An ergodic dynamical system $(\Omega,\phi^t,\mu)$ is such that the time average $\bar{f}$ of every function $f\in L_1(\Omega,\mu)$ equal the space average $\langle f \rangle_\mu$, i.e. the system densely cover all the phase space ($\mu$-almost everywhere). Another equal condition of ergodicity is that the only invariant sets ($\phi^t(B)=B$) are the trivial ones (no way of partitioning the phase space), or looking at the $L_1$ space, the only invariant functions ($f\circ\phi^t=f$) are the constant functions. Moreover, we have a stronger property than ergodicity, namely mixing that implies the former. A system is said to be mixing if $$\mu(\phi^{-t}(A)\cap B)\rightarrow\mu(A)\mu(B),\qquad\text{as }t\rightarrow\infty.$$ At the end we have the recurrence, i.e. the system pass through all the points of the phase space infinitely many times. My question arise because in my mind for a system that approaches the equilibrium, there exist a time $T>0$ such that for all $t>T$ the system will spend its future time in a smaller region of the phase space, i.e. there exist a partition of $\Omega$ and then the system cannot be ergodic.
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1$\begingroup$ I think one often considers the phase space given that the system is in equilibrium and tries to see if this equilibrium system is ergodic, or am I missing something here? $\endgroup$– DanuCommented Aug 29, 2014 at 14:02
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$\begingroup$ Ok, but for definition an ergodic system cannot evolves towards equilibrium. Or i am missing something? $\endgroup$– yngablCommented Aug 29, 2014 at 14:09
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$\begingroup$ I think one can technically say that an ergodic system will not be capable of forever maintaining equilibrium at all instances, but I think this may be irrelevant because of averaging (I really don't know!)... Also, this question is related, please read the last section of my answer. $\endgroup$– DanuCommented Aug 29, 2014 at 14:11
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2$\begingroup$ I am not sure I understand your concern. I assume that your dynamical system is measure-preserving (as is the relevant setting for application to stat. mech.). In that case, it is clear that the system does not "spend its future time in a smaller region of the phase space", at least not as measured with $\mu$. Maybe you'd enjoy reading this very nice paper, which although not discussing specifically the issue you have, might well clarify things for you. $\endgroup$– Yvan VelenikCommented Aug 29, 2014 at 14:29
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1$\begingroup$ What is your definition of equilibrium? In physics an ergodic system is only "in equilibrium" during most of its dynamic evolution. Indeed, the recurrence theorem guarantees that all systems that start from a very special state (i.e. far from equilibrium) will, eventually, return arbitrarily close to that special state, again. $\endgroup$– CuriousOneCommented Aug 30, 2014 at 5:27
2 Answers
There are two different concepts of "state". A micro-state is a point in phase space. If we know everything there is to be knowed about the state of the system, we would know which point in phase space it occupies (at some point of time in question). This is the concept relevant to ergodicity and those trajectories you talk about and to recurrence and so on.
The other concept, relevant to "equilibrium" in thermodynamics, is when we know much less. Maybe we only know the energy, temperature, pressure, and volume, for example. Many points in phase space will satisfy these conditions, and we do not know precisely which one is occupied by the system. But suppose we know the probability that the system is in each one of these states. I.e., we know a probability distribution. That is called a "macrostate". Now some macrostates are equilibrium states and others are not. The definition is that the probability distribution does not change over time. (The definition is so totally not that the point does not move over time. Each possible point moves to another point, along a trajectory.) So there is no obstacle to an ergodic system's having an equilibrium state, i.e., an equilibrium macro-state. For example, Liouville measure restricted to a surface of constant energy is an equilibrium macrostate, called the micro-canonical distribution.
Maybe i found the source of my concern, the whole space phase have invariant measure so it's measure never change during the evolution and so i think i have a misleading idea of equilibrium approach, indeed it cannot be ''the system will spend it's future time in a set $A\subset\Omega$''.
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$\begingroup$ The whole phase space has an invariant measure. But that measure is not a probability measure, so this measure does not constitute a "macro-state". But it can be restricted to a set of fixed energy, and then it does. Or one can multiply Liouville measure by an exponential decay in the energy, so its total integral now converges. Since energy is conserved, this also yields an invariant measure, that is a probability measure, and is an equilibrium state. $\endgroup$ Commented Dec 5, 2015 at 23:00