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I am confused about phase-volume contraction in dissipative systems. Please help me catch the flaw in my understanding. From a macroscopic point of view I understand that a dynamic system tends to go to an equilibrium state or a limit cycle if its not chaotic. But now trying to understand it in phase space:

1) Consider a system in complete thermodynamic equilibrium. It is a non-dissipative system (it cannot dissipate anymore). Therefore it could be in any microstate allowed by constraints (e.g., conserved quantities) and by Liouville theorem the probability density does not change. Therefore isn't the phase space volume accessible to this system the whole phase space (allowed by constraints)?

2) Now consider a non-equilibrium system. Its location in the space has some peaked probability distribution based on the initial conditions, i.e., it occupies a small space volume, and when it approaches equilibrium the probability of finding it becomes uniform and spreads over the complete phase space. Doesn't that mean the phase volume has expanded?

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If the dissipative system has a thermodynamic equilibrium state, then in general, the set of microscopic initial conditions is larger than the set of microscopic states in the thermodynamic equilibrium state. Imagine a melting ice with a final state of water at 10°C. The initial state (some microscopic configuration corresponding to ice, or in general a set of microscopic configurations corresponding to ice) will be (almost) never visited anymore in the final equilibrium state. I say almost because mathematically it can happen, but the probability of that to happen due to statistical fluctuations is negligible in practice.

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  • $\begingroup$ Do you mean 0 degree C? I mean are we talking about an isolated system of melting ice or one that gains energy to go to 10 C. Since my posing this question a few months ago I have realized that the term "dissipative" is sort of used in two ways: 1) The system loses energy... in that case I now understand that the phase space will shrink (lower energy fewer states) 2) Constant energy but equilibration with entropy generation. I was referring to this definition in my question. In such a case, e.g., ice melting to water at 0 C constant energy $\endgroup$
    – Sankaran
    Commented Apr 9, 2013 at 18:04
  • $\begingroup$ I still think that the phase volume must be increase. Of course I am going to assume ergodicity and coarse-graining to stay away from the irreversibility paradox $\endgroup$
    – Sankaran
    Commented Apr 9, 2013 at 18:07
  • $\begingroup$ In a sense, the more you ignore about the microscopic state of the system, the less you will be able to extract energy from it. From one point of view, dissipation is the loose of information (increase of entropy) about which is the microscopic state of the system. Notice however that, when a system cools down, heat is being dissipated but its entropy is reduced! $\endgroup$
    – Juan I. Perotti
    Commented Dec 8, 2016 at 14:39
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When studying dynamical systems you consider a low-dimensional phase space that only includes macroscopic variables. In this phase space dissipation does indeed cause volume contraction - but only because the energy that has been lost has been transferred to the far bigger phase space including all the microscopic variables. In this space energy must be conserved, and volume must be constant (if viewed in fine-grained fashion) or increase (if viewed in coarse-grained fashion, corresponding to the Second Law).

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