2
$\begingroup$

I am currently thinking about the irreversibility paradox. I am not working in this area and my question is certainly not original but I couldn't see it stated in that form yet.

I can't grasp how are the trajectories of an Hamiltonian dynamic are organised in phase space. Consider a system such as an ideal gas. Here are some fact that I believe are correct, but feel contradictory.

  1. reversing velocities on the phase space preserves the Liouville measure

  2. for each trajectory, reversing velocities leads to another trajectory. On one of them the entropy is increasing while on the other one, the entorpy is decreasing. Hence both categories occupy the same volume of the phase space.

  3. the Hamiltonian dynamic preserves the Liouville measure, so when time flows, the trajectories do not get concentrated in a tiny portion of the phase space.

  4. if we follow a trajectory that decreases entropy, the tiniest deviation from it will take us back to a trajectory that increases entropy.

How can the 4th point hold if the previous ones are correct? Shouldn't we expect that trajectories that decrease entropy are surrounded by an equal amount of trajectories that increase and decrease entropy?

Improve this question
$\endgroup$
12
  • $\begingroup$ There is no paradox. In reality there are no systems with finite phase space. All real physical systems are open. We are simply approximating almost closed systems with Hamiltonians with finite phase space. It's that approximation that causes the trouble and that leads to the well known but practically useless Poincare recurrence theorem. $\endgroup$
    – FlatterMann
    Commented Jan 17 at 15:12
  • $\begingroup$ @FlatterMann you should write a detailed and full answer elaborating the details, the subject deserves it. $\endgroup$
    – hyportnex
    Commented Jan 17 at 15:20
  • $\begingroup$ @hyportnex At most I could point out in more detail that the OP has a number of conceptual problems. Hamiltonian systems don't have an entropy. Entropy change is defined as reversible heat flow over temperature. This requires that we have contact with a thermal bath, which a Hamiltonian system doesn't have.. and if we make a such a contact, then a Hamiltonian systems stops behaving like a Hamiltonian system completely because we have introduced friction. OK, maybe you are right. Somebody should write an answer explaining the conceptual problem. I am sue somebody already has. $\endgroup$
    – FlatterMann
    Commented Jan 17 at 15:33
  • $\begingroup$ @FlatterMann don't be so sure; just say yourself "I am somebody" and do it! $\endgroup$
    – hyportnex
    Commented Jan 17 at 15:45
  • 1
    $\begingroup$ There is no paradox. The mechanisms responsible for the emergence of irreversible macroscopic behavior are well understood since Boltzmann's time. $\endgroup$
    – Yvan Velenik
    Commented Jan 17 at 20:38

1 Answer 1

Reset to default
3
$\begingroup$

Before answering your question, let me just say what I mean by the entropy of an isolated Hamiltonian system. You first choose your favorite set of macrovariables, those that will be used to define the relevant notion of macrostate. You then partition phase space into subsets according to the values taken by these variables (say, up to some given precision). Then, to a given microstate $\omega$, you associate an entropy given by the logarithm of the measure of the subset it belongs to.

For more on this, see, for instance, the paper Macroscopic laws, microscopic dynamics, time's arrow and Boltzmann's entropy, J. Lebowitz, Physica A 194 (1993) 1-27, or this more recent one by the same author.

Now, let us turn to your question.

You are confusing micro- and macrostates. Suppose that the gas starts at time $0$ with all particles in the left half of the box. This corresponds to a nonequilibrium macrostate (which can be obtained by compressing the gas and inserting a removable wall to keep it there; and removing the wall at time $0$). Let us call $A$ the corresponding subset of phase space.

At positive times, the system evolves, the molecules distributing themselves in the whole box in an essentially homogeneous way, with overwhelming probability. If $A_t$ denotes the image of $A$ at time $t$, then when $t\gg 1$, the vast majority of points in $A_t$ belong to the equilibrium macrostate $B$.

Of course, the Liouville measure of $A_t$ is equal to the Liouville measure of the original set $A$. However, the measure of the set $A_t$ is ridiculously small compared to the measure of the equilibrium macrostate $B$: the points in $A_t$ comprise only a minuscule fraction of the equilibrium macrostate $B$, namely those points corresponding to microstates in which all particles were in the left half of the box at time $0$.

So, if you were to start from any point in $A_t$ and reverse all velocities, the system would return to its initial state at time $0$, displaying a decrease of entropy along the trajectory. However, if you sample a random point in the equilibrium macrostate $B$, then it will almost certainly not belong to $A_t$ and would remain at equilibrium.

So, such Hamiltonian systems are reversible in the mechanical sense: upon reversing velocities, the system follows the same trajectory in reverse. However, it is not reversible in the thermodynamic sense: taking a generic point in the equilibrium macrostate will not lead to the same evolution as selecting a point in the extremely pathological subset $A_t$.

Above, I have assumed that one has waited for the system to reach equilibrium, but the same discussion also applies if you look at what happens at a time $t$ before equilibrium has been reached: the system will find itself in another macrostate, with Liouville measure much larger than that of $A_t$. Sampling a generic point in this macrostate and reversing its velocities will lead to trajectories along which entropy increases (until equilibrium is reached), even though entropy would decrease if you do that for points belonging to $A_t$.

Improve this answer
$\endgroup$
7
  • 2
    $\begingroup$ +1 A nice discussion about these matters is also given in Jaynes' "The second law as physical fact and as human inference", in the section "What does 'reversibility' mean?". In particular, there he emphasizes the connection to experiments. $\endgroup$
    – Tobias Fünke
    Commented Jan 17 at 17:14
  • $\begingroup$ @Yvan Velenik, thank you very much for your answer. I need to meditate on it. I should probably think longer before writing but here is what comes to my mind. You say :'However, if you sample a random point in the equilibrium macrostate 𝐵, then it will almost certainly not belong to $A_t$ and would remain at equilibrium.', but how do we know the part 'and would remain at equilibrium.' ? $\endgroup$
    – Chevallier
    Commented Jan 17 at 19:50
  • $\begingroup$ Because the equilibrium macrostate occupies almost all of the phase space, so that a generic trajectory has little choice but to stay there (or go there if it starts out of equilibrium). The papers I linked probably go into quantitative estimates (I read them very long ago, so I don't remember). $\endgroup$
    – Yvan Velenik
    Commented Jan 17 at 20:03
  • $\begingroup$ @TobiasFünke Hamiltonian systems don't obey the second law. That the second law wins is because Hamiltonian systems are a poor poor approximation in the infinite time limit. Or, to put it less mildly, the infinite time limit is simply unphysical, even in most trivial cases. Newtonian gravity, for instance, is unstable for all but the most simple examples. Gravitating n-body systems evaporate long before they explore all of their phase space. $\endgroup$
    – FlatterMann
    Commented Jan 17 at 20:42
  • 1
    $\begingroup$ @YvanVelenik, thanks again, I believe that I get it now. Essentially in my reasoning I didn't to take into account the fact that high entropy values correspond to larger volumes in the phase space. $\endgroup$
    – Chevallier
    Commented Jan 17 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.