I’m reading “Why Does The World Exist”, and on page 97 it says:
Assume the billiard balls have some sort of magical force that keeps them moving around forever, are there really some configurations they will never achieve?
This was from a discussion the author was having with Richard Swinburne. There was no other context with regards to this scenario, so I’m not sure if Swinburne just made it up or if it’s a true statement.
That has to depend on what “all possible configurations” entails. If I grid up the surface into 1 foot squares, then look at which balls are in which squares, surely that would eventually occur.
If 1 foot is too large, grid it into 1 mm squares. It will take much longer, but you’ve got all the time in the world, and an infinite time after that.
If 1 mm is too large, how about 1 atomic radius? Hard to even define the position of a ball that well. But it’s still a finite number of squares, and as long as you want to achieve all the states.
Keep going smaller, and Heisenberg’s uncertainty principle eventually comes into play. You’d have to have infinite precision for where the balls are, to say that you can never achieve every state, but maybe that’s what he means.
It wouldn’t surprise me, though I can’t answer for certain.
If you think of a case with one billiard ball that is in constant motion, the presence of right angles on each edge means that it can be set on a trajectory that merely traces the same path over and over for eternity so it never achieves every position on the table.
As you add billiard balls, it’s possible to do so in a way that prevents collisions - this just repeats the single-ball repetitive motion and so there are multi-ball solutions that don’t achieve every configuration.
If you put all of the balls into random motion… there I’m not so sure. The collisions between the balls will certainly result in more of the table being traveled, with more configurations being achieved. But it wouldn’t surprise me if you still got a certain repetitive motion (perhaps repeated only after days or years) that failed to cover every possible configuration.
If ‘configurations’ is to include the vectors of each ball, as well as their positions, then there are some configurations that will never occur.
I’m going to think this out (perhaps only partially) in real-time as I type this.
Consider a simplified case, with just ONE ball rolling around. Assume it rolls with ZERO friction so it never loses momentum (and doesn’t lose any with each bounce at the edges either).
Now consider all the possible path segments that the ball can take. By “path segment” I mean one straight line segment, running from one point on the perimeter to another point on the perimeter. That is, the path a ball may follow from one bounce to the next consecutive bounce. (The two end-points of such a segment must obviously NOT be on the same edge of the table.)
Every such path may be defined by its two end-points (and the direction of the ball along that path, but I think we can ignore that for the moment). How many such paths are possible?
Lots and lots, obviously. Consider all ordered pairs (a, b) of distinct points on the perimeter. There are ℵ[sub]1[/sub] of these (if I understand my ℵ’s right).
There was another thread fairly recently in which this question came up: If the universe is “closed” (in the sense that an object moving in a “straight line” might eventually find itself back at its starting point), is it certain that an object would return to its starting point? The answer seemed to be: Not necessarily. For example, a point moving on the surface of a 3-sphere might have an “irrational” path (I think that means something like: whatever numbers you assign to describe its path include irrationals) then the path would never repeat itself.
I’m thinking that the same argument must apply to the ball rolling on the pool table.
You could (in theory) engineer cases where the ball would roll on a “rational” (repeating) path, by carefully choosing the very first path segment. But there are only ℵ[sub]0[/sub] “rational” end-points that could define such a path.
Now, from another recent thread, we discussed that, on a line segment of finite length, the ℵ[sub]0[/sub] “rational” points on that segment are so few that they contribute exactly ZERO to the total length of the segment. (This was shown to be true even if you include all the algebraic irrationals.) To get the full measure of the segment’s length – indeed, to get any non-zero length at all – you had to include the transcendental irrationals, which are so much more “numerous” as to contribute 100% to the length of the segment.
Do you see where I’m trying to go with this? If you set the ball rolling on a randomly-chosen path, I think there is a 0% probability that it will prove to be a repeating path, and a 100% chance that it will prove to be a non-repeating path. (Yet you could still find specifically-chosen paths that would repeat.) I’m also thinking that, if you had such a non-repeating path, the ball would never traverse ALL possible paths in ℵ[sub]0[/sub] time.
Okay, all of the above is entirely an informal analysis of the situation. But I think it covers the overall outline of my argument, and I’ll leave it at that. Do we have any advanced math students here who want to run with that? ( Chronos? Indistinguishable? Trinopus? etc.? Are you there?)
Assume a billiards table with a ratio of length and width as a rational number with the two sides aligned with the x and y axis respectively. Start with a ball positioned at (x, y) such that x, y e Q. Strike the ball so that it has a rational slope when looking at the coordinate plane.
Question: will there ever be a time where the ordered pair representing its position have one rational and one irrational number?
Assume the slope is irrational, will the ball ever be in a second position where both numbers of the ordered pair are rational?
I think this is the same as asking:
Let the path of the ball be: y = mx + b where slope m is rational.
For any rational x, you can compute rational y
And for any rational y, you can compute rational x.
(Since the rationals are closed under addition and multiplication.)
And also if m is irrational and b is rational can both x and y ever both be rational.
I’m still working on what what happens if the ratio l/w is irrational. I don’t think it makes a difference.
But if working with infinite sets, a 0% probability is not the same as an impossibility of the event happening like it does with finite sets.
Since the number of configurations is also infinite, I don’t find the fact that all configurations will not be reached far-fetched.
From my earlier post:
I suppose, for the record, I should question my own assumptions here: I am suggesting (based on intuition only, and we all know how reliable that is) that ℵ[sub]1[/sub] paths of a billiard ball could not be traversed in ℵ[sub]0[/sub] time. First, that’s just what I think is “obvious” but I hope the see some comment on that from someone more infinitely knowledgeable about this.
Even more so, I wonder just how long “eternity” is thought to be. Is eternity ℵ[sub]0[/sub] time or ℵ[sub]1[/sub] time? Or something else?
IANAMathematician, so bear with me, but I think the question is whether we are treating the table as “cellular” or as composed of points on a continuous plane (i.e., essentially, whether every point has Rational coordinates or Real coordinates.)
I’m also assuming that time is considered as discrete.
If time is discrete (and there are א[sub]0[/sub] “moments” in Eternity) and the table is treated as a set of points on the continuous plane (i.e., it has א[sub]1[/sub] points), then not all configurations can ever be exhausted.
However, IMLO (In my layman’s opinion) either we treat both space and time as discrete (e.g., limited by Planck’s Length and a similar analogue for time), or we treat both as continuous. In the discrete case I think it’s trivial that the number of configurations on the billiard table can be mapped to unique points on the time axis and the statement in the OP is false. I haven’t the faintest idea how to start looking at a mapping from א[sub]1[/sub] onto/into/over/whatever א[sub]1[/sub]…
I have no idea what your notation means, but my thinking is that trying to paint an area (billiard table) with a line (path of ball) is not possible as the line is one dimensional and the table is two dimensional. Is it possible to establish a one to one correspondence between a line and a rectangular area? I think no.
You’re not establishing a one-to-one correspondence, (although it is possible), but it only matters whether the path is onto. The answer is that it can be done.
Ah, but the scenario of the OP limits us to the sort of curves that can be produced by a bouncing billiard ball on a pool table (consisting of a succession of straight line segments each bounded by points on the perimeter of the table). It’s not apparent that any such succession of segments could ever produce anything like any of the Peano-like curves shown in the wiki.
That said, can it be done?
First answer in the thread…best answer in the thread!
As noted by others, it is possible to construct a pattern that repeats, and which will, thus, never attain all possible configurations. But I would say that such configurations are “infinitely” unlikely. They have to be deliberately (intelligently!) created. In practice, following ZenBeam’s process of reducing the size of the grid, and breaking the rack of balls with a “typical” opening shot, the mixing of the balls over the table is ergodic – it will encompass every possible configuration.
Quantum physics actually helps. Under QM, even if you did create a perfectly balanced, ever-repeating configuration, the uncertainty of position means that the balls would gradually drift away from that situation, and take on more and more random positions. Under QM, you cannot make one ball bounce back and forth, exactly, east-to-west and west-to-east, across the table. Eventually, uncertainty will make it veer, just slightly, north or south, and that deviation will grow.
QM also makes the question meaningless at the very, very smallest levels of ZenBeam’s grid – as ZenBeam noted. When you can no longer say whether the ball is in grid position xn,ym, since it might actually be in xn+1,ym, then “possible positions” stops being certain. Was configuration k actually attained? Best we can say is maybe.
Anyway. “Yes.”
This is a statement about something called ergodic theory. To explain it, I have to first explain something called “phase space”. For each billiard ball, you can specify four numbers that tell you its configuration: its x-coordinate, its y-coordinate, its x-velocity, and its y-velocity. This means that if there are N billiard balls on the table, you can specify the overall configuration of the system using 4N numbers (four for each ball.)
You can then view these 4N numbers as coordinates on an abstract 4N-dimensional space called phase space. Each point in this space corresponds to a configuration of the system, since it uniquely tells me the position and the velocity of each ball. As the system evolves, and the balls bounce off of each other and the walls, the corresponding point in phase space will move around. So as time goes on, the system traces out a curve in this abstract 4N-dimensional space.
The question is: does this curve get to all points in phase space?[sup]1[/sup] If that were true, then we would just have to wait long enough and the system would reach any configuration you care to name. However, this turns out to be impossible. The basic idea here is that curves are one-dimensional objects, and while (as noted above by President Johnny Gentle) it’s possible for certain curves to fill out higher-dimensional spaces, these curves don’t have the right properties for this. (The fancy word is that they’re “integral curves”, which roughly means that they’re constructed by following a smooth “flow” through phase space.) So that means that even if we extend the curve all the way out to t = ∞, there’ll still be some points in phase space that don’t get “hit”.
There are relaxed versions of this question for which the answer to the above question can be “yes”; for example, “does the system get arbitrarily close to any configuration if I give it enough time?” or “will the system return arbitrarily close to its initial configuration if I give it enough time?” The answer to the first question depends on the system; the answer to the second question is “yes”, so long as the accessible volume of phase space is bounded. Unfortunately, this text entry box is too small to contain the proofs of these remarkable statements.
[sup]1[/sup] On its face, the answer to this question is “no” — different points in phase space will correspond to different amounts of energy, and the system can’t change its energy. To make this an interesting question, you should really read “phase space” in this statement as “the region of phase space consistent with conservation of whatever quantities are conserved in this setup.”
I’m not sure whether the OP meant “configuration” in this sense (i.e., coordinates and velocity vector for each ball) or just in the sense of “set of 16 coordinates at a given point in time.”
I understand (I think) your answer to the former, but wonder whether it’s possible that the latter question also result in an unreachably large number of possible configurations
I saw what you did there 
Bizarre as it may seem, I remember discussing with my then roommate (at New College in Sarasota, Florida) a simplified version of this question as we played pool one day during our undergraduate years. My roommate was John Smillie, currently professor of mathematics at Cornell University:
http://www.math.cornell.edu/People/Faculty/smillie.html
He proved to his own satisfaction while we stood and talked that a single ball hit on a pool table in a random direction would with probability 1 enter a path that never repeats. (Remember that “with probability 1” doesn’t mean that it would always happen, just that the number of times it wouldn’t happen is infinitely swamped by the number of times that it would happen.) He proved some other things about pool ball paths at the same time. I remember jokingly calling them the Smillie Pool Theorems. I note from his publications (given in the above link) that he has since published articles about billiard flows, so he’s still interested in the subject. Maybe he can answer your question.
Only if a and b are real numbers. If they’re integers or rationals, the set of all ordered pairs of them still has cardinality ℵ[sub]0[/sub].