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What might be a simple proof showing that the time evolution of the phase space volume can't lead to splitting off of the phase space volume?

By Liouville's theorem, the total phase space volume is conserved, but does it also say that a connected phase space volume will stay connected forever under the Hamiltonian flow?

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  • $\begingroup$ Comment to the question (v1): Could you phrase more precisely the statement that you want to prove? $\endgroup$ – Qmechanic Sep 17 '14 at 13:47
  • $\begingroup$ Well, it's like this: A lump of volume in phase space if time evolved by Liouville operator preserves volume and the topology; It can never be split off in two volumes adding upto the original volume. Does this make sense? $\endgroup$ – Shuppar Sep 18 '14 at 12:18
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    $\begingroup$ Comment to the question (v1): It seems OP's question is not so much about Liouville's theorem and the conservation of phase space volume. It seems to be more about why a connected subset $R$ of phase space stays connected under time-evolution. This follows if the evolution flow is a continuous map. $\endgroup$ – Qmechanic Sep 18 '14 at 13:33
  • $\begingroup$ Yes, exactly! Can you please elaborate a little more.. I mean how is this continuous time evolution thing obvious from the form of the equation? $\endgroup$ – Shuppar Sep 19 '14 at 7:12
  • $\begingroup$ Let's say my original connected subset R is a set of initial conditions for a pencil very nearly balanced on its tip. Doesn't this split into two disconnected parts, one in which the pencil falls to the right and one in which it falls to the left? $\endgroup$ – user4552 Oct 21 '14 at 22:28
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Here's something I believe is a simple proof. Unfortunately it uses a little bit of cohomology.

Consider the canonical 2-form in extended phase space $T^*M \times \mathbb{R}$

$$\omega = \sum_{i=1}^N dq_i \wedge dp_i - dH(\vec{q},\vec{p},t) \wedge dt ,$$

where $N = dim(M)$. A function $f: M \to M$ is said to be a canonical transformation iff $f^* \omega = \omega$. Thus, for any canonical transformation,

$$\sum_{i=1}^N dq_i \wedge dp_i - dH(\vec{q},\vec{p},t) \wedge dt = \sum_{i=1}^N dQ_i \wedge dP_i - dK(\vec{Q},\vec{P},T) \wedge dT ,$$

where we defined $(\vec{Q},\vec{P},T) = f^*(\vec{q},\vec{p},t)$. This means that

$$\sum_{i=1}^N q_i dp_i - H(\vec{q},\vec{p},t) dt - \left( \sum_{i=1}^N Q_i dP_i - dK(\vec{Q},\vec{P},T) dT \right) = dG ,$$

i.e, that the tautological forms associated to the canonical 2-forms are elements of the same de Rham cohomology class. By the homotopic invariance of the de Rham cohomology,

$$\oint_\gamma \left( \sum_{i=1}^N q_i dp_i - H(\vec{q},\vec{p},t) dt \right) = \oint_{\Gamma} \left( \sum_{i=1}^N Q_i dP_i - dK(\vec{Q},\vec{P},T) dT \right) ,$$

where $\Gamma$ is the image of the curve $\gamma$ by $f$.

As movement can be considered to be a canonical transformation, we just proved that the hamiltonian evolution in extended phase space needs to relate only homotopic curves. A corollary is that if you start with a compact simply-connected set in phase space and track its evolution in time, the curve that bounds this set can only evolve to others homotopic to it. As a simply connected set boundary is never homotopic to a non-simply connected one, we just showed that hamiltonian evolution takes simply-connected sets to simply-coonected sets.

I'm really sorry if this is too technical. I'm too ignorant to offer a proof with simpler arguments (which points out how limited is my knowledge on the subject).

EDIT: Small correction on dimension.

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  • $\begingroup$ As soon as you state that the manifold id $T^*M \times \mathbb R$ ($\mathbb R$ is the line of time) you are already saying that the topology of the space of the phases $T^*M$ is constant! It seems to me that you are assuming the thesis in the hypothesis... $\endgroup$ – Valter Moretti Oct 22 '14 at 15:34
  • $\begingroup$ @ValterMoretti I might be wrong, but I think that the fact that the evolution takes place on the cotangent bundle $\times$ the axis of time has nothing to do with fixing a topology. You could have a bifurcation that would, for a given time $t$, create a whole inside a set. The process would still take place in this product manifold. I can only argue this doesn't happen because of the "proof" I gave. I'd be thrilled if you explained your argument a little deeper! $\endgroup$ – QuantumBrick Oct 22 '14 at 15:58
  • $\begingroup$ Perhaps we are interpreting the question into two different ways. My interpretation is if the phase space can change its topology. So you should think of the spacetime of phases as a foliated manifold $\cup_{t\in \mathbb R} T^*M_t$ where $M_t$ and $M_{t'}$ (different disjoint leaves) may or may not have different topology. $\endgroup$ – Valter Moretti Oct 22 '14 at 16:05
  • $\begingroup$ Actually the $T^*M_t$ could be replaced by generic symplectic manifolds without requiring that they are cotangent spaces. This allows to have compact phase spaces. In this case the topology must, in fact, be constant as I tried to explain in my answer. $\endgroup$ – Valter Moretti Oct 22 '14 at 16:09
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    $\begingroup$ @Shuppar I used something called "Cohomology Theory", in particular, I used the homotopic invariance of a cohomology (de Rham's) that's defined for differentiable manifolds. Homology/Cohomology are topics in Algebraic Topology, and they consist on studying the "holes" in surfaces based on abelian groups. My main argument is that a certain invariant in this theory, the Cohomology group, is different for a simply-connected and a non-simply connected set boundaries. It's an algebraic way of proving a topological result. Check en.wikipedia.org/wiki/De_Rham_cohomology. $\endgroup$ – QuantumBrick Oct 23 '14 at 15:35
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By the main theorem of connectedness in general topology, continuous maps preserve connectedness. Time evolution of Hamiltonian systems preserves connectedness because it is continuous. I think it is independent of from Liouville's theorem, it just requires the proving Hamiltonian time evolution is continuous.

This is just a formal way of restating @StevenMathey's answer.


Liouville's theorem states that time evolution of Hamiltonian systems is volume preserving. Volume preservation and continuity are independent properties.

Consider these simple non-Hamiltonian systems: $$ \begin{align} & && \text{volume preserving} && \text{continuous} \\ \dot{q} &= sgn(q) && + && -\\ \dot{q} &= -q && - && + \end{align} $$ Since these are 1st order systems, continuity & volume preservation can be judged simpy wrt to the configuration space, or wrt the phase space like for Hamiltonian systems.

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  • $\begingroup$ Your idea works if the phase of spaces is compact at each time. Time evolution is described by a local one parameter group. It means that the domain of time variable is not globally defined. If the space of times is not compact at every time the domain changes and you could arrange situations where there is no a common domain allowing changes of topology... $\endgroup$ – Valter Moretti Oct 22 '14 at 15:40
  • $\begingroup$ +1 I realized that I interpreted in a too restrictive way the question. If you are referring to a volume in the phase space which is contained in a common time domain of the local Hamiltonian flow, you are right. The Hamiltonian flow is continuous as a general result of continuous dependence on initial conditions of a differential equation system in normal form with continuous and locally Lipschitz in the non-temporal variable right-hand side. $\endgroup$ – Valter Moretti Oct 22 '14 at 17:46
  • $\begingroup$ Well, I think I am misinterpreting your answer. You said that the time evolution of Hamiltonian system preserves conectedness and it has nothing to do with the Liouville's theorem. Is that you want to say? But, then, is not Liouville's theorem all about the time evolution of phase points? I mean, does not it define the time evolution of the Phase points, then how can these two things be independent? $\endgroup$ – Shuppar Oct 23 '14 at 4:32
  • $\begingroup$ @Shuppar I added an explanation of the independence of volume preservation and continuity. $\endgroup$ – Daniel Mahler Oct 23 '14 at 18:55
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I can not prove anything. I hope that this hand-waving argument is satisfying for you.

Hamiltonian dynamics is a continuous process. Two neighbouring points of the phase space can only move continuously away from each other. There is no sudden jump.

The phase space can not be split in disjunct ensemble by a finite time time evolution because if this was the case then there would be a border in between the disjunct regions. This is only possible if this border is already present in the initial conditions.

Imagine you have a single closed set of initial conditions in phase space. If at some time point this set was broken in two, then it must contain some points that go on one side and some points that go on the other although they are infinitesimally close to each other at the beginning of the time evolution. This implies that you have a discontinuous jump at some point.

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  • $\begingroup$ I don't understand your border argument. Can you please explain it further; I mean what exactly you mean when you say border. You know what I mean.. Thanks! $\endgroup$ – Shuppar Sep 18 '14 at 12:23
  • $\begingroup$ If the phase space is split in two at some time, then there is a line passing through the phase space at t=0 (when it is still connected) which separates the points that will go to one side from the ones that will go to the other. Then you can start by asking where do the points that are exactly on the line go? Moreover, two points that are separated by this line can be chosen as close to each other as you want. If this line existed, they would end up far away from each other whatever their original distance. $\endgroup$ – Steven Mathey Sep 18 '14 at 12:59
  • $\begingroup$ Hmm... So in that case can you say that the initial volume was continuous? I mean, I can grasp the nub of the argument... but can it be proved somehow mathematically from the form of the equation? $\endgroup$ – Shuppar Sep 19 '14 at 7:16
  • $\begingroup$ Yes and yes, its called Liouville's theorem. $\endgroup$ – Steven Mathey Sep 19 '14 at 9:50
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    $\begingroup$ @StevenMathey: Then you can start by asking where do the points that are exactly on the line go? Can't they just stay where they are, i.e., this is an equilibrium? Moreover, two points that are separated by this line can be chosen as close to each other as you want. If this line existed, they would end up far away from each other whatever their original distance. Isn't this exactly what we expect if it's an unstable equilibrium? $\endgroup$ – user4552 Oct 21 '14 at 23:31
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REMARK. Perhaps I wrongly interpreted the question. I interpreted it as if were referred to the total volume of phase space.

The answer is negative if the question regards general changes in time of topology of the total space of phases and if you do not impose any generic restriction on the topology of the spaces, like compactness (see the final comment). This is due to basic facts much more elementary than Liouville theorem.

Consider the following situation: A particle which is free except for the fact that it is constrained to stay on the frictionless curve obtained by the intersection of a cone and a plane, whose slope depends on time along a know law. This curve passes from a circle to a hyperbola in a finite lapse of time, changing its topology. There are no difficulties in translating this problem from Newtonian mechanics to Lagrangian mechanics adopting local coordinate frames defining an atlas on a manifold which is a $2D$ manifold foliated over the axis of time $\mathbb R$, I mean $\cup_{t \in \mathbb R}Q_t$, where each leaf $Q_t$ is the configuration space at time $t$ (this overall manifold is not a fiber bundle over $\mathbb R$ consequently since the leaves are not diffeomorphic and it is not a local Cartesian product). The point is that the change of topology happens at infinity: The circle $Q_t$ becomes larger and larger as $t$ increases and, for a certain time, $t_0$, it becomes diffeomorphic to a line. Furthermore, in this model, no initial condition for a point on the curve can generate a motion which reaches the infinity, where the change of topology happens, in a finite lapse of time. So none can see the change of topology. (This strongly depends on the dynamics, no blow up solutions must be permitted.)

Since the passage from Newtonian mechanics to Lagrangian mechanics is performed in local coordinates, there is no problem in writing down the Lagrangian of the particle in local coordinates $t,q$. The spacetime of configurations, $stQ$, exists as a smooth manifold though the space of configurations $Q_t$ (which is a smooth embedded submanifold of $stQ$ for every time) changes topology in time.

The position vector in the physical $3$-space is $\vec{x}= \vec{x}(t,q)$ where the appearance of $t$ is due to the fact that the constraint change its form in time. This Lagrangian has always the form $$L(t, q, \dot{q})= \frac{1}{2} m\vec{v}^2= \frac{m}{2} \frac{\partial \vec{x}}{\partial q}\cdot \frac{\partial \vec{x}}{\partial q} \dot{q}^2 + m \frac{\partial \vec{x}}{\partial q}\cdot \frac{\partial \vec{x}}{\partial t} \dot{q} + \frac{m}{2} \frac{\partial \vec{x}}{\partial t}\cdot \frac{\partial \vec{x}}{\partial t}\:.$$ This Lagrangian is defined on a manifold which has the form $\cup_{t \in \mathbb R} Q_t \times \mathbb R$ (the last factor is the domain of $\dot q$). The topology of the sections labeled by $t$ changes in time as, again, the first homotopy group passes from $\mathbb Z$ to $1$. Since the quadratic form $\frac{m}{2} \frac{\partial \vec{x}}{\partial q}\cdot \frac{\partial \vec{x}}{\partial q}$ is non-degenerate, the Legendre transformation is well-defined: $$p = \frac{\partial \vec{x}}{\partial q}\cdot \frac{\partial \vec{x}}{\partial q} \dot{q}$$ and the Hamiltonian approach can be implemented. Each local coordinate patch $t,q, \dot{q}$ gives rise to a local coordinate patch $t,q,p$. The overall manifold, the spacetime of phases $stF$ is again of the form $\cup_{t \in \mathbb R} Q_t \times \mathbb R$ (the last factor is now the domain of $p$). As before, the topology of each space of phases $F_t = Q_t \times \mathbb R$ changes in time because, again, the first homotopy group passes from $\mathbb Z$ to $1$.

By means of a similar argument i think one could pass form a connected configuration space to a non-connected one so that the space of phases similarly passes from a connected topology to a disconnected one. Think of a free point except for the fact that it is constrained to stay on a frictionless curve $Q_t$ of the form $\subset \hskip -3pt \supset$ whose length (I mean distance from $\subset$ and $\supset$) increases in time becoming infinite in a finite lapse of time, from then on the curve separates into a pair of parallel lines. This can be described as before in a smooth spacetime of configurations of the form $\cup_{t \in \mathbb R} Q_t$.

If the space of phases $F_t$ is compact for every $t$, then no changes of topology are permitted at all. This is because compactness permits to write a global Hamiltonian flow on a sufficiently small interval $(t_0-\epsilon, t_0+\epsilon)$ of time around each fixed time $t_0$. In other words, there is a diffeomorphism $\phi : F_\tau \to F_{\tau'}$ if $\tau,\tau'\in (t_0-\epsilon, t_0+\epsilon)$. Since diffeomorphisms are homeomorphisms, the topology of $F_\tau$ and $F_{\tau'}$ must coincide. It is clear that this argument, accumulating intervals as $(t_0-\epsilon, t_0+\epsilon)$, implies that the topology of $F_\tau$ and $F_{\tau'}$ must coincide for every choice of $\tau,\tau' \in \mathbb R$.

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  • $\begingroup$ +1 I don't think this answers the original question, but it's a great answer about general topological invariants. $\endgroup$ – QuantumBrick Oct 22 '14 at 18:31

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