The stars a black hole eats come from a narrow band

A star torn apart by a black hole makes a flare that says something about the hole, and the flare’s timing depends on how its debris manages to collide. All of that begins with a star arriving at the tidal radius, and the natural question is how often that happens. The answer is not set by how many stars are near the hole. It is set by how quickly stars that were not on a disrupting orbit can be put onto one.

The reason is that a disrupting orbit is extraordinarily specific. To come within the tidal radius a star has to be moving almost straight at the hole, with almost no sideways motion, and stars already on such orbits are destroyed the first time they reach pericentre. The set of such orbits — the loss cone — is emptied once, early, and after that it is refilled only as encounters with other stars nudge a star’s angular momentum from one orbit to the next. The disruption rate is the rate of a random walk into a very small target, and almost all of the physics is in which of two bottlenecks limits it.

A cone a millionth of the sky #

A star at a distance $r$ from the hole reaches the tidal radius $r_t$ only if its angular momentum is below the value for an orbit with that pericentre. Measured as a fraction of the angular momentum of a circular orbit at $r$, squared, that threshold is

$$R_{\rm lc} \approx \frac{2GM\,r_t}{r^2\,v_c^2},$$

where $v_c$ is the circular speed. For a Sun-like star and a hole of a million solar masses, at the radius of the hole’s sphere of influence, 1.47 parsecs, $R_{\rm lc}$ is about a millionth. In terms of directions, a star there has to be moving within about a thousandth of a radian — six hundredths of a degree — of straight at the hole.

The name comes from the shape of that set in velocity space: at each point, the velocities that lead to disruption fill a narrow cone pointing at the hole. The same construction appears wherever particles are lost at a centre. In the Earth’s radiation belts, electrons whose velocities lie too close to the magnetic field line mirror so deep that they strike the atmosphere and are removed within one bounce, and the belts are drained at the rate their loss cone is refilled by scattering. A black hole’s nucleus is the gravitational version, with the scattering done by other stars.

Two ways to be limited #

Stars change their angular momenta by two-body relaxation: the accumulated effect of many weak encounters, which randomises a star’s orbit over the relaxation time $t_r$. Over one orbital period $P$, a star’s squared angular momentum, as a fraction of the circular value, wanders by about $P/t_r$. Comparing that with the size of the cone gives the one number the whole problem turns on:

$$q = \frac{P}{t_r\,R_{\rm lc}}.$$

Where $q$ is small, the cone is empty. A star’s angular momentum moves so little in one orbit that it cannot jump into the cone from outside; it has to diffuse up to the cone’s edge and is destroyed at the next pericentre after crossing it. The cone stays essentially empty, and the rate is set by how fast relaxation delivers stars to its edge — the number of stars divided by the relaxation time, reduced by a logarithm of the cone’s size.

Where $q$ is large, the cone is full. A star’s angular momentum wanders across the whole cone and beyond in a single orbit, so the cone is repopulated as fast as it is emptied, and the fraction of stars on disrupting orbits is simply $R_{\rm lc}$. The rate is then that fraction of the stars, divided by the time they take to reach pericentre — the orbital period — and relaxation no longer matters.

The logarithm in the empty regime has a simple origin. When the cone is empty, stars are being removed at its edge, so their number falls off towards the edge; but a diffusing population drains slowly, and the deficit reaches a long way into the surrounding range of angular momenta. The profile that results declines only logarithmically towards the cone, and the flux across its edge is reduced by the logarithm of the cone’s inverse size — about fourteen, for a cone of a millionth. A full cone has no such depletion, because it is repopulated faster than it can empty.

At the sphere of influence of a million-solar-mass hole the orbital period is about a hundred thousand years and the relaxation time about six billion, so a star’s squared angular momentum wanders by about two parts in a hundred thousand each orbit, sixteen times the cone. That point is in the full regime. Closer in, orbits are much shorter but the cone is much larger, and $q$ falls steeply; the figure’s boundary, where $q$ is one, is at 0.19 influence radii.

Why the band is narrow #

The two regimes push the flux in opposite directions, and that is what concentrates it.

Inside the boundary the rate is limited by relaxation, and it grows outward: there are more stars at larger radii, and inside the hole’s sphere of influence the relaxation time grows only slowly with radius. Outside the boundary the rate is limited by the cone, and it falls outward: the cone shrinks as the inverse square of radius while orbital periods lengthen, so a full cone at large radius delivers very little. A rate that climbs until one bottleneck takes over from the other peaks where they swap, and in the figure the flux peaks at 0.27 influence radii, just outside the boundary at 0.19, and falls away on both sides of it.

That is a result about where the stars that are destroyed come from, and it is robust in a way the total is not. Whatever the nucleus looks like in detail, most disrupted stars were on orbits of a few tenths of the sphere of influence before they were scattered in — a fraction of a parsec, for a hole of a million solar masses.

Lighter and heavier holes #

The mass of the hole enters twice: directly, through the tidal radius and the sphere of influence, and through the nucleus around it, since heavier holes sit in nuclei with higher velocity dispersions.

A lighter hole sits in a nucleus with a lower dispersion, which by the velocity dispersion’s role in holding a stellar system up means a denser one at the hole’s own sphere of influence, and relaxation is faster in denser, slower systems. The boundary moves inward in units of the influence radius, to 0.07, and the total rate rises to 10⁻⁴ a year.

A heavier hole sits in a hotter, sparser nucleus with slower relaxation. The boundary moves out to 0.53 influence radii, beyond the peak, so more of the delivery is in the regime limited by relaxation, and the total falls.

The trend can be estimated in a few lines. At the sphere of influence of an isothermal nucleus the density goes as $\sigma^6/M^2$, and with the dispersion rising as $M^{1/4.38}$ that is $M^{-0.63}$. The relaxation time goes as $\sigma^3/\rho$, or $M^{1.32}$, and the number of stars inside the sphere of influence goes as $M$. The empty-regime rate, stars divided by relaxation time, therefore goes as about $M^{-0.3}$.

The full calculation gives a gentler slope, $M^{-0.20}$, because the logarithm of the cone’s size and the shifting position of the peak both soften the estimate. The sign is what matters, and it is the same as the sign of every other selection in this subject. Lighter holes tear apart stars more often, and above the capture ceiling heavier ones swallow them without a flare. A survey of flares is weighted towards the lightest holes twice over.

What a million-solar-mass hole eats in a Hubble time #

At 7.4·10⁻⁵ a year, a nucleus like the one in the first figure destroys a star about every fourteen thousand years. That is too rare to wait for in any one galaxy, which is why the flares are found by surveys that watch hundreds of thousands of galaxies at once.

Over ten billion years it is a large number. Seven hundred thousand stars would be disrupted, and with about half of each star bound to the hole, several hundred thousand solar masses would be offered to a hole of a million — a substantial fraction of its mass, even allowing for most of the returning debris being blown away rather than swallowed. For the lightest holes, whose rates are highest and whose masses are smallest, disruptions are among the candidates for how they grew at all.

The measured rate, from optical surveys, is a few times 10⁻⁵ per galaxy per year. Calculations like this one have generally come out higher, by factors of a few, and the gap has narrowed as surveys have found fainter flares. For a model with a single dispersion, a single stellar mass and a spherical nucleus, agreement to that level is better than it has a right to be. The count comes from surveys that watch the whole of the visible sky every few nights and now find tens of these flares a year; divided by the number of galaxies in which they could have been seen, it is a rate per galaxy.

The Milky Way’s own hole, of about four million solar masses, would on the same model destroy a star every eighteen thousand years or so, and a century of watching was never likely to catch one. Within its sphere of influence a hole in this model is surrounded by about twice its own mass in stars — the influence radius is, by construction, where the stellar mass enclosed becomes comparable to the hole’s — so over ten billion years a million-solar-mass hole destroys roughly a third as many stars as it has neighbours inside that radius, and the supply has to be replenished from beyond it.

The logarithm nobody can pin down #

The relaxation time contains the Coulomb logarithm, $\ln\Lambda$, which counts the range of encounter distances that contribute to scattering. It is a logarithm of a ratio of the largest to the smallest relevant distance, and for a nucleus it is somewhere between ten and twenty, with no way of fixing it better from first principles — the same logarithm that dynamical friction carries, and with the same difficulty.

A smaller logarithm means slower relaxation, which pushes the boundary outward and lowers the rate. But the rate falls by less than the logarithm does: a third less relaxation gives a rate lower by 28 per cent, not 33, because the stars delivered from the full regime do not depend on relaxation at all.

The heavier holes are more sensitive: the rate at 3·10⁷ solar masses falls by about a third, and at 10⁵ by under a quarter. That follows from the flux figures. Round the heavier holes more of the delivery is inside the boundary, where relaxation is the bottleneck and the logarithm enters directly; round the lighter ones more comes from the full regime, which does not care.

A cusp only the lighter holes have had time to grow #

Inside the sphere of influence the figures assume a density rising towards the hole as the radius to the power −7/4. That profile is not arbitrary. It is the steady state a population of stars reaches round a massive point mass once relaxation has had time to act — the same relaxation that lets a star cluster boil itself away. Stars diffuse in energy, are destroyed or swallowed at the centre, and the distribution settles into the one that carries no net flow of energy through it. John Bahcall and Richard Wolf derived it in 1976, and it is the reason the rate inside the boundary stays finite.

A steady state needs time, and the time is the relaxation time at the sphere of influence. For a hole of a million solar masses that is about six billion years, less than the age of the universe, so a relaxed cusp is a reasonable expectation. The relaxation time grows faster than the hole’s mass — as its 1.3 power, by the estimate above — so for a hole of ten million solar masses it exceeds a hundred billion years and the cusp cannot have formed, while for a hole of a hundred thousand it is a few hundred million years and the cusp would have formed many times over.

The calculation is most secure where the rates are highest. For the heavier holes on these figures the nucleus has not had time to reach the profile assumed, and the true inner density could be lower, or, if stars formed there recently, higher. The Milky Way’s centre, the one nucleus near enough to count its stars individually, poses exactly this question: its old stars are distributed less steeply towards the hole than a relaxed cusp would require, and why is still argued.

The loss cone itself was emptied long before any of this. Stars that happened to be on disrupting orbits when the hole formed were destroyed within an orbital period — about ten thousand years at a few tenths of the sphere of influence — and every flare since has been a star that relaxation carried in from outside.

What the model leaves out #

Real nuclei are not spherical. In a flattened or triaxial nucleus, stars on some orbits have their angular momentum changed by the smooth potential as well as by encounters, and it can carry them into the cone far faster than relaxation. Rates in such nuclei can exceed the spherical estimate by large factors.

Stars are not all the same. A realistic population has many low-mass stars and a few massive ones, and the massive ones relax the others faster; stellar remnants sink towards the centre. Both change the relaxation time by factors of a few.

Correlated encounters. Close to the hole, orbits are nearly Keplerian and the torques between them add coherently for long stretches — resonant relaxation — which can change angular momenta faster than ordinary relaxation, though for most of the band that supplies disruptions it is not the dominant process.

And a second hole. A binary black hole at the centre of a nucleus scatters stars into its loss cone with great efficiency for a while, briefly raising the rate by orders of magnitude, and depletes the stars on low-angular-momentum orbits in doing so — the same shortage of such stars that stalls the binary’s own shrinking.

A rate that belongs to the nucleus, not the hole #

The answer to how often a black hole tears a star apart is, in the end, a statement about the nucleus more than about the hole: how many stars there are at a few tenths of its sphere of influence, and how fast they relax. A hole of given mass in a dense, recently formed nucleus should flare far more often than the same hole in an old, relaxed, sparse one, and that is a prediction of the whole of this picture that is easy to state and hard to calculate.

Still open: why flares prefer galaxies that recently stopped forming stars #

The galaxies in which tidal disruption flares are found are not a fair sample. They are strongly overrepresented among post-starburst galaxies — systems that formed stars vigorously until a few hundred million years ago and then stopped, which sit in the thinly populated region between star-forming and quiescent galaxies. A burst that built a dense central star cluster, a recent merger that left the nucleus flattened, or a second black hole still sinking to the centre would each raise the rate, and the preference is one of the few observations able to tell which of the omissions above matters most.