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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.

Write $f=1-2M/r$ in the Schwarzschild metric. For the stationary Killing vector field $V={\partial }_t$, its dual is $V^{♭}=-fdt$ and the Papapetrou electromagnetic field isOutside the Schwarzschild event horizon, use the static orthonormal coframe $e^{\hat{0}}=\sqrt{f}dt$, $e^{\hat{r}}=dr/\sqrt{f}$, $e^{\hat{\theta }}=rd\theta$, $e^{\hat{\varphi }}=r\sin \theta d\varphi$. Then $F=(2M/r^2)e^{\hat{0}}\wedge e^{\hat{r}}$. With the electric field convention $E_{\hat{i}}=F_{\hat{i}\hat{0}}$ from the preceding part,Thus the stationary Killing vector field gives a Coulomb electric field, directed inward in this convention. If the charge is normalized by $E_{\hat{r}}=Q/r^2$, its charge is $Q=-2M$. A convention with $F\mapsto -F$ reverses that charge; the invariant content is a radial monopole electric field of magnitude $2M/r^2$.

For the axial Killing vector field $V={\partial }_{\varphi }$, $V^{♭}=r^2{\sin }^2\theta d\varphi$, givingIn the same static orthonormal coframe, this is a purely magnetic field withAt large $r$ these are the spherical components of $2{\mathbf{e}}_z$. This is therefore the regular magnetic test field asymptotic to a uniform axial field, distorted by the Schwarzschild black hole. Its net magnetic monopole flux is zero because the integral of $\cos \theta$ over a sphere vanishes. Both electromagnetic fields are regular at the future event horizon: in Ingoing Eddington-Finkelstein coordinates, the electric field is $(2M/r^2)dv\wedge dr$, and the magnetic two-form already contains only regular spatial coordinates. Static observers themselves cease to exist on the event horizon.

Write $M=-a^2$ with $a>0$. On a constant-time slice, the proper distance from the axis to radius $r$ and the circumference at that radius areHence $C/s\to 2\pi a$. A smooth axis in a Riemannian metric requires the Euclidean limit $C/s\to 2\pi$, so regularity forces $a=1$, or $M=-1$. For $a\ne 1$ the conical singularity of negative-mass BTZ geometry has deficit angle $2\pi (1-a)$; for $a>1$ this is an angular excess. Even integer $a>1$ is singular with the stipulated angular period $2\pi$: a multiple covering does not make the tip a smooth point.

At $M=-1$, set $r=\ell \sinh \chi$. The metric tensor becomeswhich is regular at $\chi =0$ and describes Anti-de Sitter spacetime with nonperiodic time. For every $M<0$, $f=a^2+r^2/{\ell }^2$ stays positive. An outward radial null geodesic obeys $dt/dr=1/f$, and reaches the timelike conformal boundary in finite coordinate time, since ${\int }_0^{\infty }dr/f=\pi \ell /(2a)$. Thus no event horizon hides the conical axis. The geometry is nakedly singular except at $M=-1$. Its local curvature away from the axis is that of Anti-de Sitter spacetime; the argument is a global angular-identification test, not curvature blowup.

An event horizon is a null hypersurface. For its spherical defining function $H(v,r)=r-r_H(v)$, the inverse radial metric tensor has $g^{vv}=0$, $g^{vr}=1$ and $g^{rr}=f=1-2m(v)/r$. The normal's null condition isIt follows that the event horizon follows the outgoing radial null geodesics of the Vaidya metric:This ordinary differential equation also describes other outgoing null hypersurfaces; the global event horizon is selected by its future boundary condition, not by this local equation alone.

In the final Schwarzschild spacetime, an outgoing ray with $r>2M_1$ escapes to future null infinity, whereas one with $r<2M_1$ cannot do so and ends at the Schwarzschild singularity. Their boundary is the outgoing null geodesic $r=2M_1$. Since the final region continues indefinitely, the boundary of the causal past of future null infinity there is exactly that Schwarzschild event horizon. Equivalently, integrating the outgoing equation backward from this final boundary selects the complete event horizon. ThusThe original Penrose diagram below shows the right-hand exterior and black-hole interior appropriate to ingoing coordinates. The green band is the ingoing accretion interval; the dashed blue curve is the spherical apparent horizon $r=2m(v)$. Outgoing null geodesics escaping to future null infinity lie to the right of the red event horizon; those to its left terminate on the future spacelike singularity. In the final stationary region the two horizons coincide. The red event horizon already lies outside $r=2M_0$ before the ingoing matter arrives, because its position is fixed by the future escape criterion.

Let $\Delta M=M_1-M_0$ and retain only terms first order in $\mu /M_0$ and $\rho /M_0$. The outgoing event horizon equation expands toHere ${\kappa }_0$ is the initial Schwarzschild surface gravity. The integrating factor $e^{-{\kappa }_{0}v}$ gives $d(e^{-{\kappa }_{0}v}\rho )/dv=-e^{-{\kappa }_{0}v}\mu /(2M_0)$. Future stationarity imposes $\rho =2\Delta M$ for $v>v_0$ and removes the growing homogeneous solution. Integrating from $v$ to infinity yields the teleological response of a Vaidya event horizon:The integral is finite because $\mu$ is bounded. Integrating by parts also givesFor $v>v_0$, this reduces to $2\Delta M$. For $v<0$, the first form becomesConsequently $r_H\to 2M_0$ as $v\to -\infty$. The positive constant $C$ shows that this event horizon is already expanding in the initially vacuum region. For monotone accretion, $\rho -2\mu \ge 0$, so it lies outside the instantaneous apparent horizon to first order. All displayed radius corrections are understood up to $O((\Delta M{)}^2/M_0)$.

The stress-energy tensor is already first order, so its energy flux can be evaluated on the unperturbed Schwarzschild event horizon $r=2M_0$, with the infinity-normalized Killing vector field $\xi ={\partial }_v$. On this background $g^{rv}=1$, and henceThe conserved stress-energy current from a Killing vector is $J^a=-T^a{}_b{\xi }^b$. Its radial component is negative, so the positive inward energy flux is $-J^r$. Using the directed null-surface measure $r_H^{2}dvd\Omega$ givesThe initial and final event horizon areas are $A_0=16\pi M_0^2$ and $A_1=16\pi (M_0+\Delta M{)}^2$. Thus $\Delta A=32\pi M_0\Delta M+O((\Delta M{)}^2)$. With Schwarzschild surface gravity ${\kappa }_0=1/(4M_0)$, the Physical-process first law of black-hole mechanics follows:The flux is Killing energy, including gravitational redshift, rather than the energy measured by a sequence of static observers arbitrarily close to the event horizon.

Work initially in units $G=c=\hslash =k_B=1$ and consider gravitational collapse settling to a nonextremal Schwarzschild black hole. Classically, outgoing rays sufficiently close to the forming event horizon suffer an arbitrarily large redshift. Quantum mechanically, the meaning of positive frequency at past null infinity differs from that at future null infinity; a Bogoliubov transformation between these two mode descriptions can therefore mix creation operators and annihilation operators. Hawking's argument computes that mixing and obtains a thermal occupation, even when the incoming state has no particles.

To see why the frequency mixing is universal, use final Schwarzschild time $t$, the Schwarzschild tortoise coordinate $r_{\ast }$, and retarded time $u=t-r_{\ast }$. Near the final event horizon, $r_{\ast }\sim (2\kappa {)}^{-1}\log (r-2M)$, with Schwarzschild surface gravity $\kappa =1/(4M)$. On a fixed advanced-time section, $u=v-2r_{\ast }$, so $r-2M$ is proportional to $e^{-\kappa u}$. Tracing a ray back through the regular collapsing region maps this small separation smoothly to its separation from the last escaping incoming ray. If $U$ is an affine incoming null coordinate and $U_H$ labels that limiting ray, the Hawking exponential ray map isThe logarithm at a simple horizon root fixes the exponent. Smooth propagation through the earlier collapse changes $A$ and phases but not the leading exponential redshift.

For a massless bosonic test scalar field, an outgoing mode of frequency $\omega >0$ at future null infinity behaves as $e^{-i\omega u}$. Trace it backward in the high-frequency geometric optics approximation. Its relevant incoming dependence is, up to normalization and a phase,Here $\Theta$ is the Heaviside step function. Incoming rays with $U>U_H$ do not escape. This truncated logarithmic phase is not purely positive frequency with respect to $U$. Decomposing it into incoming positive-frequency solutions $e^{-i{\omega }^{\prime }U}$ and negative modes $e^{+i{\omega }^{\prime }U}$ gives the Bogoliubov transformation. With $x=U_H-U$, the two relevant Fourier transform integrals areThe former is the positive-frequency coefficient and the latter is the negative-frequency coefficient; the Klein-Gordon inner product introduces the same frequency normalization into their modulus ratio. By the defining integral of the Gamma function and analytic continuation in its Laplace parameter,For a complex number $z$ in the right half-plane, $|z^{-1-ia}|=|z{|}^{-1}{e}^{a\arg z}$. As $ϵ\downarrow 0$, the two arguments tend to $+\pi /2$ and $-\pi /2$. This gives the thermal ratio of Hawking Bogoliubov coefficients:The regulator fixes the branches and is essential to the sign of the thermal exponent.

Expand the outgoing annihilator as $b=\int d{\omega }^{\prime }({\alpha }_{\omega {\omega }^{\prime }}^{\ast }{a}_{{\omega }^{\prime }}-{\beta }_{\omega {\omega }^{\prime }}^{\ast }{a}_{{\omega }^{\prime }}^{†})$, including a complete set of incoming channels. The canonical identities for a bosonic Bogoliubov transformation express $[b,b^{†}]=1$. For normalized outgoing wave packets, they give $\int (|\alpha {|}^2-|\beta {|}^2)d{\omega }^{\prime }=1$. In the incoming Fock vacuum, the outgoing mean of the particle number operator is $N_{\omega }=\int |\beta {|}^{2}d{\omega }^{\prime }$. Combining the two identities with the thermal ratio, in the late-time narrow-frequency packet limit, yieldsWave packets avoid interpreting the divergent normalization of infinitely long continuum modes as a finite particle count. The temperature is the Hawking temperature, and the occupation is the zero-chemical-potential Bose-Einstein distribution. Restoring constants, the Schwarzschild black hole temperature is $T_H=\hslash c^3/(8\pi GM{k}_B)$ for physical mass $M$.

The preceding thermal occupation describes the horizon-originating channel before exterior scattering. The exterior angular and curvature potential partly reflects it. For a scalar field, the number flux at future null infinity is consequentlywhere the greybody factor ${\Gamma }_{\ell }$ is the transmission probability of a partial wave. Multiplication by $\omega$ gives its energy flux. Thus the asymptotic spectrum has the thermal denominator but is not an exact blackbody spectrum. For fermionic fields the canonical anticommutation relations replace the minus sign in the occupation denominator by a plus sign, at the same Hawking temperature.

The relevant state is the collapse vacuum for Hawking radiation: no incoming thermal bath at past null infinity, and regular short-distance behavior in freely falling coordinates at the forming event horizon. It predicts an outgoing flux, unlike equilibrium with a thermal bath on an eternal black hole. The derivation uses quantum field theory in curved spacetime on a prescribed classical geometry; it does not require outgoing particles to follow forbidden classical paths out of the interior. Positive energy carried to infinity is accompanied, in the semiclassical description, by negative Killing-energy flux into the black hole. Including that flux in slow backreaction decreases its mass, giving black-hole evaporation.

The approximation applies to the late radiation of a large nonextremal black hole while its surface gravity changes slowly on a time scale ${\kappa }^{-1}$. Backward propagation involves very high locally measured frequencies, so the calculation assumes the usual regular short-distance quantum-field state. It determines the leading radiation and temperature, not the Planck-scale endpoint of black-hole evaporation. The exponential horizon redshift forces positive/negative frequency mixing with a thermal ratio, giving outgoing Hawking radiation at $T_H=\kappa /(2\pi )$.

For nonzero $a$, $\Sigma =0$ requires $r=0$, $\theta =\pi /2$. Oblate Cartesian coordinates satisfy $x^2+y^2=(r^2+a^2){\sin }^2\theta$, $z=r\cos \theta$, so this is the Kerr ring singularity of radius $|a|$. Curvature components and generic invariants diverge there. The whole $r=0$ surface is not singular: away from the ring it is a disk through which the extension continues to a sheet with $r<0$. This oblate radial parameter is not a nonnegative Euclidean distance.

In $G=c=1$ units, $\boxed{J=Ma}$. Horizon candidates solve $\Delta =0$, giving $r_{\pm }=M\pm \sqrt{M^2-a^2}$. For $J^2>M^4$, equivalently $|a|>M$, there are no real roots. The over-rotating Kerr solution has no event horizon hiding its naked ring singularity. The term black hole in this regime names the Kerr family, not a hidden-singularity black hole.

The future domain of dependence $D^{+}(\Sigma )$ consists of points through which every past-inextendible causal curve meets the partial Cauchy hypersurface $\Sigma$. Data on $\Sigma$ determine evolution there for suitable hyperbolic equations. Its future boundary is the Cauchy horizon $H^{+}(\Sigma )$; one precise definition is $\overline{D^{+}(\Sigma )}\setminus I^{-}(D^{+}(\Sigma ))$.

For a nonextremal Reissner-Nordstrom spacetime, $0<|Q|<M$, the metric function is $F=1-2M/r+Q^2/r^2$, with $r_{\pm }=M\pm \sqrt{M^2-Q^2}$. The outer event horizon lies at $r_{+}$. The future trapped region leads toward the inner null surface $r_{-}$, a future Cauchy horizon for bridge data on the illustrated $\Sigma$. Beyond it some past causal curves arrive from other analytic blocks or timelike singularities without meeting $\Sigma$. The CP diagram repeats these analytic blocks; $r=0$ is a timelike curvature singularity.

Near the inner horizon, arbitrarily late exterior radiation is compressed into finite infaller proper time. Its frequency is exponentially blueshifted, with scale $e^{|{\kappa }_{-}|v}$, so even a decaying flux can have diverging measured stress. Generic ingoing and outgoing perturbations produce mass inflation, growing internal mass and curvature. The smooth RN Cauchy horizon is expected to be unstable to back-reaction. This is a generic instability argument, not a universal theorem from every specially tuned test trajectory; the approximation of a harmless test particle becomes inconsistent.

A null geodesic congruence is a smooth family of null geodesics filling a region without intersections there. For affine tangent $k$, the null expansion is the trace of the optical tensor: $\theta =d(\log A)/d\lambda$ for an infinitesimal pencil's area. Positive expansion means spreading; negative expansion means focusing.

For hypersurface-orthogonal generators in four dimensions, the Null Raychaudhuri equation readsThe null twist vanishes. The null energy condition implies $R_{ab}{k}^a{k}^b\ge 0$ through the Einstein field equations. An initial ${\theta }_0<0$ would force a focal point within affine distance at most $2/|{\theta }_0|$. A future-complete generator cannot develop such a point while remaining on an achronal event horizon. Thus $\boxed{\theta \ge 0}$ under the energy and global predictability/future-completeness hypotheses of Hawking's area theorem. Quantum energy-condition violations or failure of those global assumptions remove the conclusion.

The event horizon is the global boundary $\partial J^{-}({\mathcal{I}}^{+})$ and depends on the whole future. An apparent horizon on a chosen spacelike slice is the outermost marginally outer trapped surface, ordinarily ${\theta }_{\mathrm{out}}=0$, ${\theta }_{\mathrm{in}}<0$. It is quasi-local and slice-dependent; its tube need not be null.

For a shell arriving at $v=0$, the final horizon is $r=2M$. Before arrival, outgoing flat-space rays obey $dr/dv=1/2$. Tracing this horizon backward gives $r=2M+v/2$ until its beginning at $r=0$, $v=-4M$. No apparent horizon exists in that earlier flat region, explicitly illustrating the event horizon's dependence on future collapse.