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Suppose $n$ air molecules (infinitesimal points) are bouncing around in a unit $d$-dimensional cube, with perfectly elastic wall collisions. Let $k=n^{\frac{1}{d}}$. For example, in 3D, $d=3$, with $n=64$, $k=4$. In analogy with the lonely runner conjecture, say that a molecule $m$ is lonely if it is at least distance $\frac{1}{k}$ from every other molecule, $\frac{1}{4}$ in the example.

I am not entirely certain this $\frac{1}{k}$ distance is the appropriate generalization, but let me proceed nevertheless.

Q1. Assuming the molecules start at random positions, and have distinct speeds and distinct, irrational-multiples-of-$\pi$ initial direction vector angles w.r.t. the cube edges—and so never collide—is it the case that each molecule is lonely at some time? Or more generally, what is the largest distance $\lambda$ for which is it guaranteed that each molecule is at least $\lambda$ from every other molecule at some time?

Q2. And an obverse question: Under the same assumptions, what is the smallest subcube volume $v$ in which it is guaranteed that all molecules will at some time simultaneously reside?

Perhaps $v=(\frac{1}{k-1})^d$ is the value to beat for Q2?

These questions lay behind this recent question. I was wondering about the chance that all air molecules would spontaneously cluster in a corner of a room. :-) I am unclear on whether the circle in the lonely runner conjecture, and reflecting cubes above, are sufficiently analogous to transfer insights from one to the other.


Addendum. The questions are answered in the comments, by Anthony Quas & Evan Jenkins. Q1: Each molecule will become arbitrarily lonely (up to the diagonal length of the box). Q2: The molecules will cluster arbitrarily densely. The reason, to quote Evan, is that "any nonempty open condition on particle configurations will be obtained." As Will Sawin's comment makes clear, the nondegeneracy conditions must be more carefully crafted to obtain lonely-runner-like complications.

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    $\begingroup$ Why put up walls? Have them run around in a 3-torus. You might get quicker answers that way. $\endgroup$ Commented Jul 25, 2014 at 0:46
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    $\begingroup$ No such condition (on the angles) can prevent collisions: we can still shift parallel to the coordinate axes. However, just like the runners, we can just let the molecules ignore each other when they do collide. $\endgroup$ Commented Jul 25, 2014 at 0:48
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    $\begingroup$ If they're randomly placed, then they will almost surely cluster somewhere sooner or later. It's easiest to work on the torus, but completely equivalent because you can unfold $2^d$ copies of the cube to make 1 copy of the torus. $\endgroup$ Commented Jul 25, 2014 at 1:16
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    $\begingroup$ @JosephO'Rourke: Consider your $n$ particles in a $d$-dimensional cube instead as a single particle in the $nd$-dimensional unit torus (an unfolding of $2^{nd}$ cubes). Provided the $nd$ coordinates of the velocity vector are linearly independent over $\mathbb{Q}$, it follows that the $\mathbb{Q}$-span, and hence the $\mathbb{Z}$-span, of this vector is dense in the torus. This implies that any nonempty open condition on particle configurations will be obtained. $\endgroup$ Commented Jul 25, 2014 at 2:25
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    $\begingroup$ Precisely which non degeneracy condition you choose is critical. In the lonely runner conjecture, the velocities are required to be distinct, but not linearly independent over $\mathbb Q$. I would guess the right approach is to require the velocities be pairwise distinct in each coordinate axis (or just one?). One might also consider what notion of distance is appropriate - $L^1$ norm? $L^2$ norm?. $\endgroup$
    – Will Sawin
    Commented Jul 25, 2014 at 2:45

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