Exercises — Principle of least action — Hamilton's principle derivation

KYA: $T$ motion ki energy hai, $V$ stored energy. Kinetic: $T=\frac{1}{2}m{\dot{x}}^2$. Potential: height fixed hai aur koi spring nahi, isliye $V=0$ (koi bhi constant chalega; constant Euler–Lagrange ko kabhi affect nahi karta kyunki uske derivatives zero hote hain). $L=T-V=\frac{1}{2}m{\dot{x}}^2.$ YE KYUN MATTER KARTA HAI: yahi free particle hai — sabse simple test case. $L$ haath mein aane ke baad baaki har step mechanical hai.

KYA: ek number bob ki position uske circular arc par fix kar deta hai — downward vertical se swing angle $\theta$. Toh $q=\theta$. YE COORDINATE KYUN: rod rigid hai, isliye radius kabhi nahi badalta; sirf angle free hai. $(x,y)$ ki jagah $\theta$ use karne ka matlab hai ek coordinate do ki jagah plus ek constraint.

Neeche figure padhna. Dashed vertical line reference hai: ye pivot se arc ke lowest point (open circle) tak seedha neeche point karti hai. Solid rod us dashed line se angle $\theta$ banata hai. Red segment lowest point se upar bob ki height $h$ measure karta hai. Geometry se notice karo: pivot lowest point se $\ell$ upar baithta hai, jabki bob pivot se $\ell \cos \theta$ neeche baithta hai — toh bob lowest point se $\ell -\ell \cos \theta$ upar hai. Woh difference exactly red segment hai: $h=\ell -\ell \cos \theta =\ell (1-\cos \theta ).$ Figure par extremes check karo: $\theta =0$ par rod dashed line ke saath lie karta hai, red segment zero ho jaata hai ($h=0$); swing karo aur red segment badhta hai, toh $h$ badhta hai. Sahi hai.

$V=mgh=mg\ell (1-\cos \theta ).$

Jhooth. Condition hai $\delta S=0$ — yaani action $S=\int Ldt$ mein first-order change har allowed wiggle ke liye vanish karta hai. Isse action stationary hota hai, zaruri nahi minimal. Ye minimum, saddle point, ya (conjugate point ke baad) maximum ho sakta hai. Honest naam hai principle of stationary action.

Step (KYA/KYUN): equation ko jo do pieces chahiye unhe compute karo. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$ (derivative treating $\dot{x}$ as a slot), toh $\frac{d}{dt}(m\dot{x})=m\ddot{x}$. $\frac{\partial L}{\partial x}=0$ (koi $x$ appear nahi karta). Euler–Lagrange: $m\ddot{x}-0=0\Rightarrow \ddot{x}=0$. Do baar integrate karo: $\dot{x}=v_0$ (const), $x(t)=x_0+v_{0}t$. Seedhi line, uniform speed. ✅

Orientation flag: $y$ positive upward ke saath, gravitational potential energy height ke saath badhti hai, isliye $V=+mgy$ aur $L=T-V=\frac{1}{2}m{\dot{y}}^2-mgy$. Yahi sign convention hai jo answer ko downward acceleration ke roop mein deta hai. $\frac{\partial L}{\partial \dot{y}}=m\dot{y}\Rightarrow \frac{d}{dt}=m\ddot{y}$. $\frac{\partial L}{\partial y}=-mg$. EL: $m\ddot{y}-(-mg)=0\Rightarrow m\ddot{y}=-mg\Rightarrow \ddot{y}=-g$. ✅ Free fall: minus sign ka matlab hai acceleration downward point karta hai (decreasing $y$ ki taraf), exactly jaisa gravity ko karna chahiye.

$\dot{x}$ piece (KYUN): $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$ — sirf $\frac{1}{2}m{\dot{x}}^2$ term mein $\dot{x}$ hai; uska slot-derivative $m\dot{x}$ hai. Phir $\frac{d}{dt}(m\dot{x})=m\ddot{x}$. $x$ piece (KYUN): $\frac{\partial L}{\partial x}=-kx$. Reason: $L$ ko term by term $x$ ke respect mein differentiate karo. Kinetic term $\frac{1}{2}m{\dot{x}}^2$ mein koi bare $x$ nahi, isliye ye $0$ contribute karta hai. Potential term $-\frac{1}{2}k{x}^2$ power rule se differentiate hota hai ($\frac{d}{dx}{x}^2=2x$): $-\frac{1}{2}k\cdot 2x=-kx$. Yehi poora $\partial L/\partial x$ hai. EL: $m\ddot{x}-(-kx)=0\Rightarrow m\ddot{x}+kx=0\Rightarrow \ddot{x}=-\frac{k}{m}x=-{\omega }^{2}x$, with $\omega =\sqrt{k/m}$. Numbers: $\omega =\sqrt{8/2}=\sqrt{4}=2\text{rad/s}$.

KYUN $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$: bob ki speed apne arc par $\ell \dot{\theta }$ hai (arc length $=\ell \theta$, differentiate karo). Use square karo, aadha karo, $m$ se multiply karo. $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \frac{d}{dt}=m{\ell }^2\ddot{\theta }$. KYUN $\partial L/\partial \theta =-mg\ell \sin \theta$: kinetic term mein koi bare $\theta$ nahi, isliye ye $0$ deta hai. $L$ ka potential part hai $-V=-mg\ell (1-\cos \theta )=-mg\ell +mg\ell \cos \theta$. $\theta$ ke respect mein differentiate karo: constant $-mg\ell$ se $0$ milta hai, aur $\frac{d}{d\theta }\cos \theta =-\sin \theta$, toh $mg\ell \cos \theta$ se $-mg\ell \sin \theta$ milta hai. EL: $m{\ell }^2\ddot{\theta }-(-mg\ell \sin \theta )=0\Rightarrow m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0\Rightarrow \boxed{\ddot{\theta }=-\frac{g}{\ell }\sin \theta }$. Small $\theta$ ke liye, $\sin \theta \approx \theta$, jisse $\omega =\sqrt{g/\ell }$ milta hai.

Naya Lagrangian $L^{\prime }=T-V-C$. Euler–Lagrange mein hume $\partial L^{\prime }/\partial q$ aur $\partial L^{\prime }/\partial \dot{q}$ chahiye. $\frac{\partial L^{\prime }}{\partial \dot{q}}=\frac{\partial L}{\partial \dot{q}}$ (constant ka zero derivative hota hai). $\frac{\partial L^{\prime }}{\partial q}=\frac{\partial L}{\partial q}$ (same reason). Dono pieces identical hain, isliye equation identical hai. YE KYUN MATTER KARTA HAI: sirf potential energy mein differences ka physical meaning hota hai — aap $V$ ka zero jahan convenient ho wahan rakh sakte ho.

Added piece hai $g\equiv \frac{df}{dt}=\frac{\partial f}{\partial q}\dot{q}+\frac{\partial f}{\partial t}$. Action mein uska contribution hai ${\int }_{t_1}^{t_2}\frac{df}{dt}dt=f(q(t_2),t_2)-f(q(t_1),t_1)$ — ek pure boundary term. Kyunki endpoints $q(t_1),q(t_2)$ fixed hain, yeh boundary value kisi bhi path variation ke respect mein constant hai. Iska variation zero hai: $\delta \int gdt=0$. Isliye $\delta S^{\prime }=\delta S$, aur true path — woh jo $\delta S=0$ banata hai — unchanged hai. YE KYUN MATTER KARTA HAI: Lagrangian unique nahi hai; yahi freedom Hamiltonian Mechanics mein gauge choices aur canonical transformations ke neeche hai.

Agar $q$ absent hai toh $\frac{\partial L}{\partial q}=0$. Euler–Lagrange ban jaata hai $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=0\Rightarrow \frac{\partial L}{\partial \dot{q}}=\text{const}.$ Conserved quantity $p\equiv \frac{\partial L}{\partial \dot{q}}$ generalized momentum hai. YE KYUN MATTER KARTA HAI: ek symmetry ($q$ par kuch depend nahi karta, yaani $q$ shift karne se physics unchanged rehti hai) ek conservation law produce karta hai. Yahi Noether's Theorem ka seed hai.

KYUN woh $T$: bead ki velocity ka ek radial part hai $\dot{r}$ (bahar slide karna) aur ek tangential part $r\Omega$ (rotation se carry hona). Speed${}^2={\dot{r}}^2+(r\Omega {)}^2$. $L=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\Omega }^2$. $\frac{\partial L}{\partial \dot{r}}=m\dot{r}\Rightarrow m\ddot{r}$. $\frac{\partial L}{\partial r}=mr{\Omega }^2$ ($\frac{1}{2}m{r}^2{\Omega }^2$ ko power rule se $r$ ke respect mein differentiate karo: $\frac{1}{2}m{\Omega }^2\cdot 2r$). EL: $m\ddot{r}-mr{\Omega }^2=0\Rightarrow \boxed{\ddot{r}={\Omega }^{2}r}$. Interpretation: bead outward accelerate karta hai — jaana-pehchana centrifugal push, yahan bina kisi force diagram ke, purely $r^2{\Omega }^2$ term se aata hai jo ek inverted potential ki tarah act karta hai.

Euler–Lagrange har coordinate ke liye ek baar apply hota hai (har $q_i$ ko apni equation milti hai). $x$ ke liye: $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$, $\frac{\partial L}{\partial x}=0\Rightarrow m\ddot{x}=0$. $y$ ke liye: $\frac{\partial L}{\partial \dot{y}}=m\dot{y}$, $\frac{\partial L}{\partial y}=-mg\Rightarrow m\ddot{y}=-mg$. Toh $\ddot{x}=0$ (uniform horizontal drift) aur $\ddot{y}=-g$ (vertical free fall) — projectile parabola. YE KYUN MATTER KARTA HAI: kyunki $x$ cyclic hai ($L$ mein koi bare $x$ nahi), $p_x=m\dot{x}$ conserved hai — horizontal momentum, exactly jaisi expect ki jaati hai.

Neeche figure padhna. Horizontal black line do media ke beech boundary hai. Point $A=(0,a)$ top medium mein baitha hai (light speed $v_1$), point $B=(d,-b)$ bottom medium mein (light speed $v_2$). Red bent line ray hai: ye $A$ se seedha boundary par crossing point $(x,0)$ tak travel karti hai, phir seedhe $B$ tak. Ek number jo hum choose kar sakte hain woh hai $x$, jahan ye cross karta hai. Dashed vertical line crossing point se boundary ke normal (perpendicular) hai. ==Angle of incidence ${\theta }_1$ red incoming ray aur us dashed normal ke beech measure kiya jaata hai; angle of refraction ${\theta }_2$== red outgoing ray aur normal ke beech. Yahi do marked angles hain jo algebra pin down karega.

Time = (distance/speed) har medium mein: $\mathcal{T}(x)=\frac{\sqrt{x^2+a^2}}{v_1}+\frac{\sqrt{(d-x{)}^2+b^2}}{v_2}.$ Stationary ka matlab $\frac{d\mathcal{T}}{dx}=0$: $\frac{x}{v_1\sqrt{x^2+a^2}}-\frac{(d-x)}{v_2\sqrt{(d-x{)}^2+b^2}}=0.$ Ab figure mein do right triangles padho: upper triangle mein ${\theta }_1$ ke opposite side (normal se $A$ tak horizontal run) $x$ hai aur hypotenuse red segment $\sqrt{x^2+a^2}$ hai, isliye $\frac{x}{\sqrt{x^2+a^2}}=\sin {\theta }_1$. Isi tarah lower triangle mein $\frac{d-x}{\sqrt{(d-x{)}^2+b^2}}=\sin {\theta }_2$. Substitute karne par, $\frac{\sin {\theta }_1}{v_1}=\frac{\sin {\theta }_2}{v_2}.✓$ YE KYUN MATTER KARTA HAI: wahi variational logic jo mechanics mein chala, optics mein bhi chalta hai — ek crossing point poore path ki wiggle ka role play karta hai. Yahi kinship hai jiski wajah se Feynman Path Integral baad mein dono ko unify karta hai.

Orientation flag: $y$ ko positive downward lo (badhta $y$ = start ke neeche zyada gehra). Ye L2.2 ke upward-positive convention se ulta hai — hum ise yahan isliye choose karte hain taaki start ke neeche ka drop simply $+y$ ho aur speed formula mein koi stray minus sign na ho. Horizontal endpoint $x=x_B$ hai. KYA minimize karein: total travel time $\mathcal{T}=\int \frac{ds}{v}$, jahan $ds$ curve ke saath arc length ka chota piece hai aur $v$ wahan bead ki speed. Arc-length piece (KYUN): graph $y(x)$ ke liye, ek small step $dx$ curve ko $dy=y^{\prime }dx$ raise karta hai jahan $y^{\prime }=dy/dx$. Pythagoras se slanted step $ds=\sqrt{d{x}^2+d{y}^2}=\sqrt{1+y^{\prime 2}}dx$ hai. Speed piece (KYUN): energy conservation ke saath $y$ downward measure kiya gaya — bead ek height $y$ gira hai, potential ko kinetic mein convert kar raha hai: $\frac{1}{2}m{v}^2=mgy\Rightarrow v=\sqrt{2gy}$. Put them together: $\boxed{\mathcal{T}[y]={\int }_0^{x_B}\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}dx},F(y,y^{\prime })=\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}.$ Yahan $x$ woh role play karta hai jo mechanics mein time $t$ ne play kiya, aur $y(x)$ path $q(t)$ ka role play karta hai; integrand $F(y,y^{\prime })$ Lagrangian $L(q,\dot{q})$ ka role play karta hai. YE KYUN MATTER KARTA HAI: ye precisely ek Calculus of Variations problem hai — wahi Euler–Lagrange engine ise solve karta hai. Kyunki $F$ mein koi explicit $x$ nahi hai, Beltrami identity ($F-y^{\prime }\partial F/\partial y^{\prime }=\text{const}$) apply hoti hai aur cycloid deti hai. (Us ODE ko solve karna agla step hai; yahan hume sirf correct functional aur integrand $F$ chahiye tha.)

KYA test karte hain. True path $x_{\text{true}}(t)$ pehle se $\delta S=0$ satisfy karta hai (first variation vanish hota hai). Ise classify karne ke liye hum second variation ${\delta }^{2}S$ examine karte hain — $S$ mein leading change jab hum ek small endpoint-vanishing wiggle $\eta (t)$ add karte hain. Perturbed action ko second order tak likhne par, $S[x_{\text{true}}+\eta ]=S[x_{\text{true}}]+\underset{=0}{\underbrace{\delta S}}+\frac{1}{2}{\delta }^{2}S+\cdots$ Path minimum hai agar ${\delta }^{2}S>0$ har allowed $\eta$ ke liye, aur saddle agar koi $\eta$ ${\delta }^{2}S<0$ banata hai. KYUN oscillator turn karta hai. $L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}m{\omega }^2{x}^2$ ke liye milta hai ${\delta }^{2}S={\int }_{t_1}^{t_2}(m{\dot{\eta }}^2-m{\omega }^2{\eta }^2)dt.$ ${\dot{\eta }}^2$ term ko $\eta$ ke liye parts se integrate karo (boundary mar jaata hai kyunki $\eta (t_1)=\eta (t_2)=0$): ${\delta }^{2}S=-{\int }_{t_1}^{t_2}m\eta (\ddot{\eta }+{\omega }^2\eta )dt.$ Iska sign operator $\ddot{\eta }+{\omega }^2\eta$ se govern hota hai. Ek nonzero wiggle $\eta$ jo dono endpoints par vanish kare aur $\ddot{\eta }+{\omega }^2\eta =0$ satisfy kare, exactly ek sine hai $\eta \propto \sin (\omega (t-t_1))$ jo zero return karti hai — aur pehli baar ye kar sakti hai jab $\omega (t_2-t_1)=\pi$, yaani half period ke baad $t_2-t_1=\pi /\omega$. Woh instant conjugate point hai. Teen regimes padh ke.

• $t_2-t_1<\pi /\omega$: koi allowed wiggle ${\delta }^{2}S$ ko negative nahi kar sakti, isliye ${\delta }^{2}S>0$ sab $\eta$ ke liye ⇒ path genuinely minimum hai. KYUN: restoring "spring cost" $m{\omega }^2{\eta }^2$ itne chote span par kinetic penalty $m{\dot{\eta }}^2$ se zyada nahi ho sakta.
• $t_2-t_1=\pi /\omega$ (conjugate point): special sine ${\delta }^{2}S=0$ banata hai — second variation bilkul flat ho jaata hai, exact borderline mark karta hai.
• $t_2-t_1>\pi /\omega$: ab ek wiggle exist karta hai jo ${\delta }^{2}S<0$ banata hai ⇒ stationary path saddle hai, minimum nahi. KYUN: itne lambe span par potential term us shape ke liye jeet jaata hai, isliye ek competing path action ko lower karta hai. YE KYUN MATTER KARTA HAI. Yahi L1.3 ke peeche concrete proof hai: "least action" ek misnomer hai — conjugate point ke baad true path stationary hai lekin least nahi.

$V^{\prime }$ ko $V$ ka derivative uske single argument $u=x_1-x_2$ ke respect mein likho. Har particle ke liye Euler–Lagrange: $x_1$: $\frac{d}{dt}(m_1{\dot{x}}_1)-\frac{\partial L}{\partial x_1}=0$. Kyunki $\frac{\partial L}{\partial x_1}=-V^{\prime }(x_1-x_2)\cdot \frac{\partial u}{\partial x_1}=-V^{\prime }(x_1-x_2)$, ye deta hai $m_1{\ddot{x}}_1=-V^{\prime }(x_1-x_2)$. $x_2$: $\frac{\partial L}{\partial x_2}=-V^{\prime }(x_1-x_2)\cdot \frac{\partial u}{\partial x_2}=+V^{\prime }(x_1-x_2)$ (chain rule sign flip karta hai kyunki $x_2$ $-x_2$ ke roop mein enter karta hai), deta hai $m_2{\ddot{x}}_2=+V^{\prime }(x_1-x_2)$. Do equations of motion add karo: $\frac{d}{dt}(m_1{\dot{x}}_1+m_2{\dot{x}}_2)=-V^{\prime }(x_1-x_2)+V^{\prime }(x_1-x_2)=0$. Toh $p=m_1{\dot{x}}_1+m_2{\dot{x}}_2=\text{const}$. ✅ KYUN kaam karta hai: dono particles ko same amount se shift karne par ($x_i\to x_i+c$) $x_1-x_2$ — aur isliye $L$ — unchanged rehta hai. Woh translation symmetry hi momentum conservation hai, Noether's Theorem ka statement miniature mein.

Yahan har problem ek hi move hai: $L=T-V$ banao, ise $\frac{d}{dt}{\partial }_{\dot{q}}L={\partial }_{q}L$ ko do, ek equation per coordinate — aur physics padho (motion, conservation, optics, minimality).