Visual walkthrough — Hamilton's equations of motion

2.1.12 · D2 · Physics › Analytical Mechanics › Hamilton's equations of motion

Hum sirf ek cheez assume karte hain jo aap Lagrangian Mechanics se jaante ho: ek system ka ek function $L$ hota hai jise Lagrangian kehte hain, aur uski motion Euler–Lagrange rule follow karti hai. Baaki sab hum naye sire se banate hain.

Step 1 — Ek curve, aur us curve ki slope

KYA. Ek position $q$ aur time $t$ fix karo, aur sirf velocity $\dot{q}$ (position kitni tezi se change ho rahi hai, "kitna fast") ko vary karne do. Tab $L$ ek ordinary curve ban jaati hai: horizontal axis $\dot{q}$, vertical axis $L$.

KYUN. Hamilton jo bhi karta hai woh sab is curve ki slopes ka ek trick hai. $\dot{q}$ ko kisi aur cheez se trade karne se pehle, hume dekhna hoga ki $L$ versus $\dot{q}$ ka curve aakhir hota kya hai. Ek free particle ke liye $L=\frac{1}{2}m{\dot{q}}^2$, toh yeh ek seedha upward bowl (parabola) hai.

PICTURE. Pale-yellow curve dekho. Us par ek chosen point par, chalk-blue line tangent hai — woh sirf curve ko graze karti hai. Uski steepness (rise over run) woh number hai jise hum abhi naam dene wale hain.

Step 2 — Curve par ek point ko do tarike se padhna

KYA. Bowl-shaped curve ko har point $(\dot{q},L)$ list karke describe kiya ja sakta hai — YA har tangent line list karke: uski slope $p$ aur woh tangent vertical axis par kahan milti hai.

KYUN. Yahi Legendre Transform ka gehra idea hai (dekho Legendre Transform). Ek convex curve apni tangent lines ke family se poori tarah pin ho jaati hai. Toh "velocity $\dot{q}$ par height $L$" ki jagah hum kisi ko "slope $p$ aur intercept" de sakte hain. Koi information nahi jaati — humne sirf kaunsa variable haath mein rakhein woh $\dot{q}$ se $p$ mein badal diya.

PICTURE. Blue tangent line, extend karne par, vertical axis ko negative height par cross karti hai. Origin se us intercept tak ki positive doori ko hum apni nayi quantity $H$ kehte hain.

Step 3 — Velocity kyun gayab hoti hai: tangent ko slide hote dekho

KYA. $\dot{q}$ ko thoda sa nudge karo. Curve height $dL=pd\dot{q}$ (slope times step) se change hoti hai. Saath hi $p\dot{q}$ change hota hai $\dot{q}dp+pd\dot{q}$ se. Subtract karo, jaisa $H$ demand karta hai.

KYUN. Hum chahte hain $H$ $\dot{q}$ ko bilkul bhool jaaye. Chaliye cancellation hote dekhen, sirf claim na karein.

PICTURE. Do nearby tangent lines: jab $\dot{q}$ right mein $d\dot{q}$ step karta hai, toh extra bit $pd\dot{q}$ jo curve climb karti hai woh exactly woh extra bit hai jo term $p\dot{q}$ climb karta hai. Woh ek doosre ko erase kar dete hain, sirf woh change bachta hai jo $p$ ke move karne se aata hai.

$dH=\underset{\text{from }p\text{ moving}}{\underbrace{\dot{q}dp}}+\underset{\text{cancels}\to }{\underbrace{pd\dot{q}}}-\underset{\leftarrow dL\text{ piece}}{\underbrace{pd\dot{q}}}-\frac{\partial L}{\partial q}dq-\frac{\partial L}{\partial t}dt.$

Beech ke do terms equal aur opposite hain kyunki $\partial L/\partial \dot{q}=p$ by definition hai (Step 1). Velocity ka differential chala gaya.

Step 4 — Baaki bachi slopes ko naam dene ke liye Euler–Lagrange daalo

KYA. Step 3 se do pieces bachti hain: $dq$ wala term aur $dp$ wala term. Euler–Lagrange humein $dq$ term ko relabel karne deta hai.

KYUN. Abhi $dq$ ka coefficient $-\partial L/\partial q$ hai — Lagrangian language mein likha hai. Hum chahte hain ise momentum language mein likhen taaki dono equations symmetric lagein.

PICTURE. Euler–Lagrange kehta hai: "momentum-slope ka time mein change karne ka rate, $L$ ki position-slope ke barabar hai." Chalk-pink arrow follow karo $p$ ke time-derivative se $\partial L/\partial q$ tak.

$\dot{p}\equiv \frac{dp}{dt}=\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\stackrel{\text{E–L}}{=}\frac{\partial L}{\partial q}.$

Toh Step 3 ka leftover ban jaata hai $dH=\dot{q}dp-\dot{p}dq-\frac{\partial L}{\partial t}dt.$

Ab har term sirf $q$, $p$, aur $t$ mein bolta hai. Achha — yeh match karta hai jo $H$ depend kar sakta hai.

Step 5 — Same $dH$ ke liye do expressions match karo

KYA. Hamare paas same chhote change $dH$ ke liye do formulas hain. Ek jo humne abhi derive kiya (Step 4). Doosra woh jo calculus hamesha kehta hai $(q,p,t)$ ke function ke liye.

KYUN. Agar do expressions har independent wiggle $dq$, $dp$, $dt$ ke liye same $dH$ ke barabar hain, toh har wiggle par baithe coefficient alag-alag match hone chahiye. Isi tarah hum equations padhte hain.

PICTURE. Do rows line up karo. Upar $dp$ ka coefficient neeche $dp$ ke coefficient se match hona chahiye; aise hi $dq$ aur $dt$ ke liye. Matching colours pairings dikhate hain.

Step 6 — Minus sign optional kyun nahi hai: flow ko swirl karna hi hai

KYA. Phase space mein velocity field draw karo (axes: $q$ horizontal, $p$ vertical). Har point par state velocity $(\dot{q},\dot{p})=({\partial }_{p}H,-{\partial }_{q}H)$ se move karti hai.

KYUN. Minus gradient ka ek component $90^{\circ }$ flip kar deta hai. Plus-plus rule ki tarah $H$ ke lowest point ki taraf flow downhill karne ki jagah (jo sab kuch ek point par collapse kar deta), state $H$ ke constant contours ke around flow karti hai. Motion perpetual hai, energy conserve hoti hai — yeh Phase Space and Liouville's Theorem ka seed hai.

PICTURE. Left panel: galat plus-plus field — har arrow pit ki taraf jaata hai, motion mar jaati hai. Right panel: sahi Hamilton field — arrows $H$ ke contours ke circle karte hain, closed loop trace karte hain.

Step 7 — Degenerate & edge cases (koi gap mat chhoodo)

KYA & KYUN & PICTURE saath mein, kyunki har ek neeche ke same figure ka ek special reading hai.

• Free particle, $V=0$. $H=p^2/2m$ mein $q$ andar nahi hai, toh $\dot{p}=-{\partial }_{q}H=0$: momentum constant hai. Phase-space lines horizontal hain (top panel) — state fixed height par right slide karta hai.
• Equilibrium par (${\partial }_{q}H=0$ aur ${\partial }_{p}H=0$). Dono velocities vanish: state ek fixed point hai aur kabhi move nahi karta (centre dot).
• Curve convex nahi (${\partial }^{2}L/\partial {\dot{q}}^2=0$, inflection hai). Slope $p$ ab uniquely ek $\dot{q}$ nahi pick karta — Step 2 ka Legendre step invert nahi ho sakta. Yeh singular case hai jahan poori Hamiltonian recipe stall ho jaati hai (bottom panel: do velocities ek slope share karti hain).
• $L$ explicitly time-dependent hai. Tab $\partial L/\partial t\ne 0$, toh $\partial H/\partial t\ne 0$ aur $H$ drift karta hai: $dH/dt=\partial H/\partial t$. Energy conserve nahi hoti even though do canonical equations unchanged hold karte hain.

Ek-picture summary

Is page par sab kuch ek journey hai: curve → uski slope → intercept → $\dot{q}$ nikalo → differentials match karo → circulate karo. Final board ise compress karta hai.

Active-Recall