Worked examples — Hamilton's principle — least action

4.10.15 · D3 · Maths › Advanced Topics (Elite Level) › Hamilton's principle — least action

Shuru karne se pehle, ek reminder us recipe ka jo tum har baar use karoge:

Yahaan $\dot{q}$ ka matlab hai "$q$ ka time ke saath change ki rate" aur $\partial$ (curly d) ka matlab hai "differentiate karo, doosre slots ko frozen constants maanke". Bas yahi notation chahiye.

Scenario matrix

Is topic ka har problem in cells mein se kisi ek mein aata hai. Neeche ke examples mein har ek ke saath [Cell A] jaisa tag hai taaki tum dekh sako ki poora grid cover ho raha hai.

Hum force slot ke har sign ko bhi cover karte hain: positive (Ex 2), negative (Ex 1, 4), zero (Ex 3, 5, 6).

Example 1 — Bowl-shaped potential (spring) · [Cell A]

Forecast: padhne se pehle guess karo — mass ko oscillate karna chahiye; agar hum spring stiffen karein (bada $k$), toh frequency upar jaayegi ya neeche?

1. Momentum slot. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$. Yeh step kyun? Sirf $\frac{1}{2}m{\dot{x}}^2$ term mein $\dot{x}$ hai; differentiate karne par $m\dot{x}$ milta hai — ordinary momentum.
2. Force slot. $\frac{\partial L}{\partial x}=-kx$. Yeh step kyun? Sirf $-\frac{1}{2}k{x}^2$ mein $x$ hai; iska slope $-kx$ hai. Minus sign ka matlab hai force $x=0$ ki taraf wapas kheenchti hai — ek bowl (positive curvature $V^{\prime \prime }=+k>0$).
3. E–L assemble karo. $\frac{d}{dt}(m\dot{x})-(-kx)=m\ddot{x}+kx=0$. Yeh step kyun? Yeh recipe ki line 4 hai, word-for-word.
4. Physics padho. $\ddot{x}=-\frac{k}{m}x=-{\omega }^{2}x$ jahan $\omega =\sqrt{k/m}$. Solution $x(t)=A\cos (\omega t+\varphi )$.

Verify: $\omega =\sqrt{8/2}=\sqrt{4}=2\text{rad/s}$. Units: $\sqrt{(\text{N/m})/\text{kg}}=\sqrt{({\text{kg⋅m/s}}^2/\text{m})/\text{kg}}=\sqrt{1/{\text{s}}^2}={\text{s}}^{-1}$ ✓. Stiffer spring ⇒ bada $k$ ⇒ zyada $\omega$ ✓ (forecast se match karta hai).

Example 2 — Hill-shaped potential (unstable) · [Cell B]

Forecast: $V$ ka sign flip ho gaya. Kya mass ab bhi oscillate karega, ya bhaag jaayega?

1. $V$ ko $L$ se read karo. Hume $L=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}k{x}^2$ diya gaya hai. Recipe ke $L=T-V$ se match karo: kinetic part $T=\frac{1}{2}m{\dot{x}}^2$ hai, toh jo bacha woh $-V$ ke barabar hona chahiye. Yaani, $-V=+\frac{1}{2}k{x}^2$, isliye $V=-\frac{1}{2}k{x}^2$. Yeh step kyun? Potential woh nahi hai jo $L$ mein minus sign ke baad aata hai — yeh minus the non-kinetic part hai. Kyunki $L$ mein $+\frac{1}{2}k{x}^2$ hai, algebra force karta hai $V=-\frac{1}{2}k{x}^2$: curvature $V^{\prime \prime }=-k<0$, ek hilltop.
2. Momentum slot. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$. Yeh step kyun? Recipe line 3, pehla half — kabhi skip mat karo. Sirf $\frac{1}{2}m{\dot{x}}^2$ mein $\dot{x}$ hai; iska derivative ordinary momentum $m\dot{x}$ hai (Ex 1 jaisa hi, kyunki kinetic energy same hai).
3. Force slot. $\frac{\partial L}{\partial x}=+kx$. Yeh step kyun? Sirf $+\frac{1}{2}k{x}^2$ mein $x$ hai; iska slope $+kx$ hai. Plus sign ka matlab hai force origin se door dhakelta hai — step 1 mein mila hilltop iske saath consistent hai.
4. E–L assemble karo. $\frac{d}{dt}(m\dot{x})-(+kx)=m\ddot{x}-kx=0\Rightarrow \ddot{x}=+\frac{k}{m}x$. Yeh step kyun? Positive coefficient ka matlab acceleration baahir ki taraf point karti hai — Ex 1 se ulta sign.
5. Physics padho. $\ddot{x}={\omega }^{2}x$ ke solutions hain $x=A{e}^{\lambda t}+B{e}^{-\lambda t}$ jahan $\lambda =\sqrt{k/m}$ — exponential growth, oscillation nahi. $e^{\lambda t}$ kyun na $\cos$? Hume ek aisi function chahiye jiska second derivative plus ek constant times itself ho. $\cos$ minus deta hai; exponential $e^{\lambda t}$ deta hai ${\lambda }^2{e}^{\lambda t}$ — same sign. Yahi precisely woh tool hai jo "second derivative = $+$ itself" ka jawaab deta hai.

Verify: $\lambda =\sqrt{4/1}=2{\text{s}}^{-1}$. Check: $x=e^{2t}\Rightarrow \ddot{x}=4e^{2t}=4x$ aur ${\omega }^{2}x=4x$ ✓. Runaway confirm hua — yeh ek unstable equilibrium hai; least action tab bhi hold karta hai, par extremum ek saddle hai, minimum nahi.

Example 3 — Constant force (linear potential) · [Cell C]

Forecast: constant force ka matlab constant acceleration. Kaunsa classic formula nikalne ki tumhe expect hai?

1. Momentum slot. $\frac{\partial L}{\partial \dot{y}}=m\dot{y}$. Yeh step kyun? Sirf kinetic term $\frac{1}{2}m{\dot{y}}^2$ hi $\dot{y}$ par depend karta hai; iska derivative ordinary momentum $m\dot{y}$ hai.
2. Force slot. $\frac{\partial L}{\partial y}=-mg$ (constant — force slot $y$ par depend nahi karta; yahaan yeh ek nonzero constant hai, hamara "constant-force" case). Yeh step kyun? Sirf $-mgy$ mein $y$ hai; iska slope constant $-mg$ hai, neeche ki taraf kheenchne wala weight.
3. Assemble karo. $\frac{d}{dt}(m\dot{y})-(-mg)=m\ddot{y}+mg=0\Rightarrow \ddot{y}=-g$. Yeh step kyun? Ek linear potential $mgy$ ka slope constant hota hai, isliye acceleration constant hai.
4. Do baar integrate karo. $\dot{y}=-gt$, $y=y_0-\frac{1}{2}g{t}^2$. Yeh step kyun? Ek constant ke antiderivatives familiar kinematic equations dete hain — E–L unhe exactly reproduce karta hai.

Verify: $y=0$ set karo: $20=\frac{1}{2}(10)t^2=5t^2\Rightarrow t^2=4\Rightarrow t=2\text{s}$. Units: $\sqrt{\text{m}/({\text{m/s}}^2)}=\sqrt{{\text{s}}^2}=\text{s}$ ✓.

Example 4 — Angular coordinate: the pendulum · [Cell D]

Forecast: koi forces resolve kiye bina, kya ek single angle saari geometry carry kar sakta hai?

1. Geometry se $T,V$ banao. Bob ki speed $=\ell \dot{\theta }$ (arc length rate), isliye $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$. Lowest point ke upar height $=\ell (1-\cos \theta )$, isliye $V=mg\ell (1-\cos \theta )$. Constant drop karo: $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta$. Yeh step kyun? Figure dekho: amber dotted arc bob ka path dikhata hai; iska length rate hi speed hai. Cyan vertical height dikhata hai. Koi tension nahi, koi force components nahi.
2. Momentum slot. $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }$ (yeh angular momentum hai). Yeh step kyun? Sirf kinetic term $\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$ mein $\dot{\theta }$ hai; differentiate karne par $m{\ell }^2\dot{\theta }$ milta hai, angular momentum.
3. Force slot. $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta$ ($0<\theta <180°$ ke liye negative: neeche wapas kheenchta hai — ek restoring torque). Yeh step kyun? Sirf $mg\ell \cos \theta$ mein $\theta$ hai; $\cos \theta$ ka derivative $-\sin \theta$ hai, jo $-mg\ell \sin \theta$ deta hai.
4. Assemble karo. $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0\Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta$. Yeh step kyun? Yeh recipe ki line 4 hai; $m{\ell }^2$ se divide karne par angular acceleration isolate ho jaata hai.

Verify: $\theta =30°=\pi /6$ par, $\sin 30°=0.5$, isliye $\ddot{\theta }=-9.8/1\times 0.5=-4.9{\text{rad/s}}^2$. Minus sign keh raha hai yeh $\theta =0$ ki taraf wapas accelerate kar raha hai ✓. Units ${\text{m/s}}^2÷\text{m}={\text{s}}^{-2}$, angular acceleration ke liye sahi ✓.

Example 5 — Cyclic coordinate: conserved momentum · [Cell E]

Forecast: E–L ka kya hota hai jab ek coordinate $L$ mein kabhi appear hi nahi karta (sirf iska rate karta hai)?

1. Missing coordinate dhundo. $L$ mein $\dot{\varphi }$ hai par $\varphi$ nahi. Aise coordinate ko hum cyclic (ya ignorable) kehte hain. Yeh step kyun? Agar $\varphi$ absent hai, toh $\frac{\partial L}{\partial \varphi }=0$ — $\varphi$ ka force slot exactly zero hai.
2. $\varphi$ ke liye E–L likho. $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\varphi }}\right)-0=0\Rightarrow \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\varphi }}\right)=0$. Yeh step kyun? Zero force slot ka matlab momentum slot time mein constant hai — yeh ek conservation law hi hai, Noether's Theorem ka beej.
3. Constant identify karo. $\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }=p_{\varphi }=$ const. Yeh angular momentum hai. Yeh step kyun? Sirf $\frac{1}{2}m{r}^2{\dot{\varphi }}^2$ term mein $\dot{\varphi }$ hai; iska derivative $m{r}^2\dot{\varphi }$ hai, jise step 2 ne abhi prove kiya ki constant hai.

Verify: $p_{\varphi }=m{r}^2\dot{\varphi }=3\times 2^2\times 5=3\times 4\times 5=60{\text{kg⋅m}}^2/\text{s}$. Units: ${\text{kg⋅m}}^2/\text{s}$ = angular momentum ✓. (Ex 1–4 se contrast karo jahan force slot nonzero tha, isliye momentum badla.)

Example 6 — Degenerate case: the free particle · [Cell F]

Forecast: $V=0$ hone par sab kuch "kinetic" hai. Kaunsi single path total minimize karti hai?

1. Force slot zero hai. $\frac{\partial L}{\partial x}=0$ ($L$ mein koi $x$ nahi — yeh degenerate case hai). Yeh step kyun? $L$ sirf $\dot{x}$ par depend karta hai, kabhi $x$ par nahi, isliye iska $x$-derivative zero ho jaata hai.
2. E–L. $\frac{d}{dt}(m\dot{x})=0\Rightarrow \dot{x}=$ const. Constant velocity ⇒ straight line $x(t)=2t$ (slope $6/3=2$). Yeh step kyun? Zero force slot ⇒ momentum constant ⇒ uniform speed. Figure ki straight cyan line true path hai.
3. True path ka action. $\dot{x}=2$, isliye $S_{\text{true}}={\int }_0^3\frac{1}{2}m(2{)}^{2}dt=\frac{1}{2}m\cdot 4\cdot 3=6m$. Yeh step kyun? Constant $\dot{x}=2$ hone par integrand constant hai, isliye integral bas value $\times$ length hai.
4. Ek wiggly rival ka action. Try karo $x_w(t)=2t+\sin \left(\frac{\pi t}{3}\right)$ (same endpoints: $\sin 0=\sin \pi =0$). Phir ${\dot{x}}_w=2+\frac{\pi }{3}\cos \left(\frac{\pi t}{3}\right)$ aur $S_w={\int }_0^3\frac{1}{2}m{\dot{x}}_w^{2}dt$. Yeh step kyun? Cross-term $\int 2\cdot \frac{\pi }{3}\cos dt$ ek half-period mein zero integrate hota hai, bacha hua extra positive $\int (\frac{\pi }{3}\cos {)}^2$ — pure added cost.
5. Yeh genuine minimum kyun hai (saddle nahi)? Kisi bhi rival ko $x=\bar{x}+\eta$ likho jahan $\eta (0)=\eta (3)=0$. Action ka difference exactly $S-S_{\text{true}}=\frac{1}{2}m{\int }_0^3{\dot{\eta }}^{2}dt$ hai. Yeh step kyun? $\eta$ mein linear term zero ho jaata hai (E–L hold karta hai), aur second variation $\frac{1}{2}m\int {\dot{\eta }}^{2}dt\ge 0$ hai — squares ka sum, isliye yeh kabhi negative nahi ho sakta. Is liye true path yahaan ek strict minimum hai, kisi bhi interval length ke liye, sirf "short" walo ke liye nahi. Free particles ek aisa case hai jahan minimum globally guaranteed hai; "short interval" caveat (conjugate points) sirf tab kaatata hai jab potential nearby paths ko wapas saath la sake, jo flat $V=0$ kabhi nahi karta.

Verify: $S_w=6m+\frac{1}{2}m{\int }_0^3(\frac{\pi }{3}\cos \frac{\pi t}{3}{)}^{2}dt=6m+\frac{1}{2}m\cdot \frac{{\pi }^2}{9}\cdot \frac{3}{2}=6m+\frac{m{\pi }^2}{12}\approx 6m+0.822m>6m$. Straight line jeet jaati hai ✓ — aur step 5 prove karta hai yeh true minimum hai, kyunki $\frac{1}{2}m\int {\dot{\eta }}^2\ge 0$ hamesha.

Example 7 — Limiting behaviour: small-angle pendulum · [Cell G]

Forecast: tiny swings ke liye, kya nonlinear pendulum Example 1 ka linear oscillator ban jaata hai?

1. $\sin \theta$ approximate karo. Small $\theta$ ke liye, $\sin \theta \approx \theta$ (arc aur iska chord coincide karte hain). Yeh step kyun? $\sin \theta =\theta -\frac{{\theta }^3}{6}+\dots$; jaise $\theta \to 0$, leading term $\theta$ dominate karta hai. Yeh woh limit hai jo ek mushkil equation ko aasaan bana deta hai.
2. Linearize karo. $\ddot{\theta }\approx -\frac{g}{\ell }\theta$. Yeh exactly $\ddot{\theta }=-{\omega }^2\theta$ hai jahan $\omega =\sqrt{g/\ell }$ — bilkul Ex 1 ke spring jaisi shape. Yeh step kyun? Hum SHM template recognize karte hain; iska solution angular frequency $\omega$ ka cosine hai.
3. Period. $T_{\text{period}}=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{\ell }{g}}$. Yeh step kyun? $\cos (\omega t)$ ke liye ek full cycle time $2\pi /\omega$ leta hai — period ki definition.

Verify: $T=2\pi \sqrt{1/9.8}=2\pi \times 0.3194\approx 2.007\text{s}$. Units $\sqrt{\text{m}/({\text{m/s}}^2)}=\text{s}$ ✓. Amplitude → 0 hone par nonlinear system linear tak limit karta hai ✓.

Example 8 — Word problem: bead on a parabolic wire · [Cell H]

Forecast: sentence ko $T$ aur $V$ mein translate karo. Kaunsa single number ($x$) bead ko pin karta hai?

1. Constraint ko coordinates mein daalo. Wire par, $y=\frac{1}{2}a{x}^2$, isliye time mein differentiate karne par $\dot{y}=ax\dot{x}$. Bead ki speed${}^2={\dot{x}}^2+{\dot{y}}^2={\dot{x}}^2+a^2{x}^2{\dot{x}}^2={\dot{x}}^2(1+a^2{x}^2)$. Yeh step kyun? Bead curve par rehne ke liye forced hai, isliye $y$ free nahi hai — yeh $x$ ka function hai. Ek coordinate kaafi hai, bilkul figure ke single pinned bead ki tarah.
2. Energies. $T=\frac{1}{2}m{\dot{x}}^2(1+a^2{x}^2)$ aur $V=mgy=\frac{1}{2}mga{x}^2$. Isliye $L=T-V=\frac{1}{2}m{\dot{x}}^2(1+a^2{x}^2)-\frac{1}{2}mga{x}^2$. Yeh step kyun? Kinetic energy $\frac{1}{2}m(\text{speed}{)}^2$ hai step 1 use karke; potential weight times height $y=\frac{1}{2}a{x}^2$ hai.
3. Momentum slot. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}(1+a^2{x}^2)$. Yeh step kyun? Kinetic term ko $\dot{x}$ mein differentiate karo, $x$ ko frozen maanke — factor $(1+a^2{x}^2)$ saath aa jaata hai.
4. Force slot. $\frac{\partial L}{\partial x}=m{\dot{x}}^2{a}^{2}x-mgax$. Yeh step kyun? $x$ dono mein appear karta hai — kinetic term (through $1+a^2{x}^2$) aur potential mein; dono differentiate karo. Pehla piece $\frac{1}{2}m{\dot{x}}^2\cdot a^2\cdot 2x$ se aata hai, doosra $-\frac{1}{2}mga\cdot 2x$ se.
5. E–L assemble karo. $\frac{d}{dt}[m\dot{x}(1+a^2{x}^2)]-(m{\dot{x}}^2{a}^{2}x-mgax)=0$. Time-derivative expand karo: $m\ddot{x}(1+a^2{x}^2)+2m{a}^{2}x{\dot{x}}^2-m{a}^{2}x{\dot{x}}^2+mgax=0$, yaani $m\ddot{x}(1+a^2{x}^2)+m{a}^{2}x{\dot{x}}^2+mgax=0$. Yeh step kyun? Recipe line 4. Momentum slot par product rule ${\dot{x}}^2$ terms produce karta hai; unme se ek kinetic force-slot term ke saath cancel ho jaata hai, clean exact equation chhodke.
6. $x=0$ ke paas small oscillations. Small $x$ ke liye: $a^2{x}^2\to 0$ aur $x{\dot{x}}^2\to 0$ (higher order), jo $m\ddot{x}+mgax=0\Rightarrow \ddot{x}=-gax$ deta hai. Yeh step kyun? Small swings higher-order-in-$x$ terms drop kar dete hain — Ex 7 jaisi same limiting move — SHM deti hai jahan $\omega =\sqrt{ga}$.

Verify: $\omega =\sqrt{ga}=\sqrt{9.8\times 2}=\sqrt{19.6}\approx 4.427\text{rad/s}$. Units $\sqrt{({\text{m/s}}^2)(1/\text{m})}=\sqrt{{\text{s}}^{-2}}={\text{s}}^{-1}$ ✓ (yahaan $a$ ke units $1/\text{m}$ hain).

Example 9 — Exam twist: two coupled coordinates · [Cell I]

Forecast: do coordinates hone par, hume kitne E–L equations milenge, aur woh aapas mein kaise baat karte hain?

1. $x_1$ ke liye E–L. Momentum slot $\frac{\partial L}{\partial {\dot{x}}_1}=m{\dot{x}}_1$; force slot $\frac{\partial L}{\partial x_1}=+k(x_2-x_1)$. Isliye $m{\ddot{x}}_1=k(x_2-x_1)$. Yeh step kyun? Ek E–L equation per coordinate — parent ka "one per generalized coordinate" rule. $-\frac{1}{2}k(x_2-x_1{)}^2$ par $x_1$ mein chain rule: $-k(x_2-x_1)\cdot (-1)=+k(x_2-x_1)$ milta hai.
2. $x_2$ ke liye E–L. Momentum slot $\frac{\partial L}{\partial {\dot{x}}_2}=m{\dot{x}}_2$; force slot $\frac{\partial L}{\partial x_2}=-k(x_2-x_1)$. Isliye $m{\ddot{x}}_2=-k(x_2-x_1)$. Yeh step kyun? Same potential, par $x_2$ mein differentiate karne par $-k(x_2-x_1)\cdot (+1)=-k(x_2-x_1)$ milta hai — inner sign flip ho jaata hai.
3. Relative coordinate mein decouple karo. $u=x_2-x_1$ (stretch) lo. Eq 1 ko eq 2 se subtract karo: $m({\ddot{x}}_2-{\ddot{x}}_1)=-k(x_2-x_1)-k(x_2-x_1)=-2ku$, isliye $m\ddot{u}=-2ku\Rightarrow \ddot{u}=-\frac{2k}{m}u$. Yeh step kyun? Subtract karne par shared centre-of-mass motion cancel ho jaati hai, relative coordinate mein ek clean SHM expose hoti hai — frequency $\omega =\sqrt{2k/m}$.
4. Centre-of-mass check. Eq 1 aur eq 2 add karo: $m({\ddot{x}}_1+{\ddot{x}}_2)=+k(x_2-x_1)-k(x_2-x_1)=0$: centre of mass constant velocity par drift karta hai (ek conservation jo cyclic case Ex 5 ki echo karta hai). Yeh step kyun? Add karne par equal-and-opposite spring terms cancel ho jaate hain, pair par koi net force nahi bachti.

Verify: $\omega =\sqrt{2\times 6/1}=\sqrt{12}\approx 3.464\text{rad/s}$. Do coordinates → do equations ✓; factor $2$ isliye aata hai kyunki dono masses ek spring share karte hain ✓. Units $\sqrt{(\text{N/m})/\text{kg}}={\text{s}}^{-1}$ ✓.

Active recall