For a system with $n$ degrees of freedom (DOF), one has to deal with $2n$ independent coordinates ($2n$ dimensional phase space), of position $q$ and $\dot{q}$ in Lagrangian formulation, or independent coordinates of $q$ and generalized momentum $p$ in the Hamiltonian formulation.

We remind the reader that if a system with $n$ DOF exhibits at least $n$ globally defined integrals of motion (first integrals), where all such conserved variables are in (Poisson) involution with one another, then the system is (Liouville) integrable.

Furthermore a system with $n$ DOF can at most have $2n-1$ globally defined integrals of motion. A system will generically have $2n$ locally defined constants of motion. We will only be interested in integrals of motion that are globally defined.

Now coming to the famous case of the 2D double pendulum, with weightless rigid wires attaching the two masses, having the lengths $\ell_1$ and $\ell_2$, the generalized coordinates here are given by the two angles that each mass makes with the vertical, denoted respectively by $\theta_1$ and $\theta_2.$

It is rather straightforward to show then that under constant gravity field, the Lagrangian is given by:

$$L~=~T-V~=~\frac{1}{2}(m_1+m_2)\ell_1^2\dot{\theta_1}^2+\frac{1}{2}m_2 \ell_2^2 \dot{\theta_2}^2+m_2 \ell_1 \ell_2 \dot{\theta_1} \dot{\theta_2} \cos(\theta_1 - \theta_2)+(m_1+m_2)g\ell_1\cos\theta_1 + m_2g\ell_2\cos\theta_2.$$

From here calculating the Euler-Lagrange differential equations, one obtains a coupled 2nd-order ordinary differential equation that can be solved only numerically for $\theta_1(t)$ and $\theta_2(t).$

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Knowing that one integral of motion here is the total energy $E$, and that angular momentum component orthogonal $L_z$ to the plane of motion is also a integral of motion independent of $E$. Unfortunately, they do not Poisson commute.

  1. Are there any other integrals of motion to be found here?

  2. Just by looking at the Lagrangian, as given above, how can we show the system is not integrable, at least at a conceptual level? (we just want to predict, by reasoning what is conserved, and what quantities are not first integrals here).

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    $\begingroup$ 2. Again, I'm not sure but I don't think it is easy to prove... This Phys.SE question might be of interest. $\endgroup$
    – anderstood
    Oct 21, 2014 at 15:03
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    $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Oct 21, 2014 at 20:30
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    $\begingroup$ Comment to the question (v4): The total external torque $\tau_z=\frac{dL_z}{dt}$ (produced by gravity) around the suspension point, is rarely zero. Hence $L_z$ is not an integral of motion in the presence of gravity. $\endgroup$
    – Qmechanic
    Oct 22, 2014 at 19:31
  • $\begingroup$ Look at math.stackexchange.com/q/1682368 and the green picture in myphysicslab.com/pendulum/double-pendulum/… and the comment in the former web page about KAM theory. It all suggests that there is another conserved quantity. I have no idea what it is though. Maybe if one studied KAM theory, one could figure it out. $\endgroup$ Nov 15, 2016 at 16:38
  • $\begingroup$ These seem relevant: 1. V. Salnikov, arxiv.org/abs/1303.4904 Hm. Very short. 2. T. Stachowiak & W. Szuminski, arxiv.org/abs/1511.01850 Hm their eq. (2.1) looks different. $\endgroup$
    – Qmechanic
    Apr 3, 2018 at 9:12

2 Answers 2


To start things off I'd say that noting the $L_z$ component is conserved seems to mean pretty much nothing, since you're considering the motion as restricted to the $\mathcal{X}\mathcal{Y}$ plane. If you had assumed the motion along the $\mathcal{Z}$ axis to be possible, then we'd be talking about the spherical double pendulum instead of the planar one (which is the case, since the lagrangian has two degrees of freedom).

The energy will be conserved because the system is autonomous (time-independent). Notice also that an integrable autonomous system with $n$ freedoms has $n$ conserved quantities, one of them being the energy. So, since our system has two degrees of freedom, there is one constant of motion missing for it to be integrable. Differentiating $L$ with respect to $\dot{\theta}_1$ and $\dot{\theta}_2$ will give us the canonical momenta of the system, but notice that the derivatives of $L$ with respect to $\theta_1$ and $\theta_2$ are not zero. Thus, the canonical momenta are not conserved quantities. Also, the total energy of the system doesn't factor as a sum of individual energies, since the Lagrangian has mixed terms in it. You cannot extract any other quantity from the Lagrangian which should be conserved, since mechanics talks about the conservation of momenta and energy (and sometimes their projections). Since there are not as much conserved quantities as there are degrees of freedom, the system is non-integrable.

P.S.: In one dimension we can clearly see that (using simple examples as the harmonic oscillator or the simple pendulum) some systems do not conserve their momenta. Even though, if they're autonomous then they're integrable, because they have one degree of freedom and one conserved quantity: energy.

EDIT.: Since the question is directly oriented to "is it possible to obtain, from the Lagrangian, an answer about integrability?", then (as suggested in the comments) I'll sketch something on Noether's theorem. The theorem relates Lie groups to invariant quantities, so it's just one more way to find conserved quantities. It basically says that if you find a transformation that leaves the Lagrangian invariant, then there's a conserved quantity associated to that transformation. As an example, when the transformation reduces to a translation, invariance of the Lagrangian implies momentum conservation; much the same way, if the transformation is a rotation, then invariance implies the conservation of angular momentum along the axis of rotation. So this is basically a way of using symmetries of the Lagrangian to obtain conservation laws (knowing Noether's theorem content well is very important to a clear understanding of many concepts of Quantum Mechanics and Quantum Field Theory). I'm not being quantitative because proving the double pendulum's Lagrangian is not an invariant is tedious, even thought it should be somehow obvious by just looking at it.

  • $\begingroup$ @Phonon Thanks! What I'm saying is indeed that the system is coupled. Compare with the case of two masses glued by a spring and glued by springs on two walls: the total energy can be factored as a sum of energies (the system is not coupled), so the total energy gives, alone, two conserved quantities: the energies of each of the particles. I just thought that "the energy can't be factored" came more naturally to the discussion. $\endgroup$ Oct 21, 2014 at 19:53
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    $\begingroup$ yeah thought so :), for completeness maybe you could add a word or two on the point of view using Noether's theorem, and how the lack of symmetry in the system is already indicative of lack of conserved quantities. (if you want, just a suggestion), would be cherry on top :) $\endgroup$
    – Ellie
    Oct 21, 2014 at 19:58
  • $\begingroup$ @Phonon I agree it would be nice to say something about Noether's theorem qualitatively, but the quantitative part doesn't attract me. The lack of symmetry of the system would be evident by proving there is no group action associated with the momenta that would leave the Lagrangian invariant, and I'd need some paper to do calculations of a result the Lagrangian itself already tells me won't give me nothing. What do you suggest? $\endgroup$ Oct 21, 2014 at 20:12
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    $\begingroup$ @user929304 1) We call a system autonomous if the hamiltonian does not depend explicitly on time. It'll always depend implicitly, since Mechanics is the study of movement. 2) We get stuck as soon as we notice our equations of motion cannot be integrated, and this is due to non-integrability. 3) By the Euler Lagrange equation, $d_t \partial_{\dot{\theta_1}} L = \partial_{\theta_1} L$, that is, if the Lagrangian depends explicitly on the generalized coordinates, than the canonical momentum can't be conserved. Feel free to ask any other questions you have! $\endgroup$ Oct 22, 2014 at 18:26
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    $\begingroup$ @QuantumBrick no problem, knew you'd be interested in this. Well anyway, it was more out of curiosity for me, my own field of study has actually kinda nothing to do with this (condensed matter physics). Hope you make good progress with your thesis, don't hesitate asking if you ever think I can help out in any way. Cheers $\endgroup$
    – Ellie
    Nov 18, 2014 at 17:55

It is hard to "see" the integrability or non-integrability of a Lagrangian or Hamiltonian. One method to prove non-integrability is Melnikov's method, and for the physical pendulum in 2D this has been done in the paper "Melnikov's method applied to the double pendulum" by Holger Dullin, Zeitschrift für Physik B, 93:521-528, 1994, https://doi.org/10.1007/BF01314257

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    $\begingroup$ Since you are the author of the paper, perhaps you could flesh this answer out more - the answer is supposed to stand alone, not just be a pointer to a paper. $\endgroup$
    – Jon Custer
    Jul 31, 2019 at 21:14
  • $\begingroup$ OK Jon, about Melnikov's method: It requires that you have an integrable case that has a separatrix. Perturbing the integrable case you need to show that the separatrix splits and has transverse intersections of stable and unstable manifolds. This proves that the system is chaotic, and has no integral other than the energy. To apply this to the double pendulum you need to consider the physical double pendulum (i.e. with extended bodies with moments of inertia), so that an integrable limiting case can be found, e.g. where one pendulum is just a rotor. $\endgroup$ Sep 15, 2019 at 12:05

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