Ships die of tides, not of curvature

The Newtonian version

Gravity at distance $r$ from a mass $M$ is $g(r) = GM/r^2$. Two ends of a body, a length $L$ apart along the line of fall, feel different pulls. For small $L$ the difference is the slope of $g$ times $L$:

$$\Delta g = \left|\frac{dg}{dr}\right| L = \frac{2GM}{r^3}\,L$$

Across the line of fall it goes the other way. Two points at the same distance $r$ from the centre but $L$ apart sideways are both pulled toward the centre, along lines that converge. Each pull points inward by a small angle $L/2r$ off the line through the middle, so the two points are squeezed together at

$$\Delta g_\perp = -\frac{GM}{r^3}\,L$$

So a falling body is stretched along its line of fall at rate $2GM/r^3$ and squeezed in each of the two directions across at rate $GM/r^3$. Add the three: $2 - 1 - 1 = 0$. To first order the body's volume does not change. It is stretched one way and squeezed the other two, by exactly the amounts that keep the volume. That is spaghettification.

The relativistic version

General relativity says the same thing in other words. Two neighbouring free-fall paths, separated by a small vector $\xi$, drift apart or together by the equation of geodesic deviation:

$$\frac{D^2 \xi^a}{d\tau^2} = -R^a{}_{bcd}\,u^b\,\xi^c\,u^d$$

Here $R$ is the Riemann tensor, the full description of curvature, $u$ is the four-velocity of the free-falling observer and $\tau$ is that observer's own time. Pirani showed in 1956 what this means physically: the Riemann tensor is the tidal field, the thing you would measure by watching free particles near you accelerate apart.

For the Schwarzschild solution and an observer falling radially, the equation in the observer's own frame reads

$$\frac{d^2\xi^{\hat r}}{d\tau^2} = +\frac{2GM}{r^3}\,\xi^{\hat r}, \qquad \frac{d^2\xi^{\hat\theta}}{d\tau^2} = -\frac{GM}{r^3}\,\xi^{\hat\theta}, \qquad \frac{d^2\xi^{\hat\phi}}{d\tau^2} = -\frac{GM}{r^3}\,\xi^{\hat\phi}$$

These are the Newtonian numbers exactly. Two facts make that possible. The three rates sum to zero because the empty space outside a mass has no Ricci curvature, the same statement as the volume staying fixed. And in Schwarzschild a boost along the radial direction leaves all three components unchanged, so it does not matter how fast you are falling or where you started: a radial faller feels $2GM/r^3$. The formula holds just as well inside the horizon, where $r$ is no longer a distance in the ordinary sense but a measure of the sphere's area.

A body moving sideways is different, because a sideways boost does mix the components. At the closest point of a pass the motion is all sideways, at speed $v$ with Lorentz factor $\gamma$, and boosting the static tidal field gives

$$\frac{d^2\xi^{\hat r}}{d\tau^2} = \frac{2GM}{r^3}\left(1 + \frac{3}{2}\frac{\gamma^2 v^2}{c^2}\right)\xi^{\hat r}, \qquad \frac{d^2\xi^{\perp}}{d\tau^2} = -\frac{GM}{r^3}\left(1 + 3\,\frac{\gamma^2 v^2}{c^2}\right)\xi^{\perp}$$

for the stretch along the radius and the squeeze out of the orbital plane, while the squeeze along the direction of motion stays at $GM/r^3$. The three still sum to zero. So the radial formula undershoots, but only a little at these speeds. At closest approach the slow pass moves at 1,519 km/s, which raises the stretch by 0.0039 percent; even the fast pass gains only 0.87 percent. The Newtonian estimates above are adequate at these speeds.

Why the horizon tide goes as one over M squared

Put $r = r_s = 2GM/c^2$ into $2GM/r^3$:

$$\frac{2GM}{r_s^3} = \frac{2GM\,c^6}{8\,G^3 M^3} = \frac{c^6}{4\,G^2 M^2}$$

For Vesper that is 7.2 × 10⁷ m/s² per metre, the number in the table. The scaling has a physical reading. The horizon's volume is $(4\pi/3) r_s^3$, so the mean density inside it is $\bar\rho = 3M/(4\pi r_s^3)$, and

$$\frac{2GM}{r_s^3} = \frac{8\pi}{3}\,G\bar\rho$$

The tide at a horizon is set by the average density inside it. The radius grows as $M$, the volume as $M^3$, and so the density falls as $1/M^2$. A bigger hole has a lower average density by this definition. For M87* the mean density inside the horizon works out to 0.44 kg/m³, about a third of the density of air at sea level (1.23 kg/m³).

The yardstick mass

Set the stretch across a length $L$ at the horizon equal to a limit $a$ and solve for the mass:

$$\frac{L\,c^6}{4\,G^2 M^2} = a \quad\Longrightarrow\quad M = \frac{c^3}{2G}\sqrt{\frac{L}{a}}$$

With $L = 2$ m and $a = 98.1$ m/s², that gives 14,496 Suns. The limit is this page's own yardstick, and the mass depends on it only as $a^{-1/2}$, so even a tenfold change in what a body can bear moves the answer by about a factor of three.

Why a faster pass is so much worse

On a hyperbolic pass with speed $v_\infty$ far away, turning through an angle $\delta$, the closest approach is

$$r_p = \frac{GM}{v_\infty^2}\left(\frac{1}{\sin(\delta/2)} - 1\right)$$

For a fixed turn, $r_p \propto v_\infty^{-2}$. The tide goes as $r_p^{-3}$, so for a fixed turn it goes as $v_\infty^{6}$. The speed ratio 15.0 to the sixth power is 1.1 × 10⁷, the factor found above. Reducing the arrival speed also reduces the tidal load: every halving of arrival speed makes the tide 64 times gentler.

The Roche limit and the tidal radius

A star holds itself together with its own gravity, $Gm/R^2$ at the surface, where $m$ is its mass and $R$ its radius. A hole's tide across the star is $2GMR/r^3$. Set the two equal and the distance inside which the tide wins is, up to a factor near one,

$$r_t \sim R\left(\frac{M}{m}\right)^{1/3}$$

This is the tidal radius. Édouard Roche's careful treatment, for a fluid body held together only by its own gravity and orbiting so that it always shows the same face, lets the body stretch into a football shape and puts the limit at 2.44 times this distance; that is the Roche limit. A star plunging in on a single pass is not in that steady state, and the tidal radius is the usual estimate for it. A body with real strength, like a ship, holds out until the tide stress passes what its materials can bear, which can be much closer.

For a Sun-like star near Vesper, $r_t$ is 1,592,755 km, or 44,942 horizon radii. The star is shredded far out in space and never reaches the horizon. Ask when the tidal radius shrinks to the horizon, $R\,(M/m)^{1/3} = 2GM/c^2$, and the answer is

$$M = \left(\frac{R\,c^2}{2G}\right)^{3/2} \frac{1}{\sqrt{m}}$$

With the Sun's radius and mass, that is about 1.1 × 10⁸ Suns. A heavier hole lets a Sun-like star reach the horizon whole. A lighter one tears it apart first. The boundary is usually called the Hills mass, after Jack Hills, who in 1975 proposed that stars torn apart inside a massive black hole's Roche limit could be what powers quasars.

Stars being shredded

Martin Rees worked out in 1988 what happens when a star in the centre of a nearby galaxy meets a hole of a million to a hundred million Suns. The star is torn apart outside the horizon; some of the debris is flung away at high speed, and the rest falls back and is swallowed, making a bright flare. Observed flares fade over months to years, and Rees's estimate tied them to holes of this size in the nuclei of nearby galaxies. The first flares were found in archival searches of the ROSAT X-ray sky survey in the 1990s, and surveys in visible light have since made them a class of their own. In 2021 Suvi Gezari's review counted 56 reported tidal disruption events. They are rare, and each is a view of a quiet black hole waking up for a few months.

The mass boundary matters here. Above the 1.1 × 10⁸ Suns found in the fold above, the estimate says a Sun-like star falls in whole and nothing flares. But that estimate is Newtonian, and for the heaviest holes the star is torn close to the horizon, where relativity changes the tide. In 2012 Michael Kesden followed stars on orbits tilted every which way around spinning holes and evaluated the relativistic tidal tensor at each one's closest approach. He found, in line with earlier work, that a fast spin raises the heaviest hole able to disrupt a Sun-like star to about 7 × 10⁸ Suns, some 6 times the simple estimate above. Spin lets a star on the right orbit swing closer without crossing the horizon.

What is still argued

• What makes the light. In the events seen so far, the ultraviolet and optical light and the soft X-rays come from different places, and the ratio between them swings widely from one event to the next. What powers each component is still being worked out. Gezari's census also finds that the fading of the light follows the hole's mass as the fallback of debris would predict, but the rise time does not.
• Tides along real paths. The formula on this page is exact for a radial faller around a hole that does not spin. For a spinning hole, or any path that is not radial, you need the full tidal tensor along the path. In 1983 Jean-Alain Marck showed that the special symmetry of the Kerr solution lets you write down, in closed form, a frame carried along any free-fall path around a spinning hole, and the tidal tensor in that frame; Kesden's calculation rests on it. These calculations refine the tidal forces for each trajectory.

Why is Ember not torn apart?

The tidal radius from the first fold applies to the Vigil's own star. Ember, a dwarf with 0.70 of the Sun's mass and 0.68 of its radius, would be pulled apart by Vesper inside 0.0082 AU, 2.6 times its own radius. It circles at 1.5 AU, 184 times farther out.

The more useful measure for a star on a steady orbit is its Roche lobe: the region, shaped like a teardrop pointing at the companion, inside which the star's own gravity holds its gas. Peter Eggleton's 1983 formula gives the lobe's size from the mass ratio alone. Ember's radius is 1.2 percent of its lobe's, and it would only start spilling gas toward the hole if the two were closer than 0.018 AU. Close binaries in which a star does overflow onto a black hole are among the brightest X-ray sources in the sky. Ember is far from that, so nothing streams from the star to Vesper, and the hole stays dark until the Vigil feeds it.

What did the seed feel on arrival?

The seed needed no close pass of either body. It stopped on its own fusion engine, burning for decades far from both (Stopping from 0.1c); the engine's push peaked at 0.061 m/s², the same on every part of the hull, so it stretched nothing. It then settled on Kindling, Ember's close planet. A star is a gentle tidal body: its mass is spread through a ball far wider than a stellar hole's horizon, so its pull changes slowly with distance. Over a 100 m hull, Ember's tide at Kindling's orbit is 1 × 10⁻⁸ m/s², 5 × 10⁸ times under the seed's limit.

The figures here treat each pass as a clean turn around one body. A real path from the end of the seed's burn onto Kindling's orbit runs through the fields of both Ember and Vesper, which is a three-body problem. Eventide's plan includes a search for such a path that keeps the tide under 5 m/s² over 100 m design; as of 2026 it has not been run.

Further reading

1. Eggleton (1983), Approximations to the radii of Roche lobes, The Astrophysical Journal 268, 368A simple formula, good to about one percent, for the size of a star's Roche lobe at any mass ratio.
2. Pirani (1956; republished 2009), On the physical significance of the Riemann tensor, General Relativity and GravitationThe Riemann tensor read as the relative acceleration of neighbouring free particles, which is the tidal field.
3. Marck (1983), Solution to the equations of parallel transport in Kerr geometry; tidal tensor, Proceedings of the Royal Society AA frame carried along any free-fall path around a spinning hole, in closed form, and the tidal tensor written in it.
4. Hills (1975), Possible power source of Seyfert galaxies and QSOs, NatureStars broken up inside a massive black hole's Roche limit as the engine of quasars and Seyfert nuclei; the swallowed-whole boundary bears his name.
5. Rees (1988), Tidal disruption of stars by black holes of 10⁶ to 10⁸ solar masses in nearby galaxies, NaturePart of a shredded star's debris escapes and the rest is swallowed, making a bright flare.
6. Kesden (2012), Tidal-disruption rate of stars by spinning supermassive black holes, Physical Review DRelativistic tides at closest approach, and how spin raises the heaviest hole that can disrupt a Sun-like star.
7. Gezari (2021), Tidal Disruption Events, Annual Review of Astronomy and AstrophysicsA census of 56 observed events and what they show.
8. Event Horizon Telescope Collaboration (2019), First M87 Event Horizon Telescope Results. I. The Shadow of the Supermassive Black Hole, The Astrophysical Journal LettersThe image of M87*'s shadow, and the mass this page uses for it.