Clocks near the horizon

The Schwarzschild metric determines both the clock rate and the acceleration required to hold a fixed radius. It leans on the picture of the horizon from What is a black hole?.

Slow clocks from the metric

For a hole that does not spin, the spacetime outside the horizon is the Schwarzschild solution. Write $f(r) = 1 - r_s/r$, with $r_s = 2GM/c^2$. The interval between nearby events is

$$ds^2 = -f\,c^2\,dt^2 + \frac{dr^2}{f} + r^2\,d\Omega^2$$

Here $t$ is time on a clock far from the hole, and $r$ is defined so that a sphere at $r$ has area $4\pi r^2$. For two nearby events on one clock's path, $-ds^2/c^2$ is the square of the time that clock reads between them, $d\tau^2$. A hovering clock does not change $r$, $\theta$ or $\varphi$, so $dr = d\Omega = 0$ and

$$d\tau = \sqrt{f}\,dt$$

That is the clock rate, $\sqrt{1 - r_s/r}$, and it vanishes at $r = r_s$. This follows by evaluating the metric along a path at fixed $r$.

Gravitational redshift

A photon moving through this spacetime keeps one number fixed along its path, the energy measured at infinity, $E_\infty = -p_t$. That holds because the metric does not depend on $t$. A hovering observer at radius $r$ has four-velocity with only a time part, $u^t = 1/\sqrt{f}$ (so that $g_{tt}(u^t)^2 = -1$ in units with $c = 1$), and the energy they measure is $E = -p_\mu u^\mu = E_\infty/\sqrt{f(r)}$. Light sent from a hoverer at $r_e$ to a hoverer at $r_o$ is therefore received at

$$\frac{\nu_o}{\nu_e} = \sqrt{\frac{f(r_e)}{f(r_o)}}$$

A deeper emitter has a smaller $f$, so its light arrives lower in frequency. This is the same fact as the slow clock: the light's wave crests are ticks of the emitter's clock, and those ticks are spaced out by $1/\sqrt{f}$ in far time. Far from the hole, $\sqrt{1 - r_s/r} \approx 1 - GM/rc^2$, so

$$\frac{\nu_o}{\nu_e} \approx 1 - \frac{GM}{c^2}\left(\frac{1}{r_e} - \frac{1}{r_o}\right)$$

For a receiver a height $h$ above the emitter on a planet of radius $R$, with $h \ll R$, the bracket is $h/R^2$, and $GM/R^2$ is the surface gravity $g$. The fractional shift becomes

$$\frac{\Delta\nu}{\nu} = \frac{g\,h}{c^2}$$

This is the formula behind every Earth experiment in the next fold. Our arithmetic with standard gravity: a rise of 22.6 m shifts a frequency by only 2.5 × 10⁻¹⁵, a rise of 0.33 m by 3.6 × 10⁻¹⁷.

The thrust, from the proper acceleration

Hovering is not a free fall, so a hoverer feels a push. Keep $c = 1$. The acceleration of a worldline is the part of its change in four-velocity that is not just parallel transport:

$$a^\mu = u^\nu \nabla_\nu u^\mu = \frac{du^\mu}{d\tau} + \Gamma^\mu_{\nu\lambda}u^\nu u^\lambda$$

The hoverer's four-velocity is constant in these coordinates, so the first term is zero. Only the connection term survives, and for $\mu = r$ it is $\Gamma^r_{tt}(u^t)^2$. From $\Gamma^r_{tt} = -\tfrac12 g^{rr}\,\partial_r g_{tt}$, with $g^{rr} = f$ and $g_{tt} = -f$, we get $\Gamma^r_{tt} = \tfrac12 f f'$, where $f' = r_s/r^2$. Then

$$a^r = \tfrac12 f f' \cdot \frac{1}{f} = \frac{r_s}{2r^2}$$

This is a component along the coordinate $r$, which is not a length. The physical acceleration is the length of the vector, $a = \sqrt{g_{rr}}\,a^r$, and $g_{rr} = 1/f$, so

$$a = \frac{r_s}{2 r^2 \sqrt{f}}$$

Restoring $c$ and using $GM = c^2 r_s/2$:

$$a = \frac{GM}{r^2}\cdot\frac{1}{\sqrt{1 - r_s/r}}$$

Newton's pull, divided by the factor that slows the clock. At $r = 3r_s$ this is 1.7 × 10¹¹ m/s²; at $1.01\,r_s$, 1.2 × 10¹³ m/s², as in the table.

The corresponding quantity referred to a distant observer remains finite. Multiply the local thrust by the clock rate to refer it to the far observer: $a\sqrt{f} = GM/r^2$, which tends to $c^4/4GM$ at the horizon. That finite number is the horizon's surface gravity. An ideal rope lowered from far away has a finite tension at its top however close to the horizon its load hangs. The local push diverges only because the local clock stops.

Circular orbits, from an effective potential

A free-falling ship follows a geodesic, and two numbers stay fixed along it, again because the metric does not change with $t$ or with $\varphi$: the energy per unit mass $E = f\,dt/d\tau$ and the angular momentum per unit mass $L = r^2\,d\varphi/d\tau$ (still $c = 1$). Put them into the metric for a path in the equatorial plane, with $-1 = g_{\mu\nu}u^\mu u^\nu$:

$$-1 = -\frac{E^2}{f} + \frac{1}{f}\left(\frac{dr}{d\tau}\right)^2 + \frac{L^2}{r^2}$$

$$\left(\frac{dr}{d\tau}\right)^2 = E^2 - V(r), \qquad V(r) = \left(1 - \frac{r_s}{r}\right)\left(1 + \frac{L^2}{r^2}\right)$$

This is a ball rolling in a one-dimensional valley. A circular orbit sits where the slope is zero, $V'(r) = 0$:

$$\frac{r_s}{r^2} - \frac{2L^2}{r^3} + \frac{3 r_s L^2}{r^4} = 0 \quad\Longrightarrow\quad L^2 = \frac{r_s\, r^2}{2r - 3r_s}$$

and then $E^2 = V(r) = 2(r - r_s)^2/\big(r(2r - 3r_s)\big)$. Two things fall out at once. First, $L^2$ blows up at $r = 3r_s/2$: no finite angular momentum holds a massive body in a circle there, and below it $L^2$ would be negative. That radius is the photon sphere, where only light circles. Second, the orbit is stable only if it sits at the bottom of the valley, $V'' > 0$. Working out $V''$ and substituting $L^2$ gives

$$r^5 V''(r) = r_s r^2\left(\frac{6r - 12 r_s}{2r - 3r_s} - 2\right) = \frac{2 r_s r^2 (r - 3r_s)}{2r - 3r_s}$$

which is positive only for $r > 3r_s$. So circular orbits are stable from three horizon radii out, unstable between one and a half and three, and absent below one and a half. That innermost stable orbit, at $3r_s$, is where a thin accretion disk around a non-spinning hole ends: gas that drifts inside it falls in.

The angular speed seen from far away is $\Omega = d\varphi/dt = (L/r^2)(f/E)$. Squaring and substituting,

$$\Omega^2 = \frac{L^2 f^2}{r^4 E^2} = \frac{r_s}{2r^3} = \frac{GM}{r^3}$$

the same law as Kepler's third, exactly, in these coordinates. The clock rate is $d\tau/dt = f/E$:

$$\left(\frac{d\tau}{dt}\right)^2 = \frac{f^2}{E^2} = \frac{2r - 3r_s}{2r} = 1 - \frac{3r_s}{2r}$$

The same result comes straight from the metric with $dr = 0$ and $d\varphi = \Omega\,dt$: $1 - r_s/r - r^2\Omega^2/c^2 = 1 - r_s/r - r_s/2r$. One term is gravity, the other is motion.

Orbiting against hovering

Comparing the two clocks at the same radius,

$$\frac{d\tau_\text{orbit}}{d\tau_\text{hover}} = \sqrt{\frac{1 - 3r_s/2r}{1 - r_s/r}}$$

which is 0.87 at $3r_s$ and 0.95 at $6r_s$. The hoverer measures the orbiter's speed with its own rulers and clock: proper distance $r\,d\varphi$ over proper time $\sqrt{f}\,dt$, so

$$\frac{v^2}{c^2} = \frac{r^2\Omega^2}{f c^2} = \frac{r_s}{2(r - r_s)}$$

and special relativity's $\sqrt{1 - v^2/c^2} = \sqrt{(2r - 3r_s)/(2(r - r_s))}$ is exactly the ratio above. Gravity slows both clocks equally. Speed slows only the orbiter. At $3r_s$ the speed is half of $c$; at the photon sphere it reaches $c$, which is why only light orbits there.

The two ships meet once per orbit, so this is a real twin experiment, and it has a twist. In flat space the twin who stays in free fall always ages most. Here the free faller ages less. In general relativity the longest-lived path between two meetings is still a free fall, but there can be several free falls joining the same two events, and the circle is not the longest. A ship tossed straight up from beside the hoverer at the first meeting, so that it falls back to meet it at the second, ages more than both.

No hovering at the horizon

At the horizon there is no hovering path at all. A four-velocity must have length $c$, and for a hoverer that needs $u^t = 1/\sqrt{f}$, which grows without bound as $f \to 0$. At $f = 0$ no slower-than-light path stays at fixed $r$. Inside the horizon $f < 0$, so $dt^2$ enters the interval with a plus sign and $dr^2$ with a minus sign: $r$ has become a time direction. A physical trajectory must then move toward smaller radius.

Measured on Earth

The slowing of a clock by gravity was first shown with gamma rays. In 1959 and 1960 Robert Pound and Glen Rebka sent gamma rays from iron-57 up and down a tower at Harvard, over a height of 22.6 m. Emitter and absorber were matched so sharply that the tiny shift of 2.5 × 10⁻¹⁵ spoiled the match, and the loss of match came out at 1.05 times the prediction, give or take 10 percent. Pound and Joseph Snider redid it in 1965 and got 0.9990 times the prediction, with an uncertainty of 0.8 percent. That is the same $g h/c^2$ as in the previous fold.

Clocks soon replaced gamma rays. In 1971 Joseph Hafele and Richard Keating carried four caesium clocks around the world on airliners, once east and once west, and compared them with the clocks of the US Naval Observatory. Eastbound they lost 59 ± 10 nanoseconds; westbound they gained 273 ± 7 nanoseconds. Both match the two effects on this page added together: height speeds a clock up, and speed slows it down, and the Earth's spin makes an eastbound airliner faster than the ground and a westbound one slower.

In 2010 a NIST team including David Wineland did it with two aluminium-ion optical clocks. They raised one by 0.33 m and watched the clocks drift apart by 4.1 × 10⁻¹⁷, with an uncertainty of 1.6 × 10⁻¹⁷. The prediction from $g h/c^2$, our arithmetic, is 3.6 × 10⁻¹⁷, inside that error bar. A shift that Pound and Rebka needed a tower to see, a clock in 2010 could see across a lab bench.

In the 2020s the same arithmetic is built into satellite navigation. Our arithmetic, from the inputs in Ashby's review: a GPS satellite's clock runs fast by 45.7 microseconds a day for sitting higher in Earth's field and slow by 7.2 for its orbital speed, and the ground clock's own ride on the spinning Earth adds 0.1, for a net of 38.6 microseconds a day. Light covers 12 km in that time, so left alone, the ranges a receiver measures would drift by that much every day. The satellite clocks are set slightly slow before launch to cancel it.

When the hole spins

A spinning hole drags space round with it. Vesper's design spin of 0.70 is a choice made for the story. illustrative The spin changes one thing that matters here. Hovering means staying put relative to the distant stars, and outside a spinning hole's horizon there is a region, the ergosphere, where the dragging is so strong that nothing can do that, however hard its engines push. At the equator the ergosphere's edge sits at $r_s$, where a non-spinning hole of the same mass would have its horizon, and for this spin the real horizon sits at 0.86 of that radius. Between the two, everything is carried round.

The replacement for the hoverer there is the zero-angular-momentum observer, or ZAMO, of James Bardeen, William Press and Saul Teukolsky (1972), working in the spacetime Roy Kerr found in 1963. A ZAMO carries no spin of its own and is swept round with the dragging of space, as close to "at rest" as the geometry allows. Its clock rate, a number called the lapse, drops to zero at the horizon, and ZAMOs exist all the way down to it.

For a horizon you can cross

The thrusts on this page shrink for a heavier hole: at a fixed number of horizon radii, the push to hover scales as one over the mass. For a hole of 6.5 × 10⁹ suns like the one in M87, 5 × 10⁸ times Vesper's mass, hovering at three horizon radii takes about 32 g: more than a person can bear for long, but a load a robot probe could be built for. A hair above the horizon it is 2,400 g. That hole's horizon has a radius of about 128 astronomical units, so wide that the tides at its edge are gentle and a falling probe crosses it whole. You can fly such a crossing yourself in M87 Descent, David Veksler's earlier simulator of a fall into a supermassive hole, and read on Ships die of tides, not of curvature why the same crossing at Vesper would kill you long before the horizon. Where an orbit stays put with Ember tugging from 1.5 AU away is on Where you can live around a hole and a star.

Still open in 2026

The slowing of clocks by gravity, and by speed, is among the best-tested predictions in physics. The apparent freezing at the horizon describes the distant observer's coordinates: an infalling ship's own clock reads a finite time on the way in, and following that clock is the job of the Dive. What is argued is whether a quantum theory of gravity leaves the horizon as quiet as general relativity says, or gives it structure of its own. frontier

Further reading

1. Pound and Rebka (1960), Apparent weight of photons, Physical Review LettersGamma rays sent up and down a tower at Harvard shift in frequency as the equivalence principle predicts.
2. Pound and Snider (1965), Effect of gravity on gamma radiation, Physical ReviewThe same experiment, redone ten times more precisely.
3. Hafele and Keating (1972), Around-the-world atomic clocks: observed relativistic time gains, ScienceCaesium clocks flown round the world eastward and westward lost and gained time as relativity predicts.
4. Chou, Hume, Rosenband and Wineland (2010), Optical clocks and relativity, ScienceAluminium-ion optical clocks see time dilation from a height change of less than a metre and from speeds of a few metres a second.
5. Ashby (2003), Relativity in the Global Positioning System, Living Reviews in RelativityHow the satellite clocks are set for gravity and for speed, and why the system would not work without it.
6. Kerr (1963), Gravitational field of a spinning mass as an example of algebraically special metrics, Physical Review LettersThe spacetime of a spinning black hole.
7. Bardeen, Press and Teukolsky (1972), Rotating black holes: locally nonrotating frames, energy extraction, and scalar synchrotron radiation, The Astrophysical JournalThe zero-angular-momentum observers of a spinning hole.
8. Event Horizon Telescope Collaboration (2019), First M87 Event Horizon Telescope results. VI. The shadow and mass of the central black hole, The Astrophysical Journal LettersThe size of the shadow of the hole in M87, and the mass it gives.