Liouville's theorem — phase space volume conservation: Exercises

This page is a workout for Liouville's theorem. You will lean on Hamilton's Equations, Phase Space, and Poisson Brackets. A few problems reach toward Statistical Mechanics — Ensembles, Canonical Transformations, Poincaré Recurrence Theorem, and Symplectic Geometry.

Before we start, one reminder of every symbol we use, in plain words:

Definition The cast of symbols

$q$ = a position coordinate. On our pictures it is the horizontal axis.
$p$ = a momentum coordinate. On our pictures it is the vertical axis.
A point $(q, p)$ = one complete "state" of the system — everything you'd need to predict its future.
$\overset{q}{˙}$ = "how fast $q$ changes per second"; $\overset{p}{˙}$ = "how fast $p$ changes per second". A dot on top always means rate of change in time.
$H$ = the Hamiltonian, usually the total energy written as a formula in $q$ and $p$ .
$ρ$ (Greek "rho") = density of points — how crowded the swarm of possible states is at a spot.
$\nabla \cdot (something)$ = the divergence, a number telling you whether a flow is spreading points apart (positive) or squeezing them together (negative). Zero means "no squeezing, no spreading".
${f, g}$ = a Poisson bracket, defined below when we first need it.

Level 1 — Recognition

L1.1

State, in one sentence each, (a) what quantity Liouville's theorem says is conserved and (b) the single geometric property of the flow that makes it true.

Recall Solution

(a) The phase-space volume of any region (equivalently the density $ρ$ measured while moving with a point, $d ρ / d t = 0$ ) is conserved. (b) The phase-space velocity field is divergence-free: $\nabla \cdot \dot{x} = 0$ . Points are neither compressed together nor spread apart.

L1.2

For the free particle $H = \frac{p ^{2}}{2 m}$ , write $\overset{q}{˙}$ and $\overset{p}{˙}$ using Hamilton's Equations, then compute the divergence $\frac{\partial q ˙}{\partial q} + \frac{\partial p ˙}{\partial p}$ .

Recall Solution

Hamilton's equations say $\overset{q}{˙} = \frac{\partial H}{\partial p} = \frac{p}{m}$ and $\overset{p}{˙} = - \frac{\partial H}{\partial q} = 0$ .

$\frac{\partial q ˙}{\partial q} = \frac{\partial}{\partial q} (\frac{p}{m}) = 0$ (no $q$ inside).
$\frac{\partial p ˙}{\partial p} = \frac{\partial}{\partial p} (0) = 0$ . Sum $= 0$ . Divergence-free ✓, so volume is conserved.

Level 2 — Application

L2.1

Free particle, $H = \frac{p ^{2}}{2 m}$ , $m = 2 kg$ . A rectangular blob at $t = 0$ has $q \in [0, 3]$ , $p \in [4, 6]$ (SI units). Find its area at $t = 0$ , and its area at $t = 5 s$ .

Recall Solution

At $t = 0$ : area $= Δ q Δ p = 3 \times 2 = 6$ . Evolution: each point moves $q \to q + \frac{p}{m} t = q + \frac{p}{2} (5) = q + 2.5 p$ . Momentum is unchanged ( $\overset{p}{˙} = 0$ ), so the vertical spread stays $Δ p = 2$ . The top edge ( $p = 6$ ) slides right by $2.5 \times 6 = 15$ ; the bottom edge ( $p = 4$ ) by $2.5 \times 4 = 10$ . The shape shears into a parallelogram, but shear along $q$ does not touch the height $Δ p$ . Area of a parallelogram $=$ base $\times$ perpendicular height. Base along $p$ is still $Δ p = 2$ ; the horizontal width of the base is still $Δ q = 3$ . $A (5) = Δ q \cdot Δ p = 3 \times 2 = 6.$ Same as $t = 0$ — Liouville confirmed. See the shear in

Figure — Liouville's theorem — phase space volume conservation

L2.2

Harmonic oscillator $H = \frac{p ^{2}}{2 m} + \frac{1}{2} m ω^{2} q^{2}$ . Directly verify divergence-free: compute $\frac{\partial q ˙}{\partial q} + \frac{\partial p ˙}{\partial p}$ .

Recall Solution

$\overset{q}{˙} = \frac{\partial H}{\partial p} = \frac{p}{m}$ , so $\frac{\partial q ˙}{\partial q} = 0$ (only $p$ inside). $\overset{p}{˙} = - \frac{\partial H}{\partial q} = - m ω^{2} q$ , so $\frac{\partial p ˙}{\partial p} = 0$ (only $q$ inside). Sum $= 0 + 0 = 0$ ✓. Notice the cancellation is trivial here because each velocity component depends only on the other coordinate — that is the generic reason Hamiltonian flows are incompressible.

Level 3 — Analysis

L3.1

Damped oscillator: $\overset{q}{˙} = \frac{p}{m}$ , $\overset{p}{˙} = - m ω^{2} q - γ p$ with friction $γ > 0$ . (a) Compute the divergence. (b) A blob has area $A_{0}$ at $t = 0$ ; give $A (t)$ . (c) With $γ = 0.4 s^{- 1}$ , $A_{0} = 10$ , find $A (t)$ at $t = 5 s$ .

Recall Solution

(a) $\frac{\partial q ˙}{\partial q} = 0$ , $\frac{\partial p ˙}{\partial p} = \frac{\partial}{\partial p} (- m ω^{2} q - γ p) = - γ$ . So $\nabla \cdot \dot{x} = - γ \neq = 0$ . Not Hamiltonian-conservative. (b) Volume obeys $\frac{d V}{d t} = \int \nabla \cdot \dot{x} d V = - γ V$ (constant divergence pulls out). This is exponential decay: $A (t) = A_{0} e^{- γ t} .$ The blob spirals into the origin attractor, area shrinking — energy is leaking out via friction. (c) $A (5) = 10 e^{- 0.4 \times 5} = 10 e^{- 2} \approx 10 \times 0.13534 = 1.3534$ . Picture the spiral-in in

L3.2

Poisson-bracket check. The Poisson bracket of two functions on phase space is defined as ${f, g} = \sum_{i} (\frac{\partial f}{\partial q _{i}} \frac{\partial g}{\partial p _{i}} - \frac{\partial f}{\partial p _{i}} \frac{\partial g}{\partial q _{i}}) .$ For the harmonic oscillator ( $N = 1$ ) take $ρ = H$ itself (a stationary density that is constant on energy shells). Show $\partial ρ / \partial t = - {ρ, H} = 0$ , so this $ρ$ is a valid equilibrium ensemble.

Recall Solution

With $ρ = H$ , compute ${ρ, H} = {H, H}$ . ${H, H} = \frac{\partial H}{\partial q} \frac{\partial H}{\partial p} - \frac{\partial H}{\partial p} \frac{\partial H}{\partial q} = 0.$ Any function bracketed with itself gives zero (the two terms are identical and subtract). So $\partial ρ / \partial t = - {ρ, H} = 0$ : the density does not change in time. Why this matters: in Statistical Mechanics — Ensembles, equilibrium ensembles are exactly the densities $ρ = ρ (H)$ that are functions of energy alone — they are stationary because they Poisson-commute with $H$ .

Level 4 — Synthesis

L4.1

A linear phase-space map advances a blob by the matrix $M = (a c b d), (q^{'} p^{'}) = M (q p) .$ The new area equals the old area times $∣ det M ∣$ . (a) State the condition on $M$ for the map to be area-preserving. (b) Is $M = (10 τ 1)$ (free-particle drift for time $τ$ , with $m = 1$ ) area-preserving? (c) Is $M = (20 0 0.5)$ area-preserving, and could it come from a Hamiltonian flow?

Recall Solution

(a) Area-preserving $⟺ ∣ det M ∣ = 1$ . In Symplectic Geometry the sharper condition for a genuine canonical map is $det M = + 1$ (a symplectic matrix); this both preserves area and orientation. (b) $det = 10 τ 1 = 1 \cdot 1 - τ \cdot 0 = 1$ . Area-preserving ✓ — this is exactly the shear of Example 1, for any $τ$ . (c) $det = 2 \times 0.5 - 0 = 1$ . Area is preserved (it stretches by $2 \times$ along $q$ and squeezes by $2 \times$ along $p$ — a perfect trade). And since $det = + 1$ it can arise from a Hamiltonian flow: this is the local picture of chaotic stretch-and-fold — a filament forms while area stays fixed.

L4.2

Two canonical maps are applied one after the other, $M_{2}$ then $M_{1}$ . Given $det M_{1} = 1$ and $det M_{2} = 1$ , show the composite still preserves area, and explain why this guarantees Liouville holds for the whole evolution, not just tiny steps.

Recall Solution

The composite map is the matrix product $M_{1} M_{2}$ . Determinants multiply: $det (M_{1} M_{2}) = det M_{1} \cdot det M_{2} = 1 \times 1 = 1.$ Area preserved ✓. The chain of reasoning: any finite-time evolution is a limit of many infinitesimal Hamiltonian steps, each with $det = 1$ . Multiplying any number of unit determinants gives $1$ . So volume is conserved over any time interval, not merely for instantaneous rates. This is the group structure of canonical transformations: they form a group under composition, and area-preservation is preserved by the group operation.

Level 5 — Mastery

L5.1

Poincaré recurrence. A Hamiltonian system is confined to a phase-space region of finite total volume $V_{tot}$ (e.g. fixed energy shell, bounded). Argue, using only Liouville's theorem, that almost every starting point must eventually return arbitrarily close to where it began. Then explain why a damped system (L3.1) is exempt.

Recall Solution

Take a tiny neighbourhood $U$ around a starting point, with volume $V (U) > 0$ . Let the flow for time $T$ be the map $ϕ_{T}$ . By Liouville, $ϕ_{T}$ preserves volume: every image $ϕ_{T} (U), ϕ_{2 T} (U), ϕ_{3 T} (U), \dots$ has the same volume $V (U)$ . Now suppose, for contradiction, these images never overlapped. Then infinitely many disjoint sets each of volume $V (U) > 0$ sit inside the finite region $V_{tot}$ . Their total volume would be $\infty \cdot V (U) = \infty > V_{tot}$ — impossible. So two images must overlap: $ϕ_{i T} (U) \cap ϕ_{j T} (U) \neq = \emptyset$ for some $i > j$ . Apply $ϕ_{- j T}$ (also volume-preserving): $U \cap ϕ_{(i - j) T} (U) \neq = \emptyset$ . Hence some point of $U$ returns to $U$ after time $(i - j) T$ . Shrinking $U$ shows return can be arbitrarily close. This is the Poincaré Recurrence Theorem. Damped exemption: in L3.1 volume is not preserved; blobs shrink toward the attractor. Images can pile up onto the same shrinking set, so the "disjoint boxes fill up the space" argument fails. No volume floor ⇒ no forced recurrence.

L5.2

Ensemble entropy is constant. The Gibbs entropy is $S = - k_{B} \int ρ ln ρ d Γ$ , where $d Γ$ is the phase-space volume element. Show that under Hamiltonian flow $\frac{d S}{d t} = 0$ , i.e. fine-grained entropy never changes. (Use $d ρ / d t = 0$ and volume conservation.)

Recall Solution

Move to coordinates that flow with the fluid (comoving frame). Two Liouville facts:

Along a trajectory $\frac{d ρ}{d t} = 0$ , so $ρ$ carried by a moving point is constant; hence $f (ρ) = - ρ ln ρ$ is also constant along that trajectory.
Volume is conserved: the little cell $d Γ$ each point carries keeps the same size. The integral $S = - k_{B} \int f (ρ) d Γ$ is a sum of (constant value $f (ρ)$ ) $\times$ (constant cell size $d Γ$ ) over the moving cells. Nothing in the sum changes, so $\frac{d S}{d t} = 0.$ Interpretation: microscopic (fine-grained) entropy is frozen — mechanics is reversible. The Second Law's entropy increase comes from coarse-graining (blurring the filaments that Liouville stretches out), not from the exact flow. This is a cornerstone of Statistical Mechanics — Ensembles.

Recall Final self-test (cover the answers)

Which determinant value marks an area-preserving linear map? ::: $∣ det M ∣ = 1$ (symplectic: $det = + 1$ ). Why does a damped system escape Poincaré recurrence? ::: Its flow is not volume-preserving, so images can pile onto a shrinking attractor — no volume floor to force overlaps. What is frozen by Liouville that makes fine-grained Gibbs entropy constant? ::: The comoving density $ρ$ ( $d ρ / d t = 0$ ) and the carried volume element $d Γ$ .