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In units $c=1$, a receiver on the hyperbolic motion worldline $t=a^{-1}\sinh (a\tau )$, $x=a^{-1}\cosh (a\tau )$ has retarded null coordinate $p=t-x=-a^{-1}{e}^{-a\tau }$. A right-moving light ray emitted at $(t_0,L)$ has $p=t_0-L$, so it meets the receiver only if $t_0<L$. The photon wave covector is proportional to $(1,-1)$, and contraction with the receiver's four-velocity gives its measured frequency as ${\nu }_0(\dot{t}-\dot{x})={\nu }_0{e}^{-a\tau }$. The intersection condition gives the second displayed equality. Emission at $t_0=L$ lies on the Rindler horizon and never reaches the receiver at finite proper time; $L$ is the supremum of receivable emission times, not an attained latest time.

A null coordinate is a coordinate function $u$ whose gradient is null, $g^{ab}{\partial }_{a}u{\partial }_{b}u=0$. Its level sets are null hypersurfaces. Pairs of such coordinates straighten radial null directions in two-dimensional orbit metrics; a monotone reparametrization of either function preserves its null level sets. Examples include the retarded and advanced null coordinates and the radial Kruskal–Szekeres coordinates.

Use the active sign convention for the linearized coordinate gauge transformation, $h_{\mu \nu }\mapsto h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$. The paper's prescribed generator has precisely this sign; using the passive convention would require its negative. Direct partial differentiation givesThe middle terms cancel when the indices are symmetrized. ThereforeTo test both equations (3), put $H(u)=\tilde{h}(u)={\eta }^{\kappa \sigma }{\tilde{h}}_{\kappa \sigma }(u)$, the trace of the original metric perturbation. Since $n$ is a null vector, $\hat{h}=0$ and its trace-reversed metric perturbation equals $\hat{h}$. Differentiating the result of part (a) givesIn the second identity, the term with $n^2$ is zero. ThusIn general $\hat{h}$ does not satisfy (3). On an open interval of $u$, both equations hold exactly when $H^{\prime \prime }=0$ there: the first equation forces this for $u\ne 0$, and smoothness gives it at $u=0$ as well. A linearized coordinate gauge transformation preserves the linearized vacuum Einstein equations, but only a residual gauge symmetry of linearized gravity preserves the imposed Lorenz gauge in linearized gravity.

For an explicit failure, use the null coordinates of part (c) and take only ${\tilde{h}}_{uv}={\tilde{h}}_{vu}=u^2/4$ nonzero. Then $H=u^2$, the original field obeys (3), but $\varphi =uv/2$ gives ${\partial }^{\mu }{\overline{\hat{h}}}_{\mu u}=u$ and $\square {\hat{h}}_{uu}=2$.

With $u=-t+z$ and $v=t+z$, the Minkowski metric and the specified null vector becomeThe Minkowski wave operator in null coordinates is $\square =4{\partial }_u{\partial }_v+{\partial }_x^2+{\partial }_y^2$. All components of the original metric perturbation still depend only on $u$, since this coordinate change is linear. The result of part (a), differentiated once, readsIt follows that ${\tilde{h}}_{vv}^{\prime \prime }={\tilde{h}}_{vx}^{\prime \prime }={\tilde{h}}_{vy}^{\prime \prime }=0$ and ${\tilde{h}}_{uv}^{\prime \prime }=H^{\prime \prime }/4$. Taking the trace givesExpanding the scalar $\varphi ={\tilde{h}}_{AB}^{\prime \prime }{x}^A{x}^B/2$ in these null coordinates therefore yieldswhere the coefficients are explicitlyThe quadratic transverse term is a harmonic quadratic plane-wave profile. In particular ${\partial }_v\varphi =c$, ${\partial }^{\mu }{\overline{\hat{h}}}_{\mu \nu }=2c{n}_{\nu }$ and $\square \varphi =4c^{\prime }$, consistently reproducing the failure of wave gauge in part (b).

Continue with the active sign convention for the linearized coordinate gauge transformation. Choose local functions $A,B,C,F$ of $u$ by integration so thatFor example, for any fixed $u_0$ one may takeA generator, specified as a covector with components in the null coordinates $(u,v,x,y)$, isIndeed, with $\delta h_{AB}={\partial }_A{\chi }_B+{\partial }_B{\chi }_A$, the mixed components cancel:All components without a $u$ index vanish as well, andSince $n^{♭}=du$, only the $uu$ component needs changing. ThusYes, $\mathring{h}$ satisfies both equations (3). Its trace is zero, ${\partial }_v\psi =0$ gives zero divergence, and the harmonic quadratic plane-wave profile givesThe two surviving functions encode the two gravitational wave polarizations; constant and linear transverse terms have been removed by the explicit linearized coordinate gauge transformation.

First use the original null coordinates $u=t+m/\lambda$, $v=t-m/\lambda$. They obey $u-v=2m/\lambda$, and the radial metric tensor becomesApply the given substitutions $u=\tan (U/2)$ and $v=-\cot (V/2)$. The identitiesgive the analytic extension of a spacetimeThe poles of the individual tangent and cotangent substitutions are now ordinary coordinate lines wherever $\cos ((U-V)/2)\ne 0$. Set $T=(U+V)/2$ and $X=(U-V)/2$. Allowing these coordinate functions to continue across every such line givesThis is the universal cover of two-dimensional anti-de Sitter spacetime of radius $m$, times a two-sphere of radius $m$: the simply connected Bertotti-Robinson spacetime. In defining this maximal analytic extension, $T$ ranges over the whole real line without periodic identification. A periodic identification would instead create closed timelike curves and would be an additional global choice.

For clarity, the original patch is embedded by the branches $-\pi <U<\pi$ and $0<V<2\pi$. The inequality $u>v$ is equivalent on these branches to $|U-V|<\pi$, so the original patch occupiesThe first spacetime extension in part (a) occupies the larger band $0<T-X<2\pi$, with $|X|<\pi /2$. Its finite-$v$ Killing horizon is the interior line $T+X=\pi$. The edges $T-X=0,2\pi$ of that band are further regular Poincaré horizons; continuing through them produces the full conformal strip of two-dimensional anti-de Sitter spacetime.

The original coordinate $\lambda$ has a smooth continuation throughout the strip:Consequently the complete set of zero-level Killing horizons isThese are two alternating families of parallel null lines; different families meet only at the excluded conformal boundaries, never inside the spacetime. The continuation of the original Killing vector field isThe field never vanishes in the interior, including on the Killing horizons.

To obtain the Penrose diagram, suppress the two-sphere and multiply by the conformal factor ${\cos }^{2}X/m^2$. The resulting radial metric tensor is $-d{T}^2+d{X}^2$, so the diagram is an infinitely tall strip of width $\pi$, with null lines at 45 degrees and timelike edges $X=\pm \pi /2$. Generically,At the special edge points where the two families of Killing horizons end, the displayed formula is a $0/0$ limit and $\lambda$ has no unique limiting value. Thus the infinity labels on the diagram refer to the open edge segments, not those vertices. Each interior point of the diagram represents a two-sphere of fixed radius $m$; neither edge is a centre where that two-sphere shrinks.

Finally, these timelike edges are conformal boundaries, not omitted regular points of a further spacetime extension. One way to establish maximality is geodesic completeness. The two-dimensional anti-de Sitter spacetime factor embeds locally as $Z\cdot Z=-m^2$ in a flat space with metric signature $(-,-,+)$. Along an affinely parametrized geodesic with constant $ϵ=\dot{Z}\cdot \dot{Z}$, the ambient equation isIts oscillatory, linear or hyperbolic solutions for negative, zero or positive $ϵ$ exist for all real affine parameters and lift to the universal cover. The two-sphere is also geodesically complete, and geodesics of the product split into geodesics of its factors. Hence $\tilde{g}$ is geodesically complete. A proper regular spacetime extension would furnish a geodesic reaching its boundary in finite affine parameter, contradicting that completeness. This proves that the displayed simply connected analytic spacetime is maximal.

Introduce conformal timeIt satisfies $d\eta /dt=\mathrm{sech}t$ and $\cos \eta =\mathrm{sech}t$, so the metric becomes $d{s}^2={\sec }^2\eta (-d{\eta }^2+d{\xi }^2)$. Choose null coordinatesThus $f(u,v)=\sec [(v-u)/2]$ is positive on $-\pi <v-u<\pi$. Choosing the two null coordinates with these orientations gives the requested positive coefficient of $dudv$.

There is a global-topology distinction in the source: unrestricted $\xi \in \mathbb{R}$ gives the universal cover of two-dimensional de Sitter spacetime. The usual two-dimensional de Sitter spacetime has $\xi$ periodic modulo $2\pi$. Indeed its standard embedding is $X^0=\sinh t$, $X^1=\cosh t\cos \xi$, $X^2=\cosh t\sin \xi$, and is unchanged by $\xi \mapsto \xi +2\pi$. The local metric and the null-coordinate calculation apply to either choice.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $T_H=\kappa /(2\pi )$.

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.

Use the specified hyperbolic trajectory. The emitted right-moving photon follows $x=L+t-t_0$, so its retarded null coordinate is $p=t_0-L$. Interception at astronaut proper time $\tau$ requiresThe astronaut has velocity $v=\tanh (a\tau )$ relative to the star and $\gamma =\cosh (a\tau )$. The collinear relativistic Doppler effect givesThis is the Doppler shift for a uniformly accelerated receiver. A finite reception time exists precisely when $t_0<L$, so the limiting emission time is $L$, approached from below. At $t_0=L$ the photon lies on the Rindler horizon and never reaches the astronaut. If reception is restricted to $\tau \ge 0$, the full emission range is $L-1/a\le t_0<L$; for emission after $t=0$, it is $0\le t_0<L$.

For the standard right-wedge hyperbola and a stationary source at $x=L$ emitting rightmoving photons, the retarded null coordinate is $p=t-x=-e^{-a\tau }/a$. Equating it to the emission value $t_e-L$ gives the displayed reception relation. Differentiating yields the Doppler frequency ratio $\nu /{\nu }_0=d{t}_e/d\tau =e^{-a\tau }$. The limiting emission time $L$ is a supremum: emission exactly at that time is never received at finite proper time. These expressions use $c=1$ and acceleration continued indefinitely.