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Choose an auxiliary null vector $N$ with $U\cdot N=-1$, parallel-transported along each affine ray. The screen-space projector and its induced positive metric areThey annihilate $U,N$, obey $P^2=P$, and have trace two in four spacetime dimensions. Since both $U$ and $N$ are parallel along $U$, ${\nabla }_{U}P=0$. Equivalently, the screen represents the two-dimensional quotient $U^{\perp }/\mathrm{span}(U)$.

Define the optical matrix of a null congruence byFor a transverse deviation, its screen projection $\zeta =PZ$ satisfies ${\nabla }_U{\zeta }^a={\hat{B}}^a{}_b{\zeta }^b$. Decompose the matrix asThese are the null expansion, null shear and null twist. Null expansion gives the fractional change of a small transverse area, $\theta =d\log (dA)/d\lambda$. Null shear is symmetric trace-free distortion; null twist is the antisymmetric rotation part. This null expansion convention uses the full trace, not half the trace. The auxiliary screen fixes representatives, while the transverse optical content is intrinsic to the null geodesic congruence.

Use signed velocities in Skaro's inertial frame; the returning velocity $w$ must be negative. For the first two rockets, the Lorentz transformation of infinitesimal separations is $dx={\gamma }_u(d{x}^{\prime }+ud{t}^{\prime })$, $dt={\gamma }_u(d{t}^{\prime }+ud{x}^{\prime }/c^2)$. Dividing with $d{x}^{\prime }/d{t}^{\prime }=v^{\prime }$ gives the relativistic velocity-addition formulaFor the third rocket, its signed velocity in the second frame is $-w^{\prime \prime }$, so another application givesSince $0<v<c$ and $0<w^{\prime \prime }<c$, the denominator $1-v{w}^{\prime \prime }/c^2$ is positive. The third rocket returns from positive $x$ to Skaro if and only if $w<0$, henceEquality leaves it stationary in Skaro's frame, so it never returns from $x=2L$; a smaller $w^{\prime \prime }$ leaves it travelling outward.

Let $O$ denote departure, $A$ boarding the second rocket, $B$ boarding the third, and $C$ reunion with Skaro. In coordinates $(t,x)$, their positions on the worldline areIndeed, the first two coordinate durations are $L/u$ and $L/v$, and the returning leg has duration $-2L/w>0$. Equivalently, their $(ct,x)$ coordinates are obtained by multiplying each displayed time by $c$; these are the coordinates used for the spacetime diagram.

The diagram uses representative velocities satisfying the return condition; its event labels apply to the general formulas above. Skaro's elapsed time isOn a constant-velocity leg the traveller's proper time is $\Delta t/{\gamma }_V=\Delta t\sqrt{1-V^2/c^2}$, by time dilation. Neglecting the specified brief accelerations, Davros therefore ages byEvery contribution is positive because $w<0$. Each square-root factor is smaller than one, so $\Delta \tau <T$: Davros returns younger than a companion who remained on Skaro. The elapsed times compare the same departure and reunion events; they cannot be obtained by applying one time-dilation factor to the whole journey, whose worldline has three different velocities.

Choose a Euclidean string worldsheet and the usual normalization of the Polyakov action. For a closed oriented worldsheet $\Sigma$, the string nonlinear sigma model isHere ${\epsilon }^{12}=1$ is the alternating density, not the tensor with components $1/\sqrt{g}$. The imaginary coefficient in the Kalb–Ramond field term is the Euclidean continuation; in Lorentzian signature this term is real. $G$ is the spacetime metric and $\Phi$ the dilaton.

All three terms have worldsheet diffeomorphism invariance, and are covariant under spacetime coordinate transformations when $X$, $G$, $B$ and $\Phi$ are transformed together. Under a Weyl transformation $g_{ab}\mapsto e^{2\omega }{g}_{ab}$, $\sqrt{g}{g}^{ab}$ is invariant in two dimensions, so the metric coupling has classical Weyl invariance. The antisymmetric coupling is independent of the worldsheet metric, so it too has classical Weyl invariance.

The curvature coupling requires care. An infinitesimal Weyl transformation gives ${\delta }_{\omega }(\sqrt{g}{R}^{(2)})=-2\sqrt{g}{\Delta }_g\omega$. Integration by parts on the closed worldsheet therefore gives the Weyl variation of a nonconstant dilaton coupling:A constant dilaton coupling is topological and Weyl invariant; a general varying dilaton coupling is not separately classically Weyl invariant. Its classical variation participates in canceling the quantum worldsheet Weyl anomaly, producing the improved background equations below.

To relate these background couplings to the flat-space bosonic string mass spectrum, expand $G_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }(X)$ and express $h$ and $B$ as superpositions of plane waves. The perturbation of the action is a sum of integrated string vertex operators of the formThis is the vertex of ${\alpha }_{-1}^{\mu }{\tilde{\alpha }}_{-1}^{\nu }|p⟩$ in the massless first closed-string level. Its conformal weights are $(1+{\alpha }^{\prime }{p}^2/4,1+{\alpha }^{\prime }{p}^2/4)$, so marginality gives $p^2=0$. The primary operator conditions give transversality, and the string-state gauge redundancy removes longitudinal components. The symmetric trace-free polarization tensor is the graviton and perturbs $G$; its antisymmetric part is the Kalb–Ramond field and perturbs $B$. The scalar physical polarization is the dilaton. At the level of background fields its vertex involves the metric trace and the curvature improvement, so an arbitrary off-shell metric trace should not simply be identified with the physical dilaton without this distinction.

The spacetime gauge invariance of the antisymmetric coupling isIndeed its action is $S_B=i(2\pi {\alpha }^{\prime }{)}^{-1}{\int }_{\Sigma }{X}^{\ast }B$. Its variation isfor a closed worldsheet, by Stokes theorem. In components this is the total derivative ${\partial }_a({\epsilon }^{ab}{\Lambda }_{\nu }(X){\partial }_b{X}^{\nu })$: the term containing ${\partial }_a{\partial }_b{X}^{\nu }$ vanishes by antisymmetry. Thus only the Kalb-Ramond field strength $H$ can enter the local spacetime equations. With boundaries, a compensating boundary gauge-field transformation is needed.

Separate the dilaton into a constant ${\Phi }_0$ and a varying part. The Gauss-Bonnet theorem givesfor a connected closed orientable genus-$g$ worldsheet. Its Euclidean path-integral weight containsConsequently the sum over worldsheet topologies is the string genus expansion ${\sum }_{g\ge 0}{g}_s^{2g-2}{\mathcal{A}}_g$, with each coefficient integrating over that genus's worldsheet moduli. Each extra handle costs $g_s^2$, which is the loop-counting factor of string perturbation theory. With the usual normalization of $n$ external closed-string vertices, the connected amplitude carries $g_s^{2g-2+n}$. This dilaton Euler-characteristic weighting is distinct from the expansion in ${\alpha }^{\prime }$ of a fixed-worldsheet string nonlinear sigma model.

Finally quantize that two-dimensional theory. The background fields are its couplings. The background field expansion of a string sigma model, $X=x+\xi$ in target-space normal coordinates, contains curvature vertices quadratic in the fluctuation $\xi$. Contracting their two fluctuation indices produces the Ricci tensor multiplying $\partial x^{\mu }\bar{\partial }{x}^{\nu }$. Logarithmic short-distance contractions renormalize $G$, $B$, and $\Phi$, giving functional sigma-model beta functions. Gauge independence requires vanishing of the worldsheet Weyl anomaly, including the classical improvement from the varying dilaton, not merely an arbitrary choice of renormalization scale.

For clarity, in conventional leading-order normalization the improved coefficients areHere $H^2=H_{\mu \nu \rho }{H}^{\mu \nu \rho }$. The constant term in the dilaton coefficient is the matter-plus-ghost central charge deficit. At $D=26$, setting these coefficients to zero gives the leading spacetime field equationsFor the last equation, take the trace of the first, $R-H^2/4+2{\nabla }^2\Phi =0$, and combine it with ${\overline{\beta }}^{\Phi }=0$; this verifies the coefficients and signs. These are equivalently the equations from the leading string-frame massless effective actionHigher worldsheet orders give higher-derivative ${\alpha }^{\prime }$ corrections, whereas additional handles give string-loop corrections controlled by $g_s$. Thus quantum consistency of a two-dimensional string nonlinear sigma model determines the dynamics of its spacetime background fields. These leading coefficient conventions agree with the primary string-theory lectures' background-field calculation.

A linearized coordinate gauge transformation changes the identification of points of the perturbed spacetime with its Minkowski spacetime background. With $x^{\prime i}=x^i+ϵ{\xi }^i$, the same physical metric has perturbationThese changes are coordinate freedom, not extra physical gravitational wave polarizations. Raising indices with $\eta$, define $h={\eta }^{ij}{h}_{ij}$ and the trace-reversed metric perturbation ${\bar{h}}_{ij}=h_{ij}-{\eta }_{ij}h/2$. Then

The first-order Levi-Civita connection is ${\Gamma }^{(1)k}{}_{ij}=\frac{1}{2}{\eta }^{k\ell }({\partial }_i{h}_{j\ell }+{\partial }_j{h}_{i\ell }-{\partial }_{\ell }{h}_{ij})$. Contracting the Riemann curvature tensor in the stated convention givesPut $A_j={\partial }^i{\bar{h}}_{ij}$. The corresponding Einstein tensor isUnder the linearized coordinate gauge transformation, $A_j^{\prime }=A_j-\square {\xi }_j$. Choosing a solution of $\square {\xi }_j=A_j$ imposes Lorenz gauge in linearized gravity, $A_j^{\prime }=0$. Thus the vacuum Einstein field equations reduce toThere remains residual gauge symmetry of linearized gravity with $\square {\xi }_j=0$.

For the transverse profiles, write both symmetric off-diagonal entries as $h_{xy}=h_{yx}=h_{\times }(t-z)$ and the diagonal entries as $h_{xx}=-h_{yy}=h_{+}(t-z)$. The trace-reversed metric perturbation has zero trace and therefore equals $h$. Its divergence vanishes: its only nonzero index directions are $x,y$, while its coefficients depend only on $t-z$. Also $(-{\partial }_t^2+{\partial }_z^2)F(t-z)=0$ for every twice differentiable profile $F$. These are consequently plane gravitational waves in linearized gravity in transverse-traceless gauge, with the two independent gravitational wave polarizations $h_{+}$ and $h_{\times }$.

To calculate the detector response, use a parallel-propagated orthonormal frame along the centre of a short nonrotating stick. To first order its clock reads $t$, and the tidal curvature isFor a sliding-bead gravitational wave detector, the stick constrains transverse motion but supplies no axial restoring force. Projecting geodesic deviation along an $x$-directed stick givesHere $L$ is the proper separation and $L_0=L(t_0)$, with $\dot{L}(t_0)=0$. Integrating with those actual initial conditions yieldsFor beads at rest before an incident wave, both initial profile terms vanish and the familiar result is $\boxed{L(t)=L_0[1+ϵh_{+}(t-z_0)/2]+O({ϵ}^2)}$. The cross polarization produces a transverse tidal force, balanced by the guide, and no axial first-order displacement in this orientation. In transverse-traceless gauge the same plus response can be read from the proper line element, $d\ell =\sqrt{1+ϵh_{+}}dx$; coordinate distances alone are not measured distances. When a cross polarization is present, keeping the guide physically nonrotating may require transverse coordinate adjustments, which do not change the axial answer at this order.

For a $z$-directed stick, $R_{z0z0}=0$ and the wave is transverse. Therefore beads initially at rest retain their proper separation to first order: $L(t)=L_0+O({ϵ}^2)$. The tidal calculation is the usual linear detector limit, with bead spacing small compared with the wavelength; it does not treat a macroscopic rigid body as exactly rigid over arbitrary relativistic distances.

Assume $M>0$ and use geometric units. The stationary Killing vector field $K={\partial }_t$ gives the conserved Killing energy $E=-g(K,U)=f\dot{t}$. For radial motion, normalization of the four-velocity gives $-f{\dot{t}}^2+f^{-1}{\dot{r}}^2=-1$. Combining these identities yieldsAt rest at infinity, $f\to 1$ and $\dot{r}\to 0$, so $E=1$. Choose the inward branch $\dot{r}=-\sqrt{2M/r}$. Its proper time from the Schwarzschild event horizon to the curvature singularity isIn ordinary units this is $4GM/(3c^3)$. The endpoint $r=0$ is not a regular centre of the spacetime. This is the Schwarzschild horizon-to-singularity proper time for fall from infinity, rather than a universal time for all infall energies.

For this geodesic congruence, set $w(r)=\sqrt{2M/r}$. Its four-velocity and metric dual areThe right side is an exact one-form because $w/f$ depends only on $r$. Thus $U_a=-{\partial }_a\tau$, with free-fall proper time in Schwarzschild spacetime determined byIndeed $U^a{\partial }_a\tau =-U^a{U}_a=1$, so this scalar advances by the actual proper time on every member of the infalling geodesic congruence, with a synchronized choice of additive offsets.

Substitution of $dt=d\tau -w{f}^{-1}dr$ in the Schwarzschild metric givesUsing $f=1-w^2$, the Painlevé–Gullstrand coordinates therefore haveAll coefficients are finite at $r=2M$; the determinant of the $\tau ,r$ block is $-f-w^2=-1$. Hence the horizon is a coordinate singularity of Schwarzschild time, not a degeneracy of the metric. The new spatial slices have $d{\ell }^2=d{r}^2+r^{2}d{\Sigma }^2$, the flat Euclidean metric, and $g^{ab}{\partial }_a\tau {\partial }_b\tau =-1$ makes their normals the freely falling four-velocities. By contrast, the curvature scalar $R_{abcd}{R}^{abcd}=48M^2/r^6$ diverges at $r=0$; this genuine singularity is not removed.

For ingoing Eddington-Finkelstein coordinates, use $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=f^{-1}$, for example $r_{\ast }=r+2M\log |r/(2M)-1|$. ThenBoth constructions cross the future horizon. Their time functions have different meanings: $v=\text{constant}$ is an ingoing null hypersurface, whereas $\tau =\text{constant}$ is a flat spacelike hypersurface orthogonal to the infallers. Thus the former is adapted to ingoing null geodesics, and the latter to a timelike free-fall clock congruence.

For the Kruskal diagram, choose Kruskal–Szekeres coordinates with $U_K=-e^{-(t-r_{\ast })/(4M)}$, $V_K=e^{(t+r_{\ast })/(4M)}$ in the right exterior and extend them across its horizon. They obeyThe four regions of Kruskal spacetime are I, the right exterior $(U_K<0,V_K>0)$; II, the future black hole $(U_K>0,V_K>0)$; III, the other exterior $(U_K>0,V_K<0)$; and IV, the past white hole $(U_K<0,V_K<0)$. The null axes are horizons, their intersection is the bifurcation two-sphere, and the spacelike boundaries $U_K{V}_K=1$ are the future and past curvature singularities. A point in this radial diagram represents a symmetry two-sphere.

The domain of the ingoing Painleve-Gullstrand chart is region I together with region II and their common future horizon. More precisely, it is $V_K>0$, $U_K{V}_K<1$, or $r>0$ with finite $\tau$. To verify the boundary rather than infer it from the locally regular metric, put $\rho =\sqrt{r/(2M)}$ and writeThis follows by subtracting $r_{\ast }$ from the preceding radial primitive, and $S$ is finite at $r=2M$. Thus $V_K=\exp [(\tau -S(r))/(4M)]>0$ throughout the chart. Conversely every point with $V_K>0$ and $r>0$ gives a finite $\tau$. The past horizon $V_K=0$, the bifurcation sphere, region III and region IV are absent. Reversing the sign of the radial term produces the outgoing chart through the white hole instead.

The radial part of the Minkowski metric becomes $-dudv$, and $r=(v-u)/2$. With $u=\tan p$, $v=\tan q$ and the positive conformal factor $\Omega =2\cos p\cos q$, multiplication by ${\Omega }^2$ givesConsequently the Minkowski conformal compactification isThe spatial metric $d{R}^2+{\sin }^{2}Rd{\Sigma }^2$ is the round unit three-sphere. The extended unphysical metric is therefore that of the Einstein static universe, $\mathbb{R}\times S^3$ with unit spatial radius. Physical Minkowski spacetime occupies only the diamond-shaped causal patch of this cylinder, or a triangle after restricting to nonnegative spherical radius; it is not the whole cylinder.

The complete coordinate ranges and joint restrictions areThe polar angles have the usual coordinate degeneracies at their poles and at $r=0$. In the extended Einstein static universe one may let $T\in \mathbb{R}$ and $0\le R\le \pi$, but these are not the ranges of the original physical patch. In its interior $\Omega >0$. The line $R=0$ is the regular centre $r=0$, not a component of infinity.

The conformal completion has future timelike infinity $i^{+}=(\pi ,0)$, past timelike infinity $i^{-}=(-\pi ,0)$, and spacelike infinity $i^0=(0,\pi )$. The open sloping boundaries arenamely future null infinity and past null infinity. Each ordinary interior point of the radial diagram represents a two-sphere, and each point on either open null boundary represents a two-sphere of null directions.

A timelike geodesic is an inertial worldline $\mathbf{x}=\mathbf{b}+\mathbf{\beta }t$ with $|\mathbf{\beta }|<1$. As $t\to +\infty$, both $t-r$ and $t+r$ tend to $+\infty$, giving $p,q\to \pi /2$ and endpoint $i^{+}$. At past infinity both tend to $-\infty$, giving $i^{-}$. These endpoints are reached at infinite physical proper time, although at finite conformal coordinate time $T$. A null geodesic has $r\sim |t|$. In its outgoing future, $u$ approaches a finite retarded time while $v\to +\infty$, giving an endpoint on ${\mathcal{I}}^{+}$; in its incoming past, $v$ has a finite limit and $u\to -\infty$, giving an endpoint on ${\mathcal{I}}^{-}$. A spacelike geodesic is a straight line with spatial tangent larger than its temporal tangent, so $u\to -\infty$, $v\to +\infty$ at either end and both endpoints are $i^0$ in the compactified radial description. All these physical geodesics are complete. Only the unparametrized null geodesics are necessarily preserved as geodesics of the unphysical metric; timelike and spacelike images need not be its geodesics.

For horizons, distinguish a property of the spacetime from a property of an observer or of an initial-data hypersurface. Complete inertial observers have no particle horizon: in the flat cosmological slicing $a(t)=1$, the past light-travel distance ${\int }_{-\infty }^{t_0}dt$ is infinite. There is no finite-age initial singularity truncating communication between comoving observers. This statement does not say that an event receives signals from spacelike-separated events at the same time.

There is also no black-hole event horizon, because every event can send a future-directed null ray to future null infinity. Nevertheless accelerated timelike observers can have both future and past observer event horizons. For an entire observer worldline $\gamma$, define them by $H_{\gamma }^{+}=\partial I^{-}(\gamma )$ and $H_{\gamma }^{-}=\partial I^{+}(\gamma )$, using its complete future and past respectively. An inertial observer's chronological past and future of the whole worldline are both all of Minkowski spacetime. By contrast, for the uniformly accelerated worldline in a Cartesian spatial direction,its null coordinates obey $t-x=-a^{-1}{e}^{-a\tau }<0$ and $t+x=a^{-1}{e}^{a\tau }>0$. An event with $t-x<0$ can signal to a sufficiently late point of this worldline, whereas an event with $t-x\ge 0$ cannot; finite transverse displacements do not change that conclusion as $t+x\to \infty$. Similarly the worldline can signal to precisely the events with $t+x>0$. Thus its two observer event horizons areThey are the Rindler horizons, and their accessible common region is $x>|t|$. They do not indicate a curvature singularity or a failure of global predictability.

Finally Minkowski spacetime is a globally hyperbolic spacetime: every inextendible causal curve crosses each complete spacelike plane $t=t_0$ exactly once. Indeed $t$ is strictly monotone along future-directed causal curves, and the speed bound $|d\mathbf{x}/dt|\le 1$ prevents an inextendible curve from ending at finite $t$ by escaping to spatial infinity. Its time range is therefore all of $\mathbb{R}$. These planes are Cauchy hypersurfaces with full domain of dependence, so there are no Cauchy horizons associated with complete Cauchy data.

A Cauchy horizon is, however, defined relative to a specified initial surface. Under a literal existence reading of “admits”, non-Cauchy initial surfaces can have such horizons even in flat spacetime. For example the edgeless spacelike hyperboloid ${\Sigma }_a:t=\sqrt{a^2+r^2}$ has $D({\Sigma }_a)=\{t>r\}$. To see this, inside the future light cone $t-r$ and $t+r$ are positive and nondecreasing on future-directed causal curves. Their product $t^2-r^2$ increases past $a^2$ on every future-inextendible curve from below the hyperboloid; every past-inextendible curve from above it crosses down through $a^2$ before leaving the cone. Outside the cone there is an inextendible null curve avoiding the hyperboloid. The future light cone itself is its past Cauchy horizon. The time-reversed hyperboloid gives a future Cauchy horizon. This is the dependence of a Cauchy horizon on the initial hypersurface, not an intrinsic loss of predictability of full Minkowski spacetime. Thus a categorical claim that no choice of partial initial surface can ever have a Cauchy horizon would be too strong.

A spacetime is a smooth four-dimensional manifold equipped with a smooth Lorentzian metric, conventionally of metric signature $(-,+,+,+)$, and a choice of time orientation. The usual manifold assumptions include the Hausdorff space and second-countable space conditions. A physical model also specifies matter fields and requires the Einstein field equations and matter equations.

Diffeomorphism invariance of general relativity means that relabelling events by a smooth invertible map, while transforming the metric tensor and every matter field together, preserves the form of the equations. Passively, a coordinate change gives new components for the same geometric fields. Actively, pulling all fields back by a diffeomorphism gives another representative of the same physical geometry, subject to any prescribed boundary conditions or boundary symmetries.

The geometry and matter together carry the physical content; coordinate labels do not. A diffeomorphism need not be an isometry of a fixed metric, and unrelated metrics are not automatically physically equivalent.

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.

The second law of black-hole mechanics says that the total area of future event horizon cross-sections cannot decrease toward the future, assuming the classical Einstein field equations, the null energy condition, suitable global predictability and regularity, and here the stipulated future completeness of its generators.

A future event horizon is an achronal boundary generated by null geodesics. If a smooth generator had $\theta <0$, the null focusing theorem would produce a conjugate point to a spacelike surface within finite affine parameter. Future completeness lets the generator reach it, but beyond such a point it could no longer remain on the achronal boundary, a contradiction. Hence $\theta \ge 0$ wherever the null expansion is defined. A transported area element obeysEvery generator present on an early cut continues to a later cut without decreasing its area contribution. New generators can join at past endpoints, adding area; future endpoints in a regular spacetime are excluded. ThereforeThis argument also covers the sum of initially disconnected event horizons of the black holes in a merger. Nonsmooth crease sets require the standard regularity treatment; they do not provide a classical area-decrease loophole under these hypotheses. Quantum negative-energy flux can violate the classical null energy condition, so this is a classical black-hole area theorem, not an unconditional law for evaporating black holes.

A gravitational wave can propagate through a region with vanishing stress-energy tensor; its gravitational field is not a matter current $T_{ab}{m}^b$. Nevertheless, exact axisymmetry gives an axial Killing vector field, and the vacuum identity $d\star dm=0$ makes the Komar angular momentum equal on enclosing spacelike submanifolds in the same homology class. Applying the same identity to the spacetime tube between surrounding spacelike submanifolds at different times gives zero net flux of this charge through the tube.

Consequently gravitational waves in exact axisymmetry carry no net angular momentum about the symmetry axis. They can still carry energy and decrease the total mass. In a mode description, rotationally invariant radiative data have zero azimuthal mode number; exact axisymmetry therefore excludes the axial angular momentum flux present in nonaxisymmetric merger radiation. This does not mean there are no gravitational waves, or that radiation cannot transport angular momentum when the exact symmetry is absent. The relevant conserved charge includes any horizon contributions identified in the preceding Komar angular momentum with inner boundaries identity.

Parallel transport the auxiliary null vector $N$ and a screen basis along each generator of the null geodesic congruence. This keeps the screen-space projector fixed under the corresponding transported derivative. Use the curvature convention $[{\nabla }_c,{\nabla }_b]V_a=-R^d{}_{acb}{V}_d$.

With $B_{ab}={\nabla }_b{U}_a$, differentiation of the affine geodesic equation givesHere the first term comes from commuting the derivative past $U^c$, and the second from the curvature commutator. The constraints $B_{ab}{U}^a=B_{ab}{U}^b=0$ follow respectively from the null normalization and affine geodesic equation. Projecting onto the screen therefore yields the optical matrix equationThe trace of the tidal matrix is $R_{ab}{U}^a{U}^b$; extra terms from replacing the screen trace by the spacetime trace vanish by curvature antisymmetry. Taking the trace givesDecompose the optical tensor as $\hat{B}=\frac{1}{2}\theta I+\hat{\sigma }+\hat{\omega }$. The null shear is symmetric and trace-free and the null twist is antisymmetric. Their cross traces vanish, andThe negative sign in the last term is the identity $\mathrm{tr}({\hat{\omega }}^2)=-{\sum }_{I,J}{\hat{\omega }}_{IJ}^2$ on the positive-definite screen. Substitution proves the four-dimensional Null Raychaudhuri equation:The coefficient is $1/(D-2)$ in $D$ spacetime dimensions; the factor $1/2$ here belongs to the four-dimensional congruence in this question, not the five-dimensional geometry of the next question.

The relevant Penrose singularity theorem states: a connected time-oriented four-dimensional globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a nonempty compact orientable boundaryless trapped surface, and the null convergence condition is future null-geodesically incomplete. Under the Einstein field equations, the null energy condition supplies the curvature hypothesis. A trapped surface here is spacelike and has both future normal null expansions strictly negative. The conclusion is an incomplete null geodesic, not necessarily a divergent curvature invariant at an identifiable point.

Suppose, for contradiction, that every future null geodesic is complete. Write $S$ for the trapped surface andThe equality follows because the causal future of a compact set is closed in a globally hyperbolic spacetime. This future horismos is a closed achronal boundary. It is nonempty: the restriction of a Cauchy time function to compact $S$ has a minimum, and a point at that minimum cannot be chronologically preceded by another point of $S$.

Normalize the two future null normal directions along $S$ against a smooth future timelike field, fixing the affine scale continuously. Their initial null expansions are continuous and strictly negative. Compactness of the normalized normal bundle gives a uniform $c>0$ with ${\theta }_{\pm }\le -c$ everywhere on $S$. The preceding null focusing theorem forces a conjugate point to a spacelike surface along each normal generator within affine length $L=2/c$.

By the supplied boundary-generator result, every point of $E^{+}(S)$ is reached by an orthogonal future null geodesic which has no earlier conjugate point. No such boundary generator can remain on the boundary beyond its first focal point. Every point of $E^{+}(S)$ therefore lies inThe parameter set is compact. Future null completeness makes its geodesic flow defined throughout this common finite interval, and smooth dependence on initial data makes its image $K$ compact. Since $E^{+}(S)$ is closed and contained in $K$, it is compact. This is the compactness of the future horismos of a trapped surface step; a merely pointwise finite bound would not suffice without compactness and uniform normalization.

Choose a smooth complete timelike vector field, obtained if necessary by positive rescaling against a complete auxiliary Riemannian metric. Its inextendible integral curves of a vector field meet a chosen smooth Cauchy hypersurface $C$ exactly once. Projection along those curves defines a continuous map ${\pi }_C:E^{+}(S)\to C$. The achronal set property makes this map an injective function: two boundary points on the same timelike integral curve of a vector field would be timelike related.

An achronal boundary is a topological hypersurface without boundary, even at nonsmooth generator junctions; the submanifold property allowed in the question supplies this fact. Thus domain and codomain are both three-dimensional topological manifolds without boundary. Invariance of domain makes ${\pi }_C(E^{+}(S))$ open in $C$. Compactness makes it closed in the Hausdorff space $C$, and it is nonempty. Connectedness of the spacetime gives connectedness of $C$, so this image is all of $C$. That would make $C$ compact, contradicting the noncompact Cauchy hypersurface hypothesis. The completeness assumption is false, proving the theorem.

The noncompactness hypothesis and strict trapping are essential to this version. Global hyperbolicity cannot simply be omitted in the compactness and projection steps, and nonpositive initial expansion with zeros does not supply the uniform focusing bound used here.

Relative to a chosen asymptotically flat end, define the black hole region bywhere ${\mathcal{I}}^{+}$ is that end's future null infinity in a conformal completion. It consists of events unable to send a future causal signal to that infinity; its boundary is the corresponding future event horizon.

For $0<\alpha <r_0/2$, $h>0$ and $f<0$ on $r_{-}<r<r_{+}$. The gradient of $r$ has norm $g^{ab}{\nabla }_{a}r{\nabla }_{b}r=f$, so it is timelike there. Use the future extension regular in the ingoing coordinates above, reached by an exterior future ingoing null geodesic with $\dot{r}=-E\sqrt{h}<0$. This determines $\nabla r$ to be future timelike in the inter-horizon block. Every nonzero future causal tangent $V$ therefore obeysAt the future outer horizon the gradient becomes future null and gives the corresponding one-way inequality $dr/d\lambda \le 0$. No future signal can cross that horizon outward to the selected exterior. The future inter-horizon block is consequently within $\mathcal{B}$; a future-directed path can leave it only through the inner horizon, not return through the outer horizon to the selected infinity.

This statement requires a branch and an end. The maximal extension also contains time-reversed inter-horizon blocks, where future-directed curves have increasing $r$ and emerge into an exterior: they are white holes, not the selected black hole region. Thus the unqualified claim for every copy of $r_{-}<r<r_{+}$ in an arbitrary extension is false. Analytic continuation past a Cauchy horizon can also introduce other asymptotic ends; the displayed definition explicitly refers to the chosen component of infinity.

To sketch the causal structure, suppress the spacelike $\psi$ circle by its orbit-space projection. The induced three-dimensional metric tensor isThe causal projection along a spacelike circular fiber is exact: any causal curve projects to a $q$-causal curve, and every base curve has a lift $d\psi =\Omega dt$ with exactly its base norm. The diagrams therefore represent this two-dimensional orbit space, not a constant-$\psi$ slice of the three-dimensional submanifold.

There are three nonextremal block types. For $r>r_{+}$, $r_{\ast }$ spans the full real line from the outer horizon to infinity, producing an exterior diamond. For $r_{-}<r<r_{+}$, $r_{\ast }$ spans the full line with the time and space roles reversed; these diamonds contain the future black-hole or past white-hole branches. For $0<r<r_{-}$, $r_{\ast }$ is finite at zero and diverges at the inner horizon, producing a static half-diamond with a timelike singular edge. The supplied Kretschmann scalar diverges as $384{\alpha }^4{r}_0^4/r^{12}$ at $r=0$, and $f>0$ on the inner side, so that edge is a genuine timelike curvature singularity. In contrast, both horizon radii are regular in horizon-adapted coordinates.

The finite strip shows two exterior levels, a future trapped block, an inner static level with timelike singularities, and the next emerging block. Continuing through the inner Cauchy horizons repeats this pattern upward and downward in the maximal analytic extension. The inner boundary is not a spacelike Schwarzschild-type singularity. Such an ideal extension need not describe a physical collapse past its unstable inner horizon.

At $\alpha =r_0/2$, put $R=r_0/\sqrt{2}$. Then $r_{+}=r_{-}=R$, $f=(r^2-R^2{)}^2/r^4$, and the interval $r_{-}<r<r_{+}$ is empty. The lapse is positive on both sides, so there is no inter-horizon trapped diamond. The horizon is a degenerate Killing horizon, with $\kappa =0$. In the exterior $r_{\ast }\to -\infty$ as $r\downarrow R$, while inside $r_{\ast }\to +\infty$ as $r\uparrow R$; more precisely its leading pole is $-\sqrt{2}{R}^2/[4(r-R)]$. Static proper distance to the horizon is infinite. Future ingoing rays still cross it at finite affine parameter, as the regular ingoing metric tensor shows.