Hamilton-Jacobi equation: Visual walkthrough — Hinglish deep dive

2.1.17 · D2 · Physics › Analytical Mechanics › Hamilton-Jacobi equation

Step 0 — Stage: "phase space" aur "trajectory" ka matlab

Koi bhi equation se pehle, picture apne dimaag mein fix karo.

Ek degree of freedom wale mechanical system ki ek position $q$ hoti hai (cheez kahan hai) aur ek momentum $p$ (kitna motion carry kar raha hai: simple particle ke liye mass times velocity).
$q$ horizontal axis pe aur $p$ vertical axis pe plot karo. Ye plane phase space hai. Ek point $(q, p)$ system ki poori state abhi ke liye fix karta hai.
Jaise-jaise time $t$ beetha hai, point move karta hai, ek curve trace karta hai — ye trajectory hai.

Figure s01 — Phase space aur ek trajectory. Blue curve history hai; yellow dot abhi ki state hai.

Blue curve dekho: har instant pe dot hi system hai. Newton is dot ko step by step push karta. Hum ek coordinate change dhundhne wale hain jo ek naye frame mein dot ko freeze kar de.

Step 1 — Khwaish: coordinates jahan kuch move na kare

KYA. Hum bilkul naye coordinates $(Q, P)$ introduce karte hain jo same physics describe karte hain, is tarah choose kiye gaye ki in naye axes mein trajectory ek akela stationary point ho — bilkul bhi move na kare.

KYUN. Agar $\dot{Q} = 0$ aur $\dot{P} = 0$ (yaad raho dot ka matlab $d / d t$ hai, time rate of change), toh $Q$ aur $P$ sirf fixed numbers hain. Motion phir pure algebra se padhi ja sakti hai: koi differential equation baaki nahi bachti. Ye hi Hamilton-Jacobi ka poora sapna hai.

PICTURE. Same trajectory, do coordinate grids. Purane $(q, p)$ grid mein state ek curve sweep karti hai; naye $(Q, P)$ grid mein ye ek frozen dot hai.

Figure s02 — Purana moving vs naya frozen. White arrow wo canonical map hai jo sweeping curve ko ek single green dot mein bhejta hai.

$purana, moving (q, p) ⟶ naya, frozen (Q, P)$

$q, p$ — purani position aur momentum, moving dot.
$Q, P$ — nayi position aur momentum; humara goal hai ki ye constants hon.

Pure chapter ka sawaal: konsa variable change ye achieve karta hai, aur kya wo allowed hai?

Step 2 — "Allowed" matlab canonical: wo rulebook jise hum follow karna chahiye

KYA. Har variable change legal nahi hota. Ek legal change canonical transformation hota hai — ye motion ke laws ki shape ko intact rakhta hai.

KYUN. Hamilton's Equations hain $\overset{q}{˙} = \frac{\partial H}{\partial p}, \overset{p}{˙} = - \frac{\partial H}{\partial q} .$

$\overset{q}{˙}, \overset{p}{˙}$ — position aur momentum ke time rates of change (dot = $d / d t$ ).
$H$ — Hamiltonian, system ki energy $q$ aur $p$ ke function ke roop mein likhi gayi.
$\partial H / \partial p$ — "agar main $p$ ko nudge karun aur $q$ fixed rakhuun toh $H$ kitni tezi se badalti hai"; ye $\overset{q}{˙}$ ke barabar nikalta hai.
Doosri equation mein minus sign hi woh cheez hai jo phase-space flow ko swirl karaati hai na ki spread — ye Hamiltonian dynamics ki pehchaan hai.

Canonical transformation precisely woh hota hai jiske baad naye variables same-shaped equations follow karein kisi naye energy $K$ ke saath: $\dot{Q} = \frac{\partial K}{\partial P}, \dot{P} = - \frac{\partial K}{\partial Q} .$ $K$ naya Hamiltonian hai (pyaar se "Kamiltonian" bhi kahte hain).

PICTURE. Dono grids mein same swirl pattern rehta hai. Ek canonical map phase space ka ek tedha-medha re-labelling hai jo swirl kabhi nahi todta.

Figure s03 — Swirl preserve hota hai. Har blue arrow flow $(\overset{q}{˙}, \overset{p}{˙})$ hai; canonical map ko ye pattern kabhi nahi faadhna chahiye.

Step 3 — Purane aur naye ke beech bridge: ek generating function

KYA. Hum actually ek canonical map specify kaise karte hain? Ek single generating function ke through. Hum type-2 wala use karte hain, $S (q, P, t)$ — jo purani position $q$ aur nayi momentum $P$ pe depend karta hai.

TYPE-2 HI KYUN, KUCH AUR KYUN NAHI? Hum chahte hain ki naye momenta $P$ woh constants hon jo hamne Step 1 mein freeze kiye. Type-2 function $P$ ko ek independent input maanta hai, isliye ye "constant naye momenta design karne" ka natural tool hai.

TEEN RELATIONS KAHAN SE AATE HAIN. Ek canonical transformation build ki ja sakti hai ye demand karke ki purana aur naya description same action principle de. Action principle mein purane variables $p \overset{q}{˙} - H$ contribute karte hain aur naye $P \dot{Q} - K$ . In dono mein sirf kisi function ke total time-derivative ka fark ho sakta hai — warna variational principle badal jaata aur map canonical nahi rehta. Wo function $(q, P, t)$ pe depend karne wala choose karke (type-2), bookkeeping sabse clean rehti hai agar hum $\frac{d}{d t} (P Q)$ add kar dein, jo deta hai $p \overset{q}{˙} - H = - Q \dot{P} - K + \frac{d S}{d t}, S = S (q, P, t) .$ $\frac{d S}{d t} = \frac{\partial S}{\partial q} \overset{q}{˙} + \frac{\partial S}{\partial P} \dot{P} + \frac{\partial S}{\partial t}$ expand karo aur independent rates $\overset{q}{˙}$ aur $\dot{P}$ ke coefficients match karo (aur bacha hua constant piece). $\overset{q}{˙}$ match karne se $p = \partial S / \partial q$ milta hai; $\dot{P}$ match karne se $Q = \partial S / \partial P$ milta hai; leftover se $K = H + \partial S / \partial t$ :

$p = \frac{\partial S}{\partial q}, Q = \frac{\partial S}{\partial P}, K = H + \frac{\partial S}{\partial t} .$

Har ek ko ek sentence ki tarah padho:

$p = \frac{\partial S}{\partial q}$ — purana momentum $S$ ka slope hai jab tum $q$ slide karte ho ( $P, t$ fixed rakh ke). (Ye wahi relation hai jo humein derivatives se momenta dhundhne degi.)
$Q = \frac{\partial S}{\partial P}$ — naya coordinate $S$ ka slope hai jab tum naye momentum ko slide karte ho.
$K = H + \frac{\partial S}{\partial t}$ — nayi energy purani energy plus kitna $S$ khud time mein change hota hai. (Explicit-time term $\partial S / \partial t$ woh knob hai jise hum exploit karenge.)

YE GENUINE TRANSFORMATION KAB HAI? In slope relations ke invertible hone ke liye — taaki $(q, P)$ se hum actually $(p, Q)$ recover kar sakein aur ulta bhi — $S$ ka second-derivative matrix (Hessian) degenerate nahi hona chahiye: $det (\frac{\partial ^{2} S}{\partial q \partial P}) \neq = 0.$ Agar ye determinant zero hota, toh map dimensions collapse kar deta aur bona-fide canonical transformation fail ho jaata. Hum assume karte hain ki ye hold karta hai.

PICTURE. $S$ $(q, P)$ plane ke upar ek landscape ki tarah baitha hai; uske do directions mein slopes $p$ aur $Q$ ki values emit karte hain, jabki uska time mein slope extra term $K$ mein dalta hai.

Figure s04 — $S$ -landscape $p$ aur $Q$ emit karta hai. $q$ -direction mein slope $p$ deta hai; $P$ -direction mein slope $Q$ deta hai.

Step 4 — Killer demand: nayi energy ko zero hone par majboor karo

KYA. Hum ab sabse bold choice karte hain: demand karte hain ki $K \equiv 0 har jagah.$

KYUN. $K = 0$ ko Step 2 ke naye Hamilton's equations mein daalo: $\dot{Q} = \frac{\partial K}{\partial P} = 0, \dot{P} = - \frac{\partial K}{\partial Q} = 0.$

$\dot{Q} = 0$ — naya coordinate kabhi nahi badalta (dot = $d / d t$ ).
$\dot{P} = 0$ — naya momentum kabhi nahi badalta.

Dono constant hain — exactly Step 1 ka frozen dot. Hum $K$ pe available sabse strong condition impose karke trivial dynamics khareed rahe hain. Ye koi random trick nahi hai; $K = 0$ hi mathematically "naye variables ko rok do" ka spelling hai.

PICTURE. Ek flat energy surface (nayi phase space par $K = 0$ har jagah) matlab har direction mein zero slope, toh dono flow rates khatam — dot ek jagah thaak jaata hai.

Figure s05 — Flat nayi energy dot ko freeze karti hai. $K = 0$ har jagah ke saath, $K$ ki har slope zero hoti hai, isliye $\dot{Q} = \dot{P} = 0$ .

Step 5 — Equation assemble karo: rulebook ko $K = 0$ mein substitute karo

KYA. Type-2 relation lelo $K$ ke liye aur use zero set karo: $K = H + \frac{\partial S}{\partial t} = 0 ⟹ H + \frac{\partial S}{\partial t} = 0.$

Ab doosri type-2 relation use karo, $p = \partial S / \partial q$ , $H$ ke andar har momentum ko eliminate karne ke liye. Jahan bhi $H$ ne $p$ maanga, hum use $\partial S / \partial q$ de dete hain:

$H (q, \frac{\partial S}{\partial q}, t) + \frac{\partial S}{\partial t} = 0$

Term by term:

$H (q, \cdot, t)$ — energy function, jo ab momentum slot mein $S$ ki derivative le raha hai.
$\frac{\partial S}{\partial q}$ — purane momentum $p$ ki jagah khada hai.
$\frac{\partial S}{\partial t}$ — $S$ ka time-slope, jo pehle $K$ ka extra piece tha; $K = 0$ set karne se ye yahan aa gaya.

KYUN YE EK FAYDA HAI. Hum coupled ordinary differential equations ki ek system se shuru kiye the (Hamilton's equations). Humne unhe ek PDE ke badle mein diya. Ye trade tabhi worthwhile hai kyunki PDE aksar separate hoti hai (Step 7).

PICTURE. Ek flow chart: do type-2 relations $K = 0$ ke box mein daalte hain aur HJ equation bahar aati hai.

Figure s06 — HJ equation assemble ho rahi hai. Dono type-2 relations $K = 0$ ki demand mein daalte hain; HJ PDE bahar aati hai.

Step 6 — Matlab padho: $S$ action hai

KYA. Pehle, us actor ka naam lo jis se hum milne wale hain.

Ab $S (q, P, t)$ ko real trajectory ke saath follow karo aur uski total time rate of change compute karo. Chain rule deta hai $\frac{d S}{d t} = \frac{\partial S}{\partial q} \overset{q}{˙} + \frac{\partial S}{\partial t} .$

( $\partial S / \partial P$ term yahan drop ho jaata hai kyunki $P$ motion ke along ek constant hai, isliye $\dot{P} = 0$ .) Ab do facts substitute karo jo hum pehle hi prove kar chuke hain:

$\frac{\partial S}{\partial q} = p$ (Step 3),
$\frac{\partial S}{\partial t} = - H$ (Step 5, rearranged),

toh $\frac{d S}{d t} = p \overset{q}{˙} - H = L .$

$p \overset{q}{˙}$ — momentum times velocity ( $\overset{q}{˙} = d q / d t$ ).
$p \overset{q}{˙} - H$ — ye combination exactly Lagrangian $L$ hai jo abhi define kiya.

Integrate karo: $S = \int L d t + const$ .

KYUN YE MATTER KARTA HAI. $S$ koi abstract convenience nahi hai — ye action hai, wahi quantity jo Hamilton's principle mein minimize hoti hai. HJ equation action ka apna local law hai.

PICTURE. Jaise dot blue trajectory pe aage badhta hai, $S$ area $\int L d t$ accumulate karta hai; $S$ ki growth rate har instant pe $L$ hai.

Figure s07 — $S$ action accumulate karta hai. $L$ ke neeche shaded area hi $S$ ki badhti hui value hai; har instant pe uski slope $L$ ke barabar hai.

Step 7 — Degenerate/simplifying case: jab $H$ time ignore kare

KYA. Agar $H$ mein koi explicit $t$ nahi hai, toh energy conserved hai. Hum variables separate karte hain ye split guess karke $S (q, t) = W (q) - E t .$

$W (q)$ — characteristic function, sara $q$ -dependence isme hai.
$- E t$ — time mein ek saaf straight-line wala piece; $E$ conserved energy hai.
Toh $\frac{\partial S}{\partial t} = - E$ automatically aa jaata hai.

KYUN. HJ equation mein substitute karne par, do terms alag ho jaate hain: $H (q, \frac{\partial W}{\partial q}) - E = 0 ⟹ H (q, \frac{\partial W}{\partial q}) = E .$ Time gayab ho gaya; hum $W (q)$ ke liye ek ordinary differential equation ke saath bache hain. Ye tabhi kaam karta hai kyunki $H$ time-blind tha — ye hi degenerate condition hai.

PICTURE. $S$ ko $(q, t)$ ke upar ek surface ki tarah dekho: ek fixed shape $W (q)$ jo constant rate $E$ se neeche slide karta hai jaise time badhta hai — ek rigid ramp jo $t$ mein translate ho raha hai.

Figure s08 — Rigid ramp time mein slide kar raha hai. Shape $W (q)$ fixed hai; $t$ badhne se poori curve constant rate $E$ se neeche aati hai.

Step 8 — Sabse simple system pe sanity check: free particle

KYA. $H = \frac{p ^{2}}{2 m}$ lo (mass $m$ , koi force nahi). Uski position ab bhi hamara ek coordinate $q$ hai — koi naya letter nahi, free particle simply $q$ -line pe rehta hai.

Step-through (har move justified):

HJ: $\frac{1}{2 m} (\frac{\partial S}{\partial q})^{2} + \frac{\partial S}{\partial t} = 0$ — $p$ ki jagah $\partial S / \partial q$ rakho.
$S = W - E t$ separate karo: $\frac{1}{2 m} (W^{'})^{2} = E \Rightarrow W^{'} = 2 m E$ — energy conserved, time peel off karo. Yahan $W^{'} \equiv d W / d q$ .
Integrate karo: $W = 2 m E q$ , toh $S = 2 m E q - E t$ .
$P = E$ ko naya momentum lo. Naya coordinate constant hai: $Q = \frac{\partial S}{\partial E} = \frac{m}{2 E} q - t \equiv - t_{0} .$ Yahan $t_{0}$ simply integration constant hai — ye woh fixed value hai jo constant $Q$ leta hai, aur iske units time ke hain, isliye hum ise $- t_{0}$ kehte hain; physically $t_{0}$ woh instant hai jab particle $q = 0$ se guzarta hai (uska initial-time offset).
$q$ ke liye solve karo: $q = \frac{2 E}{m} (t - t_{0})$ — uniform velocity $v = 2 E / m$ , aur $t = t_{0}$ pe particle $q = 0$ pe hai. ✅

KYUN. Ye sabse humble possible test hai: free particle zaroor constant speed se straight line mein move karna chahiye, aur HJ exactly wahi reproduce karta hai. Machinery trustworthy hai.

PICTURE. Purani space mein, $q$ $t$ ke saath linearly chalta hai; naye coordinates mein pair $(Q, P) = (- t_{0}, E)$ ek single frozen dot hai — Step 1 ka promise deliver hua.

Figure s09 — Free particle recover hua. Blue line constant speed pe $q (t)$ hai; red dot $t_{0}$ mark karta hai jahan $q = 0$ ; green square frozen naya state hai.

Ek-picture summary

Upar sab kuch ek single arrow diagram mein simat jaata hai: do type-2 relations se, $K = 0$ demand ke through, HJ equation tak, aur bahar frozen constants $Q, P$ plus recovered trajectory.

Figure s10 — Poori derivation ek picture mein. Type-2 relations → $K = 0$ ki demand → HJ PDE → ( $S$ action hai) → frozen $Q, P$ → algebra se trajectory.

Recall Feynman retelling — poora walk plain words mein

Tumhare paas ek dot hai jo ek swirl mein ghoom rahi hai (phase space). Tum chahte ho ek jaaduyi goggles ka pair jo dot ko bilkul still kar de. Goggles ek canonical re-labelling hai — allowed tabhi hai jab ye swirl intact rakhe. Unhe banane ke liye tum ek landscape function $S$ invent karte ho: uski slopes automatically purana momentum bata deti hain ( $p = \partial S / \partial q$ ) aur naya position ( $Q = \partial S / \partial P$ ), jab tak landscape degenerate na ho (uska curvature determinant non-zero ho). Phir tum sabse bold wish karte ho: naya energy exactly zero ho. Zero energy ka matlab zero slope har jagah, matlab dot move nahi kar sakta — frozen, exactly jaisa chahte the. Wo wish likhne par, "purani energy plus $S$ ka time-slope zero ke barabar hai," aur har momentum ko $S$ ki slope se swap karne par, tumhe ek equation milti hai: Hamilton-Jacobi equation. Naya momentum $P$ bas ek dial ki tarah saath chalta hai jo tum set karte ho. Aur punch line: wo landscape $S$ wahi action nikalta hai jo tumhe pehle se pata thi — uski growth rate Lagrangian $L = p \overset{q}{˙} - H$ hai. Ek PDE solve karo, frozen constants ko waapas follow karo, aur poori motion algebra se nikal aati hai. Ye hi woh jaaduyi map hai jo tumhare liye maze chalti hai.

Recall Rapid self-test

Naye coordinates mein kya cheez $\dot{Q} = \dot{P} = 0$ banati hai? ::: $K = 0$ ki demand, kyunki $\dot{Q} = \partial K / \partial P$ aur $\dot{P} = - \partial K / \partial Q$ .
Konsi type-2 relation $H$ ko PDE mein badal deti hai? ::: $p = \partial S / \partial q$ .
HJ PDE solve hote waqt $P$ kya role karta hai? ::: Ek held-fixed parameter (ek dial); har value se ek solution milta hai, aur $P$ conserved naya momentum ban jaata hai.
Generating function ek real canonical map define karne ke liye kya condition chahiye? ::: Non-degenerate Hessian, $det (\partial^{2} S / \partial q \partial P) \neq = 0$ .
$S$ physically kya hai? ::: Real path ke along action $\int L d t$ , jahan $L = p \overset{q}{˙} - H$ .
$W$ kab exist karta hai aur wo kaunsi equation solve karta hai? ::: Jab $H$ time-independent ho; $H (q, \partial W / \partial q) = E$ .
Free particle ka $q (t)$ answer kya hai, aur $t_{0}$ kya hai? ::: $q = 2 E / m (t - t_{0})$ ; $t_{0}$ integration constant hai, woh instant jab particle $q = 0$ se guzarta hai.

Aage kahan jaayein: Action-Angle Variables is machinery ko periodic motion pe apply karta hai, aur Schrodinger Equation woh hai jo HJ ban jaata hai jab tum $S$ ko ek exponential ke andar rehne do.