Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 73 / 1 / ii / Solution

Write ${\mu }_g=\gamma m$ for the gravitational parameter of the relative Kepler orbit, and use specific orbital energy and momenta per unit reduced mass. The Hamiltonian isIf physical momenta rather than these specific momenta are used, multiply the actions by the reduced mass. This only introduces a constant factor into the phase space measure and leaves the eccentricity fractions unchanged. The formulas in the question adopt the specific-action convention.

Here is an explicit local canonical transformation. A separated complete integral of the Hamilton-Jacobi equation isChoose branches along an orbit and take $p_j=\partial S/\partial q_j$, $l=\partial S/\partial L$, $g=\partial S/\partial G$, $h=\partial S/\partial H$. Thenwhose exterior derivative proves preservation of the canonical symplectic form. Thus these variables are canonical, and the absolute Jacobian determinant is one. The angular separation gives $G=|\mathbf{r}\times \mathbf{v}|$ and $H=G\cos i$, including the sign of the angular momentum along the reference axis.

For clarity, the radial action can be evaluated rather than assumed. Let $r=a(1-e\cos u)$, with $u$ the eccentric anomaly. Then $p_r=\sqrt{{\mu }_g/a}e\sin u/(1-e\cos u)$, andIn the integral, use $e^2{\sin }^{2}u/(1-e\cos u)=1+e\cos u-(1-e^2)/(1-e\cos u)$ and ${\int }_0^{2\pi }(1-e\cos u{)}^{-1}du=2\pi /\sqrt{1-e^2}$. The polar action is $J_{\theta }=G-|H|$, while the signed azimuthal action is $J_{\varphi }=H$. The separated angular coordinates have the usual Delaunay variables interpretation: $g$ is the argument of periapsis, and $h$ is the longitude of ascending node. In particular,with a suitable origin at periapsis, so $l$ is the mean anomaly. We have therefore recoverednot merely assigned names to the new coordinates.

For nondegenerate bound Kepler orbits, the ranges areThe boundaries describe circular, coplanar or radial degeneracies, where some angles cease to be defined; they have zero measure in the present calculation. The actions are canonical coordinates with nested bounds, rather than three independently unrestricted Cartesian coordinates. The square of $H$ alone would miss the distinction between prograde and retrograde inclinations.

By the canonical transformation, the population measure becomesAt fixed $L$, the condition $e^{\prime }<e$ is $L\sqrt{1-e^2}<G<L$. Integrating all three angles and then $H$ givesConsequently the Delaunay phase-volume proof of the thermal eccentricity distribution yieldsThe PDF's $e^2$ dependence must be interpreted as a cumulative distribution function, not an eccentricity probability density function or the value of the original six-dimensional phase-space distribution function. A finite value of $F(E)$ at every energy also does not suffice to make the total population finite: $F=1$ on all bound energies gives the divergent integral ${\int }_0^{\infty }{L}^{2}dL$. The fractions are well defined at each fixed energy, over a finite allowed energy range, or for an energy distribution with a finite total integral.

A Boltzmann distribution is therefore unnecessary. Any normalizable energy-only distribution has the same eccentricity marginal, although such a distribution is a collisionless steady state by Jeans theorem. The marginal alone does not even prove stationarity: for $0<|\eta |<1$,is a positive time-dependent solution of the Collisionless Boltzmann equation and has exactly the same eccentricity marginal after integration over $l$. An observed $e^2$ cumulative law establishes neither thermal relaxation nor the age of the system or Universe.