Two-Body Relaxation Time

A gravitating stellar system is an N-body problem in disguise, and its entire theoretical machinery hinges on one question: does the discreteness of the stars matter? On short timescales every star coasts through the smooth mean field of all the others, on an orbit set by the global potential — the collisionless picture underlying the collisionless Boltzmann equation and Jeans theorem. But the potential is not truly smooth. Mass is carried by discrete point masses, and that graininess makes each star's energy and angular momentum drift, in a slow random walk, away from their mean-field values. The two-body relaxation time is the clock of that drift: the timescale over which the summed gravitational nudges of individual stars change a star's velocity by an amount comparable to itself. The decisive result, made quantitative by Chandrasekhar in 1942, is that relaxation is driven not by rare close collisions but by the cumulative effect of countless distant, weak deflections. Each faraway star tugs the test star by a vanishingly small angle, yet there are so many of them, and their contributions accumulate as a random walk, that the summed effect eventually rivals the star's own kinetic energy. Integrating over impact parameter produces a logarithmic weighting — the Coulomb logarithm lnΛ — cut off at the system size and at the distance where weak deflections give way to strong scattering. The upshot is one of the most important numbers in astrophysics, the ratio t_relax/t_cross ∼ N/(8 lnN): it grows with the number of stars, so bigger systems are more nearly collisionless. A galaxy of 10¹¹ stars is collisionless for the age of the Universe; a globular cluster of 10⁵–10⁶ stars relaxes within a few Gyr and shows every collisional signature — mass segregation and gravothermal core collapse.

Key takeaways

The two-body relaxation time $t_{\mathrm{relax}}$ is the timescale over which accumulated small-angle gravitational deflections change a star's velocity by order itself, erasing its initial orbit.
Distant, weak encounters dominate over rare strong ones: their cumulative effect scales as the Coulomb logarithm $\ln\Lambda$ , with $\Lambda\sim N$ .
The central estimate is $t_{\mathrm{relax}}\simeq \dfrac{0.1\,N}{\ln N}\,t_{\mathrm{cross}}$ , many crossing times when $N$ is large.
Systems with $t_{\mathrm{relax}}\gg$ age are collisionless (galaxies, dark-matter halos) and obey the collisionless Boltzmann equation; those with $t_{\mathrm{relax}}\lesssim$ age are collisional (globular and open clusters, galactic nuclei).
Relaxation drives equipartition, mass segregation, gravothermal core collapse, and evaporation — the secular evolution of dense stellar systems.
Key idea: relaxation is a velocity-space diffusion driven by the summed weak deflections of many distant encounters — the kicks add in quadrature, so $t_{\mathrm{relax}}$ is how long the cumulative random walk takes to reach order $v^2$ , longer than the crossing time by a factor $\sim N/\ln N$ .

The core insight

Two-body relaxation is a velocity-space diffusion process driven by the summed weak deflections of many distant encounters: the perturbations add in quadrature, so $t_{\mathrm{relax}}$ measures how long the cumulative random walk in velocity takes to reach order $v^2$ — and it exceeds the crossing time by the factor $\sim N/(8\ln N)$ .

Where this lives today

Two-body relaxation is the engine behind the most active problems in dense stellar dynamics. Direct N-body integrators (NBODY6++GPU) and Monte-Carlo codes (MOCCA, CMC) evolve million-body clusters over a Hubble time by resolving exactly this graininess, and their predictions — core-collapse times, mass functions, binary-burning — are matched against Gaia proper motions and HST photometry of Galactic globulars. Relaxation-driven segregation funnels the heaviest remnants inward, so a cluster's stellar-mass black holes sink, pair up, and merge or are ejected: the dynamical channel for the binary-black-hole mergers LIGO–Virgo–KAGRA detects. In galactic nuclei, relaxation refills the "loss cone" of orbits that plunge toward the central black hole, setting the rate of tidal disruption events and of the extreme-mass-ratio inspirals targeted by LISA; near Sgr A* the two-body process is supplemented by faster resonant relaxation of the near-Keplerian orbits. The same physics, run in reverse, sets the particle counts cosmological N-body simulations need to avoid spurious numerical relaxation of their dark-matter halos.

The physical picture

Picture one star ploughing through the sea of all the others. Every star it passes tugs it sideways by a minute angle. Any single tug is negligible, but the tugs point in random directions, so the star's velocity performs a random walk. Relaxation is complete when the accumulated wander in velocity has grown as large as the velocity itself — when the star has, in effect, forgotten the orbit it started on. The first figure fixes this: a single distant encounter delivers a transverse kick δv⊥ = 2Gm/(bv), and many such kicks pile up incoherently into a diffusing velocity.

Figure

A distant field star of mass m deflects the test star by δv⊥ = 2Gm/(bv); because successive kicks point in random directions the velocity performs a random walk, and a single crossing adds Δv⊥²/v² ≈ 8 lnΛ/N, so relaxation is complete after N/(8 lnΛ) crossings

Why do weak encounters win? A distant encounter is weak but common. The number of field stars at impact parameter b grows as b db, while each contributes a kick scaling as 1/b in velocity, hence 1/b² in energy. The product is db/b, whose integral is a logarithm — every decade of impact parameter contributes equally, and that flat, scale-free weighting is the origin of the Coulomb logarithm lnΛ. The second figure carries the payoff: the crossings needed to relax, N/(8 lnN), against star count N, and where real systems fall.

Figure

The relaxation time in crossing times, t_relax/t_cross = N/(8 lnN), versus the number of stars N: an open cluster with N ≈ 10³ relaxes in about 20 crossings, a globular with N ≈ 3×10⁵ in a few thousand, and a galaxy with N ≈ 10¹¹ in ~5×10⁸ crossings — vastly more than a Hubble time, so galaxies are collisionless

The scaling t_relax ∼ (N/lnN) t_cross says discreteness matters less as N grows: a smooth potential is a good approximation precisely because N is enormous, so galaxies are collisionless not because gravity is weak but because graininess averages out. Relaxation is also thermodynamic — it drives equipartition, ½ m⟨v²⟩ constant across masses, so heavy stars slow and sink (mass segregation) while light ones speed up and escape (evaporation) — and because a self-gravitating system has negative heat capacity, it does not settle into a stable isothermal state but runs away into gravothermal core collapse.

The core equations

The headline result compresses the whole subject into one ratio. It says the number of crossing times a system needs to relax is set almost linearly by its star count, softened only by the slowly varying Coulomb logarithm.

t_{\mathrm{relax}} \simeq \frac{0.1\,N}{\ln N}\, t_{\mathrm{cross}} \qquad (1)

where: $t_{\mathrm{relax}}$ — two-body relaxation time; $t_{\mathrm{cross}}\sim R/v$ — crossing time; $N$ — number of stars; $\ln N$ — Coulomb logarithm with $\Lambda\sim N$ . Reading: the coefficient $0.1\approx 1/8$ is order-unity; the physics lives in the $N/\ln N$ growth that makes big systems collisionless. A single encounter is a hyperbolic Kepler orbit, and the exact deflection of the relative velocity depends only on the impact parameter, the relative speed, and the total mass. This is the microscopic event whose accumulation is relaxation.

\tan\!\left(\frac{\theta}{2}\right) = \frac{G(m_1+m_2)}{b\,v_0^2} \qquad (2)

where: $\theta$ — scattering angle of the relative velocity; $b$ — impact parameter; $v_0$ — relative speed at infinity; $m_1,m_2$ — the two masses. Reading: strong deflection ( $\theta\to 90^\circ$ ) needs small $b$ ; distant passages barely bend the orbit. For the distant passages that dominate relaxation, the deflection is small and the transverse velocity kick follows from the impulse approximation — treat the trajectory as a straight line and integrate the perpendicular force.

|\delta v_\perp| \simeq \frac{2Gm}{b\,v_0}\qquad (b \gg b_{90}) \qquad (3)

where: $\delta v_\perp$ — perpendicular velocity change per encounter; $m$ — perturber mass; $b_{90}$ — right-angle-deflection impact parameter. Reading: the kick falls off only as $1/b$ , so distant stars still contribute — the slow fall-off is why they add up. The boundary between weak and strong scattering is the 90° impact parameter, obtained by setting $\theta=90^\circ$ in equation (2). It is the natural inner cutoff of the relaxation integral.

b_{90} = \frac{G(m_1+m_2)}{v_0^2} \qquad (4)

where: $b_{90}$ — impact parameter giving a right-angle deflection. Reading: inside $b_{90}$ the weak-kick formula (3) fails; such close passages are rare and handled separately. Summing the weak kicks in quadrature over one crossing gives the fractional velocity diffusion, and its reciprocal is the number of crossings to relax. This is the quantitative heart of the subject.

\frac{\Delta v_\perp^2}{v^2} \simeq \frac{8\ln\Lambda}{N},\qquad n_{\mathrm{relax}} = \frac{N}{8\ln\Lambda},\qquad \Lambda \equiv \frac{b_{\max}}{b_{\min}}\approx\frac{R\,v^2}{Gm}\sim N \qquad (5)

where: $\Delta v_\perp^2$ — summed mean-square transverse kick per crossing; $n_{\mathrm{relax}}$ — crossings to relax; $\Lambda$ — Coulomb argument, $b_{\max}\sim R$ , $b_{\min}\sim b_{90}$ . Reading: the whole many-body sum collapses to $N$ over a logarithm, because the diffusion is a velocity random walk. For quantitative work on real clusters the estimate is sharpened into Spitzer's median relaxation time, expressed in locally measurable quantities — velocity dispersion and mass density — rather than the global $N$ and $R$ .

t_{\mathrm{relax}} = \frac{0.34\,\sigma^3}{G^2 m\,\rho\,\ln\Lambda},\qquad t_{\mathrm{evap}} \approx 100\,t_{\mathrm{relax}} \qquad (6)

where: $\sigma$ — one-dimensional velocity dispersion; $\rho$ — stellar mass density; $m$ — mean stellar mass; $t_{\mathrm{evap}}$ — evaporation time. Reading: relaxation is fast where density is high and dispersion low, so it runs fastest in cluster cores.

Derivation AThe transverse kick from a distant passage

The goal is to derive the single-encounter velocity kick $\delta v_\perp = 2Gm/(bv)$ that seeds everything else, using the impulse approximation valid for the distant, weak passages that dominate relaxation. We treat the test star as moving in a straight line at constant speed past a stationary perturber of mass $m$ , integrate the transverse gravitational force over the whole passage, and find the result depends on impact parameter only as $1/b$ — the slow fall-off that lets faraway stars matter.

Step 1. Place the perturber at the origin and let the test star move along the line $x=vt$ at perpendicular distance $b$ , so its separation is $r=\sqrt{b^2+v^2t^2}$ . The gravitational acceleration has magnitude $Gm/r^2$ ; only its component perpendicular to the motion survives the integral, because the along-track component is antisymmetric in time and cancels. Why: to first order the deflection is entirely transverse — the parallel tug speeds the star up on approach and slows it equally on recession.

Step 2. The transverse acceleration is $a_\perp = (Gm/r^2)\,(b/r) = Gm\,b/(b^2+v^2t^2)^{3/2}$ . Integrate over the entire passage: $\delta v_\perp = \int_{-\infty}^{\infty} \dfrac{Gm\,b\,\mathrm{d}t}{(b^2+v^2t^2)^{3/2}}$ . Substituting $u=vt/b$ reduces this to $\dfrac{Gm}{bv}\int_{-\infty}^{\infty}\dfrac{\mathrm{d}u}{(1+u^2)^{3/2}}$ . Why: the substitution strips the physical scales out front, leaving a pure dimensionless integral to evaluate once and for all.

Step 3. The remaining integral is standard: $\int_{-\infty}^{\infty}(1+u^2)^{-3/2}\,\mathrm{d}u = \left[u/\sqrt{1+u^2}\right]_{-\infty}^{\infty}=2$ . Hence $\delta v_\perp = 2Gm/(bv)$ , valid whenever the deflection is small, i.e. $b\gg b_{90}$ . Why: the factor of 2 is exact and the $1/b$ scaling is the crux — energy kicks go as $\delta v_\perp^2\propto 1/b^2$ , which when weighted by the $b\,\mathrm{d}b$ number of perturbers gives the logarithmic integrand.

Go deeper · Derivation B · 1 minRelaxation time from summed weak encounters

Derivation BRelaxation time from summed weak encounters

The goal is to derive the central result $n_{\mathrm{relax}} = N/(8\ln\Lambda)$ , hence $t_{\mathrm{relax}} = [N/(8\ln\Lambda)]\,t_{\mathrm{cross}}$ , by adding the weak kicks of equation (3) in quadrature over a single traversal. We count the encounters a test star suffers in one crossing from the projected surface density of stars, integrate their squared kicks over impact parameter, normalize by $v^2$ with the virial relation, and invert to find how many crossings relaxation needs.

Step 1. In one crossing the test star sees the whole cluster projected onto the sky at surface number density $N/(\pi R^2)$ . The number of field stars it passes with impact parameter between $b$ and $b+\mathrm{d}b$ is therefore $\delta n = \dfrac{N}{\pi R^2}\,2\pi b\,\mathrm{d}b = \dfrac{2N}{R^2}\,b\,\mathrm{d}b$ . Why: because kicks in different directions are uncorrelated, we track the mean-square kick $\delta v_\perp^2$ , not the vector sum, which averages to zero.

Step 2. Each encounter contributes $\delta v_\perp^2 = (2Gm/bv)^2$ . Summing over all impact parameters in one crossing gives $\Delta v_\perp^2 = \displaystyle\int_{b_{\min}}^{b_{\max}} \left(\frac{2Gm}{bv}\right)^2 \frac{2N}{R^2}\,b\,\mathrm{d}b = \frac{8N\,G^2m^2}{R^2 v^2}\int_{b_{\min}}^{b_{\max}}\frac{\mathrm{d}b}{b} = \frac{8N\,G^2m^2}{R^2 v^2}\,\ln\Lambda$ . Why: the integrand collapses to $\mathrm{d}b/b$ , and its integral is $\ln\Lambda=\ln(b_{\max}/b_{\min})$ — every decade of impact parameter contributes equally, which is the whole reason distant encounters dominate.

Step 3. Normalize by $v^2$ using the virial estimate $v^2\sim GNm/R$ , so $v^4\sim G^2N^2m^2/R^2$ . Then $\dfrac{\Delta v_\perp^2}{v^2} = \dfrac{8N\,G^2m^2}{R^2 v^4}\ln\Lambda = \dfrac{8\ln\Lambda}{N}$ . Why: virial equilibrium fixes the characteristic speed in terms of $N$ , $m$ , and $R$ , converting the dimensional prefactor into a clean $8\ln\Lambda/N$ .

Step 4. Relaxation is complete when the cumulative mean-square kick — which grows linearly with the number of crossings, being a random walk — equals $v^2$ . Inverting the per-crossing fraction gives $n_{\mathrm{relax}} = (\Delta v_\perp^2/v^2)^{-1} = N/(8\ln\Lambda)$ , so $t_{\mathrm{relax}} = n_{\mathrm{relax}}\,t_{\mathrm{cross}}$ . With $\Lambda\sim N$ this is $t_{\mathrm{relax}}\sim[N/(8\ln N)]\,t_{\mathrm{cross}}\approx 0.1\,(N/\ln N)\,t_{\mathrm{cross}}$ . Why: this reproduces Binney & Tremaine §1.2 up to the order-unity coefficient; the $N$ and $\ln\Lambda$ scalings are exact and are what physically matter.

Go deeper · Derivation C · 1 minThe Coulomb logarithm from close and distant cutoffs

Derivation CThe Coulomb logarithm from close and distant cutoffs

The goal is to explain why the relaxation integral converges only logarithmically and to pin down the cutoffs that fix $\ln\Lambda$ . The mean-square velocity change from encounters between $b$ and $b+\mathrm{d}b$ goes as $\mathrm{d}(\Delta v_\perp^2)\propto \mathrm{d}b/b$ , which diverges at both ends without cutoffs; we identify the inner and outer limits from the breakdown of the weak-kick approximation and the finite system size, and show their ratio is of order the star number.

Step 1. The integrand is $\mathrm{d}(\Delta v_\perp^2)\propto (Gm/bv)^2\,b\,\mathrm{d}b = (Gm/v)^2\,\mathrm{d}b/b$ . This $1/b$ form is the signature of the long-range $1/r^2$ force: the logarithmic integral gives no preference to any scale, so the divergences at small and large $b$ are only cut off by physics outside the idealization. Why: a short-range force would give a convergent integral dominated by close encounters; gravity's reach is what makes distant encounters collectively decisive.

Step 2. The inner cutoff is the 90° impact parameter, $b_{\min}\sim b_{90}=G(m_1+m_2)/v^2$ . Below it the deflection is order unity, the linear impulse approximation fails, and a single strong scattering — rare, and treated separately — replaces the weak-kick accounting. Why: extending the weak formula to $b\to 0$ would spuriously blow up the kick; $b_{90}$ is where "many gentle nudges" gives way to "one hard collision".

Step 3. The outer cutoff is the system size, $b_{\max}\sim R$ . Beyond it there is no uniform background of perturbers, and the collective tidal field of the whole system — already contained in the mean potential — takes over from individual encounters. Why: relaxation is about the graininess on top of the smooth field, so the largest relevant impact parameter is where the smooth field description begins.

Step 4. The ratio is $\Lambda = b_{\max}/b_{\min}\sim Rv^2/(Gm)$ , and the virial relation $v^2\sim GNm/R$ collapses it to $\Lambda\sim N$ , so $\ln\Lambda\sim\ln N$ — typically 10 to 20 for real stellar systems. Why: only the logarithm of the cutoff ratio enters, so the answer is famously insensitive to the exact choices of $b_{\min}$ and $b_{\max}$ , which is why $\ln\Lambda\approx\ln(0.4N)$ is used with confidence.

Limiting cases

The relaxation time measured against the system's age sorts every stellar system into a physical regime:

Case	Result	Meaning
Galaxies	$t_{\mathrm{relax}}\gg t_{\mathrm{age}}$	With $N\sim10^{11}$ , $t_{\mathrm{relax}}\sim10^{16}$ – $10^{17}\,\mathrm{yr}$ : strictly collisionless; the smooth-potential Jeans description is exact for all practical purposes.
Globular clusters	$t_{\mathrm{relax}}\lesssim t_{\mathrm{age}}$	With $N\sim10^{5}$ – $10^{6}$ , central $t_{\mathrm{relax}}\sim10^{8}$ – $10^{9}\,\mathrm{yr}$ : collisional; showing core collapse, mass segregation, evaporation.
Open clusters	$t_{\mathrm{relax}}\ll t_{\mathrm{age}}$	With $N\sim10^{2}$ – $10^{3}$ , $t_{\mathrm{relax}}$ is short; these sparse systems relax and dissolve within a few $10^{8}\,\mathrm{yr}$ .
Galactic nuclei and cusps	$t_{\mathrm{relax}}\sim t_{\mathrm{age}}$	Nuclear clusters and regions near a central black hole are marginally collisional, permitting loss-cone refilling and resonant relaxation.

Worked examples

1 · A galaxy is collisionless (order of magnitude)

The Milky Way disk has $N\sim10^{11}$ stars, radius $R\sim10\,\mathrm{kpc}$ , and rotation speed $v\sim200\,\mathrm{km/s}$ . Is it collisional on a Hubble time?

Step 1. Crossing time: $t_{\mathrm{cross}}\sim R/v = (3\times10^{20}\,\mathrm{m})/(2\times10^{5}\,\mathrm{m/s}) = 1.5\times10^{15}\,\mathrm{s}\approx \mathbf{5\times10^{7}\,\mathrm{yr}}$ , so $t_{\mathrm{cross}}\sim10^{8}\,\mathrm{yr}$ — consistent with the Sun's $\sim2\times10^{8}\,\mathrm{yr}$ orbital period.

Step 2. Coulomb logarithm: $\ln\Lambda\sim\ln N = \ln(10^{11}) = 25.3$ , so $8\ln\Lambda\approx 203$ .

Step 3. From equation (5): $t_{\mathrm{relax}} = \dfrac{N}{8\ln\Lambda}\,t_{\mathrm{cross}} = \dfrac{10^{11}}{203}\times10^{8}\,\mathrm{yr} \approx \mathbf{5\times10^{16}\,\mathrm{yr}}$ .

Step 4. Against the age of the Universe, $1.4\times10^{10}\,\mathrm{yr}$ , the ratio is $\sim3\times10^{6}$ : the galaxy is collisionless by six orders of magnitude — its stars have never once relaxed, which is why the collisionless Boltzmann equation and Jeans theorem describe it so well.

2 · Central relaxation time of a globular cluster (fully worked)

Take a dense globular-cluster core: one-dimensional dispersion $\sigma=10\,\mathrm{km/s}$ , mean stellar mass $m=1\,M_\odot$ , and central mass density $\rho\approx10^{4}\,M_\odot\,\mathrm{pc^{-3}}$ , with $\ln\Lambda\approx12$ . Find $t_{\mathrm{relax}}$ from Spitzer's formula.

Step 1. Convert to SI. $\sigma = 1.0\times10^{4}\,\mathrm{m/s}$ , so $\sigma^3 = 1.0\times10^{12}\,\mathrm{m^3\,s^{-3}}$ . $m = 1.99\times10^{30}\,\mathrm{kg}$ . Density: $10^{4}\,M_\odot\,\mathrm{pc^{-3}} = (2.0\times10^{34}\,\mathrm{kg})/(2.94\times10^{49}\,\mathrm{m^3}) = \mathbf{6.8\times10^{-16}\,\mathrm{kg\,m^{-3}}}$ , using $1\,\mathrm{pc}=3.086\times10^{16}\,\mathrm{m}$ .

Step 2. Assemble the denominator of equation (6): $G^2 m\,\rho\,\ln\Lambda = (6.67\times10^{-11})^2(1.99\times10^{30})(6.8\times10^{-16})(12)$ . Working left to right: $G^2 = 4.45\times10^{-21}$ ; times $m$ gives $8.86\times10^{9}$ ; times $\rho$ gives $6.02\times10^{-6}$ ; times $\ln\Lambda$ gives $\mathbf{7.2\times10^{-5}}$ (SI units of $\mathrm{m^3\,s^{-4}}$ ).

Step 3. Numerator: $0.34\,\sigma^3 = 3.4\times10^{11}$ . Divide: $t_{\mathrm{relax}} = (3.4\times10^{11})/(7.2\times10^{-5}) = 4.7\times10^{15}\,\mathrm{s}$ .

Step 4. Convert to years: $t_{\mathrm{relax}} = (4.7\times10^{15}\,\mathrm{s})/(3.16\times10^{7}\,\mathrm{s/yr}) \approx \mathbf{1.5\times10^{8}\,\mathrm{yr}}$ . This is roughly 1% of the cluster's $\sim1.2\times10^{10}\,\mathrm{yr}$ age, so the core has relaxed about a hundred times over — thoroughly collisional, consistent with the observed mass segregation and collapsed cores. Evaporation would dissolve such a cluster on $t_{\mathrm{evap}}\approx100\,t_{\mathrm{relax}}\sim1.5\times10^{10}\,\mathrm{yr}$ , a Hubble time.

Physics of the portrait

Two-body relaxation is where thermodynamics enters stellar dynamics, and it carries the name of the man who first computed it. Before 1942 the "relaxation time" of a star cluster was estimated crudely, as if stars collided like billiard balls; the realization that distant encounters dominate, and that the sum converges only logarithmically, was Chandrasekhar's.

Origins · 1 minSubrahmanyan Chandrasekhar and how this was discovered

Subrahmanyan Chandrasekhar

1910–1995 · stellar dynamics and the theory of relaxation

Subrahmanyan Chandrasekhar — "Chandra" to the field — recast stellar dynamics as a branch of statistical mechanics. In his 1942 monograph Principles of Stellar Dynamics he showed that a star crossing a cluster is deflected overwhelmingly by many weak, distant encounters rather than by rare close ones, and that summing their contributions over impact parameter yields a logarithmic factor cut off at the system size and the strong-scattering radius — the Coulomb logarithm, borrowed by analogy from a charged particle in a plasma. His 1943 papers went further, treating a star's velocity as a quantity diffusing under a Fokker–Planck equation, and introducing dynamical friction: the systematic drag a massive body feels as it stirs a wake in the background stars, the companion of random-walk diffusion. That framework remains the language of the collisional evolution of clusters and galactic nuclei. Already famous for the Chandrasekhar mass of white dwarfs (1983 Nobel Prize), he treated stellar dynamics as one of several complete subjects he built and left, each in a definitive book.

History

1942

— Chandrasekhar's Principles of Stellar Dynamics founds the weak-encounter theory of relaxation

1943

— Chandrasekhar formulates dynamical friction and the Fokker–Planck picture of velocity diffusion

1968

— Lynden-Bell and Wood identify the gravothermal catastrophe driving core collapse

1980

— Cohn's Fokker–Planck models follow a cluster core all the way to collapse

1987

— Spitzer's monograph systematizes the collisional evolution of globular clusters

Going further

Texts. Binney & Tremaine, Galactic Dynamics, 2nd ed., §1.2 (relaxation and the Coulomb logarithm) and §7.4 (Fokker–Planck and collisional evolution) · Spitzer, Dynamical Evolution of Globular Clusters (1987) · Chandrasekhar, Principles of Stellar Dynamics (1942) · Carroll & Ostlie, An Introduction to Modern Astrophysics, §25.1 · Mo, van den Bosch & White, Galaxy Formation and Evolution, §5.4.

Provenance. The weak-encounter derivation and the $N/(8\ln\Lambda)$ result follow Binney & Tremaine §1.2, using the surface-density counting that yields the coefficient 8 cleanly; the impulse-approximation kick $\delta v_\perp=2Gm/(bv)$ and the deflection angle $\tan(\theta/2)=G(m_1+m_2)/(bv_0^2)$ are the standard two-body results; the Spitzer median relaxation time $t_{\mathrm{relax}}=0.34\,\sigma^3/(G^2 m\rho\ln\Lambda)$ and $t_{\mathrm{evap}}\approx100\,t_{\mathrm{relax}}$ are from Spitzer (1987); the Coulomb-logarithm cutoffs and $\Lambda\sim N$ trace to Chandrasekhar (1942). A verifier should recompute: the impact-parameter integral yielding $\ln\Lambda$ ; the ratio $t_{\mathrm{relax}}/t_{\mathrm{cross}}=N/(8\ln N)$ ; the galaxy estimate ( $\sim5\times10^{16}\,\mathrm{yr}$ ); and the globular-cluster Spitzer value ( $\sim1.5\times10^{8}\,\mathrm{yr}$ ). Order-unity prefactors are approximate; the scalings in $N$ and $\ln\Lambda$ are exact.

Prerequisites. Gravitational Two-Body Problem · Gravitational N-Body Problem · Distribution Function and Phase Space · Virial Theorem in Stellar Systems. Leads to. Collisionless Boltzmann Equation · Jeans Theorem · Galactic Potential Models.

Ask the tutor: Want to see how the crossing-time and Spitzer estimates of $t_{\mathrm{relax}}$ compare for a specific cluster, or how relaxation feeds into core collapse and the Fokker–Planck equation? Ask and we will work it through.

Why is a bigger system more collisionless, when it has more stars to bump into?

It is counterintuitive but exact. Adding stars does add perturbers, but for a self-gravitating system in virial balance it also raises the characteristic speed. Each star's fractional velocity change per crossing scales as $8\ln\Lambda/N$ , so it falls almost as $1/N$ : doubling the star count nearly halves the relative disturbance any one star suffers per orbit. That is why a galaxy of $10^{11}$ stars, despite containing vastly more potential collision partners, is far more nearly collisionless than a cluster of $10^{5}$ .

What exactly is the Coulomb logarithm, and why "Coulomb"?

It is the logarithm of the ratio of the largest to the smallest relevant impact parameter, $\ln\Lambda=\ln(b_{\max}/b_{\min})$ , that emerges when you integrate the weak-encounter kicks over impact parameter. The name is borrowed from plasma physics: a charged particle scattering off other charges through the long-range Coulomb force produces exactly the same $\mathrm{d}b/b$ integrand and the same logarithmic cutoff structure, because gravity and electrostatics share the $1/r^2$ form. In stellar systems $b_{\max}\sim R$ (the system size) and $b_{\min}\sim b_{90}=Gm/v^2$ (where deflections stop being weak), giving $\Lambda\sim N$ and $\ln\Lambda\approx10$ – $20$ . Its great virtue is insensitivity: only the logarithm enters, so even a factor-of-two error in a cutoff barely moves the answer.

Is the relaxation time the same everywhere in a cluster?

No — it varies strongly with position, and that variation drives the cluster's evolution. Spitzer's formula, $t_{\mathrm{relax}}\propto\sigma^3/\rho$ , makes relaxation fastest where the density is high and the velocity dispersion is low, which is the core. So a globular can have a central relaxation time of $\sim10^{8}\,\mathrm{yr}$ while its half-mass time is $\sim10^{9}\,\mathrm{yr}$ and its outskirts have barely relaxed. Because the core relaxes first, it reaches equipartition, segregates its heavy stars, and collapses long before the envelope responds. Quoted single numbers are usually the "half-mass relaxation time", a global average, but the physics happens on the shorter central timescale.

How does relaxation drive mass segregation and equipartition?

Relaxation is thermodynamic: encounters exchange kinetic energy and push the system toward equipartition, where $\tfrac12 m_i\langle v_i^2\rangle$ is the same for every mass species $i$ . To reach it, heavy stars must slow down and light stars speed up. A slower heavy star sinks deeper into the well, so massive stars and remnants concentrate toward the centre — mass segregation — while lightened low-mass stars puff outward and some reach escape speed. Because a species' relaxation time scales roughly as $1/m$ , the heaviest objects segregate on a fraction of the overall relaxation time. Full equipartition is never reached (the Spitzer instability intervenes), but the partial segregation is directly observed as centrally concentrated blue stragglers and binaries.

Why does relaxation cause core collapse instead of a comfortable equilibrium?

Because a self-gravitating system has negative heat capacity — take energy out of a bound orbit and it speeds up, like a satellite that drops to a lower, faster orbit as drag removes its energy. Relaxation lets heat flow from the hot, dense core to the cooler halo, but losing heat makes the core contract and grow hotter, steepening the gradient and driving yet more heat outward. There is no stable stopping point: the process runs away, the central density diverging in finite time. This gravothermal catastrophe, identified by Lynden-Bell and Wood in 1968, collapses a cluster core on roughly $15$ – $20$ central relaxation times, and it explains why about a fifth of Galactic globular clusters show cuspy, collapsed surface-brightness profiles.

Does two-body relaxation matter for dark-matter halos and cosmological simulations?

Physically, no — real dark-matter halos have astronomically many particles, so their relaxation time is vastly longer than the age of the Universe, and they are exquisitely collisionless (the whole premise of cold dark matter). But numerically, yes, as a hazard. An N-body simulation represents a halo with far fewer, much heavier macro-particles, so its artificial relaxation time $\propto N/\ln N$ is short. Too few particles and spurious two-body relaxation can puff out a halo's centre, erase substructure, and mimic a physical core. Modellers therefore pick particle numbers and softening lengths to keep the numerical relaxation time safely longer than the run — a direct, practical use of the formula derived here.