Liouville's theorem — phase space volume conservation: Exercises

2.1.14 · D4 · Physics › Analytical Mechanics › Liouville's theorem — phase space volume conservation

Yeh page Liouville's theorem ki practice ke liye hai. Tum Hamilton's Equations, Phase Space, aur Poisson Brackets par rely karoge. Kuch problems Statistical Mechanics — Ensembles, Canonical Transformations, Poincaré Recurrence Theorem, aur Symplectic Geometry ki taraf bhi jaati hain.

Shuru karne se pehle, ek reminder — har symbol ka plain words mein matlab:

Definition Symbols ki cast

$q$ = ek position coordinate. Humare pictures mein yeh horizontal axis hai.
$p$ = ek momentum coordinate. Humare pictures mein yeh vertical axis hai.
Ek point $(q, p)$ = system ki ek poori "state" — sab kuch jo uska future predict karne ke liye chahiye.
$\overset{q}{˙}$ = " $q$ kitni tezi se change ho raha hai per second"; $\overset{p}{˙}$ = " $p$ kitni tezi se change ho raha hai per second". Upar dot ka matlab hamesha time mein rate of change hota hai.
$H$ = Hamiltonian, usually total energy jo $q$ aur $p$ mein formula ke roop mein likhi hoti hai.
$ρ$ (Greek "rho") = points ki density — possible states ke swarm ki crowding ek jagah par.
$\nabla \cdot (something)$ = divergence, ek number jo batata hai ki flow points ko alag kar raha hai (positive) ya squeeze kar raha hai (negative). Zero ka matlab hai "na squeezing, na spreading".
${f, g}$ = ek Poisson bracket, neeche define kiya gaya hai jab pehli baar zarurat padegi.

Level 1 — Recognition

L1.1

Ek-ek sentence mein batao, (a) Liouville's theorem kaunsi quantity conserved kehta hai aur (b) flow ki woh single geometric property kya hai jo ise true banati hai.

Recall Solution

(a) Kisi bhi region ka phase-space volume (ya equivalently density $ρ$ jo ek point ke saath chaalte hue measure ki jaaye, $d ρ / d t = 0$ ) conserved hota hai. (b) Phase-space velocity field divergence-free hoti hai: $\nabla \cdot \dot{x} = 0$ . Points na compress hote hain, na spread hote hain.

L1.2

Free particle $H = \frac{p ^{2}}{2 m}$ ke liye, Hamilton's Equations use karke $\overset{q}{˙}$ aur $\overset{p}{˙}$ likho, phir divergence $\frac{\partial q ˙}{\partial q} + \frac{\partial p ˙}{\partial p}$ compute karo.

Recall Solution

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

$\frac{\partial q ˙}{\partial q} = \frac{\partial}{\partial q} (\frac{p}{m}) = 0$ (andar koi $q$ nahi).
$\frac{\partial p ˙}{\partial p} = \frac{\partial}{\partial p} (0) = 0$ . Sum $= 0$ . Divergence-free ✓, isliye volume conserved hai.

Level 2 — Application

L2.1

Free particle, $H = \frac{p ^{2}}{2 m}$ , $m = 2 kg$ . $t = 0$ par ek rectangular blob ka $q \in [0, 3]$ , $p \in [4, 6]$ (SI units) hai. $t = 0$ par uska area nikalo, aur $t = 5 s$ par bhi.

Recall Solution

$t = 0$ par: area $= Δ q Δ p = 3 \times 2 = 6$ . Evolution: har point move karta hai $q \to q + \frac{p}{m} t = q + \frac{p}{2} (5) = q + 2.5 p$ . Momentum unchanged rehta hai ( $\overset{p}{˙} = 0$ ), isliye vertical spread $Δ p = 2$ locked hai. Top edge ( $p = 6$ ) $2.5 \times 6 = 15$ right slide karta hai; bottom edge ( $p = 4$ ) $2.5 \times 4 = 10$ right. Shape shear hokar parallelogram banta hai, lekin $q$ ke along shear height $Δ p$ ko touch nahi karta. Parallelogram ka Area $=$ base $\times$ perpendicular height. $p$ ke along base abhi bhi $Δ p = 2$ hai; horizontal width abhi bhi $Δ q = 3$ hai. $A (5) = Δ q \cdot Δ p = 3 \times 2 = 6.$ $t = 0$ jaisa hi — Liouville confirmed. Shear dekho

Figure — Liouville's theorem — phase space volume conservation

mein.

L2.2

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

Recall Solution

$\overset{q}{˙} = \frac{\partial H}{\partial p} = \frac{p}{m}$ , isliye $\frac{\partial q ˙}{\partial q} = 0$ (andar sirf $p$ hai). $\overset{p}{˙} = - \frac{\partial H}{\partial q} = - m ω^{2} q$ , isliye $\frac{\partial p ˙}{\partial p} = 0$ (andar sirf $q$ hai). Sum $= 0 + 0 = 0$ ✓. Notice karo ki cancellation yahan trivial hai kyunki har velocity component sirf doosre coordinate par depend karta hai — yahi generic reason hai ki Hamiltonian flows incompressible hote hain.

Level 3 — Analysis

L3.1

Damped oscillator: $\overset{q}{˙} = \frac{p}{m}$ , $\overset{p}{˙} = - m ω^{2} q - γ p$ friction $γ > 0$ ke saath. (a) Divergence compute karo. (b) Ek blob ka area $t = 0$ par $A_{0}$ hai; $A (t)$ do. (c) $γ = 0.4 s^{- 1}$ , $A_{0} = 10$ ke saath, $t = 5 s$ par $A (t)$ nikalo.

Recall Solution

(a) $\frac{\partial q ˙}{\partial q} = 0$ , $\frac{\partial p ˙}{\partial p} = \frac{\partial}{\partial p} (- m ω^{2} q - γ p) = - γ$ . To $\nabla \cdot \dot{x} = - γ \neq = 0$ . Nahi hai Hamiltonian-conservative. (b) Volume yeh obey karta hai $\frac{d V}{d t} = \int \nabla \cdot \dot{x} d V = - γ V$ (constant divergence bahar aa jaata hai). Yeh exponential decay hai: $A (t) = A_{0} e^{- γ t} .$ Blob origin attractor mein spiral karta hai, area shrink hota hai — energy friction ke through leak ho rahi hai. (c) $A (5) = 10 e^{- 0.4 \times 5} = 10 e^{- 2} \approx 10 \times 0.13534 = 1.3534$ . Spiral-in

mein dekho.

L3.2

Poisson-bracket check. Phase space par do functions ka Poisson bracket aise define hota hai: ${f, g} = \sum_{i} (\frac{\partial f}{\partial q _{i}} \frac{\partial g}{\partial p _{i}} - \frac{\partial f}{\partial p _{i}} \frac{\partial g}{\partial q _{i}}) .$ Harmonic oscillator ( $N = 1$ ) ke liye $ρ = H$ hi lo (ek stationary density jo energy shells par constant hai). Dikhao ki $\partial ρ / \partial t = - {ρ, H} = 0$ , isliye yeh $ρ$ ek valid equilibrium ensemble hai.

Recall Solution

$ρ = H$ ke saath, ${ρ, H} = {H, H}$ compute karo. ${H, H} = \frac{\partial H}{\partial q} \frac{\partial H}{\partial p} - \frac{\partial H}{\partial p} \frac{\partial H}{\partial q} = 0.$ Koi bhi function khud se bracket kiya jaaye to zero deta hai (dono terms identical hain aur subtract ho jaate hain). To $\partial ρ / \partial t = - {ρ, H} = 0$ : density time mein change nahi hoti. Yeh kyun matter karta hai: Statistical Mechanics — Ensembles mein, equilibrium ensembles exactly wahi densities $ρ = ρ (H)$ hoti hain jo sirf energy ki functions hain — woh stationary hain kyunki woh $H$ ke saath Poisson-commute karti hain.

Level 4 — Synthesis

L4.1

Ek linear phase-space map ek blob ko matrix se advance karta hai $M = (a c b d), (q^{'} p^{'}) = M (q p) .$ Naya area = purana area $\times$ $∣ det M ∣$ . (a) $M$ par area-preserving hone ki condition batao. (b) Kya $M = (10 τ 1)$ (time $τ$ ke liye free-particle drift, $m = 1$ ke saath) area-preserving hai? (c) Kya $M = (20 0 0.5)$ area-preserving hai, aur kya yeh Hamiltonian flow se aa sakta hai?

Recall Solution

(a) Area-preserving $⟺ ∣ det M ∣ = 1$ . Symplectic Geometry mein genuine canonical map ke liye sharper condition hai $det M = + 1$ (ek symplectic matrix); yeh area aur orientation dono preserve karta hai. (b) $det = 10 τ 1 = 1 \cdot 1 - τ \cdot 0 = 1$ . Area-preserving ✓ — yeh exactly Example 1 ki shear hai, kisi bhi $τ$ ke liye. (c) $det = 2 \times 0.5 - 0 = 1$ . Area preserve hota hai ( $q$ ke along $2 \times$ stretch aur $p$ ke along $2 \times$ squeeze — ek perfect trade). Aur kyunki $det = + 1$ hai, yeh Hamiltonian flow se aa sakta hai: yeh chaotic stretch-and-fold ka local picture hai — ek filament banta hai jabki area fixed rehta hai.

L4.2

Do canonical maps ek ke baad ek apply hote hain, pehle $M_{2}$ phir $M_{1}$ . $det M_{1} = 1$ aur $det M_{2} = 1$ diya hua hai, dikhao ki composite abhi bhi area preserve karta hai, aur explain karo ki yeh kyun guarantee karta hai ki Liouville poori evolution ke liye hold karta hai, na sirf tiny steps ke liye.

Recall Solution

Composite map matrix product $M_{1} M_{2}$ hai. Determinants multiply hote hain: $det (M_{1} M_{2}) = det M_{1} \cdot det M_{2} = 1 \times 1 = 1.$ Area preserved ✓. Reasoning ki chain: koi bhi finite-time evolution kai infinitesimal Hamiltonian steps ka limit hai, har ek $det = 1$ ke saath. Kitne bhi unit determinants multiply karo, result $1$ hoga. To volume kisi bhi time interval mein conserved rehta hai, sirf instantaneous rates ke liye nahi. Yahi canonical transformations ka group structure hai: woh composition ke under ek group banate hain, aur area-preservation group operation se preserve hoti hai.

Level 5 — Mastery

L5.1

Poincaré recurrence. Ek Hamiltonian system finite total volume $V_{tot}$ ke phase-space region mein confined hai (jaise fixed energy shell, bounded). Sirf Liouville's theorem use karke argue karo ki almost har starting point eventually wahan wapas zaroor aa jaata hai jahan woh shuru hua tha. Phir explain karo ki ek damped system (L3.1) kyun exempt hai.

Recall Solution

Ek starting point ke aas-paas ek tiny neighbourhood $U$ lo, jiska volume $V (U) > 0$ hai. Time $T$ ke liye flow ko map $ϕ_{T}$ hone do. Liouville se, $ϕ_{T}$ volume preserve karta hai: har image $ϕ_{T} (U), ϕ_{2 T} (U), ϕ_{3 T} (U), \dots$ ka same volume $V (U)$ hota hai. Ab contradiction ke liye assume karo ki yeh images kabhi overlap nahi karte. To infinitely many disjoint sets, har ek volume $V (U) > 0$ ke saath, finite region $V_{tot}$ ke andar baith jaate hain. Unka total volume $\infty \cdot V (U) = \infty > V_{tot}$ hota — impossible. To do images overlap karni chahiye: $ϕ_{i T} (U) \cap ϕ_{j T} (U) \neq = \emptyset$ kisi $i > j$ ke liye. $ϕ_{- j T}$ apply karo (yeh bhi volume-preserving hai): $U \cap ϕ_{(i - j) T} (U) \neq = \emptyset$ . Isliye $U$ ka koi point time $(i - j) T$ ke baad $U$ mein wapas aa jaata hai. $U$ ko chhota karte jaane se dikhta hai ki return arbitrarily close ho sakta hai. Yahi Poincaré Recurrence Theorem hai. Damped exemption: L3.1 mein volume preserved nahi hota; blobs attractor ki taraf shrink karte hain. Images usi shrinking set par pile up ho sakti hain, isliye "disjoint boxes space fill karte hain" wala argument fail ho jaata hai. Koi volume floor nahi ⇒ koi forced recurrence nahi.

L5.2

Ensemble entropy constant hoti hai. Gibbs entropy hai $S = - k_{B} \int ρ ln ρ d Γ$ , jahan $d Γ$ phase-space volume element hai. Dikhao ki Hamiltonian flow ke under $\frac{d S}{d t} = 0$ , matlab fine-grained entropy kabhi change nahi hoti. ( $d ρ / d t = 0$ aur volume conservation use karo.)

Recall Solution

Un coordinates mein jao jo fluid ke saath flow karte hain (comoving frame). Do Liouville facts:

Ek trajectory ke along $\frac{d ρ}{d t} = 0$ , isliye ek moving point ke saath carried $ρ$ constant hai; isliye $f (ρ) = - ρ ln ρ$ bhi us trajectory ke along constant hai.
Volume conserved hai: har point jo chhota cell $d Γ$ carry karta hai, woh same size rakhta hai. Integral $S = - k_{B} \int f (ρ) d Γ$ moving cells par (constant value $f (ρ)$ ) $\times$ (constant cell size $d Γ$ ) ka sum hai. Sum mein kuch bhi change nahi hota, isliye $\frac{d S}{d t} = 0.$ Interpretation: microscopic (fine-grained) entropy frozen hai — mechanics reversible hai. Second Law ka entropy increase coarse-graining (un filaments ko blur karna jo Liouville stretch karta hai) se aata hai, exact flow se nahi. Yeh Statistical Mechanics — Ensembles ka ek cornerstone hai.

Recall Final self-test (answers cover karo)

Kaunsi determinant value ek area-preserving linear map mark karta hai? ::: $∣ det M ∣ = 1$ (symplectic: $det = + 1$ ). Damped system Poincaré recurrence se kyun bachta hai? ::: Uska flow volume-preserving nahi hai, isliye images ek shrinking attractor par pile up ho sakti hain — overlaps force karne ke liye koi volume floor nahi. Liouville kya freeze karta hai jo fine-grained Gibbs entropy constant banata hai? ::: Comoving density $ρ$ ( $d ρ / d t = 0$ ) aur carried volume element $d Γ$ .