# Liouville's Theorem & Flows in Phase Space for Particle in a Box

A Hamiltonian system of $$100$$ interacting oxygen atoms, each of mass $$16$$ $$m_p$$, is confined within a cubical box of sides $$1 m$$. The average initial speed of each particle is $$300 ms^{-1}$$. Estimate the timescale for the system to return close to its initial state so that each particle is within $$0.1 cm$$ of its initial location and with a momentum vector $$\mathbf{p}$$ satisfying $$|\mathbf{p} − \mathbf{p}_{initial}| < 7.8 × 10^{-26} kg m s^{-1}$$.

Liouville's theorem in classical mechanics states that for a Hamiltonian system, the volume of a phase space element $$\Delta V = \Delta q_1\Delta q_2\Delta q_3...\Delta q_n\Delta p_1\Delta p_2\Delta p_3...\Delta p_n$$ is conserved in time. In our problem, $$q_i$$ is for the position of the $$i^{th}$$ particle, and $$p_i$$ is its momentum.

Focus on just $$1$$ atom in the box instead all $$100$$ atoms. The total volume of phase space that this particle could occupy is given by $$\Delta V_{total} = \Delta q \Delta p$$, and we know $$\Delta q = 1m^3$$ and (as a rough guess) that the spread of possible momenta will be about $$2$$ orders of magnitude around its starting momentum: $$\Delta p = 100p_{initial} = 3\cdot10^4m_p\ kgms^{-1}$$.

We are tasked with working out the time it takes for the particle to return to the phase space volume of $$\Delta V_{initial} = (0.1 \cdot 10^{-2}) \cdot (7.8 \cdot 10^{-26})$$. I have a good feeling that the ratio of volumes $$\Delta V_{initial} / \Delta V_{total}$$ gives how likely the particle is to occupy that volume of phase space at some point in the future, but I'm having a lot of trouble coming up with some way to extract a "time taken" out of it.

Even for the case of a single particle in a $$2D$$ box, I don't have many ideas (any answers may wish to refer to a 2D or 1D case to distill my question down a lot). Any explanation of where to go from here would be much appreciated.

A very rough estimate - consider a single $$d$$-Dimensional particle with Hamiltonian $$H = \sum_i \frac{p_i^2}{2m}+V(x)$$ where $$V(x)$$ defines the walls. The box has linear dimension $$L$$ and the particle initially has momentum of magnitude $$p$$. The available phase space is a cube of volume $$L^d$$ producted with (by energy conservation) a sphere of volume $$p^{d-1} \delta p$$ where we thicken the sphere for convenience and drop a dimensionless constant (that may be rather large).
By Lioville's theorem a blob of phase space of dimensions $$\Delta x^d \Delta p^d$$ retains its volume. There are $$N_{\rm cells} \sim \left(\frac{L}{\Delta x}\right)^d \left(\frac{p}{\Delta p}\right)^{d-1} \frac{\delta p}{\Delta p}$$ total cells of this size within the available phase space and the particle travels through one cells roughly every $$\Delta t = \frac{m\Delta x}{p} .$$ The recurrence time is then crudely given by $$T_{\rm Rec} = N_{\rm cells}\Delta t \sim \left( \frac{Lp}{\Delta x\Delta p} \right)^{d-1} \frac{Lm}{p} \frac{\delta p}{\Delta p}$$ the first factor is exponentially large in the dimension $$d$$ and thus the recurrence time will be very large. The second factor is roughly the system crossing time. The third factor includes the undefined quantity $$\delta p$$ which we used to blur the size of the constant energy sphere. A more careful treatment would I suppose restrict to the $$2d-1$$-dimensional surface properly but I suspect one can set $$\Delta p \sim \delta p$$ and then drop this as an $$\cal{O}(1)$$ quantity.