Double pendulum

The double pendulum here is two equal point masses on two equal weightless rods, with no friction. Both angles are measured from straight down. In units where $m = l = g = 1$, time is counted in $\sqrt{l/g}$, and the energy of height, counted from the pendulum hanging at rest, is

Here you can let the pendulum go from any pose at rest, release a thousand nearly identical pendulums together, and see which starts can, and which do, send an arm over the top.

By default 1 000 pendulums are let go together, their lower arms spread over a millionth of a radian. The thicker lines are yours, and the trail follows your lower bob. The inset shows the energy, which stays the same while it passes between height and motion, and the two angles over time, with a red dot each time an arm passes the top.

The equations of motion are exact, and nothing in them is random: the same start always gives the same motion. Yet from the default start, (120°, 60°), the farthest lower bob is more than 0.1 arm lengths from yours by $t \approx 33.5$, and the bundle soon spreads into clearly different motions. More precision buys little. For this start, a tiny difference grows roughly $e$-fold every three time units while it stays small, so spreading the lower arms over $10^{-9}$ radian instead only delays the split to about 56. From a small swing, (30°, 30°), the same bundle hardly separates through $t = 300$.

Small swings are orderly. For small angles the equations become linear, and they have two normal modes, in which both arms keep one rhythm:

In the slow mode the arms swing the same way, in the fast one against each other, and any small swing is a mix of the two.

Energy draws a fence. Without friction the total energy $E'$ stays the same and the energy of motion is never negative, so the pendulum can only take poses with $V' \le E'$. For the lower arm to point straight up, $V'$ must be at least 2, and for the upper arm at least 4: the two marks on the inset’s energy bar. A pendulum let go from rest has the energy of its starting pose, so

That is a proof, not a simulation. Between 2 and 4 only the lower arm may go over, and above 4 both may, but enough energy only allows it; it does not mean that an arm will. The blue dashed line on the map of starts marks the poses exactly as high as yours.

Map of starts shows every start from rest at once, the upper arm’s angle across and the lower arm’s up, with the hanging pose in the middle. Coloured cells show when either arm first passes the top. Uncoloured cells outside the hatching have not flipped by the time shown, which runs up to $t = 60$. The hatched region is where energy forbids it. Away from the hatching the colours lie in broad bands. Near it they break into fine grain, where neighbouring starts go over at very different times. The grain comes from sensitivity to the start, so the time shown for a late flip is only approximate.

The map shows only the first flip, and it is not a map of where the motion is chaotic. Chaos does not need an arm to go over: in the page’s tests many starts below energy 2 already behave chaotically.