Worked examples — Hamilton-Jacobi equation

2.1.17 · D3 · Physics › Analytical Mechanics › Hamilton-Jacobi equation

Symbols aane se pehle, chaar plain-word reminders (neeche use hone wale har symbol inhi mein se ek hai):

• $H$ = energy jo coordinates $q$ aur momenta $p$ mein express ki gayi ho (Hamiltonian).
• $S$ = Hamilton's principal function; kisi coordinate mein iska slope wahan ki momentum hoti hai: $p_i=\partial S/\partial q_i$. Physically $S$ action hai $\int Ldt$.
• $W$ = $S$ ka time-independent part jab energy conserved ho: $S=W-Et$.
• $P$ = ek naya canonical momentum jo canonical transformation se milta hai. HJ ka poora point yahi hai ki ye $P$ constants of the motion hote hain. Neeche har 1-DOF example mein hum conserved energy ko hi woh role dete hain, likhte hain $P=E$: energy ek constant hai, isliye woh machine ko dene ke liye sabse natural "new momentum" hai.

Woh ek recipe jo hum har baar repeat karte hain:

1. $H(q,\partial S/\partial q,t)+\partial S/\partial t=0$ likho.
2. Agar $H$ mein explicit $t$ nahi hai: $S=W-Et$ rakho, jisse $H(q,\partial W/\partial q)=E$ milta hai.
3. Har $\partial W/\partial q=W^{\prime }$ ke liye solve karo, integrate karke $W$ nikalo.
4. Constant energy ko naya momentum $P=E$ choose karo, phir naya coordinate $Q=\partial S/\partial P=\text{const}$ set karo — yeh constant equation hi trajectory hai.
5. $q(t)$ ke liye invert karo; check karo.

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Scenario matrix

HJ method ke har problem ka ek cell hota hai. Neeche ke examples is tarah choose kiye gaye hain ki milke woh sab cells cover kar lein.

Cell	Kya alag hai	Example
A. Separable, unbounded 1-DOF	motion infinity tak jaati hai; $E$ kisi bhi sign ka momentum	Ex 1 (free particle, dono directions)
B. Separable, bounded 1-DOF	turning points, oscillation; $p$ ka sign flip hota hai	Ex 2 (oscillator, full period + dono branches)
C. Degenerate / zero input	$E=0$ ya vanishing force — limiting case	Ex 3 ($E=0$ free particle; constant-force limit)
D. Constant force (linear potential)	non-quadratic, ek taraf unbounded, doosri taraf bounded	Ex 4 (gravity mein particle)
E. Multi-DOF additive separation	$S=\sum W_i(q_i)$; har coordinate ka apna constant	Ex 5 (2-D free particle)
F. Non-Cartesian / central force	$r,\theta$ mein separation; angular momentum ek constant ke roop mein	Ex 6 (central force, radial equation)
G. Real-world word problem	tumhe story se $H$ banana hota hai	Ex 7 (frictionless slope par skier)
H. Exam twist — time-dependent $H$	$S=W-Et$ forbidden; $t$ term rakhna zaroori	Ex 8 (time-varying push wala particle)

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Ex 1 — Free particle, dono directions (Cell A)

Forecast: padhne se pehle guess karo — ek free particle ko straight line dena chahiye $x=vt+x_0$ jahan $v=\pm \sqrt{2E/m}$. Method kaunsa sign dega?

Figure dono solutions plot karti hai: red line right-going particle ($+v$) hai aur black line left-going ($-v$). Same energy, opposite slope — yahi is cell ki kahani hai, aur steps 1–4 ise build karte hain.

1. HJ likho, time separate karo. $\frac{1}{2m}(W^{\prime }{)}^2=E$, isliye $W^{\prime }=\pm \sqrt{2mE}$. Yeh step kyun? Energy conserved hai (koi explicit $t$ nahi), isliye $S=W-Et$ aur PDE sirf slope $W^{\prime }=p$ ke liye ek algebraic equation ban jaata hai. $\pm$ hi is cell ka poora point hai: same energy par $p$ kisi bhi sign ka ho sakta hai — momentum square direction kho deta hai, hum yahan ise restore karte hain.
2. Integrate karo. $W=\pm \sqrt{2mE}x$, isliye $S=\pm \sqrt{2mE}x-Et$. Kyun? Slope ko integrate karne se function rebuild hota hai; sign batata hai particle kis direction mein ja raha hai.
3. Constant new coordinate. $P=E$ lo (energy conserved new momentum hai). Tab $Q=\frac{\partial S}{\partial E}=\pm \sqrt{\frac{m}{2E}}x-t=-t_0$. Kyun? Humne $K=0$ demand ki thi, isliye Hamilton's equations se $\dot{Q}=\partial K/\partial P=0$: $Q$ literally time mein change nahi kar sakta. Time mein frozen cheez ek constant ke barabar hoti hai, jise hum $-t_0$ name dete hain — aur woh equation hi disguise mein trajectory hai.
4. Invert karo. $\pm \sqrt{m/2E}x-t=-t_0$ ko $x$ ke liye rearrange karne par milta hai $x=\pm \sqrt{\frac{2E}{m}}(t-t_0)$. Kyun? Hum position as a function of time chahte hain, $x(t)$, isliye $x$ isolate karte hain. Bracket $(t-t_0)$ dikhata hai ki $t_0$ woh instant hai jab particle origin se guzarta hai.

Verify: velocity $\dot{x}=\pm \sqrt{2E/m}$, isliye $H=\frac{1}{2}m{\dot{x}}^2=E$ ✅. $E=8\text{ J},m=1\text{ kg}$ lo: $v=\pm 4\text{ m/s}$. Units: $\sqrt{\text{J}/\text{kg}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$ ✅. Dono signs figure mein dono lines hain (right-going red mein, left-going black mein).

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Ex 2 — Harmonic oscillator, full period, dono branches (Cell B)

Forecast: bounded matlab $x$ sirf wahin reh sakta hai jahan $E\ge \frac{1}{2}m{\omega }^2{x}^2$. Compute karne se pehle turning points aur period guess karo.

Figure potential $V(x)$ (black parabola) ko energy line $E$ se cut hote dikhata hai; axis par red band allowed region $|x|\le x_{\max }$ hai, aur do red dots jahan $V=E$ hain turning points hain. Compute karte waqt us band ko dekhte rehna — woh exactly wahi range hai jo algebra allow karta hai.

1. Slope solve karo. $W^{\prime }=p=\pm \sqrt{2m(E-\frac{1}{2}m{\omega }^2{x}^2)}$. Kyun? $H(x,p)=E$ ko $p$ ke liye solve karo. Square-root tab hi real hai jab $E-\frac{1}{2}m{\omega }^2{x}^2\ge 0$, yaani $|x|\le x_{\max }=\sqrt{2E/(m{\omega }^2)}$. Woh do equalities turning points hain — figure mein red allowed region ki edges.
2. $p$ ka sign branch name karta hai. Right jaate waqt ($x$ increasing) $+$ use hota hai; $x_{\max }$ hit karne ke baad particle turn karta hai aur $-$ use hota hai. Yahi bounded case ki subtlety hai jise cell B dikhane ke liye exist karta hai.
3. Constant coordinate. $P=E$ ke saath, $Q=\frac{\partial S}{\partial E}=\int \frac{mdx}{\sqrt{2m(E-\frac{1}{2}m{\omega }^2{x}^2)}}-t=\text{const}.$ Kyun? Same $K=0$ logic; $\partial /\partial E$ integral ke andar act karta hai.
4. Integral karo. $x=x_{\max }\sin \theta$ substitute karne par milta hai $\frac{1}{\omega }\arcsin (x/x_{\max })=t+\text{const}$. $\arcsin$ kyun? Kyunki integrand exactly ek arcsine ka derivative hai — phase space mein ek circle ki geometry. Action-Angle Variables mein yahi integral angle variable ban jaata hai.
5. Invert karo. $x(t)=x_{\max }\sin (\omega t+\varphi )$, amplitude $x_{\max }=\sqrt{2E/(m{\omega }^2)}$. Kyun? Pichli line ko $x$ ke liye solve kiya dono sides ka sine lekar (sine arcsine ko undo karta hai); bacha hua constant phase $\varphi$ ban jaata hai. Isse position time ka explicit function milti hai.
6. Period. Ek full loop (right branch aur left branch) phase $2\pi$ cover karta hai: $T=2\pi /\omega$. Kyun? Sine repeat hota hai jab uska argument $\omega t$ $2\pi$ advance karta hai; uske liye elapsed time $T=2\pi /\omega$ hai. Dono branches (jaana aur aana) milke exactly ek aisa repeat banate hain.

Verify: $x=x_{\max }$ par, $\dot{x}=x_{\max }\omega \cos (\pm \pi /2)=0$ — particle turning point par ruk jaata hai ✅. Energy: $\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2{x}_{\max }^2=E$ ✅. Numbers $m=1,\omega =2,E=8$: $x_{\max }=\sqrt{16/4}=2$, $T=\pi \approx 3.1416$ ✅.

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Ex 3 — Degenerate input: $E=0$ aur vanishing case (Cell C)

Forecast: zero energy ka matlab usually "still baithna" hota hai, lekin kya method dono systems ke liye agree karta hai?

1. (a) Free particle, $E=0$. $W^{\prime }=\sqrt{2m\cdot 0}=0\Rightarrow p=0$, isliye $\dot{x}=p/m=0$: particle kahin bhi rest mein hai. Kyun? $E=0$ sirf kinetic energy ke saath $p=0$ force karta hai. Yeh Cell A ka degenerate edge hai — velocity zero ho jaati hai.
2. (b) Oscillator, $E=0$. $x_{\max }=\sqrt{2\cdot 0/(m{\omega }^2)}=0$, isliye sirf allowed point $x=0$ hai. Kyun? Allowed region $|x|\le x_{\max }$ ek single point tak shrink ho jaati hai — oscillator apne well ke bottom par baitha rehta hai, $x(t)\equiv 0$.

Verify: (a) $x(t)=x_0$ constant se $\dot{x}=0$, $H=0$ ✅. (b) $x_{\max }=0\Rightarrow x(t)=0$, aur $H=0$ ✅. Dono degenerate limits Ex 1–2 ke $E\to 0$ ke saath consistent hain (amplitude aur speed dono $\to 0$).

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Ex 4 — Constant force / linear potential (Cell D)

Forecast: yeh 1-D mein projectile motion hai — expect karo $z(t)$ quadratic hoga, time mein ek parabola. Energy mein downward unbounded lekin upar throw karne par ek top turning point hoga.

Figure resulting height $z(t)$ ko time mein red downward parabola ke roop mein plot karta hai, jahan black dot peak ko mark karta hai jahan particle momentarily ruka hota hai. Steps 1–4 exactly wahi curve derive karte hain; step 1 ke turning point tak pahunchte waqt peak ko phir dekhna.

1. Slope. $W^{\prime }=p=\pm \sqrt{2m(E-mgz)}$. Kyun? $H=E$ ko $p$ ke liye solve karo (momentum isolate karo, kyunki $W^{\prime }=p$ hamesha hamara pehla target hota hai). Real tabhi jab $z\le E/(mg)$ — highest reachable height $z_{\max }=E/(mg)$, single turning point (upar bounded, neeche unbounded): linear potential ki signature, aur figure mein peak.
2. Constant coordinate. $P=E$ ke saath, $Q=\frac{\partial S}{\partial E}=\int \frac{mdz}{\sqrt{2m(E-mgz)}}-t=\text{const}.$ Kyun? Kyunki $K=0$ force karta hai $\dot{Q}=0$, isliye $Q$ frozen hai aur ek constant ke barabar hai. $S=\int W^{\prime }dz-Et$ ko new momentum $E$ se differentiate karne par $\partial /\partial E$ integral ke andar aata hai aur $-Et$ term se ek $-t$ girta hai.
3. Integrate karo. $\int \frac{mdz}{\sqrt{2m(E-mgz)}}=-\frac{1}{g}\sqrt{\frac{2(E-mgz)}{m}}$. Kyun? Direct substitution $u=E-mgz$; $\sqrt{u}$ antiderivative appear hota hai.
4. Constant set karo, solve karo. $-\frac{1}{g}\sqrt{\frac{2(E-mgz)}{m}}-t=-t_0$. Square karke rearrange: $z(t)=\frac{E}{mg}-\frac{g}{2}(t-t_0{)}^2.$ Kyun? Hum $z$ ko $t$ ka function chahte hain, isliye square-root isolate karo, remove karne ke liye square karo, phir $z$-mein-linear equation solve karo. Subtracted $(t-t_0{)}^2$ term hi trajectory ko figure ki parabola mein bend karta hai.

Verify: $\dot{z}=-g(t-t_0)$, $\ddot{z}=-g$ — constant downward acceleration ✅ (Newton ka $F=-mg$). $t=t_0$ par, $z=E/(mg)=z_{\max }$ aur $\dot{z}=0$: peak ✅. Numbers $E=20\text{ J},m=1\text{ kg},g=10{\text{ m/s}}^2$: $z_{\max }=2\text{ m}$ ✅.

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Ex 5 — Multi-DOF additive separation (Cell E)

Forecast: har direction ek independent free particle hai, isliye guess karo ki $S$ ek $x$-part plus $y$-part mein split hoga, har ek apna constant carry karta hua.

1. Split assume karo. $S=W_x(x)+W_y(y)-Et$. Kyun? Kyunki $H$ independent pieces ka sum hai, ansatz "har coordinate ke liye ek $W$" PDE ko alag equations mein tod deta hai — multi-DOF HJ ki core trick. Yahan $W_x^{\prime }\equiv d{W}_x/dx$ aur $W_y^{\prime }\equiv d{W}_y/dy$.
2. Separate karo. $\frac{1}{2m}(W_x^{\prime }{)}^2+\frac{1}{2m}(W_y^{\prime }{)}^2=E$. Set karo $\frac{1}{2m}(W_x^{\prime }{)}^2={\alpha }_x$ (ek constant). Tab $\frac{1}{2m}(W_y^{\prime }{)}^2=E-{\alpha }_x\equiv {\alpha }_y$. ${\alpha }_x$ kyun introduce karo? Sirf $x$ wala term sab $x,y$ ke liye ($y$+const) term ke barabar hota hai tabhi jab har side constant ho. Woh constants ${\alpha }_x,{\alpha }_y$ new momenta hain — exactly woh "$n$ constants" jo parent note ne warn kiye the.
3. Integrate karo. $W_x=\sqrt{2m{\alpha }_x}x$, $W_y=\sqrt{2m{\alpha }_y}y$.
4. Trajectory. $Q_x=\partial S/\partial {\alpha }_x=\sqrt{\frac{m}{2{\alpha }_x}}x-t=\text{const}$ se milta hai $x=\sqrt{2{\alpha }_x/m}(t-t_0)$; similarly $y=\sqrt{2{\alpha }_y/m}(t-t_1)$. Kyun? Har $\alpha$ ek new momentum hai, isliye har $Q=\partial S/\partial \alpha$ frozen hai ($\dot{Q}=0$) aur ek constant ke barabar hai; un do constant equations se $x$ (aur $y$) isolate karne par dono uniform-motion lines milti hain.

Verify: speeds $v_x=\sqrt{2{\alpha }_x/m}$, $v_y=\sqrt{2{\alpha }_y/m}$ se milta hai $\frac{1}{2}m(v_x^2+v_y^2)={\alpha }_x+{\alpha }_y=E$ ✅. Numbers $m=1,{\alpha }_x=2,{\alpha }_y=3$: $v_x=2,v_y=\sqrt{6}\approx 2.449$, total $E=5$ ✅.

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Ex 6 — Central force, $r,\theta$ mein separation (Cell F)

Forecast: $\theta$ kabhi $H$ mein appear nahi karta (sirf $p_{\theta }$), isliye guess karo ki uska momentum conserved hai — woh constant angular momentum hona chahiye.

1. Ansatz. $S=W_r(r)+W_{\theta }(\theta )-Et$, jahan $p_r=W_r^{\prime }\equiv d{W}_r/dr$, $p_{\theta }=W_{\theta }^{\prime }\equiv d{W}_{\theta }/d\theta$. Kyun? Same additive-separation idea jaise Ex 5, ab polar coordinates mein.
2. $\theta$ isolate karo. $\theta$-dependence sirf $p_{\theta }^2/r^2$ ke through aata hai. Kyunki $H$ mein $\theta$ itself nahi hai, $p_{\theta }=W_{\theta }^{\prime }$ constant hona chahiye; ise $\ell$ (angular momentum) kaho. Isliye $W_{\theta }=\ell \theta$. Yeh step kyun? $H$ mein absent coordinate cyclic hoti hai; uska momentum automatically conserved hota hai — HJ mein yeh separation constant ke roop mein fall out hota hai.
3. Radial equation. $p_{\theta }=\ell$ substitute karo: $\frac{1}{2m}[(W_r^{\prime }{)}^2+\frac{{\ell }^2}{r^2}]+V(r)=E\Rightarrow W_r^{\prime }=\pm \sqrt{2m(E-V(r))-\frac{{\ell }^2}{r^2}}.$ Kyun? $\theta$ handle ho gaya, sirf $r$ bacha — effective potential $V_{\text{eff}}=V+{\ell }^2/(2m{r}^2)$ ke saath ek 1-DOF problem. $\pm$ yahan matter karta hai: $+$ root outward ja rahe particle ($r$ increasing) hai, $-$ root inward ja rahe particle ($r$ decreasing) ka; ek bound orbit dono use karta hai baar-baar, har radial turning point par sign flip karta hai jahan root zero ho jaata hai.
4. Result. $S=\pm \int \sqrt{2m(E-V)-{\ell }^2/r^2}dr+\ell \theta -Et$; do constants $(E,\ell )$ new momenta hain.

Verify (special case $V=0$, closest-approach turning point). $V=0$ ke liye turning point ($W_r^{\prime }=0$) $r_0=\ell /\sqrt{2mE}$ par hai. Numbers $\ell =6,m=1,E=8$: $r_0=6/\sqrt{16}=1.5$ ✅. Us radius par saari energy angular hai: ${\ell }^2/(2m{r}_0^2)=36/(2\cdot 2.25)=8=E$ ✅.

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Ex 7 — Real-world word problem: frictionless slope (Cell G)

Forecast: energy conservation kehta hai $v=\sqrt{2gh}$, angle se independent. Kya HJ wahi reproduce karega?

Figure scene set karta hai: red dot skier hai jo black slope par angle $\alpha$ ke saath partially neeche aa gayi hai, vertical height $h$ mark hai. Jo coordinate hum use karte hain, $s$, slope ke saath start se run karta hai — step 1 is picture ko Hamiltonian mein turn karta hai.

1. $H$ banao. Slope ke along distance par drop $\sin \alpha$ hai, isliye height $=h-s\sin \alpha$ aur potential $V=mg(h-s\sin \alpha )$. Isliye $H=\frac{p_s^2}{2m}+mg(h-s\sin \alpha )=E$. Yeh step kyun? Word problem ka sabse mushkil hissa $H$ likhna hota hai: natural coordinate choose karo ($s$ incline ke along, jaise drawn hai) aur energy use mein express karo.
2. Start conditions $E$ fix karte hain. Top par rest mein: $s=0,p_s=0\Rightarrow E=mgh$. Kyun? $E$ ek number hai jo initial state se set hota hai.
3. $W$ ka slope. $W_s^{\prime }=p_s=\sqrt{2m(E-mg(h-s\sin \alpha ))}=\sqrt{2m^{2}gs\sin \alpha }$ ($E=mgh$ use karke). Kyun? $H=E$ ko $p_s$ ke liye solve karo; $mgh$ beautifully cancel ho jaata hai, sirf drop $s\sin \alpha$ bachta hai.
4. Speed. $v=p_s/m=\sqrt{2gs\sin \alpha }$. Vertical drop $d=s\sin \alpha$ hai, isliye $v=\sqrt{2gd}$.

Verify: ek slope ke bottom par jiska foot height $0$ par hai, total vertical drop $d=h$, jisse milta hai $v=\sqrt{2gh}$ — angle cancel ho jaata hai, energy conservation se match ✅. Numbers $g=10,h=5$: $v=\sqrt{100}=10\text{ m/s}$ ✅. Units $\sqrt{({\text{m/s}}^2)\cdot \text{m}}=\text{m/s}$ ✅.

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Ex 8 — Exam twist: explicitly time-dependent $H$ (Cell H)

Forecast: kyunki energy conserved nahi hai (force time-dependent work karta hai), Ex 1–7 wali trick forbidden hai. Expect karo $p$ $\frac{1}{2}b{t}^2$ ki tarah badhe.

1. Shortcut kyun fail karta hai. $H$ mein $t$ explicitly hai, isliye $\partial S/\partial t\ne -\text{const}$: split $S=W(x)-Et$ constant energy assume karta hai aur yahan allowed nahi hai. Exam exactly yahi trap set karta hai.
2. Full HJ. $\frac{1}{2m}(\frac{\partial S}{\partial x}{)}^2-btx+\frac{\partial S}{\partial t}=0$. Kyun? Honest time term rakho — koi conserved energy nahi, isliye full time-dependent PDE se kaam lena hoga.
3. $x$ mein linear ansatz. Try karo $S=f(t)x+g(t)$ taaki $p=\partial S/\partial x=f(t)$. Yeh shape kyun? Force $x$ mein uniform hai (har jagah same), isliye momentum sirf time par depend karna chahiye; $x$ mein first-degree function sabse simple $S$ hai jiska $x$-slope $x$-independent ho.
4. Match karo. Substitute karo: $\frac{f^2}{2m}-btx+f^{\prime }x+g^{\prime }=0$. $x$-term collect karo: $f^{\prime }=bt\Rightarrow f=\frac{1}{2}b{t}^2+p_0$. Baaki: $g^{\prime }=-f^2/(2m)$. Kyun? Equation sab $x$ ke liye hold karni chahiye, isliye $x$ ka coefficient aur $x$-free part alag-alag zero hone chahiye — ek equation $f$ ke liye, ek $g$ ke liye. Yahan $p_0$ integration constant hai ($t=0$ par momentum).
5. Momentum aur velocity. $p=f(t)=\frac{1}{2}b{t}^2+p_0$ se aur $\dot{x}=p/m$: $\dot{x}=\frac{1}{2m}b{t}^2+\frac{p_0}{m}.$ Kyun? Type-2 relation directly $p$ deta hai; $m$ se divide karne par momentum velocity ban jaata hai, kyunki is kinetic term ke liye $p=m\dot{x}$.
6. Position ke liye integrate karo. $\dot{x}$ ko time mein ek baar integrate karke: $x(t)=x_0+\frac{p_0}{m}t+\frac{b}{6m}{t}^3.$ Kyun? Hum $x(t)$ chahte the, aur velocity $dx/dt$ hai; $t^2\to t^3/3$ integrate karne par cubic milta hai, $x_0$ starting position ke saath. Badhta $t^3$ term us force ka fingerprint hai jo time ke saath khud badh rahi hai.

Verify: Newton kehta hai $m\ddot{x}=F=bt$, isliye $\ddot{x}=bt/m$. Hamare $x(t)$ se: $\ddot{x}=\frac{b}{m}t$ ✅. Numbers $b=6,m=1,p_0=0,x_0=0$: $x(t)=t^3$, isliye $x(2)=8$ aur $\ddot{x}(t)=6t\Rightarrow F=6t$, $bt$ se match ✅.

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Wrap-up: har cell fill ho gaya

Recall Kaun sa example kaun sa scenario cover karta hai?

Unbounded 1-DOF, $p$ ke dono signs ::: Ex 1 (Cell A) Bounded oscillation, turning points, period ::: Ex 2 (Cell B) Degenerate zero-energy limit ::: Ex 3 (Cell C) Linear potential (constant force) ::: Ex 4 (Cell D) Multi-DOF additive separation ::: Ex 5 (Cell E) Central force, cyclic angle, angular momentum constant ::: Ex 6 (Cell F) Real-world story jahan H banana hota hai ::: Ex 7 (Cell G) Explicitly time-dependent H — no S=W−Et ::: Ex 8 (Cell H)