The Yukiad, Perpetual Motion and Me

yukiadIt’s a bit of an odd feeling to be reading a novel and stumble upon yourself as a character. Well, at least a well-disguised version of yourself. The novel is Victor Snaith’s The Yukiad, and the character is a large Scotsman named Pans who tugs at his earrings when he becomes agitated. I am neither Scottish, nor earringed, nor particularly large, but I suspect that Pans, viewed through the haze of poetic license, is I.

When we meet Pans, he is hovering over a glass contraption—a perpetual motion machine, really—consisting of a circular tube containing several colored beads, which travel around the tube, some clockwise, some counterclockwise, all at the same speed, bouncing off each other in perfectly elastic collisions whenever they collide. Pans is currently tugging at his earrings so hard as to cause some concern for the integrity of his earlobes, as he ponders the following question:

But wull tha’ aver gut bark to weer tha’s started, at a’, at a’?

Well, okay, maybe I’m not Pans. Maybe I’m the character Sherloch Humes, a “trim but rather wrinkled gentleman in worsteds”, who calculates for Pans’s benefit that “the configuration of beads is guaranteed to have exactly replicated itself by the year two thousand and nineteen”. I believe that I am the inspiration for one of these characters and that the mathematician Leonid Vaserstein (who is neither Scottish nor wrinkled) is the inspiration for the other, and here is why:

Many years ago, for reasons I can no longer remember, Vaserstein and I posed ourselves the following question: Suppose several point particles, all of different colors, are located on a circle, with some traveling clockwise and others counterclockwise, all at the same speed. Whenever two particles collide, they instantly bounce off each other and proceed in the opposite of their original directions, still at the same speed. Take a snapshot of this system at, say, 12PM. Must there be some time in the future when another snapshot of the system will look identical? In other words, does the history of the system repeat itself?

As I said, I can no longer remember how we came up with this question, but I do remember that it triggered a vigorous discussion. We were riding in a car at the time, myself in the passenger seat and Vaserstein in the rear. Victor Snaith, who some years later was to write The Yukiad, was driving.

The solution—which I’m pretty sure Vaserstein contributed more to than I did—is beautiful and elegant and crystal clear once you see it. Do you see it?

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8 Responses to “The Yukiad, Perpetual Motion and Me”


  1. 1 1 Bennett Haselton

    Spoiler alert: solution given (I think).

    Consider the point particles to be identical (i.e. ignore their colors). Then two particles colliding and moving apart are equivalent to the two particles passing each other while each continuing at the same speed. So obviously, ignoring the colors, you just have particles looping around the circle, and you have infinitely many times at which the pattern repeats itself, where the interval between them is the time taken for each particle to make a trip around the circle.

    Now, fill the colors back in. There are only finitely many ways that the colors can be assigned to the particles. That, by itself, doesn’t quite prove that your initial pattern I must repeat itself. But it proves that *some* pattern P must repeat itself.

    Once you know that P repeats itself, look at the first occurrence of P and rewind time backwards to your initial pattern I. Since the perpetual motion machine is completely deterministic (forwards and backwards), you can start at the second occurrence of P and work backwards, and through exactly the same sequence of events, you must hit on a second occurrence of I.

  2. 2 2 Dave

    I think the answer is that it certainly could repeat itself. If you start with 2 particles only. A blue particle and a red particle sitting at the top of the circle at 12pm and they start moving at 1 loop per hour in opposite directions, every half hour they would meet at the bottom of the circle, collide, then make their way back up to the top where they meet again at the top of ever hour. This will continue indefinately and you could even tell what time past the hour it is if you take a snapshot and knew the direction they were moving in at the time the photo was taken.

    Repeat but with 2 particles at the top and 2 at the bottom, they would collide every 15 minutes and reverse their track repeating the pattern again and again.

    So obviously works with a perfectly symettrical start. I just don’t have the imagination to work out what happens in non symmetrical situations.

    My gut feel is that given enough time you get repetition but then my understanding that pi never gets a repetitive pattern of decimals bites against that theory.

  3. 3 3 Jon Shea

    I suspect that the configuration will repeat once every nl / v time units, where n is the number of beads, v is there velocity, and l is the circumference of the loop.

    It helps to think about the problem in the reference frame that is corotating with, say, the counterclockwise beads. This makes it the same as a problem where some beads are stationary and some beads are moving clockwise.

  4. 4 4 Nick

    I think the more profound question is whether the result would obtain if you endow the beads with spherical volume and allow them to move freely in three dimensions (i.e. not constricted to the torus’ central ring), bouncing off walls and so forth. Even in the two-bead case I’m not sure you could guarantee the recurrence of initial conditions with anything other than contrived starting orientations.

  5. 5 5 xan

    The answer is an emphatic yes!

    Suppose the balls all are traveling at 1 revolution/minute.

    The first insight is that, ignoring colors, the problem is identical to a situation in which the balls are passing through each other instead of bouncing off. From this, we see that after 1 minute, there will be _some_ ball in each position that there was originally a ball, going in all the same directions as the original setup. That’s the coolest thing about this problem: you can have whatever crazy spacing you want, but it turns out to be irrelevant. From here on out, the problem has only to do with the _order_ of the balls around the circle.

    Now, since balls are _not_ passing through each other, their order around the circle is preserved at all times. After a minute, they may have jointly rotated around the circle, but they are still in the same order as before. Call this 1-minute rotation R. Each time we apply R, it does the exact same rotation, shuffling the balls around from place to place until they come back to where they started. If you know a bit of (abstract) algebra, you are probably convinced of the answer by now. But for those who haven’t, some examples will make it clear.

    First, suppose there are 3 balls. Number their starting positions 1,2,3. If R takes the ball in position 1 back to position 1, then it also must take 2 to 2 and 3 to 3, meaning that after a minute, everything is as it was.
    Alternatively, R could take 1 to 2, in which case it must take 2 to 3 and 3 to 1. We can represent this by the cycle R=(1 2 3) (read “1 goes to 2 goes to 3 goes to 1″). Applying R twice gives R*R=(1 2 3)*(1 2 3) = (1 3 2), which is to say, letting the system run for 2 minutes will take ball 1 to position 3 and so forth. Applying R three times, we get back to where we started. To convince yourself of this, it helps to draw a picture and track what’s happening.

    If you have more balls, say 6, you will get a cycle like (1 2 3 4 5 6) or (1 3 5)(2 4 6) (read “1 goes to 3 goes to 5 goes to 1, and 2 goes to 4 goes to 6 goes to 2″). Convince yourself that no matter what, you will always come back to where you started if you do it 6 times. (Of course, you may get back sooner, as in the second case).

    Generalizing this, we see that for n balls, a snapshot taken after n minutes will always be identical to the starting position. What’s more, the balls will be going their original directions, ready to repeat the whole thing over again.

  6. 6 6 xan

    As an afterthought, we don’t actually need the order of the balls being preserved. No matter what, we’d get some cycle like (1 4 2 3 6 5) which would take everything back to where it started after 6 minutes. But the above may be easier for people to see, fwiw.

  7. 7 7 David Grayson

    Good job, Bennett Haselton: you were the first poster, you were correct, AND you were the most concise one here.

  8. 8 8 Jonathan

    I am curious what led you and Prof. Vaserstein to pose this question. It obviously has a very interesting and nifty solution, but without knowing the answer, it does not seem at all clear that the question will lead to anything interesting. Did you consider other similar questions, e.g. without the constraint that all balls have the same speed, before deciding that only this one had an interesting answer? (Or maybe that would be interesting too?)

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