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An analytic extension embeds a spacetime isometrically as a proper open region in a larger analytic Lorentzian manifold with a nondegenerate metric tensor. A divergence of coordinate components does not obstruct extension if a new chart makes all metric coefficients analytic and the determinant nonzero. For a static spherical metric with an analytic lapse $f$ and a positive simple horizon radius, ingoing coordinates replace $-fd{t}^2+d{r}^2/f$ by $-fd{v}^2+2dvdr$. This removes the coordinate singularity at $f=0$; transverse metric coefficients must also remain regular. An extension across such a horizon is distinct from extending through a curvature singularity.

A coordinate singularity is a failure of a coordinate description which disappears in a regular chart. The horizon singularity of the Schwarzschild metric in $(t,r)$ coordinates is removable; its $r=0$ curvature singularity is not.

For $0<|Q|<M$, a Reissner-Nordstrom spacetime has an outer event horizon and an inner Cauchy horizon. Between them $r$ is timelike; inside the inner horizon $r$ is spacelike and the $r=0$ curvature singularity is timelike. The exact analytic extension contains repeated exterior blocks beyond the Cauchy horizons. Small perturbations produce severe inner-horizon blueshift and mass inflation, so that exact continuation is not a robust prediction for collapse. At $|Q|=M$ the horizons merge into a degenerate Killing horizon; for $|Q|>M$ the spherical electrovacuum solution instead has a naked singularity.

The Kretschmann scalar is the scalar tensor contraction of two copies of the Riemann curvature tensor. In Schwarzschild spacetime, with metric coefficient $1-2M/r$ and $G=c=1$, it is $48M^2/r^6$. Divergence of this scalar along a curve approaching a proposed regular point proves that there is no twice continuously differentiable, nondegenerate metric extension through that point: such an extension would have continuous finite curvature components and hence a finite scalar contraction. This gives a coordinate-independent test of a curvature singularity. Boundedness of this single scalar does not prove regularity.

For $V=1-2M/r^2+Q^2/r^4$, an orthonormal-frame curvature calculation gives $(V^{\prime \prime }{)}^2+6(V^{\prime }/r{)}^2+12((1-V)/r^2{)}^2$. Expansion gives the displayed invariant, diverging at zero radius. It is finite at every positive root of V. Thus horizon coordinate singularities are distinct from the central curvature singularity.

The Kruskal spacetime is the maximal analytic extension of the positive-mass vacuum Schwarzschild spacetime. It contains two exterior regions, a black hole, and a white hole. Kruskal–Szekeres coordinates remove both horizon coordinate singularities, while $r=0$ remains a curvature singularity.

A maximal analytic extension of an analytic spacetime is an analytic spacetime extension admitting no further proper analytic extension as a regular spacetime of the same dimension. It is different from a maximal Cauchy development, which can end at a smooth Cauchy horizon. Additional global choices or quotients can affect uniqueness. The simply connected extension of positive-mass Schwarzschild spacetime through both horizons is the Kruskal spacetime; its curvature singularities remain boundaries.

For $M>0$ and $Q^2>2M^2$, the physical exterior starts at $r=b>2M$. Throughout it, $f>0$, so the putative zero of $f$ is not in the exterior and no regular horizon shields the curvature singularity. The normal to $r=b$ has positive limiting squared norm $g^{rr}=f(b)>0$, identifying a timelike singular boundary. Outgoing radial null geodesics obey $dr/dt=f(r)>0$ and escape from arbitrarily close to that boundary. HenceAt $Q^2=2M^2$, the singular boundary coincides with $f=0$ and is null. There is no regular shielding horizon of positive area; this marginal singular extremal case must be separated from both the regular black hole and the supercritical timelike naked singularity. If “naked” is used simply to mean unshielded by a regular horizon, equality belongs on that side of the distinction, but its causal boundary is different.

The usual positive-mass restriction matters. For $M<0$, $f(r)>0$ at every positive radius and there is no positive-radius horizon; the central zero-area boundary at $r=0$ is singular and naked. For $Q=0$ its angular sectional curvature is $2M/r^3$, which diverges there. For $Q\ne 0$, $C=r(r-b)$ with $b<0$, and the angular sectional curvature $[1-f(C^{\prime }{)}^2/(4C)]/C$ also diverges as $r\downarrow 0$. The displayed family is undefined at $M=0$ with nonzero $Q$; its uncharged zero-mass limit is flat.

Take $M>0$, set $b=Q^2/M$, and denote the squared areal radius by $C(r)=r(r-b)$. The positive-area exterior has $r>b$. The stationary norm is $-f(r)$, where $f(r)=1-2M/r$, so its candidate Killing horizon is at $r=2M$. There it has areaA regular horizon must have $b<2M$. The curvature singularity then lies strictly inside the horizon, and ingoing coordinates extend smoothly across it because $C(2M)>0$. In the future interior $f<0$, the gradient of $r$ is timelike and future causal curves move to decreasing $r$, rather than back to infinity. Thus the regular black hole condition isThe equality case has zero horizon area and a curvature-singular null limiting surface. It is often included in the family as a singular extremal solution, but it is not a regular black hole. If that limiting nomenclature is used, the family condition is $Q^2\le 2M^2$ with this essential qualification. This geometry has the same spherical form as the magnetically charged dilaton black hole; no unprinted field equation is needed for the horizon comparison.

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.

Let $f(r)=1-2M/r+Q^2/r^2$. A positive-radius horizon requires a root of $r^2-2Mr+Q^2=0$, givingFor the usual asymptotically flat Reissner-Nordstrom spacetime, the black-hole conditions areThe strict inequality gives a nonextremal outer event horizon at $r_{+}$; equality gives an extremal black hole. If $M<|Q|$ with $M>0$, there is no horizon and the central curvature singularity is naked. Nonpositive $M$ cannot supply a positive-radius black-hole horizon; $M=Q=0$ is Minkowski spacetime. For $Q=0$, the zero root $r_{-}=0$ is the singularity, not an additional regular horizon.

In the static coordinates the vector $k={\partial }_t$ has constant components and every metric component is independent of $t$. Its Lie derivative therefore satisfiesFor the torsion-free Levi-Civita connection, ${\mathcal{L}}_{k}g=2{\nabla }_{(a}{k}_{b)}$, proving the Killing equation directly. This also normalizes the Killing vector field to unit time translation at infinity.

At $r=r_{+}$, $k^2=-f=0$. Use the regular advanced chart $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=f^{-1}$, in whichOn the horizon, $k_a=(dr{)}_a$. Thus the Killing vector field is both null and normal to the horizon, and its orbits generate that Killing horizon. The normal lines of a null hypersurface are null geodesics, possibly with nonaffine parameter. To see this without assuming a separate geodesic property, the Killing equation givesRestriction to the future horizon therefore givesThe acceleration is proportional to the tangent, which is exactly the geodesic equation with a nonaffine parameter. An affinely parametrized tangent can be obtained by rescaling $k$ along each generator; that rescaled vector need not itself extend to a spacetime Killing vector field. At a nonextremal bifurcation surface, $k$ vanishes, so a nonzero generator tangent must be described in a regular affine parametrization there.

On the past branch, use the retarded chart with radial-time cross term $-2dudr$. Now $k_a=-(dr{)}_a$ at the horizon, so the same calculation yields the opposite signed nonaffinity. Since $r_{+}$ is constant and the spacetime is spherically symmetric, $f^{\prime }(r_{+})/2$ is constant over the future horizon. This gives the two signs in the horizon relation without assuming their value beforehand.

Factor $f=(r-r_{+})(r-r_{-})/r^2$. Its derivative at the outer horizon gives the Reissner-Nordstrom horizon surface gravityIt is positive for $M>|Q|$, and at $Q=0$ reduces to the Schwarzschild surface gravity $1/(4M)$. For $M=|Q|>0$, the root is double and $\kappa =0$. Thus the horizon remains present when its surface gravity vanishes: it is a degenerate Killing horizon, and $k$ itself has affine parametrization along its nonzero horizon generators. The strictly positive-constant description applies only to the nonextremal case; the zero value is the legitimate extremal limit.

Suppress the angular three-spheres and use radial null coordinates $u=t-r_{\ast }$, $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=1/V$. Compactifying these null coordinates puts radial light rays at 45 degrees in the Penrose diagram. Simple zeros of V give logarithmic ends of the tortoise coordinate and are crossed by regular null charts. The sign of V distinguishes static blocks, where r is spacelike, from inter-horizon blocks, where r is timelike.

The center is a genuine curvature singularity. An orthonormal-frame calculation gives the Kretschmann scalar of a five-dimensional charged black holeIt diverges at zero radius and is finite at positive horizons. Since the normal to a constant-r hypersurface has squared norm V, the singular boundary is spacelike when V is negative near zero and timelike when V is positive there.

For $Q=0$, V is negative inside the sole horizon and the maximal extension is Schwarzschild-like: two asymptotically flat exteriors, a future black-hole triangle ending at a spacelike singularity, and a past white hole triangle beginning at a spacelike singularity. The two pairs of horizon branches meet at a regular bifurcation sphere.

For $|Q|>M$, V is positive for every positive r. The radial conformal domain is a single exterior with a timelike naked singularity at its left boundary and the usual past/future null infinities on its outer boundaries. There is no horizon. These two cases are shown together below; red zigzags denote curvature singularities.

For $0<|Q|<M$, the exterior $r>r_{+}$ and deep interior $0<r<r_{-}$ are static, while $r_{-}<r<r_{+}$ is a trapped or time-reversed band. The outer roots are event horizons for a chosen exterior, and the inner roots are Cauchy horizons. The center is timelike because $V\sim Q^2/r^4>0$ there. In the exact analytic extension, crossing successive inner and outer horizon branches produces an infinite succession of exterior, trapped, inner-static and white hole blocks. The following finite strip specifies the repetition; it does not terminate the spacetime at the drawing's upper or lower edge.

For $|Q|=M$, $V=(1-M/r^2{)}^2$ is nonnegative on both sides of $r_h=\sqrt{M}$. There is no inter-horizon trapped band. Near the horizon, $r_{\ast }\sim -M/[4(r-r_h)]$, so the horizon is an infinite throat end of each static coordinate block. It is null and degenerate, without a regular bifurcation surface. The extension consists of alternating exterior and inner-static blocks, with timelike singularities in the latter. The separate conformal blocks below, together with their future-to-past horizon gluing rule, describe the full repeated extension. The open throat endpoints are ideal boundaries, not additional events where horizon branches meet.

Thus the neutral, strictly subextremal charged, extremal and overcharged cases all have distinct causal structures. These drawings describe the maximal analytic stationary metrics. The additional angular dimension changes areas and curvature falloff, but does not change these radial causal block types.

The physical spherical coordinates cover $t\in \mathbb{R}$, $r\in [0,\infty )$, $\theta \in [0,\pi ]$, and $\varphi \in [0,2\pi )$ with periodic identification. The center and the polar axes are the usual coordinate singularities, rather than physical curvature singularities; a regular polar chart restricts $r>0$ and $0<\theta <\pi$. For the null coordinates,Thus the physical metric tensor isThe conformal transformation followed by $p=\arctan u$, $q=\arctan v$ gives $du=(1+u^2)dp$, $dv=(1+v^2)dq$, andConsequently $d{s}^2=4dpdq-{\sin }^2(q-p)d{\Sigma }^2$. Finally $T=p+q$, $R=q-p$ converts $4dpdq$ to $d{T}^2-d{R}^2$, yieldingThe ranges must retain their joint constraints:Equivalently, $-\pi <T-R<\pi$ and $-\pi <T+R<\pi$, with $R\ge 0$. The angles retain their original ranges and identifications. Extending $R$ to negative values would double-count the radial polar geometry. The conformal factor isThe unphysical metric tensor extends onto the Einstein static universe, while physical Minkowski spacetime occupies the triangular region in its radial Penrose diagram. The edge $R=0$ is the regular center and belongs to the physical region away from its endpoints.

The conformal boundary consists of the following infinities. Future null infinity has $q=\pi /2$, hence $T+R=\pi$ with $0<R<\pi$, and corresponds to finite $u$ with $v\to +\infty$. Past null infinity has $p=-\pi /2$, hence $T-R=-\pi$, and corresponds to finite $v$ with $u\to -\infty$. Future timelike infinity is $(T,R)=(\pi ,0)$, past timelike infinity is $(-\pi ,0)$, and spacelike infinity is $(0,\pi )$. Although angles label points on null infinity, their two-spheres collapse at these three endpoint locations in the compactified picture.

For the geodesic endpoints in Minkowski conformal compactification, one can check the limits directly. A timelike inertial trajectory is $\mathbf{x}=\mathbf{b}+\mathbf{w}t$ with $|\mathbf{w}|<1$. At $t\to +\infty$ both $t-r$ and $t+r$ tend to $+\infty$, so $p,q\to \pi /2$ and the trajectory ends at future timelike infinity. At $t\to -\infty$ both tend to $-\infty$, giving past timelike infinity. This includes a stationary inertial worldline with bounded $r$.

For a null inertial trajectory, $|\mathbf{w}|=1$. As $t\to +\infty$, $r=t+\mathbf{w}\cdot \mathbf{b}+o(1)$, so $u$ has a finite limit while $v\to +\infty$; its future endpoint lies on future null infinity. As $t\to -\infty$, $r=-t-\mathbf{w}\cdot \mathbf{b}+o(1)$, giving a finite limiting $v$ and $u\to -\infty$, hence a past endpoint on past null infinity. Radial null geodesic paths have $T\pm R$ constant. A ray passing through the center changes its angular direction; in the angle-suppressed radial diagram it reflects at $R=0$ rather than terminating there.

A spacelike geodesic is also a straight line, now parametrized so that $|d\mathbf{x}/d\lambda |>|dt/d\lambda |$. At either end, $r>|t|$ asymptotically, $u\to -\infty$, $v\to +\infty$, and the endpoint is spacelike infinity. A radial spacelike line at $t=0$ has two opposite-angle halves which project onto the same horizontal interval in the diagram.

The conformal compactification endpoints occur at finite $T$ but at infinite physical proper time or affine parameter; Minkowski spacetime is geodesically complete. The conformal preservation of null geodesic paths does not preserve an affine parameter: if $\lambda$ is physically affine, an unphysical affine parameter satisfies $d\tilde{\lambda }/d\lambda ={\Omega }^2$ up to a constant. Along a complete null ray this can have a finite integral at infinity. Timelike and spacelike physical geodesic paths are not generally geodesics of the rescaled metric tensor, so their drawn curves must not be interpreted as freely falling unphysical observers.

Finally, every physical event can send an outgoing null ray to future null infinity, so $J^{-}({\mathcal{I}}^{+})$ is all of Minkowski spacetime. Its black-hole region and intrinsic event horizon are empty. The null edges in the diagram are infinity, not horizons concealing part of the physical manifold. Complete inertial observers likewise have no observer event horizon. Accelerated observers restricted to a Rindler wedge can have a Rindler horizon, but that is observer-dependent and does not contradict the horizon-free Minkowski spacetime. The conformal diagram compactifies infinity; it does not introduce a physical horizon.

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.

There is only one zero of $f$, at $r=2m$, and the Kruskal–Szekeres coordinates give four quadrants. Both exterior quadrants have $r>2m$, while the future and past interior quadrants have $0<r<2m$. The tortoise coordinate is finite at $r=0$: for $q>0$, $d{r}_{\ast }/dr\sim -\sqrt{q}{r}^{1/2}/(2m)$, and for $q=0$ it is $-r/(2m)$. Throughout each interior $g^{rr}=f<0$, so constant-$r$ hypersurfaces are spacelike. The assumed curvature singularity at $r=0$ therefore forms a spacelike future boundary in the black-hole region and a spacelike past boundary in the white-hole region. At infinity, $f\to 1$ and $R/r\to 1$, giving two standard asymptotically flat ends.

The resulting Penrose diagram is the same four-region causal diagram as the maximal Schwarzschild spacetime:

Each exterior has its own past null infinity, future null infinity, spatial infinity and timelike infinities. The future and past event horizons intersect on the bifurcation surface. The diagram describes the eternal analytic solution; a spacetime formed by collapse need not contain its white-hole region or second exterior.

Unlike a nonextreme Reissner-Nordstrom black hole, this electrically charged solution has one horizon and spacelike singularities, with no inner Cauchy horizon. A nonextreme Reissner-Nordstrom solution has two distinct horizons, including an inner Cauchy horizon, and a timelike singularity; its ideal maximal extension contains an infinite sequence of exterior and interior blocks. Charge by itself does not determine this causal structure: the metric and field equations of the gravitational theory matter.