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I have seen several authors mentioning that KAM theory contradicts the Ergodic hypothesis. Unfortunately, the authors do not elaborate on this. I have some background in KAM theory but very little in Ergodic theory. Could somebody explain the argument?

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"KAM Theory contradicting the Ergodic hypothesis" really just means that in a typical nearly integrable Hamiltonian system with finitely many degrees of freedom - one of those systems for which the KAM Theorem holds, there can be no ergodicity. So KAM doesn't hold on ergodic systems, and the systems for which we can use KAM theory are not ergodic.

This seems as if it would be devastating to basically all of thermodynamics and statistical mechanics, as a number of topics studied in these theores assume that a nearly integrable hamiltonian system satisfies ergodicity if limited to potential energy hypersurfaces. Ergodicity however seems to be a reasonable assumption for very large n, though despite there not being a possibility of "true" ergodicity.

So why are these two domains disjoint? Well, KAM states that in a hamiltonial system with n degrees of freedom, there are typically n quasiperiodic tori with positive Liouville measure in every potential energy hypersurface. Since these tori are distinct, and they have positive measure in a hypersurface, this hypersurface can be decomposed into disjoint sets that are invariant and positively measured, so while the ergodic hypothesis would claim that a path spends time in this torus proportional to it's measure, it is quasiperiodic in some torus and therefore the time it spends in that torus is not proportional to the torus's measure.

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I don't think the statement "KAM contradicts the Ergodic theory hypothesis" is correct or even mathematically meaningful. The Ergodic hypothesis (or Birkhoff's Ergodic theorem) loosely states that for ergodic systems (ergodic systems is rigorously defined, see wikipedia), time average of the iterates of a randomly chosen point will equal the space average of iterates all over the set where the ergodic measure is non-zero. Intuitively, it says that a typical trajectory will sample all points of the phase space (in the limit of infinite time), and the fraction of time-spent in each neighborhood will be same as (ergodic) measure of that neighborhood.

The KAM theorem is used in the context of NON-ERGODIC systems, and especially, nearly integrable systems with a small perturbation away from complete integrability. It states that under suitable hypothesis, for a large set of initial conditions, the trajectory in a nearly-integrable system will still have quasiperiodic behavior, i.e. it will be confined to a measure-zero subset of the full-phase space.

In some sense, I can see that why people will make the false statement you mentioned. But the point is that the systems to which KAM is typically applicable (nearly integrable systems) are very different from where the Ergodic hypothesis applies (ergodic systems).

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You've basically just said that it doesn't make sense to ask why nearly integrable systems aren't ergodic because nearly integrable systems are very different from ergodic ones. –  Sam DeHority Feb 1 '13 at 0:10
    
Yes!Once you define these two types of systems, it becomes clear why it doesn't make sense to apply Ergodic hypothesis to system where KAM type methods work. –  nonlinearism Feb 1 '13 at 0:17
    
There's nothing to say that the ergodic hypothesis isn't true for systems that aren't ergodic though. And that's the application of KAM that he's asking for. –  Sam DeHority Feb 1 '13 at 0:19
    
So he wants to know why nearly integrable systems don't display ergodicity. –  Sam DeHority Feb 1 '13 at 0:20

Given an open subset $B$ of $\mathbb R^n$, let us consider on $B^n\times\mathbb T^n$ the symplectic form $$\Omega=\sum_{i=1}^ndq_i\wedge dp_i,$$ where $p=(p_1,\ldots,p_n)$ are coordinates on $B$, and $q=(q_1,\ldots,q_n)$ are the angular coordinates on $\mathbb T^n$.

Let us assign an Hamiltonian function $H(q,p)=H_0(p)+\varepsilon H_1(q,p)$, satisfying the non-degeneration condition $$\det\left(\frac{\partial^2 H_0}{\partial q\partial q}\right)\neq 0$$ The associated Hamilton equations are $$\dot p=-\varepsilon\frac{\partial H_1}{\partial q},\quad \dot q=\omega_0(p)+\varepsilon\frac{\partial H_1}{\partial p}$$ where $\omega_0(p):=\dfrac{\partial H_0}{\partial p}.$

  1. The un-perturbated case ($\varepsilon=0$) the Hamiltonian flow, on the invariant $n$-toruses $p=\mathrm{constant}$, is quasi-periodic with quasi-frequencies $\omega_0(p)$. $$(t,(q,p))\mapsto(q+t\omega_0(q),p)\tag{*}$$ For any $p_0$ in a certain open everywhere dense subset $B_0$ of $B$, the components of $\omega(p_0)$ are $\mathbb Q$-linear indipendent, so each integral curve is dense on the invariant $n$-torus $p=p_0$, which is called non-resonant.

  2. The pertubated case ($\varepsilon<<1$) Kolmogorov showed that, in the Lebesgue-measure theoretic sense, almost all the non-resonant $n$-toruses with the flow(*) are only slightly deformed.

Therefore the Hamiltonian flow on the energy hyperurface $H=\mathrm{constant}$ is not ergodic, while, between its ergodic components, there are almost all the invariant $n$-toruses.

All this is just a rewording of section 21 in Arnold, Avez, "Problemes Ergodiques De La Mecanique Classique".

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Ergodic hypothesis more or less states the the Hamiltonian flow restricted to each energy surface is ergodic for the Liouville measure.

Now if you take an integrable Hamiltonian system with a 2n dimensional phase space, action angle coordinates theorem tells you that there exists n first integral of motions $I_i$ (such that dI_i are independent) and some angle coordintes $\theta_i$ defined on $S^1$ for i=1,...,n such that under the coordinate change $(x,p) \rightarrow (\theta(x,p),I(x,p))$ the Hamiltonian flow still satisfies the Hamilton-Jacobi equations for the new Hamiltonian function in these coordinates. But since $I_i$ are first integrals it means that $\frac{dI_i}{dt}=0$ (I will drop subscript i from now on) and if you look at the form of Hamiltonian equations than you get that the new Hamiltonian does not depend on $\theta$ but it depends on $I$. And so $I(t) = cnst$ and $\theta(t)= \theta(0) + I(0)t$.

Now note that one of the first integrals will be the energy say $I_1$. It is obvious that an integrable system does not satisfy the ergodic hyphothesis. This is because there is a subset of union of torii of positive measure inside $I_1 =cnst$ such that the union is invariant but has positive measure (not 1). But this is in itself not enough to say that ergodic hyphothesis will not hold for "all most all" systems since integrable systems can in theory be destroyed by small perturbations (and people in dynamical systems are interested in statements that hold generically but not for all examples). But the KAM theorem says that even if you do this perturbation, there will remain a positive measure set of invariant torii such that restricted to each Torus the hamiltonian flow is conjugate to an rationally independent translation. So you have an open set of Hamiltonian flows (dont remember what is the smoothness degree of the perturbation but lets say for simplicity in $C^{\infty}$ topology) where the ergodic hyp. fails.

In this case each torus represents a level set of the form $I=cnst$ and in particular is a subset of $I_1=cnst$. Therefore restricted to each energy surface the Hamiltonian flow is non-ergodic. Yet restricted to each $I=cnst$ the Hamiltonian flow is still ergodic since it is conjugate to rationally independent translation on the torus. Whether if this has any physical significance or not I have not understood yet. Also it is provocative to think that in action angle coordinates, constants of motion represent the momentum. Now since the new Hamiltonian $H$ only depends on $I$, then the level sets $I =cnst$ somehow represent energy surfaces in this new coordinates and somehow this seems to what Ergodic hyp. claims.

p.s: I must note that there is some finiteness problem of using Liouville measure directly. I think this makes more sense when there is a compact energy surface which you restrict to your self and then on this surface, using the Liouville measure, you can construct another invariant and finite measure that gives full measure to this energy surface (which is to no suprise called a microcanonical distribution).

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