A ring of clocks settles into a twist

A long enough chain never locks found that oscillators with different natural frequencies, each pulled only by its neighbours, lock into a common frequency only if the coupling can carry the accumulated difference of rhythms across every link. That essay ended on the case it made trivial. If the oscillators are identical, there is no difference to carry, every running sum is nought, and a chain locks at any coupling. But a ring of identical oscillators — the chain with its ends joined — can lock in more than one way, because its angles can wind round it. This essay starts identical oscillators on a ring from random phases and asks which way they lock.

The equations are those of the chain with the natural frequencies removed, since a common frequency can be subtracted by watching from a rotating frame. Oscillator $j$ on a ring of $N$ is pulled towards its two neighbours:

$$\frac{d\theta_j}{dt} = \sin(\theta_{j+1} - \theta_j) + \sin(\theta_{j-1} - \theta_j),$$

with $\theta_{N+1} = \theta_1$. The fully synchronised state, every angle equal, is a solution and it is stable. It is not the only stable solution, and in a random start it is not even the likeliest outcome on a ring of any size.

Locked, and not in step #

The hero figure shows a ring of forty that has stopped changing. Every oscillator runs at the same frequency — in the rotating frame, none of them moves — so the ring has locked. But the angles are not equal. Walking round the ring from any oscillator to the next, the angle advances by the same amount, $2\pi \times 2/40$, so that after forty steps it has turned twice and come back to the start. This is a twisted state, $\theta_j = 2\pi q j / N$ with $q = 2$, and it is an exact solution of the equations: every oscillator is pulled forward by its next neighbour and back by its previous one by exactly the same amount, so the pulls cancel.

A twisted state is the strangest kind of agreement. Each oscillator is in close agreement with its neighbours, a ninth of a radian apart; the ring as a whole has no common phase at all. The Kuramoto order parameter of a crowd of clocks that falls into step — the length of the average of all the arrows — is exactly nought for every twisted state with $q \ne 0$, because the arrows point evenly in every direction. The figure measures it as $2 \times 10^{-16}$, rounding error. By the measure used for the all-to-all crowd, the locked ring is perfectly incoherent.

The integer $q$ is the ring’s winding number: the number of full turns the angle makes as the ring is walked once round, found by adding up the differences of angle between neighbours, each taken between minus and plus half a turn. It is a topological quantity in the sense of the loops that never cross themselves: it can change only if two neighbours pass through a phase difference of exactly half a turn, which a smooth evolution does only at isolated moments.

Which twist a random start chooses #

Started from independent random phases, a ring settles into some twisted state, and the winding number it chooses varies from start to start.

On a ring of 20, 42 per cent of six hundred random starts end fully synchronised, with $q = 0$; 45 per cent end with $q = \pm 1$, and the rest with $q = \pm 2$. On a ring of 40, synchrony takes 39 per cent and the winding numbers spread to $\pm 3$. On a ring of 60, synchrony takes only 28 per cent of 250 starts. In each case the distribution of $q$ is close to a bell curve centred on nought, and its variance is close to $0.03$ times the ring’s length — $0.033 N$, $0.030 N$, $0.036 N$ for the three rings — so its width grows like $\sqrt N$. A ring four times as long spreads its outcomes twice as wide, and the share that end in full synchrony falls like $1/\sqrt N$.

The bell shape has a simple origin. A random start already has a winding number of its own — the sum of $N$ independent phase differences, each spread evenly over half a turn either way — and that sum has a bell-shaped distribution with variance $N/12$, by the central limit theorem. The dynamics then resolves the largest disagreements by letting some neighbouring pairs slip through half a turn, which changes the winding number, and what survives is narrower: about $0.03 N$ rather than $N/12 \approx 0.083 N$. Daniel Wiley, Steven Strogatz and Michelle Girvan made these measurements in 2006 and found the same Gaussian spread of basin sizes, with a width proportional to $\sqrt N$, and an argument from the random start’s own winding that predicts the shape if not the exact constant.

A winding number cannot be unwound smoothly #

The reason a twist persists is the same reason a loop that cannot be pulled tight stays caught on a hole. Read the ring’s phases as a map from the ring of oscillators to the circle of phases: oscillator $j$ is sent to the point $e^{i\theta_j}$. If neighbouring phases always differ by less than half a turn, the map can be drawn as a continuous loop on the circle, joining each oscillator’s point to its neighbour’s by the shorter arc, and the winding number is that loop’s degree — the number of times it goes round. A continuous deformation of a loop cannot change its degree; only a jump, when some neighbouring pair is exactly half a turn apart and the shorter arc flips sides, can.

So the dynamics has walls in it. The space of all ring states is divided into regions, one for each winding number, by the states in which some neighbouring pair is half a turn apart; a trajectory can cross a wall only by passing through such a state, and once every pair is well inside half a turn, no wall is near. Each twisted state sits in the middle of its own region, and the synchronised state is the twisted state of winding number nought. What a random start does in its first few time units is decide which region it will be in when the walls stop being crossed.

The open chain of a long enough chain never locks has no such walls, since its ends are free and its angles cannot go round. That is the whole difference between the two shapes for identical oscillators. A chain’s phase differences can always relax to nought, so a chain of identical clocks always ends in step; a ring’s phase differences must add to a whole number of turns, and the number is fixed once the walls are no longer crossed.

Twists up to a quarter turn per link #

Not every winding number can be the end of a random start, because not every twisted state is stable.

Linearising the equations about the $q$-twisted state, every neighbour difference is the same angle $\phi = 2\pi q/N$, and a small disturbance of oscillator $j$ is pulled back by each neighbour at the rate $\cos\phi$ — the same factor whose sign separated the stable locked orbit from the unstable one in how a lock comes apart. The disturbances decay as the modes of a ring’s Laplacian, at rates $4\cos\phi\,\sin^2(\pi m/N)$ for $m = 1, \ldots, N - 1$, all of the same sign as $\cos\phi$. So the twisted state is stable exactly when $\cos(2\pi q/N) > 0$: when neighbouring phases differ by less than a quarter turn. On a ring of 24 that allows $q$ from $-5$ to $5$, and the figure confirms it: each of $q = 0$ to $5$, nudged and relaxed, returns to itself; $q = 7$ slips to $1$, and $q = 8$ to $11$ slip all the way to synchrony. At $q = 6$, a quarter turn per link exactly, the cosine is nought and the linear argument is silent; the nudged state drifts and slips to $q = 4$.

A ring of $N$ therefore has about $N/2$ stable twisted states, and every one of them is a locked state in which every oscillator agrees with its neighbours. The largest stable twists are rarely chosen from random starts, since the bell curve of winding numbers has width only about $0.17\sqrt N$, but they exist, and a ring placed in one stays there.

The twist is chosen early #

The winding number can change only when some neighbouring pair passes through half a turn apart. The last figure watches it during the relaxation.

All the changes happen at the start. A random initial state has many neighbouring pairs nearly half a turn apart, and within the first few time units those disagreements are resolved one way or the other, each resolution either keeping or changing the winding number by one. After that every neighbouring difference is well inside half a turn, the winding number is frozen, and the ring spends the rest of its relaxation — hundreds of time units on a ring of sixty, since a twist relaxes diffusively along the ring — straightening out the angles into the exact twisted state with the winding number it has already chosen. Which twist a ring ends in is decided in a moment; how long it takes to get there is decided by the ring’s length.

That separation of time scales is the reason the basins are bell-shaped. The fate of a random start is sealed by local events — a handful of near-half-turn pairs scattered round the ring, each resolved independently — and the winding number is their sum. A sum of many small independent contributions is bell-shaped whatever their individual shapes, which is the central limit theorem appearing in a deterministic system through the randomness of its start.

Feeling more of the ring #

Coupling each oscillator to more neighbours changes the balance. With $k$ neighbours on each side, the pull towards agreement acts over longer distances, and a twist that is gentle locally becomes costly at longer range.

With one neighbour on each side, 36 per cent of three hundred random starts on a ring of forty reach full synchrony. With two, 56 per cent; with six, 77 per cent; with ten, 97 per cent; and from twelve on — each oscillator feeling three tenths of the ring — every start synchronises. Wiley, Strogatz and Girvan found that the twisted states lose their stability altogether when $k/N$ exceeds about $0.34$, so that above that connectivity synchrony is the only stable state and every start must reach it. The measurement here finds every start synchronising slightly below that value, which is consistent: below it the twisted states still exist and are stable, but their basins have become too small for two hundred random starts to land in.

The comparison with the all-to-all crowd is now complete in one direction. When every oscillator feels every other, $k/N$ is as large as it can be, the twisted states are all unstable, and identical oscillators always synchronise. When each feels only its nearest neighbours, synchrony is one stable state among about $N/2$, reached by a share of random starts that falls like $1/\sqrt N$. In between there is a threshold of connectivity, about a third, above which the network behaves for this purpose like a complete one.

What the simulations cannot show #

The basin measurements are samples — six hundred, five hundred and two hundred and fifty starts — and the variances they give, $0.030$ to $0.036$ times $N$, scatter by more than their own sampling error suggests, partly because the rings are short and partly because the bell curve is an approximation whose tails are cut off by the stability limit at $|q| < N/4$. Whether the variance per oscillator converges to a definite constant as $N$ grows, and what that constant is, the figures do not settle. Wiley, Strogatz and Girvan gave a heuristic for the shape and measured the width; an exact value derived from the dynamics is not known.

The relaxations themselves are numerical. Each ring was integrated by Euler’s method with a step of a tenth, for a time proportional to the square of its length — about $0.6N^2$, long enough for the slowest mode of the twist to decay — and a run was counted only if every neighbouring difference ended within a thousandth of the exact twisted state’s. A gradient flow like this one is forgiving of the method, since every step lowers the same energy, but a step too large could carry a pair across a wall that the true flow would not have crossed, and so change a winding number; the orbit a computer draws is the standing warning that a computed trajectory is a nearby trajectory, not the trajectory. Halving the step on sixty starts of a ring of forty changed the winding number of one of them: a start that passed close to a wall, where the two step sizes resolved a near-half-turn pair in opposite directions. The basin figures therefore carry an error of that order on top of their sampling error, about one start in sixty, which is small beside the spread being measured and not nothing.

The stability criterion, on the other hand, is exact for the nearest-neighbour ring, and the perturbation experiment only confirms it. What the experiment cannot show is what happens exactly at the boundary, $q = N/4$, where the linearisation is degenerate: there the outcome depends on the nonlinear terms and on the disturbance, and the slip to $q = 4$ in the figure is one outcome among several possible.

Still open: the basins in higher dimensions #

The ring is one-dimensional and its twisted states are classified by a single integer. A grid of identical oscillators on a torus has twisted states classified by two integers, one for each direction, and a sphere or a random network has no such classification at all: their locked states other than synchrony are not twists but irregular patterns, and the questions of how many there are, which are stable and how large their basins are have no general answers. For random networks the question of whether identical oscillators always synchronise has been asked as a graph property — a network is globally synchronising if synchrony is the only stable state from almost every start — and it has been proved for random graphs down to nearly the density at which they become connected, while for structured sparse networks such as rings it fails, as the figures here show, and which structures fail is not known in general.

There is a sharper question on the ring itself. The figures show synchrony becoming certain somewhere near $k/N \approx 0.3$; the twisted states’ loss of stability at $0.34$ is a linear calculation. Between the two lies the question of how the basin of synchrony grows with $k/N$ — whether it reaches the whole space only at the stability threshold or earlier — and it is a question about the global geometry of a high-dimensional flow, of the kind two chaotic orbits made to agree met for rings of chaotic maps, that local analysis does not answer.

Agreement that goes round #

A ring of identical clocks, each adjusting only to its neighbours, always locks, and most of the time it locks wrong: into a twist in which every clock agrees closely with the clocks beside it and the ring as a whole points in every direction at once. The twist is chosen in the first moments from the random start, by a handful of local resolutions that add up to a bell-shaped winding number, and then it is frozen in, because no smooth evolution can unwind a whole turn of phase. Coupling over longer distances makes the twists unstable and synchrony certain, at a connectivity of about a third of the ring.

The open chain failed to lock because local coupling could not carry an accumulated difference of natural frequencies. The ring of identical clocks locks every time and fails to synchronise because local coupling cannot see a twist: every link is content, and the disagreement is global, stored in a topological integer that no link can measure.