Hamilton-Jacobi equation: Exercises — Hinglish deep dive

2.1.17 · D4 · Physics › Analytical Mechanics › Hamilton-Jacobi equation

Shuru karne se pehle, ek shared vocabulary reminder, kyunki har problem isko use karta hai:

Definition Woh symbols jo hum har jagah reuse karte hain

$q$ — ek generalized coordinate: system ka abstract "position" variable. Neeche har concrete exercise mein $q$ ek physical naam pehanta hai — 1D horizontal motion ke liye hum likhte hain $q = x$ (ek line ke saath ek length), vertical free fall ke liye $q = z$ (ek height), aur 2D problem mein do coordinates hain $q_{1} = x$ aur $q_{2} = y$ . Ye $q$ ki tarah ka same kind of object hain; sirf letter badalta hai picture se match karne ke liye.
$H$ — purana Hamiltonian, woh machine jo ek position $q$ aur momentum $p$ leta hai aur total energy return karta hai.
$L$ — Lagrangian, kinetic minus potential energy, $L = T - V$ . Yeh woh quantity hai jiska time-integral action hai; hume isko sirf yeh kehne ke liye chahiye ki $S$ physically kya hai (agla bullet dekhein). Aur detail Action and Hamilton's Principle mein.
$K$ — variables $(q, p) \to (Q, P)$ change karne ke baad naya Hamiltonian (the "Kamiltonian"). $K = 0$ set karna hi poora trick hai: yeh naye coordinates $Q$ aur naye momenta $P$ ko constant bana deta hai. Canonical Transformations dekhein.
$S (q, t)$ — Hamilton's principal function, woh unknown jise hum solve karte hain. Yeh sach wali path par action $\int L d t$ ke barabar hai. Yeh generating function hai jo purane aur naye variables ko link karti hai: $K = H + \partial S / \partial t$ .
$W (q)$ — Hamilton's characteristic function: $S$ ka time-stripped hissa. Yeh tabhi exist karta hai jab $H$ mein explicit time na ho (energy conserved ho), aur tab $S (q, t) = W (q) - E t$ . Iska slope momentum bhi deta hai: $p = \partial W / \partial q$ . Yeh sawaal ka jawaab deta hai "action kaisa dikhta hai jab hum steady $- E t$ drift factor out kar lein?" Hum isko almost har exercise mein use karte hain.
$\frac{\partial S}{\partial q}$ — padho "kitni tezi se $S$ badalta hai agar main $q$ ko nudge karun jabki time freeze karun". HJ rule kehta hai yeh momentum $p$ ke barabar hai.
$\frac{\partial S}{\partial t}$ — "kitni tezi se $S$ badalta hai jab time tick kare aur $q$ frozen ho". $K = 0$ ke saath yeh $- H$ ke barabar hai.
Notation guard: lowercase $p$ = purana momentum; uppercase $P$ = naya (constant) momentum; $α$ = ek separation constant, jise hum phir ek naye momentum $P$ se identify karte hain. Toh $α$ aur $P$ ek hi object hain, do hats pehne hue.

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

Level 1 — Recognition

Exercise 1.1 (L1)

$H = \frac{p ^{2}}{2 m} + V (x)$ wale 1D system ke liye (yahan coordinate $q = x$ hai), Hamilton-Jacobi equation likho bina solve kiye.

Recall Solution

Kya karna hai: har $p$ ban jaata hai $\partial S / \partial x$ , aur hum $\partial S / \partial t$ append karte hain. Kyun: ye generating-function relations $p = \partial S / \partial x$ aur $K = H + \partial S / \partial t$ se aate hain, defining demand $K = 0$ ke saath (new Hamiltonian $K$ vocabulary box mein define kiya gaya tha). $K$ ko kill karne par bachta hai: $\frac{1}{2 m} (\frac{\partial S}{\partial x})^{2} + V (x) + \frac{\partial S}{\partial t} = 0.$ Yahi answer hai. Koi integration nahi chahiye — sirf recognition.

Exercise 1.2 (L1)

Batao kaunsa principal function hai aur kaunsa characteristic function, aur woh ek equation jo unhe link karta hai.

Recall Solution

Principal $= S (q, t)$ — time carry karta hai.
Characteristic $= W (q)$ — iske andar koi time nahi (vocabulary box mein define kiya gaya hai).
Link (sirf tab valid jab $H$ time-independent ho, toh energy $E$ conserved ho): $S (q, t) = W (q) - E t .$ Minus $E t$ kyun: agar $\partial S / \partial t = - E$ hai toh $H (q, \partial W / \partial q) = E$ , bilkul wahi conserved-energy ka statement.

Exercise 1.3 (L1)

HJ equation kis order ki PDE hai aur kya yeh linear hai ya nonlinear? Ek sentence mein karan do.

Recall Solution

Yeh first order hai ( $S$ ke sirf first derivatives appear hote hain) aur nonlinear hai (momentum $\partial S / \partial q$ , $H$ mein typically square hoke enter karta hai, jaise $(\partial S / \partial x)^{2}$ , aur square linear nahi hota).

Level 2 — Application

Exercise 2.1 (L2)

Free particle, $H = \frac{p ^{2}}{2 m}$ jisme coordinate $q = x$ hai. Time-independent HJ equation solve karo $W (x)$ ke liye aur poora $S (x, t)$ likho.

Recall Solution

Step 1 — restricted HJ. $S = W - E t$ ke saath: $\frac{1}{2 m} (W^{'})^{2} = E$ . Step 2 — slope solve karo. $W^{'} = \pm 2 m E$ (rightward motion ke liye $+$ root lo; $-$ root leftward motion deta hai — dono legal branches hain). Step 3 — integrate karo. $W = 2 m E x + C$ . $C$ kyun drop karte hain: $C$ ek pure additive constant hai. Kyunki $S$ ke sirf derivatives hi physics mein enter karte hain ( $p = \partial S / \partial x$ , $Q = \partial S / \partial E$ ), $S$ mein constant add karna kuch measurable nahi badalta. Hum $C = 0$ bina loss of generality ke set karte hain — solution poora general rehta hai. Step 4 — assemble. $S = 2 m E x - E t$ . Parent note ke Example A se match karta hai.

Exercise 2.2 (L2)

2.1 continue karte hue, $E$ ko new momentum $P = E$ se identify karo. Constant new coordinate $Q = \partial S / \partial E$ compute karo aur $x (t)$ solve karo.

Recall Solution

Step 1. $Q = \frac{\partial S}{\partial E} = \frac{\partial}{\partial E} (2 m E x - E t) = \frac{m}{2 E} x - t .$ Kyun: $\partial_{E} 2 m E = 2 m \cdot \frac{1}{2 E} = m /2 E$ . Step 2. $Q$ constant hai (kyunki $K = 0 \Rightarrow \dot{Q} = 0$ ); isko $- t_{0}$ bolo, toh $m /2 E x = t - t_{0}$ . Step 3. $x (t) = \frac{2 E}{m} (t - t_{0})$ — uniform velocity $v = 2 E / m$ . Sanity check: $\frac{1}{2} m v^{2} = E$ . ✅

Exercise 2.3 (L2)

Free fall mein ek particle, $H = \frac{p ^{2}}{2 m} + m g z$ , jahan coordinate height $q = z$ hai. HJ equation likho aur separate karo $W^{'} (z)$ paane ke liye.

Recall Solution

Step 1 — HJ: $\frac{1}{2 m} (\partial_{z} S)^{2} + m g z + \partial_{t} S = 0$ . Step 2 — separate $S = W (z) - E t$ . Yeh separation kyun allowed hai: Hamiltonian $H = \frac{p ^{2}}{2 m} + m g z$ mein koi explicit $t$ nahi hai, toh energy $E$ conserved hai. Jab bhi $H$ time-independent ho, $S$ ki poori $t$ -dependence sirf ek steady drift $- E t$ ho sakti hai (yahi precisely $\partial_{t} S = - E$ kehta hai). Isko peel karne par ek purely spatial function $W (z)$ bachti hai — yeh wahi time-separation move hai jo har exercise mein aur Separation of Variables mein use hota hai. $\partial_{t} S = - E$ substitute karne par: $\frac{1}{2 m} (W^{'})^{2} + m g z = E .$ Step 3 — slope solve karo: $W^{'} (z) = \pm 2 m (E - m g z)$ . Notice karo $W^{'} \to 0$ turning height $z = E / (m g)$ par jahan saari energy potential hai — momentum wahan vanish karta hai, exactly jaise physics demand karta hai. $W^{'}$ ko phir integrate karne par ek additive constant aata hai jise hum drop karte hain (sirf derivatives matter karte hain, jaise 2.1 mein).

Level 3 — Analysis

Exercise 3.1 (L3)

Harmonic oscillator, $H = \frac{p ^{2}}{2 m} + \frac{1}{2} m ω^{2} x^{2} = E$ (coordinate $q = x$ ). $W^{'} = \pm 2 m (E - \frac{1}{2} m ω^{2} x^{2})$ se shuru karke $x (t)$ explicitly derive karo, substitution dikhate hue.

Recall Solution

Step 0 — do branches. $H = E$ solve karne par slope milta hai $W^{'} = \pm 2 m (E - \frac{1}{2} m ω^{2} x^{2})$ . $+$ branch particle ko right move karta dikhata hai ( $+ A$ ki taraf), $-$ branch left. Ye dono turning points $x = \pm A$ par milte hain aur swap karte hain jahan $W^{'} = 0$ . Ek poora oscillation in do branches ka stitched version hai — exactly L2 ka lesson. Neeche hum $+$ branch integrate karte hain; $-$ branch returning half deta hai aur dono milkar ek continuous $sin$ banate hain. Step 1 — constant coordinate ko ek explicit lower limit ke saath well-posed banao. Pehle $W$ ko ek fixed reference point $x_{0}$ se genuine definite integral ki tarah likho (ek constant jo $E$ par depend nahi karta): $W (x, E) = \int_{x_{0}}^{x} 2 m (E - \frac{1}{2} m ω^{2} x^{'2}) d x^{'} .$ Fixed lower limit choose karna exactly wahi hai jo "additive constant drop karna" matlab hai: $x_{0}$ shift karna sirf $W$ mein ek $x$ -independent constant add karta hai, jo kabhi koi physics affect nahi karta (sirf $S$ ke derivatives enter karte hain). Kyunki lower limit $x_{0}$ $E$ -independent hai aur upper limit $x$ bhi $E$ -independent hai, differentiation under the integral sign ab legitimate hai: sirf integrand $E$ par depend karta hai. $P = E$ ke saath, $Q = \frac{\partial S}{\partial E} = \int_{x_{0}}^{x} \frac{\partial}{\partial E} 2 m (E - \frac{1}{2} m ω^{2} x^{'2}) d x^{'} - t = \int_{x_{0}}^{x} \frac{m d x ^{'}}{2 m ( E - \frac{1}{2} m ω ^{2} x ^{'2} )} - t .$ $\partial_{E}$ se $m$ kyun neeche aata hai: $\partial_{E} 2 m (E - \dots) = \frac{2 m}{2 2 m ( \dots )} = \frac{m}{2 m ( \dots )}$ . Step 2 — substitute $x^{'} = \frac{2 E}{m ω ^{2}} sin θ$ . Tab radicand ban jaata hai $2 m E cos^{2} θ$ aur $d x^{'} = 2 E / m ω^{2} cos θ d θ$ . Integral collapse ho jaata hai: $\int \frac{m \cdot 2 E / m ω ^{2} c o s θ d θ}{2 m E c o s θ} = \frac{1}{ω} \int d θ = \frac{θ}{ω} .$ Step 3 — substitution undo karo: $θ = arcsin (x \frac{m ω ^{2}}{2 E})$ , toh (fixed lower-limit contribution ko constant $Q$ mein absorb karte hue) $Q + t = \frac{1}{ω} arcsin (\dots) + const$ . Step 4 — invert ( $- ω Q + const \equiv ϕ$ set karo): $x (t) = \frac{2 E}{m ω ^{2}} sin (ω t + ϕ)$ . Amplitude $A = 2 E / (m ω^{2})$ energy $E$ se poori tarah fixed hai — neeche figure dekhein.

Neeche ka figure yeh solution plot karta hai taaki tum dekh sako algebra ne kya produce kiya: horizontal axis time $t$ hai, vertical axis position $x (t)$ hai. Lavender curve $x (t) = A sin (ω t + ϕ)$ hai. Do coral dashed lines amplitude $\pm A$ mark karti hain — classical turning points jahan Step 0 ke $W^{'}$ ke do branches (the $+$ aur $-$ roots) milte hain aur momentum $p$ vanish karta hai. Mint dot ek zero crossing $x = 0$ mark karta hai, jahan speed maximum hai aur saari energy kinetic hai. Picture padhna: energy $E$ set karta hai wave kitni tall hai (amplitude), jabki $ω$ set karta hai kitni fast wiggle karti hai.

Figure s01 — 1D harmonic oscillator ka HJ solution. Horizontal axis: time $t$ . Vertical axis: position $x (t) = A sin (ω t + ϕ)$ (lavender). Coral dashed lines: amplitude $\pm A = \pm 2 E / (m ω^{2})$ , woh turning points jahan momentum $p = 0$ aur saari energy potential hai. Mint dot: ek zero crossing $x = 0$ , jahan speed maximal hai aur saari energy kinetic hai. Energy $E$ wave ki height control karta hai; $ω$ uski frequency control karta hai.

Exercise 3.2 (L3)

Us oscillator ke liye, verify karo ki amplitude $A = 2 E / (m ω^{2})$ maximum displacement par exactly energy $E$ carry karta hai.

Recall Solution

Maximum displacement par saari energy potential hai ( $p = 0$ ): $\frac{1}{2} m ω^{2} A^{2} = \frac{1}{2} m ω^{2} \cdot \frac{2 E}{m ω ^{2}} = E . ✅$ HJ construction self-consistent hai: algebra ne jo amplitude spat out kiya woh precisely wahi classical turning point hai jo figure mein coral dashed lines se mark hai.

Exercise 3.3 (L3)

Numerical check. $m = 2 kg$ , $ω = 3 s^{- 1}$ , $E = 27 J$ wala oscillator. Amplitude $A$ aur maximum speed $v_{m a x}$ nikalo.

Recall Solution

Amplitude: $A = \frac{2 E}{m ω ^{2}} = \frac{2 \cdot 27}{2 \cdot 9} = 3 \approx 1.732 m$ . Max speed ( $x = 0$ par saari energy kinetic): $\frac{1}{2} m v_{m a x}^{2} = E \Rightarrow v_{m a x} = 2 E / m = 27 = 33 \approx 5.196 m/s$ . Cross-check: $v_{m a x} = ω A = 3 \cdot 3 = 33$ . ✅ Consistent.

Level 4 — Synthesis

Exercise 4.1 (L4)

Separable 2D problem. $H = \frac{p _{x}^{2}}{2 m} + \frac{p _{y}^{2}}{2 m} + \frac{1}{2} m ω^{2} x^{2}$ ( $x$ mein oscillator, $y$ mein free; do generalized coordinates $q_{1} = x$ aur $q_{2} = y$ hain). Additive separation $W = W_{x} (x) + W_{y} (y)$ use karke HJ equation ko do independent 1D pieces mein split karo. Separation constants identify karo.

Recall Solution

Step 1 — restricted HJ: $\frac{1}{2 m} (W_{x}^{'})^{2} + \frac{1}{2} m ω^{2} x^{2} + \frac{1}{2 m} (W_{y}^{'})^{2} = E$ . Step 2 — additive kyun? Kyunki $H$ ek $x$ -only part aur $y$ -only part ka sum hai. $W = W_{x} + W_{y}$ guess karne par har $\partial W / \partial x = W_{x}^{'}$ sirf $x$ par depend karta hai (Separation of Variables). Step 3 — regroup: $function of x \frac{1}{2 m} (W_{x}^{'})^{2} + \frac{1}{2} m ω^{2} x^{2} = function of y E - \frac{1}{2 m} (W_{y}^{'})^{2} .$ $x$ ki function $y$ ki function ke barabar sab $x, y$ ke liye tabhi hogi jab dono ek constant ke barabar hon. Isko $α_{x}$ bolo. Step 4 — do 1D equations: $\frac{1}{2 m} (W_{x}^{'})^{2} + \frac{1}{2} m ω^{2} x^{2} = α_{x}, \frac{1}{2 m} (W_{y}^{'})^{2} = E - α_{x} \equiv α_{y} .$ Separation constants: $α_{x}$ ( $x$ -oscillator mein energy) aur $α_{y} = E - α_{x}$ (free $y$ -motion mein energy). Hamari notation guard follow karte hue, hum in constants ko new momenta se identify karte hain: $P_{1} = α_{x}, P_{2} = α_{y}$ — same objects, do hats. Yeh directly Action-Angle Variables se link karta hai, jahan aise constants actions ban jaate hain.

Exercise 4.2 (L4)

4.1 mein, $y$ -piece completely solve karo aur $y (t)$ do.

Recall Solution

Step 1: $(W_{y}^{'})^{2} = 2 m α_{y} \Rightarrow W_{y}^{'} = \pm 2 m α_{y} \Rightarrow W_{y} = 2 m α_{y} y$ (rightward drift ke liye $+$ branch lete hain; additive constant pehle ki tarah drop kiya). Step 2 — conjugate coordinate: $α_{y}$ ko new momentum $P_{2}$ maano, toh $Q_{y} = \partial S / \partial α_{y}$ . $S$ ka relevant piece $W_{y} - α_{y} t$ hai ( $E t$ ka $y$ -share), toh $Q_{y} = \frac{m}{2 α _{y}} y - t = const .$ Step 3 — invert: $y (t) = \frac{2 α _{y}}{m} (t - t_{0})$ — ek free particle speed $2 α_{y} / m$ par drift karta hua, exactly jaise expected. Oscillator part Exercise 3.1 follow karta hai $E \to α_{x}$ ke saath.

Level 5 — Mastery

Exercise 5.1 (L5)

Quantum mechanics se bridge. $ψ = e^{i S /ℏ}$ ("action se bana wave") time-dependent Schrodinger Equation $i ℏ \partial_{t} ψ = - \frac{ℏ ^{2}}{2 m} \partial_{x}^{2} ψ + V ψ$ (coordinate $q = x$ ) mein insert karo. Dikhao ki $ℏ \to 0$ par tum HJ equation recover karte ho, aur leftover term identify karo.

Recall Solution

Step 1 — $ψ = e^{i S /ℏ}$ ke derivatives compute karo. Chain rule use karke (aur second $x$ -derivative ke liye, do factors par product rule use karke jo yeh create karta hai): $\partial_{t} ψ = \frac{i}{ℏ} (\partial_{t} S) ψ, \partial_{x} ψ = \frac{i}{ℏ} (\partial_{x} S) ψ,$ $\partial_{x}^{2} ψ = \partial_{x} [\frac{i}{ℏ} (\partial_{x} S) ψ] = [\frac{i}{ℏ} \partial_{x}^{2} S + (\frac{i}{ℏ} \partial_{x} S)^{2}] ψ = [- \frac{1}{ℏ ^{2}} (\partial_{x} S)^{2} + \frac{i}{ℏ} \partial_{x}^{2} S] ψ .$ $\partial_{x}^{2} ψ$ mein do terms notice karo: exponent ko do baar differentiate karne ka square, aur slope ko ek baar differentiate karne ka $\partial_{x}^{2} S$ — yeh doosra term is exercise ka poora point hai. Step 2 — substitute karo aur $ψ$ se divide karo (har jagah nonzero): $i ℏ \cdot \frac{i}{ℏ} \partial_{t} S = - \frac{ℏ ^{2}}{2 m} [- \frac{1}{ℏ ^{2}} (\partial_{x} S)^{2} + \frac{i}{ℏ} \partial_{x}^{2} S] + V .$ Step 3 — simplify. Left side: $i ℏ \cdot \frac{i}{ℏ} = i^{2} = - 1$ , toh yeh $- \partial_{t} S$ hai. Right side, $- \frac{ℏ ^{2}}{2 m}$ distribute karo: $- \partial_{t} S = \frac{1}{2 m} (\partial_{x} S)^{2} - \frac{i ℏ}{2 m} \partial_{x}^{2} S + V .$ Step 4 — rearrange HJ shape mein (sab ek side): $\frac{1}{2 m} (\partial_{x} S)^{2} + V + \partial_{t} S = \frac{i ℏ}{2 m} \partial_{x}^{2} S .$ Step 5 — $ℏ \to 0$ lo. Right-hand side $ℏ$ ke proportional hai, toh limit mein vanish ho jaata hai, bachta hai $\frac{1}{2 m} (\partial_{x} S)^{2} + V + \partial_{t} S = 0$ — exactly the Hamilton-Jacobi equation. Leftover $\frac{i ℏ}{2 m} \partial_{x}^{2} S$ first quantum correction hai (WKB expansion ka seed). Toh HJ wahi hai Schrödinger equation ka classical (short-wavelength) limit — Hamilton ne wave mechanics ek century pehle anticipate ki thi.

Exercise 5.2 (L5)

Design problem. Tumhe bataya gaya hai ki desired trajectory $x (t) = a t$ (constant velocity $a$ ) hai ek free particle ke liye. Reverse-engineer karo woh principal function $S (x, t)$ jo ise generate kare, aur confirm karo ki $p = \partial S / \partial x$ sahi momentum deta hai.

Recall Solution

Step 1 — energy choose karo. Constant speed $a$ matlab $E = \frac{1}{2} m a^{2}$ . Step 2 — free-particle $S$ use karo: $S = 2 m E x - E t = (ma) x - \frac{1}{2} m a^{2} t$ , kyunki $2 m E = m^{2} a^{2} = ma$ . Step 3 — momentum check karo: $p = \partial S / \partial x = ma$ . ✅ Sahi: constant momentum $ma$ speed $a$ ke liye. Step 4 — constant coordinate check karo: $Q = \partial S / \partial E$ . $E = \frac{1}{2} m a^{2}$ ke saath, general form differentiate karna better hai: $Q = m /2 E x - t = x / a - t$ . $Q = 0$ set karne par $x = a t$ milta hai. ✅ Design target trajectory reproduce karta hai.

Exercise 5.3 (L5)

Numerical mastery. Free particle, $m = 4 kg$ , prescribed speed $a = 5 m/s$ . $E$ , momentum $p$ , aur $S$ ko $x = 10 m, t = 2 s$ par evaluate karo.

Recall Solution

$E = \frac{1}{2} m a^{2} = \frac{1}{2} \cdot 4 \cdot 25 = 50 J$ .
$p = ma = 4 \cdot 5 = 20 kg\cdotm/s$ .
$S = p x - E t = 20 \cdot 10 - 50 \cdot 2 = 200 - 100 = 100 J\cdots$ . Sab consistent hai $2 m E = 2 \cdot 4 \cdot 50 = 400 = 20 = p$ ke saath. ✅

Recall Final self-test (hide and answer)

HJ ki order aur linearity? ::: First order, nonlinear.
2D oscillator-plus-free system ke liye kitne separation constants? ::: Do: $α_{x}$ aur $α_{y} = E - α_{x}$ .
Schrödinger→HJ mein $ℏ \to 0$ se pehle kaunsa term bachta hai? ::: $\frac{i ℏ}{2 m} \partial_{x}^{2} S$ .
$m = 2, ω = 3, E = 27$ wale oscillator ka amplitude? ::: $3 \approx 1.732 m$ .

ψ = e^{i S /ℏ}

ka classical limit kahan rehta hai?

ℏ \to 0

mein, exactly HJ equation deta hua.

2D separable system ke liye, naye momenta kya hain?

Separation constants

P_{1} = α_{x}, P_{2} = α_{y} = E - α_{x}