I was always assuming that a double pendulum does not have any non-trivial integrals of motion besides the Hamiltonian, but after asking this question I tried to simulate the motion of the pendulum, and I was surprised.

The motion of a double pendulum is described by 4 differential equation, and if there were no non-trivial integrals of motion, one would expect the trajectory to cover a 3D manifold (corresponding to constant energy) in the 4D phase space. What I discovered, was that the motion is constrained to a 2D surface (see the picture, which shows the projection of the trajectory of $(\theta_1, \theta_2, \dot{\theta_1}, \dot{\theta_2})$ with initial position $(1,0,0,0)$ onto a 3D space and then onto the 2D screen). It may be not very easy to see from this picture, but I am positive it is a 2D manifold (after rotating it for a while). So, is there an extra integral of motion or is there not one? I saw some people mentioning local integrals in a physics post, but I am not sure what to make of it with respect to this result. Could someone educate me why the motion is constrained to a 2D manifold, when one naively would expect it to be constrained to a 3D manifold? enter image description here



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