Is a random box of gas predictable after 20 seconds?

A little (perhaps pedantic) point of observation:

 "there's no reason information should be preserved for 20 seconds" - A sort of aside on this comment. For an idealised classical mechanical (reversible) system the information will be preserved forever. Chaos, roughly speaking, is moving the same information into different significant figures (or relations between them). So an initial uncertainty in the 20th decimal place soon becomes an uncertainty in the first significant figure as the information moves about. All that information you had about the leading digits of precision describing the initial state is still there in some sense, it has been mapped into some bizarre constraints connecting different figures deep behind the decimals, and is practically useless.

So, in this sense the question (I think) is whether that remaining information you have tells you anything about which side of the box will have more molecules at t=20. So, instead of a 50/50 guess, does the information let you get to 60/40 or whatever.

My feeling is that it must be almost worthless. Lets say that information takes you from a 50/50 guess to a  50+s/50-s guess. My intuition is that if we plotted this "s" value as a function of time it is likely an exponential decay, and that 20 seconds feels like a very long time compared to the timescales involved in the molecular motion. At t=0  s will be very close to 50 (only if the perturbed molecule is within one angstrom of the dividing line between left and right will s be less than 50 at t=0). But at t=20 it has undergone many half-lives. So its probably 10^{-big number} after 20 seconds.

The prediction that the information would be significant implies an assumption that s does not exponentially decay with time, but is described by some other function (maybe a constant). So I think the core of the dispute might different assumptions on the shape of the s(t) function.

You could instead ask whether or not the observer could predict the location of a single particle p0, perhaps stipulating that p0 isn't the particle that's randomly perturbed.

My guess is that a random 1 angstrom perturbation is enough so that p0's location after 20s is ~uniform. This question seems easier to answer, and I wouldn't really be surprised if the answer is no?

Here's a really rough estimate: This says 10^{10} s^{-1} per collision, so 3s after start ~everything will have hit the randomly perturbed particle, and then there are 17 * 10^{10} more collisions, each of which add's ~1 angstrom of uncertainty to p0. 1 angstrom is 10^{-10}m, so the total uncertainty is on the order of 10m, which means it's probably uniform? This actually came out closer than I thought it would be, so now I'm less certain that it's uniform.

This is a slightly different question than the total # of particles on each side, but it becomes intuitively much harder to answer # of particles if you have to make your prediction via higher order effects, which will probably be smaller.

Standing waves of pressure (a.k.a. sound resonances) are macroscopic patterns with some persistence. Their amplitudes would be really low (≈kT of sound energy per resonance) in a random initial configuration, but if you knew every particle, you would still know the starting amplitude and phase of each standing wave, and could project it out, and one particle won’t appreciably affect the starting amplitude and phase I think.

So, do the standing waves have enough persistence to matter? How much do those standing waves decay after 20 seconds? I dunno. I know how to calculate all the frequencies, but I have no idea how to calculate the Q-factors—at least, not off the top of my head.

I'm also not sure which standing waves (if any?) bear on left-right difference in particle count.

I think the term "forecaster" is perhaps confusing here and it would be more clear to say "what fraction of the time is the final configuration in terms of left/right the same under a tiny random perturbation".

That is, let $\theta$ be the fixed initial random configuration and let $is\_left\left(\theta \right)$ be whether after 20 seconds the final configuration (when starting from $\theta$) is to the left. (Note that this is a fully deterministic quantity.) Then, let $m(\theta ,X)$ pertub a particle in the state $\theta$ as described given the random variable $X$.

Now, we care about the value of:

$$\underset{X}{E}\left[is\_left\right(m(\theta ,X)\left)\right]$$

(In particular whether it is very close to $0.5$ or is $>0.51$ or $<0.49$.)

Minor nitpicks: -I read "1 angstrom of uncertainty in 1 atom" as the location is normally distributed with mean <center> and SD 1 angstrom, or as uniformly distributed in solid sphere of radius 1 angstrom. Taken literally, though, "perturb one of the particles by 1 angstrom in a random direction" is distributed on the surface of the sphere (particle is known to be exactly 1 angstrom from <center>). -The answer will absolutely depend on the temperature. (in a neighborhood of absolute zero, the final positions of the gas particles are very close to the initial positions.) -The answer also might depend on the exact starting configuration. While I think most configurations would end up ~50/50 chance after 20 seconds, there are definitely configurations that would be stably strongly on one side.

Nothing conclusive below, but things that might help: -Back-of-envelope calculation said the single uncertain particle has ~(10 million * sqrt(temp in K)) collisions /sec. -If I'm using MSD right (big if!) then at STP, particles move from initial position only by about 5 cm in 20 seconds (cover massive distance, but the brownian motion cancels in expectation.) -I think that at standard temp, this would be at roughly 1/50 standard pressure?

Does this question, as posed, admit the possibility that the answer could be "it depends on what the initial random configuration is"? I suspect that's what it is: there are states that robustly result in the left side having more particles than the right after 20s.