Suppose we have a bunch of spherical billiard balls rolling around on an infinite plane. Suppose there is no friction and the collisions are elastic. They don’t feel the influence of gravity or any other force except the collisions. At least one ball is moving. Can they ever return to their original positions and velocities?
1) If only positions should match, then the answer is yes. Just roll two identical balls at each other.
2) If both positions and velocities should match, then the answer is no. Here’s a sketch of a proof:
Assume without loss of generality that some balls have nonzero velocity along the X axis. Let’s define the following function of time: take all balls having nonzero X velocity, and take the lowest X coordinate of their centers. A case analysis shows that the function can change from increasing to decreasing, but never the other way. Therefore it cannot be periodic. But since it’s determined from the configuration, that means the configuration can’t be periodic, QED.
Note that the case analysis is tricky because the function can be discontinuous. The interesting cases are when a ball’s X velocity becomes zero due to a collision, or (more subtly) two balls with only Y velocity gain X velocity due to an off-center collision. But I think the statement about monotonicity still holds.
Yeah I thought about those two cases as well, but I agree that they are correct. Perhaps we could make the proof a bit simpler by picking the X direction to be one that the balls never travel perpendicular too (although in fact I can’t even think of a proof that such a direction exists).
The full answer is: they cannot return if there are only finitely many balls, but they can if there are infinitely many.
Let’s first assume that there are finitely many balls. As Thomas pointed out, we can assume that the center of mass is fixed. Let’s consider R defined to be the distance from the center of mass to the furthest ball and call that furthest ball B (which ball that is might change over time). R might be decreasing at the start—we might start with B going towards the center of mass. But if R decreased forever then we would know that they never return to their starting location (since R would be different)! So at some point it must become at least as large as it was at the start. At that point either the derivative of R is 0 or it is positive. In either case, R must increase forever onwards—which again shows it can’t return to its original starting point. Why is it always increasing from that point onwards? Well, the only way for the ball B to turn around and start heading back towards the center is if there is another ball further away than it to collide with it. But that can’t be, since B is the furthest out ball! (Edit: I see now that this is essentially equivalent to cousin_it’s argument.)
For infinitely many balls, you can construct a situation where they return to their original position! We’re going to put a bunch of balls on a line (you don’t even need the whole plane). In the interval [0,1], there’ll be two balls with initial velocity heading in towards each other at unit speed, with one ball at the left edge of the interval and one ball at the right. Then do the same thing for each interval [k,k+1]. When you let them go, each pair in each interval will collide and then be heading outwards with unit interval. Then they’ll collide at the boundary with the next interval with the ball from the next interval. That sets them back at the starting position. I.e. all balls collide first with their neighbor on one side, then their neighbor on the other side, setting them back to their starting position.
Possibly not a rational answer (so possilbly not living up to the less wrong philosophy!) but given the assumption of an infinite plane I would think the probability is vanishingly small of returning to the original position and velocity.
Something would need to constrain the vectors taken to prevent any ball from taking off in some direction that could be described as “away from the group”. Perhaps that could be understood be be on a path for which the the path of no other ball can possible intersect. At that point this ball can will never change it’s current velosity and never return to it’s oiginal position.
I cannot offer a proof that such a condition must eventually occur in your experimnt but my intuition is that it will. If so that vanishing small probablity that everyting return to some orginal state goes to zero.
Two balls which are orbiting around its common mass center, they do return to a previous position. If there is no gravity, a finite bunch of rolling balls will never again return to the present state. Never again.
If the center of gravity moves, it moves with the velocity v. So it will be in the position r+v*t after some time t. Now it’s at the position r. A different position of gravity (mass) center means different position. For the whatever finite t.
In the case when the gravity center doesn’t move, you can divide the composition into two sub-compositions, where both gravity centers do move. If only one had moved, then the combined gravity center would move and we would have the solved case above.
But if both gravity centers move, they can either move apart and never collide—in which case they will both have different position vectors lately - or they will collide. In that case, they will reverse their directions after the elastic collision and we have a solved case then.
But if both gravity centers move, they can either move apart and never collide—in which case they will both have different position vectors lately—or they will collide. In that case, they will reverse their directions after the elastic collision and we have a solved case then.
I’m not convinced by this bit. Usually we can calculate the results of an elastic collision by using both conservation of energy and conservation of momentum. But we can’t know the energy of the sub-compositions based just on the velocity of their centres of mass. They will also have some internal energy. So we can’t calculate the results of the collision.
Then, both gravity centers travel with a certain velocity each and will collide. How can they return back here? If they will reverse both velocities after the collision. Then, they can return. But with the opposite velocities, and therefore this will not be the same state.
Since they always move in lines, there will be no another collision. And therefore no return to the present state.
Doesn’t matter how many collisions will happen, the momentum conservation will hold. Even if only two small balls of each gravity center will collide, the sum of momentums of that two gravity centers will remain the same. Doesn’t matter which partition of balls we chose, only that is the same before and after the collision.
After some collisions have happened and the two parts are heading away from each other the two parts could still overlap and then some more of their balls could collide. This could lead to the two parts heading back together.
EDIT: Clarified some things.
Suppose we have a bunch of spherical billiard balls rolling around on an infinite plane. Suppose there is no friction and the collisions are elastic. They don’t feel the influence of gravity or any other force except the collisions. At least one ball is moving. Can they ever return to their original positions and velocities?
1) If only positions should match, then the answer is yes. Just roll two identical balls at each other.
2) If both positions and velocities should match, then the answer is no. Here’s a sketch of a proof:
Assume without loss of generality that some balls have nonzero velocity along the X axis. Let’s define the following function of time: take all balls having nonzero X velocity, and take the lowest X coordinate of their centers. A case analysis shows that the function can change from increasing to decreasing, but never the other way. Therefore it cannot be periodic. But since it’s determined from the configuration, that means the configuration can’t be periodic, QED.
Yep, that looks pretty airtight to me. Well done!
Note that the case analysis is tricky because the function can be discontinuous. The interesting cases are when a ball’s X velocity becomes zero due to a collision, or (more subtly) two balls with only Y velocity gain X velocity due to an off-center collision. But I think the statement about monotonicity still holds.
Yeah I thought about those two cases as well, but I agree that they are correct. Perhaps we could make the proof a bit simpler by picking the X direction to be one that the balls never travel perpendicular too (although in fact I can’t even think of a proof that such a direction exists).
That wouldn’t help with the first case, because a ball can stop completely. Let’s keep the proof as it is :-)
The full answer is: they cannot return if there are only finitely many balls, but they can if there are infinitely many.
Let’s first assume that there are finitely many balls. As Thomas pointed out, we can assume that the center of mass is fixed. Let’s consider R defined to be the distance from the center of mass to the furthest ball and call that furthest ball B (which ball that is might change over time). R might be decreasing at the start—we might start with B going towards the center of mass. But if R decreased forever then we would know that they never return to their starting location (since R would be different)! So at some point it must become at least as large as it was at the start. At that point either the derivative of R is 0 or it is positive. In either case, R must increase forever onwards—which again shows it can’t return to its original starting point. Why is it always increasing from that point onwards? Well, the only way for the ball B to turn around and start heading back towards the center is if there is another ball further away than it to collide with it. But that can’t be, since B is the furthest out ball! (Edit: I see now that this is essentially equivalent to cousin_it’s argument.)
For infinitely many balls, you can construct a situation where they return to their original position! We’re going to put a bunch of balls on a line (you don’t even need the whole plane). In the interval [0,1], there’ll be two balls with initial velocity heading in towards each other at unit speed, with one ball at the left edge of the interval and one ball at the right. Then do the same thing for each interval [k,k+1]. When you let them go, each pair in each interval will collide and then be heading outwards with unit interval. Then they’ll collide at the boundary with the next interval with the ball from the next interval. That sets them back at the starting position. I.e. all balls collide first with their neighbor on one side, then their neighbor on the other side, setting them back to their starting position.
Possibly not a rational answer (so possilbly not living up to the less wrong philosophy!) but given the assumption of an infinite plane I would think the probability is vanishingly small of returning to the original position and velocity.
Something would need to constrain the vectors taken to prevent any ball from taking off in some direction that could be described as “away from the group”. Perhaps that could be understood be be on a path for which the the path of no other ball can possible intersect. At that point this ball can will never change it’s current velosity and never return to it’s oiginal position.
I cannot offer a proof that such a condition must eventually occur in your experimnt but my intuition is that it will. If so that vanishing small probablity that everyting return to some orginal state goes to zero.
Two balls which are orbiting around its common mass center, they do return to a previous position. If there is no gravity, a finite bunch of rolling balls will never again return to the present state. Never again.
I’m assuming no gravity (and that at least one ball is moving). Do you have a proof for your assertion?
Sure.
If the center of gravity moves, it moves with the velocity v. So it will be in the position r+v*t after some time t. Now it’s at the position r. A different position of gravity (mass) center means different position. For the whatever finite t.
In the case when the gravity center doesn’t move, you can divide the composition into two sub-compositions, where both gravity centers do move. If only one had moved, then the combined gravity center would move and we would have the solved case above.
But if both gravity centers move, they can either move apart and never collide—in which case they will both have different position vectors lately - or they will collide. In that case, they will reverse their directions after the elastic collision and we have a solved case then.
Well, that’s an approximate proof.
I’m not convinced by this bit. Usually we can calculate the results of an elastic collision by using both conservation of energy and conservation of momentum. But we can’t know the energy of the sub-compositions based just on the velocity of their centres of mass. They will also have some internal energy. So we can’t calculate the results of the collision.
Do we agree, that if there will be no collision, and both gravity centers move, that they will never return to the present position?
Yes.
Then, both gravity centers travel with a certain velocity each and will collide. How can they return back here? If they will reverse both velocities after the collision. Then, they can return. But with the opposite velocities, and therefore this will not be the same state.
Since they always move in lines, there will be no another collision. And therefore no return to the present state.
What do you mean by “the” collision? If each part has several balls then there will be multiple collisions.
Doesn’t matter how many collisions will happen, the momentum conservation will hold. Even if only two small balls of each gravity center will collide, the sum of momentums of that two gravity centers will remain the same. Doesn’t matter which partition of balls we chose, only that is the same before and after the collision.
After some collisions have happened and the two parts are heading away from each other the two parts could still overlap and then some more of their balls could collide. This could lead to the two parts heading back together.
No, that’s impossible. However you choose to divide this set of balls and however they later collide, both impulses are still conserved.