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.
The system is chaotic, so the uncertainty increases exponentially with each collision. Also atoms are only about 1 angstrom wide, so the first unpredictable collision p0 makes will send it flying in some random direction, and totally miss the 2nd atom it should have collided with, instead probably hitting some other atom.
However, the position of a single particle p0 after 20 seconds is not uniform. It simply doesn’t have time to diffuse far enough; with 2*10^11 collisions and mean free path 70 nm, it will travel a total displacement of 70 nm *sqrt(2*10^11) = 3 cm. Even if the box contains only 1 mol of gas, lower than atmospheric pressure, I think it should only travel 20 cm on average.
It could still be that the total uncertainty in many particles adds up to the majority side being unpredictable.
Doing it for one particle seems like it would be harder than doing it for all particles, since even if you are highly uncertain about each individual particle, in-aggregate that could still produce a quite high confidence about which side has more particles. So my guess is it matters a lot whether it’s almost uniform or not.
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.
The system is chaotic, so the uncertainty increases exponentially with each collision. Also atoms are only about 1 angstrom wide, so the first unpredictable collision p0 makes will send it flying in some random direction, and totally miss the 2nd atom it should have collided with, instead probably hitting some other atom.
However, the position of a single particle p0 after 20 seconds is not uniform. It simply doesn’t have time to diffuse far enough; with 2*10^11 collisions and mean free path 70 nm, it will travel a total displacement of 70 nm * sqrt(2*10^11) = 3 cm. Even if the box contains only 1 mol of gas, lower than atmospheric pressure, I think it should only travel 20 cm on average.
It could still be that the total uncertainty in many particles adds up to the majority side being unpredictable.
Doing it for one particle seems like it would be harder than doing it for all particles, since even if you are highly uncertain about each individual particle, in-aggregate that could still produce a quite high confidence about which side has more particles. So my guess is it matters a lot whether it’s almost uniform or not.