Page 272 of No-Nonsense Classical Mechanics sketches why paths in phase space can never intersect:

Problem: It seems to me this reasoning only implies that paths can never "strictly" intersect, in the sense that two points in phase space pass through the same point (at $t$), and then split into two distinct paths.

Question: Can't two paths in phase space merge into one single path? It seems to avoid this argument entirely.

Rigor: To make this more rigorous: suppose for sake of argument that two paths "strictly" intersect. Let the first path be $(q_1, p_1)$ and the second path be $(q_2, p_2)$. Suppose at time $t$ there is an intersection: $q_1 = q_2$ and $p_1 = p_2$ (I'm abusing notation here by referring to $q$ and $p$ as both paths and points). Since there is an intersection between two distinct paths, then $dq_1/dt ≠ dq_2/dt$. Yet according to Hamilton's equations: $dq_1/dt = ∂H/∂p_1 = ∂H/p_2 = dq_2/dt$, which is a contradiction. But notice this argument doesn't work when two paths "merge" into one path, since in that case we couldn't say that $dq_1/dt ≠ dq_2/dt$ at the point that the paths merge.

Thus it seems entirely consistent with Hamilton's equations that two paths in phase space could merge into one path.


3 Answers 3


Your logic is correct. Often in classical mechanics, though, we restrict our attention to conservative forces, in which case the dynamics have time-reversal symmetry, that is, the system should have a unique phase-space path whether you go forwards or backwards. For merging paths, if the system is at a point past the merger and you reverse time, it's not clear which branch the system will take when it gets to the fork. On the other hand, merges are possible if there is dissipation. Imagine a system subject to friction with no driving - all starting conditions will lead to the same limit point, with the system at rest, which is to say that they will merge (at least, given infinite time).

There's another problem with finite-time mergers, however - at least one of the merging paths has to have a kink at the point of merger. That's unphysical, because it corresponds to an infinite time derivative in either position or velocity (so an infinite velocity or infinite acceleration). Still, that might be close enough if the time scales on which dissipation occurs are short so we can allow for "approximately infinite" forces that still only impart finite impulses, e.g. if your system is made up of balls of putty that stick together in almost-instantaneous inelastic collisions.

  • 3
    $\begingroup$ The part about there needing to be a kink in the phase space trajectory where there might be a merger is incorrect. Mathematically, there does not need to be a discontinuity in any variable, or even any of their derivatives (first, second, etc.). $\endgroup$
    – Buzz
    Sep 11, 2020 at 2:46
  • 1
    $\begingroup$ FWIW, dissipative systems usually don't have a Hamiltonian formulation, cf. e.g. this Phys.SE post. $\endgroup$
    – Qmechanic
    Sep 11, 2020 at 11:41
  • $\begingroup$ @Buzz Hmm, I guess you're right. Yet the system on the two merging paths behave differently an infinitesimal distance away from the merger (or from each other), and then behave the same just after - that seems like it would involve a discontinuity in the forces with respect to either the positions or velocities (e.g. a friction force bringing an object to rest and then ceasing to act), though I can't show it rigorously off the top of my head. $\endgroup$
    – pwf
    Oct 2, 2020 at 16:25

TL;DR: The solutions to a first-order ODEs never intersect transversally as OP already noted. The absence of merging/splitting paths (i.e. intersecting tangentially) is guaranteed by the local uniqueness of first-order ODEs. A sufficient condition is that the evolutionary vector field $X_H$ should be Lipschitz continuous, cf. the Picard-Lindelöf theorem.


  1. Here is a 2D phase space example of 2 merging/splitting paths $$ q(t)~=~\pm \frac{t^2}{2}, \qquad p(t)~=~\pm t,\tag{1} $$ in the origin (0,0). Let the Hamiltonian be $$ H~=~\frac{p^2}{2}-|q|.\tag{2} $$ Hamilton's equations read $$ \dot{q}~=~p, \qquad \dot{p}~=~{\rm sgn}(q), \tag{3}$$ which have the solutions (1).

  2. In the above example 1 the 2 paths are meeting head-on. It is possible to change the time-direction of one of the paths so that the 2 paths
    $$ q(t)~=~\pm \frac{t^2}{2}, \qquad p(t)~=~ t, \tag{4} $$ are merging/splitting from the same direction. Namely consider the modified Hamiltonian $$ H~=~\frac{p^2}{2}{\rm sgn}(q)-|q| \tag{5} $$ instead. Hamilton's equations then read $$ \dot{q}~=~p~{\rm sgn}(q), \qquad \dot{p}~=~{\rm sgn}(q)-p^2\delta(q), \tag{6}$$ which have the solutions (4).

  3. If you allow one of the paths to be a constant path, then another example is Norton's dome, cf. e.g. this Phys.SE post.


You are right I think. The idea of attractors is similar to this. In case of attractors, you can start from any initial state and yet eventually evolve to a unique final state. From what I understand, it is only under some very special conditions when systems with vastly different initial condition tend to evolve into a unique final state. But that is not always true. It happens only when a good amount of information is lost in the evolution process and the phase-space curves merge into one unique solution. I think you are specifically asking about this case. In that case, my answer will be yes, it is possible for two phas-space curves can merge after some time. But the system can also diverge from one another. I would suggest you to take a look at Hydrodynamic attractors in phase space for a general idea and also at Exact solutions and attractors of higher-order viscous fluid dynamics for Bjorken flow for some insight into the possible conditions when phase space curves of systems with different initial conditions may or may not merge into one.

  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – ohneVal
    Sep 11, 2020 at 16:14
  • $\begingroup$ Thanks for the comment. I have made changes. Hope it makes things a bit clearer now. $\endgroup$ Sep 15, 2020 at 5:01

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