The last circle to break

The circle map of the essays on mode-locking loses energy: its orbits settle onto attracting cycles, and that is why its plateaus exist. Its most famous relative does not. Take a rotor spinning freely, and once per turn give it a kick whose strength depends on its angle. Write $x$ for the angle as a fraction of a turn and $p$ for the momentum, and the state after each kick is

$$p' = p + \frac{k}{2\pi}\sin 2\pi x, \qquad x' = x + p'.$$

This is the standard map, written down by Boris Chirikov in 1969 and studied under that name since 1979. It preserves area — a patch of states is carried to a patch of the same area, as the laws of mechanics require — so nothing is attracted anywhere, and the question it poses is different in kind from the circle map’s. It is not where do orbits settle but how far can they wander.

At $k = 0$ the answer is not at all: the momentum never changes, each orbit goes round at its own constant speed $p$, and the square of states is filled with horizontal lines, each one a circle traced out forever. The whole subject is what happens to those circles when the kick is switched on.

Circles that survive, and circles that do not #

At $k = 0.6$ most orbits still trace curves that run right across the square: the flat lines of $k = 0$, bent into waves. They are invariant circles — curves the map carries onto themselves, on which an orbit goes round forever at a fixed average speed. Between them sit islands, chains of loops round periodic orbits, where the speed has locked to a fraction as the circle map’s does. By $k = 0.9716$ the spanning circles have almost all gone, and the islands are surrounded by a thin chaotic layer. At $k = 1.3$ the orbits that are not trapped on islands wander up and down through a sea of scattered points.

The reason spanning circles matter is topological. The map preserves area and is continuous, and a curve that runs all the way across the square divides it into a region above and a region below. An orbit that starts below can never cross — the map carries the region below onto itself — so a single spanning circle bounds the momentum of every orbit beneath it for all time. As long as one survives, no orbit can drift from low momentum to high. When the last one breaks, orbits can wander without limit, and the rotor can be kicked up to any speed.

Which circle is the last? The classical result about the survival of circles is the theorem of Andrey Kolmogorov, Vladimir Arnold and Jürgen Moser — KAM theory — from the decade after 1954. It says that for a small enough kick most circles survive, deformed but intact, and that the survivors are those whose speed is badly approximated by fractions. A circle whose speed is close to a fraction $p/q$ is close to resonance with the periodic orbits of that fraction, and the kick destroys it; a circle whose speed stays far from every fraction resists.

The number hardest to approximate #

“Far from every fraction” has a precise measure, and one number is the extreme. A number’s best approximations by fractions are the convergents of its continued fraction, the fractions that beat every smaller one, and how good they are depends on how large the continued fraction’s terms are: a large term means an unusually good approximation. The number whose continued-fraction terms are as small as possible — every one equal to one — is the golden mean,

$$\frac{\sqrt5 - 1}{2} = \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cdots}}} = 0.6180\ldots,$$

and its convergents are ratios of consecutive Fibonacci numbers, $1/2, 2/3, 3/5, 5/8, \dots$, each only as good an approximation as it is forced to be. In the sense that matters for resonance, the golden mean is the most irrational number, and John Greene conjectured in 1979 that the last spanning circle to break is the one whose speed is the golden mean.

The figure sweeps starting momenta along the map’s line of symmetry at $k = 0.9716$ and records the average advance of each orbit. Long flat stretches are islands — the widest, at an advance of one half, belongs to the orbit of period two, which a twist map must have for every fraction between its boundaries. Ragged stretches are chaotic orbits, whose “average advance” over three thousand steps is whatever they happened to do. And at one starting momentum the advance is the golden value: the orbit is on the golden circle, which at this kick still exists and is about to break.

Periodic orbits that approximate the circle #

Greene’s method for locating the breakup is the essay’s central idea, and it is indirect in a way that makes it practical. An invariant circle is an infinite object, hard to find numerically and harder to certify. A periodic orbit is finite: $q$ points that the map carries round in a cycle, advancing $P$ turns in $q$ steps. And the Fibonacci periodic orbits — speeds $2/3, 3/5, 5/8, \dots, 55/89$ — approach the golden circle as the fractions approach the golden mean.

Each periodic orbit has a residue, Greene’s measure of its stability: linearise the map round the orbit, take the trace of the resulting matrix, and set

$$R = \frac{2 - \operatorname{trace}}{4}.$$

An orbit with $0 < R < 1$ is surrounded by small loops — it is at the centre of an island — and one with $R$ outside that range is unstable, with orbits peeling away from it. Greene’s criterion is about the sequence of residues of the approximating orbits. If the golden circle exists and is smooth, the orbits nestle against it and their residues shrink to nothing. If it has broken, the orbits that used to approximate it are in chaos and their residues grow without bound. And at the exact moment of breakup, the residues neither shrink nor grow: they settle at a fixed value, about a quarter.

The figure computes each orbit — by starting on the map’s symmetry line and adjusting the momentum until the orbit closes after $q$ steps — and then its residue. At $k = 0.9$ the residues fall from $0.2$ to under $0.001$ by the orbit $55/89$: the circle exists. At $k = 0.95$ they fall more slowly. At $k = 1.02$ they climb to eighteen: the circle is gone. And at $k = 0.9716$ they sit at $0.248, 0.256, 0.254$ through the longest orbits computed — the critical value.

What a residue measures #

The residue is a compact way of reading one number off a two-by-two matrix. Linearising the map round a periodic orbit gives a matrix $M$ that says how a small displacement from the orbit is carried round it once. Because the map preserves area, $\det M = 1$, so its two eigenvalues multiply to one, and everything about it is decided by their sum, the trace.

If the trace lies strictly between $-2$ and $2$, the eigenvalues are a complex pair on the unit circle: the directions the matrix leaves alone are not real, and a small displacement is rotated round the orbit rather than stretched. The orbit is elliptic — the centre of an island of loops — and nearby orbits stay nearby. If the trace is beyond $\pm 2$, the eigenvalues are real, $\lambda$ and $1/\lambda$, one stretching and one squeezing; the orbit is hyperbolic, and nearby orbits leave. The residue $R = (2 - \operatorname{trace})/4$ is just the trace rescaled so that the elliptic range becomes $0 < R < 1$.

Periodic orbits come in pairs, one elliptic and one hyperbolic for each fraction — the elliptic one at the centre of an island chain, the hyperbolic one at the pinch between neighbouring islands — which is why the residues computed here alternate in sign: the symmetry line picks up one kind for some fractions and the other kind for others. The figures plot the size. What matters is how the size moves along the Fibonacci sequence, and the three behaviours — shrinking, holding, exploding — are three answers to whether the orbits are converging onto a smooth curve, onto a critical one, or onto nothing.

Reading off the critical strength #

The criterion turns into a number by asking, for each orbit, at what kick its residue reaches a quarter.

The crossings alternate about a limit and close in on it: $0.9739$ for $3/5$, $0.9690$ for $5/8$, then $0.9711, 0.9703, 0.9718, 0.9712, 0.9714$ and $0.9716$ for $89/144$. Greene’s own calculation, carried to orbits with hundreds of thousands of points in high precision, gave $k_c = 0.971635\ldots$, and the eight crossings here are already within a few ten-thousandths of it. That the crossings alternate is a trace of the Fibonacci structure: the approximations $F_{n-1}/F_n$ themselves alternate above and below the golden mean.

The rate of convergence is also a property of the critical point. The gaps between successive crossings shrink by roughly a constant factor each time, the way distances between period doublings shrink by Feigenbaum’s constant, and extrapolating that geometric shrinking from eight terms already lands on $0.9716$. Getting the next digits honestly needs longer orbits, and longer orbits need more precision: the orbit $89/144$ is computed here in ordinary double precision, and at the critical kick its instability multiplies round-off errors by factors that eventually swamp sixteen digits. Greene’s longest orbits needed arithmetic far beyond that, which is the practical reason the number is known to six places and not to sixty.

A rotor, a beam of particles, and the asteroids #

The standard map is a toy, and it earns its place by being the local model of every system in which a periodic kick meets a free rotation. A charged particle circulating in an accelerator is kicked once a turn by the focusing magnets; its deviation from the design orbit evolves by a map of exactly this shape, and the invariant circles are what keep the beam in the pipe. Designers of storage rings speak of the dynamic aperture, the region of starting deviations bounded by the outermost surviving circle, and particles outside it wander off in the chaotic sea within minutes.

An asteroid orbiting the Sun between Mars and Jupiter is kicked by Jupiter each time the two pass, and asteroids whose periods are close to simple fractions of Jupiter’s — three to one, five to two — sit at resonances. The Kirkwood gaps in the belt, first noticed in 1866, are at those resonances, where the circles broke long ago and chaotic wandering has carried asteroids onto orbits that cross Mars or the Earth. And the ball that stays outside the table is kept from escaping by the same kind of invariant curve, round a convex table smooth enough for KAM theory to apply. In each case the question is the one the standard map poses cleanly: which motions are fenced in by surviving circles, and which have nothing left to stop them.

The circle as it breaks #

Drawing the $89/144$ orbit as a curve shows what the circle looks like on its way out. At $k = 0.5$ it is a single smooth wave. At $k = 0.9$ it is steeper, flattened at the top and bottom where it passes near the islands of the neighbouring fractions. At $k = 0.9716$ the flat stretches have developed wrinkles — small oscillations where the circle is squeezed between islands on either side. At the critical kick the golden circle is still a continuous curve that divides the square, but it has lost the smoothness it had at small $k$, and renormalisation theory describes it as self-similar: magnify it round its most squeezed point by a fixed factor and the same wrinkled shape reappears.

The wrinkling has a precise meaning. On an invariant circle the map is a rigid rotation by the golden mean in different coordinates, and the change of coordinates is the thing that loses smoothness. At small kicks it is analytic, as smooth as a function can be; as the kick grows its derivatives grow; and at the critical kick it is still continuous but no longer smooth, stretching some arcs of the circle enormously relative to others. The rise-and-fall figure is a crude reading of that: it doubles between $k = 0.5$ and the breakup, from $0.16$ to $0.33$ along one orbit of 144 points.

Why the circle map and the rotor meet at the golden mean #

The golden mean has now appeared at the critical point of two quite different maps. For the dissipative circle map, at $K = 1$ the unlocked parameters have dimension about 0.87, and the golden rotation number is where the self-similar structure is measured most cleanly. For the area-preserving standard map, the golden circle is the last barrier to break. In both cases the reason is the same: the golden mean is the rotation number least disturbed by resonances, because its continued fraction has the smallest possible terms, and so it is the last to be reached by the resonances that spread as a nonlinearity grows.

Robert MacKay made this precise in 1983 with a renormalisation scheme for the standard map: an operator that takes a map at one scale near the golden circle to a map at the next scale, with a fixed point that is the critical golden circle. The residue a quarter, the rate at which the crossings converge and the self-similar wrinkling are all properties of that fixed point, and they turn out to be universal — the same for any area-preserving twist map, not just this one — in the same way that Feigenbaum’s constant is the same for every map that doubles its period.

What the pictures cannot show #

That a circle exists. The phase portraits are orbits of 260 steps from sixty starting points; a curve drawn by an orbit looks like a circle whether or not it is one, and a thin chaotic layer looks like a curve at this resolution. Nothing in the portraits proves a circle survives, and nothing in them shows that the circles at $k = 0.6$ are the deformed horizontal lines of $k = 0$ rather than new curves — the continuity from one kick to the next is KAM theory’s, not the picture’s.

Greene’s criterion itself. The link between residues and the existence of the circle is a conjecture Greene supported numerically, and it has been proved only in parts: that a smooth circle forces the residues to zero is a theorem under suitable hypotheses; that residues tending to infinity forces the circle to be absent is established in some settings and not in general. The figures compute residues; the criterion is what gives them meaning.

The exact threshold. Eight crossings reach $0.9716$ to four places, which agrees with Greene’s value. They do not prove that the threshold is Greene’s number; they show that the approximating orbits behave as the criterion predicts, near that number.

Still open: proving the number #

Greene’s $0.971635$ is a numerical discovery, and what has been proved lies on either side of it. Robert MacKay and Ian Percival proved in 1985, by a computer-assisted argument that finds orbits crossing every would-be barrier, that no spanning invariant circle exists for $k > 63/64 = 0.984375$. In the other direction, computer-assisted versions of KAM theory, pushed by Alessandra Celletti, Luigi Chierchia, Rafael de la Llave and others, have proved the golden circle’s existence for kicks up to around nine tenths — below Greene’s value, with a gap that later methods have narrowed but not closed.

So the threshold is known to lie between those proved bounds, and it is believed to be $0.971635\ldots$ exactly, with the critical circle described by MacKay’s renormalisation fixed point. That fixed point’s existence has itself been established only with computer assistance, and a proof that the golden circle is the last circle to break — that no circle with any other rotation number outlives it — has not been given. The picture of the breakup is complete, detailed and universal, and the theorem that would make it more than a very well supported picture is still missing.