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The phase trajectory of a one-dimensional simple harmonic oscillator is a closed one (In particular, it's an ellipse). Is closed phase trajectory a generic feature of any periodic motion at least in one-dimension? If it is, is there a simple proof of it?

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    $\begingroup$ Consider the meaning of 'periodic'. Though we sometimes generalize it for cases like damped harmonic motion. $\endgroup$ – dmckee Aug 28 '17 at 15:59
  • $\begingroup$ @dmckee Do you want to me define periodic motion? By periodic motion, I mean that the particle describes the same path again and again and reaches the same point of its path at fixed intervals of time. This is not true for damped harmonic motion. $\endgroup$ – SRS Aug 28 '17 at 16:03
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    $\begingroup$ @SRS So what does "closed trajectory" mean for you? $\endgroup$ – JMac Aug 28 '17 at 16:17
  • $\begingroup$ @JMac By closed trajectory I mean closed phase trajectory. Title edited. $\endgroup$ – SRS Aug 28 '17 at 16:18
  • $\begingroup$ Are you more concerned about whether the phase trajectories are closed, or whether they're ellipses? $\endgroup$ – Michael Seifert Aug 28 '17 at 19:52
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No. At least if I understand your question correctly.

Say, a simple harmonic oscillator has a $\cos$ position curve. Then the velocity curve is going to be $dp/dt$ which will be $\sin$, and so you get your ellipse when you plot them on a phase diagram.

But now consider saw-tooth motion. Then the velocity curve switches from a constant positive, to a constant negative ($dp/dt$ is constant, because the saw-tooth is simply a straight line).

If you plot position against velocity, you end up with a box (or two parallel lines, with an instantaneous jump between them). So closed figure, but not an ellipse.

Furthermore, while a closed loop is necessary, there's no reason that the loop needs to be simple. It can overlap with itself. @stafusa's answer adds the condition that the path be deterministic, but this condition, too, can be relaxed.

Considered constant speed motion moving from $-2\rightarrow1\rightarrow-1\rightarrow2\rightarrow2$. Here the phase diagram is a box that does a kind of loop on its way over.

Specifically, it's a path of straight lines going from (in $(p,v)$ space):

$(-2,1)\rightarrow(1,1)\rightarrow(1,-1)\rightarrow(-1,-1)\rightarrow(-1,1)\rightarrow(2,1)\rightarrow(2,-1)\rightarrow(-2,-1)\rightarrow(-2,1)$

Note that the line segment from $(-1,1)\rightarrow(1,1)$ is covered twice in this loop.

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    $\begingroup$ I don't know why this answer was downvoted. This is a reasonably valid example. $\endgroup$ – SRS Aug 28 '17 at 16:30
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    $\begingroup$ Real system describe continuous trajectories in phase space. The discontinuities of this system are idealizations that don't exist for real system. $\endgroup$ – dmckee Aug 28 '17 at 20:57
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    $\begingroup$ @dmckee, it's indeed a limit case, but a continuous approximation of it also makes Xorile's point that the closed curves must not be ellipses (I think a reference to a previous, less clear, version of the OP). Besides, an abrupt change in speed is commonplace in simple classical systems (think collisions). $\endgroup$ – stafusa Aug 29 '17 at 5:58
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    $\begingroup$ @SRS A real ball flexes. A real wall deforms. Then they rebound, but it takes finite time. Real systems have continuous phase trajectories. $\endgroup$ – dmckee Aug 29 '17 at 15:04
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    $\begingroup$ @dmckee, this is true, but the shape will be approximately a square in this condition. I read the earlier version of the question regarding whether the closed curve was required to be elliptical, and the answer to that is no, which this example shows. $\endgroup$ – Dr Xorile Aug 29 '17 at 16:44
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Yes.

  • A phase space trajectory of a smooth system$^1$ has to be a continuous curve.

  • For it to be called "periodic", the movement has to repeat itself, both velocity and position: i.e., it must come back to the same spot in phase space.

  • For a deterministic system, the current condition determines uniquely its future evolution, and there can not be "crossroads".

All of that together mean that, once a phase space trajectory comes back to where it was started, it's forced by determinism to run over the same path again, which forms a closed loop, given continuity.

enter image description here

Source: Reartes, W., Actas del XII Congreso Dr. Antonio A. R. Monteiro (2013), 2014, pp. 98$-$103 (PDF file).

Notice that pendulum rotations can be considered either divergent or periodic. They're divergent when the phase space is the plane, and periodic when it's a cylinder ($\theta = 0$ and $\theta=2\pi$ identified), which means that also rotations are closed curves (over the cylinder) when considered periodic.

As for a formal proof, most texts on ODE's and dynamical systems will have some. For example Hirsch-Smale-Devaney chapters 2 and 3, or the first six pages of Lebovitz textbook chapter on dynamical systems.

$^1$ Dr Xorile's answer shows this condition can be relaxed.

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  • $\begingroup$ According to Goldstein, there are two types of periodic motions- librations and rotations. For rotations, the phase trajectories are not closed because only the momentum variable returns to the same value at the same point of its path again and again but not position. I don't understand why. @stafusa $\endgroup$ – SRS Oct 9 '17 at 8:40
  • $\begingroup$ It's a choice to identify $\theta=0$ with $\theta=2\pi$: if you think that, after going around once, the pendulum is back at the same position, then you're making this identification and the phase space trajectory is closed. Goldstein chooses to let $\theta$ grow indefinitely $-$ that's a valid choice, but one that I see as not truly compatible with calling the movement "periodic". $\endgroup$ – stafusa Oct 9 '17 at 18:35

protected by Qmechanic Aug 28 '17 at 20:08

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