# Random N-body problem

Suppose there are $N$ unit-mass particles whose initial positions are uniformly distributed in a unit-radius disk. Each particle is assigned a randomly oriented initial velocity vector $v_i$ of length $1$ (green below). Added: Robert Israel's incisive comment suggests that I should also stipulate that $\sum_i v_i = 0$. Then the particles act upon one another via inverse-square gravity.

Dots show initial positions inside unit disk. Green vectors: initial velocity; red vectors: final velocity.

Q. What is the probability that $k \ge 1$ of the $N$ particles remain within a disk of some radius $R \ge 1$ forever?

In the illustration above, $N=8$ and $R=3$. (But I do not trust my crude simulations.)

I am wondering if the answer is: zero, independent of $k$ and $R$ and the gravitational constant? Perhaps the answer differs for points in $\mathbb{R}^2$ vs. points in $\mathbb{R}^3$ (or $\mathbb{R}^d$, $d > 3$)?

• I bet cosmologists thought about this problem in considering origins of the universe. Gerhard "Into The Universality Of It" Paseman, 2017.06.16. – Gerhard Paseman Jun 16 '17 at 22:47
• Indeed, for d=3, I submit that your post is evidence that for some large N and some large k less than N, the probability is greater than zero. Gerhard "Witnessing The Meaning Of Life" Paseman, 2017.06.16. – Gerhard Paseman Jun 16 '17 at 22:52
• If you're talking about a fixed disk of radius $R$, the probability should be $0$. The centre of mass undergoes uniform motion, with velocity the initial average of the velocity vectors, and that average will almost surely be nonzero. This isn't quite conclusive, e.g. it is possible that in the limit various clusters of particles will move away with certain velocities while one cluster has average velocity $0$, but it should be intuitively clear that an average cluster velocity of exactly $0$ has probability $0$. – Robert Israel Jun 16 '17 at 23:07
• Maybe a better fix would be to ask if the $k$ particles stay in $B_R(C(t))$, where $C(t)=a+bt$ is the center of mass. In the current version, it's not clear to me if there is a natural interpretation of picking the velocities "at random." – Christian Remling Jun 17 '17 at 1:13
• @ChristianRemling Not at all even if $N=3$. The energy bounds alone do not preclude the possibility that two particles will go into a very low energy configuration and the third one will escape. – fedja Jun 17 '17 at 2:01

What is the probability that $k \ge 1$ of the $N$ particles remain within a disk of some radius $R \ge 1$ forever?

If the gravitational constant is small enough, then the probability is definitely less than one, since it's easy for one of the particles to be initially headed away from the others at a high enough speed to escape.

For $N\le 3$, the probability appears to be nonzero given a high enough gravitational constant. For $N=1$, the probability is 1. For $N=2$, we have Keplerian orbits. For $N=3$, there are figure-eight configurations known as 3-body choreographies that are believed to be stable in the KAM sense, and there seem to be solutions of this type for a variety of potentials, including $-1/r$. Stability against small perturbations means that you have bound states that occupy a region of phase space with nonvanishing volume, which implies nonzero probability.

For $N\ge4$, this appears to be an open problem. In general, we expect such systems to be unbound for thermodynamic reasons. This is because for large $N$ the probability of a given state for a particular particle falls off like $e^{-E/T}$, where $E$ is the kinetic energy and $T$ is the temperature. Since there is no bound on $E$, any system like this is generically expected to evaporate its particles off into the surrounding space. This is really just a phase space argument. There is infinite phase space out there at large distances. Such evaporation is in fact observed in galaxies and globular clusters, where we see stars shooting off with anomalously high velocities.

So all such systems should be considered guilty until proven innocent, i.e., we expect them to be unbound by default unless we have some tricky way of constructing a specific example and proving that it's bound. (Numerical simulation doesn't work, because these systems are normally chaotic.) To prove that the probability of being bound is nonzero, we also need to prove that it remains bound under a small perturbation. For $N\ge 4$, I did not come across any mention of any examples that are known to be stable, but I could certainly be missing relevant work.

By the way, physically the "right" way to state the problem in two dimensions is probably to use a $1/r$ force, not a $1/r^2$ force, since we expect Gauss's law to hold for fundamental reasons.

Gerhard Paseman says:

This would be more about the far future than the distant past. The relevant tool is general relativity, not Newtonian mechanics. The thermodynamic equilibrium state of a gravitationally interacting system according to classical GR is basically a black hole, although it can be more complicated than that depending on the large-scale geometry and topology. You don't get stable orbiting states with a purely gravitational interaction, because gravitational radiation sucks energy away.

• Thanks for your knowledgeable response! For $N=3$, wouldn't those special $8$-configurations occur with probability zero? Of course I am more interested in $\mathbb{R}^3$ than $\mathbb{R}^2$. – Joseph O'Rourke Jun 18 '17 at 23:08
• @JosephO'Rourke: If I'm interpreting the scholarpedia article correctly, then a figure-eight configuration "looks to be" stable for $N=3$ for small perturbations. That means that you have bound states that occupy a region of phase space with nonvanishing volume. That is, stable <=> nonzero probability. From their wording, it sounds like stability hasn't actually been rigorously proved. – Ben Crowell Jun 18 '17 at 23:12
• "For N≥4, this appears to be an open problem." I tagged it as an open problem. Thanks. – Joseph O'Rourke Jun 20 '17 at 10:38

The probability of a bounded motion is neither 0 nor 1. (I am assuming that the initial total momentum is 0). Indeed, for the case of two bodies, we have an explicit solution, and it shows that there is an inequality for the initial data which distinguished bounded motion from unbounded one.

For arbitrary number of bodies, there is an open set of initial positions when the motion is bounded (think of something like a solar system), and also an open set of initial positions which give unbounded motion.

• Alexandre, could you please give some details for the case of three bodies? The case of two bodies is clear, since it is integrable. By the way, I forgot, is it proven by now the the solar system is stable? (I remember Arnold had some ideas of how to prove this using KAM theory but not sure if he was successful). – aglearner Jun 17 '17 at 9:31
• The case when one mass is large, second much smaller, and the third much smaller than the second, so that the first is in the center, second moves approximately on Kepler orbit and the third (satellite) around the second, I suppose this case is stable. Like for example Sun-Mars-Fobos. – Alexandre Eremenko Jun 17 '17 at 10:09
• I agree, such case should be doable by KAM theory. But in the case under consideration the masses are equal... – aglearner Jun 17 '17 at 10:30
• "there is an open set of initial positions when the motion is bounded": May I ask: Does this open set have volume a positive fraction of the volume of the space of all initial positions? – Joseph O'Rourke Jun 18 '17 at 23:11
• Why is the solar system bounded? Isn't it expected that the solar system is unstable and will eject planets, say every few billion years? I can believe that there are stable solar systems, but why would every small perturbation also be bounded, wouldn't you expect to be able to get a planet ejected with arbitrarily small perturbations? – Douglas Zare Jun 22 '17 at 22:47

Update: There's nothing obviously preventing (negative) energy from bleeding from a pair of close bodies to others, so this answer is wrong as stated.

At least for sufficiently large $N$, the case of equal masses reduces to the case of unequal masses to give a positive probability of bounded motion. This is because two bodies that are sufficiently close together relative to other masses can act as a body of twice the mass, since they will be energetically bound. Thus if one can build a solar system out of unequal masses, one can build a stable solar system out of equal masses.

Edit: Since two masses are stable, there is a positive probability of stability for any $N$ by iteratively replacing one mass by two.

• What does "energetically bound" mean in mathematics? – Douglas Zare Jun 18 '17 at 7:45
• It means that (1) treating the two bodies as a single body at the center of mass, the system is stable in a neighborhood of the current orbits under small perturbations (2) it would violate conservation of energy for the two bodies to move far enough apart to create more than a small perturbation. This leaves out some details, since small perturbations needs to be restricted to zero mean somehow; I'll leave the remaining details to the reader. :) – Geoffrey Irving Jun 18 '17 at 18:48
• By "stable," you mean for all small perturbations and all time, not just a time so long that physicists don't care about it? How do you prove that? It seems to me that the local system can gain or lose small amounts of energy from the rest of the system. How do you know that these small amounts don't add up to enough to cause the bodies to move apart after a long time? I don't find that plausible, actually. Do you have a mathematical reference? – Douglas Zare Jun 18 '17 at 19:04
• Yep, you're right that this isn't plausible: energy can certainly bleed slowly from one pair to others. I'll edit to note this answer is wrong. Thanks! – Geoffrey Irving Jun 18 '17 at 19:11