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This is in context of classical Newtonian physics. Consider a system of n different point mass particles. Initially all are spread around on one plane. No particle possess any velocity to begin with. There is no external force acting on this system. Each particle has some mass. The only force to consider here is gravitation between any two particles. When two particles collide, lets say they form a lump with mass equal to sum of the two. This now behaves like one single particle.

My first assumption is, the centre of mass of this system, throughout time won't move. Right?

Given a sufficient amount of time, will everything collapse into one single particle at the centre of mass?

Or it's possible to see a system where few particles are rotating in one direction and other in other direction, so that angular momentum is conserved and net angular momentum is still zero?

If yes, can you give one simple example with points, mass and initial position which will achieve this kind of state.

EDIT: On similar note, can we have say 3 particles which are attracted to each other but never collide, something like a simple harmonic motion?

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Your particles are points, nevertheless they collide inelastically. That is some contradiction. What about conservation of energy? Where does the gravitational energy go when the particles make up a lump? – Georg Jul 20 '11 at 13:49
@georg lets say kinetic energy on collision is transformed to heat. New formed particle is more hot. – Ankush Jul 20 '11 at 15:51
For the three body problem, simpler but not simple, read in wiki: . Quote: "The 3-body system is the simplest mechanical system that allows for unstable trajectories and therefore probability, in the case of classical mechanics. In the case of gravitating masses, one of the questions of the 3-body problem is: For some given class of initial conditions, what is the probability that during some time t, two particles get close enough, providing the energy that would allow the third particle to leave the system?" – anna v Jul 20 '11 at 17:30
up vote 1 down vote accepted

My first assumption is, the centre of mass of this system, throughout time won't move. Right?

Yes, by conservation of linear momentum.

Or it's possible to see a system where few particles are rotating in one direction and other in other direction, so that angular momentum is conserved and net angular momentum is still zero?

Even with only three bodies, solutions can be found where they escape to infinity (with respect to the center of mass). This requires a binary system to be formed by energy conservation. See figure 10.b at page 10 of this paper for an example.

EDIT: On similar note, can we have say 3 particles which are attracted to each other but never collide, something like a simple harmonic motion?

This is also possible. See "Periodic Solution Standish" at page 6 of the same paper.

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Thanks. This paper is good. – Ankush Jul 21 '11 at 15:47

Well, the planetary system is an example, though it is not isolated as in your problem.

What you are requesting is a simple solution to a many body problem. It does not exist. Deterministic chaos is one way to study such problems, also numerical solutions in computers.

For example here is such a study:

Onset of Secular Chaos in Planetary Systems: Period Doubling & Strange Attractors

Also here , from their abstract:

Over the last two decades, there has come about a recognition that chaotic dynamics is pervasive in the solar system. We now understand that the orbits of small members of the solar system—asteroids, comets, and interplanetary dust—are chaotic and undergo large changes on geological time scales. Are the major planets' orbits also chaotic? The answer is not straightforward, and the subtleties have prompted new questions

Numerical solutions are used in the virtual planetaria.

So as genneth has said angular momentum and momentum will be conserved, but that is the only sure statement for an N-body gravitational system as the one you postulate. (same is true in N body systems for other forces too). Since you posit initial conditions at rest the total system will have 0 momentum and angular momentum, but some individual orbits will appear as not all solutions will be head on collisions. Rather the opposite, unless there is one very large mass particle to start with at the centre of mass.

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This doesn't answer the question: it's completely ignoring the assumption that all the particles are initially at rest. – Peter Shor Jul 20 '11 at 14:32
It does not make a difference if they are at rest or not. That is the constants entering the numerical solution. You start a random distribution of N particles at rest, the larger ones will attract the smaller ones more, angular momentum will appear, but it is a many body problem and has to be treated with the tools of many body interactions. – anna v Jul 20 '11 at 14:58
@anna I don't get it. So you are saying it's possible to have an initial distribution of N bodies such that they never form a lump. Right? – Ankush Jul 20 '11 at 16:03
@Ankush Yes, depending on the constants of the system, distances and masses ( momenta and angular momenta you posit as 0). Different velocities will arise on different particles and it depends on the exact numerical solution ( similar to the planetarium numerical solutions) . Once there are velocities, orbits arise and angular momenta internal to the system,the total over all momentum and angular momentum 0 as you require. – anna v Jul 20 '11 at 17:11
@geneth the difference between a theorist's acceptable solution and an experimentalists ( which I am) In the comments it says explicitly that the method cannot be used practically as it converges too slowly to be useful. – anna v Jul 25 '11 at 12:53

If you are thinking about largeish N (hundreds to millions or more), then it resembles a star cluster. There is the issue raised by more than one other poster, that if they are point particles, they can't combine/stick. Close graviational interactions resemble a sort of collision. Some particles will leave collisions with enough kinetic energy to escape from the system (i.e. fly away, and maintain a nonzero velocity in the limit where time goes to infinity), so you get a clump of gravitationally bound particles, and occasionally a particle acquires enough kinetic energy to escape from the collective graviational embrace.

In a truly 2D universe, the configuation will stay 2D, but in any real 3D situation any infinitessimal motion in the third dimension will cause the system to spreadout over three dimensions. I believe it reaches a longterm quasi-steady state that resembles a globular cluster, but remember that occasionally a particle will escape, removing energy (and energy per mass) from the cluster. Also the net momementum and angular momentum of the bound particles, equals zero, but nothing constrains these two subsets individually to have zero momentum or angular momentum, so the cluster might end up with nontrivial momentum and angular momentum.

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It sounds like you've got a bunch of particles with $1/r^2$ attractive forces which (perfectly) inelastically collide. If so, the momentum and angular momentum of the system will be conserved, so the centre of mass will not move, and there will be no net rotation. A single "clump" in the middle is certainly a stable situation, and I think there cannot be others --- they will necessarily have finite angular momentum, though a proof escapes me.

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A stable final situation with no angular momentum would be: two particles are left orbiting each other, while three more particles escape off to infinity in various directions, such that the total angular momentum and total momentum are both zero. I have no idea if it is possible to achieve such a configuration, but you can certainly have such configurations with no net momentum or angular momentum. – Peter Shor Jul 20 '11 at 14:34
@genneth I'm not convinced... I feel there can be some initial distribution of these particles resulting into some binary system or something revolving each other but not collapsing. Dare to prove this wrong? – Ankush Jul 20 '11 at 15:56
@Ankush Try modifying the mass parameters of this simulator to 1, 10, 30. This solution can be transformed into a proof by carefully bounding the integration errors (I think). – mmc Jul 21 '11 at 3:37
@mmc this simulator is awesome... thanks for sharing – Ankush Jul 21 '11 at 16:31

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