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For $g=q+C(d\psi +A{)}^2$ with $C>0$, projection of every causal curve is causal for $q$, since $q(\dot{\gamma },\dot{\gamma })\le g(\dot{\gamma },\dot{\gamma })\le 0$. Conversely, the lift defined by $d\psi =-A$ along any given base curve has the same norm as that curve. Thus the orbit-space metric $q$ describes the projected causal structure exactly. A Penrose diagram suppressing that circle is a quotient diagram, not the metric of a constant-$\psi$ slice.

For $r_{-}<r_{+}$, an ingoing chart with $v=t+r_{\ast }$ has metric $-Vd{v}^2+2dvdr+R^{2}d{\Omega }_2^2$. In its future interior, $\nabla r={\partial }_v+V{\partial }_r$ is future timelike, so every future causal curve decreases $r$ and cannot reach future null infinity. The singularity $r=r_{-}$ is spacelike. A Kruskal–Szekeres coordinates extension has the same qualitative Penrose diagram as Schwarzschild spacetime, with two exterior regions, a future black-hole interior and a past white hole interior. The past interior has the same radial interval but is not a black hole region; specifying the radial interval alone does not identify the future component. For $r_{-}>r_{+}$, the physical exterior $r>r_{-}$ has $V>0$ everywhere and outgoing null rays escape from arbitrarily near a timelike naked singularity.

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.

This radial spacetime diagram uses Kruskal–Szekeres coordinates to display both exteriors, the black hole and the white hole of maximally extended Schwarzschild spacetime. Each point represents a symmetry two-sphere. Radial null geodesics have slopes $\pm 1$ in the displayed coordinates; the horizons are $T_K=\pm X_K$ and the singularities satisfy $T_K^2-X_K^2=1$ in the usual normalization. A further compactification gives the corresponding Penrose diagram.

For four-dimensional Minkowski spacetime, set $p=\arctan ((t-r)/L)$ and $q=\arctan ((t+r)/L)$, with $L>0$, then $T=p+q$, $R=q-p$. Multiplication by the square of the conformal factor $2\cos p\cos q/L$ yields the displayed metric on $R\ge 0$, $|T|+R<\pi$. Its radial Penrose diagram is a triangle: the timelike edge $R=0$ is the ordinary centre, while the two sloping edges are future null infinity and past null infinity. A signed Cartesian spatial coordinate in two dimensions gives a full diamond instead.

Null infinity is the ideal boundary reached by asymptotically escaping null geodesics in a suitable conformal completion. It has future and past components. A finite location on a Penrose diagram need not represent a finite affine parameter.

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 the causally unwrapped universal cover of Anti-de Sitter spacetime, with curvature radius $L$. In global coordinates,The ordinary radial Penrose diagram is a half-strip: the left edge $\rho =0$ is a regular center and the right edge $\rho =\pi /2$ is a timelike conformal boundary. For displaying signed spatial motion, the figure doubles the radial diameter to $-\pi /2<\rho <\pi /2$. Its two ends represent antipodal spatial directions of the higher-dimensional boundary; they are not a claim that the full AdS boundary has two disconnected components.

The vertical edges are spatial infinity at every finite global time, forming the timelike boundary $\mathbb{R}\times S^{d-2}$. The arrows mark the future and past timelike ends $T\to \pm \infty$, conventionally $i^{+}$ and $i^{-}$. There are no separate null-infinity edges like Minkowski ${\mathcal{I}}^{\pm }$: radial null rays satisfy $dT=\pm d\rho$ and reach the timelike boundary in finite global time. Their physical affine parameter is nevertheless infinite there. The drawing is a finite excerpt of an infinitely tall strip, not an identification of its top and bottom. On the original hyperboloid global time is periodic and there are closed timelike curves; unwrapping it is essential to the usual physical maximally extended AdS spacetime.

In ordinary spherical coordinates, $t\in \mathbb{R}$, $r\in [0,\infty )$, $\theta \in [0,\pi ]$ and $\varphi \in [0,2\pi )$ with periodic identification. The regular coordinate chart excludes the origin and polar axes; the endpoint ranges include the usual spherical identifications. With null coordinates $u=t-r$ and $v=t+r$, both are real and their joint domain is $v\ge u$. The inverse relations giveThe arctangent coordinates obey $-\pi /2<p\le q<\pi /2$. Since $du={\sec }^{2}pdp$ and $dv={\sec }^{2}qdq$, the rescaled metric isNow $T=p+q$, $R=q-p$, so $4dpdq=d{T}^2-d{R}^2$. Thus the Minkowski conformal compactification hasThe full coordinate ranges are linked, not independent: $T\in (-\pi ,\pi )$ and $R\in [0,\pi )$ with $|T|+R<\pi$. Inverse coordinates are $p=(T-R)/2$, $q=(T+R)/2$, $t=[\tan q+\tan p]/2$, $r=[\tan q-\tan p]/2$. The conformal factor isThe unphysical metric extends as the unit-radius Einstein static universe on $\mathbb{R}\times S^3$, where $0\le R\le \pi$ with the endpoint two-spheres collapsed. Physical Minkowski spacetime occupies the triangular radial domain $R\ge 0$, $|T|+R<\pi$; the edge $R=0$ is the ordinary spatial center, not an infinity or physical boundary.

In the closure, the infinite-coordinate limits have locationsThese are future timelike infinity, past timelike infinity, spacelike infinity, future null infinity and past null infinity. For example ${\mathcal{I}}^{+}$ has $v\to +\infty$ with finite $u$, while ${\mathcal{I}}^{-}$ has $u\to -\infty$ with finite $v$. A diagram point with $0<R<\pi$ stands for a two-sphere, and each null boundary is topologically $\mathbb{R}\times S^2$; their three corners are points after the angular collapse. On the null boundaries $\Omega =0$ and $d\Omega \ne 0$, whereas the three corners have a degenerate zero gradient. Thus a compactifying metric can be smooth there even though those infinity points are not ordinary smooth null-boundary points.

The geodesic endpoints in Minkowski conformal compactification can be established directly. A physical inertial timelike geodesic has $t=t_0+E\tau$, $\mathbf{x}={\mathbf{x}}_0+\mathbf{p}\tau$, with $E>0$ and $E^2-|\mathbf{p}{|}^2=1$. At future infinity $u,v\to +\infty$, so $p,q\to \pi /2$ and the curve ends at $i^{+}$. At past infinity both tend to $-\infty$, giving $i^{-}$. Every complete inertial timelike path has these endpoints, regardless of velocity.

For a future-directed null geodesic, $t=t_0+E\lambda$, $\mathbf{x}={\mathbf{x}}_0+E\mathbf{n}\lambda$ with $|\mathbf{n}|=1$. As $\lambda \to +\infty$, $v\to +\infty$ but $u$ has a finite limit; the endpoint lies on ${\mathcal{I}}^{+}$. As $\lambda \to -\infty$, $u\to -\infty$ but $v$ has a finite limit, giving an endpoint on ${\mathcal{I}}^{-}$. A conformal transformation preserves these null paths as unparametrized geodesics. The null affine parameter under a conformal rescaling obeys $d\tilde{\lambda }/d\lambda ={\Omega }^2$; here ${\Omega }^2=O({\lambda }^{-2})$ at the null endpoints, so the unphysical affine parameter reaches them in finite time, although the physical affine parameter is infinite.

A physical spacelike geodesic has $t=t_0+a\lambda$, $\mathbf{x}={\mathbf{x}}_0+\mathbf{b}\lambda$, with $|\mathbf{b}|>|a|$. At either end $r>|t|$ asymptotically, $u\to -\infty$ and $v\to +\infty$, so both ends approach $i^0$. Timelike and spacelike physical geodesic paths are not generally geodesics of the rescaled metric. Their physical proper time or length is infinite, while $\Omega =O(|\lambda {|}^{-2})$ gives finite unphysical proper time or length near the endpoints. Infinity at finite diagram coordinates does not imply incompleteness of physical Minkowski geodesics.

The original Penrose diagram suppresses the angular spheres. The displayed physical paths illustrate their distinct endpoint types; the two sloping boundaries are null infinity, while the left edge is the regular center.

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.

The positive conformal factor leaves the causal structure unchanged. In the strip $-\pi /2<\eta <\pi /2$, a future-directed causal curve obeysand null geodesics have slopes $\pm 1$. Its past and future conformal boundaries $\eta =\mp \pi /2$ are spacelike. A finite conformal time interval limits the range of light signals: the observer at $\xi =0$ can receive a signal from $(\eta ,\xi )$ before future infinity exactly when $|\xi |<\pi /2-\eta$. Equality describes its observer-dependent cosmological event horizon.

For a finite Penrose diagram of the unwrapped spatial coordinate, set $U=\arctan u$, $V=\arctan v$, $X=(U+V)/2$, and $Y=(V-U)/2$. Then the metric is conformal to $-d{Y}^2+d{X}^2$; light rays remain at $45$ degrees, and the boundaries are the images of $v-u=\pm \pi$. The figure also shows the spatially periodic de Sitter cylinder so that the global distinction is explicit.

Both the covering spacetime and the de Sitter cylinder are globally hyperbolic. To see this directly in the covering spacetime, suppose an inextendible future-directed causal curve had a future limiting conformal time strictly below $\pi /2$. The slope bound makes $\xi$ Cauchy as that time is approached, so the curve has a limiting point inside the spacetime and can be extended, a contradiction. The past endpoint similarly approaches $-\pi /2$. Thus every constant-$\eta$ line is crossed exactly once and is a Cauchy hypersurface. On the cylinder the same argument uses compactness of the spatial circle. Equivalently causal diamonds between finite events are bounded and closed within the strip, hence compact.

Initial field data on a complete Cauchy hypersurface determine a unique solution for a well-posed hyperbolic Cauchy problem, with the relevant constraints imposed for constrained systems. Spacelike conformal infinity does not require additional timelike-boundary conditions. An observer's cosmological horizon limits which data that observer can see; it does not destroy global uniqueness from data on the whole hypersurface.

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.

For the printed metric tensor, first use the global Minkowski time $T=\sinh t$. SetThenRemoving the positive conformal factor yields the ordinary Penrose diagram of two-dimensional Minkowski spacetime: a diamond with null future/past infinity on its sloping edges, timelike infinity at the top/bottom, and spatial infinity at the left/right corners. Null geodesics have slope $\pm 1$; future/past causal cones are ordinary Minkowski cones. There is no physical boundary or curvature singularity at a finite original coordinate value.

For the alternative spatial-lapse metric tensor identified in part (a), set $\rho =\arctan (\sinh x)$. Then $dx=\sec \rho d\rho$ and $\cosh x=\sec \rho$, soIts universal-cover conformal diagram is an infinitely tall strip with two timelike conformal boundaries, the conformal strip of two-dimensional anti-de Sitter spacetime. Null geodesics reach either boundary in finite coordinate time, though the boundary is not part of the physical manifold and the physical null affine parameter diverges there. The strip and diamond have different causal structures; no conformal coordinate change can identify their global causal properties.

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.

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.

The original spherical coordinates have $t\in \mathbb{R}$, $0\le r<\infty$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$, with $\varphi$ periodic. The angular coordinates degenerate at the polar axis and all angles identify at $r=0$; a regular spherical chart uses $r>0$, $0<\theta <\pi$ and suitable overlapping angular patches.

The retarded time $u=t-r$ and advanced coordinate $v=t+r$ both range over $\mathbb{R}$, with the joint restriction $u\le v$. Their inverse relations are $t=(u+v)/2$ and $r=(v-u)/2$, soSet $p=\arctan u$, $q=\arctan v$. ThenThe conformal factor is $\Omega =2\cos p\cos q>0$. Since $(\tan q-\tan p)\cos p\cos q=\sin (q-p)$, multiplication by ${\Omega }^2$ givesFinally $T=p+q$ and $R=q-p$ give $4dpdq=d{T}^2-d{R}^2$. HenceThis is a region of the Einstein static universe, with positive time and negative spatial signature. The exact joint coordinate domain isor equivalently $R\ge 0$ and $|T|+R<\pi$. The regular radial chart has $R>0$; $R=0$ is the ordinary centre. The angular ranges are unchanged. It is important not to treat the stated individual ranges of $T,R$ as independent. The inverse map and conformal factor provide a useful check: