Visual walkthrough — Principle of least action — Hamilton's principle derivation

2.1.19 · D2 · Physics › Analytical Mechanics › Principle of least action — Hamilton's principle derivation

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Step 0 — Kirdar ka parichay (use karne se pehle draw karo)

Kisi bhi calculus se pehle, stage par har object se milte hain.

KYA. Humare paas ek chalti cheez hai (ek bead, ek ball) jo ek number se describe hoti hai jo time ke saath badalta hai, jise $q(t)$ kehte hain. Ise padho "cheez ki position time $t$ par". Start time $t_1$ par yeh height $q_1$ par hai; end time $t_2$ par yeh $q_2$ par hai. Yeh do dots pakke hain — humein diye gaye hain, choose nahi kiye.

KYUN. Har doosra symbol jo hum milenge woh $q(t)$ ke upar bana hai. Agar tum ek "position vs time" graph par ek curve picture kar sako, tumhare paas poori foundation hai.

PICTURE. Horizontal axis time $t$ hai, vertical axis position $q$. Solid curve sahi path hai; do bade dots $(t_1,q_1)$ aur $(t_2,q_2)$ par fixed endpoints hain.

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Step 1 — Lagrangian har path ko ek "score-rate" deta hai

KYA. Path ke har instant par hum ek number attach karte hain: $L(q,\dot{q},t)=T-V$ Yahan $T=\frac{1}{2}m{\dot{q}}^2$ kinetic energy hai (motion ki energy, badi hoti hai jab slope $\dot{q}$ steep ho) aur $V(q)$ potential energy hai (position ki energy, pahaad par badi hoti hai). Inका difference Lagrangian hai.

• $L$ — ek instant par score-rate.
• $T$ — motion cost, speed $\dot{q}$ ke saath badhti hai.
• $V$ — position cost, height $q$ ke saath badhti hai.
• Minus sign — "kinetic credit minus potential debt".

KYUN yeh combination aur $T+V$ kyun nahi? Hum woh rule dhundh rahe hain jo Newton ko reproduce kare. Pata chalta hai ki sirf $T-V$ karta hai (parent note check karta hai). Abhi ke liye $T-V$ ko ek labelled machine ki tarah treat karo jo hum apne output se justify karenge.

PICTURE. Ek sample path ke along time ke against plot ki gayi do energy curves: $T$ teal mein, $V$ plum mein, aur unka difference $L=T-V$ burnt orange mein. Dekho kaise $L$ wahan dip karta hai jahan ball chadhti hai (high $V$) aur wahan utha hai jahan tezi se chalti hai (high $T$).

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Step 2 — Poore path par score add karo: the action

KYA. Poore safar par score-rate sum (integrate) karo: $S[q]={\int }_{t_1}^{t_2}L(q,\dot{q},t)dt$

• $S[q]$ — action: ek poore path ke liye ek number. Square brackets $[q]$ warn karte hain ki yeh ek function khata hai, value nahi.
• ${\int }_{t_1}^{t_2}\dots dt$ — "start se end tak har instant par add karo". Geometrically, $L$-vs-$t$ curve ke neeche ka area.

KYUN integrate karo? Ek instant nahi bata sakta kaunsa path best hai — tumhe poora safar judge karna hoga. Integral exactly "total accumulated score" hai.

PICTURE. Step 2 ki orange $L$ curve ke neeche ka area, shaded — woh shaded area is path ke liye $S$ hai.

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Step 3 — Path wiggle karo: family $q_{\epsilon }(t)$

KYA. Sahi path $q(t)$ lo aur use ek chhoti wiggle se nudge karo: $q_{\epsilon }(t)=q(t)+\epsilon \eta (t),\eta (t_1)=\eta (t_2)=0$

• $\eta (t)$ — ek arbitrary smooth "bump" shape (nudge ki direction).
• $\epsilon$ — ek tiny knob jo control karta hai kitna nudge.
• $\eta (t_1)=\eta (t_2)=0$ — wiggle dono ends par zero par pinned hai, kyunki endpoints fixed data hain jise hum move nahi kar sakte.

KYUN wiggle karo? Yeh poochhne ke liye ki "kya yeh path special hai?", hum test karte hain ki chhote detours score ko change karte hain ya nahi. Agar koi chhota detour $S$ ko first order tak change nahi karta, toh path stationary hai — nature ki choice.

PICTURE. Sahi path (solid) aur alag-alag $\epsilon$ ke liye kai wiggled paths $q_{\epsilon }$ (dashed), saare do endpoints par ek saath pinched. Dashed aur solid ke beech vertical gap $\epsilon \eta (t)$ hai.

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Step 4 — Path-hunt ko ek one-variable calculus problem mein badlo

KYA. Wiggled family ko action mein daalo; ab $S$ sirf knob $\epsilon$ par depend karta hai: $S(\epsilon )={\int }_{t_1}^{t_2}L(q+\epsilon \eta ,\dot{q}+\epsilon \dot{\eta },t)dt$ Sahi path $\epsilon =0$ hai. "Stationary" ka matlab hai ki $S$ versus $\epsilon$ ke graph ka $\epsilon =0$ par flat tangent ho: ${\frac{dS}{d\epsilon }|}_{\epsilon =0}=0$

• $S(\epsilon )$ — ab ek number ka ordinary function, handle karna aasaan.
• $\frac{dS}{d\epsilon }$ — iska slope; $\epsilon =0$ par zero set karna ordinary "flat point" calculus hai.

KYUN. Humne ek dara dene wali "saari functions par search" ko familiar "dhundho jahan slope zero hai" mein convert kiya one-variable calculus se. Yahi Calculus of Variations ki master trick hai.

PICTURE. $S(\epsilon )$ ki ek parabola-jaisi curve; iska lowest point exactly $\epsilon =0$ par hai ek horizontal red tangent line ke saath — flat means stationary.

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Step 5 — Chain rule: score nudge par kaise react karta hai

KYA. Integral ke andar differentiate karo. $L$ ko $\epsilon$ sirf apne do slots $q+\epsilon \eta$ aur $\dot{q}+\epsilon \dot{\eta }$ se feel hota hai: $\frac{dS}{d\epsilon }={\int }_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\eta +\frac{\partial L}{\partial \dot{q}}\dot{\eta }\right)dt$

• $\frac{\partial L}{\partial q}$ — $L$ kitna change hota hai agar tum sirf position nudge karo. $\eta$ se multiply kiya kyunki position slot $\eta$ se shift hua.
• $\frac{\partial L}{\partial \dot{q}}$ — $L$ kitna change hota hai agar tum sirf slope nudge karo. $\dot{\eta }$ se multiply kiya kyunki slope slot $\dot{\eta }$ se shift hua (wiggle ka derivative).

KYUN chain rule? Yeh exact tool hai jo jawab deta hai "agar mere inputs thoda hilein, output kitna hilega?" — precisely humara sawaal. Note karo $\dot{\eta }$, $\eta$ se independent nahi hai: yeh uska time-derivative hai. Yeh link aage matter karta hai.

PICTURE. Ek chhota box $L$ labelled, do input wires ke saath: $q$-wire $\eta$ se wiggled, $\dot{q}$-wire $\dot{\eta }$ se wiggled; do contribution arrows $dS/d\epsilon$ mein add up hote hain.

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Step 6 — Integration by parts: $\dot{\eta }$ ko $\eta$ ke liye trade karo

KYA. Doosra term $\dot{\eta }$ carry karta hai, lekin pehla $\eta$ carry karta hai. $\eta$ factor out karne ke liye humein $\dot{\eta }\to \eta$ convert karna hoga. Integration by parts exactly yahi karta hai: ${\int }_{t_1}^{t_2}\frac{\partial L}{\partial \dot{q}}\dot{\eta }dt=\underset{=0\text{(ends pinned)}}{\underbrace{{\left[\frac{\partial L}{\partial \dot{q}}\eta \right]}_{t_1}^{t_2}}}-{\int }_{t_1}^{t_2}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\eta dt$

• Boundary chunk $[\frac{\partial L}{\partial \dot{q}}\eta {]}_{t_1}^{t_2}$ — do ends par evaluate kiya, jahan $\eta =0$ (Step 3 ka pin!) — toh yeh khatam ho jaata hai.
• Bachne wala integral ab $\eta$ carry karta hai ($\dot{\eta }$ nahi), factor karne ke liye ready.

KYUN. Yeh poori derivation ka pivot hai. Sirf tab jab dono terms ek hi $\eta$ share karein tab hum unhe combine kar sakte hain. Aur Step 4 ka pin hi boundary term ko saaf tarike se vanish karta hai.

PICTURE. Boundary term do endpoints par chhoti arrows ke roop mein draw kiya jo zero par collapse ho jaati hain (kyunki $\eta =0$ wahan hai), middle integral intact chodh kar.

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Step 7 — $\eta$ factor karo, phir use arbitrary hone do

KYA. Step 5 aur Step 6 combine karo: ${\frac{dS}{d\epsilon }|}_0={\int }_{t_1}^{t_2}\left[\frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\right]\eta (t)dt=0$ Kyunki $\eta (t)$ koi bhi bump hai jo hum chahein, bracket har instant par zero hona chahiye. (Agar yeh kahin positive hota, wahan ek bump rakho aur integral positive ho jaata — contradiction.) Yahi Fundamental Lemma of the Calculus of Variations hai.

• Bracket $[\cdots ]$ — "residual force" jo path feel karta hai.
• $\eta (t)$ — kisi bhi instant ko probe karne ke liye free, isliye bracket kahin chhup nahi sakta.

KYUN. $\eta$ ki freedom hi hai jo ek integral equation ko pointwise law mein upgrade karti hai jo har time par valid ho.

PICTURE. Ek localized bump $\eta$ (ek "test poke") ek time par curve ke neeche baitha; agar bracket wahan nonzero hota, shaded product cancel nahi hota — bracket ko har jagah zero hone par majboor karta hai.

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Step 8 — Edge cases: kya ho agar bracket saaf minimum na ho?

KYA. $\delta S=0$ kehta hai stationary, minimum nahi. Teen flavours exist karte hain:

1. Minimum — valley bottom; chhote trips, free particle. Har wiggle $S$ badhata hai.
2. Saddle — nudge direction mein flat lekin kuch wiggles $S$ badhate hain aur kuch ghataate hain. "Focal point" ke baad lambe time par common.
3. Degenerate / free-endpoint — agar ek endpoint pinned nahi hai, $\eta \ne 0$ wahan, Step 6 ka boundary term survive karta hai aur ek extra natural boundary condition deta hai $\frac{\partial L}{\partial \dot{q}}=0$ us end par.

KYUN. Jo reader sirf valley picture dekha hoga woh galat maanenge ki "least" literal hai. Yeh nahi hai — parent note mein mistakes dekho. Saddle aur free-end cases cover karne ka matlab hai koi scenario tumhe surprise nahi kar sakta.

PICTURE. Teen miniature $S$-vs-$\epsilon$ curves side by side: valley (minimum), true saddle (inflection-flat), aur free-endpoint case jahan pin release ho gaya aur boundary arrow ab vanish nahi hota.

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Ek-picture summary

Upar sab kuch, ek canvas par: pinned endpoints, sahi path apne wiggled cousins ke saath, score $S$ area ke roop mein, $S(\epsilon )$ parabola $\epsilon =0$ par flat hoti hui, aur boxed result jo woh force karta hai.

Recall Poore walkthrough ki Feynman retelling

Ek graph picture karo jahan tum ho vs time. Tumhe ek dot par start karna hai aur doosre dot par finish karna hai — woh fixed hain. Beech mein tum koi bhi wiggly raasta le sakte ho. Har route ko hum ek "score" dete hain: har moment par hum tumhari energy-of-moving minus energy-of-height lete hain, aur un sab chhote numbers ko poore safar par add karte hain. Woh total action hai.

Ab ek game khelo: sahi route lo aur use thoda jiggle karo, do dots pinned rakho. Kya score change hota hai? Nature ke chosen route ke liye, jawab hai nahi — first order tak nahi. Score ek flat spot par baitha hai, jaise bowl ka bottom. Hum "poora route jiggle karo" ko ordinary calculus mein convert karte hain ek knob $\epsilon$ introduce karke: knob-zero par score-versus-knob curve flat hai. Crank ghoomana (chain rule, phir integration by parts, aur pinned ends leftover boundary bit khatam karte hain) poori cheez ko ek demand par squeeze kar deta hai: ek certain bracket har instant par zero hona chahiye. Woh bracket, zero set karke, hi Euler–Lagrange equation hai — aur agar tum kinetic-minus-potential energy daalo, $F=ma$ nikalta hai. Ek flat score ke baare mein ek lazy rule, ek bowl ke roop mein draw kiya, mechanics ban jaata hai.

Recall Quick self-check

Step 6 mein boundary term kyun vanish hota hai? ::: Kyunki wiggle $\eta$ dono fixed endpoints par zero par pinned hai, isliye $[\frac{\partial L}{\partial \dot{q}}\eta {]}_{t_1}^{t_2}=0$. Ek integral equation ko har instant par law mein kya upgrade karta hai? ::: $\eta$ ki freedom (Fundamental Lemma) — bracket kahin bhi nonzero nahi ho sakta. Kya $\delta S=0$ ka matlab minimum hai? ::: Nahi — iska matlab stationary hai: minimum, saddle, ya maximum teeno qualify karte hain.

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See also: Euler–Lagrange Equation · Lagrangian Mechanics · Hamiltonian Mechanics · Noether's Theorem · Newton's Second Law