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Suppose there are N non-interacting classical particles in a box, so their state can be described by the $\{\mathbf{x}_i(t), \mathbf{p}_i(t) \}$. If the particles are initially at the left of the box, they can eventually occupy the whole box according to the Newtons law. In this case, we call the gas expand and this process is irreversible.

Nevertheless, Newtons laws also have time-reversal symmetry, so we should be able to construct an initial condition such that particles occupy the whole box (i.e. not all particles in the left chamber) will all move to the left chamber. Here are the simple questions:

  1. How to select the initial condition $\{\mathbf{x}_i(0), \mathbf{p}_i(0) \}$ if you can solve the set of equation of motion.

  2. Why the gas expansion is irreversible even though you can select the initial condition above.

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4 Answers 4

up vote 5 down vote accepted

1/ OK, let's start from an initial condition where all the particles are made to fit a tiny little corner of the room and their initial velocities are chosen randomly, according to a Maxwell-Boltzmann distribution for instance. As we let the system evolve, the gas will expand, that is true because it corresponds to the behaviour the Maxwell-Boltzmann predicts, that of an ideal gas near equilibrium and therefore the gas will try to fill the room.

Now, suppose at some moment we stop the evolution of the system, for instance after the gas has filled the entire room. Now, if we reverse all the velocities of the particles but do not change their positions, it is clear that by construction, we have found a solution that will evolve backwards to the corner where the gas was originally. So, it is indeed possible to find an initial configuration and in fact several initial configurations that satisfy your requirement. (But it takes a Maxwell Demon to realize them in practice.)

2/ Now, why doesn't that invalidate the macroscopic irreversibility of the system? Well, we should look at how many microstates corresponding to the macrostate "box entirely filled" do go back to the corner. Say the box has volume $V$. Now, assuming we have $N$ particles, the total number of microscopic configurations compatible with the box filled will be proportional to $V^N$. (I am omitting the velocities in this analysis, they will only make the analysis less simple to follow.)

How many of these configurations go back to the corner? Well, to compute that, remember that we can easily construct the solutions which go back to the corner by taking the configurations that start from the corner and expand to fill the entire room and then reverse the velocities. So, let's say we call the volume of the corner $W$. So the amount of trajectories that go back to the corner will be proportional to $W^N$. So, the fraction of trajectories that go back to the corner among the trajectories that fill the entire volume will be $W^N/V^N$. Since $W/V<1$ and $N$ is a large number, of the order of the Avogadro number, you can see why you never actually observe these trajectories in real life.

But, if you make your system small enough, less than 10 particles, and your corner is a half of the room, then the probability is about $1/2^{10} \approx 0.001$, I'd say it is worth the wait. ;p

OK, I wanted to add something to this explanation. The apparant paradox that is proposed in the OP is originally attributed to Loschmidt. The paradox Marek is talking about which is related to the ergodicity of the system is due to Zermelo, they are different paradoxes and require different answers.

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The argument for the second question isn't quite correct because you left time out. It's true that there are $W^N$ initial conditions starting in the corner but these will give you lot more configurations for different durations of evolution. Actually, complete opposite is true. The generic configuration in the box is ergodic (the strongly non-ergodic portion of them is negligible). Because the phase space is compact it's then clear that any configuration will visit the corner sooner or later. The reason why we don't observe it is that it's actually later than sooner :-) –  Marek Dec 11 '10 at 19:23
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Time is not important, only the relative sizes of the phase space volumes is important. I concentrated on the configurational part because that's the most easily explained and suffices to drive the point home. Second, I'm well aware that ergodicity of the system implies that the system will visit the corner again sooner or later (I implied it at the very end of the second part.) I could have added an extra computation of how long exactly it would take for a typical configuration to get in the vicinity of the corner state. But that is not the crux of the argument. –  Raskolnikov Dec 11 '10 at 20:10
    
It is not essential to the argument because I can start in a special state like the one constructed in 1, and then the time to reach the corner will be very small. So the real point is in explaining why such trajectories are rare. –  Raskolnikov Dec 11 '10 at 20:12
    
@Raskolnikov: all right then. I think I'll agree that just to give an impression of where irreversibility enters the picture it's enough to point out the probability nature of the problem. –  Marek Dec 11 '10 at 21:22
    
I added an extra comment in relation to your objection. Can't find a wiki page for the Zermelo paradox, except in french. I'll look for another page. –  Raskolnikov Dec 11 '10 at 21:25

The short answer is that it is not technically irreversible. If you wait some huge amount of time, the gas will de-expand, as per the Poincaré recurrence theorem. The problem with this is that the amount of time for a system to evolve from a low-entropy configuration to a high entropy configuration is very small, while you would be waiting several ages of the universe for the process to reverse itself.

Also note how the recent Maxwell Demon experiment has exploited the subatomic reversibility of macroscopically irreversible processes to generate useable work. (Ignore the overstatements about 'converting information to energy').

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If you take the thermodynamics limit, the relative probability of low-entropy state tends to zero. And it should called irreversible. –  hwlau Dec 13 '10 at 9:05
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@hwlau: but the thermodynamic limit is an abstraction. Any system (aside from perhaps the whole universe) has a finite number of particles. Certainly for a closed system involving a finite number of gas particles in a finite volume, the Poincaré theorem applies, no? –  Jerry Schirmer Dec 15 '10 at 1:03

Trivially, we could set the initial conditions so that particles all have velocities pointing directly towards one corner with speeds proportional to their distances from the corner. I don't think this is what you were looking for, though, since it doesn't look like the typical configurations of state space you'd expect.

One thing to keep in mind (in addition to Jerry and Rasklonikov's answers) is that any realistic macroscopic system is not in going to be perfectly isolated. Even if you could reach in and adjust the positions and velocities of every particle, the particles would interact with the edges of the box somewhat randomly because the box is presumably sitting in some sort of heat bath. These unpredictable interactions would mess up any carefully-calculated dance of the particles. Even if you set up the initial conditions correctly, in practice you still wouldn't get the particles to go back to the corner.

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If the particles move according to Newtons laws then the system is reversible. Newton laws conserve entropy. To add irreversibility to you need to modify Newton laws, e.g. adding a irreversible term arising from collisions with the box and accounting the dissipation of the box with its surrounds. This irreversible term will generate entropy for any given initial condition $\{\mathbf{x}_i(0), \mathbf{p}_i(0)\}$.

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