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The trajectory of a body in phase space $(x, v)$ is plotted above. I have observed there to be many crossings, or singular points, in the graph, but do not comprehend their meaning. Were an initial condition $(x_0, v_0)$ placed exactly at one of them, one should be then uncertain as to which path the body will next take–– this is quite strange to me, for I had considered that, in Classical Mechanics, the position and the velocity were always sufficient to determine the state and subsequent motion of the system. What meaning is there, then, to these crossings?

Edit: This phase portrait is one of a classical body whose position is given by $x(t) = 3\cos(3t+π/5) + 4\cos(4t+π/8)$. I shall assume $(x_0, v_0)$ to be sufficient to determine the motion of the system for any time which succeeds $t=0$.

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  • $\begingroup$ Can you specify the system this phase portrait belongs to? Side notes: the correct variables of a phase portrait are momentum and position, not velocity and position (the canonical momentum may not be proportional to the velocity e.g. if there is an electromagnetic field). Also, if the potential is time-dependent, then the tuple $(x, p)$ does not predict the future of the system. Otherwise, crossing points of trajectories in phase space are fixed points of the motion and the trajectories connecting them are called separatrices and separate qualitatively different motions. $\endgroup$ Jul 17 at 17:05
  • $\begingroup$ The phase portrait is of a mere classical body, in which it is certain that the potential is time-independent and in which the momentum is always proportional to the velocity. Its equation of motion, in particular, is given by the superposition of two cosines $3cos(3t+π/5) + 4cos(4t+π/8)$. Also, I do wish to inquire what is meant by a "fixed point of the motion"? $\endgroup$ Jul 17 at 17:11
  • $\begingroup$ If the potential is time-dependent then the interpretation of the phase-portrait becomes much more subtle and crossing points of the phase-trajectory are no longer immediately special. $\endgroup$ Jul 17 at 17:18
  • $\begingroup$ But, supposing the potential were independent of time, then would I say the crossing points to be those which bound different motions of the body, which the body could pursue should it have a suitable initial condition? $\endgroup$ Jul 17 at 17:23
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It is a statement that you find in textbooks that lines in the phase plane do not cross. Let's consider the simplest case, a system with 1 degree of freedom, so the phase space is 2-dimensional, (x.p). If, as one often assumes, the system conserves energy, then in particular energy is conserved along each line in the phase portrait. Different lines correspond to different energies; hence, they cannot intersect.

There are exceptions to this, though. Consider the double-well oscillator $ H[x,p] = {1\over 2} p^2 + V(x)$ with $V(x) = - {1\over 2} \omega^2 x^2 + {1\over 4}\lambda x^4$. If you plot the phase portrait for this potential, you will find that there is a cusp at $x = 0, p=0$. And, that is easy to understand. If you place the particle at the top of the hill, $(x=0)$, it will be stationary. With a small perturbation, the particle will run down into the well on one side or the other. So it is really true in this case that, if the particle is at $(0,0)$, we are uncertain about the subsequent evolution. This does not violate any mathematical theorem, because mathematically you can assume that the velocity is absolutely zero. But that situation is not physically realizable.

What is going on in this case, though, is different. The motion in Vera's question, $x(t) = 3 \cos(3t + \pi/5) + 4\cos(4t + \pi/8)$, is the solution of a system of two harmonic oscillators $H[x_1,x_2,p_1,p_2] = {1\over 2} p_1^2 + {1\over 2} p_2^2 +V(x_1,x_2)$, with $V(x_1,x_2) = {1\over 2} \omega_1^2 x_1^2 + {1\over 2} \omega_2^2 x_2^2$ and $\omega_1 = 3, \omega_2 = 4$, with the identification $x = x_1 + x_2$. This is an energy-conserving system, but it has a 4-dimensional phase space. The phase-space trajectories will not intersect in 4 dimensions. That is a consequence of energy conservation. However, the projections of these trajectories onto a 2-dimensional plane can certainly intersect. Crossings of trajectories in a projection of the full phase space have no meaning and do not raise any problems of principle. They indicate only that energy is being passed from the visible degrees of freedom to the hidden degrees of freedom.

There are other realizations of Vera's solution that involve time-dependent potentials or time-dependent forces. Here too, I would argue, the trajectories seem to cross because the true phase space has extra dimensions that are being ignored. In nature, all systems conserve energy. But, to have the constraints of energy conservation, you need to consider all of the relevant degrees of freedom, including those that create the time-dependent forces.

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    $\begingroup$ The "Cannot intersect" parts needs a bit of explanation IMO (+1 nevertheless): There may be phase lines that have the same energy (e.g. ones related by symmetries), that can approach fixed points of the system in a way that the phase portrait line for that energy has a kink (e.g. the limiting trajectories of a pendulum between oscillation, in other words the separatrix), of course they don't cross in that they approach the fixed points, but still, kinks in the trajectories for given energies, that look like crossing phase lines in the phase portrait, are possible. $\endgroup$ Jul 17 at 18:44
  • $\begingroup$ Sebastian: You are correct. This is exactly the situation in paragraph 2 of my answer. I thought that just giving this simple example makes the general point clear. $\endgroup$ Jul 17 at 18:59
  • $\begingroup$ You are right, I somehow overlooked that part. $\endgroup$ Jul 17 at 20:13
  • $\begingroup$ I thank you sincerely for your answer & do wish to at last ask what had been meant by a "fixed point" of a sytem. $\endgroup$ Jul 17 at 20:22
  • $\begingroup$ This is an excellent answer, but it’s incorrect that one does not know the evolution of the double well oscillator that starts at (0,0). That point is a equilibrium: a system placed there will stay there. Of course, this is not possible in practice, but at the level of the maths this is the case. $\endgroup$
    – Andrea
    Jul 17 at 21:26

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