KAM theory and the Ergodic hypothesis

Question

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?

Answer 1

"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.

Answer 2

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).

Answer 3

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".