Jeans Theorem

Give a galaxy its statistics and you stop tracking a hundred billion stars: you replace them with a smooth distribution function $f(\mathbf{x},\mathbf{v},t)$ , the phase-space density of stars in each box of position and velocity. Over a galaxy's age the graininess of that swarm almost never matters — two-body encounters take millions of times the age of the Universe to reshuffle orbits — so $f$ flows along orbits in the smooth mean-field potential $\Phi(\mathbf{x})$ , obeying the collisionless Boltzmann equation $\mathrm{d}f/\mathrm{d}t=0$ . Jeans theorem turns this into a construction kit for equilibrium models. Its logic is almost embarrassingly short. In a steady state the equation says $f$ never changes as you ride along an orbit; an integral of motion is, by definition, a quantity that never changes along an orbit. So the two are one idea in different clothes: any steady-state $f$ is a function of the integrals, and vice versa. The theorem is about building blocks — which functions of the six phase-space coordinates may appear in an equilibrium $f$ . The answer: only the integrals of motion, and among those only the isolating ones that cut phase space cleanly enough to label an orbit. That one restriction organizes equilibrium galactic dynamics: isotropic spheres are $f(E)$ , anisotropic spheres $f(E,L)$ , flattened disks need a hidden third integral, and the cleanest modern models are $f(\mathbf{J})$ in the three actions. Take velocity moments of any of them and out drop the Jeans equations, so this theorem is the microscopic bedrock beneath the machinery used to weigh galaxies and clusters.

Key takeaways

Jeans theorem: any steady-state solution of the collisionless Boltzmann equation depends on phase space only through integrals of motion, and any function of such integrals is a steady-state solution.
An integral of motion $I(\mathbf{x},\mathbf{v})$ is constant along every orbit; only isolating ones — those confining an orbit to a lower-dimensional surface — can be independent arguments of $f$ .
In a spherical potential $E$ and $\mathbf{L}$ are isolating, so $f=f(E,\mathbf{L})$ ; an isotropic model collapses to $f=f(E)$ , the basis of Eddington's inversion formula.
The strong form: when almost all orbits are regular, $f$ is a function of the three actions, $f=f(J_1,J_2,J_3)$ — the framework used to fit Gaia data on the Milky Way.
The theorem applies because galaxies are collisionless: a galaxy's two-body relaxation time is roughly $6\times10^{16}$ yr, about four million times the age of the Universe.
Key idea: in a steady state the collisionless Boltzmann equation makes $f$ constant along each orbit, and integrals of motion are exactly the phase-space functions constant along orbits, so steady-state distribution functions are functions of the isolating integrals alone.

The core insight

In a steady state the collisionless Boltzmann equation makes $f$ constant along each orbit, and integrals of motion are precisely the phase-space functions constant along orbits; therefore every steady-state distribution function is a function of the isolating integrals alone.

Where this lives today

Jeans theorem is the load-bearing assumption behind how modern surveys weigh the Galaxy and its neighbours. Gaia has measured positions and velocities for over a billion Milky Way stars, and the action-based codes that fit them — Agama, Galpy — write the distribution function directly as $f(\mathbf{J})$ , the purest realization of the strong theorem, because actions stay meaningful as the potential slowly evolves. For external galaxies, Schwarzschild's orbit-superposition method builds an equilibrium by weighting a library of orbits, each defined by its integrals of motion; this is the technique that delivered the mass of the supermassive black hole in M87, later imaged by the Event Horizon Telescope. In the Milky Way's dwarf spheroidal satellites, spherical Jeans modelling on an assumed $f(E,L)$ turns line-of-sight velocities into a dark-matter density, feeding the cored-versus-cusped debate. The open frontier is where the theorem is weakest: real galaxies harbour resonant and chaotic orbits, bars and arms are not steady, and quantifying how much of phase space is genuinely regular — and so how well $f=f(\mathbf{J})$ can fit it — is active research.

The physical picture

Picture a single star's orbit threading the six-dimensional phase space of positions and velocities. The steady-state collisionless Boltzmann equation says the value of $f$ never changes as you ride along that orbit — $f$ is painted onto orbits, one colour per orbit. Which coordinates stay fixed along an orbit? Exactly the integrals of motion. So the integrals are the natural labels, and specifying $f$ means giving one number per orbit as a function of its integrals. The first figure shows this in the surface-of-section view, where each orbit is a closed loop that $f$ decorates with a single value.

Figure

Orbits foliate phase space and f is painted on them: in the r–v_r plane each bound orbit traces a closed invariant loop, and the steady-state condition df/dt = 0 forces f to take one constant value all the way around each loop, so f = f(E, L) — a density over loops, not over points

Now zoom out. The integrals foliate phase space into nested surfaces: energy alone slices the six-dimensional space into five-dimensional shells, and each extra isolating integral cuts them thinner, until a full set of three shrinks the shells to three-dimensional invariant tori and $f$ becomes a density over the space of tori — one value per set of three actions. Which symmetry the potential has decides how far this slicing goes, and so which distribution functions are allowed. That hierarchy — chaotic, spherical, axisymmetric, integrable — is the theorem's real content, and the second figure lays it out as a ladder from weakest to strongest.

Figure

Symmetry sets which integrals are isolating, and that sets the allowed form of f: a chaotic potential leaves only energy so f = f(E); a spherical one adds angular momentum, f = f(E, L); an axisymmetric one adds L_z and a non-classical third integral; a fully integrable one has three actions and a regular orbit is a 3-torus in the 6-dimensional phase space

The distinction between an isolating and a non-isolating integral is the difference between a map coordinate and a tangled thread: an isolating integral cuts phase space cleanly, like a contour line, while a non-isolating one winds so densely across the energy surface that it never separates neighbouring orbits and cannot appear in $f$ . Energy is always isolating; whether anything else is depends on the potential.

The core equations

The starting point is the equation of motion for the whole swarm at once. In the collisionless limit — no close encounters, only the smooth potential — the phase-space density is conserved along the flow, Liouville's theorem specialised to a mean field.

\frac{\mathrm{d}f}{\mathrm{d}t}=\frac{\partial f}{\partial t}+\mathbf{v}\cdot\nabla f-\nabla\Phi\cdot\frac{\partial f}{\partial\mathbf{v}}=0 \qquad (1)

where: $f(\mathbf{x},\mathbf{v},t)$ — phase-space distribution function; $\Phi(\mathbf{x})$ — smooth mean-field potential; $\mathbf{v}$ — velocity; $\nabla,\ \partial/\partial\mathbf{v}$ — spatial and velocity gradients. Reading: the total derivative following a star is zero, so $f$ is conserved along orbits. An integral of motion is any phase-space function that is likewise constant along orbits. Carrying no explicit time dependence, its total derivative is built from the same convective operator as the steady-state Boltzmann equation.

\frac{\mathrm{d}I}{\mathrm{d}t}=\mathbf{v}\cdot\nabla I-\nabla\Phi\cdot\frac{\partial I}{\partial\mathbf{v}}=0 \qquad (2)

where: $I(\mathbf{x},\mathbf{v})$ — a phase-space function constant along every orbit. Reading: an integral obeys exactly the operator equation a steady-state $f$ must obey — that coincidence is the whole theorem. Setting the explicit time dependence to zero leaves the steady-state condition: the same operator acting on $f$ must vanish, so $f$ is constant along orbits, exactly as an integral is.

\frac{\partial f}{\partial t}=0\ \Longrightarrow\ \mathbf{v}\cdot\nabla f-\nabla\Phi\cdot\frac{\partial f}{\partial\mathbf{v}}=0 \qquad (3)

where: $\partial f/\partial t$ — explicit time dependence, zero in equilibrium. Reading: $f$ and every integral $I$ are annihilated by the identical operator. Comparing (2) and (3) gives the theorem in one line: $f$ can depend on phase space only through quantities annihilated by that same operator — the isolating integrals.

f=f\big(I_1(\mathbf{x},\mathbf{v}),\ldots,I_n(\mathbf{x},\mathbf{v})\big) \qquad (4)

where: $I_1,\ldots,I_n$ — the isolating integrals admitted by the potential. Reading: any function of the integrals the symmetry allows is a valid equilibrium; nothing else is. In a spherical potential the symmetry hands us the integrals for free: energy from time-translation invariance, the full angular-momentum vector from rotational invariance.

E=\tfrac{1}{2}v^2+\Phi(r),\qquad \mathbf{L}=\mathbf{r}\times\mathbf{v} \qquad (5)

where: $E$ — specific orbital energy; $\mathbf{L}$ — specific angular-momentum vector; $r$ — galactocentric radius. Reading: so $f=f(E,\mathbf{L})$ , and every orbit is planar because $\mathbf{L}$ fixes its plane. If the model has no preferred direction and equal dispersion in all velocity components, dependence on $\mathbf{L}$ drops out — the isotropic sphere. Integrating $f(E)$ over velocities gives the density.

f=f(E),\qquad \rho(r)=4\pi\int_0^{\sqrt{-2\Phi}} f(E)\,v^2\,\mathrm{d}v,\quad E=\tfrac12 v^2+\Phi \qquad (6)

where: $f(E)$ — distribution function of energy alone; $\rho(r)$ — mass density; $\Phi$ — potential (negative, vanishing at infinity). Reading: this forward map from $f(E)$ to $\rho(r)$ inverts uniquely. That inversion is Eddington's formula: given a spherical density and its self-consistent potential, it returns the one isotropic distribution function reproducing them, if the result is non-negative.

f(\mathcal{E})=\frac{1}{\sqrt{8}\,\pi^2}\left[\int_0^{\mathcal{E}}\frac{\mathrm{d}^2\rho}{\mathrm{d}\Psi^2}\frac{\mathrm{d}\Psi}{\sqrt{\mathcal{E}-\Psi}}+\frac{1}{\sqrt{\mathcal{E}}}\left(\frac{\mathrm{d}\rho}{\mathrm{d}\Psi}\right)_{\Psi=0}\right] \qquad (7)

where: $\Psi=-\Phi$ — relative potential (positive); $\mathcal{E}=-E$ — relative energy (positive for bound stars). Reading: $f\ge0$ everywhere is a non-trivial constraint that rules some density profiles out as isotropic models. The strong form replaces the ad hoc list of integrals with the canonical actions of a regular orbit — the cleanest arguments $f$ can have.

f=f(J_1,J_2,J_3) \qquad (8)

where: $J_i$ — the three actions, canonical isolating integrals of a regular orbit. Reading: when almost every orbit is regular this is the most general steady-state $f$ ; actions are adiabatic invariants, so the model survives slow changes of $\Phi$ .

Derivation AJeans theorem as a statement of sufficiency and necessity

The goal is to prove both halves of the theorem. Sufficiency: a one-line chain rule shows any function built from integrals satisfies the steady-state collisionless Boltzmann equation. Necessity: when almost all orbits are regular, any steady-state $f$ is forced to be a function of the isolating integrals, because constancy along an orbit plus dense filling of an invariant torus leaves no other freedom. Work in the smooth mean-field potential $\Phi(\mathbf{x})$ where orbits obey $\dot{\mathbf{x}}=\mathbf{v}$ and $\dot{\mathbf{v}}=-\nabla\Phi$ .

Step 1. Write the steady-state condition explicitly: with $\partial f/\partial t=0$ , equation (1) reduces to $\mathbf{v}\cdot\nabla f-\nabla\Phi\cdot\partial f/\partial\mathbf{v}=0$ , the plain assertion that $f$ takes one value all along a single orbit. Why: a steady state cannot let the local density drift as the star it describes moves, so equilibrium and constant-along-orbits are synonyms.

Step 2. Prove sufficiency. Let $f=f(I_1,\dots,I_n)$ with each $I_k$ an integral. The chain rule gives $\mathrm{d}f/\mathrm{d}t=\sum_k(\partial f/\partial I_k)\,\mathrm{d}I_k/\mathrm{d}t=0$ , since every $\mathrm{d}I_k/\mathrm{d}t=0$ . Why: no property of $f$ beyond differentiability was used, so the construction manufactures equilibria wholesale — a practical model-building recipe.

Step 3. Set up necessity. Suppose almost all orbits are regular, so through a generic point passes one invariant torus labelled by a full set of isolating integrals (the actions), and a non-resonant orbit winds around its torus, passing arbitrarily close to every point. Why: the torus, not the instantaneous point, carries an orbit's identity, and dense winding promotes constancy along the curve to constancy over the whole torus.

Step 4. Close the argument. A steady-state $f$ is constant along the orbit; since the orbit fills its torus densely and $f$ is continuous, $f$ is constant over the whole torus. Distinct tori are labelled precisely by the isolating integrals, so $f$ varies only from torus to torus — a function of those integrals alone. Why: continuity turns a statement about a measure-zero curve into one about the whole torus it covers.

Step 5. Caveat. On a resonant torus the orbit closes and does not fill the torus, so $f$ may vary across it; in a chaotic region only $E$ is isolating, so the clean form fails. The tidy conclusion holds when such regions have negligible volume. Why: the strong theorem is a statement about a particular potential, not a universal law — integrability must be checked.

Go deeper · Derivation B · 1 minEddington inversion from a spherical density to the isotropic distribution function

Derivation BEddington inversion from a spherical density to the isotropic distribution function

The goal is to show that the $f(E)$ ansatz Jeans theorem permits for an isotropic sphere is not merely allowed but uniquely determined by the density. We write the density as a velocity integral of $f$ , change variables so it becomes an Abel transform, and invert that transform to recover $f$ from $\rho$ and its self-consistent potential. Use the relative potential $\Psi=-\Phi$ and relative energy $\mathcal{E}=-E=\Psi-\tfrac12 v^2$ , so bound stars have $\mathcal{E}>0$ .

Step 1. Start from equation (6) in relative variables: $\rho(\Psi)=4\pi\int_0^{\sqrt{2\Psi}} f\big(\Psi-\tfrac12 v^2\big)\,v^2\,\mathrm{d}v$ . Because the model is isotropic, $f$ depends on speed only through $\mathcal{E}$ , and integrating over the velocity sphere leaves $\rho$ a function of $\Psi(r)$ alone. Why: isotropy collapses a six-dimensional object to a one-dimensional relation between two functions of radius.

Step 2. Change the integration variable from $v$ to $\mathcal{E}$ . With $\mathcal{E}=\Psi-\tfrac12 v^2$ one has $v=\sqrt{2(\Psi-\mathcal{E})}$ and $v^2\,\mathrm{d}v=-\sqrt{2(\Psi-\mathcal{E})}\,\mathrm{d}\mathcal{E}$ , so $\rho(\Psi)=4\pi\int_0^{\Psi} f(\mathcal{E})\sqrt{2(\Psi-\mathcal{E})}\,\mathrm{d}\mathcal{E}$ . Why: the square-root kernel is the fingerprint of an Abel integral equation, the one transform with a clean closed-form inverse.

Step 3. Differentiate once in $\Psi$ : $\tfrac{1}{\sqrt{8}\pi}\,\mathrm{d}\rho/\mathrm{d}\Psi=\int_0^{\Psi} f(\mathcal{E})/\sqrt{\Psi-\mathcal{E}}\,\mathrm{d}\mathcal{E}$ , the standard Abel equation with a $(\Psi-\mathcal{E})^{-1/2}$ kernel. Why: differentiating trades the awkward $+\tfrac12$ power for the canonical $-\tfrac12$ Abel form whose inversion is tabulated.

Step 4. Apply the Abel inversion and differentiate once more: $f(\mathcal{E})=\tfrac{1}{\sqrt{8}\pi^2}\,\mathrm{d}/\mathrm{d}\mathcal{E}\int_0^{\mathcal{E}}(\mathrm{d}\rho/\mathrm{d}\Psi)/\sqrt{\mathcal{E}-\Psi}\,\mathrm{d}\Psi$ , the boundary-term form of equation (7). The recovered $f$ is unique but only physical if $f(\mathcal{E})\ge0$ everywhere. Why: the positivity check converts a formal inversion into a real constraint — it says which densities cannot be built from an isotropic velocity distribution. *As a concrete check: the Plummer profile $\rho\propto\Psi^5$ passes through this machinery to give $f\propto\mathcal{E}^{7/2}$ , worked out in full in the examples below.*

Go deeper · Derivation C · 1 minWhy a galaxy is collisionless and the theorem applies

Derivation CWhy a galaxy is collisionless and the theorem applies

The goal is to justify the assumption that makes everything above legitimate — that stars flow under a smooth potential rather than scattering off one another — by estimating the two-body relaxation time and comparing it with the age of the system. We build it from the cumulative effect of many weak deflections, express it as a multiple of the crossing time, and read off, system by system, whether Jeans theorem is exact enough to use.

Step 1. Add up the weak kicks. A star passing another at impact parameter $b$ and speed $v$ gains a transverse kick $\delta v_\perp\sim 2Gm/(bv)$ ; summing $(\delta v_\perp)^2$ over all impact parameters gives a per-crossing change $\Delta v^2/v^2\sim 8\ln\Lambda/N$ , with $N$ the number of stars and $\ln\Lambda\approx\ln N$ the Coulomb logarithm. Why: gravity is long-ranged, so the orbit random-walks under many tiny nudges from distant stars, and the logarithm records that both very close and very distant passages contribute only weakly.

Step 2. Convert to a relaxation time. The orbit is deflected by order itself after $n_{\text{relax}}=v^2/\Delta v^2\sim N/(8\ln N)$ crossings, so $t_{\text{relax}}\approx\dfrac{0.1\,N}{\ln N}\,t_{\text{cross}}$ with $t_{\text{cross}}$ the crossing time. Why: relaxation is slow precisely because $N$ is enormous and only its logarithm appears in the denominator.

Step 3. Read the verdict. For a galaxy $N\sim10^{11}$ , so $t_{\text{relax}}$ is roughly $4\times10^{8}$ crossing times — far longer than the age of the Universe — and Jeans theorem is essentially exact. For a globular cluster $N\sim10^{5}$ the factor is only a few hundred, so its core relaxes within a Hubble time and the idealization breaks. Why: the same formula both licenses the theorem for galaxies and warns where it fails. *The scaling $t_{\text{relax}}\approx0.1\,N\,t_{\text{cross}}/\ln N$ is applied to both a galaxy and a globular cluster in the worked examples below.*

Limiting cases

How many isolating integrals survive is decided entirely by the symmetry of the potential, and that number in turn fixes the most general form $f$ may take:

Case	Result	Meaning
Spherical potential	$f=f(E,\mathbf{L})$	Energy and all three components of $\mathbf{L}$ are isolating and every orbit is planar; reduces to $f(E,L)$ if anisotropic, or isotropic $f(E)$
Axisymmetric potential	$f=f(E,L_z,I_3)$	$E$ and axial $L_z$ are classical isolating integrals; a non-classical third integral $I_3$ exists for regular orbits and is required for flattened models
Generic / chaotic potential	$f=f(E)$	If energy is the only isolating integral, only $f(E)$ is permitted; richer forms would need integrals that do not exist as global isolating quantities
Integrable / action-angle	$f=f(J_1,J_2,J_3)$	When three actions exist, any steady-state $f$ is a function of them — the cleanest and most portable form

Common trapEvery conserved quantity is an integral, but not every integral can appear in f

The tempting slip is to assume a full set of three isolating integrals always exists, so one may freely write $f=f(I_1,I_2,I_3)$ in any potential. It is not so. Energy is always an integral, but in a generic non-integrable potential the only isolating integral may be $E$ itself. Other conserved quantities are non-isolating: they wind densely over the energy surface and never separate neighbouring orbits, so they cannot serve as arguments of $f$ . Assuming three isolating integrals exist is an assumption about the specific potential — verified by orbit integration or action-angle analysis — not a guarantee. The surprise runs the other way too: many realistic axisymmetric potentials admit an effective third integral with no known analytic expression, which is why flattened anisotropic models work at all.

Worked examples

1 · Two-body relaxation time — why a galaxy is collisionless but a globular cluster core is not (fully worked)

Estimate the relaxation time of a bright galaxy ( $N\approx10^{11}$ stars, half-light radius $\approx30$ kpc, typical speed $\approx200$ km/s) and of a globular cluster ( $N\approx10^{5}$ , radius $\approx10$ pc, speed $\approx10$ km/s). Compare each with the $13.8$ Gyr age of the Universe and decide where Jeans theorem is exact.

Step 1. Crossing time of the galaxy. With $R=30\ \mathrm{kpc}=9.3\times10^{20}$ m and $v=2\times10^{5}$ m/s, $t_{\text{cross}}=R/v=4.6\times10^{15}$ s $\approx\mathbf{1.5\times10^{8}\ yr}$ — about $150$ Myr per crossing.

Step 2. Relaxation factor. With $\ln N=\ln(10^{11})=25.3$ , the multiplier $0.1\,N/\ln N\approx\mathbf{4\times10^{8}}$ crossing times — four hundred million orbits away.

Step 3. Galaxy relaxation time. Multiplying, $t_{\text{relax}}\approx4\times10^{8}\times1.5\times10^{8}\ \mathrm{yr}\approx\mathbf{6\times10^{16}\ yr}$ , or $6\times10^{16}/1.38\times10^{10}\approx\mathbf{4\times10^{6}}$ Hubble times. Two-body encounters are utterly negligible, so $\mathrm{d}f/\mathrm{d}t=0$ and Jeans theorem hold to superb accuracy.

Step 4. The globular cluster for contrast. Here $t_{\text{cross}}\approx10\ \mathrm{pc}/10\ \mathrm{km/s}\approx10^{6}$ yr and $\ln N=11.5$ , so the multiplier is $0.1\times10^{5}/11.5\approx870$ and $t_{\text{relax}}\approx\mathbf{0.9\ Gyr}$ — a fraction of the cluster's age, so its core does relax, driving mass segregation and core collapse. A galaxy is collisionless to one part in a million; a globular cluster core is not — the same formula draws the line.

2 · The Plummer sphere — an isotropic f(E) from Eddington inversion (fully worked)

The Plummer model has potential $\Phi(r)=-GM/\sqrt{r^2+a^2}$ and density $\rho(r)=\dfrac{3M}{4\pi a^3}\left(1+r^2/a^2\right)^{-5/2}$ . Find the isotropic distribution function $f(\mathcal{E})$ that Jeans theorem permits and Eddington inversion determines, and give its exponent and coefficient.

Step 1. Express density as a power of the relative potential. With $\Psi=-\Phi=GM/\sqrt{r^2+a^2}$ one has $\sqrt{r^2+a^2}=GM/\Psi$ , and substituting into $\rho$ gives $\rho=\dfrac{3a^2}{4\pi G^5M^4}\,\Psi^5$ — a pure fifth power, $\rho\propto\Psi^{5}$ .

Step 2. Use the Abel power-law shortcut. For an isotropic sphere, $\rho(\Psi)=4\pi\sqrt2\int_0^\Psi f(\mathcal{E})\sqrt{\Psi-\mathcal{E}}\,\mathrm{d}\mathcal{E}$ ; if $f=C\mathcal{E}^{p}$ , the Beta-function integral gives $\rho\propto\Psi^{p+3/2}$ . Matching to $\rho\propto\Psi^{5}$ requires $p+\tfrac32=5$ , so $p=\tfrac72$ and $f\propto\mathcal{E}^{7/2}$ .

Step 3. Fix the coefficient. Carrying the Beta function $B(\tfrac92,\tfrac32)=7\pi/256$ through $\rho=4\pi\sqrt2\,C\,B(\tfrac92,\tfrac32)\,\Psi^5=\tfrac{7\sqrt2\pi^2}{64}C\,\Psi^5$ and equating to the Step 1 prefactor $\tfrac{3a^2}{4\pi G^5M^4}$ gives $C=\dfrac{24\sqrt2}{7\pi^3}\dfrac{a^2}{G^5M^4}$ . Hence

f(\mathcal{E})=\frac{24\sqrt{2}}{7\pi^3}\,\frac{a^2}{G^5M^4}\,\mathcal{E}^{7/2} \qquad (9)

Step 4. Sanity-check. The exponent $\tfrac72>0$ , so $f>0$ for every bound energy $\mathcal{E}>0$ : the Plummer model passes the Eddington positivity test and is a legitimate isotropic equilibrium. It is manifestly a function of $E$ alone — exactly the form Jeans theorem allows for an isotropic sphere — and reproduces the tabulated Binney–Tremaine coefficient. $f(\mathcal{E})\propto\mathcal{E}^{7/2}$ with prefactor $24\sqrt2/7\pi^3=0.156$ — a clean, positive, self-consistent $f(E)$ .

Physics of the portrait

Jeans theorem carries the name of the physicist who first treated a star system as a fluid in phase space — and who was corrected on its subtlest point half a century later.

Origins · 1 minJames Jeans and how this was discovered

James Jeans

1877–1946 · kinetic theory and stellar dynamics

James Hopwood Jeans was an English physicist whose instinct was to import the methods of gas kinetics into astronomy. Between 1913 and 1916 he took Boltzmann's kinetic equation, stripped it of collisions — because stars, unlike molecules, almost never touch — and applied the resulting collisionless equation to the streaming motions of stars that Kapteyn and others had begun to measure. In his 1915 star-streaming paper he argued that a steady-state stellar system must have a distribution function expressible through the integrals of its orbits — the statement now called Jeans theorem. It launched stellar dynamics as a quantitative subject, but Jeans stated it too loosely: he did not distinguish integrals that genuinely label orbits from those that wind uselessly through phase space. That gap was closed in 1962 by Donald Lynden-Bell, whose sharply titled paper insisted that only isolating integrals may appear in $f$ — the modern strong form. Jeans himself had by then moved on to gravitational collapse (the Jeans length and mass that bear his name in a different context), a celebrated dispute with Eddington over stellar structure, and popular books that made him a famous scientific voice of the interwar years. The theorem endures as the microscopic charter of galactic dynamics: every equilibrium galaxy model, from Eddington's clusters to today's action-based Milky Way, is a function of integrals of motion because Jeans said it had to be.

History

1915

— Jeans casts star-streaming as the collisionless Boltzmann equation and argues that a steady-state f depends only on integrals of motion

1916

— Eddington inverts a spherical density profile to obtain the isotropic f(E) of a stellar system

1960

— Contopoulos finds a numerical third integral confining orbits in an axisymmetric galactic potential

1962

— Lynden-Bell sharpens the theorem, insisting that only isolating integrals may appear in f

2008

— action-based f(J) modelling of the Milky Way matures through torus mapping and Gaia-era dynamics

Going further

Texts. Binney and Tremaine, Galactic Dynamics, 2nd ed., sections 4.1–4.4 (Boltzmann equation, integrals of motion, weak and strong Jeans theorems, Eddington inversion) and 1.2 (relaxation time) · Carroll and Ostlie, An Introduction to Modern Astrophysics, galactic-dynamics chapter · Mo, van den Bosch and White, Galaxy Formation and Evolution, chapter 5 · Dodelson, Modern Cosmology (collisionless framing). Key results to keep in mind: (1) the chain-rule sufficiency proof turns on each bracket equalling $\mathrm{d}I_k/\mathrm{d}t$ ; (2) Eddington's formula follows from an Abel transform of the velocity integral; (3) the Plummer result $\rho\propto\Psi^5\Rightarrow f\propto\mathcal{E}^{7/2}$ , with coefficient $24\sqrt2/7\pi^3=0.156$ , is checked numerically above; and (4) $t_{\text{relax}}\approx0.1\,N\,t_{\text{cross}}/\ln N$ gives $\approx6\times10^{16}$ yr for a galaxy and $\approx0.9$ Gyr for a globular cluster, using standard order-of-magnitude astronomical inputs.

Prerequisites. Collisionless Boltzmann Equation · Distribution Function and Phase Space · Orbits in Spherical Potentials · Gravitational N-Body Problem. Leads to. Jeans Equations · Action-Angle Variables · Orbits in Axisymmetric and Triaxial Potentials · Galactic Potential Models.

Ask the tutor: Ask me to show why an isotropic sphere collapses to $f(E)$ , or to invert a density profile of your choice with the Eddington formula and check whether it stays positive.

What is the difference between the weak and strong Jeans theorems

The weak (general) theorem is the easy half: any function of integrals solves the steady-state collisionless Boltzmann equation, and any steady-state $f$ is expressible through integrals. It follows from a one-line chain rule and holds in any potential. The strong theorem is sharper: when almost all orbits are regular — confined to invariant tori by three isolating integrals — $f$ is a function of just those three, equivalently the three actions. The strong form licenses concrete models like $f(\mathbf{J})$ , but it assumes the potential is integrable, and it fails wherever chaos occupies a non-negligible slice of phase space.

Why must a steady-state f depend only on integrals of motion

Because a steady state cannot let the phase-space density drift as the stars it describes move. The condition $\mathrm{d}f/\mathrm{d}t=0$ says $f$ takes one value along a single orbit, and integrals of motion are by definition the functions constant along orbits. So $f$ and the integrals are annihilated by the same operator, and $f$ can depend on the coordinates only in the combinations the integrals form; any other dependence would make $f$ vary along an orbit, contradicting equilibrium. The theorem is just this identification, dressed up.

What exactly makes an integral isolating rather than non-isolating

An isolating integral confines an orbit to a lower-dimensional surface in a way you can use to label it — like a contour line, it separates one orbit from its neighbours. A non-isolating integral, though conserved, threads so densely through the energy surface that its level sets are space-filling: orbits with different values lie arbitrarily close, so it cannot distinguish them or appear as an argument of $f$ . Energy is always isolating; angular momentum is isolating in a spherical potential; whether a useful third integral is depends on the detailed shape of the potential, and answering that for real galaxies is much of modern orbit analysis.

Does Jeans theorem require the system to be spherical

No. The theorem is completely general — a steady-state $f$ is a function of whatever isolating integrals the potential admits, in any geometry. Spheres are just the simplest case, where symmetry hands you energy and the full angular-momentum vector for free. An axisymmetric potential keeps $E$ and $L_z$ and gains a non-classical third integral, so $f=f(E,L_z,I_3)$ ; a triaxial or integrable potential uses the three actions; only a chaotic potential shrinks the theorem back to $f(E)$ alone. Sphericity is a convenience, not a requirement.

How does Jeans theorem connect to the Jeans equations

They are the microscopic and macroscopic faces of the same physics. Jeans theorem says which distribution functions are valid equilibria — functions of the isolating integrals. The Jeans equations are the velocity moments of the collisionless Boltzmann equation: continuum relations for the mean velocities and the stress (velocity-dispersion) tensor, without needing $f$ in full. Because a Jeans-theorem $f$ depends only on integrals, their symmetries imprint on the stress tensor — an $f(E,L^2)$ model yields a diagonal but radially anisotropic dispersion tensor that fixes the anisotropy $\beta(r)$ used in mass estimates. The theorem guarantees a consistent $f$ exists; the Jeans equations work with its moments.

Why is the collisionless assumption safe for galaxies but not always for star clusters

It comes down to the two-body relaxation time, roughly $0.1\,N/\ln N$ crossing times. A galaxy has about $10^{11}$ stars, so its relaxation time is some $6\times10^{16}$ years — four million times the age of the Universe — and encounters have had no time to matter, making the collisionless Boltzmann equation and Jeans theorem essentially exact. A globular cluster has only $10^5$ stars and a short crossing time, so its core relaxes in under a billion years, driving mass segregation and core collapse; there one must add a collision term (Fokker–Planck). The line is set almost entirely by $N$ .