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In my statistical mechanics lecture, it was claimed that a volume of phase-space cannot be split into two separate volumes as time evolves.

I suspect that this is a topological fact that I am not familiar with, and I believe it amounts to the requirement that the image of a connected subset remains connected under time evolution generated by the Hamiltonian. Is this a consequence of the trajectories in phase space being continuous and unique, or is there another requirement on the time-evolution for this to hold? How would one go about showing this explicity?

Furthermore, since Liouville's theorem essentially states that the phase space density behaves like an incompressible fluid, does this imply that a connected volume of incompressible fluid will remain connected as well? If so, can this be physically justified?

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First of all we fix the general setup.

Let $M$ be the space of phases, I assume that all structures I am henceforth describing are $C^\infty$ (a $C^2$ Hamiltonian would be pretty sufficient for many issues actually, $C^3$ for the validity of Liouville theorem).

The Hamiltonian flow $\Phi$ is defined as follow. $$I_a \ni t \mapsto \Phi_t(a) \in M$$ is the maximal solution of Hamilton equations with initial condition $\Phi_0(a) = a \in M$ and the open interval $I_a\subseteq \mathbb{R}$ is the (maximal) temporal domain of that solution. Hence, the domain of $(t,a) \mapsto \Phi_t(a)$ is $$\Gamma := \bigcup_{a\in M} I_a \times \{a\} \subseteq \mathbb{R} \times M\:.$$ It is possible to prove that $\Gamma$ is open. Finally from the uniqueness theorem for solution of evolution equations the one-parameter group of local diffeomorphisms relations are valid, $$\Phi_0(a) = a\:, \quad \Phi_t(\Phi_u(a)) = \Phi_{t+u}(a)\tag{1}$$ the latter is true for $t,u\in \mathbb{R}$ and $a\in M$ such that the left-hand side is defined.

Now consider a ``volume'' $V\subset M$, i.e., an open connected subset. In general, there is no guarantee that its evolution $V_t:= \Phi_t(V) $ is defined for all $t\in \mathbb R$. However, using the fact that $\Gamma$ is open, one easily proves that, if $V$ is a sufficiently small (open connected) neigborhood of a state $a$ and $|t| <T$, for some sufficiently small constant $T>0$, then $\Phi_t(V)$ is well defined and open.

The map $\Phi_t : V \to \Phi_t(V)=:V_t$ is differentiable ad admits $\Phi_{-t}: V_t \to V$ as inverse. Since both maps are differentiable they are a fortiori continuous. Hence they are homeomorphisms between $V$ to $V_t$ and thus they preserve all topological properties.

In particular, $V_t$ is connected if $V$ is connected (actually to prove it it is sufficient that $\Phi_t$ is continuous, but here all topological properties are also preserved).

Final remarks.

  1. In some physically relevant cases $\Gamma = \mathbb{R} \times M$ so that the latter in (1) is valid for every $a\in M$ and all $t,u\in \mathbb{R}$ and $\Phi_t:M \to M$ is a global diffeomorphism. Sufficient conditions assuring it are that every maximal solution of Hamilton equations $\gamma : I_a \to M$ is included in a compact set (that may depend on the solution) $K_\gamma \subseteq M$. That is the case if, for instance the Hamiltonian does not depend on time exicitly and its level sets are compact. In these cases no restrictions exist on $V$ and $t$ to define $V_t$.

  2. The Liouville theorem has nothing to do with the preservation of conectedness, as what I said above is true for every dynamical system, even on a manifold $M$ with odd dimension, where no Hamiltonian formulation is possible (in the sense of symplectic formulation).

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