A particle on a line that can spell any word

Most demonstrations of chaos in gravitating systems need several dimensions to work in. Poincaré’s original result used the restricted three-body problem, and it was so hard to draw that he confessed he would not try. It has become conventional to think that a system must have room to move in, that chaos comes from having many directions and competing frequencies. The Sitnikov problem takes nearly all of that away. It has two stars, equal in mass. Their orbit is a fixed known ellipse. A third body of no mass sits on a straight line passing through their centre of mass at right angles to their orbit, and by symmetry it stays on that line for ever.

There is one coordinate and one velocity. In all of orbital dynamics there is hardly a smaller problem that is not simply solved. Kirill Sitnikov introduced it in 1960, in order to show something quite different: that a body can move away and return, with no upper bound on how far it goes, and yet not escape, a behaviour whose possibility had been an open question since Chazy’s classification. And then in 1973 Jürgen Moser used this example for a lecture course and showed that its motion could reproduce any sequence at all. The result is among the strongest statements about randomness in a deterministic system. It requires a coordinate that a pencil and a ruler could plot.

One coordinate and a clock #

Choose units with the two stars’ total mass 1, their semi-major axis of separation 1, and their period $2\pi$. Each star is at a distance $r(t)$ from the centre, with $r=\tfrac12(1-e\cos E)$ and $E$ the eccentric anomaly. At height $z$ on the axis, the particle feels the two stars’ pull along the axis, and the components in the plane cancel. So

$$\ddot z = -\frac{z}{\left(z^2+r(t)^2\right)^{3/2}}.$$

That is the whole system. For $e=0$ the distance is constant at a half and there is no explicit time in the equation, so energy is conserved and the motion can be integrated in elliptic functions. A particle with less than the escape energy goes up and down in a periodic oscillation, and each crossing of the plane is at the same speed. For $e\ne0$ the stars move in and out, $r(t)$ is a known periodic function of time, and that one external clock is all that is added. The particle has a clock of its own, which depends on how high it goes. These two clocks now interact.

The picture does not look like anything very dramatic. There is a smooth curve going through zero, up and down. Yet the only way to say what it will do next, as its excursions grow and shrink, is to compute it. Nothing simple links one excursion to the next, and a tiny change in launch alters the entire later sequence.

The map of crossings #

The natural way to look at a system with one coordinate and a clock is to reduce it to a map. Each crossing of the plane is described by two numbers: the speed $|\dot z|$ there, and the binary’s phase at that moment. The equation of motion then takes one pair into the next pair, and because the equations are Hamiltonian, the map conserves area. Everything that the particle does is coded in the sequence of points.

For a circular binary this gives a stack of straight lines, as expected. Each orbit keeps one speed, and only the phase changes, by an amount that depends on the speed. Those are the invariant curves of an integrable system, and the map is simply a shear.

Allow the stars to swing in and out and the picture breaks where the shear is greatest. Near the plane the orbits are short and the particle sees an averaged binary, so the motion is close to the integrable case and the curves survive. High up, where the speeds approach the escape value, the time for an excursion gets very long and very sensitive to the launch, and a small change in the arrival phase changes everything downstream.

The escape curve shown is what the two stars’ instantaneous potential would give, taking the binary’s distance as frozen at that moment. Whether a real launch escapes depends also on how the binary moves during the climb. The curve is therefore a guide and not a boundary, and that imprecision hides the most important fact about the map.

A time that takes every value #

Fix the phase of the launch and increase the speed. The particle goes higher and takes longer to come back. At some speed it does not come back. Near that speed, the time to return is rising without limit.

This is the essential ingredient. The return time is a continuous function of launch speed, and it goes to infinity at an escape boundary. So every sufficiently large value is attained, and for each large whole number $s$ there is a narrow band of launch speeds whose excursion lasts about $s$ binary periods. Those bands get closer and closer together as they approach the boundary.

Now take into account how the arrival depends on the launch. Stretch the launch speed across one of those narrow bands, and the outgoing conditions spread from one end to the other. Far up, the particle is moving very slowly and any change in energy from the swinging binary, however small, makes a large change in time. So each band is stretched over a huge range of arrival phases, and then all the way across the map again. The map takes a thin strip and stretches it right across the set of possible conditions, a number of times. There lies the classic mechanism of a horseshoe: stretch and fold, and the set of points that survive is a Cantor set whose addresses are sequences of symbols.

Moser’s sequence #

Moser’s theorem says exactly what that structure yields. Define $s_k$ as the integer part of the time between crossing number $k$ and the next, in binary periods. For small enough non-zero $e$, there is an integer $m$ such that every sequence $\ldots,s_{-1},s_0,s_1,\ldots$ with all $s_k\ge m$ is the itinerary of a genuine orbit. The sequence may be finite on one side and end in escape, or it may end in capture from infinity, or be bi-infinite.

Stated without symbols: choose any list of large numbers. Roll dice for them if desired, or write them out from the digits of pi. There is an orbit that does those excursions in that order.

The figure draws one such orbit, and it should be said what kind of thing it is and is not. The integers in the word are small, the eccentricity moderate, and the word short. It was found by a numerical search, launching a thousand trial orbits and zooming in on those that matched the longest beginning of the word, so it does not illustrate the theorem’s hypotheses. What it shows is what the claim looks like when one tries it.

The search also shows something of its own. Each additional letter in the word took the launch box down by a factor of about four. The six-letter word requires launch conditions set to about one part in a thousand, in each of two numbers. A twelve-letter word would require about a part in a million. That is what the determinism of the system looks like from outside. An orbit’s whole future is in its starting numbers, but the amount of its future that is available from a given number of digits is limited, and each further symbol of itinerary costs a fixed number of bits. Taking one more symbol costs one more digit, approximately.

Why this is randomness and not just complication #

A random process can be defined by exactly this property: that a sequence of its outcomes could be any sequence. A coin tossed repeatedly gives heads and tails with no constraints, and knowing what has come out so far gives no information about the next. The Sitnikov itinerary is a list that has this structure. Any large sequence is possible, including all of those that are incompressible.

So there are orbits of this system that are as complex as any random sequence is, in the sense that the shortest description of them is essentially the list itself. The theorem says that no finite formula could cover all orbits, because if it could, then there would be a formula generating a random sequence. It is a stronger and less ambiguous statement than saying that the problem is unsolvable, which has several distinct senses, and it does not require any measure of the chaos such as a Lyapunov exponent.

There is an important qualification. The set of orbits realised in this way has measure zero in the space of initial conditions. It is a Cantor set, a dust of orbits. The theorem says that orbits exist with any itinerary, not that a randomly chosen launch has an itinerary that looks random. Near the plane there remain regular islands. How much of the phase space has chaotic motion of positive probability has been explored numerically and is a harder question than the existence of a horseshoe.

A bridge between Chazy’s final motions #

The original reason for studying the problem is an interesting one. In 1922 Jean Chazy classified how a three-body system can behave at very late times: it can come apart hyperbolically, it can leave a binary and a body on a parabolic escape, and so on. A possibility he could not eliminate was one he called oscillatory: a body goes to arbitrarily great distances but keeps returning, with $\limsup|z|=\infty$ and $\liminf|z|<\infty$. Without an example of it, it was unclear whether it could exist in Newtonian mechanics.

Sitnikov’s example answers that. A sequence $s_k$ that grows without bound gives a particle whose excursions have no upper limit and which keeps coming back. Moser’s symbolic dynamics contains that as one sequence among others. For the same reason, an orbit can be captured from infinity, spend some long time bound to the stars, and then leave for ever. That was once believed impossible by an argument that seemed sound, since energy exchange in a two-body encounter is time-reversible and capture seemed to demand a third body already there. Here the third body is the binary’s phase. This shows why the lightest body leaves in general and can also come back.

What the numerical picture leaves out #

The integrator has a limited memory. These curves were calculated by a fourth-order Runge–Kutta method at a fixed step, and such an integrator is not symplectic and adds a small drift in energy. Long integrations of the map must take that into account. For excursions of a few periods the effect is small, but a figure of this kind could not show the fine structure near an escape boundary, because that structure is finer than the numerical error at any given precision.

The bodies are not massless and are not on a line. The real three-body problem has no such axis. The Sitnikov configuration is a symmetric special case which is exact only if the particle has no mass and the two stars are precisely equal. Small asymmetries break the invariant line, and a particle displaced from it would leave. So a real planet or dust grain is not going to do this; it is an exact model of an idealised system, not a description of an object in the sky.

The theorem’s hypotheses are not those of the figures. The theorem’s minimum integer depends on $e$ and grows as $e$ goes to zero, as the driving is weaker. For large eccentricities, proofs exist only under additional numerical verification. None of the figures tests the existence of all sequences. They show one sequence, for one case.

A section is not a history. The maps take a two-dimensional cross section, hiding the time between points, and those may be long. Showing the map at all assumes that the transition from one crossing to the next is unique, which fails at escape.

The same structure in a real sky #

Despite the artificial geometry, the mechanism has an important counterpart. A comet or a long-period body on a nearly parabolic orbit around the Sun and a giant planet has a return time that depends sharply on its energy, which diverges at parabolic escape, and it receives a small kick at each passage that depends on the planet’s phase. This is the Keplerian map used to model the long-term evolution of comets, and it has the same stretching property as the Sitnikov map: a small change in energy at the top of a very long orbit produces a large change in return time and hence in the phase at the next passage.

For those bodies the dynamics really is chaotic and is studied as a random walk in energy, as with the way a comet’s binding energy is changed at each passage. It is the same physical structure, slow motion far away with a long-period clock set by the energy, and it is why an orbit near the edge of escape behaves so differently from one deeper in. The same divergence appears in the stickiness near a chaotic region’s boundary and in the way a Lyapunov time sets how far ahead such a body can be forecast.

The same structure also sits behind the behaviour of stars around a pair of massive black holes, which sling a star out at a rate determined by repeated close encounters, and behind the binaries which give up energy to third stars in a dense cluster. In each case a long excursion comes between short and violent exchanges, and in each the exchange depends on a clock’s phase at the moment of arrival. The clean case on a line is what makes this structure provable.

Why one equation can hold this #

The surprising thing about the system is how little it takes. Consider what Poincaré would have required: the conservative structure, a separatrix, and a perturbation that crosses it transversely. Sitnikov’s problem has a separatrix between returning and escaping orbits in the plane of $z$ and $\dot z$, the parabolic orbit with zero energy at infinity. The periodic pull of the binary’s in-and-out motion breaks it, so that the stable and unstable manifolds of the point at infinity intersect transversely. That transverse crossing, which can be calculated from the Melnikov integral for small $e$, immediately gives the horseshoe through the Smale–Birkhoff theorem, and hence the symbolic sequence.

Everything therefore follows from whether one particular integral is non-zero, and that can be checked by hand. The most complicated possible motion is implied by a calculation with a single, smooth integrand. That is a fact about mathematics as much as physics, and the sequence that emerges is what might be called the signature of a homoclinic tangle: an orbit that goes out and gets back to the same place a countable number of times.

How far the digits reach #

A last way of stating the result concerns what can be known in practice. An orbit on this line has a whole future determined by two numbers, and a finite-precision copy of those numbers determines only a finite part of that future. Each further symbol costs a fixed number of digits. So a calculation carried out with fifteen significant figures, as an ordinary computer does, has a definite horizon, beyond which it is no longer following any particular orbit at all, only some nearby one. This is the sense in which the problem joins a system with a probability for its one future. Nothing random is happening, and yet the only useful statements at long range are statistical ones.

Still open: how large the regular part is #

Moser’s theorem, as stated, concerns a Cantor set of zero measure, at small eccentricity. A much more practical question is how large a share of the launch conditions are actually chaotic, at any eccentricity, and whether a set of chaotic orbits with positive measure exists. Numerical sections suggest that large chaotic zones exist with a non-trivial area. But proving positive measure for a chaotic zone in a nearly integrable system is one of the hard open problems of Hamiltonian dynamics, in this case as in the standard map. Numerical results cannot resolve it, since they do not see sets that are small and they cannot tell an extremely long transient from a true invariant set.

What is more, the map’s boundary with escape has structure at every scale, and the rate at which orbits are lost through that boundary is not a simple exponential. The practical consequence would be for long-period dynamics in comet clouds and in loosely bound binaries. Its mathematical expression is the same question of how much of a Hamiltonian system’s phase space is chaotic, and it remains open even when the system has one coordinate.