Worked examples — Action-angle variables — integrable systems

2.1.18 · D3 · Physics › Analytical Mechanics › Action-angle variables — integrable systems

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Scenario matrix

Compute karne se pehle, har cell list karte hain jisme ek 1-degree-of-freedom periodic system aa sakta hai. Ek cell = ek qualitatively alag phase-space loop ya ek alag question type.

Cell	Naam	Isme kya alag hai	Example
A	Smooth well (libration)	Loop ek closed oval hai; $p$ dono signs $\pm$ leta hai	Ex 1 (SHO)
B	Hard walls (libration)	$\Vert p\Vert$ constant hai, loop ek rectangle hai	Ex 2 (box)
C	Power-law well	Loop na oval hai na rectangle; $\omega$, $E$ pe depend karta hai	Ex 3 (quartic)
D	Pendulum barrier se neeche	Aage-peeche swing karta hai — libration, same as A but with $\sin$	Ex 4
E	Pendulum barrier se upar	Poora round ghoomta hai — rotation, ek alag contour	Ex 5
F	Attractive $1/r$ (bound orbit)	Turning points $r_{\min },r_{\max }$ wala loop; ek real astronomy word problem	Ex 6
G	Degenerate / zero input	Zero energy, zero amplitude — kya recipe is limit mein survive karti hai?	Ex 7
H	Exam twist	Ek aisa relation jo lagta hai trajectory chahiye par aslaan nahi chahiye	Ex 8

Ab hum har cell ko baari-baari lete hain. "Forecast" padho aur scroll karne se pehle guess karo — wahi pe learning hoti hai.

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Ex 1 — Cell A: smooth well (harmonic oscillator)

Phase-space picture (ellipse kyun?). Equation $\frac{p^2}{2mE}+\frac{q^2}{2E/m{\omega }_0^2}=1$ ek ellipse ki equation hai — ek squashed circle. Iske do half-widths (jinhe semi-axes kehte hain, center se edge tak ki distance har axis ke along) hain:

$a=\sqrt{\frac{2E}{m{\omega }_0^2}}(\text{along }q),b=\sqrt{2mE}(\text{along }p).$

Step 1 — $H=E$ ko $p$ ke liye solve karo. $p=\pm \sqrt{2mE-m^2{\omega }_0^2{q}^2}.$ Yeh step kyun? $J$ ko $p$ chahiye $q$ ke function ke roop mein; energy equation rearrange karna hi woh jagah hai jahan $p$ appear hota hai. $\pm$ oval ki top aur bottom halves hain (figure mein blue = $+$, red = $-$).

Step 2 — Area bina integrate kiye nikalo. Ellipse ka area $\pi ab$ hota hai (circle $\pi r^2$ ka area, $a/r$ ek direction mein aur $b/r$ doosri direction mein stretch kiya). To: $\oint pdq=\pi ab=\pi \sqrt{\frac{2E}{m{\omega }_0^2}}\sqrt{2mE}=\frac{2\pi E}{{\omega }_0}.$ Yeh step kyun? $\oint pdq$ hi enclosed area hai — ek known formula integral se behtar hai.

Step 3 — $2\pi$ se divide karo aur invert karo. $J=\frac{1}{2\pi }\cdot \frac{2\pi E}{{\omega }_0}=\frac{E}{{\omega }_0}\Rightarrow E={\omega }_{0}J.$

Step 4 — Frequency ke liye differentiate karo. $\omega =\frac{dE}{dJ}={\omega }_0.$ Yeh step kyun? $\omega =\partial H/\partial J$; kyunki $E=H(J)={\omega }_{0}J$ ek straight line hai, iska slope ${\omega }_0$ hai — energy se independent. Yahi woh special property hai jo oscillator ko isochronous banati hai (kisi bhi amplitude pe same period).

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Ex 2 — Cell B: hard walls (particle in a box)

Phase-space picture. Walls ke beech koi force nahi, isliye $|p|=\sqrt{2mE}$ constant hai. Particle right mein $+p$ pe move karta hai, wall se hit hota hai, instantly $-p$ pe reverse hota hai, wapas aata hai. $(q,p)$ plane mein yeh ek rectangle hai: top edge $+p$ par, bottom edge $-p$ par, walls par vertical jumps se juda.

Step 1 — Momentum. $p=\sqrt{2mE}$, constant. Kyun? Koi potential nahi matlab saari energy kinetic hai.

Step 2 — Area = poora loop (there-and-back). $\oint pdq=\underset{\text{right jaate waqt, }+p}{\underbrace{pL}}+\underset{\text{wapas aate waqt, }-p\times -dq}{\underbrace{pL}}=2pL.$ Yeh step kyun? Ek poora cycle out aur back hai — box do baar traverse hota hai. Bottom edge ek doosra positive $pL$ contribute karta hai kyunki wahan $p$ aur $dq$ dono sign flip karte hain (negative times negative positive hota hai). Yeh classic factor-of-2 trap hai.

Step 3 — Action aur inversion. $J=\frac{1}{2\pi }2pL=\frac{pL}{\pi }=\frac{L\sqrt{2mE}}{\pi }\Rightarrow E=\frac{{\pi }^2{J}^2}{2m{L}^2}.$

Step 4 — Frequency aur period. $\omega =\frac{dE}{dJ}=\frac{{\pi }^{2}J}{m{L}^2}=\frac{\pi v}{L},T=\frac{2\pi }{\omega }=\frac{2L}{v},$ $v=p/m$ use karke. Yeh step kyun? Spring se alag, $E\propto J^2$ ek parabola hai, isliye iska slope $\omega$ $J$ ke saath badhta hai — tez particle zyada baar bounce karta hai.

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Ex 3 — Cell C: power-law well (quartic oscillator)

Idea — hard integral ke bajay scaling. Hume messy integral ki zaroorat nahi; hume sirf yeh chahiye ki $J$ kaise scale karta hai $E$ ke saath. Turning point $q_{\max }$ likho jahan $p=0$: $\alpha q_{\max }^4=E$, to $q_{\max }=(E/\alpha {)}^{1/4}$.

Step 1 — Integral ko rescale karo. $J=\frac{1}{2\pi }\oint \sqrt{2m(E-\alpha q^4)}dq.$ $q=q_{\max }x$ substitute karo taki $x$ ek fixed range pe run kare jo $E$ se independent ho: $J=\frac{1}{2\pi }\oint \sqrt{2mE}\sqrt{1-x^4}{q}_{\max }dx=C\sqrt{E}{E}^{1/4}=C{E}^{3/4},$ jahan $C$ saare $E$-independent constants collect karta hai aur $q_{\max }\propto E^{1/4}$. Yeh step kyun? Har $E$ ko integral ke bahar kheenchna ek unsolvable integral ko pure power law mein badal deta hai — frequency ke liye hume bas yahi chahiye.

Step 2 — Invert karo. $J\propto E^{3/4}\Rightarrow E\propto J^{4/3}.$

Step 3 — Frequency. $\omega =\frac{dE}{dJ}\propto J^{1/3}\propto E^{1/4}.$ Yeh step kyun? $\omega$ energy ke saath badhta hai: spring se stiffer well ($q^4$, $q^2$ se tez badhta hai) particle ko jaldi turn around karata hai, isliye yeh zyada amplitude pe tezi se oscillate karta hai. Period shrinks — shayad tumhare guess ke opposite.

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Ex 4 — Cell D: pendulum barrier se neeche (libration)

Pendulum ke do regimes. Angle $\varphi$ seedha neeche se measure hota hai. Agar total energy $E<mg\ell$ (top par energy) hai, to bob top tak nahi pahunch sakta; woh aage-peeche swing karta hai — ek libration loop (closed oval), hamara Cell D. Agar $E>mg\ell$ hai to woh poora ghoom jaata hai — Cell E (agla example). Dividing energy $E=mg\ell$ separatrix hai.

Step 1 — Small-angle limit. Chote $\varphi$ ke liye, $\cos \varphi \approx 1-{\varphi }^2/2$, to: $H\approx \frac{p^2}{2m{\ell }^2}+\frac{1}{2}mg\ell {\varphi }^2-mg\ell .$ Yeh step kyun? Bottom ke paas $\cos$ well parabolic lagta hai — yeh exactly Ex 1 ka harmonic form hai mass $M=m{\ell }^2$ aur stiffness $k=mg\ell$ ke saath.

Step 2 — Analogy se frequency padho. Ek oscillator $H=\frac{p^2}{2M}+\frac{1}{2}M{\Omega }^2{\varphi }^2$ ke liye hume $M{\Omega }^2=k$ chahiye: ${\Omega }^2=\frac{k}{M}=\frac{mg\ell }{m{\ell }^2}=\frac{g}{\ell }\Rightarrow {\omega }_0=\sqrt{\frac{g}{\ell }}.$ Yeh step kyun? Harmonic case hum Ex 1 mein pehle hi solve kar chuke hain ($\omega ={\omega }_0$, energy-independent); uss par map karna saara kaam bachata hai.

Step 3 — Bottom ke paas action. Ex 1 ka result $J=E^{\prime }/{\omega }_0$ use karke (jahan $E^{\prime }=E+mg\ell$ well bottom se measured hai): $J\approx \frac{E+mg\ell }{\sqrt{g/\ell }}.$

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Ex 5 — Cell E: pendulum barrier se upar (rotation)

Alag contour kyun (is cell ka poora point). Libration mein (Ex 4) $\varphi$ maximum tak pahunchta hai, rukta hai, aur wapas aata hai — loop apne aap mein band hota hai, $p$ dono signs leta hai. Rotation mein, $p$ kabhi sign nahi badalta: bob same direction mein chalda rehta hai, aur $\varphi$ har ghoom mein poora $2\pi$ badhta hai. "Ek period" ek poori revolution hai, isliye contour $\varphi :0\to 2\pi$ ek baar run karta hai, there-and-back nahi.

Step 1 — Momentum single-signed rehta hai. $p=+\sqrt{2m{\ell }^2(E+mg\ell \cos \varphi )}>0\text{for all }\varphi (\text{kyunki }E>mg\ell ).$ Yeh step kyun? $E>mg\ell$ guarantee karta hai ki bracket hamesha positive hai — bob kabhi kinetic energy khatam nahi karta, isliye kabhi rukta nahi. Koi $\pm$ nahi; bas $+$.

Step 2 — Action integral, single sweep. $J=\frac{1}{2\pi }{\int }_0^{2\pi }\sqrt{2m{\ell }^2(E+mg\ell \cos \varphi )}d\varphi .$ Yeh step kyun? Ek poori turn $=0\to 2\pi$ exactly ek period hai, isliye yahan koi factor of 2 nahi (Ex 2 ke box se compare karo, jisko yeh chahiye tha). Yeh sabse common mistake hai: rotation aur libration alag contours use karte hain.

Step 3 — High-energy limit (fast spinning). Jab $E\gg mg\ell$, to $\cos \varphi$ koi khaas matter nahi karta: $J\approx \frac{1}{2\pi }{\int }_0^{2\pi }\sqrt{2m{\ell }^{2}E}d\varphi =\sqrt{2m{\ell }^{2}E}=|p|,E\approx \frac{J^2}{2m{\ell }^2},$ to $\omega =\frac{dE}{dJ}=\frac{J}{m{\ell }^2}=\frac{p}{m{\ell }^2}=\dot{\varphi }$ — frequency bas spin rate hai. Kyun? Ek fast rotor gravity ignore karta hai; yeh ek free rotor ki tarah behave karta hai jiska "period" ek revolution hai.

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Ex 6 — Cell F: gravity word problem (bound orbit ka radial action)

Step 1 — Radial turning points. $p_r=0$ set karo: bob ka radial motion wahan reverse hota hai jahan saari radial kinetic energy khatam ho jaati hai. $E=\frac{L^2}{2m{r}^2}-\frac{k}{r}$ ke do roots $r_{\min },r_{\max }$ hain — perigee aur apogee. Inke beech, $r$ ek 1-DOF well ki tarah librate karta hai.

Step 2 — Radial action (standard contour integral). $J_r=\frac{1}{2\pi }\oint p_{r}dr=\frac{1}{2\pi }\oint \sqrt{2m(E+\frac{k}{r}-\frac{L^2}{2m{r}^2})}dr=-L+k\sqrt{\frac{m}{-2E}}.$ Yeh step kyun? Yeh ek known residue integral hai; hamare liye important hai result, jo $E$ ko cleanly isolate karta hai.

Step 3 — Energy ke liye invert karo. $E$ solve karo: $J_r+L=k\sqrt{\frac{m}{-2E}}\Rightarrow E=-\frac{m{k}^2}{2(J_r+L{)}^2}.$ Yeh step kyun? $E$, $J_r$ aur $L$ pe sirf sum $J_r+L$ ke zariye depend karta hai — ek hint ki dono frequencies equal honge (ek degeneracy jo $1/r$ ke liye special hai).

Step 4 — Frequency. $\omega =\frac{\partial E}{\partial J_r}=\frac{m{k}^2}{(J_r+L{)}^3}.$ Kyunki $E$ sirf $J_r+L$ pe depend karta hai, ${\omega }_r={\omega }_{\text{angular}}$ — radial aur angular periods coincide karte hain, isliye orbit ek fixed ellipse mein close ho jaati hai (koi precession nahi). Kyun? Yeh equality exactly isliye hai ki Kepler orbits drift nahi karte — pure $1/r$ law ki ek hallmark.

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Ex 7 — Cell G: degenerate / zero input (kya recipe survive karti hai?)

Step 1 — Oscillator $E\to 0$ par. Ex 1 se, $J=E/{\omega }_0\to 0$ aur $\omega ={\omega }_0$ (constant). Yeh step kyun? Loop area zero tak shrink hota hai (origin par ek point), to $J\to 0$; lekin shape same aspect ratio ki ellipse rehti hai, isliye frequency unchanged hai. Ek point ka bhi ek well-defined frequency hota hai — infinitesimally chote swings ki limiting frequency. Koi blow-up nahi.

Step 2 — Box $E\to 0$ par. Ex 2 se, $J=L\sqrt{2mE}/\pi \to 0$ aur $\omega =\pi v/L$ jahan $v=\sqrt{2E/m}\to 0$, to $\omega \to 0$, $T=2L/v\to \infty$. Yeh step kyun? Zero speed wala particle kabhi bounce complete nahi karta — period diverge karta hai. Yeh correct, physical degenerate limit hai: koi motion nahi, koi period nahi.

Step 3 — Lesson. Recipe boundary par robust hai: spring ke liye degenerate limit finite hai (isochronous), box ke liye diverge karta hai (speed-controlled). Koi $0/0$ nonsense produce nahi hoti kyunki hum $J(E)$ aur $\omega (E)$ ke limits lete hain jo areas se build hue the, jo smoothly vanish karte hain.

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Ex 8 — Cell H: exam twist (kuch bhi solve kiye bina frequency)

Step 1 — Trajectory dhundho mat. Action–angle exist karne ka poora karan yeh hai ki $\omega =dE/dJ$ ko sirf $J-E$ relation chahiye. Yeh step kyun? Parent ka magic line: frequency area-vs-energy curve se aati hai, kabhi $\dot{q},\dot{p}$ integrate karke nahi.

Step 2 — Invert karo. $J=a{E}^{2/3}\Rightarrow E=(J/a{)}^{3/2}=a^{-3/2}{J}^{3/2}.$

Step 3 — Differentiate karo. $\omega =\frac{dE}{dJ}=\frac{3}{2}{a}^{-3/2}{J}^{1/2}.$ $E$ ke terms mein rewrite karo $J=a{E}^{2/3}$ use karke, to $J^{1/2}=a^{1/2}{E}^{1/3}$: $\omega =\frac{3}{2}{a}^{-3/2}\cdot a^{1/2}{E}^{1/3}=\frac{3}{2a}{E}^{1/3},T=\frac{2\pi }{\omega }=\frac{4\pi a}{3}{E}^{-1/3}.$ Yeh step kyun? $T\propto E^{-1/3}$ energy ke saath decrease karta hai — ek stiffening well (Ex 3 ke quartic ka mild version jaisa). Tumne sirf algebra se solve kar diya; "impossible-looking" problem do lines mein tha.

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Recall Self-test (answers cover karo)

$E$ just above $mg\ell$ wala pendulum — kaun sa cell? ::: Cell E (rotation), lekin barely — separatrix ke paas period diverge karta hai. Box ko factor of 2 kyun milta hai lekin whirling pendulum ko nahi? ::: Box librate karta hai (out-and-back, $\pm p$); whirling pendulum rotate karta hai (ek sweep $0\to 2\pi$, single-signed $p$). Spring ki frequency $E$ se independent hai. Kaun si matrix property yeh cause karti hai? ::: $E={\omega }_{0}J$ ek straight line hai, isliye iska slope $\omega =dE/dJ$ constant hai.

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Connections

• Parent topic — recipe
• Hamiltonian mechanics — har example ek Hamiltonian $H(q,p)$ hai
• Canonical transformations aur Generating functions — wo machinery jo $(q,p)$ ko $(\theta ,J)$ mein turn karti hai
• Hamilton–Jacobi equation — jahan se $W(q,J)$ aata hai
• Poisson brackets aur Liouville–Arnold theorem — multi-DOF (torus) generalization
• Adiabatic invariants — kyun $J$ slow changes mein preserve hota hai (Ex 7 ka shrinking loop)
• Bohr–Sommerfeld quantization — Ex 1 ka $E={\omega }_{0}J$ SHO spectrum ban jaata hai
• KAM theorem — kya bachta hai jab ye integrable systems perturb hote hain