More reasons a binary's eccentricity should be ~0.6
In my Oct. 28, 2010, post, I explain that if all the original particles have the same angular momentum, and if their mean orbital energy is zero, and if the orbital energies follow a Boltzmann distribution, then, if all the bound (i.e., negative orbital energy) particles agglomerate without loss of orbital energy, the eccentricity will be 0.64655. Particles are likelier to agglomerate when they intersect with similar velocities; at similar velocities, rather little orbital energy is lost. Some of the orbital energy lost when the particles agglomerate, might be recoverable (e.g. orbital energy of one particle about the other). Yet some orbital energy will be lost, and this will cause the final eccentricity to be < 0.64655.
If the apses of the original particles are distributed symmetrically in the plane, then the final eccentricity (the eccentricity after all particles are agglomerated into one) must be zero, because there is no preferred direction. On the other hand, Barbarossa's aphelion aligns with the galactic center, so, plausibly the original particles all aligned with the galactic center for the same reason; let's suppose that the original particles did have the same aphelion direction (and the same angular momentum vector).
The simplest particle interaction, is that of point masses, as discussed in the dynamics chapter of most freshman college physics texts. Particles with the same aphelion direction and angular momentum vector, have the same latus rectum segment, and for different eccentricities, intersect only at the ends of the latus rectum. At the latus rectum, the potential energy is the same for all particles because the position is the same; the energy of transverse velocity is the same because the angular momenta are the same; only the energies of radial velocity differ, proportional to eccentricity^2. The eccentricity is proportional to the radial velocity; so the mean eccentricity, will be the eccentricity of the agglomeration. This mean eccentricity, e, can be found as the ratio of two integrals:
Numerator: integral of d(e^2)*exp(-e^2)*e, from 0 to 1
Denominator: integral of d(e^2)*exp(-e^2), from 0 to 1
(The exponential function gives a Boltzmann distribution, with mean corresponding to eccentricity = 1.) Integrating the numerator numerically (use the substitution u = e^2), I find
eccentricity = 0.59948... . The eccentricity might be higher than this if some particles in hyperbolic orbits agglomerate also, on their one-time pass through the system; this describes Barbarossa (e = 0.6106 according to the orbit I fit to the sky surveys last year).
