Hamilton's principle — least action

4.10.15 · Maths › Advanced Topics (Elite Level)

Action KYA hai?

$T-V$ kyun aur $T+V$ kyun nahi? Energy conservation ($T+V$) true path par constant rehti hai — yeh paths ko distinguish karne ki koi information carry nahi karti. Difference $T-V$ woh hai jo time ke saath "balance" hota hai: system kinetic ko potential se trade karta hai aur us trade-off ka integral extremize hota hai.

Equation of motion KAISE derive karein (scratch se)

Hum jaanna chahte hain: kaun sa $q(t)$ $\delta S$ ko zero karta hai? True path ko perturb karo: $q(t)\to q(t)+\epsilon \eta (t),\eta (t_1)=\eta (t_2)=0,$ jahaan $\eta$ ek arbitrary smooth "wiggle" hai jo fixed endpoints par vanish karti hai (WHY wahan zero? endpoints fixed hain — humein unhe vary karne ki permission nahi hai).

Action $\epsilon$ ka ek function ban jaata hai: $S(\epsilon )={\int }_{t_1}^{t_2}L(q+\epsilon \eta ,\dot{q}+\epsilon \dot{\eta },t)dt.$

Stationary ka matlab ${\frac{dS}{d\epsilon }|}_{\epsilon =0}=0$ hai. Integral ke neeche differentiate karo (chain rule):

$\frac{dS}{d\epsilon }{|}_0={\int }_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\eta +\frac{\partial L}{\partial \dot{q}}\dot{\eta }\right)dt.$

Yeh step kyun? Har path-component $L$ ka ek derivative pick up karta hai; $\eta$ $q$-slot multiply karta hai, $\dot{\eta }$ $\dot{q}$-slot multiply karta hai.

Ab $\dot{\eta }$ ko $\eta$ mein freee karne ke liye doosre term ko integrate by parts karo: ${\int }_{t_1}^{t_2}\frac{\partial L}{\partial \dot{q}}\dot{\eta }dt=\underset{=0(\eta \text{ ends par vanish karta hai})}{\underbrace{{\left[\frac{\partial L}{\partial \dot{q}}\eta \right]}_{t_1}^{t_2}}}-{\int }_{t_1}^{t_2}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\eta dt.$

Yeh step kyun? Boundary term mar jaata hai kyunki $\eta (t_1)=\eta (t_2)=0$ — exactly yahi wajah hai ki fixed endpoints matter karte hain. Wapas substitute karo:

$\frac{dS}{d\epsilon }{|}_0={\int }_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right)\eta (t)dt=0.$

Yeh har wiggle $\eta$ ke liye hold karna chahiye. Fundamental Lemma of Calculus of Variations se (agar $\int f\eta dt=0$ sabhi $\eta$ ke liye, toh $f\equiv 0$), bracket vanish karna chahiye:

WHY yeh Newton recover karta hai: $L=\frac{1}{2}m{\dot{x}}^2-V(x)$ ke liye: $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$, toh $\frac{d}{dt}(m\dot{x})=m\ddot{x}$; aur $\frac{\partial L}{\partial x}=-V^{\prime }(x)=F$. Euler–Lagrange deta hai $m\ddot{x}-F=0\Rightarrow F=m\ddot{x}$. Newton's law free mein nikal aata hai.

Worked examples

Steel-manned mistakes

Active recall

Connections

• Euler–Lagrange equation — $\delta S=0$ ka differential consequence.
• Calculus of Variations — general machinery (functionals, first variation, fundamental lemma).
• Lagrangian Mechanics — generalized coordinates geometry ko automatic banate hain.
• Noether's Theorem — $L$ ki symmetries ⇒ conserved quantities.
• Hamiltonian Mechanics — $L$ ka Legendre transform jisse $H=T+V$ milta hai.
• Fermat's Principle — optics ka analogue: light stationary time ka path leta hai.
• Newton's Laws — ek special case ke roop mein recover hota hai.

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