Worked examples — Principle of least action — Hamilton's principle derivation

2.1.19 · D3 · Physics › Analytical Mechanics › Principle of least action — Hamilton's principle derivation

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The scenario matrix

Kuch bhi solve karne se pehle, har case class ka naam rakhte hain jo yeh topic tumpe throw kar sakta hai. Har row ek "flavour" ki problem hai; aakhri column batata hai ki neeche kaun sa worked example use cover karta hai.

#	Case class	Kya cheez ise alag banati hai	Covered by
A	No potential ($V=0$)	$\partial L/\partial q=0$ ⟹ ek conserved momentum	Ex 1
B	Constant force ($V$ linear in $q$)	$\partial L/\partial q=\text{const}$ ⟹ constant acceleration	Ex 2
C	Restoring force ($V$ quadratic)	oscillation, stiffness ka sign matter karta hai	Ex 3
D	Sign flip / degenerate stiffness ($k<0$ or $k=0$)	runaway vs. free motion — limiting behaviour	Ex 4
E	Curvilinear / non-Cartesian coordinate ($q=\theta$)	kinetic energy nahi hai $\frac{1}{2}m{\dot{q}}^2$; nayi geometry	Ex 5
F	Cyclic (ignorable) coordinate	$L$ mein $q$ nahi hai ⟹ Noether's Theorem conserved quantity	Ex 6
G	Explicit time dependence ($L$ mein $t$ hai)	energy conserved nahi; forced system	Ex 7
H	Real-world word problem	words ko translate karo → $T$, $V$, $L$ mein	Ex 8
I	Exam twist: free endpoint	boundary term nahi marta ⟹ natural boundary condition	Ex 9

Ab hum A → I order mein chalenge. Page ke neeche har numeric answer machine-checked hai.

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Ex 1 — Case A: free particle (no potential)

Step 1. Do partial derivatives nikalo. $\frac{\partial L}{\partial x}=0,\frac{\partial L}{\partial \dot{x}}=m\dot{x}.$ Yeh step kyun? Euler–Lagrange ko exactly yeh do ingredients chahiye. Yahan $L$ mein koi $x$ nahi hai, sirf $\dot{x}$ hai — yahi "no potential" ka matlab hai.

Step 2. $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0$ mein plug karo: $\frac{d}{dt}(m\dot{x})-0=0\Rightarrow m\ddot{x}=0\Rightarrow \ddot{x}=0.$ Yeh step kyun? $\partial L/\partial x$ zero hone ka matlab hai ki momentum $m\dot{x}$ ki time-derivative zero hai — momentum conserved hai. Yeh Noether's Theorem ka pehla taste hai: $L$ mein $x$ nahi ⟹ $x$-momentum conserved.

Step 3. Integrate karo: $\dot{x}=\text{const}=v$. $x(0)=0$, $x(3)=6$ ke saath: $v=\frac{6-0}{3}=2\text{m/s}.$ Yeh step kyun? Constant velocity hi ek aisi motion hai jisme $\ddot{x}=0$ ho; fixed endpoints constant ko pin kar dete hain.

Step 4. Momentum $p=m\dot{x}=2\times 2=4\text{kg⋅m/s}$.

Verify: Units: $\text{kg}\cdot \text{m/s}$ ✅. Sanity: koi force nahi ⟹ straight uniform motion. Yeh constant-speed path $\int \frac{1}{2}m{\dot{x}}^{2}dt$ ko stationary banati hai (yahan sach mein short times ke liye minimum — koi bhi wobble ${\dot{x}}^2$ ka average badha deta hai), exactly jaisa Calculus of Variations promise karta hai. Answer $p=4$.

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Ex 2 — Case B: falling body (constant force)

Step 1. $\frac{\partial L}{\partial \dot{y}}=m\dot{y}$, isliye $\frac{d}{dt}(m\dot{y})=m\ddot{y}$. Yeh step kyun? Yeh Euler–Lagrange ka "momentum rate" term hai.

Step 2. $\frac{\partial L}{\partial y}=-mg$. Yeh step kyun? Potential debt $-mgy$ ko differentiate karo; linear potential ek constant slope deta hai — yahi constant force ki pehchaan hai.

Step 3. EL: $m\ddot{y}-(-mg)=0\Rightarrow m\ddot{y}=-mg\Rightarrow \ddot{y}=-g=-10{\text{m/s}}^2$. Yeh step kyun? Mass cancel ho jata hai — bhari aur halki dono cheezein ek jaisi girti hain, aur minus "neeche ki taraf" batata hai.

Step 4. Origin par rest se, $y(t)=-\frac{1}{2}g{t}^2$. $t=2$ par: $y=-\frac{1}{2}(10)(4)=-20\text{m}$.

Verify: $\frac{1}{2}g{t}^2$ ke units: $({\text{m/s}}^2)({\text{s}}^2)=\text{m}$ ✅. Sign negative = neeche gira ✅. Answer $y(2)=-20\text{m}$, jo Newton's Second Law $m\ddot{y}=-mg$ ko recover karta hai.

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Ex 3 — Case C: simple harmonic oscillator (restoring force)

Step 1. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}(m\dot{x})=m\ddot{x}$. Yeh step kyun? Phir se momentum-rate term.

Step 2. $\frac{\partial L}{\partial x}=-kx$. Yeh step kyun? Quadratic potential differentiate hoke linear restoring force deta hai — oscillation ki pehchaan. Minus $x=0$ ki taraf wapas kheenchta hai.

Step 3. EL: $m\ddot{x}+kx=0\Rightarrow \ddot{x}=-\frac{k}{m}x=-{\omega }^{2}x$ jahan $\omega =\sqrt{k/m}$. Yeh step kyun? $\ddot{x}=-{\omega }^{2}x$ se match karke inspection se frequency read ho jaati hai.

Step 4. $\omega =\sqrt{8/0.5}=\sqrt{16}=4\text{rad/s}$; period $T_{\text{per}}=\frac{2\pi }{\omega }=\frac{2\pi }{4}=\frac{\pi }{2}\approx 1.571\text{s}$.

Verify: $\sqrt{k/m}$ ke units: $\sqrt{(\text{N/m})/\text{kg}}=\sqrt{1/{\text{s}}^2}=\text{1/s}$ ✅. Stiffer $k$ ⟹ bada $\omega$ ⟹ faster ✅. Answers $\omega =4$, $T_{\text{per}}=\pi /2$.

Figure s01 — teen stiffness regimes. Yeh plot $V(x)=\frac{1}{2}k{x}^2$ potential draw karta hai $k>0$ ke liye (lavender, ek ghati: motion oscillate karti hai), $k=0$ ke liye (mint, bilkul flat: free particle), aur $k<0$ ke liye (coral, ulta hill: motion bhaag jaati hai). Iska pedagogical kaam hai $k$ ka sign ko landscape ki shape ke roop mein visible banana — stable minimum versus unstable maximum — jo exactly woh distinction hai jo Ex 4 algebraically exploit karta hai.

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Ex 4 — Case D: degenerate & sign-flipped stiffness (limiting behaviour)

Step 1. Ex 3 ka EL equation kisi bhi $k$ ke liye hold karta hai: $m\ddot{x}+kx=0$. Yeh step kyun? Machine reuse karo; sirf number $k$ badalta hai.

Step 2 — case (a), $k=0$. Tab $m\ddot{x}=0\Rightarrow \ddot{x}=0$: oscillator degenerate hokar Ex 1 ka free particle ban jaata hai. Yeh step kyun? Yeh limiting value $k\to 0$ hai — SHM answer $\omega =\sqrt{k/m}\to 0$ ka matlab hai "infinitely long period", yaani bilkul oscillation nahi.

Step 3 — case (b), $k<0$. $k=-|k|$ likho: $\ddot{x}=-\frac{k}{m}x=+\frac{|k|}{m}x={\gamma }^{2}x$ jahan $\gamma =\sqrt{|k|/m}$. Yeh step kyun? $x$ par positive coefficient (flipped sign se) sines ko exponentials mein badal deta hai: $x(t)=A{e}^{\gamma t}+B{e}^{-\gamma t}$ — solution bhaag jaata hai. Isliye $\delta S=0$ stationary deta hai, minimum nahi: hilltop par action ek saddle/maximum hai, jo parent ke pehle "mistake" callout se match karta hai.

Step 4. (b) ke liye numeric growth rate: $\gamma =\sqrt{2/0.5}=\sqrt{4}=2{\text{s}}^{-1}$.

Verify: (a) $\omega =\sqrt{0/m}=0$ ✅ degenerate. (b) $\gamma =2$, to displacement $e^{2t}$ ki tarah badhta hai — unstable ✅. Yeh edge case exactly woh hai jahan "least action" ek jhooth hai aur "stationary action" sach hai.

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Ex 5 — Case E: simple pendulum (curvilinear coordinate)

Step 1 — angle mein kinetic energy. Bob radius $\ell$ ke circle par move karta hai; uski speed $v=\ell \dot{\theta }$ hai. To $T=\frac{1}{2}m{v}^2=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2.$ Yeh step kyun? Yahi curvilinear coordinate ka poora point hai: angle ke per unit mein distance $\ell$ hai, to $\theta$ ke liye "effective mass" $m{\ell }^2$ hai — ek moment of inertia. Figure dekho: arc length $\ell \theta$ hai.

Step 2 — potential energy. Lowest point se height $\ell (1-\cos \theta )$ hai, to $V=mg\ell (1-\cos \theta )$. Yeh step kyun? Bob ko uthane mein potential debt lagti hai; circle ki geometry height set karti hai.

Step 3 — assemble karo aur EL apply karo. $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2-mg\ell (1-\cos \theta )$.

• $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \frac{d}{dt}=m{\ell }^2\ddot{\theta }$.
• $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta$.

EL: $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0\Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta .$ Yeh step kyun? Wahi chhe-step machine; sirf symbols badal gaye kyunki geometry badal gayi.

Step 4 — small angle limit. Chhote $\theta$ ke liye, $\sin \theta \approx \theta$, to $\ddot{\theta }=-\frac{g}{\ell }\theta =-{\omega }^2\theta$, $\omega =\sqrt{g/\ell }$. $\omega =\sqrt{10/2.5}=\sqrt{4}=2\text{rad/s}.$

Verify: Units $\sqrt{({\text{m/s}}^2)/\text{m}}=\text{1/s}$ ✅. Structure Ex 3 ke SHM se match karta hai jisme $g/\ell$ ki role $k/m$ jaisi hai ✅. Answer $\omega =2\text{rad/s}$.

Figure s02 — pendulum geometry. Yeh diagram Ex 5 mein use har symbol earn karta hai: lavender rod length $\ell$ ka, coral bob $m$, mint angle $\theta$ dashed vertical se measure kiya hua, butter-coloured arc jiska length $\ell \theta$ hai (to speed $v=\ell \dot{\theta }$), aur height $\ell (1-\cos \theta )$ mark karne wala vertical drop. Iska pedagogical role hai yeh dikhana ki kyun $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$ aur $V=mg\ell (1-\cos \theta )$ hain, inhe likhne se pehle — geometry hi derivation hai.

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Ex 6 — Case F: cyclic coordinate & ek conserved quantity

Step 1. $\varphi$-Euler–Lagrange equation test karo: $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\varphi }}\right)-\frac{\partial L}{\partial \varphi }=0$. Yeh step kyun? Hume specifically $\varphi$ coordinate ke liye do ingredients banana hai.

Step 2. $\frac{\partial L}{\partial \varphi }=0$ (cyclic!) aur $\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }$. Yeh step kyun? Cyclic (ignorable) coordinate woh hota hai jo $L$ mein absent ho; uska $\partial L/\partial q$ vanish karta hai.

Step 3. Tab EL read karta hai $\frac{d}{dt}(m{r}^2\dot{\varphi })=0\Rightarrow p_{\varphi }=m{r}^2\dot{\varphi }=\text{const}.$ Yeh step kyun? Zero force-slot ⟹ conjugate momentum conserved hai. $p_{\varphi }$ angular momentum hai — yeh Noether's Theorem concrete form mein: rotational symmetry ↔ conserved angular momentum.

Step 4. $p_{\varphi }=m{r}^2\dot{\varphi }=1\times 2^2\times 3=12{\text{kg⋅m}}^2/\text{s}.$

Verify: Units $\text{kg}\cdot {\text{m}}^2/\text{s}$ = angular momentum ✅. Answer $p_{\varphi }=12$. (Hamiltonian Mechanics se link: $p_{\varphi }$ exactly woh momentum hai jo tum Legendre transform mein le jaoge.)

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Ex 7 — Case G: explicitly time-dependent Lagrangian (forced system)

Step 1. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}(m\dot{x})=m\ddot{x}$. Yeh step kyun? Momentum-rate term.

Step 2. $\frac{\partial L}{\partial x}=F_0\cos (\Omega t)$. Yeh step kyun? $F_{0}x\cos (\Omega t)$ ko $x$ ke w.r.t. differentiate karne par $t$ frozen maana jaata hai — yeh time-varying driving force expose karta hai.

Step 3. EL: $m\ddot{x}-F_0\cos (\Omega t)=0\Rightarrow m\ddot{x}=F_0\cos (\Omega t).$ Yeh step kyun? Yeh Newton hai ek oscillating external force ke saath. Kyunki $L$ explicitly $t$ par depend karta hai, total energy conserved nahi hai — energy drive ke zariye andar-bahar pump hoti hai.

Step 4. $t=\frac{\pi }{6}$ par: drive phase $\Omega t=2\cdot \frac{\pi }{6}=\frac{\pi }{3}$ hai, aur $\cos \frac{\pi }{3}=\frac{1}{2}$ hai, to $\ddot{x}\left(\frac{\pi }{6}\right)=\frac{F_0}{m}\cos \left(\frac{\pi }{3}\right)=\frac{4}{1}\times \frac{1}{2}=2{\text{m/s}}^2.$ Yeh step kyun? EL equation ko $m$ se divide karo aur diye gaye instant par cosine evaluate karo — acceleration simply driving force ko track karta hai.

Verify: Units $\text{N/kg}={\text{m/s}}^2$ ✅. Answer $\ddot{x}\left(\frac{\pi }{6}\right)=2{\text{m/s}}^2$. Ex 6 (symmetry ⟹ conservation) se compare karo is example se (broken time-symmetry ⟹ energy conservation nahi).

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Ex 8 — Case H: real-world word problem (Atwood machine)

Step 1 — pure system ke liye ek coordinate. String unhe bandha hai: agar $m_1$ $x$ neeche gira, to $m_2$ $x$ upar gaya, aur dono $\dot{x}$ speed se chalte hain. Kinetic energy: $T=\frac{1}{2}(m_1+m_2){\dot{x}}^2.$ Yeh step kyun? Constraint do objects ko ek degree of freedom $x$ mein collapse kar deta hai — Lagrangian bookkeeping ki elegance.

Step 2 — potential energy. Neeche ko positive $x$ maano aur height ka zero pulley par rakho. Tab $m_1$ $x$ neeche gaya hai (height $-x$, PE $=-m_{1}gx$) jabki $m_2$ $x$ upar gaya hai (height $+x$, PE $=+m_{2}gx$). Total: $V=-m_{1}gx+m_{2}gx=(m_2-m_1)gx.$ Yeh step kyun? Height changes equal aur opposite hain; isliye potential debt sirf weights ke difference par depend karti hai. Yeh woh piece hai jo naive "sirf $T$ likho" attempt mein missing hota hai — koi $V$ nahi, koi forces nahi.

Step 3 — assemble karo aur EL apply karo. $L=T-V=\frac{1}{2}(m_1+m_2){\dot{x}}^2-(m_2-m_1)gx.$

• $\frac{\partial L}{\partial \dot{x}}=(m_1+m_2)\dot{x}\Rightarrow \frac{d}{dt}=(m_1+m_2)\ddot{x}.$ Kyun? combined mass ke liye momentum-rate term.
• $\frac{\partial L}{\partial x}=-(m_2-m_1)g=(m_1-m_2)g.$ Kyun? linear potential ka derivative — net driving weight.

EL: $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0\Rightarrow (m_1+m_2)\ddot{x}-(m_1-m_2)g=0.$

Step 4 — acceleration solve karo. $\ddot{x}=\frac{m_1-m_2}{m_1+m_2}g=\frac{3-1}{3+1}\times 10=\frac{2}{4}\times 10=5{\text{m/s}}^2.$

Verify: $m_1>m_2$ hone se $\ddot{x}>0$: bhaari side neeche jaayega ✅. Magnitude $5<g=10$ kyunki dono masses ko saath accelerate karna padta hai ✅. Units ${\text{m/s}}^2$ ✅. Answer $\ddot{x}=5{\text{m/s}}^2$, jo Newton's Second Law ka standard Atwood result match karta hai.

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Ex 9 — Case I: exam twist — ek free endpoint

Step 1 — path vary karo, boundary term rakhte hue. Parent ke integration by parts ki tarah, lekin endpoints discard na karo: $\frac{dS}{d\epsilon }{|}_0={\left[\frac{\partial L}{\partial \dot{x}}\eta \right]}_{t_1}^{t_2}+{\int }_{t_1}^{t_2}\left[\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}\right]\eta dt=0.$ Yeh step kyun? Parent wali same machine — lekin ab hum boundary term phenkne nahi denge, kyunki $\eta (t_2)$ ab forced zero nahi hai.

Step 2 — do demands alag karo. Interior integral arbitrary $\eta$ ke liye zero hona chahiye (usual EL equation deta hai), aur bachha hua boundary piece apne aap zero hona chahiye. Fixed start par $\eta (t_1)=0$ us end ko maar deta hai; free end par $\eta (t_2)$ arbitrary hai, to uska coefficient zero hona chahiye: ${\frac{\partial L}{\partial \dot{x}}|}_{t_2}=0\Rightarrow m\dot{x}(t_2)=0.$ Yeh step kyun? Jab bhi koi factor ek arbitrary quantity ko multiply karta hai ek sum mein jo zero hone par force ho, to woh factor khud zero hona chahiye. Yeh natural boundary condition hai jiske baare mein parent ka aakhri "mistake" callout ne warning diya tha.

Step 3 — interpret karo. $m\dot{x}(t_2)=0$ kehta hai bead free end par zero momentum ke saath pahunchta hai — woh coast karke ruk jaata hai. Physically, jab finish ko kuch pin nahi kar raha, to sabse aasaan path $\int \frac{1}{2}m{\dot{x}}^{2}dt$ ko bilkul na chalke minimize karta hai: $\dot{x}\equiv 0$. Yeh step kyun? Interior EL equation abhi bhi $\ddot{x}=0$ deta hai (constant velocity); free-end condition phir us constant ko zero velocity par pin kar deta hai.

Step 4 — resulting motion ka numeric check. $x(t_1)=0$ aur $\dot{x}=0$ everywhere ke saath, $x(t)=0$ sab $t$ ke liye, aur free end par momentum $p(t_2)=m\dot{x}(t_2)=0$ hai.

Verify: Natural boundary condition deta hai $p(t_2)=0$. $x(0)=0$ ke liye poora path $x\equiv 0$ hai, momentum $0$ ✅. Ex 1 se contrast karo, jahan fixed finish ne $p=4\ne 0$ force kiya tha: same $L$, alag boundary data, alag physics. Answer: $p(t_2)=0$.

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Recall Kaun sa case ek

conserved quantity deta hai, aur kyun? $L$ mein missing coordinate (cyclic) ::: uska conjugate momentum $\partial L/\partial \dot{q}$ conserved hota hai — yeh Noether's Theorem hai (Ex 6, aur Ex 1 ka momentum).

Recall "Least action" ek misnomer kyun hai, aur kaunse example mein dikhta hai?

Ex 4(b): negative stiffness ke saath action ek saddle/maximum hai, isliye sirf stationary $\delta S=0$ sahi hai, minimum nahi. ::: correct.

Recall Integration by parts mein boundary term kab survive karta hai?

Jab ek endpoint free ho (Ex 9) ::: to natural boundary condition milti hai $\partial L/\partial \dot{q}=0$ wahan.

Yeh bhi dekho: Fermat's Principle (stationary action ka optics twin) aur Feynman Path Integral (jahan sab paths contribute karte hain, aur stationary wala dominate karta hai).