Hamilton-Jacobi equation — Hinglish notes

2.1.17 · Physics › Analytical Mechanics

1. Setup: canonical transformations (KISE lean on karte hain)

Hum type-2 generating function $F_{2} (q, P, t) \equiv S (q, P, t)$ use karte hain, jo purane coordinates $q$ aur nayi momenta $P$ par depend karta hai. Iske defining relations hain: $p_{i} = \frac{\partial S}{\partial q _{i}}, Q_{i} = \frac{\partial S}{\partial P _{i}}, K = H + \frac{\partial S}{\partial t},$ jahan $K (Q, P, t)$ naya Hamiltonian hai (yaani "Kamiltonian").

2. HJ equation ko scratch se derive karna (KAISE)

Goal: $S$ isko is tarah choose karo ki naya Hamiltonian zero ho jaaye: $K \equiv 0$ .

$K = 0$ kyun help karta hai? Nayi variables mein Hamilton's equations hain: $\dot{Q}_{i} = \frac{\partial K}{\partial P _{i}}, \dot{P}_{i} = - \frac{\partial K}{\partial Q _{i}} .$ Agar $K = 0$ har jagah ho, toh $\dot{Q}_{i} = 0$ aur $\dot{P}_{i} = 0$ : saari nayi coordinates aur momenta constants hain. Motion solved.

Yeh step kyun? Hum deliberately trivial dynamics "khareed" rahe hain $K$ par sabse strong possible condition lagaake.

Ab $K = H + \partial S / \partial t$ mein $K = 0$ impose karo: $H (q_{1}, \dots, q_{n}, \frac{\partial S}{\partial q _{1}}, \dots, \frac{\partial S}{\partial q _{n}}, t) + \frac{\partial S}{\partial t} = 0.$

Yeh step kyun? Humne $p_{i} = \partial S / \partial q_{i}$ (type-2 relation) ko $H$ mein substitute kiya, har momentum ko $S$ ki derivative mein badal diya. Result hai yeh celebrated equation:

3. $S$ physically kya hai?

Yeh step kyun? Humne $\partial S / \partial q_{i} = p_{i}$ aur $\partial S / \partial t = - H$ use kiya, phir $\sum p_{i} \overset{q}{˙}_{i} - H = L$ recognize kiya (Legendre transform). Yahi deep payoff hai: HJ action, canonical transformations, aur PDEs ko unify karta hai.

4. Time-independent case: $t$ ka separation

Agar $H$ mein koi explicit time dependence nahi hai, toh energy conserved hai. Separation try karo: $S (q, P, t) = W (q, P) - E t .$

Yeh step kyun? Plug in karne par, $\partial S / \partial t = - E$ , toh HJ equation ban jaati hai: $H (q, \frac{\partial W}{\partial q}) = E .$

5. Worked Example A — Free particle (1D)

Hamiltonian: $H = \frac{p ^{2}}{2 m}$ .

Step 1. HJ likho: $\frac{1}{2 m} (\frac{\partial S}{\partial x})^{2} + \frac{\partial S}{\partial t} = 0.$ Kyun? $p \to \partial S / \partial x$ replace karo.

Step 2. $S = W (x) - E t$ separate karo: $\frac{1}{2 m} (W^{'})^{2} = E \Rightarrow W^{'} = 2 m E .$ Kyun? Energy conserved hai, toh time part alag karo.

Step 3. Integrate karo: $W = 2 m E x$ , toh $S = 2 m E x - E t .$

Step 4. Naya coordinate constant hai: $E$ ko naya momentum $P = E$ lo. Toh: $Q = \frac{\partial S}{\partial E} = \frac{m}{2 E} x - t = const \equiv - t_{0} .$ Kyun? $Q$ construction se constant hai ( $K = 0$ ).

Step 5. $x$ ke liye solve karo: $x = \frac{2 E}{m} (t - t_{0})$ — uniform velocity $v = 2 E / m$ . ✅ Exactly free particle.

6. Worked Example B — Harmonic oscillator (1D)

$H = \frac{p ^{2}}{2 m} + \frac{1}{2} m ω^{2} x^{2} = E .$

Step 1. $W^{'} = 2 m (E - \frac{1}{2} m ω^{2} x^{2})$ . Kyun? $W^{'} = \partial W / \partial x$ ke liye $H (x, W^{'}) = E$ solve karo.

Step 2. $S = \int 2 m (E - \frac{1}{2} m ω^{2} x^{2}) d x - E t$ .

Step 3. Conserved naya coordinate: $P = E$ ke saath, $Q = \frac{\partial S}{\partial E} = \int \frac{m d x}{2 m ( E - \frac{1}{2} m ω ^{2} x ^{2} )} - t = const .$ Kyun? Integral ke andar differentiate karo; $\partial / \partial E$ $m$ le aata hai.

Step 4. Integral evaluate hota hai (sub $x = 2 E / m ω^{2} sin θ$ ) is tarah: $\frac{1}{ω} arcsin (x \frac{m ω ^{2}}{2 E}) = t + const .$

Step 5. Invert karo: $x (t) = \frac{2 E}{m ω ^{2}} sin (ω t + ϕ)$ . ✅ Familiar oscillation jiska amplitude $E$ se set hota hai.

7. Common mistakes

8. Active recall

Recall Quick self-test (hide karke jawab do)

Naye Hamiltonian $K$ par kaun si condition HJ equation produce karti hai? → $K = 0$ .
$S$ physically kya hai? → action $\int L d t$ .
$n$ DOF ke complete integral mein kitne constants hote hain? → $n$ (plus ek trivial additive constant).
$H$ mein $p_{i}$ ki jagah kya aata hai? → $\partial S / \partial q_{i}$ .

Hamilton-Jacobi equation poori tarah kaise likhi jaati hai?

H (q, \partial S / \partial q, t) + \partial S / \partial t = 0

Naye Hamiltonian

K

par kaun si condition HJ equation generate karti hai?

K \equiv 0

, jisse saari nayi coordinates aur momenta constant ho jaati hain.

Hamilton's principal function

S

physically kya hai?

Action

\int L d t

actual trajectory par evaluate kiya gaya.

Kaun sa generating-function type use hota hai aur uske relations kya hain?

Type-2

F_{2} = S (q, P, t)

, jisme

p_{i} = \partial S / \partial q_{i}

Q_{i} = \partial S / \partial P_{i}

K = H + \partial S / \partial t

Hamilton's characteristic function

W

kya hai aur kab defined hoti hai?

W (q)

from

S = W - E t

, defined jab

H

time-independent ho; satisfies

H (q, \partial W / \partial q) = E

n

degrees of freedom ke liye, complete integral mein kitne independent constants hote hain?

n

non-trivial constants, nayi momenta se identify kiye jaate hain

P_{i} = α_{i}

Free particle ke liye

S

kya hai?

S = 2 m E x - E t

K = 0

motion ko trivial kyun bana deta hai?

Kyunki

\dot{Q} = \partial K / \partial P = 0

aur

\dot{P} = - \partial K / \partial Q = 0

, toh saari nayi variables constant ho jaati hain.

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho tumhe ek complicated maze cross karni hai. Newton ka tarika: deewarein feel karo aur step by step push karo. Hamilton-Jacobi ka tarika: pehle ek magic map banao jahan har point par pehle se ek arrow likha ho jo bolta hai "yahan jao." Ek baar woh perfect map ( $S$ ) ban jaaye, tum kabhi nahi sochte — bas arrows follow karo aur maze khud solve ho jaata hai. Magic map banana mushkil hai (ek bada equation hai), lekin uske baad chalana free hai. Arrows momenta hain, aur map ki "height" har point tak pahunchne ki total mehnat (action) hai.

9. Connections

Canonical Transformations — HJ constant variables ka canonical transformation hai.
Hamilton's Equations — starting point; HJ $2 n$ ODEs ko ek PDE se replace karta hai.
Action and Hamilton's Principle — $S$ action hai; HJ woh PDE hai jo action obey karta hai.
Separation of Variables — HJ solve karne ka practical method integrable systems ke liye.
Action-Angle Variables — periodic motion ke liye $W$ se directly banaye jaate hain.
Schrodinger Equation — QM ka classical limit HJ tak reduce hota hai ( $ψ \sim e^{i S /ℏ}$ ).