Consider the following idealized motions

(i) The motion of a bob attached to a spring on a horizontal frictionless table,

(ii) the motion of a pendulum with an arbitrary amplitude without air resistance,

(iii) the motion of the earth around the Sun,

(iv) the motion of a ball dropped from a height that hits the ground elastically so that there is no loss of energy into heat and every time it bounces back to the same height.

The only commonality between these motions is that all of them are conservative systems and all of them are periodic. Though the mathematical forms of the potentials very different, all are conservative systems without dissipation. From these examples, my strong hunch is that a necessary condition for periodicity must be that the system must be conservative. Apart from that, all these motions are examples of motion that are either effectively one-dimensional motion or two-dimensional planar motion.

  • Is being conservative the only necessary condition for a motion to be periodic?

  • Is being conservative also the sufficient condition for a motion to be periodic?

Conditions for periodic motions in classical mechanics is a pre-existing answer. The accepted answer is packed with jargons (integrable motion, regular motion, quasi-periodic motion etc) and is unintelligible for me.

  • $\begingroup$ Motion in any number of dimensions is periodic if it has a definite non-zero time period. $\endgroup$
    – Sam
    Feb 3, 2020 at 14:29
  • 1
    $\begingroup$ there are conservative systems that are non periodic, imagine a billiard ball in this situation plus.maths.org/content/chaos-billiard-table $\endgroup$
    – user65081
    Feb 3, 2020 at 14:48
  • $\begingroup$ @Wolphramjonny Two cases (both left and right) of the second diagram on the left are periodic. Which diagram are you talking about? Thanks for the reference. $\endgroup$ Feb 3, 2020 at 14:54
  • $\begingroup$ It’s definitely not sufficient. An object under no force need not exhibit periodic behaviour. $\endgroup$ Feb 3, 2020 at 15:08
  • 2
    $\begingroup$ The 2D double pendulum is a conservative system which can be chaotic and therefore aperiodic for some sets of starting conditions. $\endgroup$ Feb 3, 2020 at 19:52

3 Answers 3


There's no relation between being conservative and being periodic:

Even a simple damped harmonic oscillator can display periodic motion as well: namely, that of rest.

There are a few analytical methods for figuring out whether a periodic orbit exists or not, but most often one has to resort to numerics.


A periodic motion is a trajectory in phase space which visits the same phase space point more then once. Any system in which there are trajectories in phase space that visit the same point more than once must be conservative (because if a non-conservative system can revisit a point in its phase space then by going round a loop in phase space in one direction or the other you could gain energy). Therefore the only systems that allow periodic motions are conservative systems.

However, not all motions in conservative systems are periodic. For example, the two-body problem has periodic solutions, but it also allows non-periodic hyperbolic trajectories.


I'm not quite sure what you mean with "conservative". But I guess you mean there must be no loss of energy.

Very generalized: If you lose energy, your system can never be in the same state again, unless you have a constant source of energy attached to the system, eg a driven pendulum.

Edit: I missed something when reading the question the first time.

No, the system only being conservative is not sufficient to make it periodic. Throw a ball into space, it will fly away and not come back. But the other way around it works. Conservatitivism is a requirement for periodicity, which i previously thought was the question.

  • $\begingroup$ so the answer is no, as your own example says $\endgroup$
    – user65081
    Feb 3, 2020 at 14:36
  • $\begingroup$ the answer on what? $\endgroup$
    – Leviathan
    Feb 3, 2020 at 14:40
  • $\begingroup$ that being conservative is not a necessary or sufficient condition $\endgroup$
    – user65081
    Feb 3, 2020 at 14:41
  • $\begingroup$ depends on what you interpret as the system. the motor that drives your system is part of the system. to make it short: the system has to be conservative. in real life applications losses can be overcome by attaching sources of energy. $\endgroup$
    – Leviathan
    Feb 3, 2020 at 14:44

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