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If we assume that a fluid is a continuum then if we have for example a cup of tea and we stir the fluid then there will be a point in the fluid that is on the same location before and after the stirring.

Now, a practical fluid is not a continuum although it is for many practical situation in fluid mechanics.

Does this influence the statement much? Is there for example a small box where the particle will be in (which is somewhat in the same magnitude as the size of the particles) or does the complete statement break down?

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up vote 4 down vote accepted

You can substitute the discretized version of the Brouwer fixed point theorem (which can be used to prove the usual version): Sperner's lemma. This isn't a complete answer but it should point the way. I believe the idea is to use colors to indicate the possible directions a particle travels, although I haven't thought about the details.

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To apply Sperner's lemma to Brouwer's fixed-point theorem, you have to let the size of the polytopes tend to zero, which breaks down in a real-world context. I would be surprised if there were any kind of approximate fixed-point result that would apply to a real cup of tea. – TonyK May 29 '11 at 17:56
Yes, there are problems. Sperner's lemma will let you find a point such that nearby points are being stirred in every possible direction, or something like that, but with a fixed triangulation that doesn't necessarily tell you anything about the point itself. I agree that the basic issue is that it's unclear what reasonable notion of continuity might be preserved in the discrete setting. – Qiaochu Yuan May 29 '11 at 18:36
Sperner's lemma allows you to find almost-fixed points, points that are mapped with epsilon distance to itself. What you can not guarantee is that the point is close to a real fixed point. – Michael Greinecker Jan 4 '12 at 15:07

I imagine that after stirring a cup of ideal tea twenty or thirty times, some points that were initially less than an atomic radius apart will become separated by more than half the diameter of the cup. Which makes Brouwer's fixed-point theorem physically meaningless.

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I don't see how this answers the question. Why does this preclude the possibility that some points that were initially less than an atomic radius apart are still less than an atomic radius apart? – Qiaochu Yuan May 29 '11 at 16:28
It doesn't preclude the possibility, but it is also possible that all molecules that were initially close are now widely separated. After a certain time (even without stirring, just diffusion) the molecules in the tea have essentially been permuted randomly. Then we basically have a version of the "Problem of Derangements". If we divide up the liquid into cells such that each cell contains one molecule, the probability that at least one molecule is back in its initial cell is approximately $1 - 1/e$. – Robert Israel May 29 '11 at 17:03
@Qiaochu: The possibility is not precluded by my argument. But Brouwer's theorem relies on the displacement function being continuous, which is not the case at the molecular level. – TonyK May 29 '11 at 17:48
@Robert: However you divide the liquid up into cells based on the initial positions of the tea molecules, the cells in the final state will certainly (OK, almost certainly) not contain one molecule each. Suppose we re-pose the question, to ask: What is the probability that there exists at least one molecule which is closer to its initial position than to the initial position of any other molecule? Do we still get $1 - 1/e$ ? Or is that a hard question? – TonyK May 29 '11 at 18:51

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