Exercises — Hamilton's principle — least action

Pieces ka matlab: $T=\frac{1}{2}m{\dot{x}}^2$ motion ki energy hai (half mass times speed-squared). "Free" ka matlab kuch bhi nahi kheenchta, isliye $V=0$. $L=T-V=\frac{1}{2}m{\dot{x}}^2-0=\frac{1}{2}m{\dot{x}}^2.$ Momentum slot: $L$ ko differentiate karo $\dot{x}$ ko variable maanke, $\frac{\partial L}{\partial \dot{x}}=m\dot{x}(\text{yeh momentum }p\text{ hai}).$ Force slot: $L$ ko differentiate karo $x$ ko variable maanke. Kyunki $x$ appear hi nahi karta, $\frac{\partial L}{\partial x}=0.$ Jo coordinate $L$ mein appear nahi karta use cyclic kehte hain — uska momentum conserved hoga. Dekho Noether's Theorem.

Spring potential hai $V=\frac{1}{2}k{x}^2$ (stored energy stretch ke square ke saath badhti hai). Kinetic energy $T=\frac{1}{2}m{\dot{x}}^2$. $L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2.$ $\frac{\partial L}{\partial \dot{x}}=m\dot{x},\frac{\partial L}{\partial x}=-kx.$ Minus sign spring ka $x=0$ ki taraf pull back karna hai — ek restoring tendency, random sign nahi.

(b) $S={\int }_{t_1}^{t_2}Ldt$ sahi hai: action Lagrangian ka time ke upar accumulation hai. (a) galat variable par integrate karta hai — action ek "cost per unit time" sum karta hai, isliye hum $dt$ integrate karte hain, $dx$ nahi. (c) sirf values ka difference hai; yeh beech ka poora safar ignore karta hai, aur "action" exactly wohi capture karne ke liye bana hai.

E–L kehta hai $\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x}=0$. Problem 1.1 se: $\partial L/\partial \dot{x}=m\dot{x}$ aur $\partial L/\partial x=0$. Toh $\frac{d}{dt}(m\dot{x})-0=0\Rightarrow m\ddot{x}=0\Rightarrow \ddot{x}=0.$ Iska matlab: acceleration zero hai, isliye $\dot{x}=\text{const}$ (constant velocity), aur ek baar aur integrate karne par $x(t)=x_0+vt$ — position–time plot mein ek straight line. Yeh Newton's Laws se match karta hai: no force ⇒ no acceleration.

Slots (Problem 1.2 se): $\partial L/\partial \dot{x}=m\dot{x}$, $\partial L/\partial x=-kx$. $\frac{d}{dt}(m\dot{x})-(-kx)=0\Rightarrow m\ddot{x}+kx=0.$ $m$ se divide karo: $\ddot{x}=-\frac{k}{m}x$. $\ddot{x}=-{\omega }^{2}x$ se compare karte hue, $\boxed{\omega =\sqrt{k/m}}.$ Restoring sign $-kx$ woh $-{\omega }^{2}x$ ban jaata hai jo oscillation produce karta hai.

Kinetic energy ke liye real speed chahiye. Bead horizontally aur vertically dono taraf move karta hai: $v^2={\dot{x}}^2+{\dot{y}}^2={\dot{x}}^2+(2x\dot{x}{)}^2={\dot{x}}^2(1+4x^2).$ Toh $T=\frac{1}{2}m{\dot{x}}^2(1+4x^2)$. Potential gravity hai: $V=mgy=mg{x}^2$. $L=\frac{1}{2}m{\dot{x}}^2(1+4x^2)-mg{x}^2.$ Momentum slot: $\frac{\partial L}{\partial \dot{x}}=m\dot{x}(1+4x^2)$. Time-derivative (product rule — dono $\dot{x}$ aur $x$ vary karte hain): $\frac{d}{dt}[m\dot{x}(1+4x^2)]=m\ddot{x}(1+4x^2)+m\dot{x}\cdot 8x\dot{x}=m\ddot{x}(1+4x^2)+8mx{\dot{x}}^2.$ Force slot: $\frac{\partial L}{\partial x}=\frac{1}{2}m{\dot{x}}^2\cdot 8x-2mgx=4mx{\dot{x}}^2-2mgx.$ E–L (force slot subtract karke): $m\ddot{x}(1+4x^2)+8mx{\dot{x}}^2-(4mx{\dot{x}}^2-2mgx)=0.$ $\boxed{\ddot{x}(1+4x^2)+4x{\dot{x}}^2+2gx=0}$ Padho isko: $x=0$ ke paas wire almost flat hai, $4x^2\approx 0$, aur yeh reduce ho jaata hai $\ddot{x}\approx -2gx$ mein — small oscillations, bilkul spring jaisi. Method ki power: humne wire ki normal force kabhi resolve nahi ki; $x$ choose karke aur $T,V$ likhke saari geometry ho gayi.

Speed: bob radius $\ell$ ka circle trace karta hai, isliye uski speed $\ell \dot{\theta }$ hai aur $T=\frac{1}{2}m(\ell \dot{\theta }{)}^2=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$. Height: pivot se height measure karte hue, bob $-\ell \cos \theta$ par hota hai (lowest jab $\theta =0$). Toh $V=mg(-\ell \cos \theta )=-mg\ell \cos \theta$. $L=T-V=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta .$ Slots: $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }$, aur $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta$. E–L: $\frac{d}{dt}(m{\ell }^2\dot{\theta })-(-mg\ell \sin \theta )=0\Rightarrow m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0$. $\boxed{\ddot{\theta }=-\frac{g}{\ell }\sin \theta }.$ Small angles: $\sin \theta \approx \theta$ ($\sin$ ka graph line $y=\theta$ ke paas $0$ ke paas chipka rehta hai), isliye $\ddot{\theta }\approx -\frac{g}{\ell }\theta$, jo $\omega =\sqrt{g/\ell }$ deta hai.

String tension koi work nahi karta: yeh hamesha string ke along point karta hai, bilkul perpendicular bob ki motion ke (jo circle ke along hoti hai). Energy methods kabhi bhi forces ko sirf energies $T$ aur $V$ ke through dekhte hain; ek force jo koi energy store nahi karti aur koi work nahi karta woh $L$ mein kabhi enter nahi karta. Constraint "length fixed hai" single coordinate $\theta$ use karke pehle se bake in ho gayi thi. Yahi reason hai generalized coordinates itne powerful hain — Lagrangian Mechanics constraint forces ko completely sidestep karta hai.

$\varphi$ equation dekho. $\varphi$ momentum slot hai $\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }$. $\varphi$ force slot hai $\frac{\partial L}{\partial \varphi }=0$ kyunki $\varphi$ $L$ mein appear hi nahi karta (sirf $\dot{\varphi }$ karta hai; $V$ sirf $r$ par depend karta hai). $\varphi$ ke liye E–L: $\frac{d}{dt}(m{r}^2\dot{\varphi })-0=0\Rightarrow m{r}^2\dot{\varphi }=\text{const}.$ Woh constant angular momentum $p_{\varphi }$ hai. Kyun: $\varphi$ ek cyclic coordinate hai ($L$ mein absent), aur E–L phir kehta hai uski momentum ki time-derivative zero hai. Yeh Noether's Theorem miniature mein hai: $L$ ki rotational symmetry ($\varphi$ rotate karne par $L$ unchanged rehta hai) hi conservation of angular momentum hai.

E–L set up karo. Yahan "coordinate" $y$ hai aur "time" $x$ hai, isliye E–L padhta hai $\frac{d}{dx}\left(\frac{\partial L}{\partial y^{\prime }}\right)-\frac{\partial L}{\partial y}=0.$ $L=\frac{\sqrt{1+y^{\prime 2}}}{v(y)}$ ke liye har slot compute karo:

Pehle sign convention. Hum $x,y$ neeche measure karte hain (positive = aur neeche). Gravity positive work karta hai jab mass descend karta hai, isliye potential energy depth ke saath decrease honi chahiye: $V_1=-m_{1}gx$ aur $V_2=-m_{2}gy$. Yahan se har minus sign aata hai — "PE girne par drop hoti hai" se forced hai, free choice nahi. Constraint apply karo. Rope inextensible hai: $x+y=\ell$ (constant), isliye $y=\ell -x$ aur differentiate karne par, $\dot{y}=-\dot{x}$ (jab $m_1$ drop karta hai, $m_2$ same rate par rise karta hai). Kinetic energy ($\dot{y}=-\dot{x}$ substitute karo, aur $(-\dot{x}{)}^2={\dot{x}}^2$): $T=\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{y}}^2=\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{x}}^2=\frac{1}{2}(m_1+m_2){\dot{x}}^2.$ Potential ($y=\ell -x$ substitute karo): $V=-m_{1}gx-m_{2}g(\ell -x)=-(m_1-m_2)gx\underset{\text{constant}}{\underbrace{-m_{2}g\ell }}.$ Kyun constant drop kiya ja sakta hai — dikhaya gaya, assert nahi kiya. $V$ mein constant $c$ add karne par $L$ $-c$ se change hota hai. Dono slots mein uska contribution vanish ho jaata hai: $\frac{\partial (-c)}{\partial \dot{x}}=0$ aur $\frac{\partial (-c)}{\partial x}=0$ (constant ki derivative zero hoti hai). Toh yeh E–L equation ko affect hi nahi kar sakta — hum ise $L$ mein rakhte hain lekin yeh koi role play nahi karta. $L=\frac{1}{2}(m_1+m_2){\dot{x}}^2+(m_1-m_2)gx-m_{2}g\ell .$ Slots: $\frac{\partial L}{\partial \dot{x}}=(m_1+m_2)\dot{x}$, $\frac{\partial L}{\partial x}=(m_1-m_2)g$. E–L: $(m_1+m_2)\ddot{x}-(m_1-m_2)g=0$, isliye $\boxed{\ddot{x}=\frac{(m_1-m_2)}{m_1+m_2}g}.$ Sanity check: agar $m_1=m_2$ toh acceleration zero hai (balanced); agar $m_2=0$ toh $+g$ hai ($m_1$ ki free fall). Yeh classic Atwood machine result hai, bina ek bhi free-body diagram ke derive kiya gaya.

One-line argument: E–L ki derivation ne sirf kabhi action ka first-order change zero force kiya, ${\frac{dS}{d\epsilon }|}_{\epsilon =0}=0$; lekin vanishing first derivative ek minimum, maximum, ya saddle mark karta hai — isliye physics sirf stationarity guarantee karta hai, minimality nahi. Ek concrete non-minimum case. Spring par mass consider karo (ya koi bhi oscillator) period $T_0=2\pi /\omega$ ke saath. Do points ke beech true path ko nearby paths se compare karo:

• Chhote time interval $t_2-t_1<T_0/2$ ke liye, true path action ka genuine local minimum hai.
• Jab interval ek conjugate point cross karta hai (yahan $t_2-t_1=T_0/2$, half a period, jahan saare nearby oscillator paths refocus hote hain), true path ek saddle ban jaati hai: kuch neighbouring paths action lower karti hain jabki kuch raise karti hain. Yeh abhi bhi $\delta S=0$ satisfy karta hai, lekin ab minimum nahi raha. Yehi light ke liye ek focus se guzarne par hota hai aur orbits ke liye unke focal point ke baad. Isliye honest naam hai principle of stationary action.

Extra action hai ${\int }_{t_1}^{t_2}\frac{d}{dt}fdt=f(q(t_2),t_2)-f(q(t_1),t_1),$ jo sirf fixed endpoints par depend karta hai. Jab hum path $\eta (t_1)=\eta (t_2)=0$ ke saath vary karte hain, yeh endpoint term bilkul change nahi hota, isliye $\delta S^{\prime }=\delta S$. Same variation ⇒ same E–L equation. Kyun matter karta hai: Lagrangian unique nahi hai — tum koi bhi total time-derivative add kar sakte ho ("gauge freedom") physics change kiye bina. Yahi reason hai alag textbooks alag-dikhne wale $L$'s likhte hain jo dynamics par agree karte hain, aur yeh Hamiltonian Mechanics aur gauge theory ke central hai.

Momentum slot: $\frac{\partial L}{\partial \dot{x}}=0$ (koi $\dot{x}$ hai hi nahi). Force slot: $\frac{\partial L}{\partial x}=-V^{\prime }(x)$. E–L: $\frac{d}{dt}(0)-(-V^{\prime }(x))=0\Rightarrow V^{\prime }(x)=0.$ Kya hua: koi equation nahi hai ki $x$ time mein kaise evolve karta hai — E–L sirf demand karta hai ki $x$ $V$ ke critical point par baithe (ek algebraic condition, differential nahi). Koi inertia nahi, theory degenerate hai: yeh motion predict nahi kar sakta, sirf equilibrium. Yahi reason hai genuine kinetic term $T$ ($\dot{q}$ mein quadratic) hi E–L equation ko ek second-order equation of motion banata hai jo well-posed initial-value problem ke saath ho.

$\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}(m\dot{x})=m\ddot{x}$. Aur $\frac{\partial L}{\partial x}=-V^{\prime }(x)$. E–L: $m\ddot{x}-(-V^{\prime }(x))=0\Rightarrow m\ddot{x}=-V^{\prime }(x)$. Kyunki force hai $F=-V^{\prime }(x)$ (force potential par downhill point karta hai), yeh hai $m\ddot{x}=F$, Newton's Laws. Replacement: "$F=ma$ har instant par hold karta hai" ban jaata hai "poora path $\int Ldt$ ko stationary banata hai." Ek local, instant-by-instant law ek single global optimisation ke roop mein re-express hoti hai. Dono identical trajectory dete hain — lekin variational form fields, relativity, aur quantum mechanics mein cleanly transfer hoti hai, jahan "force" define karna mushkil hota hai.