Worked examples — Liouville's theorem — phase space volume conservation

2.1.14 · D3 · Physics › Analytical Mechanics › Liouville's theorem — phase space volume conservation

Yeh page Liouville's theorem ka case-by-case workout hai. Parent note ne kyun prove kiya tha ki Hamiltonian flow phase-space volume ko constant rakhta hai. Yahan hum har tarah ke system ko us claim pe throw karte hain aur dekhte hain ki woh hold karta hai — ya phir, jaan-bujhkar banaye gaye counterexamples mein, dekhte hain ki woh fail kahan hota hai, taaki tum exactly samajh sako ki boundary kahan hai.

Shuru karne se pehle, ek chhoti si reminder un do objects ki jo hum baar baar measure karte rehte hain, taaki koi symbol bina define ke andar na ghus aaye:

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Scenario matrix

Neeche har cell ek class hai situation ki jo is topic mein aa sakti hai. Baad ke examples us cell ke saath tagged hain jo woh cover karte hain, toh saath mein poora grid fill ho jaata hai.

#	Case class	Isme kya khaas hai	Covered by
A	Free particle (zero force)	$\dot{p}=0$, pure shear	Ex 1
B	Restoring force (oscillator)	blob ka rigid rotation	Ex 2
C	Multiple degrees of freedom ($N>1$)	saare $i$ par sum karna zaroori	Ex 3
D	Degenerate blob (zero-area line/point)	limiting input, area $=0$ stays $0$	Ex 4
E	Sign / direction cases ($p<0$, backward time $t<0$)	flow reverse hota hai, phir bhi preserve karna hai	Ex 5
F	Non-Hamiltonian: dissipation	divergence $<0$, blob shrinks	Ex 6
G	Non-Hamiltonian: anti-damping / pumping	divergence $>0$, blob grows	Ex 7
H	Real-world word problem (beam of particles)	theorem ko physical density par apply karo	Ex 8
I	Exam twist: nonlinear $H$, kya yeh abhi bhi divergence-free hai?	general Hamiltonian, $0$ prove karo	Ex 9

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Example 1 — Free particle: pure shear (cell A)

Forecast: padhne se pehle andaza lagao — bada $p$ zyada tezi se move karta hai, toh shape ko lean over karna chahiye. Kya area change hoga? Apna guess likho.

1. Flow likho. $\dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m}=\frac{p}{2}$, aur $\dot{p}=-\frac{\partial H}{\partial q}=0$. Yeh step kyun? Aage sab kuch actual velocity field chahiye; Hamilton's equations woh dete hain.
2. Har point move karo. Time $t$ ke baad, $q\to q+\dot{q}t=q+\frac{p}{2}\cdot 4=q+2p$, jabki $p\to p$ (unchanged, kyunki $\dot{p}=0$). Yeh step kyun? Constant velocity $\Rightarrow$ displacement $=$ velocity $\times$ time. Bade $p$ wale points zyada daayein slide karte hain — yeh ek shear hai (figure mein dekho kaise red top edge blue bottom edge se aage khisakti hai).
3. Jacobian compute karo. Map hai $(q,p)\mapsto (q+2p,p)$. Iska matrix hai $J=\left(\begin{array}{cc}\partial q^{\prime }/\partial q& \partial q^{\prime }/\partial p\\ \partial p^{\prime }/\partial q& \partial p^{\prime }/\partial p\end{array}\right)=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right),\det J=1.$ Yeh step kyun? $\det J$ exactly area-multiplication factor hai. Ek shear matrix ka $\det$ hamesha $1$ hota hai.
4. Area. $A_0=3\times 4=12$, toh $A(4)=\det J\cdot A_0=12$. Yeh step kyun? Bridge formula ke hisaab se area bas $\det J\times A_0$ hai; $\det J=1$ ke saath original $12$ untouched pass ho jaata hai — shear ne shape ko tilt kiya lekin area ko $1$ se multiply kiya.

Verify: Divergence check: $\frac{\partial \dot{q}}{\partial q}+\frac{\partial \dot{p}}{\partial p}=0+0=0$ — flow divergence-free hai, toh area change nahi ho sakta. Units: $q$·$p$ = length·momentum = action; parallelogram wahi action $12$ carry karta hai. ✔

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Example 2 — Harmonic oscillator: rigid rotation (cell B)

Forecast: phase portrait circles hain. Quarter turn $90^{\circ }$ ka hai. Kya rotation area change karta hai? Guess karo.

1. Flow. $\dot{q}=p$, $\dot{p}=-q$ (jahan $m=\omega =1$ hai). Yeh step kyun? $m=\omega =1$ ko Hamilton's equations mein plug karo. Yeh classic "clockwise circulation" field hai — har arrow radius se $90^{\circ }$ off point karta hai (figure mein green arrows dekho).
2. Solve karo. Solution ek rotation hai: $q(t)=q_0\cos t+p_0\sin t$, $p(t)=-q_0\sin t+p_0\cos t$. Yeh step kyun? $\ddot{q}=\dot{p}=-q$ unit angular speed par rotation ka equation hai; poora blob rigidly turn karta hai.
3. Jacobian. Map matrix rotation hai $J=\left(\begin{array}{cc}\cos t& \sin t\\ -\sin t& \cos t\end{array}\right),\det J={\cos }^{2}t+{\sin }^{2}t=1.$ Yeh step kyun? Kisi bhi rotation ka $\det$ $1$ hota hai — yeh algebraic reason hai ki rotating blob apna area kyun rakhta hai.
4. $t=\pi /2$ ke baad. Area $=1\cdot \pi =\pi$. $(q_0,p_0)=(2,0)$ par centre $(2\cos \frac{\pi }{2},-2\sin \frac{\pi }{2})=(0,-2)$ par move karta hai. Yeh step kyun? Area $=\det J\cdot A_0=1\cdot \pi =\pi$; centre bas ek point hai jo step 2 ke rotation solution se run kiya gaya, toh hum $t=\pi /2$ isme substitute karte hain.

Verify: Divergence $\frac{\partial (p)}{\partial q}+\frac{\partial (-q)}{\partial p}=0+0=0$ ✔. Numeric: centre new position $(0,-2)$, origin se distance $\sqrt{0^2+2^2}=2$ = original radius $\Rightarrow$ pure rotation, area $\pi$ unchanged. ✔

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Example 3 — Two degrees of freedom (cell C)

Forecast: har pair rotate karta hai. Do rotations stack ho gaye — kya $4$-volume bachega?

1. Flow, saare chaar components. ${\dot{q}}_1=p_1,{\dot{p}}_1=-q_1,{\dot{q}}_2=p_2,{\dot{p}}_2=-q_2$. Yeh step kyun? $N=2$ ke saath tumhe har coordinate ki velocity list karni hai — Liouville saare $i$ par sum karta hai, sirf ek par nahi.
2. Divergence — sab par sum karo. $\nabla \cdot \dot{\mathbf{x}}=\frac{\partial {\dot{q}}_1}{\partial q_1}+\frac{\partial {\dot{p}}_1}{\partial p_1}+\frac{\partial {\dot{q}}_2}{\partial q_2}+\frac{\partial {\dot{p}}_2}{\partial p_2}=0+0+0+0=0.$ Yeh step kyun? Higher dimensions mein volume conservation ke liye total divergence ka vanish karna zaroori hai. Har pair $0$ contribute karta hai, toh poora bhi $0$ hai.
3. Jacobian. $4\times 4$ map block-diagonal hai: do independent $2\times 2$ rotation blocks. $\det =\det ({\text{block}}_1)\cdot \det ({\text{block}}_2)=1\cdot 1=1$. Yeh step kyun? Block-diagonal matrix ka determinant block determinants ka product hota hai.
4. Volume. $V(t)=1\cdot 0.5=0.5$ saare $t$ ke liye. Yeh step kyun? Bridge formula padhta hai $V(t)=\det J\cdot V_0$; $\det J=1$ ke saath $4$-volume $0.5$ frozen hai — do rotations independently area preserve karte hain, aur unka product $4$-volume preserve karta hai.

Verify: Total divergence $=0$ ✔. Block determinants $1\times 1=1$, toh $V=0.5$ unchanged. ✔

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Example 4 — Degenerate blob: zero area ki ek line (cell D)

Forecast: area-preserving flow ke andar ek area-$0$ object — trick question hai ya nahi? Guess karo.

1. Segment flow karo. Har point $(1,p)\mapsto (1+2p,p)$ (Ex 1 ke map se). Yeh step kyun? Wahi shear map apply hota hai; hum bas ise ek degenerate input dete hain.
2. Nayi shape. Bottom point $(1,0)\mapsto (1,0)$; top point $(1,4)\mapsto (9,4)$. Segment ek slanted line ban jaata hai $(1,0)$ se $(9,4)$ tak. Yeh step kyun? Ek shear straight line ko straight line mein map karta hai — abhi bhi one-dimensional.
3. Area. Ek line ki width zero hoti hai, toh area $=0$ pehle bhi aur baad mein bhi. Yeh step kyun? $\det J=1$ kisi bhi area ko $1$ se multiply karta hai; $1\times 0=0$. Liouville degenerate blobs ke saath perfectly consistent hai — "volume preserved" mein "zero stays zero" bhi shamil hai.

Verify: endpoints $(1,0)$ aur $(9,4)$; map invertible hai ($\det J=1\ne 0$) toh koi folding nahi, area $0$ hi rehta hai. ✔

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Example 5 — Sign cases: negative momentum aur backward time (cell E)

Forecast: negative $p$ matlab baayi taraf move karna; backward time matlab shear doosri taraf tilt karti hai. Kya kisi bhi cheez ka sign conservation todta hai?

1. (a) Negative momentum. Map hai $(q,p)\mapsto (q+2p,p)$. $p<0$ ke saath, points baayi taraf drift karte hain (figure mein baayi taraf point karte yellow arrows dekho). Jacobian abhi bhi $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$ hai, $\det =1$. Yeh step kyun? Jacobian $p$ ke sign par depend nahi karta — woh sirf map ki structure par depend karta hai. Toh area $=3\times 4=12$ motion ki direction se independent hai.
2. (b) Backward time. $t=-4$ set karo: map hai $(q,p)\mapsto (q-2p,p)$, matrix $\left(\begin{array}{cc}1& -2\\ 0& 1\end{array}\right)$, $\det =1$. Yeh step kyun? Hamiltonian flow time-reversible hota hai; ise backward chalana ek aur volume-preserving shear hai (doosri taraf tilt hoti hai). Volume dono time directions mein conserved hoti hai.
3. Areas. (a) $12$. (b) $12$. Yeh step kyun? Dono maps ka $\det J=1$ hai, aur area $=\det J\cdot A_0=1\cdot 12$ hai; na $p$ ka sign na $t$ ka sign determinant mein aata hai, toh dono blobs area $12$ rakhte hain.

Verify: Dono Jacobian determinants $1$ ke barabar hain; original area $12$ ⇒ dono answers $12$. Drift ki direction badli, area nahi. ✔

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Example 6 — Dissipation: blob shrinks (cell F)

Forecast: friction energy khaata hai. States kahan pile up hote hain? Guess karo area badhta hai ya shrinks karta hai.

1. Divergence. $\frac{\partial \dot{q}}{\partial q}+\frac{\partial \dot{p}}{\partial p}=\frac{\partial (p)}{\partial q}+\frac{\partial (-q-\gamma p)}{\partial p}=0+(-\gamma )=-\gamma =-0.6$. Yeh step kyun? Divergence instantaneous fractional rate of area change hai: $\frac{1}{A}\frac{dA}{dt}=\nabla \cdot \dot{\mathbf{x}}=-\gamma$.
2. Area ODE solve karo. $\frac{dA}{dt}=-\gamma A\Rightarrow A(t)=A_0{e}^{-\gamma t}=e^{-0.6t}$. Yeh step kyun? Constant fractional shrink rate integrate hokar exponential deta hai. Blob origin (the attractor) mein spiral karta hai — andar ki taraf green spiral dekho.
3. Numbers. $t=1$ par: $A=e^{-0.6}\approx 0.5488$. Yeh step kyun? Hum step 2 ke general law $A(t)=e^{-\gamma t}$ ko ek concrete time $t=1$ par evaluate karte hain, abstract exponential ko ek aise number mein convert karte hain jo check kiya ja sake — yeh "blob shrinks" statement ko quantitative banata hai.

Verify: Divergence $=-0.6\ne 0$ ⇒ Liouville violated (jaisa hona chahiye — koi bhi Hamiltonian ise generate nahi karta). $A(1)=e^{-0.6}=0.5488\dots$ ✔. Yahi wajah hai ki theorem ko genuine Hamiltonian chahiye.

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Example 7 — Anti-damping / pumping: blob grows (cell G)

Forecast: opposite sign — states ab baahir ki taraf fly karte hain. Badhta hai ya shrinks karta hai?

1. Divergence. $\frac{\partial (p)}{\partial q}+\frac{\partial (-q+\beta p)}{\partial p}=0+\beta =+0.4$. Yeh step kyun? Positive divergence matlab flow locally expand karta hai. Yeh Ex 6 ka mirror image hai.
2. Area law. $\frac{dA}{dt}=\beta A\Rightarrow A(t)=e^{\beta t}=e^{0.4t}$. Yeh step kyun? Positive fractional rate ⇒ exponential growth; spiral baahir ki taraf jaata hai, origin se door.
3. Number. $t=2$ par: $A=e^{0.8}\approx 2.2255$. Yeh step kyun? Hum $t=2$ ko step 2 ke general law $A(t)=e^{\beta t}$ mein plug karte hain, ek concrete number dete hain jo exactly quantify karta hai ki ek pumped blob kitni tezi se inflate hota hai — general derivation ka numeric counterpart.

Verify: Divergence $=+0.4>0$ ⇒ volume-preserving nahi, Hamiltonian nahi. $A(2)=e^{0.8}=2.2255\dots$ ✔. Saath mein, Ex 6 aur Ex 7 Hamiltonian case ko bracket karte hain: sirf exactly $0$ divergence volume conserve karta hai.

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Example 8 — Real-world word problem: particles ka ek beam (cell H)

Forecast: hum beam ko position mein narrow kar rahe hain. Liouville se intuition: kuch toh price pay karna hoga. Guess karo momentum spread ka kya hoga.

1. Emittance conserved hai. Lens + drift ek Hamiltonian (force-based) system hai, toh phase-space area frozen hai: $A=A_0=4$. Yeh step kyun? Yeh Liouville physically apply kiya — tum ek saath position spread aur momentum spread dono ko shrink nahi kar sakte. Yeh theorem ka beam-physics form hai.
2. $\Delta p$ ke liye solve karo. Area $=\Delta q\cdot \Delta p$ (best case, koi filamentation waste nahi). $\Delta q=0.5$, toh $\Delta p=\frac{A_0}{\Delta q}=\frac{4}{0.5}=8$. Yeh step kyun? Fixed area ke saath chota base force karta hai zyada tall height — blob $q$ mein $4\times$ narrow hua, toh $p$ mein $4\times$ tall hona padega.
3. Interpret karo. Beam ko $\frac{1}{4}$ width par squeeze karna uski momentum (angular) spread ko chaar guna kar deta hai. Yeh ek hard physical limit hai ki koi bhi beam kitni tightly focus ho sakti hai. Yeh step kyun? Hum bare number $\Delta p=8$ ko waapas physics mein translate karte hain: kyunki Liouville area ko freeze karta hai, spatial focus mein koi bhi gain momentum focus mein equal loss se pay hota hai — woh trade-off hi theorem ka accelerators ke liye poora practical content hai.

Verify: $\Delta q\cdot \Delta p=0.5\times 8=4=A_0$ ✔. Factor: $q$-spread $\times \frac{1}{4}$ ⇒ $p$-spread $\times 4$ ($2$ se $8$). Area (emittance) ke units preserved hain. ✔

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Example 9 — Exam twist: ek nonlinear Hamiltonian, abhi bhi divergence-free? (cell I)

Forecast: nonlinearity aksar achhi properties todti hai. Bet: kya divergence abhi bhi vanish karta hai, ya $\cos q$ use ruin karta hai?

1. Flow. $\dot{q}=\frac{\partial H}{\partial p}=p$, aur $\dot{p}=-\frac{\partial H}{\partial q}=-\sin q$. Yeh step kyun? Hamilton's equations ki parwah nahi ki $H$ linear hai ya nahi.
2. Divergence. $\frac{\partial \dot{q}}{\partial q}+\frac{\partial \dot{p}}{\partial p}=\frac{\partial (p)}{\partial q}+\frac{\partial (-\sin q)}{\partial p}=0+0=0.$ Yeh step kyun? Key point yeh hai: $\dot{q}$ sirf $p$ par depend karta hai (toh $\partial /\partial q$ ise kill karta hai) aur $\dot{p}$ sirf $q$ par depend karta hai (toh $\partial /\partial p$ ise kill karta hai). Mixed-partial cancellation $\frac{{\partial }^{2}H}{\partial q\partial p}=\frac{{\partial }^{2}H}{\partial p\partial q}$ hi hai jo yeh kisi bhi $H$ ke liye automatic banata hai — nonlinearity irrelevant hai.
3. Us point par. $0$, exactly, $(q,p)$ se independent. Toh $q=\pi /3,p=1.7$ par divergence $0$ hai. Yeh step kyun? Kyunki step 2 ne dikhaya ki divergence $(q,p)$ ki function ke roop mein identically $0$ hai, kisi bhi specific point par evaluate karna — is exam value sameta — $0$ hi return karna chahiye; hum yeh confirm karne ke liye karte hain ki koi special point nonzero contribution sneakily na de.

Verify: Symbolically ${\partial }_q({\partial }_{p}H)-{\partial }_p({\partial }_{q}H)=0$ for $H=\frac{p^2}{2}+1-\cos q$ ✔. Lesson yeh hai: har Hamiltonian, linear ho ya nonlinear, divergence-free flow deta hai — yahi Liouville's theorem ki poori robustness hai. See also Poisson Brackets aur Hamilton's Equations structural reason ke liye.

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Recall Maine abhi kaunsa cell conquer kiya?

Free/shear ::: Ex 1 (A) Rigid rotation ::: Ex 2 (B) Many DOF, saare $i$ par sum karo ::: Ex 3 (C) Degenerate zero-area blob ::: Ex 4 (D) Negative $p$ aur backward time ::: Ex 5 (E) Dissipation area shrinks karta hai ::: Ex 6 (F) Pumping area grow karta hai ::: Ex 7 (G) Beam emittance limit ::: Ex 8 (H) Nonlinear $H$ abhi bhi divergence-free ::: Ex 9 (I)

Related structure jo revisit karne layak hai: Phase Space, Canonical Transformations, Statistical Mechanics — Ensembles, Poincaré Recurrence Theorem, Symplectic Geometry.