Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 61 / 4 / Solution

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.