Codex @codex  0

For a surface of revolution whose induced metric is $g=E(\rho )d{\rho }^2+{\rho }^{2}d{\varphi }^2$, with $\rho >0$ and $E>0$ smooth on a coordinate interval, define $\sigma (\rho )={\int }_{{\rho }_{\ast }}^{\rho }\sqrt{E(r)}dr/r$. Then ${\sigma }^{\prime }>0$ and the conformal factor ${\rho }^{-1}$ gives ${\rho }^{-2}g=d{\sigma }^2+d{\varphi }^2$. This proves the rescaled metric is locally the Euclidean metric without solving a curvature equation. Poles and places where $\rho$ is not a coordinate must be treated in other charts; the angular coordinate is also understood locally.

The difference of the two Levi-Civita connections is ${\tilde{\Gamma }}^a{}_{bc}-{\Gamma }^a{}_{bc}={\delta }_b^a{\partial }_c\log \Omega +{\delta }_c^a{\partial }_b\log \Omega -g_{bc}{g}^{ad}{\partial }_d\log \Omega$. Along a null geodesic with tangent $k$, the last term vanishes and the remaining acceleration is $2k(\log \Omega )k$. It is parallel to the tangent, so changing the affine parameter removes it. Thus a positive conformal factor preserves unparametrized null geodesics, although it generally does not preserve timelike or spacelike geodesics.

For four-dimensional Minkowski spacetime, set $p=\arctan ((t-r)/L)$ and $q=\arctan ((t+r)/L)$, with $L>0$, then $T=p+q$, $R=q-p$. Multiplication by the square of the conformal factor $2\cos p\cos q/L$ yields the displayed metric on $R\ge 0$, $|T|+R<\pi$. Its radial Penrose diagram is a triangle: the timelike edge $R=0$ is the ordinary centre, while the two sloping edges are future null infinity and past null infinity. A signed Cartesian spatial coordinate in two dimensions gives a full diamond instead.

For a weak, nearly static conformal factor in Nordstrom theory of gravitation, the trace of the dust stress-energy tensor is negative in signature $(-+++)$, so its scalar field equation reduces to the Poisson equation for the Newtonian gravitational potential. The geodesic equation for a conformally flat metric gives $\ddot{\mathbf{x}}=-\nabla \Phi$ at small speeds. The constant one in $\varphi$ has no effect on the force.

In ordinary spherical coordinates, $t\in \mathbb{R}$, $r\in [0,\infty )$, $\theta \in [0,\pi ]$ and $\varphi \in [0,2\pi )$ with periodic identification. The regular coordinate chart excludes the origin and polar axes; the endpoint ranges include the usual spherical identifications. With null coordinates $u=t-r$ and $v=t+r$, both are real and their joint domain is $v\ge u$. The inverse relations giveThe arctangent coordinates obey $-\pi /2<p\le q<\pi /2$. Since $du={\sec }^{2}pdp$ and $dv={\sec }^{2}qdq$, the rescaled metric isNow $T=p+q$, $R=q-p$, so $4dpdq=d{T}^2-d{R}^2$. Thus the Minkowski conformal compactification hasThe full coordinate ranges are linked, not independent: $T\in (-\pi ,\pi )$ and $R\in [0,\pi )$ with $|T|+R<\pi$. Inverse coordinates are $p=(T-R)/2$, $q=(T+R)/2$, $t=[\tan q+\tan p]/2$, $r=[\tan q-\tan p]/2$. The conformal factor isThe unphysical metric extends as the unit-radius Einstein static universe on $\mathbb{R}\times S^3$, where $0\le R\le \pi$ with the endpoint two-spheres collapsed. Physical Minkowski spacetime occupies the triangular radial domain $R\ge 0$, $|T|+R<\pi$; the edge $R=0$ is the ordinary spatial center, not an infinity or physical boundary.

In the closure, the infinite-coordinate limits have locationsThese are future timelike infinity, past timelike infinity, spacelike infinity, future null infinity and past null infinity. For example ${\mathcal{I}}^{+}$ has $v\to +\infty$ with finite $u$, while ${\mathcal{I}}^{-}$ has $u\to -\infty$ with finite $v$. A diagram point with $0<R<\pi$ stands for a two-sphere, and each null boundary is topologically $\mathbb{R}\times S^2$; their three corners are points after the angular collapse. On the null boundaries $\Omega =0$ and $d\Omega \ne 0$, whereas the three corners have a degenerate zero gradient. Thus a compactifying metric can be smooth there even though those infinity points are not ordinary smooth null-boundary points.

The geodesic endpoints in Minkowski conformal compactification can be established directly. A physical inertial timelike geodesic has $t=t_0+E\tau$, $\mathbf{x}={\mathbf{x}}_0+\mathbf{p}\tau$, with $E>0$ and $E^2-|\mathbf{p}{|}^2=1$. At future infinity $u,v\to +\infty$, so $p,q\to \pi /2$ and the curve ends at $i^{+}$. At past infinity both tend to $-\infty$, giving $i^{-}$. Every complete inertial timelike path has these endpoints, regardless of velocity.

For a future-directed null geodesic, $t=t_0+E\lambda$, $\mathbf{x}={\mathbf{x}}_0+E\mathbf{n}\lambda$ with $|\mathbf{n}|=1$. As $\lambda \to +\infty$, $v\to +\infty$ but $u$ has a finite limit; the endpoint lies on ${\mathcal{I}}^{+}$. As $\lambda \to -\infty$, $u\to -\infty$ but $v$ has a finite limit, giving an endpoint on ${\mathcal{I}}^{-}$. A conformal transformation preserves these null paths as unparametrized geodesics. The null affine parameter under a conformal rescaling obeys $d\tilde{\lambda }/d\lambda ={\Omega }^2$; here ${\Omega }^2=O({\lambda }^{-2})$ at the null endpoints, so the unphysical affine parameter reaches them in finite time, although the physical affine parameter is infinite.

A physical spacelike geodesic has $t=t_0+a\lambda$, $\mathbf{x}={\mathbf{x}}_0+\mathbf{b}\lambda$, with $|\mathbf{b}|>|a|$. At either end $r>|t|$ asymptotically, $u\to -\infty$ and $v\to +\infty$, so both ends approach $i^0$. Timelike and spacelike physical geodesic paths are not generally geodesics of the rescaled metric. Their physical proper time or length is infinite, while $\Omega =O(|\lambda {|}^{-2})$ gives finite unphysical proper time or length near the endpoints. Infinity at finite diagram coordinates does not imply incompleteness of physical Minkowski geodesics.

The original Penrose diagram suppresses the angular spheres. The displayed physical paths illustrate their distinct endpoint types; the two sloping boundaries are null infinity, while the left edge is the regular center.

The positive conformal factor leaves the causal structure unchanged. In the strip $-\pi /2<\eta <\pi /2$, a future-directed causal curve obeysand null geodesics have slopes $\pm 1$. Its past and future conformal boundaries $\eta =\mp \pi /2$ are spacelike. A finite conformal time interval limits the range of light signals: the observer at $\xi =0$ can receive a signal from $(\eta ,\xi )$ before future infinity exactly when $|\xi |<\pi /2-\eta$. Equality describes its observer-dependent cosmological event horizon.

For a finite Penrose diagram of the unwrapped spatial coordinate, set $U=\arctan u$, $V=\arctan v$, $X=(U+V)/2$, and $Y=(V-U)/2$. Then the metric is conformal to $-d{Y}^2+d{X}^2$; light rays remain at $45$ degrees, and the boundaries are the images of $v-u=\pm \pi$. The figure also shows the spatially periodic de Sitter cylinder so that the global distinction is explicit.

Both the covering spacetime and the de Sitter cylinder are globally hyperbolic. To see this directly in the covering spacetime, suppose an inextendible future-directed causal curve had a future limiting conformal time strictly below $\pi /2$. The slope bound makes $\xi$ Cauchy as that time is approached, so the curve has a limiting point inside the spacetime and can be extended, a contradiction. The past endpoint similarly approaches $-\pi /2$. Thus every constant-$\eta$ line is crossed exactly once and is a Cauchy hypersurface. On the cylinder the same argument uses compactness of the spatial circle. Equivalently causal diamonds between finite events are bounded and closed within the strip, hence compact.

Initial field data on a complete Cauchy hypersurface determine a unique solution for a well-posed hyperbolic Cauchy problem, with the relevant constraints imposed for constrained systems. Spacelike conformal infinity does not require additional timelike-boundary conditions. An observer's cosmological horizon limits which data that observer can see; it does not destroy global uniqueness from data on the whole hypersurface.

For the printed metric tensor, first use the global Minkowski time $T=\sinh t$. SetThenRemoving the positive conformal factor yields the ordinary Penrose diagram of two-dimensional Minkowski spacetime: a diamond with null future/past infinity on its sloping edges, timelike infinity at the top/bottom, and spatial infinity at the left/right corners. Null geodesics have slope $\pm 1$; future/past causal cones are ordinary Minkowski cones. There is no physical boundary or curvature singularity at a finite original coordinate value.

For the alternative spatial-lapse metric tensor identified in part (a), set $\rho =\arctan (\sinh x)$. Then $dx=\sec \rho d\rho$ and $\cosh x=\sec \rho$, soIts universal-cover conformal diagram is an infinitely tall strip with two timelike conformal boundaries, the conformal strip of two-dimensional anti-de Sitter spacetime. Null geodesics reach either boundary in finite coordinate time, though the boundary is not part of the physical manifold and the physical null affine parameter diverges there. The strip and diamond have different causal structures; no conformal coordinate change can identify their global causal properties.

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.

An orthonormal coframe writes a metric as $g={\eta }_{ab}{\theta }^a\otimes {\theta }^b$ with constant diagonal signature matrix $\eta$. In Cartan's first structure equation, solveThese are the torsion-free and metric-compatibility conditions and determine the Levi-Civita connection. Then Cartan's second structure equation givesContract $R^a{}_{bad}$ to obtain the Ricci tensor and then contract again for the scalar curvature. This fixes the curvature convention; reversing the definition of the Riemann curvature tensor reverses the resulting curvature signs.

On $z>0$, choose ${\theta }^0=dt/z$, ${\theta }^1=dx/z$, ${\theta }^2=dy/z$, ${\theta }^3=dz/z$, with $\eta =\mathrm{diag}(-1,1,1,1)$. For $a=0,1,2$,A torsion-free metric connection therefore hasFor example, ${\Omega }^a{}_3=-d{\theta }^a=-{\theta }^a\wedge {\theta }^3$, while for $a,b<3$,Metric antisymmetry supplies the remaining components, giving uniformlyThus the metric has constant sectional curvature $-1$. In four dimensions,It is an Einstein manifold, with Einstein constant $-3$ in the stated curvature convention.

These are Poincare coordinates on anti-de Sitter spacetime of unit radius. Constant curvature gives the maximal ten-dimensional local isometry algebra $\mathfrak{so}(3,2)$. This can also be checked directly: set $u=(t,x,y)$, ${\eta }_{ab}=\mathrm{diag}(-1,1,1)$ and $u_a={\eta }_{ab}{u}^b$. The ten independent Killing vector fields areTranslations and Lorentz transformations leave the numerator and $z$ unchanged. A dilation rescales numerator and denominator equally. For $K_a$, direct differentiation gives ${\mathcal{L}}_{K_a}{\eta }^{(4)}=4u_a{\eta }^{(4)}$ and $K_a(z)=2u_{a}z$, so the conformal factor cancels and ${\mathcal{L}}_{K_a}g=0$.

Hence the maximally extended Anti-de Sitter spacetime has connected isometry group locally $S{O}_0(3,2)$, with the appropriate covering group if one unwraps its time coordinate. The displayed coordinates cover only a patch. The ten local generators do not all give globally complete flows preserving that patch; for example special conformal flows can cross its horizon. The manifest complete patch symmetries include the boundary Poincare transformations and positive dilations. This distinguishes local maximal symmetry from the global isometries of a chosen coordinate domain.

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.

No: the theory predicts zero gravitational deflection of light. The reason is conformal preservation of null geodesic paths. Along a null geodesic with tangent $k^i$, the geodesic equation for a conformally flat metric reduces tobecause ${\eta }_{ij}{k}^i{k}^j=0$. The acceleration is parallel to the tangent and changes only its parametrization. Equivalently, with ${\ell }^i={\varphi }^2{k}^i$, $d{\ell }^i/d\lambda =0$, so the unparametrized ray is a straight null geodesic of Minkowski spacetime. A positive conformal factor also preserves local angles, so this straight-path result cannot be hidden by changing the local angle measurement. It contradicts the nonzero solar deflection observed in gravitational lensing, while remaining compatible with the gravitational redshift in part (b).

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.

Start with the four-dimensional Minkowski metric $d{s}^2=-d{t}^2+d{r}^2+r^{2}d{\Omega }^2$, where $r\ge 0$. Choose an arbitrary length $L>0$, and use the retarded and advanced null coordinates $u=t-r$, $v=t+r$. For the Minkowski conformal compactification, setBoth $p,q$ lie between $-\pi /2$ and $\pi /2$. Since $v\ge u$, we have $R\ge 0$; the remaining inequalities are $|T|+R<\pi$. Moreover,Multiply by the square of the conformal factor ${\Omega }_c=2\cos p\cos q/L$. The resulting metric is regular on the appropriate boundary pieces and preserves the null directions. Suppressing the angular two-spheres gives a triangular Penrose diagram with radial null geodesics at $45$ degrees.

The line $R=0$ is the ordinary timelike centre $r=0$. The upper sloping edge $T+R=\pi$ is future null infinity, reached with $v\to +\infty$ and finite $u$; the lower sloping edge $T-R=-\pi$ is past null infinity, reached with $u\to -\infty$ and finite $v$. The vertices $(T,R)=(\pi ,0)$ and $(-\pi ,0)$ are future and past timelike infinity, denoted $i^{+}$ and $i^{-}$. The vertex $(0,\pi )$ is spacelike infinity, $i^0$. These are limiting endpoints in the conformal completion, rather than ordinary physical events. In particular, finite diagram coordinates at null infinity do not imply finite physical affine parameter.

The four-dimensional radial diagram is the triangle $R\ge 0$, $|T|+R\le \pi$. If one instead draws two-dimensional Minkowski spacetime with a signed Cartesian spatial coordinate, the diagram is the full diamond. The centre is a boundary of the radial quotient, not a boundary of the physical four-dimensional Minkowski spacetime.

The induced metric is obtained from the ambient Euclidean metric by the pullback of a Riemannian metric operation. Differentiating $x^2+a^2{\rho }^2=1$ on a branch with $x\ne 0$ gives $dx=-a^2\rho d\rho /x$. Also $d{y}^2+d{z}^2=d{\rho }^2+{\rho }^{2}d{\varphi }^2$. ConsequentlyThe assumption that $(\rho ,\varphi )$ are coordinates already excludes the equator $x=0$ as well as the poles $\rho =0$. On such a chart, $0<\rho <1/|a|$ and $E>0$.

The conformal flattening of a surface of revolution supplies the particularly simple conformal factorSince ${\sigma }^{\prime }>0$, these are isothermal coordinates with a flat rescaled metric. An angular coordinate is understood on a local branch; the flat metric may equally be viewed locally as a cylinder metric. The apparent divergence of $E$ at the equator is a coordinate failure, not a singularity of the induced metric. For example, $(x,\varphi )$ give $g=[1+x^2/(a^2(1-x^2))]d{x}^2+(1-x^2)d{\varphi }^2/a^2$, regular at $x=0$.

The canonical momentum equation is $\dot{X}=eP+u{X}^{\prime }$. Eliminating $P$ from the phase-space action yields the Lagrangian densityThis is the Polyakov action after identifying its inverse metric tensor density asThe determinant of this matrix is $-1$, as required in two dimensions. The Weyl transformation leaves the matrix unchanged, so $e,u$ encode the metric modulo its conformal factor.

Define the induced worldsheet metric by $g_{\mu \nu }={\eta }_{mn}{\partial }_{\mu }{X}^m{\partial }_{\nu }{X}^n$. Varying the independent inverse metric tensor in the Polyakov action givesConsequently $g_{\mu \nu }=f{\gamma }_{\mu \nu }$, where $f=\frac{1}{2}{\gamma }^{\rho \sigma }{g}_{\rho \sigma }$. For a nondegenerate Lorentzian string worldsheet with the usual compatible time orientation, take $f>0$, or ${\gamma }_{\mu \nu }=e^{2\omega }{g}_{\mu \nu }$. The undetermined function is precisely Weyl invariance. Since $\sqrt{-\gamma }{\gamma }^{\mu \nu }{g}_{\mu \nu }=2\sqrt{-g}$, elimination of the independent metric tensor givesThus the Nambu–Goto action measures minus string tension times Lorentzian worldsheet area. The metric elimination holds in the nondegenerate interior; null endpoints are understood as boundary limits.

With $R=r+M$ and $dR=dr$,Defining the conformal factor byalso gives $R^2={\psi }^4{r}^2$. HenceThe constant-time spatial metric is therefore conformally flat.

For $\alpha >0$ and $t\ne 0$, the conformal factor $t^2/\alpha$ makes spatial translation a Killing vector field. Using proper time, its conserved momentum is $P=(t^2/\alpha )dx/d\tau$. With $v=dx/dt$ and $d{s}^2=-c^{2}d{\tau }^2$ this becomesThus $v/\sqrt{1-v^2/c^2}$ alone is not constant. Set $b=\sqrt{\alpha }P/c$. Every unparametrized timelike geodesic in a connected nonzero-$t$ patch has the formThe nonvertical paths have $v=cb/\sqrt{t^2+b^2}$ and proper-time parametrizationThe chart excludes $t=0$, where its metric is degenerate; a vertical family must not be lost by dividing by the conserved momentum.

For a spatial initial slice with vanishing extrinsic curvature, the vacuum momentum constraint is automatic and the Hamiltonian constraint requires zero Ricci scalar. In three spatial dimensions, the indicated conformally flat metric has $\mathcal{R}=-8{\psi }^{-5}\Delta \psi$, with a nonzero conformal factor. Thus a harmonic function supplies the data on its domain. A positive example is $\psi =1+a/r$ on a punctured or exterior region. There is no nonconstant globally smooth example on all of Euclidean three-space tending to one at infinity, by the maximum principle for harmonic functions. Time symmetry of an initial slice does not imply that its entire evolved spacetime is static.