Astronomy & Celestial Mechanics
the sky, computed
Concept

Event horizon — where it appears

The comoving distance a signal sent now will ever cover, which is finite only because the expansion accelerates. Galaxies beyond it remain visible while everything they do from now on is permanently unreachable, and it shrinks rather than grows.

Named by 8 essays across 3 fields — each of them below, with the objects they name alongside it.

Three horizons, and a light cone that bulges. Cosmic time upwards, comoving distance sideways, for the Planck 2018 cosmology. Galaxies sit still in these coordinates, so their worldlines are the vertical grey lines: comoving distance is defined to take the expansion out. The solid inner curve is the past light cone — the set of events whose light reaches here and now — and it reaches out to the particle horizon at 46.1 billion light years, which is the diagram's central number and the one that sounds impossible. The universe is 13.80 billion years old and light has travelled for 13.80 billion years, and yet the material that emitted the oldest light is now 46 billion light years away. Nothing has outrun light: the comoving distance covered is ∫c dt/a, and dividing by a scale factor that was small early on makes the integral three times ct. The dashed curve is the Hubble sphere, where the recession speed equals c, at 14.5 Gly today — and the light cone lies outside it for most of its length, which is exactly why a galaxy can be observed while receding faster than light. The outer dot-dash curve is the event horizon, at 16.7 Gly: a signal sent from here today never reaches anything beyond it.

A horizon three times larger than the age allows

The universe is 13.8 billion years old and light travels one light year a year, so the observable universe should be 13.8 billion light years across. It is 46.1, nothing has outrun light, and the discrepancy is a piece of arithmetic rather than a paradox.

cosmology · Horizons Beyond redshift 1.87, what a galaxy does today will never be seen. For a galaxy at each redshift, the cosmic time of the last event on it that will ever be visible from here — not the last that has arrived, but the last that ever arrives, integrated to infinite future time. The horizontal line is the present, 13.8 billion years. A galaxy below redshift 1.87 has its curve above that line: its entire future will be seen from here, arriving ever more slowly and ever more redshifted, so it never quite disappears. A galaxy above redshift 1.87 has its curve below the line, and that is the whole content of an event horizon: what such a galaxy is doing today will never be seen, ever, by anybody here. Only a finite slice of its history is coming, and when the last of that light arrives the object stops changing. The redshift at which the curve crosses is 1.87, the comoving distance there is 16.7 billion light years against a particle horizon of 46.1, and the ratio of the volumes says that 95 per cent of the galaxies now observable are already beyond reach. Superluminal recession is not what does this. Everything past about redshift 1.5 has always been receding faster than light and is seen perfectly well, because the Hubble sphere grows to meet the photon; what closes the horizon is that with a cosmological constant the comoving Hubble radius stops growing and begins to shrink, so a photon that has not already been overtaken never will be.

The galaxies that are already out of reach

With a cosmological constant the comoving distance a photon can ever cover converges, so there is a redshift beyond which light leaving today never arrives. It is 1.87, and about ninety-five per cent of the galaxies now visible are past it — which superluminal recession has nothing to do with.

cosmology · Horizons Two lengths that cross at 1.1·10⁸ solar masses, above which nothing is seen. The radius at which a star of 1 solar radius and 1 solar mass is pulled apart by a black hole, and the hole's own horizon, both against the hole's mass and both on logarithmic axes. The tidal radius is the star's own radius times the cube root of the mass ratio, so it climbs with a slope of one third; the horizon is proportional to the mass, so it climbs with a slope of one. Two lines of different slope cross once, and this pair crosses at 1.14·10⁸ solar masses. Below that the star is torn apart outside the horizon, half of it is thrown out and half falls back, and the fallback is visible for months. Above it the star crosses the horizon while it is still a star, is swallowed whole, and produces no flare at all. The consequence is the reason these events are worth watching: a flare that is seen is an upper limit on the mass of the hole that made it, obtained without resolving anything, and it is the only such limit available for a hole that is not currently accreting.

A flare that puts a ceiling on a mass

A star torn apart by a black hole lights up for a year. The tidal radius grows as the cube root of the hole's mass and the horizon grows as the mass itself, so above about a hundred million suns the star is swallowed whole and nothing is seen — which makes the existence of a flare a measurement.

galaxies · Tidal disruption From six gravitational radii to one. The radius of the innermost stable circular orbit against the dimensionless spin a = Jc/GM², in units of GM/c², for orbits prograde and retrograde with the hole's rotation. Both curves are the Bardeen–Press–Teukolsky expression and are checked at the three places it has exact values: 6 at zero spin, and 1 and 9 at the extremal limit. The separation is the observable consequence of frame dragging — space near the hole is itself circulating, so an orbit going the same way can stay closer before it becomes unstable, and one going the other way cannot come as close as a non-rotating hole allows. The prograde branch is required to fall and the retrograde branch to rise at every step drawn, which is a claim about the direction of the effect rather than about its size. The marked spin of 0.998 is not the extremal value but the equilibrium a hole fed by a thin disc actually reaches, because photons emitted by the disc are preferentially captured on retrograde orbits and spin the hole down again. Nothing here depends on what the hole is made of: two numbers fix the whole geometry, and this figure is the first of them holding still while the second moves.

The second number a black hole has

A black hole in equilibrium is described by its mass and its spin, and nothing else. The mass decides how strongly it pulls. The spin decides how much light a kilogram of infalling matter can emit before it disappears — and between the two extremes that figure changes by a factor of seven.

gravitation · Black hole spin The same history, on the clock that straightens light. Conformal time upward against comoving distance sideways, for the Planck 2018 cosmology. Conformal time is ∫dt/a, which is exactly the comoving distance light covers, so on these axes every photon moves at forty-five degrees — at every epoch, whatever the expansion is doing. That single property turns every curved thing in the ordinary space-time diagram into a straight one. The universe began at η = 0 and is now at η = 46.1 Gly of conformal time; it will ever accumulate only 62.8, because the integral ∫dt/a converges once Λ dominates, and that finite ceiling is the whole reason an event horizon exists. The particle horizon is the 45° line from the origin and the event horizon is the 45° line back from the ceiling, so the two horizons that were curves are now the two edges of one light cone drawn twice. The shaded wedges are the past light cones of two points on the last scattering surface, at conformal time 0.914 Gly and comoving distance 45.2 Gly from here. They do not overlap. Two points on that surface separated by more than 2η_rec were never in causal contact, which subtends 2.31° on the sky, and the microwave sky therefore contains about 9,805 patches that have no common past and the same temperature to one part in a hundred thousand. That is the horizon problem, and in these coordinates it is a statement about whether two triangles intersect.

The clock on which light travels in straight lines

Cosmic time makes light cones bulge and horizons curve. There is another time coordinate on which a photon's worldline is a forty-five degree line at every epoch, and on it the horizon problem stops being a piece of arithmetic and becomes a question about whether two triangles overlap.

cosmology · Horizons Whether an event horizon exists at all, against one number. The comoving event horizon today — the distance a signal sent now will ever cover — against the equation of state of the dark energy, for a flat universe with the measured matter density. The curve runs away at w = −1/3 and does not exist above it: that is where the expansion stops accelerating, and in a universe that does not accelerate the integral ∫da/a²E diverges and every galaxy is eventually reachable, however far away. Below −1/3 the horizon is finite and shrinks as w falls, because a more negative equation of state makes the dark energy density grow with time rather than stay constant. At the cosmological constant's w = −1 the horizon is 16.7 billion light years against a particle horizon of 46.1, so 4.7 per cent of the volume now observable is still reachable. The band is the measured −1.03 ± 0.03. What the figure is for is the asymmetry in what the measurement still allows: two sigma toward zero puts the horizon at 17.5 Gly and two sigma the other way at 14.7, and the shape of the curve means that the closer the true value sits to −1/3 the more violently the answer moves. The reachable fraction is not a robust number in the way the particle horizon is.

Whether there is a horizon at all

An event horizon exists precisely when the expansion accelerates, and its size is not a smooth function of how much. The integral that defines it runs away as the equation of state approaches minus a third, so two sigma either way on a measured number are two very different futures.

cosmology · Horizons Two horizons, one formula, and a factor of two. Horizon temperature against horizon radius, on logarithmic axes, for the two kinds of horizon this collection has. The upper line is a de Sitter horizon at T = ħc/2πk_BR and the lower is a Schwarzschild horizon at T = ħc/4πk_BR — the same expression with the same constants, differing by exactly two, and both falling as one over the radius so that a bigger horizon is a colder one. The cosmological horizon today has a radius of 4,451 megaparsecs and a temperature of 2.65e-30 K, which is thirty orders of magnitude below the microwave background and will never be measured by anything. A solar-mass black hole sits at 6.2e-8 K, and a black hole as cold as the sky would weigh 2.32e+22 solar masses — of the order of the mass inside the observable universe, which is not a coincidence, since both numbers are c³/GH up to factors of order one. The factor of two between the two lines is the one place the analogy is not exact, and it is not a convention: it comes from the periodicity of the Euclidean time coordinate, which is 8πGM/c³ for a black hole and 2π/H for de Sitter space. The entropies, by contrast, agree exactly.

Two horizons that differ only in who is inside

A black hole's horizon and the cosmological one share an entropy formula exactly and differ in temperature by precisely a factor of two. The factor of two is the whole of the difference, and what it encodes is which side of the surface the observer stands on.

cosmology · Horizons The horizon and the orbit that cannot come back, against the hole's spin. Two radii round a rotating black hole, in gravitational radii GM/c², against its spin a from −1 (an orbit against the rotation) to +1 (with it), for orbits in the equatorial plane. The lower curve is the horizon, 1 + √(1 − a²), which is the same whichever way the orbit goes. The upper one is the marginally bound orbit, 2 − a + 2√(1 − a): the closest a body falling in from far away can pass and still escape back out. For a hole with no spin the horizon is at 2 and the marginally bound orbit at 4 — twice as far out. With maximal spin the marginally bound orbit comes in to 1.09 for a prograde orbit (a = 0.998) and moves out to 5.83 for a retrograde one. A star whose tidal radius lies inside this curve is swallowed whole, so the curve, not the horizon, is the line a disruption flare is measured against — and it moves by a factor of 5.3 with the spin.

The line a star is swallowed at is not the horizon

A star torn apart by a black hole makes a flare, and a star swallowed whole makes nothing, so the heaviest hole that can produce a flare is a measurement. That ceiling is set not at the horizon but at the closest orbit from which infalling matter can still come back out — twice as far out for a hole that does not spin, and moved by a factor of five by the hole's rotation.

galaxies · Tidal disruption

Named alongside it

The objects these essays reach for when they reach for this one.

Comoving distanceConformal timeDe sitter spaceObservable universeParticle horizonCausal contactCosmological constantHubble sphereLight coneAccretion discBlack hole spinEddington limit

All concepts