The line a star is swallowed at is not the horizon

The heaviest black hole that can tear a star apart is found by setting two lengths equal. One is the tidal radius, the distance at which the hole’s pull across the star exceeds the star’s own gravity; it grows as the cube root of the hole’s mass. The other is the distance at which the star is lost to the hole before it can be disrupted; it grows as the mass itself. Two lengths with different slopes cross once, and above the crossing no flare can be produced.

The first calculation of that crossing took the second length to be the horizon, and the horizon is the natural first guess. It is also the wrong line. A star does not have to cross the horizon to be lost. It has only to come close enough that the orbit it is on can no longer carry it back out, and for a star falling in from far away that happens well outside the horizon. The correct line is the marginally bound orbit, and it is twice as far out for a hole that does not rotate — and for a hole that does, it moves by a factor of five depending on which way the star arrives.

Why a star can be lost outside the horizon #

The reason is a property of orbits near a black hole that has no counterpart in Newtonian gravity. Round an ordinary mass, the angular momentum of an orbiting body builds a wall in its effective potential: the closer the body comes, the faster it has to swing sideways, and the centrifugal term rises without limit, so any body with angular momentum turns round before it reaches the centre.

General relativity adds a term to that potential that grows faster than the centrifugal one as the distance shrinks, and near the hole it wins. The wall no longer rises for ever; it reaches a maximum and falls away inside it. A body arriving with too little angular momentum to clear the top of the wall does not bounce. It goes over the top and falls in, horizon or no horizon.

For a body that begins at rest far away the energy is exactly that of an escaping body, and the question is what angular momentum makes the top of the wall reachable. The answer, for a hole that does not spin, is the orbit whose closest approach is four gravitational radii. Any star whose pericentre would be smaller than that is captured. Four gravitational radii is twice the Schwarzschild radius, and it is the relevant line for disruption because the star has to survive to its pericentre to be torn apart there.

A star’s orbit is a parabola to a part in a million #

Using the orbit of a body that starts at rest at infinity is not an idealisation of convenience; it is very nearly exact, and the reason is where the stars come from.

A star is put onto a disrupting orbit by the slow accumulation of small gravitational nudges from other stars, and the orbits that deliver it are the nearly radial ones that the stellar population has to keep resupplying. Most of that resupply happens near the radius at which the hole’s gravity first dominates the galaxy’s, a parsec or more out for a hole of a few million solar masses, where the star’s orbital energy per unit mass is of the order of the square of the local stellar velocity dispersion — for a hundred kilometres per second, a ten-millionth of $c^2$. At a pericentre of a few gravitational radii the kinetic energy per unit mass is a sizeable fraction of $c^2$. The star’s binding to the hole, measured against the energy it has when it arrives, is therefore a few parts in ten million. To the hole, every star it disrupts is arriving from infinity, and the marginally bound orbit is the line exactly.

Spin moves the line, and direction matters #

A rotating hole drags the space around it. An orbit that goes round in the same sense as the rotation is carried along, needs less angular momentum of its own to stay out, and can come closer before it is lost. An orbit that goes the other way is working against the drag and is lost further out.

That is the asymmetry the first figure draws. The horizon of a spinning hole shrinks with spin, from two gravitational radii to one, whichever way anything is moving round it, because it is a property of the hole alone. The marginally bound orbit splits: for a star arriving with the rotation it falls from four gravitational radii to 1.09 at the highest spin astrophysical holes are expected to reach, and for a star arriving against it, it rises to 5.83. The same hole swallows a star arriving from one direction at five times the distance it lets another star, arriving from the other, pass close and escape.

For a star arriving out of the equatorial plane the line lies between the two extremes, which is why the figures here are for equatorial orbits: they bracket every other case.

The ceiling, redrawn #

Replacing the horizon with the marginally bound orbit changes the heaviest hole that can make a flare, and the spin then changes it again.

The factor of 2.8 between the horizon ceiling and the correct one is simple arithmetic once the lengths are right. The tidal radius goes as $M^{1/3}$ and the capture radius as $M$, so their ratio goes as $M^{-2/3}$; moving the capture radius out by a factor of two therefore lowers the crossing mass by $2^{3/2}$, which is 2.83. That is the whole of the correction for a hole that does not spin, and it applies to every ceiling quoted from the horizon.

The spin then spreads the ceiling over a factor of twelve for the same star: from 2.3·10⁷ solar masses, for a star arriving against the rotation of a maximally spinning hole, to 2.84·10⁸ for one arriving with it. Because the ceiling depends on the capture radius to the power $-3/2$, a factor of 5.3 in radius becomes a factor of twelve in mass.

A flare that should not exist #

The consequence runs in the useful direction. If a flare is seen from a galactic nucleus whose hole is known, from the motions of stars and gas around it or from the galaxy’s velocity dispersion, to be heavier than the non-spinning ceiling, the flare is not an inconsistency. It says that the hole spins, that the star arrived going the same way round, or that the star was larger than the Sun — and the three can be told apart.

Such a case has been argued. An exceptionally luminous transient discovered in 2015, first read as the most powerful supernova ever seen, was later interpreted as a tidal disruption by a hole of around a hundred million solar masses or more, in a galaxy whose central mass placed it above the ceiling for a non-spinning hole and a Sun-like star. If that interpretation is right, the event itself is a measurement that the hole spins rapidly and that the disrupted star arrived nearly aligned with that spin. The interpretation is contested, and what is not contested is the logic: above four times ten to the seven solar masses, a Sun-like star cannot be disrupted by a hole that does not spin.

A larger star lifts the ceiling, and a smaller one lowers it #

The disrupted star is the other free parameter, and it moves the ceiling by more than the spin does.

A swollen star has a far larger tidal radius for the same mass, so it is disrupted by much heavier holes: the non-spinning ceiling rises by a factor of 22.6 for a factor of eight in radius, because the ceiling goes as the star’s radius to the power 3/2. A star that has left the main sequence and begun to climb the giant branch, its envelope expanding because its core contracts, can be torn apart by a hole of nearly a billion solar masses, and around a maximally spinning hole by one of six billion.

The commonest stars in any galaxy are low-mass dwarfs, and they are the ones whose ceiling is lowest. That means the population of flares from the heaviest holes should be dominated by the rarer evolved stars and by spinning holes, and the flares from lighter holes by ordinary dwarfs. A catalogue of flares sorted by host mass is therefore also, in principle, a catalogue of which stars and which spins were available at each mass — though with the handful of events known above the non-spinning ceiling, that is a statement about the future.

The ceiling is a density #

The two stars above changed both radius and mass, and the ceiling moved by 22.6 for one and by exactly one half for the other. Both numbers follow from a single fact about the crossing, which is clearer written out than drawn. The tidal radius is $R_\star (M/m_\star)^{1/3}$ and the capture radius is $k\,GM/c^2$, with $k$ equal to 4 for a hole that does not spin. Setting them equal and solving for the hole’s mass:

$$M_{\rm ceiling} = \left(\frac{c^2}{k\,G}\right)^{3/2} \left(\frac{R_\star^3}{m_\star}\right)^{1/2} \;\propto\; k^{-3/2}\,\bar\rho_\star^{\,-1/2}.$$

The star enters only through its mean density. The half-solar star is four times denser than the Sun, and its ceiling is half the Sun’s; the swollen star is 512 times less dense, and its ceiling is $\sqrt{512}$, or 22.6, times higher.

There is a picture behind the formula. The tidal radius is, by construction, the distance at which the hole’s mass spread through a sphere of that radius would have the star’s own mean density — the star is torn apart where it is no denser than the hole’s gravity averaged over its orbit. The mean density inside the capture radius falls as $M^{-2}$, because the radius grows as the mass. So the ceiling is simply the hole whose mean density inside its capture line equals the star’s. For the Sun that density is 1.41 grams per cubic centimetre, and a hole of 4.04·10⁷ solar masses holds exactly that density inside four gravitational radii. A black hole can disrupt a star only if, in this sense, it is less dense than the star.

The densest stars need the lightest holes #

That makes a prediction about the densest ordinary stars, and it is a sharp one.

The mass is chosen because it is where the white dwarf mass distribution peaks. At that mass a white dwarf is about three hundred thousand times denser than the Sun, and its ceiling is lower by the square root of that: 522 times, from 4.04·10⁷ solar masses to 7.74·10⁴. No supermassive black hole can disrupt one. Even a maximally spinning hole, with the star arriving along its rotation, stops producing white dwarf flares above about half a million solar masses.

That range — tens of thousands to a few hundred thousand solar masses — is the range of the intermediate-mass black holes whose existence is argued largely from indirect evidence. A tidal disruption of a white dwarf would be direct evidence of one, and it would look quite different from the flares of Sun-like stars: the debris of so compact a star returns to pericentre within about an hour rather than over a month or two, and the tidal compression can be strong enough to ignite nuclear burning in the star as it passes. The degeneracy that holds a white dwarf up also makes the heaviest ones the smallest, so the ceiling falls further as the white dwarf’s mass rises, the opposite of what a ceiling proportional to stellar mass would suggest.

The first ceiling, and what it assumed #

The horizon version of the argument is still worth drawing, because the difference between it and the corrected one is exactly the physics this essay adds.

The horizon crossing assumes that a star is safe until it passes the one surface from which nothing returns. What it leaves out is that the orbit, not the surface, is what the star is following, and an orbit can be committed to falling in while still outside. The same distinction appears wherever general relativity is compared with its Newtonian limit near a compact object: the innermost stable circular orbit of an accretion disc lies outside the horizon for the same reason, and it moves with spin in the same way, from six gravitational radii for a non-rotating hole to one for a maximally rotating one on the prograde side. That circular orbit is how a hole’s spin is measured from the light of its disc, where the spin turns out to be one length in disguise. The marginally bound orbit is the parabolic counterpart of that circular one, and the capture of a star is the parabolic counterpart of the inner edge of a disc.

What the lines leave out #

The star is a point until it arrives. The tidal radius is computed for a star of uniform response, and a real star is centrally concentrated, so it is fully disrupted only somewhat inside the radius drawn here and partially stripped outside it. Detailed simulations move the effective tidal radius by a factor of order one, which moves every ceiling by the same factor to the power 3/2.

The orbits are equatorial. A star arriving at an angle to the hole’s spin plane sees a capture radius between the prograde and retrograde values, so every real ceiling lies between the extremes drawn, and averaging over random arrival directions gives a typical ceiling for a spinning hole that is higher than the non-spinning one but well below the prograde extreme.

The tidal radius is Newtonian. Close to a heavy hole the tidal field itself is modified by relativity, and for a star disrupted at a few gravitational radii that modification is not small. It strengthens the tide slightly and does not change the ordering or the size of the spin effect.

And the ceiling says nothing about the flare’s brightness. A star disrupted just outside the capture line is disrupted deep in the hole’s potential, where the debris orbits precess strongly and the returning streams collide close in. Whether that makes the flare brighter or earlier is a separate question, and it is not the one the ceiling answers.

What a ceiling is #

A ceiling in this subject is not a limit on what a black hole can do; it is a limit on what can be seen, and every such limit is only as good as the line it is drawn against. Moving that line from the horizon to the marginally bound orbit lowers the ceiling by a factor of 2.8 for every star, and making the hole spin spreads it by a factor of twelve. Neither change requires any new observation. They require only taking the orbit the star is on, rather than the surface it might eventually reach, as the thing that decides its fate — and noticing that what the hole compares against the star, in the end, is a density.

Still open: why a flare from a light hole is late #

The debris of a disrupted star does not fall straight in. It returns to pericentre on elongated orbits and swings back out, and the relativistic precession of those orbits decides where the outgoing stream crashes into the stream still falling. Round a heavy hole the swing is large and the collision prompt and close; round a light one the swing is a few degrees and the streams may not meet until far out, which suggests that the lightest holes, whose orbital clocks are fastest, may be the slowest to light up.