# Dropping three bodies

Consider the usual three-body problem with Newtonian $1/r^2$ force between masses. Let the three masses start off at rest, and not collinear. Then they will become collinear a finite time later by a theorem I proved some time ago. (See the papers "Infinitely Many Syzygies" and "The zero angular momentum three-body problem: all but one solution has syzygies" available on my web site or the arXivs.) Let $t_c$ denote the first such time.

Write $r_{ij} (t)$ for the distance between mass $i$ and mass $j$ at time $t$.

Question 1. For general masses $m_i >0$, is it true that the "moment of inertia" $I = m_1 m_2 r_{12}^2 + m_2 m_3 r_{23}^2 + m_1 m_3 r_{13}^2$ monotonically decreases over the interval $(0, t_c)$?

Question 2. If the masses are all equal and if the initial side-lengths satsify $0 < r_{12}(0) < r_{23} (0)< r_{13} (0)$
is it true that these inequalities remain in force: $0 < r_{12} (t) < r_{23} (t) < r_{13}(t)$ for $0 < t < t_c$? In other words: if the triangle starts off as scalene (not isosceles, and having nonzero area) does it remain scalene up to collinearity?

Motivation: The space of collinear triangles, consisting of triangles of zero area, acts like a global Poincare section for the zero-angular momentum, negative energy three-body problem. To obtain some understanding of the return map from this space to itself the "brake orbits"-- those solutions for which all velocities vanish at some instant -- seem to play an organizing role. Answering either questions would yield useful information about brake orbits.

Aside: I suspect that if the answers to either question is yes for the standard $1/r^2$ force, then it is also yes for any attractive "power law" $1/r^a$ force between masses, any $a > 0$.

added, Sept 20, 2010. The bounty is for an answer to either question 1 or 2.

I've made partial progress toward 2 using variational methods (direct method of the calculus of variations). I can prove that if a syzygy is chosen anywhere in a neighborhood of binary collision (so $r_{12}(t_c) = \delta$, small, $r_{23} (t_) = r_{13}(t_c) + \delta$) then there exists a brake orbit solution arc ending in this syzygy and satisfying the inequality of question 2. The proof suggests, but does not prove, that the result holds locally near isosceles, meaning for brake initial conditions in a neighborhood of isosceles brake initial conditions ( so
$r_{13} (0) = r_{12} (0) + \epsilon$). If I had uniqueness [modulo rotation and reflection] of brake orbits with specified syzygy endpoints, then my proof would yield a proof of this local version of the alleged theorem.
Unfortunately, my proof does not exclude the possibility of more than one orbit ending in the chosen syzygy, one of which violates the inequality.

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I just wanted to chime in to say that this is perhaps the most "gangster" question title on MO. –  Harrison Brown Feb 3 '10 at 6:40
I think the title of the post might be a bit misleading since it could be taken to imply a fourth body. In other words, one could read that as dropping three bodies in a single gravitational field which is a different problem than the one you are proposing (which is a fascinating problem, by the way). –  Ian Durham Feb 3 '10 at 12:32
Can you answer these questions for infinitesimal perturbations of an isosceles configuration? If so, perhaps it would help to restrict the cases on the boundary of those which satisfy those conditions, e.g., ones which become isosceles or singular at the moment of collinearity. –  Douglas Zare Feb 3 '10 at 16:18
@Harrison: As long as Stringer isn't dropping bodies, he's not our problem. –  Sheikraisinrollbank Sep 16 '10 at 10:56
"they will become collinear a finite time later": You hardly need this, but I think this is a remarkable theorem! –  Joseph O'Rourke Sep 20 '10 at 23:00