# Infinite sequence of nested, falling, colliding spheres

Imagine an infinite collection of nested, concentric spheres, of radius 1, $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, and so on. Suppose they are somehow suspended in space, fixed on their common center $x$. Then the outermost sphere is "released" from its center, and falls vertically under the influence of gravity, while all the other spheres remain "pinned" with their centers on $x$. Next, the top interior of the $r=1$ sphere collides with the top exterior of the $r=\frac{1}{2}$ sphere, knocking it loose from $x$ via a perfectly elastic collision, sending it downward. And so on.

Essentially my question is: What happens? It would be pleasing to understand the behavior of this system without resorting to explicit calculation of all the interactions. Assume the spheres are made of some homogenous, thin material so that their weight is proportional to their surface area (or circumference if you'd prefer to drop down to $\mathbb{R}^2$). I cannot see intuitively the sequence of collisions and overall behavior, and I have not yet tried careful calculations. Perhaps there is a line of reasoning that demystifies the apparent complexities...?

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At least you could say that the system will behave symmetrically around the "gravitational axis", so I guess the only difference between the 2D and 3D-cases are the ratios of the masses of the different circles/spheres. – TMM Sep 7 '11 at 1:24
@Thijs: Yes, the only dependence on dimension is the mass ratios, just as you observe. – Joseph O'Rourke Sep 7 '11 at 1:26

I don't think your system is well-specified. Consider the following simpler variant on the same theme:

Infinitely many ideal, identical billiard balls occupy a 1D world. The initial distance between successive balls (excluding the diameter of the balls themselves) is 1, 1/2, 1/4, 1/8 and so forth.

At time t=0, we give the first ball a push so it starts moving with speed 1 towards the infinite line of balls. A cascade of collisions happen, each transferring 100% of the energy and momentum from one ball to the next. By time 2, infinitely many collisions have happened, and all balls are now at rest and never move again! You can write down parametric equations for the position of each ball at any time, and see that locally the rules for ideal billiard balls are satisfied at every time. But somehow the energy and momentum we put into the system has disappeared.

Even more disturbing, the standard billiard-ball rules are time symmetric. So we can run the experiment backwards -- if we have an infinite line of balls sitting, all at rest, it is completely according to the rules if they spontaneously start an infinite cascade of collisions that end with ball 0 being ejected at some unpredictable speed, at an unpredictable moment in time.

It's not clear that your system of spheres has a failure mode that's quite as dramatic as this one. The fact that total mass is finite may conceivably prevent any momentum shenanigans. But according to statistical mechanics one should expect that the available energy will eventually drain away into micro-vibrations of ever smaller spheres as it tries to distribute itself evenly over all degrees of freedom. And the rules are still time-symmetric, so there are probably valid histories where energy arises spontaneously and ends up affecting the larger shells.

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In this example, you're assuming the balls are perfectly rigid so that a collision instantly starts the whole ball moving. In reality this can't happen: a disturbance can't propagate faster than the speed of light (and in fact I think it may only go at the speed of sound in the ball). – Robert Israel Sep 7 '11 at 3:10
@Robert: Incisive point, which I certainly never considered! It makes sense to first analyze assuming instant transfer of energy upon collision, and then see if finite propagation alters the behavior. – Joseph O'Rourke Sep 7 '11 at 10:51
@Henning: Fascinating analysis! Suppose there were only $n$ spheres rather than infinitely many. Then even for very large $n$, the system is well-specified, and some behavior ensues. Then letting $n \rightarrow \infty$ should indicate the behavior of the full collection. What is unclear to me now is whether any sphere velocity becomes unbounded as $n \rightarrow \infty$, which would demand a relativistic analysis. – Joseph O'Rourke Sep 7 '11 at 10:59
@Joseph, yes, you can try to let $n\to\infty$. However, non-determinism in the infinite case suggests to me that the system is probably chaotic, so I wouldn't expect the behavior under $n\to\infty$ to converge towards a limit -- beyond the constraints implied by macroscopic laws such as how the center of mass moves. – Henning Makholm Sep 7 '11 at 11:37
@Robert, yes, my example is unphysical. (So is Joseph's assumption that we have infinitely many separate objects in a finite space). The point was just that this kind of infinite ensembles of simple interacting laws can have aggregate behavior that is very different from -- and less reasonable than -- finite ones. – Henning Makholm Sep 7 '11 at 11:41

I worked out the details for the 2D circles case, where the mass of ring $i$ is twice the mass of ring $i+1$ (because the latter is half the radius). When ring $i$, traveling at velocity $v_i$, collides with stationary ring $i+1$, their velocities after the collision are $v'_i= \frac{1}{3} v_i$ and $v'_{i+1}= \frac{4}{3} v_i$, a simple consequence of conservation of momentum and energy.† (This incidentally corrects some miscalculations in the original answers.) Because $v_{i+1}$ then increases a bit under acceleration, we know that $v_k$ grows at least as fast as $\left( \frac{4}{3} \right)^k$.

So, if I've calculated correctly, assuming a 1-meter radius outer ring, it would take fewer than 64 nested rings to reach the speed of light. :-) Which means that, answering my original question ("What happens?") requires an analysis that incorporates the laws of special relativity.

† Just for the record, here is the derivation of the velocities just before and just after a collision. Let the $i$-th ring have mass $m$, and so the $(i+1)$-st ring has mass $\frac{1}{2}m$. Conservation of momentum: $$m v_i + \frac{1}{2} m v_{i+1} = m v'_i + \frac{1}{2} m v'_{i+1} \;.$$ Because $v_{i+1} = 0$, this reduces to: $v_i=v'_i+\frac{1}{2} v'_{i+1}$. Conservation of (kinetic) energy: $$\frac{1}{2} m v_i^2 + \frac{1}{4} m v_{i+1}^2 = \frac{1}{2} m (v'_i)^2 + \frac{1}{4} m (v'_{i+1})^2 \;,$$ which reduces to $v_i^2 = (v'_i)^2 + \frac{1}{2} (v'_{i+1})^2$. Solving these two equations simultaneously gives $v'_i = \frac{1}{3} v_i$ and $v'_{i+1} = \frac{4}{3} v_i$.

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Making many extremely unrealistic assumptions: After a collision, the second sphere will move 4 times as fast as the first (1/4 the mass) and the distance between it and the next sphere wil be halved, hence the time between each consecutive collision will be an eighth the time for the previous collision. It both spheres can feel gravity once released, both will accelerate downwards at the same rate, so the relative speeds will be the same as if there were no gravity and the first sphere were set off at an initial speed x m/s. $\sum_0^\infty 1/(8^{n})=1/9$ so it will take r/9 seconds for the collision to reach the centre (r is the radius of the first sphere), and the 'centre' will now be a distance of r/2 from the top so it will take 2r/9 seconds for the outer sphere to be hit again. Now the 'centre' will be 1/4 of the distance from the bottom. So first it takes r/3 seconds between when the outer sphere is hit, and after that the intervals between when the outer sphere is hit are 4r/9 seconds. In addition, the spheres will be constantly accelerating downwards.

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Thanks for this analysis! Do you see any sphere velocity becoming arbitrarily large? If so, a relativistic analysis might be necessary. – Joseph O'Rourke Sep 7 '11 at 11:01
"After a collision, the second sphere will move 4 times as fast as the first (1/4 the mass)" -- this assumes that the entire momentum of the outer sphere was transferred to the inner one. But its kinetic energy will then be 4 times the initial energy of the outer sphere, which cannot be right. – Henning Makholm Sep 7 '11 at 11:43