Cosmic Horizons and Spacetime Curvature — Epoche C2
The escape-velocity picture, and the coincidence that keeps it alive The Schwarzschild radius $R_s = 2GM/c^2$ — the coordinate radius of the event horizon of a static, uncharged black hole of mass $M$, with $G$ the gravitational constant and $c$ the speed of light in vacuum — is almost always introduced by an argument that gets the right number for the wrong reasons, and that argument is the source of the picture this essay is written against: a horizon as the surface on which the escape speed reaches $c$. John Michell in 1784, and Laplace a decade later, computed the size of a body from which light could not escape by setting the Newtonian escape speed $v_{\rm esc} = \sqrt{2GM/r}$ equal to $c$ and solving. The result is $r = 2GM/c^2$, numerically identical to the horizon radius that Schwarzschild's 1916 solution of the field equations delivers. For the Sun the arithmetic runs $2 \times (6.674 \times 10^{-11}) \times (1.989 \times 10^{30}) / (8.988 \times 10^{16}) = 2.95 \times 10^{3}$ metres, the familiar 2.95 km. The agreement is a coincidence in the strict sense that no step of the Newtonian argument survives translation. Three failures matter, and each of them points to what a horizon really is. A dark star has no causal boundary. In the Newtonian picture, light emitted radially outward from just inside the critical surface rises, decelerates and falls back, so an observer hovering just outside could intercept it during its ascent. What Michell described is a bound trajectory, not a one-way membrane; the relativistic horizon admits no future-directed causal curve at all from inside to arbitrarily large radius. The symbol $r$ does not mean what the derivation assumes. In the Schwarzschild line element $r$ is not a measured distance from a centre; it is defined by the area of the sphere of symmetry through the point, $r \equiv \sqrt{A/4\pi}$. Inside $R_s$ that coordinate is timelike, so decreasing $r$ is as unavoidable as the passage of time is outside, and hovering at fixed radius is not an available motion at all. The horizon is a null surface, and an escape-speed surface is timelike. A timelike hypersurface can be crossed in both directions; that is what makes it timelike. The Schwarzschild horizon is generated by null geodesics that neither reach infinity nor fall in, but remain on it for infinite affine parameter. Written out, the exterior geometry is $$ds^2 = -\left(1-\frac{R_s}{r}\right)c^2\,dt^2 + \left(1-\frac{R_s}{r}\right)^{-1}dr^2 + r^2\,d\Omega^2,$$ with $d\Omega^2$ the unit two-sphere metric. At $r = R_s$ the coefficient of $dt^2$ vanishes and that of $dr^2$ diverges — an artefact of the chart, as the Eddington–Finkelstein and Kruskal coordinates show by covering the surface smoothly. The invariant that would have to blow up for a genuine singularity is the Kretschmann scalar $R_{abcd}R^{abcd} = 48G^2M^2/(c^4 r^6)$, which at $r = R_s$ is finite and, for a large hole, tiny. For the black hole at the centre of M87, whose mass the Event Horizon Telescope Collaboration (2019) put at $6.5 \times 10^{9}$ solar masses, $GM/c^2 = 9.60 \times 10^{12}$ m, the horizon sits at $r = 1.92 \times 10^{13}$ m, and the tidal acceleration across a two-metre body there is of order $2GM\ell/r^3 = 2 \times (8.63 \times 10^{29}) \times 2 / (7.07 \times 10^{39}) \approx 4.9 \times 10^{-10}\ \mathrm{m\,s^{-2}}$, about five parts in $10^{11}$ of Earth's surface gravity. Whatever makes that surface a horizon, it is not the strength of the curvature there. Three inequivalent definitions, and what each of them needs Since the escape-speed criterion has failed, what replaces it? Three definitions are in use; they coincide in the simplest cases and come apart in general, and keeping them apart is most of what this subject demands. The event horizon is defined globally. Let $\mathscr{I}^+$ denote future null infinity, the conformal boundary at which outgoing light rays terminate in an asymptotically flat spacetime, and let $J^-(\mathscr{I}^+)$ be the set of events from which some future-directed causal curve reaches it. The event horizon is the boundary $\partial J^-(\mathscr{I}^+)$, and the black hole is what lies outside $J^-(\mathscr{I}^+)$. This is the definition Hawking and Ellis (1973) develop in their treatment of gravitational collapse, and it carries two heavy commitments. It requires asymptotic flatness, so that the conformal boundary exists at all — already a reason to doubt that the notion transfers unchanged to cosmology. And it is teleological: to decide whether the event you are now at lies inside, one must know the entire future of the spacetime. If a spherical null shell collapses onto an existing black hole, the event horizon begins to expand before the shell arrives, at a place where the curvature is still exactly that of the smaller hole, because the generators that will belong to the final horizon are already fixed by what is going to happen. A horizon can move through effectively flat space in anticipation. The apparent horizon is the quasi-local repair for that non-locality, and rests on Penrose's (1965) notion of a closed trapped surface: a compact spacelike two-surface whose two families of future-directed null normals both have negative expansion, $\theta_+ The Killing horizon is defined by symmetry: a null hypersurface on which a Killing vector field, the generator of a continuous isometry, becomes null. In Schwarzschild the time-translation field $\partial_t$ has norm $-(1 - R_s/r)c^2$, which vanishes at $r = R_s$. Hawking's rigidity theorem, given in Hawking and Ellis (1973), states that the event horizon of a stationary, analytic, asymptotically flat solution is a Killing horizon; that theorem is what makes the three notions agree for the stationary black holes of the textbooks, and its hypotheses are exactly what fails in dynamical situations. That settles the black-hole cases. The decisive counterexample to the whole curvature story is different, and it is the reason the