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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    $\begingroup$ I just wanted to chime in to say that this is perhaps the most "gangster" question title on MO. $\endgroup$ Feb 3, 2010 at 6:40
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    $\begingroup$ 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). $\endgroup$
    – Ian Durham
    Feb 3, 2010 at 12:32
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    $\begingroup$ 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. $\endgroup$ Feb 3, 2010 at 16:18
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    $\begingroup$ @Harrison: As long as Stringer isn't dropping bodies, he's not our problem. $\endgroup$ Sep 16, 2010 at 10:56
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    $\begingroup$ "they will become collinear a finite time later": You hardly need this, but I think this is a remarkable theorem! $\endgroup$ Sep 20, 2010 at 23:00

1 Answer 1


I believe your first proof is trivial since any 3 points may always be placed on a plane and since their gravity will attract along the plane they will remain on it in perpetuity. Then any 2 points say $a$ and $b$ can be placed on a line and their centre of gravity plotted on the line.

Since the centre of gravity of $a$ and $b$ is on this line, this line must accelerate across the plane towards $c$ until $c$ crosses the line. For $c$ never to reached $ab$ would require the entire system to be generating a state of its own perpetual acceleration opposing the direction from $c$ to its nearest point on $ab$, and creating a perpetual angular acceleration around the centre of gravity of $a$ and $b$; a contradiction since there is no fourth mass with respect to which the system $abc$ could be accelerating.

Moving on to the monotonic decrease in momentum, if we choose $c$ such that $c$ is the body which passes between the other two then holding the line $ab$ as a reference point, $a$ and $b$ must move towards each other along $ab$ so $I_{ab}$ monotonically decreases, and $c$ moves towards some point between $a$ and $b$ until it passes between them so again $I_{ca}+I_{cb}$ must decrease monotonically.

Re the final part, the triangles will remain scalene until $t_c$ but will not necessarily do so in perpetuity. This is because it will be the body which is closest to the other two which passes between them. If we draw a scalene triangle, exactly one of the vertices connects the shortest two sides. This body (say $a$) will move between the other two and since it is closer to both of the other two than they are to each other, it will approach both of them faster than they close the distance between each other.

Furthermore due to the inverse square law it will approach the body nearest to it faster than it approaches the other, and therefore there is no means by which the ordering of their distances can change before $a$ reaches the line between $b$ and $c$, which is what would be required to create an isosceles or equilateral triangle from a scalene one.

  • $\begingroup$ Thanks Robert for the feedback. My ``first proof'' result (about existence of collinear instants) is known to be false when the total energy is positive: simply send each mass outward faster than its `escape velocity'. But you do not use negativity of energy in your argument. Can you think of a way that negativity of energy might enter into your thinking on this situation? $\endgroup$ May 16, 2017 at 16:56
  • $\begingroup$ @RichardMontgomery if I interpret your meaning correctly, one or more masses are moving outwards faster than escape velocity would negate my argument that "this line ($ab$) must accelerate across the plane towards $c$ until $c$ crosses the line. In stating that, I assumed the bodies start out stationary. Although they would still accelerate together they would do so having started out on divergent paths, and the relative acceleration of $ab$ towards $c$ would be asymptotic towards some permanently divergent path; insufficient to halt their divergence. Is that what you meant? $\endgroup$ May 16, 2017 at 17:12
  • $\begingroup$ @RichardMontgomery re-reading your question and the matters you're touching upon, perhaps this is the answer you are looking for. If you were to define in some way the divergent paths which the bodies move asymptotically towards, I believe this would encode some of the energy information similarly to the ways the collinear triangles do for negative energy states. $\endgroup$ May 16, 2017 at 17:30
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    $\begingroup$ I will have time to think about what you wrote more carefully next week. Your first proof' is heuristically good. The problem with turning it into a rigorous proof is that the line connecting any two of the bodies keeps moving. How do you know it is not moving away' from the third body faster than the third body is approaching it? That is true, for at least one decomposition ab; c, but since by fixing a line you are working in a non-inertial reference plane, I do not think you've proved it , even when starting all three bodies at rest. $\endgroup$ May 18, 2017 at 3:54
  • $\begingroup$ @RichardMontgomery A good point! That would require it to have attained escape velocity by drawing energy off the other two. I'm not an expert and you almost certainly know the answer to this, but I think it would have to encircle them both first to do that, also completing the proof. $\endgroup$ May 18, 2017 at 10:03