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Imagine a ball bouncing in a box for a long time. We know, there is a certain path it can go to bounce off infinitely (see the image). If it gets to this state, it will never be able to get back again.

Now, I suppose, it is as probable for the ball to reach one point in the system (with a random starting position), as to reach any else. And I suppose, there is the same probability for it to come from any direction. With these assumptions, the system unstopably tends to get to the regular state.

I have heard in a pop-sci show, that we could theoretically perfectly detect the past from the data we have today (the locations of all the particles with their characteristics). But we can't really find out how had it all begun if we see a system like on the animation below.

So will the information get lost after infinite time? bouncing ball in a box

  • Sorry for this inappropriate image, there is a bigger probability for the ball to reach the middles of the lines than the corners, let's ignore the bouncing of the walls)
  • Note that even though 0.0...1 equals 0, the stable point does exist as any else
  • Let's assume the ball gets all it's energy back...I'm interested in something a bit else
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  • $\begingroup$ Nice GIF! Like it. $\endgroup$ – Omar Nagib May 10 '16 at 13:51
  • $\begingroup$ Just to be sure...is that a simulation or a cartoon? $\endgroup$ – gatsu May 10 '16 at 14:34
  • $\begingroup$ I have wondered about this also. Orbiting bodies are classified as elliptical but if there are tidal forces involved the orbit will tend to go perfectly circular if nothing else interferes. As Orion says "after a system finds one of the common and easy to find states, it'll be very unlikely for it to arrive back to the very specific (low entropy) state." What I don't understand is why is the elliptic orbit a lower entropy than the perfectly circular orbit? $\endgroup$ – Bill Alsept May 10 '16 at 15:23
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    $\begingroup$ Possible duplicate of Difference between irreversible and entropy? $\endgroup$ – honeste_vivere May 22 '16 at 19:09
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If you can define a perfectly closed system which eventually arrives back at its starting state, that system is reversible. A reversible system is not irreversible, and can never "become" irreversable without changing the system.

However, in the real world, there are limits which come into play. The first is that the systems are never actually isolated like that. There are always interactions with the world outside of the system, even if that is nothing more than emitting energy into a cold environment. This is also important because a truly reversible system cannot do work. See Laplace's Daemon for how that works out.

The second issue comes with the challenge of measuring the state of such a reversible system. If, at any point, you need to describe a variable in the state of the system with a random variable, information about the system has been lost. This occurs with any real physical measurement (we have no ways of perfectly measuring anything to an infinite degree of accuracy, nor could we write down the result of such a measurement). It also occurs in the interpretations of quantum mechanics.

If said random variable is a state in a chaotic system, the entire state of the system quickly becomes unknown, "erasing" the information that would be needed to return to that state, unless you could somehow measure the exact value of said random variable. It doesn't matter if that random variable is the product of measurement error or quantum mechanics or any other reason.

We actually have no way of proving the laws of thermodynamics. It is entirely possible that, outside of our corner of the universe, there is a reflective construct which was perfectly constructed to capture the state of every value "forgotten" by us, and feed it right back to us to allow us to undo the big bang. However, we have little reason to believe this is actually the case.

This gets more interesting when the measuring devices is part of the system, as happens in quantum mechanics with interactions. In these systems, the measuring device's behavior has to be modeled as part of the system, and laws like Godel's Incompleteness Theorem make it very hard for the system to prove that a particular measurement/interaction is reversible.

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Let's take the 1D equivalent of your problem for simplicity: a particle bouncing back and forth along a segment, reversing its velocity every time it hits the boundaries of the segment.

If we know perfectly the initial state of the particle, i.e. its position and velocity at time $0$, $(x(0),v(0))$, we will know exactly what the motion of the particle will be. In the absence of external forces, its equation of motion between a collision and the other will be (let's take positive velocity for simplicity)

$$x(t)=x(0) + v(0) t$$

We will then know the exact position of the particle at any time, an will be able in principle to revert its state to get back to the state at $t=0$.

But what if we don't know the exact value of, say, the initial velocity? Let's introduce an error $\delta v$ on the initial velocity. We will then have

$$x(t)=x(0)+ (v(0)\pm \delta v) t = x(0) + v(0)t \pm \delta v \ t$$

As you can see, the error gets larger and larger at times passes (it increases linearly with time). After some time $t$, we will only be able to say that the particle is in some segment of length $2 \delta v \ t$. At a certain time $t^*$, this number will become equal to the length $L$ of the whole segment where the motion is taking place:

$$2 \delta v \ t^* = L \rightarrow t^* = \frac L {2 \delta v}$$

We will then be only able to say that the particle is "somewhere inside the box", but won't be able to tell where: every position is equiprobable! (i.e. the probability distribution function of the variable $x$ is uniform).

Since it is impossible to know exactly both $x(0)$ and $v(0)$ (because our instruments are not good enough, or, if the instruments are perfect, because of the uncertainty principle), the situation in real life will always be the latter: after some time you will lose all the information about the position of the particle!

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  • $\begingroup$ Ok, so there are many variations that lead to one variation: The ball is somewhere inside the box, aren't there? So, indeed, it is possible for the system to become irreversible from this point, isn't it? $\endgroup$ – foggy Oct 1 '16 at 12:34
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If you assume that the ball is in the state you describe, and will continue to run through the same path over and over again, then you assume, that your system is deterministic, that means, from one position and momentum of the ball, you will be able to calculate its path for all the times that will come after that. In your special case: If the time goes on, and the path of the ball never changes, then you can also turn the time back, and the path of the ball never changes: You can see some kind of periodicity in the path of the ball, after a Time $T$ it is at the same point again, and the trajectorie repeats. But that means (because the laws of physics allow you to calculate the trajectory for everytime before and everytime after) that you can also go "back in Time" by the Time $T$ and the ball will be at the same spot, and so on and so on.

This should show you that there is no possibility for the ball to randomly reach the state you described. If the ball is in the state, then it has

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Every deterministic system with chaotic trajectories is losing information on account of infinitely small deviations leading to completely different results. In your case, if the ball eventually settles into a stable orbit after some time, no matter where it started, then computing backwards would amplify any uncertainty so that after a few cycles, you know nothing about the past.

This is precisely the second law of thermodynamics. Theoretically, knowing the state of the system PERFECTLY, you could compute the past and the future just following physical laws. However, every time the system finds one of the common and easy to find states, it'll be very unlikely for it to arrive back to the very specific (low entropy) state. And trying to predict where it came from is the same as trying to put it back into that state. You miss by a femtometer in position of one of the particles, and instead of putting all the water back into the glass, you just spill it a different way.

People talk about entropy as "measure of disorder", and chaos theory studies how deterministic laws can lead to extreme sensitivity to initial conditions and thus to a system that will "lose information" (increase its entropy) when you leave it running. At the end, it's all just "it's easier to break things than to mend them". It's a law more general than the universe.

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  • $\begingroup$ Thank you, I have just one question outside of this topic: Could the system work in the Newtonian universe (If no quantum effects were discovered)? Or is there some fundamental problem, that allow such system work only in the "quantum" universe? $\endgroup$ – foggy May 10 '16 at 16:46
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    $\begingroup$ Statistical physics can be (and was) done also for classical systems, and as it's a purely mathematical concept, the same formalism can be applied on any sufficiently complex system. The only "catch" is that classical systems have infinite degrees of freedom, so the absolute scale of the entropy isn't defined (there's no smallest chunk of phase space). There are many arguments that show that the world as we know it wouldn't be stable, but statistical mechanics does work even for hard marbles (colloidal systems, for instance). $\endgroup$ – orion May 11 '16 at 6:35

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