Lagrangian L = T − V

2.1.4 · Physics › Analytical Mechanics

Lagrangian KIYA HAI?

Minus sign dhyaan se dekho. Total energy hoti hai $E=T+V$; Lagrangian hota hai $T-V$. Yeh dono ek cheez nahi hain, aur yahi fark poori baat hai (Mistakes section dekho).

$T-V$ KYUN? (First principles se Derivation)

Hum yeh dikhana chahte hain ki $L=T-V$ ke saath action minimize karna Newton ka law $F=ma$ reproduce karta hai. Agar karta hai, toh $T-V$ ka choice justified hai, arbitrary nahi.

Step 1 — Action define karo. $S={\int }_{t_1}^{t_2}L(q,\dot{q},t)dt$ Yeh step kyun? Humein ek aisa single number chahiye jo poori path ko measure kare taaki hum pooch sakein "kaun si path best hai." Ek scalar ko time pe integrate karne se ek path ke liye ek number milta hai.

Step 2 — Action ko stationary hone ka demand karo. Ek true path $q(t)$ aur ek nearby varied path $q(t)+\delta q(t)$ lo jisme fixed endpoints hon ($\delta q(t_1)=\delta q(t_2)=0$). Stationarity ka matlab hai $\delta S=0$ first order tak. Fixed endpoints kyun? Hum paths ko ek hi start aur end events ke beech compare karte hain; sawaal route ka hai, destination ka nahi.

Step 3 — $S$ ko vary karo. $\delta S={\int }_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot{q}}\delta \dot{q}\right)dt$ Kyun? $L$ ka uske arguments mein first-order Taylor expansion.

Step 4 — Doosre term ko parts se integrate karo. $\delta \dot{q}=\frac{d}{dt}\delta q$ use karte hue: $\int \frac{\partial L}{\partial \dot{q}}\frac{d}{dt}\delta qdt={\left[\frac{\partial L}{\partial \dot{q}}\delta q\right]}_{t_1}^{t_2}-\int \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\delta qdt$ Boundary term vanish ho jaata hai kyunki endpoints par $\delta q=0$ hai. Parts se kyun? Har $\delta q$ ko uski derivative ke bahar laane ke liye taaki hum usse factor kar sakein.

Step 5 — Factor karo aur fundamental lemma apply karo. $\delta S={\int }_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right)\delta qdt=0$ Kyunki $\delta q$ arbitrary hai, bracket zero hona chahiye: $\boxed{\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}=0}$ Yahi hai Euler–Lagrange equation.

Step 6 — 1D mein ek particle ke liye $L=T-V$ plug in karo. $T=\frac{1}{2}m{\dot{x}}^2$ aur $V=V(x)$ ke saath: $\frac{\partial L}{\partial \dot{x}}=m\dot{x},\frac{d}{dt}(m\dot{x})=m\ddot{x},\frac{\partial L}{\partial x}=-\frac{dV}{dx}=F$ Toh Euler–Lagrange equation deta hai: $m\ddot{x}-F=0⟹F=m\ddot{x}.$ Yeh exactly Newton recover karta hai. Isliye $L=T-V$ hai: yeh woh unique combination hai jiska stationary-action condition $ma=-\nabla V$ reproduce karta hai. Agar hum $T+V$ choose karte, toh velocity term aur force term ka same sign hota aur physics galat ho jaati.

Ise USE kaise karein (recipe)

1. Generalized coordinates $q_i$ choose karo jo system ki symmetry/constraints ka faayda uthaayen.
2. Un coordinates mein $T$ aur $V$ likho.
3. $L=T-V$ banao.
4. Har $q_i$ ke liye Euler–Lagrange equation apply karo.

Worked examples

Common mistakes (Steel-manned)

Active recall

Connections

• Euler–Lagrange Equation — upar derive ki gayi equation of motion.
• Principle of Least Action — variational principle $\delta S=0$ jo $L$ ko justify karta hai.
• Generalized Coordinates and Constraints — constraint forces kyun vanish ho jaate hain.
• Generalized Momentum and Conservation — $p_i=\partial L/\partial {\dot{q}}_i$; cyclic coordinates.
• Hamiltonian H = T + V — $L$ ka Legendre transform; energy yahaan wapas aati hai.
• Newton's Second Law — woh special case jismein $L$ reduce ho jaata hai.

Concept Map