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A bifurcate Killing horizon consists of two intersecting null Killing horizons of the same Killing vector field, which vanishes on their spacelike intersection, the bifurcation surface. In regular Kruskal coordinates the horizons are $U=0$ and $V=0$, and their intersection is $U=V=0$. The displayed field is tangent and normal to each horizon branch and zero on the intersection.

The null expansion, null shear and null twist of a congruence tangent to the generators of a Killing horizon all vanish there. If $k$ is its Killing generator and $\ell =fk$ is an affine generator, the screen-space projector kills every term proportional to $k$. The symmetric part of the projected derivative of $\ell$ therefore vanishes by the Killing equation. The antisymmetric part vanishes by Frobenius theorem, since $\ell$ is hypersurface-orthogonal. These conclusions extend through a regular bifurcation surface by continuity.

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.

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.

Set $r_H=2m$. Since $f^{\prime }(r_H)=2\kappa$, the integrand defining the tortoise coordinate has a simple pole:where $h$ is analytic near $r_H$. Define the Kruskal–Szekeres coordinates in the right exterior by $U=-e^{-\kappa u}$ and $V=e^{\kappa v}$. ThenThe right-hand side has positive nonzero derivative at $r_H$. Continuing this relation analytically, rather than retaining an absolute value, makes $r$ an analytic function of $UV$ through both signs. The Kruskal extension across a simple static horizon has metricIts radial coefficient tends to $-2e^{-2\kappa h(r_H)}/\kappa$, finite and nonzero, while $R(r_H{)}^2=4m^2\cosh \alpha$. Therefore both null surfaces $U=0$ and $V=0$ are regular horizon branches, intersecting at a smooth two-sphere $U=V=0$.

The stationary Killing vector field becomesIt is null, tangent and normal on each horizon branch and vanishes on their intersection. This proves that $r=2m$ is a bifurcate Killing horizon, with bifurcation surface of area $\boxed{A_H=16\pi m^2\cosh \alpha }$. The exponential definitions specify one exterior; allowing all signs of $U,V$ supplies the remaining quadrants of the extension.

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.

Put $r_{+}=\ell \sqrt{M}$ and $f=(r^2-r_{+}^2)/{\ell }^2$. The apparent failure of the nonrotating BTZ black hole coordinates at $r=r_{+}$ is a coordinate singularity. Define the tortoise coordinate and ingoing time byThe Ingoing BTZ coordinates giveAll coefficients are analytic at $r_{+}>0$, and the determinant is $-r^2$, so the metric tensor extends nondegenerately through that radius. The Killing vector field $K={\partial }_v$ agrees with ${\partial }_t$ in the exterior. On $r=r_{+}$, $K^2=-f=0$ and $K_a=(dr{)}_a$. It is therefore both normal and tangent to this null hypersurface, proving that it is a Killing horizon.

One may include both horizon branches and their bifurcation surface explicitly. Set $\kappa =r_{+}/{\ell }^2$, $u=t-r_{\ast }$ and, initially in the exterior, $U=-e^{-\kappa u}$, $V=e^{\kappa v}$. ThenThis analytic extension of a spacetime is regular near $UV=0$, and $K=\kappa (V{\partial }_V-U{\partial }_U)$ extends analytically. The surfaces $U=0$ and $V=0$ form the extended Killing horizon at $r=r_{+}$; $K$ vanishes only at their regular bifurcation circle.

Away from a bifurcation set, the normal to a Killing horizon is its nonzero Killing vector field $K$. Any affine tangent to the same generators has the form $k=hK$. In the optical tensor, derivatives of $h$ contribute a factor of $K_a$ and disappear under the screen-space projector. Consequentlyby the Killing equation. The trace and trace-free symmetric part are therefore zero: $\theta =0$ and ${\sigma }_{ab}=0$.

For the antisymmetric part, locally the horizon normal is proportional to a gradient. The restriction of its dual one-form to every tangent direction on the horizon is zero. Taking its exterior derivative and evaluating on two screen tangents therefore gives zero as well; equivalently this is the hypersurface-orthogonality consequence of the Frobenius theorem. Since both screen indices are tangential, $B_{[ab]}=0$. Thus the optical scalars on a Killing horizon satisfyNo affine normalization of the Killing vector field itself was assumed. At a regular bifurcation surface, $K$ vanishes but the smooth null generators do not; the conclusions for their optical quantities follow by continuity along each horizon branch.

The two identities. Set $F_{\beta \gamma }={\nabla }_{\beta }{K}_{\gamma }$. The Killing equation says $F_{\beta \gamma }=-F_{\gamma \beta }$. The Levi-Civita connection has symmetric lower connection indices, so they cancel in the antisymmetric difference:Use normalized antisymmetrization, soPut $a^{\sigma }=K^{\tau }{\nabla }_{\tau }{K}^{\sigma }$. Contracting with $F^{\beta \gamma }$, the second term is $a^{\gamma }{F}_{\gamma \alpha }$ and the third is the same after using antisymmetry. ThusRearranging proves the Killing derivative contraction identity

Restriction to a Killing horizon. On a Killing horizon, the generator $K$ is normal to the horizon as well as tangent to its null generators. Consequently $K_{[\alpha }{\nabla }_{\beta }{K}_{\gamma ]}=0$ there. One can see this without assuming hypersurface orthogonality away from the horizon: if the horizon is locally $\Phi =0$, write its dual one-form as $K=fd\Phi +\Phi q$ near it; then $K\wedge dK=0$ on $\Phi =0$. By the definition of surface gravity, $a^{\sigma }=\kappa K^{\sigma }$ on the horizon, and lowering that relation gives $K^{\sigma }{\nabla }_{\sigma }{K}_{\alpha }=\kappa K_{\alpha }$. The contraction identity becomesAway from points where $K$ vanishes, cancel a nonzero component of $K_{\alpha }$; at a regular bifurcation surface extend by continuity. ThereforeThis is surface gravity from the Killing derivative; it concerns the horizon, not every point in the exterior.

Static spherical metric. Choose the Killing vector $K={\partial }_t$. Its dual one-form is $-Adt$, and the first identity immediately giveswith all other components zero. HenceUsing the horizon identity and the simultaneous simple zeros,Here the positive value is chosen for a regular outer Killing horizon with $A,B>0$ on the exterior side. This is surface gravity of a static spherical horizon. A different normalization $K\mapsto cK$ rescales surface gravity by $c$; when an asymptotically flat normalization is wanted, the time coordinate is chosen so $A\to 1$ at infinity.

Euclidean regularity and temperature. Let $\delta =r-r_{+}$, $a=A^{\prime }(r_{+})$ and $b=B^{\prime }(r_{+})$. On the exterior side, define the proper radial coordinate $R=2\sqrt{\delta /b}$. After Wick rotation, the near-horizon metric isThe radial-time plane is a polar plane with angular coordinate $\vartheta =\kappa \tau$. Its circumference-to-radius ratio is $\kappa \Delta \tau$; the conical singularity disappears precisely whenThis is the Euclidean black-hole regularity condition. If a signed surface gravity convention is used, the period uses $|\kappa |$.

In a thermal quantum field theory, imaginary-time periodicity is the inverse-temperature condition expressed by the KMS condition. Therefore smoothness identifies the Hawking temperaturein units $\hslash =k_B=1$, or $T_H=\hslash \kappa /(2\pi k_B)$ when $\kappa$ is measured as an inverse time. It provides the thermal interpretation of surface gravity in black-hole thermodynamics. The regular Euclidean construction describes an equilibrium thermal state; the collapse calculation of Hawking radiation in the last solution supplies the outgoing spectrum. The simple-zero hypothesis excludes an extremal black hole, for which this polar-plane argument changes.

Use geometrized units and metric signature $(-+++)$, with $M>0$. In the exterior of Schwarzschild spacetime, put $f=1-2M/r$. The Schwarzschild tortoise coordinate satisfiesThe retarded and advanced null coordinates $u=t-r_{\ast }$ and $v=t+r_{\ast }$ then give $d{s}^2=-fdudv+r^{2}d{\Omega }^2$. The logarithmic divergence of $r_{\ast }$ suggests exponentiating these null coordinates. In the right exterior define the Kruskal–Szekeres coordinatesTheir product eliminates $t$:Since $du=-4MdU/U$ and $dv=4MdV/V$, the Schwarzschild metric becomesHere $r$ is an implicitly defined function of $UV$. The derivative of the right side with respect to $r$ is $-r{e}^{r/(2M)}/(4M^2)$, which is nonzero at $r=2M$. The inverse function theorem therefore makes $r(UV)$ smooth across that surface, and the coefficient of $dUdV$ tends to $-16M^2/e$. Thus the Schwarzschild event horizon is a coordinate singularity of the original chart, while this Lorentzian metric remains regular there.

Extend the Kruskal–Szekeres coordinates to all real $U,V$ with $UV<1$. The signs give two exterior regions, $U<0<V$ and $V<0<U$, a future black hole region $U,V>0$, and a past white hole region $U,V<0$. The event horizons are $U=0$ or $V=0$, intersecting at the bifurcation surface. The boundary $UV=1$ has $r=0$ and is a genuine Schwarzschild singularity, as the Kretschmann scalar $48M^2/r^6$ diverges there.

Finally, $T=(V+U)/2$ and $X=(V-U)/2$ give $UV=T^2-X^2$ and a radial metric proportional to $-d{T}^2+d{X}^2$. Hence radial null geodesics have slopes $\pm 1$, the event horizons are $T=\pm X$, and the singular boundaries are $T^2-X^2=1$. This constructs the maximal Kruskal extension; a black hole produced by collapse need not contain the second exterior or the white hole of that eternal extension.

A spacetime is geodesically complete if every maximal geodesic has an affine parameter ranging over all of $\mathbb{R}$. For timelike geodesics this is equivalent to unbounded proper time in both directions. An extendible geodesic segment can be prolonged in the same spacetime; an inextendible geodesic cannot. A finite coordinate endpoint need not imply finite affine parameter.

For the Kruskal spacetime, use $U,V$ withIn the right exterior $U<0<V$. A truncated ray $U=U_0<0$, $V_1<V<V_2$ is an extendible geodesic of radial null type: neither artificial endpoint is a spacetime boundary. A future ray $U=U_0>0$ in the black hole reaches $UV=1$, hence $r=0$, and is inextendible geodesic and future null-geodesically incomplete. Its Killing energy $E\ne 0$ gives $dr/d\lambda =-|E|$, so the Schwarzschild singularity occurs at finite affine parameter. The maximal continuation toward the past supplies the other half of this same null geodesic.

There is no inextendible, complete radial timelike geodesic in positive-mass Kruskal spacetime. This requested example is impossible as printed. For a radial timelike geodesic, the conserved Killing energy and normalization giveIf $E^2<1$, $r$ has at most one turning point, a maximum $r_{\max }=2M/(1-E^2)$; a maximal trajectory runs from the white hole Schwarzschild singularity to the Schwarzschild singularity. If $E^2\ge 1$, there is no finite turning point; one end can lie at infinity but the other reaches $r=0$. The exceptional $E=0$ trajectory through the bifurcation surface also reaches $r=0$ in both time directions. Near $r=0$,whose integral is finite. Constant-$r$ radial timelike curves are accelerated, not geodesics.

Two plausible repairs have different meanings. Removing “radial” permits a complete circular timelike geodesic at $r=R>6M$, with nonzero angular momentum and proper time ranging over $\mathbb{R}$. Replacing “timelike” by “null” permits a complete horizon null geodesic: $U=0$, $V\in \mathbb{R}$, with $V$ an affine parameter. The Penrose diagram shows both repairs explicitly, together with the two valid requested examples; the circular trajectory is only a radial projection and is labelled as nonradial.

The horizontal boundaries are the Schwarzschild singularities, diagonal dashed lines are the Killing horizons, and outer diagonal edges are null infinity.

Use the subextremal four-dimensional Reissner-Nordstrom spacetime, with $0<|Q|<M$. Its static radial function isTake the time-symmetric two-ended bridge through the outer bifurcation surface. Its initial data have $K_{ij}=0$ andon each $r\ge r_{+}$ end. The bridge is smooth: in proper radial distance $s$, $r-r_{+}=f^{\prime }(r_{+})s^2/4+O(s^4)$. Each asymptotically flat end is an infinite Riemannian distance away. Hence the spatial Riemannian manifold is complete and admits no proper same-dimensional smooth isometric extension as a connected spatial manifold.

Its maximal Cauchy development includes the two exteriors and the adjacent future and past regions between $r_{+}$ and $r_{-}$. It ends at inner Cauchy horizons, not at a curvature singularity. Since the simple root at $r_{-}$ is removable in horizon-penetrating coordinates and all curvature invariants are finite there, the exact solution extends across these Cauchy horizons. The extension is no longer globally determined by the given initial data.

The shaded region is the maximal Cauchy development; dashed upper and lower null edges are inner Cauchy horizons. The displayed neighboring diamonds illustrate smooth continuation, rather than the entire infinite extension.

The strong cosmic censorship conjecture concerns generic admissible initial data, in a specified extension regularity. Exact charged spherical data are exceptional. Perturbations can produce mass inflation at the inner Cauchy horizon, obstructing suitably regular extensions; the precise conjecture depends on the matter model and whether extensions are required to be $C^0$, $C^2$, or another regularity. Thus this exact extendible example does not refute a generic strong cosmic censorship conjecture.

For an Einstein-Maxwell example the gravitational triple $(\Sigma ,h,K)$ must be accompanied by electromagnetic initial data. One may take zero magnetic field and the smooth radial electric flux $Q/r^2$ through the bridge. The charges at the two ends have opposite signs when measured with outward normals. This is an example in the electrovacuum theory, not vacuum initial data.

Let $k$ generate the Killing horizon and let $\ell =fk$ be an affinely parametrized generator wherever $k\ne 0$. Choose an auxiliary null vector $n$ with $n\cdot \ell =-1$ and the screen-space projector $q_{ab}=g_{ab}+{\ell }_a{n}_b+n_a{\ell }_b$. The optical tensor is the screen projection $B_{ab}=q_a{}^c{q}_b{}^d{\nabla }_d{\ell }_c$.

Since the projector annihilates $k$, the symmetric part isby the Killing equation. Its trace is the null expansion and its trace-free symmetric part is the null shear, so both vanish. Also $\ell$ is normal to the null hypersurface. Frobenius theorem gives ${\ell }_{[a}{\nabla }_b{\ell }_{c]}=0$, which implies $q_a{}^c{q}_b{}^d{\nabla }_{[d}{\ell }_{c]}=0$: the null twist vanishes. These arguments use derivatives tangent to the horizon and hence apply to any smooth null geodesic congruence containing its generators, independently of its extension off the horizon. At a regular bifurcation surface, use the smooth affine generator and continuity. ThusThis is the vanishing of the optical scalars on a Killing horizon.

The Physical-process first law for a rotating black hole states that a small flux of matter through an initially and finally stationary horizon obeysin units $G=1$. Let $k^a$ be an affinely parametrized horizon tangent with affine parameter $\lambda =0$ on a bifurcation surface. Constancy of the surface gravity from the Zeroth law of black-hole mechanics and Gaussian null coordinates give the horizon generatorTo first order about a stationary horizon, the squared expansion and shear in the Null Raychaudhuri equation are second order. The Einstein field equations reduce it toThe final stationary condition $\theta (\infty )=0$ givesSince $d(\delta A)/d\lambda =\int \theta dA$, reversing the order of integration yieldsThe stress-energy current from a Killing vector gives the horizon Killing-energy fluxwhich proves the stated law.