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For bound Kepler orbits, the Delaunay variables obey $0<G<L$, $-G<H<G$, and three periodic angles. A canonical transformation preserves phase space volume. At fixed $L$, integrating $H$ supplies the measure $2GdG$. Since $G=L\sqrt{1-e^2}$, the fraction with orbital eccentricity less than $e$ is $L^{-2}{\int }_{L\sqrt{1-e^2}}^{L}2GdG=e^2$. Thus the probability density function is $2e$, independently of any normalizable energy-only phase-space distribution function. A Boltzmann distribution is unnecessary. Even the time-dependent density $F(E(L))[1+\eta \cos (l-n(L)t)]$, $|\eta |<1$, has this same marginal because integrating the mean anomaly removes its oscillatory part; an eccentricity marginal alone does not imply equilibrium.

For bound Kepler orbits with gravitational parameter $\mu$, integrating the Delaunay variables over $0<G<L$, $-G<H<G$ and their three full angle periods gives the displayed population. A globally defined thermal eccentricity distribution requires this integral to converge, or an explicit finite action range. Merely having finite $F(E)$ at every energy does not suffice: $F=1$ yields an infinite integral at large $L$. At each fixed energy, or on every finite normalized action band, the cumulative eccentricity fraction remains $e^2$.

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.

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.

For a bound Kepler orbit, the semimajor axis is half the long diameter of the ellipse; if $r_p,r_{mathrma}$ are its apsidal distances, $a=(r_p+r_{mathrma})/2$. The orbital eccentricity is $e=(r_{mathrma}-r_p)/(r_{mathrma}+r_p)$, with $0\le e<1$ and semiminor axis $b=a\sqrt{1-e^2}$. The gravitating centre is a focus, displaced by $ae$ from the ellipse's geometric centre.

Choose a reference plane, a positive normal and a fixed reference direction in that plane. The ascending node is where the orbit crosses toward positive normal displacement. The longitude of ascending node $\Omega$ is the reference-plane angle from the fixed direction to that node. The argument of periapsis $\omega$ is the angle from the ascending node to periapsis, measured in the orbital plane in the direction of motion. The orbital inclination $i$ is the angle between the reference normal and the orbital angular-momentum normal, with $0\le i\le \pi$. The diagram separates the ellipse's shape from these orientation angles.

Let $\mu =GM$ and use actions per unit test-particle mass. In the prograde spherical-action chart underlying the stated Hamiltonian, $J_{\varphi }=L_z$ and $J_{\theta }=L-L_z$, so $J_{\theta }+J_{\varphi }=L$. The radial action then satisfies $J_r+L=\sqrt{\mu a}$, consistently with $H_K=-\mu /(2a)$. DefineIf $(w_r,w_{\theta },w_{\varphi })$ are the old angles, a type-two generating function isDifferentiation with respect to the old angles recovers $J_r,J_{\theta },J_{\varphi }$, and differentiation with respect to the new actions gives the canonical transformationIts triangular integer action matrix has determinant one, so it preserves the angle periodicities. In the new coordinates,These are the Delaunay variables:With the usual choices of angle origins, $w_1=\ell$ is mean anomaly, $w_2=\omega$ is argument of periapsis, and $w_3=\Omega$ is longitude of ascending node. The generator $\mathcal{G}=L$ rotates the orbit within its plane, whereas $\mathcal{H}=L_z$ rotates that plane about the reference normal; this explains the angular identification. The fast angle obeys Kepler's equation $\ell =\mathcal{E}-e\sin \mathcal{E}$, where $\mathcal{E}$ is eccentric anomaly.

For retrograde orbits, signed $J_{\varphi }=L_z$ gives $J_{\theta }=L-|L_z|$ and the spherical Hamiltonian uses $J_r+J_{\theta }+|J_{\varphi }|$. The stated plus-sign form is therefore a prograde chart, not a global signed-action formula. The Delaunay variables still use $\mathcal{H}=L_z$; on the retrograde branch the generating function is $(\mathcal{L}-\mathcal{G})w_r+(\mathcal{G}+\mathcal{H})w_{\theta }+\mathcal{H}{w}_{\varphi }$. At $e=0$, periapsis is undefined, and at $i=0$ or $\pi$ the node is undefined: these are angle-chart degeneracies, not physical singularities.

For the perturbed planet orbit, the radial and vertical frequencies mean small oscillations about a circular equatorial orbit of cylindrical radius $R$. Write $r=\sqrt{R^2+z^2}$ and retain the stated quadrupole potential. On the equator,At fixed axial specific angular momentum $L_z$, the radial effective potential is $\varphi (R,0)+L_z^2/(2R^2)$. Its second derivative at circular balance $L_z=R^2{\Omega }_c$ gives the radial epicyclic frequencyReflection symmetry decouples vertical motion to first order. Expanding the potential in $z$ givesso the vertical epicyclic frequency isThe circular orbit is radially stable only for $R^2>3{\alpha }^2$. For $\alpha \ne 0$ in that domain, $\nu >{\Omega }_c>\kappa$. Vertical node crossings repeat at frequency $\nu$ while azimuth advances at ${\Omega }_c$, so the line of nodes has nodal precession rate ${\Omega }_c-\nu <0$. Radial periapsis passages repeat at $\kappa$, giving apsidal precession rate ${\Omega }_c-\kappa >0$. To leading quadrupole order, with $n=\sqrt{\mu /R^3}$,This is oblate-quadrupole epicyclic precession. Here $\varpi$ denotes the nearly equatorial longitude of periapsis, not $\omega$ alone.

The precession signs above have a small-eccentricity, small-inclination domain. They are not a universal claim for every inclined Kepler ellipse. To see the distinction with the same Delaunay variables, average the perturbation over an unperturbed orbit. Since $z/r=\sin i\sin (\omega +f)$ and $dt=r^{2}df/\mathcal{G}$, the term linear in $e\cos f$ integrates to zero and the ${\sin }^2(\omega +f)$ average in the $r^{-3}$ integral is $1/2$. ThusDifferentiating at fixed other actions yields the first-order secular ratesTheir nearly equatorial sum reproduces the prograde $\dot{\varpi }$. For a polar orbit, however, the node is fixed and $\dot{\omega }<0$. This supplies a concrete counterexample to extending the requested forward periapsis precession beyond the intended epicyclic regime.