Exercises — Lagrangian L = T − V

WHAT: dono energies ko naam do. WHY: $L=T-V$ ko dono chahiye. Ek chalte mass ki kinetic energy $T=\frac{1}{2}m{\dot{x}}^2$ hoti hai — ye "moving-energy" hai. Kyunki motion horizontal hai, gravity ka potential $V=mgy$ kabhi nahi badalta, isliye hum $V=0$ lete hain (constant potential equation of motion mein contribute nahi karta). $L=T-V=\frac{1}{2}m{\dot{x}}^2-0=\boxed{\frac{1}{2}m{\dot{x}}^2}$ WHAT IT LOOKS LIKE: ek flat wire par sliding bead; koi cheez ise push nahi karti, isliye $L$ purely kinetic hai.

Ek stretched/compressed spring energy store karti hai: $V=\frac{1}{2}k{x}^2$. Kinetic term wahi $T=\frac{1}{2}m{\dot{x}}^2$ hai. $L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2.$ WHY sign aisa hai: spring energy subtract hoti hai kyunki $L=T-V$; wo subtraction restoring force $-kx$ ko sahi sign ke saath produce karegi.

Ek generalized coordinate koi bhi independent number hota hai jo configuration fix karta hai. Bob ek rigid rod se bandha hai, isliye $x=\ell \sin \theta$ aur $y=-\ell \cos \theta$ independent nahi hain — ye constraint $x^2+y^2={\ell }^2$ satisfy karte hain. Matlab do Cartesian numbers minus ek constraint $=$ ek true degree of freedom, exactly $\theta$ se capture hota hai. $\theta$ choose karne se constraint automatically gayab ho jaata hai aur rod ka tension kabhi appear nahi karta.

$L=\frac{1}{2}m{\dot{y}}^2-mgy$. Step 1 (momentum term): $\frac{\partial L}{\partial \dot{y}}=m\dot{y}$. WHY: ye generalized momentum $p_y$ hai. Step 2 (uski rate): $\frac{d}{dt}(m\dot{y})=m\ddot{y}$. Step 3 (position term): $\frac{\partial L}{\partial y}=-mg$. Step 4 (EL): $m\ddot{y}-(-mg)=0\Rightarrow \boxed{\ddot{y}=-g}$. WHAT IT LOOKS LIKE: acceleration neeche point karti hai (negative), free fall se match karta hai. ✓

$\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}=m\ddot{x}$. Aur $\frac{\partial L}{\partial x}=-kx$. EL: $m\ddot{x}-(-kx)=0\Rightarrow m\ddot{x}+kx=0$. Standard form $\ddot{x}=-{\omega }^{2}x$ se compare karne par ${\omega }^2=k/m$ milta hai, isliye $\boxed{\omega =\sqrt{k/m}}$. Numeric check: agar $m=2\text{ kg}$, $k=8\text{ N/m}$, to $\omega =\sqrt{8/2}=2\text{ rad/s}$.

$L=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\varphi }}^2)-V(r)$. WHAT: notice karo ki $\varphi$ $L$ mein appear nahi karta (sirf $\dot{\varphi }$ karta hai). Aisa coordinate cyclic ya ignorable kehlata hai. WHY it matters: $\frac{\partial L}{\partial \varphi }=0$, isliye $\varphi$ ke liye EL equation ban jaata hai $\frac{d}{dt}\frac{\partial L}{\partial \dot{\varphi }}=0$ — matlab $\frac{\partial L}{\partial \dot{\varphi }}$ time mein constant hai. Compute karo: $\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }$. Ye conserved hai — ye angular momentum $\ell =m{r}^2\dot{\varphi }$ hai. WHAT IT LOOKS LIKE: jab bhi koi coordinate $L$ mein missing ho, uska momentum ek conserved constant hota hai — Noether ka idea miniature mein.

$\frac{\partial L^{\prime }}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}=m\ddot{x}$; $\frac{\partial L^{\prime }}{\partial x}=+kx$. EL: $m\ddot{x}-kx=0\Rightarrow \ddot{x}=+\frac{k}{m}x$. WHY unphysical: acceleration equilibrium se door point karti hai ($x$ ke same sign mein). Solutions hain $e^{\pm t\sqrt{k/m}}$ — exponential blow-up, oscillation nahi. Ek real spring wapas kheenchti hai. Ye concrete proof hai ki sirf $L=T-V$ sahi physics deta hai.

Bob speed $=\ell \dot{\theta }$, isliye $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$. Gravity potential $=mg\ell (1-\cos \theta )$; spring potential $=\frac{1}{2}\kappa {\theta }^2$. Total $V=mg\ell (1-\cos \theta )+\frac{1}{2}\kappa {\theta }^2$. $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2-mg\ell (1-\cos \theta )-\frac{1}{2}\kappa {\theta }^2.$ $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \frac{d}{dt}=m{\ell }^2\ddot{\theta }$. $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta -\kappa \theta$. EL: $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta +\kappa \theta =0$. Small angle ($\sin \theta \approx \theta$): $m{\ell }^2\ddot{\theta }+(mg\ell +\kappa )\theta =0\Rightarrow \boxed{\ddot{\theta }=-\frac{mg\ell +\kappa }{m{\ell }^2}\theta }$. WHAT IT LOOKS LIKE: SHM with $\omega =\sqrt{(mg\ell +\kappa )/(m{\ell }^2)}$ — spring pendulum ko stiffen karta hai, frequency badhata hai, exactly jaisi intuition demand karti hai.

WHAT: check karo ki $L$ explicitly time $t$ par depend karta hai ya nahi. Yahan $\Omega$ ek fixed constant hai, isliye $L$ mein explicit $t$ nahi hai — isliye Jacobi energy function (Hamiltonian-like quantity) $h=\dot{\theta }\frac{\partial L}{\partial \dot{\theta }}-L$ conserved hai. Compute karo: $h=\dot{\theta }(m{R}^2\dot{\theta })-L=\frac{1}{2}m{R}^2{\dot{\theta }}^2-\frac{1}{2}m{R}^2{\Omega }^2{\sin }^2\theta +mgR(1-\cos \theta )$. WHY ye $T+V$ nahi hai: ${\Omega }^2{\sin }^2\theta$ term $h$ mein minus sign ke saath enter karta hai, $T+V$ ke unlike jahan plus hota hai. Rotation constraint rheonomic hai (externally imposed), isliye naive $E=T+V$ conserved quantity nahi hai — $h$ hai. Yahi subtle payoff hai: "$H=T+V$" sirf time-independent, natural systems ke liye hold karta hai; yahan driven spin wo identity tod deta hai. Takeaway: conservation $L$ ki symmetry se follow karta hai (yahan time-translation), kisi memorized formula se nahi.

Level-1 skill kya hai? ::: $T$, $V$ identify karo, $L=T-V$ assemble karo. Tension pendulum/Atwood solutions mein kabhi kyun nahi aata? ::: Coordinates allowed motion ke along chalte hain, isliye ideal constraint forces koi work nahi karte aur drop out ho jaate hain. Conserved momentum ka signal kya hai? ::: $L$ mein ek coordinate missing ho (cyclic) ⇒ uska $\partial L/\partial \dot{q}$ constant hai. Critical spin of the rotating hoop? ::: ${\Omega }_c=\sqrt{g/R}$, jahan ek pitchfork bifurcation bead ko bottom se utha deta hai. $h=T+V$ kab hota hai? ::: Sirf natural, time-independent (scleronomic) systems ke liye; warna $h=\sum \dot{q}\partial L/\partial \dot{q}-L$ derive karo.