Astronomy & Celestial Mechanics
the sky, computed
Concept

Relaxation time — where it appears

How long the accumulated small deflections from distant encounters take to change a star's velocity by an amount comparable with its own. It exceeds the crossing time by roughly the number of stars over its logarithm, which makes clusters collisional and galaxies not.

Named by 10 essays across 2 fields — each of them below, with the objects they name alongside it.

7 decades of relaxation time, and a Hubble time between them. Crossing time and two-body relaxation time for 5 self-gravitating systems, against the number of bodies in each. The lower marks are the crossing time R/σ, which spans 3 decades; the upper ones are t_relax ≈ 0.1N/ln N times it, which spans 8. The horizontal rule is a Hubble time, and the same expression puts these systems on opposite sides of it. Two of them relax — an open cluster and a globular cluster — so their stars have exchanged energy, their heavy stars have sunk, and their present structure is not the one they were born with. The rest do not: a dwarf spheroidal, an elliptical galaxy, a cluster of galaxies, the elliptical missing it by a factor of 10⁶, which is why a galaxy is modelled as a smooth potential with no stars in it at all. The dependence is on N and hardly at all on anything else, because the Coulomb logarithm makes every decade of impact parameter contribute equally. A cluster that relaxes also evaporates: about a stellar mass in a hundred and forty leaves per relaxation time, so the globular cluster empties in 6.1·10¹¹ years.

How long a system takes to forget

A star crossing a galaxy is deflected by every other star, and the deflections add as a random walk. The time for that walk to change a star's energy by its own amount is a hundred million billion years for a galaxy and a billion for a globular cluster, and everything about how the two are modelled follows from which side of a Hubble time that falls on.

gravitation · Two-body relaxation 7 decades of relaxation time, and a Hubble time between them. Crossing time and two-body relaxation time for 5 self-gravitating systems, against the number of bodies in each. The lower marks are the crossing time R/σ, which spans 3 decades; the upper ones are t_relax ≈ 0.1N/ln N times it, which spans 8. The horizontal rule is a Hubble time, and the same expression puts these systems on opposite sides of it. Two of them relax — an open cluster and a globular cluster — so their stars have exchanged energy, their heavy stars have sunk, and their present structure is not the one they were born with. The rest do not: a dwarf spheroidal, an elliptical galaxy, a cluster of galaxies, the elliptical missing it by a factor of 10⁶, which is why a galaxy is modelled as a smooth potential with no stars in it at all. The dependence is on N and hardly at all on anything else, because the Coulomb logarithm makes every decade of impact parameter contribute equally. A cluster that relaxes also evaporates: about a stellar mass in a hundred and forty leaves per relaxation time, so the globular cluster empties in 6.1·10¹¹ years.

A cluster that boils itself away

A star cluster has no thermostat. Encounters between its members push a few of them above escape speed, the cluster loses them, and losing them makes it contract — which makes it hotter, which makes more of them escape. A self-gravitating system heats up as it loses energy, and the process ends by destroying the system.

gravitation · Two-body relaxation Hard below 8.3 astronomical units, soft above, and nothing settles at the line. The binding energy of a binary of two 0.7 solar-mass stars against its separation, both axes logarithmic, with the mean kinetic energy of a single cluster star at a velocity dispersion of 5 kilometres a second drawn as a level. Where the curve is above the level the binary is bound more tightly than a passing star's motion, and encounters on average take energy out of the field and put it into the pair; where it is below, they do the reverse. The crossing at 8.3 astronomical units is the hard–soft boundary, and it is the entire content of Heggie's law: hard binaries harden and soft binaries soften. The arrows are the direction each side moves, and they point away from the crossing in both directions rather than toward it. That is not a coincidence but a negative heat capacity, the same property that makes a star contract when it radiates: taking energy out of a bound pair moves it closer together and speeds it up, so a hard binary that gives energy to the cluster becomes harder still and gives more. A cluster with binaries in it therefore has a heat source that turns itself up, and the boundary drawn here is a watershed rather than an equilibrium.

A pair that heats what is trying to cool

A star cluster has a negative heat capacity, so it cannot reach equilibrium — its core contracts and gets hotter without limit. What stops it is a binary, and which way a binary exchanges energy with the stars around it is settled by one comparison of two energies.

gravitation · Binary heating The whole kick, delivered in about two encounter times. The transverse force a body feels while a mass sweeps past it on a straight line, and the velocity that force has delivered so far, both against time in units of the impact parameter divided by the relative speed. The force is the component of the inverse-square attraction perpendicular to the path, which is the impact parameter over the cube of the distance, and it is drawn at its peak value of one at closest approach. The rising curve is its running integral, scaled by twice the gravitational constant times the mass over the impact parameter and the speed. Two things are visible and both are the point. The integral of the force over all time is exactly two in these units, so the kick is exactly 2GM/bv with no free constant anywhere — an answer to a three-body-shaped question obtained without solving anything. And it arrives quickly: 71 per cent of it within a single encounter time of closest approach and 98.6 per cent within the 6 drawn, which is what licenses calling the whole thing an impulse. On the scale of anything slower, the velocity changes discontinuously.

An answer obtained along a path that was not taken

Integrate the force of a passing mass along the straight line the body would have followed if the encounter had not happened, and out comes an exact deflection with no free constant in it. The approximation is circular, it is wrong in a known direction by a known amount, and it is the reason stellar dynamics has closed forms at all.

gravitation · Impulse approximation Dynamical friction is strongest at 0.97 times the dispersion, and weaker either side. The dynamical friction on a body moving through a Maxwellian sea of stars, against its own speed in units of the square root of two times the velocity dispersion. The dashed curve is the fraction of the field moving slower than the body, which is the only part of the field that contributes: a star overtaking from behind pulls the body forward exactly as often as one being overtaken pulls it back, so the fast tail cancels out entirely. That fraction rises from nothing to one. The drag is that fraction divided by the square of the speed, and the quotient of a saturating numerator and a growing denominator has a single maximum, here at 0.97. The consequences run in both directions. A body moving much faster than the stars around it is barely slowed at all, which is why a galaxy passing through a cluster at a thousand kilometres a second does not sink; and a body already at rest with respect to the field feels nothing either, because there is no wake to be behind it. Sinking is therefore fastest in the middle of the process rather than at the start of it.

A drag that is strongest in the middle

Dynamical friction is not like air resistance. Only the stars moving slower than the body contribute at all, so the drag rises from nothing, peaks when the body has slowed to about the speed of everything around it, and falls away again — which is why a fast satellite is never captured and a slow one sinks in a hurry.

gravitation · Dynamical friction Supply and demand in a cluster's core. The rate at which hard binaries heat a star cluster's core, against the core's central density, for binary fractions of 1%, 5%, 20%, both on logarithmic axes and in units that put the balance for five per cent at one. The heating rate per binary does not depend on how hard the binary is — the energy released per encounter is proportional to its binding, and the chance of an encounter to its size, so the two cancel — and summed over a core whose radius shrinks as its density rises, the total grows as the square root of the density. The horizontal line is the demand: the energy the rest of the cluster carries away by relaxation through its half-mass radius, which the core has no say in. Each supply curve meets it at one density: 25 for 1%, 1.0 for 5%, 0.063 for 20%. The core contracts or expands until it produces exactly what is demanded, and a core rich in binaries does so at a lower density.

A core that makes what the cluster asks for

After core collapse, a star cluster's binaries heat it — but how much they heat it is not theirs to decide. The rest of the cluster carries energy outward at a rate fixed by its own size and mass, and the core contracts or swells until its binaries supply exactly that. So the core's structure follows from the halo's demand, and the cluster expands at a rate that forgets how it was born.

gravitation · Binary heating A cusp steepened into a spike by the black hole growing in it. The dark-matter density near the Galaxy's central black hole of 4.3·10⁶ solar masses, in solar masses per cubic parsec on logarithmic axes. The dashed line is the halo's NFW cusp, falling inward as one over the radius. If the black hole grew slowly at the centre of that cusp, the dark matter would have followed it inward on adiabatically shrinking orbits, steepening the cusp inside about a fifth of the black hole's sphere of influence — here 0.4 parsecs, with the sphere of influence at 1.8 — into a spike falling as r to the minus 7/3, measured off the drawn curve. Close in, the dark matter's own annihilation would cap the density at the particle mass over the annihilation rate times the Galaxy's age — for a 100 GeV particle with the thermal cross-section, 2.8·10⁸ solar masses per cubic parsec — the flat plateau. Between the spike's edge and its plateau the density rises by 6 orders of magnitude.

A cusp a black hole would sharpen

A black hole that grows slowly at the centre of a dark-matter cusp drags the dark matter in after it, turning a density that rises as one over the radius into a spike rising as the radius to the minus seven-thirds. Whatever the cusp was, the spike comes out nearly the same — and the annihilation signal it would produce is set less by the particle than by whether the spike survived the stars around it.

galaxies · Halo profiles How fast a disc forgets, by the mass of the lumps it is made of. The two-body relaxation time of a star moving at 30 km/s among perturbers of mass m, against m, for a fixed mass density of 0.015 solar masses per cubic parsec — about the density of molecular gas in the plane near the Sun — with a Coulomb logarithm of 5. The time falls as one over the perturber's mass: the same mass in heavier lumps relaxes a population proportionally faster. The orange point is the stars themselves, 0.04 solar masses per cubic parsec in half-solar-mass pieces: 4.9·10⁹ million years, 3.5·10⁵ times the age of the universe (dashed). Molecular clouds of 5·10⁵ solar masses give 13 billion years as point masses — comparable to the age of the disc. Fewer than a hundred-thousandth of the bodies, carrying less than half the mass, relax the disc 3.8·10⁵ times faster than all the stars together.

A few thousand clouds outrun a hundred billion stars

Encounters between stars relax the Galactic disc in a time hundreds of thousands of times the age of the universe, so in practice they do nothing. Put a fraction of the same mass into a few thousand molecular clouds and the time falls to the age of the disc — because relaxation counts the mass of each perturber, not their number. The disc's present thickness is the record of that, and of a bound on anything heavier hiding in the halo.

gravitation · Two-body relaxation Five clocks around the black hole at the centre of the Galaxy. Timescales against distance from a black hole of 4.3·10⁶ solar masses at the centre of a cluster whose stellar mass inside r is 10⁶ (r/1 pc)^1.5 solar masses, made of stars of 1 solar mass. Ordinary two-body relaxation (orange) is slow everywhere, 8.4·10¹⁰ years at 0.01 pc. Orbits that close are nearly Keplerian and keep their orientation until they precess — through the stars' own mass farther out, through relativity close in — and over that coherence time the torques between them add rather than random-walk. That makes scalar resonant relaxation (blue) faster than two-body relaxation inside about 0.57 pc, and vector resonant relaxation (green), which turns orbital planes and is not stopped by precession within a plane, faster still: 3.5·10⁷ years at 0.1 pc, against the 6·10⁶-year age of the young stars there. Coefficients of order one are set to one; the slopes are the content.

Torques that add because the orbits hold still

Near a massive black hole every star's orbit is almost a fixed ellipse, repeated thousands of times before it precesses. The torques the other stars exert on it therefore point the same way orbit after orbit, and they add in proportion to time instead of cancelling in a random walk — so angular momentum near the hole relaxes hundreds of times faster than ordinary encounters allow, and orbital planes faster still.

gravitation · Two-body relaxation Which populations can share a cluster's temperature. The share of a cluster's mass in a heavy population against the mass of its members relative to an average star of 0.5 solar masses, both on logarithmic scales, with Spitzer's limit: a heavy component can reach energy equipartition with the stars only below the line where its share of the mass times its members' mass ratio to the power three halves equals 0.16. Above it the heavy bodies sink, cannot slow down enough to match the stars' kinetic temperature, and decouple into a subsystem of their own. Stellar black holes, 10 solar masses and about 2% of the mass, sit at 1.8 — 11 times beyond the limit. Neutron stars sit at 0.05 and massive white dwarfs at 0.14, below it; binaries, at a tenth of the mass, sit at 0.28, just beyond it, which is part of why they gather toward the centre. Only the black holes are beyond it by an order of magnitude.

A population that cannot share the cluster's temperature

Encounters between stars push a cluster toward equal kinetic energies, so heavy bodies slow down and sink. Spitzer showed in 1969 that past a certain point they cannot slow down enough: the heavy population falls away from the rest and becomes a cluster of its own. Stellar black holes are past that point by a factor of ten, in every globular cluster that kept them — and the subsystem they form, not the stars' binaries, is what holds most clusters' cores open until the black holes have been spent.

gravitation · Binary heating

Named alongside it

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

Two-body relaxationCore collapseGlobular clusterBinary heatingCoulomb logarithmMass segregationEquipartitionImpulse approximationNegative heat capacityAdiabatic invariantClose encounterDynamical friction

All concepts