Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Math/physics teachers love to break out the double pendulum as an example of chaotic motion that is very sensitive to initial conditions. I have some questions about specific properties:

  1. For a simple double pendulum (which I'll define as two equal masses joined by two massless arms of the same length) is the highest initial energy state when both masses are completely raised vertically, as intuition would suggest?

  2. contains a plot of conditions for which the double pendulum "flips" or does not flip. My question is: does a "flipping" of either top or bottom pendulum imply subsequent chaotic motion, or are there certain energies for which we can fully predict the motion and flips? Conversely, if we are in low energies where neither top nor bottom pendulum can possibly flip, can we still have a chaotic system or do all those energies have predictable motions?

share|cite|improve this question
up vote 5 down vote accepted

It is true that the double pendulum exhibits integrable behavior, when the initial angles are very small, however, in general, it is very difficult to characterize the chaotic behavior of the double pendulum in terms of the initial angles. There are other representations which provide a clearer picture of its chaotic behavior.

The introductory section of the following article by O. A. Richter, and reference therein describe the main characteristics of the double pendulum integrability. I'll summarize here the main facts:

(Remarks: The numerical values correspond to a standard double pendulum of unit masses and unit rod lengths and unit acceleration due to gravity)

The total energy of the double pendulum is a constant of motion. The double pendulum possesses 4 equilibrium points corresponding to total energies $E= 0, 2, 4, 6$. The total energy determines the topology of the energy hypersurfaces, for $E<2$, the energy hypersurfaces are three spheres, while for $E>6$ , the energy hypersurfaces are three tori.

To further understand this point, in the case of very low energies, the system can be approximated (linearized) to an isotropic harmonic oscillator. The energy hypersurfaces are then of the form $x_1^2+x_2^2+p_1^2+p_2^2=E = const.$, while for the case of very large energies, the kinetic energy dominates and we can neglect gravity. In this case, there are two types of solutions of the equation of motion, one in which the outer rod rotates and the inner rod oscillates, and the second in which both rods rotate. The transition between the two types of solutions is determined by the value of the total angular $L$ momentum which becomes a constant of motion (due to the lack of gravity). For the standard double pendulum the transition occurs at $ L^2 = 2E$.

Both limits (small and large energies) correspond to integrable systems. This is well known, but here is a short explanation. To see that the isotropic harmonic oscillator is integrable, one needs to solve the equations of motion in polar coordinates. The polar angle just rotates with a constant angular velocity, and the radial coordinates oscillates in such a way that the trajectory has the shape of a two trip course through the donut hole drawn on the surface of a two-torus. This is the Liouville-Arnold torus (whose existence indicates the system's integrability) with respect to which the three sphere energy hypersurface is foliated.

In the high energy limit, a similar Liouville-Arnold torus exists when the inner rod oscillates and a torus generated by the two polar angles when the two angles rotate. (Here the exact solution is more difficult, see for example, the following article by Enolskii, Pronine, and Richter.

Now, since both limits of vanishing and very high energies correspond to integrable systems, the total energy also controls the system characteristics, but the dependence here is much more complicated. The transition from integrability to chaos and back as the total energy decreases from infinity to zero is described in figure 2 of Richter's article. There are a lot of details, but here are the main features: The figure correspond to the projection of the trajectories onto the plane spanned by the outer rod angle and the total angular momenta. For very large angular momentum, the projections of the trajectories are horizontal lines with constant angular momenta (which is a constant of motion). As the energy is reduced, two disjoint chaotic regions are formed, the integrable trajectories correspond to rational and irrational tori, together with stable resonances. At about E = 10.352 which corresponds to the golden winding ratio, all irrational tori vanish and a transition to global chaos occurs. The stable resonances also vanish eventually, at the low energies the resonances corresponding to the second integrable region start to appear.

share|cite|improve this answer

Here is an interactive simulation page with derivation of formulae that may enlighten you.

If the energy in the system is just the gravitational energy, then yes, the vertical initial position will have the highest potential energy which will be transmuted to kinetic when let go. Energy is conserved, kinetic + potential is constant, so the "position" helps in determining the initial energy of the system. Otherwise, if kinetic energy is supplied it will be the maximum potential + the added kinetic. What has no answer is also the "position" in this case, as it will depend on how the impulse is conveyed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.