Visual walkthrough — Hamilton's principle — least action

4.10.15 · D2 · Maths › Advanced Topics (Elite Level) › Hamilton's principle — least action

Hum is final result ka peecha kar rahe hain — lekin aapko abhi ise believe karne ki ijazat nahi hai: $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)-\frac{\partial L}{\partial q}=0.$ Last step tak upar ke har nishan ko aap ne dekha hoga.

Step 1 — Ek path sirf ek "position vs. time" chart mein ek trail hai

KYA. Ek bead ko ek wire pe slide karte hue picture karo. Har instant $t$ (waqt ka ek pal) pe bead kisi position pe baithta hai. Us position ko $q$ kaho. Jaise waqt aage badhta hai, pair $(t,q)$ ek flat chart pe ek curve trace karta hai: time left-to-right chalta hai, position bottom-to-top. Woh poora curve hi wo hai jise hum path $q(t)$ kehte hain — padho "$q$ as a function of $t$", matlab "mujhe ek time do, main tumhe bataunga bead kahan hai".

KYUN. Isse pehle ki hum "sabse achhe path" ki baat kar sakein, humein koi bhi arbitrary path dikhna chahiye. Ek path aasman mein koi formula nahi hai — yeh ek drawn curve hai jise tum point kar sakte ho.

PICTURE. Neecha wali blue curve ek path hai. Uske donoN siron pe do mote dots note karo: start $(t_1,q_1)$ aur end $(t_2,q_2)$. Yeh do dots nailed down hain — humein bataya jaata hai bead kahan shuru hota hai aur kahan khatam. Freedom sirf isme hai ki curve unke beech mein kaise wiggle karta hai.

Step 2 — Trail ka slope velocity $\dot{q}$ hai

KYA. Curve ke ek point pe zoom karo aur woh chhoti tangent line khicho jo use hug karti ho. Uski steepness — position kitni change hoti hai waqt ke chhote tick mein — bead ki velocity hai. Hum ise $\dot{q}$ likhte hain (kaho "q-dot"). Upar ka dot "rate of change with time" ka shorthand hai.

KYUN. Har physics ingredient jo hume chahiye (kinetic energy) is par depend karta hai ki bead kitni tez move karti hai, na sirf kahan hai. Speed slope mein rehti hai, isliye hume slope ka naam dena hoga. Hum yahan derivative use karte hain — na ki sirf "distance ÷ time" — kyunki slope curve ke saath point se point tak change hota hai; derivative woh tool hai jo har single point pe instantaneous steepness deta hai.

PICTURE. Usi curve pe do points: pink point pe curve steep hai isliye $\dot{q}$ bada hai; yellow point pe yeh lagbhag flat hai isliye $\dot{q}$ chhota hai. Ek hi curve, alag alag slopes.

Step 3 — Har path ko ek single number se score karo: action $S$

KYA. Har instant pe bead ke paas kinetic energy $T$ hoti hai (motion ki energy, badi jab $\dot{q}$ bada ho) aur potential energy $V$ (position ki energy, badi jab bead kahin "mehenga" baithta ho). Lagrangian banao: $L=T-V.$ Ab poore path pe slide karo, har instant pe $L$ padho, aur un saari values ko pure trip mein add karo. Woh grand total action hai: $S={\int }_{t_1}^{t_2}Ldt.$

KYUN. Hum poore paths ko ek doosre se ek number se compare karna chahte hain, isliye hume ek poori curve ko ek single score mein compress karna hoga. Woh tool jo "ek value ko continuously ek curve ke along add karta hai" woh integral $\int \cdots dt$ hai — infinitely many infinitely thin slices wala sum. Ek plain sum yeh nahi kar sakta kyunki $L$ har instant pe smoothly change hota hai; integral exactly woh continuous-sum tool hai.

PICTURE. Har time pe $L$ ki value ek height hai; action us height-curve ke neeche shaded area hai. Term by term: $S=\underset{\text{add up}}{\underbrace{{\int }_{t_1}^{t_2}}}\underset{L,\text{abhi ka score}}{\underbrace{(T-V)}}\underset{\text{har chhote time-slice per}}{\underbrace{dt}}.$

Step 4 — Ek path ko uske nudged neighbours se compare karo (wiggle $\eta$)

KYA. Candidate true path $q(t)$ lo aur ek chhoti wiggle add karo. Nudged path ko likho: $q(t)+\epsilon \eta (t).$ Yahan $\eta (t)$ (kaho "eta") koi bhi smooth bump-shape hai, aur $\epsilon$ (kaho "epsilon") ek tiny knob hai jo control karta hai ki kitna bump add karna hai. Zaroori baat yeh hai ki bump donoN siron pe zero ho jaata hai: $\eta (t_1)=\eta (t_2)=0.$

KYUN. "Best path" ka matlab hai: use kisi bhi taraf nudge karne se score improve nahi ho sakta. Ise test karne ke liye, humein demand pe neighbouring paths produce karne ka tarika chahiye — yahi kaam $\eta$ ka hai. Aur ends nailed rehne chahiye (Step 1 ne unhe thoka), isliye wiggle wahan vanish karne pe majboor hai. Yeh pinning hi woh poori wajah hai ki derivation kaam karti hai; iska dhyan rakhna.

PICTURE. Solid blue curve $q$ hai. Dashed curves $q+\epsilon \eta$ hain kuch $\epsilon$ values ke liye. Dekho kaise sab un do hi nailed dots se guzarte hain — wiggle ends pe pinch ho jaati hai.

Step 5 — Score ko $\epsilon$ mein ek curve banao, aur flat top demand karo

KYA. Nudged path ko action mein daalo. Kyunki path ab dial $\epsilon$ par depend karta hai, isliye score bhi: $S(\epsilon )={\int }_{t_1}^{t_2}L(q+\epsilon \eta ,\dot{q}+\epsilon \dot{\eta },t)dt.$ Is single number $S$ ko dial $\epsilon$ ke against plot karo. True path woh hai jahan $\epsilon =0$ is curve ke flat point pe baithta hai: ${\frac{dS}{d\epsilon }|}_{\epsilon =0}=0.$

KYUN. "Kisi bhi taraf nudging score improve nahi kar sakti" exactly translate hota hai "score-vs-dial curve ka slope $\epsilon =0$ pe zero hai". Hum derivative $\frac{dS}{d\epsilon }$ use karte hain kyunki derivative precisely woh tool hai jo detect karta hai "flat = first-order pe improve-nahi-ho-sakta". Note karo ki flat point ek valley (minimum), ek hilltop (maximum), ya saddle ho sakta hai — sab "stationary" hain. Isliye honest naam stationary action hai, least action nahi.

PICTURE. $S$ ka $\epsilon$ ke against cup-shaped curve, $\epsilon =0$ pe tangent horizontal. Chhoti green tangent line flat hai — woh flatness hi $\delta S=0$ hai.

Step 6 — Differentiate karo: $L$ ka har slot wiggle feel karta hai

KYA. $S(\epsilon )$ ko integral ke under differentiate karo. Wiggle $L$ mein do jagah enter hui — position slot (gained $\epsilon \eta$) aur velocity slot (gained $\epsilon \dot{\eta }$) — isliye chain rule humein do terms deta hai: $\frac{dS}{d\epsilon }{|}_0={\int }_{t_1}^{t_2}(\underset{\text{how }L\text{ reacts to a position nudge}}{\underbrace{\frac{\partial L}{\partial q}}}\underset{\text{the position nudge}}{\underbrace{\eta }}+\underset{\text{how }L\text{ reacts to a velocity nudge}}{\underbrace{\frac{\partial L}{\partial \dot{q}}}}\underset{\text{the velocity nudge}}{\underbrace{\dot{\eta }}})dt.$

KYUN. Hum jaanna chahte hain ki score kaise change hota hai jab hum dial ghumaate hain. Chain rule woh tool hai "several inputs mein chhoti changes se output mein chhoti change". Symbol $\frac{\partial L}{\partial q}$ (ek partial derivative) ka matlab hai "$L$ ke sirf position slot ko wiggle karo, baaki sab freeze karo" — humein yeh isliye chahiye kyunki $L$ ke several inputs hain aur hum ek waqt mein ek ki response chahte hain.

PICTURE. $L$ ko ek chhoti machine ke roop mein draw karo jiske do input dials hain, "position" aur "velocity". Wiggle donoN dials ko jiggle karti hai; har dial ki sensitivity ($\partial L/\partial q$ aur $\partial L/\partial \dot{q}$) apni jiggle ($\eta$ aur $\dot{\eta }$) se multiply hoti hai to sum ka ek term milta hai.

Step 7 — Integrate by parts karo taaki har jagah akela $\eta$ mile

KYA. Doosra term annoying hai: usme $\dot{\eta }$ hai (wiggle ka slope), jabki pehle mein bare $\eta$ hai. Unhe combine karne ke liye hume $\dot{\eta }$ ko $\eta$ mein convert karna hoga. Ek integral ke andar ek factor se dusre factor pe derivative trade karne ka tool integration by parts hai: ${\int }_{t_1}^{t_2}\frac{\partial L}{\partial \dot{q}}\dot{\eta }dt=\underset{\text{boundary term}}{\underbrace{{\left[\frac{\partial L}{\partial \dot{q}}\eta \right]}_{t_1}^{t_2}}}-{\int }_{t_1}^{t_2}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\eta dt.$ Ab pinned ends ka fayda milta hai (Step 4): $\eta (t_1)=\eta (t_2)=0$, isliye boundary term donoN ends pe $0\times (\text{stuff})=0$ hai. Yeh vanish ho jaata hai.

KYUN. Hum integration by parts sirf isliye karte hain taaki do terms ko ek single common factor $\eta$ mein unify karein. Aur endpoint-pinning hi woh cheez hai jo humein boundary term phenkne deti hai — yahi exact jagah hai jahan "fixed endpoints" kaam aata hai. Agar ends free hote, woh term bachti aur hume ek alag problem milti.

PICTURE. Boundary term har nailed dot pe evaluate hua: donoN heights zero hain kyunki wiggle wahan pinch ho jaati hai. Chhote brackets literally kuch nahi pe close hote hain.

Bachne wale pieces ko wapas saath rakh ke: $\frac{dS}{d\epsilon }{|}_0={\int }_{t_1}^{t_2}\left(\underset{\text{ise }f(t)\text{ kaho}}{\underbrace{\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}}}\right)\eta (t)dt=0.$

Step 8 — Wiggle arbitrary hai ⇒ bracket har jagah zero hona chahiye

KYA. Ab hamare paas ${\int }_{t_1}^{t_2}f(t)\eta (t)dt=0$ hai, aur yeh har allowed wiggle $\eta$ ke liye hold karna chahiye. Fundamental Lemma of the Calculus of Variations kehta hai: agar ek continuous $f$ har pinned wiggle ke saath zero area deta hai, to $f$ khud har point pe zero hai. Isliye $f(t)\equiv 0$.

KYUN. Maano $f$ kisi chhote interval par positive hota. Tab ek aisi wiggle $\eta$ choose karo jo exactly us interval ke upar ek positive bump ho aur baaki jagah zero ho. Integral positive aata, zero nahi — contradiction. Kyunki hum bump kahi bhi rakh sakte hain, $f$ kahin bhi nonzero nahi ho sakta. $\eta$ ki freedom woh crowbar hai jo equation ko integral se nikaalti hai.

PICTURE. Ek hypothetical $f$ jo zero ke upar poke karta hai, uske saath ek matching positive bump $\eta$ uske upar rakha; shaded product-area clearly positive hai — woh contradiction jo $f\equiv 0$ force karta hai.

$f(t)=0$ set karo aur sign flip karo to destination milta hai:

Ek picture mein summary

Ek figure, poora safar: ek nailed start aur end, bold mein true path, uske around pinned wiggles ka fan, score-vs-dial cup $\epsilon =0$ pe flat, aur arrow boxed E–L equation pe land karta hua.

Active recall