# Are double pendulums eventually periodic?

I've often heard it said that the motion of a double pendulum is non-periodic. (This may be related to the fact that it's a chaotic system, but I'm not sure about that.) But this does not seem possible to me, for the following reason. Let $\theta_1$ and $\theta_2$ be the angles of the two masses relative to the vertical. Then we can consider the two-dimensional phase space with a $\theta_1$ axis and a $\theta_2$ axis, and the motion of the double pendulum is a continuous curve $\gamma:[0,\infty) \rightarrow [0,2\pi]\times[0,2\pi]$. The thing is, I'm pretty sure such a curve must be self-intersecting. Because if it's not self-intersecting, then its graph would cover more and more of the codomain with time, and so I think you'd get a space-filling curve. And yet space-filling curves are always self-intersecting, so you'd get a contradiction. Thus $\gamma$ must be self-intersecting, and thus the motion of a double pendulum is always periodic.

So what's wrong with my reasoning? Or is my reasoning correct, and is the motion of a double pendulum always periodic, just with such a long period that it looks non-periodic? If so, is there a formula for the period?

• @JonCuster Well, if my reasoning is correct then the period must be finite. – Keshav Srinivasan Oct 17 '17 at 22:23
• Why are there a finite number of points in your phase space? – Jon Custer Oct 17 '17 at 22:26
• $\gamma(t)=(t^{-1},\cos(t))$ is an example of a continuous non self-intersecting curve, so you know your reasoning is false and you're missing some assumptions! – user12029 Oct 17 '17 at 23:58
• @KeshavSrinivasan my curve is nonintersecting and maps from $[1,\infty]$ into $[0,1]\times [-1,1]$. – user12029 Oct 18 '17 at 0:04
• The answers have already explained several things wrong with the premise, but it seems important to point out that your notion of a space-filling curve is not correct. For instance, on a torus $T^2 = \mathbb R^2 / (2 \pi \mathbb Z)^2$ (the space on which $\theta_1$ and $\theta_2$ live), a line of irrational slope is a curve $\mathbb R \rightarrow T^2$ which has dense image, but is not space-filling (in fact its image is measure 0). While it does approximately recur infinitely often (as in the Poincare recurrence theorem), it is an injection. – Logan M Oct 18 '17 at 0:09

Short answer: No. General trajectories of double pendulum are not periodic.

You need to distinguish between two aspects: the trajectory in the spatial coordinate system and the trajectory in phase space.

Your claim about $\gamma$ is about the first aspect and is thus false. It is perfectly okay for trajectories to intersect in the real space, and this doesn't mean the solution is periodic.

However, in the phase space it is forbidden for different trajectories to intersect (because of the uniqueness of the solution of ODEs given initial conditions). And if they do, you are correct that the dynamic is periodic. Indeed, notice that it may be that the mass travels through the same spatial point twice, but it can be with different velocities.

As @agemO suggested in a comment below, it is important to stress that although the solution is not periodic, it seem like it is getting close to there (which is probably what confuses you). Suppose for example that the mass starts from a point $(x,y)$ in the XY plane with velocity vector $(v_{x},v_{y})$. Then according to Poincare Recurrence Theorem, after some time the mass will travel as close as you want to that point with a very very similar velocity - but they are not guaranteed to be the same. In other words, the motion is as close as you want to be periodic, but it misses, and the resulting behavior is chaotic.

There is another very interesting theorem that should also be worth stating in this case. It is called Poincare Bendixson Theorem, and it states that a traped trajectory in 2D phase space must eventually repeat itself (given that the trapping region doesn't contain fixed points). But in this case the phase space is 4D and the theorem doesn't apply.

• The velocity has both direction and magnitude. The magnitude must be the same, but the direction is not. – eranreches Oct 18 '17 at 0:13
• But you have two rods. I think that gives you this freedom. – eranreches Oct 18 '17 at 0:20
• It very much does give you that freedom! For $m_1=m_2=\ell_1=\ell_2=g=1$, we have $E=\dot{\theta}_1^2 + \cos(\theta_1-\theta_2) \dot{\theta}_1\dot{\theta}_2 + \frac{1}{2}\dot{\theta}_2^2 + 2 \sin(\theta_1) + \sin(\theta_2)$, which for any given $E,\theta_1,\theta_2$ has a continuum of solutions in $\dot{\theta}_1$ and $\dot{\theta}_2$. – user12029 Oct 18 '17 at 0:27
• Note in a gravitational potential, there isn't rotational symmetry and so the double pendulum doesn't conserve angular momentum. – CDCM Oct 18 '17 at 1:54
• @KeshavSrinivasan : A simple pendulum passes through its lowest point repeatedly with various velocities. You seem to be arguing that this never happens. In a double pendulum, set the initial angles to $0$ (i.e., straight down). You seem to claim that I have only one choice of initial angular velocities, which is absurd. – Eric Towers Oct 18 '17 at 18:25

From wikipedia, the phase space of a dynamical system is defined as

... a space in which all possible states of a system are represented, with each possible state corresponding to one unique point in the phase space. For mechanical systems, the phase space usually consists of all possible values of position and momentum variables.

For the double pendulum there are two position variables $\theta_1, \theta_2$ and two momentum variables $\dot \theta_1, \dot \theta_2$. So the phase space is 4 dimensional. If the system returns to the same point in its phase space its motion will repeat, because the state of the system is identical with its previous state so its subsequent motion is identical.

The 2 dimensional space with axes $\theta_1, \theta_2$ is not the phase space for this system. It is only a projection or "shadow" of the phase space.

Each point on the 2D plane $(\theta_1, \theta_2)$ corresponds to an infinite number of points in the 4D phase space, each with a different combination of $(\dot \theta_1, \dot \theta_2)$. The system can return to the same position $(\theta_1, \theta_2)$ with a different combination of momenta $(\dot \theta_1, \dot \theta_2)$ each time. The trajectory in the 2D plane $(\theta_1, \theta_2)$ can pass through the same point an infinite number of times without retracing the same path, because each of these intersections corresponds to a different state of the system.

So it is possible for the double pendulum to never return to the same point in its 4D phase space and therefore its motion can be non-periodic.

So your argument is not quite correct because of the existence of space-filling curves and fractals and the like; to take a simple example imagine a spiral over the whole plane $\mathbb R^2$, clearly not self-intersecting, and cut out the unit circle from it, then map it according to the Möbius transform $(x, y) \mapsto \left(\frac x{x^2 +y^2}, \frac y{x^2+y^2}\right).$ This transform maps the exterior of the unit circle into the circle, but it's totally invertible and cannot cause self-intersection. (It is also conformal and some other nice things that don't really matter here.) What you would get, if you look at this like a phase space, would be a harmonic oscillator with dissipation which never allows it to quite come to rest but always saps a tiny bit of energy from the thing. One could probably find that in the 4D phase space of a double pendulum there would be weird Hamiltonians which conserve energy and yet still do this. Trajectories need never be periodic precisely.

What you can prove is that for a gobsmackingly huge number of systems, the vast majority of the "nearby" paths to these exceptional ones will nonetheless return back to "nearby" whence they came, so that in general you will see a pattern of physics from $0 < t < \tau$ and then you will see an arbitrarily close copy of that from $T < t < T + \tau$ for some big $T$. This is known as the Poincaré recurrence theorem.

• "So your argument is not quite correct because of the existence of space-filling curves and fractals and the like" Actually space-filling curves are always self-intersecting. – Keshav Srinivasan Oct 18 '17 at 0:05
• @KeshavSrinivasan But in this case, the curve doesn't actually have to be space-filling. The spiral is a good example; it will never fully fill any area, but it will not self-intersect either. Instead the distance between consecutive revolutions just grows smaller and smaller. – jpa Oct 18 '17 at 5:32

## protected by Qmechanic♦Oct 18 '17 at 5:15

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