How can bounded orbits diverge arbitrarily far at late times?

Question

While there are many definitions of chaotic dynamics, many commonly used ones require a positive Lyapunov exponent, which is defined as $$\lambda := \lim_{t \to \infty} \lim_{\delta{\bf Z}_0 \to {\bf 0}} \frac{1}{t} \ln \left( \frac{|\delta {\bf Z}(t)|}{|\delta {\bf Z}_0|} \right).$$ Here there is a dynamical map with an implied initial condition ${\bf Z}_0$ in phase space, $\delta {\bf Z}_0$ represents an initial infinitesimal shift away from that initial condition, and $\delta {\bf Z}(t)$ represents the distance at time $t$ between the time-evolved state that began at initial condition ${\bf Z}_0$ and the time-evolved state that began at initial condition ${\bf Z}_0 + \delta {\bf Z}_0$.

But - while definitions again vary - chaotic dynamics are often defined to require bounded trajectories; see here and here. This requirement is meant to exclude trivial situations like systems that just exponentially accelerate away to infinity in a straight line. Even if we don't require boundedness in the definition, clearly many chaotic systems (like the double pendulum) have bounded dynamics.

But how are bounded trajectories compatible with a positive Lyapunov exponent? If all trajectories are bounded, then isn't $\delta {\bf Z}(t)$ also bounded over all times, and therefore the logarithm is as well - in which case the expression in the limit must fall off at least as fast as $1/t$ at late $t$?

I actually have a more basic confusion about the definition of Lyapunov exponents, which involves the very definition of $\delta {\bf Z}$. As far as I know, Hamiltonian phase space does not in general have a natural distance function (nor a vector-space structure, but I assume that the vector-like notation "$\delta {\bf Z}$" is just an abuse of notation). I won't include that question here, because I think it's basically a duplicate of The natural metric of a phase space and the Lyapunov exponent. I don't fully understand the answer to that question, but I figure that the answer to that question is probably important for answering this one.

Answer 1

The key to this is that $|\delta Z|$ is infinitesimal. We're taking it as the limit approaches zero. In this sense, bounding does not cause any odd limitations like would occur with finite perturbations. An infinitesimal shift in $|\delta Z_0|$ which results in a $\delta Z(t) = 1$ would result in an infinite $\frac{|\delta Z(t)|}{|\delta Z_0|}$, so boundedness does not automatically mean this exponent is nonpositive.

Answer 2

As Cort's answer hints and you realize in your comments there, the order of the limits is important: you first take an arbitrarily close orbit and then let them evolve.

I think a numerical perspective sheds some light on the question: When calculating numerical Lyapunov exponents using methods that approximate the infinitesimals, we have to regularly normalize the distance (actually orthonormalize it, due to prototypical phase-space stretching) for the calculation to work.

As for $\delta {\bf Z}$, it lives in a flat space tangent to the phase space manifold — which doesn't mean you can ignore curvature, which in some systems is actually the origin of chaos — but does sidestep the question of a distance function for the phase space.

Answer 3

I now understand the answer. The key is the subtlety in the order of limits: somewhat unusually for physics formulas involving double limits, the order of limits is critical, and the formula no longer works if we interchange them.

I would actually frame my answer the exact opposite way as stafusa did (although we agree on the solution). I think it makes more sense to consider the two limits "from the outside in": you need to first consider a time difference $t$ of interest, and then take the limit $\delta {\bf Z}_0$ down to an infinitesimal size that depends on $t$. So the inner limit in some sense depends (exponentially) on $t$.

This dependence on $t$ is what makes the limit $\delta {\bf Z}_0 \to 0$ a little conceptually different from most infinitestimals that come up in physics. Usually, there are some local quantities - typically second derivatives that we can think of as representing some kind of curvature - that set the scale for how small we need to zoom in in order to consider a change as "infinitesimal" (i.e. we can think of the function as being locally linear). For example, for a 2D surface, the local radius of curvature (or Ricci scalar) sets the scale (far) below which we can ignore higher-order effects.

But by contrast, the limit $\delta {\bf Z}_0 \to 0$ in the definition of the Lyapunov exponent is in some sense nonlocal, because the size that you need to shrink $\delta {\bf Z}_0$ down to depends on the time difference $t$ that you plan to consider evolving out to. There is no time-independent length scale at which we can consider $\delta {\bf Z}_0$ to be "infinitesimal"; that length scale is not bounded below.