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

Question

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.

Answer 1

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.