Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 66 / 3

The factor called $N$ in the formula must be the stellar number density $n$, not a dimensionless count. Indeed ${\sigma }^3/(G^2{m}^2)$ has dimensions time per volume, so a density in the denominator is essential. If $N_{\ast }$ is the population in a representative volume $V$, then $n=N_{\ast }/V$.

For a weak encounter between equal-mass stars, let $u$ be the asymptotic relative speed and $b$ the impact parameter. In the straight-line impulse approximation, one star's transverse kick isThe other receives the opposite kick, so the relative-velocity change is $\Delta u_{\perp }=4Gm/(bu)$. Encounters in $(b,b+db)$ occur at rate $2\pi bdbnu$. Independent random kicks add in squared magnitude, givingThis is the random-walk derivation of stellar relaxation. The Coulomb logarithm in stellar dynamics is $\ln \Lambda =\ln (b_{\max }/b_{\min })$. The upper cutoff is the size or inhomogeneity scale beyond which independent local encounters are inappropriate. The lower cutoff is of order the strong-deflection scale $G(2m)/u^2$, or a larger stellar-size/softening scale if applicable. A self-gravitating system commonly has $\Lambda$ of order its total number of stars; that population count is distinct from $n$ in the diffusion rate.

To state the numerical convention explicitly, use representative encounter speed $u=\sqrt{2}\sigma$ and define the relaxation estimate from the relative-velocity diffusion rate with reference squared velocity $3{\sigma }^2$. This givesThis recovers the printed coefficient under a specified convention. There is no convention-independent exact numerical prefactor determined by a representative one-dimensional velocity alone. Using the single-star diffusion rate with the same $3{\sigma }^2$ reference instead gives four times this time; averaging over a full velocity distribution also changes the prefactor. The relaxation time conventions in stellar dynamics must therefore distinguish single-star and relative kicks. The physical scaling is $T_R\propto {\sigma }^3/(n{G}^2{m}^2\ln \Lambda )$, and the printed $N$ must be interpreted as number density.

Use $\mu =\gamma m$ for the gravitational parameter, to distinguish it from the Delaunay action called $G$. The displayed actions are per unit reduced mass. Accordingly use specific canonical momenta; with physical relative momenta, all three actions acquire a common reduced-mass factor, which leaves the eccentricity fraction below unchanged.

The canonical transformation. In spherical coordinates, separation of the Kepler Hamilton-Jacobi equation givesA local type-two generating function, continued across turning points with the appropriate momentum branches, isThe old momenta are $\partial W/\partial (r,\theta ,\varphi )$ and the new angles are $l={\partial }_{L}W$, $g={\partial }_{G}W$, $h={\partial }_{H}W$. This defines a canonical transformation, since its generating differential is $dW=p_{r}dr+p_{\theta }d\theta +p_{\varphi }d\varphi +ldL+gdG+hdH$. With standard choices of angle origins, the resulting Delaunay variables arewith $l$ the mean anomaly, $g$ the argument of periapsis, and $h$ the longitude of ascending node. In particular, the signed $H$ includes retrograde orbits; the squared equation in the PDF must not be read as $H\ge 0$.

For bound noncollision Kepler orbits, the independent ranges areThe limiting circular, radial and coplanar cases sit on boundaries where some angle charts degenerate. They have zero measure in the generic phase-volume calculation. The ranges are constrained as above, rather than a Cartesian box with three unconstrained actions.

The eccentricity count. A canonical transformation preserves phase space volume, soIntegrating the three angles contributes $(2\pi {)}^3$, and integrating $H$ gives $2G$. At fixed $L$, the condition that the orbital eccentricity be less than $e_0$ is $G>L\sqrt{1-e_0^2}$. ThusAny energy weight cancels between these integrals. If the total population is finite, or an explicit finite action band is used,This is the Delaunay phase-volume proof of the thermal eccentricity distribution. The source's $e^2$ law is a cumulative count, not a differential eccentricity density. Also, pointwise finite $F(E)$ does not alone imply finite $N_{\mathrm{tot}}$: $F=1$ gives ${\int }_0^{\infty }{L}^{2}dL=\infty$. The finite normalization of an energy-only Kepler distribution is the additional condition needed for a global probability statement; the fraction at every fixed energy is unaffected.

Why the observation does not imply relaxation or equilibrium. No Boltzmann distribution was used: every normalizable energy-only phase-space distribution function has the same eccentricity marginal, irrespective of collisional thermalization. An even stronger counterexample is a genuinely time-dependent collisionless distributionKepler motion has $\dot{l}=\nu (L)$ and constant actions and other angles, so ${\partial }_{t}f+\nu {\partial }_{l}f=0$: this is an exact nonstationary solution of the Collisionless Boltzmann equation. Integrating $l$ removes the cosine and leaves precisely the same $e^2$ cumulative count at every time. An eccentricity marginal alone cannot establish thermal relaxation, a relaxed age, or even a stationary distribution.