Question bank — Hamilton's principle — least action

False. $L=T-V$, jabki total energy $T+V$ hai. Ye sirf tab coincide karte hain jab $V=0$; warna ye $2V$ se differ karte hain.

False. Exact condition $\delta S=0$ (stationary) hai. Ek conjugate/focal point ke baad true path saddle ban jaata hai, toh ye stationary point toh hai lekin minimum nahi.

False. Fundamental Lemma ke liye ye sab smooth $\eta$ ke liye hold karna chahiye jo ends par vanish ho; ek akela $\eta$ almost kuch nahi batata.

True. $L$ mein ek total time derivative $\frac{dF(q,t)}{dt}$ add karne se $S$ sirf ek fixed boundary term se badlta hai, toh $\delta S$ aur isliye E–L equations unchanged rehte hain.

False. Yahi iska selling point hai: achhe-chosen coordinates mein $T$ aur $V$ likhne se constraint forces (jaise pendulum ki tension) automatically drop out ho jaate hain — koi force decomposition ki zaroorat nahi.

False. Ye isliye vanish karta hai kyunki endpoints fixed rakhe jaate hain; yahi cheez integration by parts mein boundary term $[\frac{\partial L}{\partial \dot{q}}\eta {]}_{t_1}^{t_2}$ ko kill karti hai, kisi infinity ki wajah se nahi.

False. $T+V$ true path ke saath same constant hai, toh ye paths ko distinguish karne ki koi information nahi deta; difference $T-V$, integrated, woh hai jo extremize hota hai.

True in spirit. $L=\frac{1}{2}m{\dot{x}}^2-V$ ke liye, E–L seedha $m\ddot{x}=-V^{\prime }$ deta hai, aur yahi principle fields aur relativity mein cleanly generalize hota hai jahan "force" awkward hai. Dekho Newton's Laws.

False. $S=\int (T-V)dt$ negative, zero, ya positive ho sakta hai depending on kinetic aur potential ke trade off par; iska sign irrelevant hai — sirf iska stationarity matter karta hai.

$\frac{d}{dt}$ missing hai. Momentum term ko time-differentiate karna zaroori hai: $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$, kyunki $p=\partial L/\partial \dot{q}$ time ke saath change hota hai.

Galat problem. Hamilton's principle $q(t_1),q(t_2)$ fix karta hai. Ends ko free karne se ek surviving boundary term reh jaata hai aur natural boundary conditions impose hoti hain — ye ek alag variational problem hai.

Misread. Stationarity sirf $\epsilon =0$ par hai: ${\frac{dS}{d\epsilon }|}_0=0$. Action phir bhi nonzero $\epsilon$ ke liye bend karta hai; ye flat sirf true path par hai.

Nahi. Partial derivative ke andar $q,\dot{q},t$ ko independent slots ki tarah treat kiya jaata hai; sirf E–L equation form karne ke baad hum yaad karte hain ki $\dot{q}=dq/dt$ aur $\frac{d}{dt}$ apply karte hain.

Zaroori nahi. Coordinate $\theta$ choose karna aur $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta$ likhna saari geometry encode karta hai; tension koi work nahi karta aur kabhi appear nahi hota. Dekho Lagrangian Mechanics.

Galat argument. $S$ ek poora path $q(\cdot )$ leta hai aur $t$ integrate karke ek number return karta hai; ye function par depend karta hai, kisi ek time par nahi.

Ye essential hai. By-parts derivative ko $\dot{\eta }$ se hatake $\frac{\partial L}{\partial \dot{q}}$ par le jaata hai, taaki har term ek hi $\eta$ multiply kare — tabhi lemma $\eta$ factor out kar sakta hai aur E–L bracket expose kar sakta hai.

Kyunki $T+V$ true motion ke saath constant hai aur kuch distinguish nahi karta; difference moment-to-moment trade measure karta hai kinetic aur potential energy ke beech, aur nature iske integral ko extremize karta hai.

True path ko sabhi nearby competitors se behtar hona chahiye. Arbitrary $\eta$ ke liye ${\frac{dS}{d\epsilon }|}_0=0$ require karna exactly wahi hai jo E–L integrand ke bracket ko har jagah vanish karne par majboor karta hai.

Ye isliye vanish hota hai kyunki $\eta (t_1)=\eta (t_2)=0$ (fixed ends). Ye poori derivation ka pivot hai — iske bina ek leftover boundary contribution clean E–L equation ko kharab kar deta.

Newton poochhta hai "abhi, yahan kaunsi force act kar rahi hai?"; Hamilton poochhta hai "kaun sa poora path best score karta hai?". Ek local instantaneous rule hai, doosra poori trajectory ke baare mein ek statement hai — phir bhi dono agree karte hain.

Ek constant shift $L$ ko constant se badlta hai, isliye $S$ ek fixed amount se badlta hai path se independent; $\delta S$ unaffected rehta hai, toh E–L equations aur motion identical hain.

Dono ek path par ek integral extremize karte hain (matter ke liye action, light ke liye optical path length/time). Fermat's Principle optics ka cousin hai: light stationary-time route leta hai jaise particle stationary-action route leta hai.

Agar $L$ kisi shift ke under (time, position, ya angle mein) nahi badlta, toh E–L matching momentum ko conserved banata hai — yahi Noether's Theorem ka content hai. E.g. $L$ jo $q$ se independent hai $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=0$ deta hai, toh $p=\partial L/\partial \dot{q}$ conserved hai.

Ek genuine minimum ordinary case ke liye: koi bhi wiggle kinetic energy add karta hai bina kisi potential ke compensate karne ke, toh $S$ strictly increase hota hai.

Coordinate cyclic/ignorable hai: $\partial L/\partial q=0$, toh $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=0$ aur conjugate momentum $p=\partial L/\partial \dot{q}$ conserved hai.

Tab ek Hamiltonian-like quantity $H=\dot{q}\frac{\partial L}{\partial \dot{q}}-L$ conserved hoti hai; scleronomic systems ke liye ye $T+V$, total energy, ke barabar hoti hai. Dekho Hamiltonian Mechanics.

Haan — wiggle argument har direction mein independently chalta hai, har coordinate ke liye ek E–L equation deta hai, kyunki ${\eta }_i$ alag-alag vary kiye ja sakte hain.

Conjugate point ke baad possible hai (e.g. sphere ke antipodes ke beech kaafi geodesics). Stationarity ek local condition hai, toh kaafi paths mein se har ek stationary ho sakta hai; extra analysis decide karta hai kaun sa minimum hai.

Sirf tab agar ye E–L satisfy karta ho. E.g. ek pendulum $\theta =0$ par hanging deta hai $\ddot{\theta }=-\frac{g}{\ell }\sin 0=0$, ek valid stationary (equilibrium) solution; $\theta =\pi$ par ye stationary hai lekin unstable.

Action integral zero ki taraf shrink hota hai aur endpoints almost coincide karte hain; true path essentially tiny straight segment hone par majboor ho jaata hai, aur stationarity trivial ban jaati hai — wiggle karne ki koi jagah nahi.

Nahi. Ye ek competitor bhi nahi hai: Hamilton's principle sirf un paths ko compare karta hai jo given $q(t_1)$ aur $q(t_2)$ share karte hain, toh ek endpoint-violating curve simply contest se bahar hai.