Action-angle variables — integrable systems: Question bank

2.1.18 · D5 · Physics › Analytical Mechanics › Action-angle variables — integrable systems

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Definition Characters, plain words mein

$q$ = position coordinate (cheez kahan hai). $p$ = uska momentum (simple cases mein mass × speed).
$H$ = Hamiltonian, woh function jo system ki total energy ko $(q, p)$ ke terms mein deta hai; dekho Hamiltonian mechanics. Time-independent systems ke liye iska numerical value conserved hota hai, aur us conserved number ko hum energy $E$ kehte hain. Toh " $H = E$ " ka matlab sirf yeh hai ki "energy function abhi constant $E$ ke barabar hai."
$J = \frac{1}{2 π} \oint p d q$ = action: woh area jo motion $(q, p)$ picture mein closed loop trace karta hai, $2 π$ se divided. Sochو "loop ka size." Kyunki $J$ loop ke zariye $E$ se bana hai, hum $E = H (J)$ likhte hain jab loop area ko $J$ ke function ke roop mein express kar dete hain.
$\oint$ ka orientation: loop us direction mein traverse hota hai jis direction mein physical motion actually jaati hai (standard oscillator loop ke liye counter-clockwise), toh enclosed area — aur isliye $J$ — positive aata hai. Doosri taraf jaane wala rotor sign flip karta hai; convention se hum $J \geq 0$ lete hain contour ko motion ke saath orient karke, ya equivalently $J = \frac{1}{2 π} \oint p d q$ use karke.
$θ$ = angle: ek clock hand jo constant rate par aage badhti hai, batati hai ki loop par tum kahan ho. Har cycle mein yeh $2 π$ se badhti hai.
$ω = \partial H / \partial J$ = frequency, woh clock hand kitni tezi se tick karti hai.
$n$ = degrees of freedom ki sankhya (independent position coordinates jo system ko chahiye); ek pendulum mein $n = 1$ hota hai, plane mein planet mein $n = 2$ .
$F_{1}, \dots, F_{n}$ = conserved quantities ( $(q, p)$ ke functions jinki values motion ke saath fixed rehti hain — energy, momenta, angular momenta). ${F_{i}, F_{j}}$ = Poisson bracket, ek "kya ye do quantities interfere karti hain?" test; agar yeh zero hai toh woh in involution hain (compatible constants). Dekho Poisson brackets.

Intuition

$\dot{J} = - \partial H / \partial θ$ kyun apply hota hai Kisi bhi canonical coordinates mein, motion Hamilton's equations follow karta hai — $(θ, J)$ pair ke liye ye padhte hain $\dot{θ} = \partial H / \partial J$ aur $\dot{J} = - \partial H / \partial θ$ . Yahi wajah hai ki humne insist kiya ki $(θ, J)$ mein transformation canonical ho (ek generating function ke zariye): canonicity exactly woh property hai jo guarantee karti hai ki Hamilton's equations apna form naye variables mein rakhein. Toh jab ek baar $H$ mein $θ$ nahi hota, $\dot{J} = - \partial H / \partial θ = 0$ immediately follow karta hai.

True ya False — justify karo

Kya

J

instantaneous momentum

p

se same cheez hai?

False.

p

loop ke saath moment-to-moment badalta hai;

J

poore loop ka area

2 π

se divided hai aur constant rehta hai. Yeh sirf special symmetric cases jaise circular orbits mein coincide karte hain jahan

J = L

hota hai.

Kya action

J

hamesha motion ke saath conserved rehta hai?

True — lekin sirf transformation ke baad, kyunki naya Hamiltonian

H (J)

mein koi

θ

nahi hai, toh Hamilton's equation

\dot{J} = - \partial H / \partial θ = 0

. Yeh construction se motion ka constant hai, luck se nahi.

Harmonic oscillator ke liye, kya frequency

ω = \partial H / \partial J

energy par depend karti hai?

False.

E = ω_{0} J

se hume milta hai

ω = ω_{0}

, ek constant. SHO special hai (isochronous); zyaadatar systems mein energy-dependent frequencies hoti hain.

Kya har bounded, periodic 1-DOF system action-angle variables se solvable hai?

True ek degree of freedom ke liye — koi bhi bounded 1-DOF motion automatically integrable hai, kyunki energy conservation akela ek closed curve trace karta hai jise tum integrate kar sakte ho.

Kya har multi-DOF Hamiltonian system integrable hai?

False. Tumhe

n

(degrees of freedom ki sankhya) independent conserved quantities in involution chahiye; most systems (double pendulum, three-body) mein yeh nahi hote aur woh chaotic hote hain. Integrable systems rare exceptions hain.

Kya

\oint d θ = 2 π

exactly ek period mein hold karta hai?

True — yahi exactly wajah hai ki

J

mein

1/2 π

factor hai, taki angle har cycle mein ek clean

2 π

se badhta hai (parent mein recall derivation dekho).

Agar ek torus par frequencies

ω_{i}

irrationally related hain, kya orbit kabhi close hoti hai?

False. Irrational ratios motion ko quasi-periodic banate hain — yeh kabhi repeat nahi hoti aur poore torus ko densely fill kar deti hai.

Kya ek integrable system ke level sets

{F_{i} = c_{i}}

hamesha spheres hote hain?

False. Liouville–Arnold theorem kehta hai ki bounded level set (jahan conserved

F_{i}

fixed values

c_{i}

lete hain) ek

n

-dimensional torus

T^{n}

(ek doughnut) hai, sphere nahi.

Error dhundho

"Swinging pendulum ke liye

J

find karne ke liye,

p d q

q_{m i n}

q_{m a x}

tak ek baar integrate karo."

Galat.

\oint

poora closed loop hai:

q_{m i n} \to q_{m a x} \to q_{m i n}

. Libration ke liye woh round trip aksar one-way integral ke versus factor of 2 deta hai.

"Kyunki energy conserved hai,

H

original

(q, p)

mein already sirf

J

par depend karta hai."

Galat. Raw

(q, p)

mein,

H

abhi bhi

q

par depend karta hai. Sirf canonical transformation ke baad

(θ, J)

mein

θ

-dependence khatam hoti hai, jisse

H = H (J)

milta hai.

J

ki units energy hain, kyunki yeh energy

E

se aata hai."

Galat.

J = \frac{1}{2 π} \oint p d q

ki units [momentum × length] = [energy × time] = angular momentum hain. Yahi wajah hai ki yeh Bohr–Sommerfeld mein

ℏ

ke saath cleanly pair karta hai.

"Angle

θ

speed up aur slow down karta hai jab particle loop ke around move karta hai."

Galat. Construction se

θ = ω t + θ_{0}

constant rate par advance karta hai.

(q, p)

mein uneven-dikhne wali motion steady winding mein repackage ho jaati hai — yahi poora payoff hai.

"Ek box mein bouncing particle ke liye, ek period box ke paar ek single trip hai."

Galat. Ek poora cycle out aur back hai, toh

\oint p d q = 2 p L

. Round-trip factor miss karna tumhara action half kar deta hai aur frequency double kar deta hai.

"Do conserved quantities hamesha matlab hai system 2 DOF mein integrable hai."

Incomplete. Tumhe unhe in involution bhi chahiye:

{F_{1}, F_{2}} = 0

. Conserved lekin non-commuting constants tumhe torus structure nahi dete.

"Action-angle variables trajectory ke liye Hamilton's equations directly solve karke find hoti hain."

Galat. Point yeh hai ki tum trajectory solve karna avoid karte ho: tum ek area

J (E)

compute karte ho,

E (J)

tak invert karte ho, aur differentiate karte ho — frequency bina kisi bhi time integration ke nikal aati hai.

\oint p d q

ka sign matter nahi karta, toh

J

negative ho sakta hai."

Galat spirit mein. Hum loop ka orientation actual motion follow karne ke liye fix karte hain (ya absolute value lete hain), toh

J \geq 0

ek genuine area measure karta hai. Blindly signed integral "doosri taraf" circulate karne wale rotor ke liye negative aa sakta hai.

Why questions

J

ki definition mein

2 π

se divide kyun karte hain?

Taaki conjugate angle

θ

har period mein ek clean

2 π

advance kare, kyunki

\oint d θ = \frac{d}{dJ} \oint p d q = \frac{d}{dJ} (2 π J) = 2 π

. Yeh "clock hand" ko neatly close karta hai.

J (E)

usually monotonic kyun hai, taki hum

E (J)

tak invert kar sakein?

Bada energy

E

bada loop enclose karta hai, isliye bada area, toh

J

E

ke saath badhta hai aur inverse

E = H (J)

single-valued hota hai. Caveat: yeh separatrix par exactly fail karta hai, jahan

J \to \infty

aur smooth

J (E)

relation toot jaata hai — toh monotonic inversion sirf us boundary se door hold karta hai.

Hum type-2 generating function

W (q, J)

kyun use karte hain kisi aur type ki jagah?

Ek type-2 generator naturally old coordinate

q

aur new momentum

J

ko apne variables ke roop mein leta hai, deliver karta hai

p = \partial W / \partial q

aur

θ = \partial W / \partial J

— exactly woh pairing jo hum chahte hain, canonical by construction.

H

ko sirf

J

par depend karana immediately dynamics kyun deta hai?

Kyunki Hamilton's equations

(θ, J)

mein tab read karti hain

\dot{J} = - \partial H / \partial θ = 0

(action frozen) aur

\dot{θ} = \partial H / \partial J = ω

= const, toh motion sirf linear winding hai — trivially solved.

ω = d E / dJ

ko "magic" kyun kehte hain?

Yeh tumhe oscillation frequency area-versus-energy relation se akela deta hai, bina kisi trajectory solving ke. Saari dynamics ek derivative mein collapse ho jaati hai.

Harmonic oscillator ka phase orbit ellipse hona

J

ko itna easy kyun banata hai?

Kyunki

\oint p d q

enclosed area hi hai, aur ek ellipse ka area

π ab

hai — ek jaana-pehchana formula. Koi integration nahi chahiye, sirf geometry.

Edge cases

Ek rotor ke liye (angle

q

monotonically

2 π

ke through badhta hai, swing back nahi karta),

\oint p d q

mein kya badalta hai?

Contour alag hai:

q

ek baar

0 \to 2 π

run karta hai bina

p

sign reverse kiye, toh koi there-and-back factor of 2 nahi hota. Orientation yahan matter karta hai — opposite taraf circulate karna sign flip karta hai, toh

J \geq 0

rakhne ke liye loop actual motion ke along lete hain.

Separatrix par (pendulum jisme top tak pahunchne ke liye exactly itni energy hai), period aur

ω

ka kya hota hai?

Period infinity tak diverge karta hai aur

ω \to 0

, kyunki motion asymptotically unstable top par stall ho jaati hai. Yahan

J (E)

monotonic nahi rehta (

J \to \infty

) aur action-angle variables exactly separatrix par break down ho jaate hain.

Genuinely free particle (unbounded, koi walls nahi) ka action kya hai?

Undefined — motion periodic nahi hai, toh koi closed loop nahi hai jो area enclose kare. Action-angle machinery ko bounded motion chahiye; ise periodic banane ke liye walls add karo (box example).

Agar ek torus par do frequencies satisfy karti hain

ω_{1} / ω_{2} = 2/3

exactly, kya orbit close hoti hai?

Haan. Ek rational ratio matlab winding finite number of turns ke baad repeat hoti hai, toh trajectory torus par ek knot mein close ho jaati hai usse fill karne ki jagah.

Jab tum small perturbation on karte ho toh ek KAM torus ka kya hota hai?

Sufficiently irrational tori survive karte hain (thoda deform hokar), jabki rational/resonant wale pehle break up hote hain — near-integrable regime jo KAM theorem describe karta hai.

Connections

Liouville–Arnold theorem — woh torus supply karta hai jis par yahan har "edge case" rehta hai.
Canonical transformations aur Generating functions — "type-2 $W (q, J)$ kyun" ke peeche ki machinery.
Poisson brackets — integrability traps mein involution test.
Adiabatic invariants aur Bohr–Sommerfeld quantization — jahan $J$ ki unit pay off karti hai.
KAM theorem — last edge-case torus ka fate.

Recall One-line self-test

Upar sab dhako: kya tum bata sakte ho ki kyun round-trip factor of 2 aata hai, kyun $ω = d E / dJ$ ko koi trajectory nahi chahiye, aur kyun most systems integrable nahi hain? ::: Round-trip = $\oint$ poora closed loop hai (out aur back); $ω = d E / dJ$ kyunki area-vs-energy already frequency encode karta hai; non-integrable kyunki $n$ commuting constants of motion almost kabhi exist nahi karte.