Delaunay phase-volume proof of the thermal eccentricity distribution ($P(e^{\prime }<e)=e^2$)

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 $e^2$.