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I would like to know if this is theoretically possible or not and if there is a way of calculating the probability of this happening.

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    $\begingroup$ Where's the momentum going to come from? $\endgroup$
    – Alan Wendt
    Commented Mar 31, 2023 at 22:09
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    $\begingroup$ @Alan Wendt: The walls of the room - Newton's 3rd law of motion predicts that as all the air in the room packs into, say, the left half, the room (and the building and the planet) are pushed to the right. - All the air molecules must have, directly or by proxy, kicked off from the right-hand wall within a few milliseconds. $\endgroup$ Commented Apr 1, 2023 at 20:33
  • $\begingroup$ Why not describe the circumstances, please? Failing that are you not relying on magic or, worse, guesswork? $\endgroup$ Commented Apr 3, 2023 at 20:21

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If we would calculate probability that air will compress "itself" by half, per time interval required for 1 molecule to travel half-room distance $L/2$,- then this "synchronized movement to particular direction" of all molecules simultaneously would be even more unlikely.

Per half-room travel molecule experiences $$n=\frac {L}{2\ell} \tag 1,$$

collisions, where $\ell$ is mean free path of molecule. Let's assume that there is $1/2$ probability that with each hit, molecule will bounce to solid angle which directs towards required room side. Then considering that there is $N/2$ molecules in half-room, probability of them spontaneously move to the target zone at once is :

$$ P_t = \left(0.5^n\right)^{\frac {N}{2}} = 0.5^{\frac {LN}{4\ell}} \tag 2$$

Now, for the sample numbers, consider some room with area $25~m^2$ which has room diameter of $L=5~m$, and mean free path of molecule in the air is $\ell = 64~nm$. Substituting this into (2) for that room gives :

$$ P_t \approx 10^{~-5,879,493~N} \tag 3$$

where $N$ is number of molecules in that room. Even without substituting $N$ into the formula it's clear enough that such probability is for all practical purposes zero.

Btw, keep in mind that if we will raise allowable time duration (make it longer) in which all molecules could "self-organize",- probability $P_t$ would be even more smaller, because with greater time duration, molecules experiences more collisions, namely :

$$ n(t)=\frac {s}{\ell} = \frac {\overline v~t}{\ell} \tag 4, $$

where $s$ - molecule accumulated travel distance, $\overline v$ is average speed of molecule. So, over time single molecule experiences more collisions and hence probability for molecule to keep required bounce solid angle becomes even smaller (probability to keep "one-directional" movement), due to independent probability addition principle. So, if you are thinking about saving your costs with not buying a vacuum/pump device,- drop it :-)

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    $\begingroup$ Its worth noting that "Let's assume that there is 1/2 probability that with each hit, molecule will bounce to solid angle which directs towards required room side. " is clearly false, since the thing it hit was probably another molecule, causing the other molecule to bounce in the wrong direction. Therefore, the odds are far far rarer than even this $\endgroup$ Commented Mar 31, 2023 at 14:38
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    $\begingroup$ I don't quite get that part: "if we will raise allowable time duration (make it longer) in which all molecules could "self-organize",- probability Pt would be even more smaller". I believe that the "infinite monkey theorem" guarantees that this will happen given an infinite amount of time. So if there is any probability > 0, then the probability is automatically 100%, no matter how small. $\endgroup$
    – Guilherme
    Commented Mar 31, 2023 at 18:26
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    $\begingroup$ "For all practical purposes zero". Well, not necessarily. If your practical purpose is working out possible ways in which a Universe might spontaneously erupt, then it's of practical interest that if you leave the system in a steady state for an infinite length of time, some highly improbable events are bound to occur eventually. $\endgroup$ Commented Apr 1, 2023 at 7:54
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    $\begingroup$ @AgniusVasiliauskas that is not true, probability that a special low entropy state will be observed actually increases with time, and after a very long time it is certainty. This is due to Poincare recurrence theorem. $\endgroup$ Commented Apr 1, 2023 at 15:09
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    $\begingroup$ @AgniusVasiliauskas it is a mathematical theorem for Hamiltonian systems with finite volume phase space. It is well-known that for mechanical systems, 2nd law holds only in statistical sense and fluctuations are possible. Universe has nothing to do with this. $\endgroup$ Commented Apr 1, 2023 at 16:04
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The probability of one molecule being in half the room is 1/2. The probability of them all being in half the room is $(1/2)^N$ where N is the total number of molecules in the room.

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    $\begingroup$ I'm not sure it makes sense to attribute a "probability" to this event. It makes more sense, to me, for this to have a "probability of happening within a certain amount of time." In other words, we should have an estimate of how often this "dice" gets re-rolled. Do you have any sense of how to estimate that? $\endgroup$
    – AXensen
    Commented Mar 30, 2023 at 12:20
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    $\begingroup$ @RogerVadim I'm pretty sure it's bigger than 100, which is really all you need to know :) $\endgroup$
    – hobbs
    Commented Mar 30, 2023 at 18:17
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    $\begingroup$ @AndrewChristensen I think both questions make sense. One is "If you observe the room at one instant, what is the probability that all the molecules will be in the western half of the room at that instant?" and one is "If you observe the room over some interval of time, what is the probability that all the molecules will be in the western half of the room at at least one instant in that interval?" Both of these are nonzero, but the former is much easier to estimate. $\endgroup$ Commented Mar 30, 2023 at 18:35
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    $\begingroup$ +1, but this assumes the molecules are all independent of each other, which isn't realistic because collisions are significant at the scale involved. My guess is that because of this the true probability will be much lower than what you'll estimate with this method. $\endgroup$
    – N. Virgo
    Commented Mar 31, 2023 at 4:18
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    $\begingroup$ @N.Virgo for a gas that can be described within the molecular chaos hypothesis this is a really good approximation. To see if this is a good approximation you just have to calculate the "excluded volume" and for air this is minuscule wrt the volume of the room. $\endgroup$
    – Quillo
    Commented Mar 31, 2023 at 12:51
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For an ideal gas, which is a good model for air at standard temperature and pressure, the the chance of finding all $N$ of the particles in the same half of the volume is $(1/2)^N$, which is tiny even for $N=1000$ let alone a typical lab scale sample where $N \sim O(10^{23})$. This probability statement can be interpreted in terms of the canonical ensemble -- i.e. it is the fraction of microstates consistent with the specified $NVT$ constraints and have all the particles in the same half of the room.


To turn this into an estimate of how long you'd have to wait, you need an estimate of how long it takes for the system to move to an independent state in the canonical ensemble. This will be governed by the timescale $\tau = L/c$ where $c$ is the speed of sound and $L$ is the characteristic size of the volume under consideration. After waiting something like thr time scale $L/c$, the system will have evolved to an effectively independent configuration of density. So overall, you'd need to wait a time $t$ something on the scale of $t \sim (L/c) 2^N$

This time scale can be precisely motivated by considering a box of side $L$. The lowest mode for density (sound) waves in this box has a frequency of $c/(2L)$. So once you wait something longer than this time, all of the modes will de-cohere relative to one another and you have an independent density configuration. One could start with the configuration of all of the gas on one side of the room and evolve the modes to see how quickly it spreads to an effectively uniform density as a concrete example of how this works.

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  • $\begingroup$ Could you put this in actual numbers for an example volume/temperature/air pressure? $\endgroup$ Commented Apr 1, 2023 at 0:40
  • $\begingroup$ @PaŭloEbermann Probability 0. Time forever. $2^{10^{27}}$ (two to the power of the number of molecules in a room of a few meters length) is uncountably huge. $L/c$ (about $10^{-2}s$) is irrelevant next to how huge it is. The difference between the number of seconds you'd have to wait and the number of lifetimes of the universe you'd have to wait is a rounding error on its power-of-10 representation. $\endgroup$
    – g s
    Commented Apr 3, 2023 at 19:43
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If there is only 1 molecule in the room, then that molecule can be either in one half of the room or in the other half. The occupancy looks like 1/0 or 0/1. In two out of two cases, all the gas in the room is in one half of the room, so the probability that all the gas in the room is in one half of the room at any given time is 2/2=1, or 100%.

If there are 2 molecules in the room, then the occupancy of the room at any given time can be 2/0, 1/1, or 0/2, so in 2 out of 3 cases all the gas is in one half of the room, and the probability is 2/3, or 67%.

With 3 molecules, the probability is 2/4, with 4 molecules it is 2/5, with 5 molecules it is 2/6, and so on. So, as long as the molecules don't influence each other (at VERY low gas density or zero pressure), the probability for N molecules is 2/(N+1).

Keep in mind that once the density of the gas increases to a level where the molecules start bouncing into each other, they will push each other in opposing directions, so the distribution of the gas in the room becomes less and less random, and besides the probability being low due to an already large number of molecules in the room, the probability further decreases exponentially with increasing gas pressure. To find a formula for this is beyond my ability at this time but the likeliness to find all molecules in the same half of the room definitely will decrease exponentially with increasing gas pressure to become infinitesimally small. This will already be the case at VERY low gas pressure close to absolute vacuum. At atmospheric pressure, the probability to find all molecules in the same half of the room will de facto be zero for that reason.

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    $\begingroup$ If the two molecules are N and O there are 4 cases (NO/-, N/O, O/N, -NO). These are just tags btw, not Nitrogen and Oxygen). Likewise 3 cases for 8, 4 for 16, etc. The probability will get smaller even more quickly than you have suggested. $\endgroup$
    – Peter
    Commented Mar 31, 2023 at 12:19
  • $\begingroup$ You are right, thanks Peter. 1 molecule: probability is 2/2. Two molecules probability is 2/4. 3 molecules probability is 2/8. N molecules probability is 2/(2^N). $\endgroup$
    – Hanno
    Commented Mar 31, 2023 at 12:57
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It is impossible for all the air to spontaneously move and occupy just one half of the room. If you started with the state of having all the molecules in just one half and perfect vacuum in the other half, all the air would spontaneously expand and fill the whole room. The odds of having this process go in reverse, without any external input, are nil, as that would break the 2nd law of thermofynamics. What you might as well bet on is having dilute combustion fumes spontaneously concentrating themselves around the ashes and then combining to reform oxygen and the original fuel material that has been burnt down.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Apr 1, 2023 at 19:27

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