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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.

This is the future endpoint of complete inertial timelike geodesics in the Minkowski conformal compactification, at $(T,R)=(\pi ,0)$. The endpoint has finite unphysical coordinate time but infinite physical proper time. Accelerated timelike observers need not have this endpoint; a Rindler horizon can accompany an observer asymptotic to a null direction instead.

Every event in complete Minkowski spacetime emits a future-directed null geodesic ending on future null infinity. Therefore its black-hole region $M\setminus J^{-}({\mathcal{I}}^{+})$ is empty and there is no intrinsic event horizon. The null edges of the Minkowski conformal compactification represent infinity, not hidden regions inside the physical spacetime. An accelerated observer can instead have an observer-dependent Rindler horizon.

An observer's future event horizon bounds the events that can ever signal to its future-inextendible worldline. Its past event horizon bounds the events reachable by signals from its past-inextendible worldline. A complete inertial observer in Minkowski spacetime has neither. The uniformly accelerated worldline $t=a^{-1}\sinh (a\tau )$, $x=a^{-1}\cosh (a\tau )$ has $I^{-}(\gamma )=\{t-x<0\}$ and $I^{+}(\gamma )=\{t+x>0\}$. Their boundaries are its two Rindler horizons. This observer dependence is distinct from a black-hole horizon defined using future null infinity.

A horizon separates regions according to a specified causal, dynamical or geometrical criterion. The criterion must be stated: an event horizon, an apparent horizon, a Killing horizon and a Cauchy horizon answer different questions. Some horizons describe a black hole, others an observer's acceleration or the universe's expansion. A vanishing coordinate metric component alone does not establish any of these definitions.

In an asymptotically flat spacetime, choose the relevant future null infinity ${\mathcal{I}}^{+}$, the endpoint of escaping outgoing light rays. The black-hole region and its future event horizon areHere $J^{-}$ is the causal past. Events in $\mathcal{B}$ cannot send a causal signal to the chosen infinity. This is a global definition: identifying the horizon requires the full future development, so a local measurement need not reveal whether an event lies on it. Smooth portions of an event horizon are null hypersurfaces generated by null geodesics. Generators may join at past caustics or crease sets, so the horizon need not be smooth everywhere. A past event horizon is defined by the time-reversed escape criterion; in the maximally extended Schwarzschild solution it bounds a white-hole region.

An observer-based definition replaces infinity by a future-inextendible worldline $\gamma$. Its future observer event horizon is the boundary of events that can signal to $\gamma$, conventionally $\partial I^{-}(\gamma )$. A past observer horizon similarly bounds events reachable from its past worldline. Such horizons depend on the observer and need not imply a black hole.

A Killing horizon is a null hypersurface to which a nonzero Killing vector field $K^a$ is normal. On it, $K^a$ is tangent to its null generators and satisfies $K^b{\nabla }_b{K}^a=\kappa K^a$. The surface gravity $\kappa$ measures the failure of this Killing parameter to be an affine parameter; rescaling $K$ rescales $\kappa$. A degenerate Killing horizon has $\kappa =0$. Stationary black-hole horizons often are Killing horizons, whereas a general evolving event horizon need not have any Killing symmetry. The condition $K^a{K}_a=0$ alone is insufficient: the null vector must also be normal to a null hypersurface.

The Schwarzschild spacetime illustrates several of these notions. In units $G=c=1$, let $f=1-2M/r$. Its metric is $d{s}^2=fd{t}^2-f^{-1}d{r}^2-r^{2}d{\Omega }^2$, and $r=2M$ is the future event horizon of the black-hole region in the appropriate extension. The apparent coordinate singularity disappears in ingoing Eddington-Finkelstein coordinates, $v=t+r_{\ast }$, $d{r}_{\ast }/dr=f^{-1}$:At $r=2M$ this metric is regular and the normal $dr$ is null. Future outgoing radial rays satisfy $dr/dv=f/2$; inside the horizon their areal radius decreases even on the outgoing family. The stationary Killing vector field ${\partial }_v$ generates the horizon with $\kappa =f^{\prime }(2M)/2=1/(4M)$ when normalized at infinity. The Kretschmann scalar is $48M^2/r^6$, finite at the horizon and divergent at $r=0$. Thus the event horizon is not the curvature singularity. The photon sphere at $r=3M$, which supports circular null orbits, is also not the horizon.

Quasi-local horizon notions use null expansions. For a closed spacelike two-surface, choose future outgoing and ingoing null normals $\ell ,n$, with $\ell \cdot n=1$ in signature $(+---)$. Each expansion is the logarithmic rate of its positive area element along the corresponding null normal. A future trapped surface has ${\theta }_{\ell }<0$ and ${\theta }_n<0$: both future light fronts initially shrink. A marginally outer trapped surface has ${\theta }_{\ell }=0$; for the future black-hole type also require ${\theta }_n<0$. On a chosen spacelike slice, the outer boundary of the future trapped region is an apparent horizon, ordinarily an outermost marginally outer trapped surface when regularity permits this characterization. It depends on the slice, and a slicing need not display an apparent horizon even when the global spacetime contains an event horizon.

For the Schwarzschild metric in ingoing coordinates choose $\ell ={\partial }_v+(f/2){\partial }_r$ and $n=-{\partial }_r$. The spherical null expansion in ingoing coordinates givesThe round spheres inside $2M$ are future trapped, while $r=2M$ is marginal. On standard horizon-penetrating slices its sections give the apparent horizon, coinciding with the stationary event and Killing horizon. This coincidence in a symmetric stationary example does not identify the notions in a general spacetime.

A trapping horizon is a hypersurface foliated by marginal surfaces. Its future outer type obeys ${\theta }_{\ell }=0$, ${\theta }_n<0$ and ${\mathcal{L}}_n{\theta }_{\ell }<0$. Unlike an event horizon, this can be diagnosed from local expansion data and their variation. The null energy condition explains its causal character. At a marginal surface, the Null Raychaudhuri equation, with the present curvature sign and vanishing null twist, givesin units $c=1$. A possible nonaffinity term vanishes here because it is proportional to the zero expansion. Write a horizon evolution vector as $V=\ell -Cn$. Preserving ${\theta }_{\ell }=0$ gives $C=({\mathcal{L}}_{\ell }{\theta }_{\ell })/({\mathcal{L}}_n{\theta }_{\ell })\ge 0$. Hence $V^2=-2C\le 0$: the future outer horizon is spacelike or null. Its area expansion is ${\theta }_V=-C{\theta }_n\ge 0$. These conclusions use the outer condition and the energy condition, not merely marginality.

For the global event horizon, the future-complete horizon focusing proof of the area theorem uses the same Null Raychaudhuri equation. Under the null energy condition, a negative generator expansion forces a future caustic in finite affine parameter. Future completeness and the usual regularity or predictability assumptions exclude such a caustic on the future event-horizon boundary. Its generator expansion is therefore nonnegative and its classical area cannot decrease. These assumptions matter: this is not an unconditional result for every null surface or every matter model.

Rotating and charged solutions provide further distinctions. A nonextremal Kerr black hole has $r_{\pm }=M\pm \sqrt{M^2-a^2}$, with $a=J/M$. Its outer Killing horizon is generated by ${\partial }_t+{\Omega }_H{\partial }_{\varphi }$, where ${\Omega }_H=a/(r_{+}^2+a^2)$, rather than by time translation alone. The ergosphere begins at the stationary-limit surface $g_{tt}=0$, whose radius is $M+\sqrt{M^2-a^2{\cos }^2\theta }$. Away from the axis this lies outside the event horizon: it forbids static observers but does not forbid escape to infinity. In Reissner-Nordstrom spacetime, the nonextremal radii are $r_{\pm }=M\pm \sqrt{M^2-Q^2}$. The outer surface is an event horizon, whereas the inner surface is a Cauchy horizon in the exact analytic extension.

A Cauchy horizon concerns predictability from initial data. The future domain of dependence $D^{+}(\Sigma )$ of a partial Cauchy surface consists of events for which every past-inextendible causal curve intersects $\Sigma$. Its future Cauchy horizon isBeyond it, causal influences can arrive without meeting the specified initial surface; data on that surface no longer determine the solution there. This is distinct from the inability to send signals to infinity. The inner horizon of the exact charged black hole supplies an example, and the definition necessarily specifies which initial surface is being used.

Flat spacetime itself has acceleration horizons. In the right Rindler wedge, set $t=\xi \sinh (a\tau )$ and $x=\xi \cosh (a\tau )$, with $\xi >0$. The Minkowski spacetime metric becomes $d{s}^2=a^2{\xi }^{2}d{\tau }^2-d{\xi }^2-d{y}^2-d{z}^2$. The null boundaries $t=x$ and $t=-x$ are Rindler horizons of the boost Killing flow. A uniformly accelerated observer cannot receive signals from one side of its future horizon, even though the curvature vanishes and inertial observers encounter no such boundary. These are observer and Killing horizons, not a black-hole event horizon defined using the full Minkowski null infinity.

In a spatially flat Friedmann-Lemaître-Robertson-Walker metric with noncompact spatial sections, radial light rays have $d\chi /dt=\pm 1/a(t)$. The particle horizon measures how far signals could have travelled since the initial cosmic time; the cosmological event horizon measures how far signals emitted now could ever travel toward a comoving observer. Their comoving and proper radii areThe respective horizon is present only if its integral is finite. The event-horizon integral uses the actual future endpoint, often $t_f=\infty$, so it is global and observer-dependent. For $a(t)=t^q$ on $0<t<\infty$, a finite particle horizon exists for $0<q<1$, while a finite event horizon exists for $q>1$. Neither integral is finite for $q=1$. In the expanding flat patch of de Sitter spacetime, $a=e^{Ht}$ and the proper event-horizon radius is $H^{-1}$. The static form $d{s}^2=(1-H^2{r}^2)d{t}^2-(1-H^2{r}^2{)}^{-1}d{r}^2-r^{2}d{\Omega }^2$ identifies it as a cosmological Killing horizon; its surface-gravity magnitude is $H$ with the central static time normalization.

Finally, the Hubble radius must not automatically be called an event horizon. For a flat expanding cosmology, the two future radial null expansions of a sphere of proper radius $R$ are ${\theta }_{\pm }=2(H\pm 1/R)$. At $R=H^{-1}$ the ingoing expansion vanishes, while outside it both expansions are positive: this is the cosmological marginal, past-trapped type rather than a future black-hole trapped surface. A dust universe with $a\propto t^{2/3}$ has such a Hubble sphere but no future event horizon. Global causal boundaries, slice-dependent trapped-surface boundaries, symmetry horizons and initial-data boundaries must therefore be distinguished, even when several coincide in a special example.

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.

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

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

The resulting Minkowski conformal compactification has a triangular radial Penrose diagram. The two sloping boundary edges areThese are future null infinity and past null infinity. The three vertices are future timelike infinity $i^{+}=(\pi ,0)$, past timelike infinity $i^{-}=(-\pi ,0)$ and spacelike infinity $i^0=(0,\pi )$. They are ideal boundary points, not events of physical Minkowski spacetime.

For a complete inertial timelike geodesic, $\mathbf{x}=\mathbf{b}+\mathbf{v}t$ with $|\mathbf{v}|<1$. As $t\to +\infty$, both $t-r$ and $t+r$ tend to $+\infty$, so $p,q\to \pi /2$ and the curve ends at $i^{+}$. As $t\to -\infty$, both tend to $-\infty$, giving $i^{-}$. These endpoints occur at infinite physical proper time.

For a complete null geodesic, write $\mathbf{x}=\mathbf{b}+\mathbf{n}t$, $|\mathbf{n}|=1$. To the future, $r=t+\mathbf{n}\cdot \mathbf{b}+O(1/t)$, so $u\to -\mathbf{n}\cdot \mathbf{b}$ is finite and $v\to +\infty$. Thus $q\to \pi /2$ with $p$ finite: the endpoint lies on ${\mathcal{I}}^{+}$. To the past, $v\to -\mathbf{n}\cdot \mathbf{b}$ remains finite while $u\to -\infty$, giving ${\mathcal{I}}^{-}$. Physical affine parameter is infinite at both ends. Along a radial null path, $dT=\pm dR$, so the diagram represents its null directions at $45$ degrees.

For a complete spacelike geodesic, take $t=t_0+a\lambda$, $\mathbf{x}=\mathbf{b}+\mathbf{w}\lambda$ with $|\mathbf{w}|>|a|$. At either end $r>|t|$ asymptotically, so $u\to -\infty$, $v\to +\infty$ and $(T,R)\to (0,\pi )$. Both ends therefore reach $i^0$, again at infinite physical affine parameter. These are the geodesic endpoints in Minkowski conformal compactification.

Only conformal preservation of null geodesic paths is guaranteed. The pictured timelike and spacelike curves are images of physical geodesics, not generally geodesics of the unphysical metric. For a null path, an unphysical affine parameter satisfies $d\tilde{\lambda }/d\lambda \propto {\Omega }^2$. Since $\Omega =O(1/|\lambda |)$ near a generic null-infinity endpoint, the unphysical parameter can converge while the physical parameter diverges. At $i^{\pm }$ and $i^0$, the usual smooth-boundary condition $d\Omega \ne 0$ degenerates; these vertices should not be treated as regular points of null infinity.

Complete Minkowski spacetime has no black-hole event horizon. Every physical event can send an outgoing null signal to ${\mathcal{I}}^{+}$, so the black-hole region $\mathcal{M}\setminus J^{-}({\mathcal{I}}^{+})$ is empty. There is likewise no particle or future event horizon for an eternal inertial observer: any finite event can eventually signal that observer. Null infinity is a conformal boundary, not a physical horizon. Uniformly accelerated observers can have Rindler horizons, but that is an observer-dependent restriction and is not a horizon of the full inertial Minkowski spacetime.

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.

Assume $qE>0$ and putThe nonzero four-velocity equations arewith $u^0(0)=c$ and $u^x(0)=0$. HenceIntegrating from the origin gives the uniformly accelerated trajectory

The light ray has $x_L=-h+ct$. At interception, $x_L=x$, soA finite solution exists exactly whenThenand, writing $H=\alpha h/c$,At $h=h_c$ interception is approached only as $t\to \infty$, and for $h>h_c$ it never occurs. The limiting backward light ray is the Rindler horizon of the accelerated particle.

Differentiating the supplied coordinates givesSubtract the squares. The cross terms cancel and ${\cosh }^2-{\sinh }^2=1$, yieldingThe spacetime is flat Minkowski spacetime in Rindler coordinates. $x>0$ covers $X>|cT|$, the right Rindler wedge, while $x<0$ covers the left wedge $X<-|cT|$; neither is the whole spacetime.

The null combinations areFor $x=C{e}^{-\alpha t}$ the first is constant; for $x=C{e}^{\alpha t}$ the second is constant. Thus part (a)'s exponentials are simply the inertial light rays $X\pm cT=\text{constant}$. Their coordinate-time asymptotes reflect the chart's Rindler horizon, not a failure of light to move at $c$ in an inertial frame.

Let $a>0$ be the constant proper acceleration away from the emitter and set $R=c^2/a$. Take the signal emission to be simultaneous, in the initial inertial frame, with the ship being at rest and its clock reading $\tau =0$. Place that initial ship event at $T=0$, $X=R$. Its hyperbolic motion isA photon emitted from $X=R-d$ at $T=0$ towards the departing ship follows $X_{\gamma }=R-d+cT$. At reception,Therefore light reception by a uniformly accelerated observer occurs afterThe same result follows in Rindler coordinates: the ship stays at $x=R$, the emitter starts at $x=R-d$, and the right-moving null ray has $x=(R-d)e^{\alpha t}$. The onboard lapse $d\tau =\alpha Rdt/c$ converts its coordinate catch-up time to the expression above.

If $d=c^2/a$, the ray lies on the limiting null horizon and never intersects it at finite proper time; if $d>c^2/a$, it never reaches the ship. The reception time diverges as the emitter approaches this Rindler horizon. For $ad/c^2\ll 1$, ${\tau }_{\mathrm{rec}}=d/c+a{d}^2/(2c^3)+\cdots$, recovering the inertial light-travel time. A different emission time or initial relative velocity would require specifying that extra datum; the formula uses the initial-rest simultaneous-emission convention stated above.

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 $0\le t_0<L$.