the horizon and divergent at $r=0$. Thus the event horizon is not the curvature singularity. The photon sphere at $r=3M$, which supports circular null orbits, is also not the horizon.

Quasi-local horizon notions use null expansions. For a closed spacelike two-surface, choose future outgoing and ingoing null normals $\ell ,n$, with $\ell \cdot n=1$ in signature $(+---)$. Each expansion is the logarithmic rate of its positive area element along the corresponding null normal. A future trapped surface has ${\theta }_{\ell }<0$ and ${\theta }_n<0$: both future light fronts initially shrink. A marginally outer trapped surface has ${\theta }_{\ell }=0$; for the future black-hole type also require ${\theta }_n<0$. On a chosen spacelike slice, the outer boundary of the future trapped region is an apparent horizon, ordinarily an outermost marginally outer trapped surface when regularity permits this characterization. It depends on the slice, and a slicing need not display an apparent horizon even when the global spacetime contains an event horizon.

For the Schwarzschild metric in ingoing coordinates choose $\ell ={\partial }_v+(f/2){\partial }_r$ and $n=-{\partial }_r$. The spherical null expansion in ingoing coordinates givesThe round spheres inside $2M$ are future trapped, while $r=2M$ is marginal. On standard horizon-penetrating slices its sections give the apparent horizon, coinciding with the stationary event and Killing horizon. This coincidence in a symmetric stationary example does not identify the notions in a general spacetime.

A trapping horizon is a hypersurface foliated by marginal surfaces. Its future outer type obeys ${\theta }_{\ell }=0$, ${\theta }_n<0$ and ${\mathcal{L}}_n{\theta }_{\ell }<0$. Unlike an event horizon, this can be diagnosed from local expansion data and their variation. The null energy condition explains its causal character. At a marginal surface, the Null Raychaudhuri equation, with the present curvature sign and vanishing null twist, givesin units $c=1$. A possible nonaffinity term vanishes here because it is proportional to the zero expansion. Write a horizon evolution vector as $V=\ell -Cn$. Preserving ${\theta }_{\ell }=0$ gives $C=({\mathcal{L}}_{\ell }{\theta }_{\ell })/({\mathcal{L}}_n{\theta }_{\ell })\ge 0$. Hence $V^2=-2C\le 0$: the future outer horizon is spacelike or null. Its area expansion is ${\theta }_V=-C{\theta }_n\ge 0$. These conclusions use the outer condition and the energy condition, not merely marginality.

For the global event horizon, the future-complete horizon focusing proof of the area theorem uses the same Null Raychaudhuri equation. Under the null energy condition, a negative generator expansion forces a future caustic in finite affine parameter. Future completeness and the usual regularity or predictability assumptions exclude such a caustic on the future event-horizon boundary. Its generator expansion is therefore nonnegative and its classical area cannot decrease. These assumptions matter: this is not an unconditional result for every null surface or every matter model.

Rotating and charged solutions provide further distinctions. A nonextremal Kerr black hole has $r_{\pm }=M\pm \sqrt{M^2-a^2}$, with $a=J/M$. Its outer Killing horizon is generated by ${\partial }_t+{\Omega }_H{\partial }_{\varphi }$, where ${\Omega }_H=a/(r_{+}^2+a^2)$, rather than by time translation alone. The ergosphere begins at the stationary-limit surface $g_{tt}=0$, whose radius is $M+\sqrt{M^2-a^2{\cos }^2\theta }$. Away from the axis this lies outside the event horizon: it forbids static observers but does not forbid escape to infinity. In Reissner-Nordstrom spacetime, the nonextremal radii are $r_{\pm }=M\pm \sqrt{M^2-Q^2}$. The outer surface is an event horizon, whereas the inner surface is a Cauchy horizon in the exact analytic extension.

A Cauchy horizon concerns predictability from initial data. The future domain of dependence $D^{+}(\Sigma )$ of a partial Cauchy surface consists of events for which every past-inextendible causal curve intersects $\Sigma$. Its future Cauchy horizon isBeyond it, causal influences can arrive without meeting the specified initial surface; data on that surface no longer determine the solution there. This is distinct from the inability to send signals to infinity. The inner horizon of the exact charged black hole supplies an example, and the definition necessarily specifies which initial surface is being used.

Flat spacetime itself has acceleration horizons. In the right Rindler wedge, set $t=\xi \sinh (a\tau )$ and $x=\xi \cosh (a\tau )$, with $\xi >0$. The Minkowski spacetime metric becomes $d{s}^2=a^2{\xi }^{2}d{\tau }^2-d{\xi }^2-d{y}^2-d{z}^2$. The null boundaries $t=x$ and $t=-x$ are Rindler horizons of the boost Killing flow. A uniformly accelerated observer cannot receive signals from one side of its future horizon, even though the curvature vanishes and inertial observers encounter no such boundary. These are observer and Killing horizons, not a black-hole event horizon defined using the full Minkowski null infinity.

In a spatially flat Friedmann-Lemaître-Robertson-Walker metric with noncompact spatial sections, radial light rays have $d\chi /dt=\pm 1/a(t)$. The particle horizon measures how far signals could have travelled since the initial cosmic time; the cosmological event horizon measures how far signals emitted now could ever travel toward a comoving observer. Their comoving and proper radii areThe respective horizon is present only if its integral is finite. The event-horizon integral uses the actual future endpoint, often $t_f=\infty$, so it is global and observer-dependent. For $a(t)=t^q$ on $0<t<\infty$, a finite particle horizon exists for $0<q<1$, while a finite event horizon exists for $q>1$. Neither integral is finite for $q=1$. In the expanding flat patch of de Sitter spacetime, $a=e^{Ht}$ and the proper event-horizon radius is $H^{-1}$. The static form $d{s}^2=(1-H^2{r}^2)d{t}^2-(1-H^2{r}^2{)}^{-1}d{r}^2-r^{2}d{\Omega }^2$ identifies it as a cosmological Killing horizon; its surface-gravity magnitude is $H$ with the central static time normalization.

Finally, the Hubble radius must not automatically be called an event horizon. For a flat expanding cosmology, the two future radial null expansions of a sphere of proper radius $R$ are ${\theta }_{\pm }=2(H\pm 1/R)$. At $R=H^{-1}$ the ingoing expansion vanishes, while outside it both expansions are positive: this is the cosmological marginal, past-trapped type rather than a future black-hole trapped surface. A dust universe with $a\propto t^{2/3}$ has such a Hubble sphere but no future event horizon. Global causal boundaries, slice-dependent trapped-surface boundaries, symmetry horizons and initial-data boundaries must therefore be distinguished, even when several coincide in a special example.

Take $M=|Q|>0$ and use $G=c=1$. The case $M=Q=0$ is flat spacetime and is not a black hole. In the nontrivial extremal Reissner-Nordstrom metric,The horizon is a double zero at $r=M$, with surface gravity $\kappa =f^{\prime }(M)/2=0$. The function $f$ is positive on both sides of the horizon. Thus the intermediate nonstatic region of a nonextremal charged hole disappears; one must not draw that region as a strip of nonzero width in the extremal Penrose diagram.

To construct the Penrose diagram, first integrate the radial tortoise coordinate:The additive constant has been chosen so that $r_{\ast }(0)=0$ in the inner block. In the exterior, $r_{\ast }\to -\infty$ at $M^{+}$ and $r_{\ast }\to +\infty$ at infinity. In the inner static region $0<r<M$, it increases from $0$ to $+\infty$. Set $u=t-r_{\ast }$ and $v=t+r_{\ast }$. The radial metric is $-fdudv$, so its radial null directions are $u=\mathrm{constant}$ and $v=\mathrm{constant}$.

Compactify each static block usingThis conformal compactification puts radial null geodesics at $45$ degrees. The exterior is a diamond: its right-hand edges are past and future null infinity, while its left-hand edges are the past and future degenerate horizons. At the left vertex, finite static $t$ and $r\to M^{+}$ give $u\to +\infty$, $v\to -\infty$. This is an end of the infinite spatial throat, not a regular bifurcation surface where the two horizon branches meet. A degenerate Killing horizon has no such bifurcation sphere. The inner block occupies the half-diamond $X>0$, because $r_{\ast }>0$ there. Its vertical edge $X=0$ is the timelike curvature singularity $r=0$; its two diagonal edges are the two horizon branches at $r=M$. Its outer vertex likewise represents an excluded infinite-throat end.

The static-coordinate compactification identifies the blocks but does not by itself establish the smooth horizon extension. For that, use ingoing Eddington-Finkelstein coordinates on the future horizon:Its radial determinant is $-1$, so it is nonsingular at $r=M$. Glue the exterior future edge, $u\to +\infty$ with $v$ finite, to the inner past edge, $u\to -\infty$ with the same finite $v$. The outgoing coordinate $u$ similarly gives $d{s}^2=-fd{u}^2-2dudr+r^{2}d{\Omega }^2$ and extends the other horizon branch. Continue adjoining these regular blocks across every extendible null edge. The construction repeats indefinitely to the future and past, producing the maximal analytic diagram with infinitely many asymptotic exterior regions and inner static regions bounded by timelike singularities. Horizons are null edges, never the excluded throat vertices. Identical edges in different copies must be glued with their time orientations preserved; they are not all the same exterior infinity.

The figure shows the two building blocks, rather than compressing the whole infinite extension into a misleading finite diagram. The red edges glue across a future horizon at finite advanced time. Outgoing coordinates extend the next edge to another exterior copy. The thin diagonal lines illustrate the radial light-cone directions.

For spatially conformally flat coordinates, compare the radial and angular parts of $f^{-1}d{r}^2+r^{2}d{\Omega }^2$ with $V(\rho {)}^2(d{\rho }^2+{\rho }^{2}d{\Omega }^2)$. The angular coefficient requires $V\rho =r$, and the radial coefficient then gives, in the exterior,Choose the integration constant to make $\rho =r-M$ and $V\to 1$ at infinity. In Cartesian coordinates $x=\rho \sin \theta \cos \phi$, $y=\rho \sin \theta \sin \phi$, $z=\rho \cos \theta$, put $\rho =\sqrt{x^2+y^2+z^2}$. The Extremal Reissner-Nordstrom isotropic coordinates giveThese coordinates have $\rho >0$ and describe the exterior only. The locus $\rho =0$ is the limit $r=M$, not an ordinary point included in this chart. An independent inner chart is obtained with $\rho =M-r$, $0<\rho <M$, giving $V=M/\rho -1$ and $W=(M/\rho -1{)}^{-1}$; it must be attached using the regular null coordinates, not by continuing the exterior radial coordinate through a Cartesian origin.

The infinite proper distance is real. Along a constant-$t$ exterior radius,It is a property of a static spacelike slice. It does not bound the proper time of a freely falling observer. Indeed, a neutral radial timelike geodesic has conserved energy per unit rest mass $E=f\dot{t}>0$ and normalization $-f{\dot{t}}^2+f^{-1}{\dot{r}}^2=-1$, givingAt the horizon $\dot{r}\to -E$, so the remaining proper time is finite. Moreover,The observer crosses the regular future horizon with finite $v$ and finite proper time. This is the Extremal Reissner-Nordstrom causal crossing, which the singular static time coordinate conceals.

There is also a direct causal test for the event horizon. Inside the future horizon, outgoing radial null curves in the ingoing chart satisfy $dr/dv=f/2$. To reach $r=M$ from below they would requireNear the horizon, $M-r\sim 2M^2/v$. A future timelike or nonradial null curve cannot increase $r$ faster than the outgoing radial null curve, since $d{s}^2\le 0$ and $\dot{v}>0$ imply $dr/dv\le f/2$ after dropping the nonnegative angular term. Hence a signal that crossed inward cannot return to the original exterior's future null infinity. Curves can encounter other blocks in the maximal analytic extension, but their infinities are not the original observer's infinity. Relative to that asymptotic end, $r=M$ is the boundary of the causal past of future null infinity: it is an event horizon.

Finally, the Reissner-Nordstrom trapped spheres test explains why the proposed observation is unsurprising. Normalize future radial null normals by $\ell ={\partial }_v+(f/2){\partial }_r$, $n=-{\partial }_r$, with $g(\ell ,n)=-1$. The round-sphere null expansions areThere is no region of strictly future-trapped round spheres, while the horizon spheres are marginal with ${\theta }_{\ell }=0$. Absence of those particular trapped surfaces does not imply absence of an event horizon. The geometry has a nonzero horizon area $4\pi M^2$ and a genuine timelike singularity at $r=0$ in its extension. The infinite static throat does not remove the black hole: infall crosses its horizon in finite proper time, and no signal returns to the same exterior infinity.

Take $M>0$ and write $f=1-2M/r$. Integrating $d{r}_{\ast }/dr=f^{-1}$ gives the Schwarzschild tortoise coordinateThe advanced and retarded Eddington-Finkelstein coordinates are respectively $v=t+r_{\ast }$ and $u=t-r_{\ast }$. Substituting $dt=dv-dr/f$ or $dt=du+dr/f$ givesBoth radial-time blocks have determinant $-1$ and smooth coefficients at $r=2M$. Consequently the Schwarzschild event horizon is a coordinate singularity of the original chart, not a degeneracy of the geometry. The Kretschmann scalar $48M^2/r^6$ is finite there, while it diverges at $r=0$. Angular coordinate degeneracies at the polar axes are removed separately by ordinary sphere charts.

The ingoing Eddington-Finkelstein coordinates extend the right exterior through its future black hole horizon. Radial null geodesics obey either $dv=0$ or $dr/dv=f/2$. Inside $r=2M$, the latter also moves towards smaller $r$ as $v$ increases, and the future-directed ingoing family has decreasing $r$. Thus neither family can escape to the exterior. The future horizon is the boundary separating points that can send signals to future null infinity from those that cannot. The retarded chart extends through the past horizon instead. Reversing the time orientation gives the white hole: future-directed signals can leave it, but exterior signals cannot enter it. These are distinct branches in the eternal extension; the white-hole branch need not occur in a spacetime formed by collapse.

For a simultaneous extension through both branches, start in the right exterior and define the Kruskal–Szekeres coordinatesTheir product and the extended metric areIndeed $du=-4MdU/U$ and $dv=4MdV/V$ transform $-fdudv$ into the displayed expression. The function of $r$ on the right of the product relation has derivative $-r{e}^{r/(2M)}/(4M^2)$, nonzero at $r=2M$, so $r$ is a smooth function of $UV$ across zero. The coefficient of $dUdV$ there is $-16M^2/e$, finite and nonzero. Allowing either sign of $U,V$ therefore gives the regular maximal extension with $UV<1$ and $r>0$.

With $T_K=(V+U)/2$ and $X_K=(V-U)/2$, the radial metric is a positive multiple of $-d{T}_K^2+d{X}_K^2$. The two exterior regions have $(U,V)=(-,+)$ and $(+,-)$. The future interior, or black hole region, has $(+,+)$; the past interior, or white hole region, has $(-,-)$. The Killing horizons are $U=0$ and $V=0$, meeting on the bifurcation surface at $U=V=0$.

For the Penrose diagram, compactify the null variables before forming time and space:This Schwarzschild conformal compactification preserves radial null directions because $dUdV={\sec }^2\bar{U}{\sec }^2\bar{V}(d{T}^2-d{X}^2)$. The ranges $|\bar{U}|,|\bar{V}|<\pi /2$ bound the diagram, and $UV<1$ cuts off its future and past tips. On $UV=1$ with $U,V>0$, $\bar{U}+\bar{V}=\pi /2$, so the future curvature singularity is $T=\pi /4$. The past curvature singularity is similarly $T=-\pi /4$. Both have $|X|<\pi /4$ and are spacelike. The horizon lines remain $T=\pm X$.

Each point of this radial Penrose diagram represents a two-sphere, except at limiting boundaries. Each exterior has its own future and past null infinity, ${\mathcal{I}}^{+}$ and ${\mathcal{I}}^{-}$. For example in the right exterior, $V\to \infty$ at fixed $U<0$ gives ${\mathcal{I}}^{+}$, whereas $U\to -\infty$ at fixed $V>0$ gives ${\mathcal{I}}^{-}$. Its spacelike infinity is $(T,X)=(0,\pi /2)$; its future and past timelike infinity are the vertices $(\pi /4,\pi /4)$ and $(-\pi /4,\pi /4)$. The left exterior has the reflected counterparts. These conformal boundary points are ideal endpoints, not finite-distance spacetime events; the physical affine or proper parameters of escaping geodesics diverge.

The horizons are regular null surfaces, while the horizontal $r=0$ boundaries are genuine curvature singularities. The central point represents the regular bifurcation surface, not a singularity. Future-directed radial causal curves have nondecreasing $U,V$, so a curve cannot go from one exterior to the other: changing the requisite signs would require one null variable to decrease. Crossing a future horizon instead leads towards the future singularity. The maximal extension has two causally disconnected exteriors, a future black-hole region and a past white-hole region.

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.

Let $r_h=\sqrt{2m}$ and $f(r)=1-r_h^2/r^2$. The geometry is the five-dimensional Schwarzschild-Tangherlini metric. Introduce a tortoise coordinate by $d{r}_{\ast }/dr=f^{-1}$. Decomposing the integrand givesAn irrelevant additive constant has been fixed. Set $v=t+r_{\ast }$ and $u=t-r_{\ast }$. Substituting $dt=dv-dr/f$ or $dt=du+dr/f$ respectively gives the ingoing Eddington-Finkelstein coordinates and outgoing Eddington-Finkelstein coordinates:The radial-time block in each has determinant $-1$, so the apparent singularity at $f=0$ has disappeared. The ingoing chart crosses the future event horizon, and the outgoing chart crosses the past horizon. Radial null geodesics in the ingoing chart obey either $v=\text{constant}$ or $dr/dv=f/2$, displaying how outward light is trapped when $r<r_h$.

For the maximal extension, the exterior radial metric is $-fdudv$. Since the surface gravity is $\kappa =f^{\prime }(r_h)/2=1/r_h$, choosein the right exterior. Their product and the extended metric areThe coefficient at the horizon is finite and nonzero, $-4r_h^2{e}^{-2}$. The product relation therefore extends smoothly through either horizon. For $r>r_h$, $UV<0$ gives two exterior regions. For $0<r<r_h$, $0<UV<1$ gives the future black-hole interior with $U,V>0$ and the past white-hole interior with $U,V<0$. The Kruskal extension of the five-dimensional Schwarzschild-Tangherlini metric has a bifurcation three-sphere at $U=V=0$.

The singularity $r=0$ corresponds to $UV=1$. It is a curvature singularity: for example, its Kretschmann scalar is $72r_h^4/r^8$. Since $g^{ab}{\partial }_{a}r{\partial }_{b}r=f<0$ in the interior, constant-$r$ surfaces are spacelike there, and the singularities are spacelike boundaries. Define $P=\arctan U$, $Q=\arctan V$, then compactified coordinates $T=(Q+P)/2$, $X=(Q-P)/2$. Radial null curves have $T\pm X=\text{constant}$. Moreover, $\tan P\tan Q=1$ gives $T=\pm \pi /4$ at the two singularities. This constructs the Penrose diagram, not just its horizon labels:

Each point away from the singular boundaries represents a three-sphere, not a two-sphere. The diagram has two asymptotically flat exteriors with separate past and future null infinities, a future black hole and a past white hole. The diagram of a collapse spacetime would omit the second exterior and the white-hole region; the maximal extension of the metric given here contains both.

For the Komar integral, normalize the stationary Killing vector field as $k={\partial }_t$, with unit norm at infinity. The Killing equation makes ${\nabla }^a{k}^b$ antisymmetric, and its divergence is proportional to $R^b{}_c{k}^c$. In a vacuum exterior this vanishes, so Stokes theorem makes the flux independent of the enclosing sphere. In the weak-field limit its radial component measures the derivative of the gravitational potential; its large-sphere flux is therefore proportional to the asymptotic gravitational mass. Normalization of $k$ is needed because rescaling it would rescale the integral.

Specify the surface orientation to fix the sign and factor. On a static exterior slice take future unit normal $u^a=f^{-1/2}({\partial }_t{)}^a$, outward radial normal $n^a=f^{1/2}({\partial }_r{)}^a$ andThen $d{\Sigma }^{tr}=-dA$ and $d{\Sigma }^{rt}=dA$, with both antisymmetric components included. Since $k_t=-f$,The three-sphere area is $2{\pi }^2{r}^3$, and $f^{\prime }=2r_h^2/r^3=4m/r^3$. ThusReversing the binormal orientation gives $+8{\pi }^{2}m$; a convention with half the antisymmetric binormal gives half this magnitude. The question does not fix those conventions, so the choice above is explicit.

For comparison with physical mass, the Komar mass normalization in higher dimensions isIn five dimensions this givesThe metric parameter $m$ has dimensions of length squared and is not itself the physical mass. An independent horizon check is $I_K=-2\kappa A_H$ with $A_H=2{\pi }^2{r}_h^3$, which reproduces the same integral. The dimension-adjusted normalization also agrees with the ADM mass; the four-dimensional Komar prefactor should not be reused unchanged.

Write $f(r)=1-2M/r$. Radial null paths satisfy $dt/dr=\pm f^{-1}$, whose integral uses the Schwarzschild tortoise coordinate $r_{\ast }=r+2M\log |r/(2M)-1|$. The incoming and outgoing families are thereforewith constants $v,u$. For an exterior outgoing ray, put $r/(2M)-1=\epsilon$. As $t\to -\infty$, $t=u+2M(1+\epsilon )+2M\log \epsilon$, so $\epsilon \sim C{e}^{t/(2M)}$ for a positive ray-dependent $C$. As $t\to \infty$, the logarithm is smaller than the linear term, giving $r\sim t$.