Worked examples — Hamilton's equations of motion

2.1.12 · D3 · Physics › Analytical Mechanics › Hamilton's equations of motion

Examples se pehle, notation ka ek reminder taaki kuch bhi unexplained na rahe:

Scenario matrix

Neeche har worked example uss cell ke saath tagged hai jo wo cover karta hai. Matrix un situations ke types ko name karti hai jinse Hamilton ki machine ko survive karna hai.

Example 1 — Baseline: gravity mein girta hua mass [C1] [C8]

1. Momentum. $p=\frac{\partial L}{\partial \dot{x}}=m\dot{x}$. Yeh step kyun? $p$ woh variable hai jo $\dot{x}$ ki jagah lega; hume $\dot{x}=p/m$ bhi invert karna hai.
2. Hamiltonian. $H=p\dot{x}-L=p\cdot \frac{p}{m}-(\frac{1}{2}m\frac{p^2}{m^2}-mgx)=\frac{p^2}{2m}+mgx$. Yeh step kyun? $\dot{x}=p/m$ substitute karne se har $\dot{x}$ khatam ho jaata hai; jo bachta hai woh sirf $(x,p)$ ka function hai. Yahan $V=mgx$ (gravity ki stored energy) hai aur yeh $T+V$ ki tarah padhta hai.
3. Canonical equations. $\dot{x}=\frac{\partial H}{\partial p}=\frac{p}{m}$, aur $\dot{p}=-\frac{\partial H}{\partial x}=-mg$. Yeh step kyun? Yeh machine ke do rules hain; doosra kehta hai ki "push" ek steady rate pe decrease hoti hai.
4. Acceleration recover karo. $\ddot{x}=\frac{\dot{p}}{m}=\frac{-mg}{m}=-g=-9.8{\text{m/s}}^2$. Yeh step kyun? $\dot{x}=p/m$ ko differentiate karke $\dot{p}$ daalne se do first-order equations wapas ek Newton equation ban jaati hain.

Example 2 — Sign flip: repulsive (inverted) spring [C2]

1. Equations padhna. $\dot{x}=\frac{\partial H}{\partial p}=p$, $\dot{p}=-\frac{\partial H}{\partial x}=-(-4x)=+4x$. Yeh step kyun? Hume minus rule literally apply karna hai — do minus signs (ek rule se, ek $V<0$ se) milke plus bante hain.
2. Case $x>0$. $\dot{p}=+4x>0$: momentum aur positive hota jaata hai, bead aur right push hoti hai. Yeh step kyun? Hum sign region-by-region check karte hain kyunki sign error sirf ek region mein reveal hogi.
3. Case $x<0$. $\dot{p}=+4x<0$: momentum aur negative hota jaata hai, bead aur left push hoti hai. Yeh step kyun? Dono regions baahir push karte hain — origin unstable hai, exactly wahi jo ek inverted potential ko karna chahiye.
4. Case $x=0$, $p=0$. $\dot{x}=0$ aur $\dot{p}=0$ dono: ek fixed point, lekin ek unstable wala (koi bhi nudge grow karta hai). Yeh step kyun? Degenerate equilibrium ko classify karna zaroori hai; yahan $\ddot{x}=\dot{p}=4x$ exponential growth $x\sim e^{2t}$ deta hai, oscillation nahi.

Neeche ki figure is potential ko ek hilltop ki tarah draw karti hai. Amber arrows dekho: woh curve pe chaar sample points pe baithe hain aur hamesha centre se door point karte hain — $x<0$ hone pe leftward, $x>0$ hone pe rightward. Yahi $\dot{p}=+4x$ ka geometric matlab hai ki iski sign $x$ jaisi hai. Amber dot top pe unstable fixed point mark karta hai: perfectly balanced, lekin zarraa sa slip bead ko dono taraf slide kara deta hai.

Example 3 — Degenerate case: free particle [C3]

1. Hamiltonian. $p=m\dot{x}$, toh $H=\frac{p^2}{2m}$ (pure kinetic, kyunki $V=0$ matlab kahin bhi koi stored energy nahi). Yeh step kyun? $V=0$ set karna sabse clean test hai ki machine "kuch nahi hota" wala case handle kar sakti hai ya nahi.
2. Equations. $\dot{x}=\frac{p}{m}$, $\dot{p}=-\frac{\partial H}{\partial x}=0$. Yeh step kyun? $H$ mein $x$ hai hi nahi, toh ${\partial }_{x}H=0$ — momentum exactly conserved hai.
3. Solve karo. $p=\text{const}=p_0$, isliye $\dot{x}=p_0/m$ constant, isliye $x(t)=x_0+\frac{p_0}{m}t$. Yeh step kyun? Constant velocity, straight line — yeh Newton ka first law formalism se nikalta hai.
4. Phase-space picture. Point $(x,p)$ ek horizontal line $p=p_0$ ke saath move karta hai, $p_0/m$ speed se right drift karta hua. Yeh step kyun? Trajectory picture karte hain taaki closed SHO ellipse ki degenerate limit dekh sakein: jaise restoring force $\to 0$ hoti hai, ellipse flat line mein khul jaata hai (Example 7 mein exactly is limit pe wapas aayenge).

Example 4 — Angular coordinate: pendulum [C4]

1. Angular momentum. $p_{\theta }=\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }$. Yeh step kyun? Ek angle ka conjugate linear momentum nahi hota — yeh angular momentum hota hai (units ${\text{kg⋅m}}^2/\text{s}$). Invert karo: $\dot{\theta }=p_{\theta }/(m{\ell }^2)$.
2. Hamiltonian. $H=p_{\theta }\dot{\theta }-L=\frac{p_{\theta }^2}{2m{\ell }^2}-mg\ell \cos \theta$. Yeh step kyun? Potential energy hai $V=-mg\ell \cos \theta$ ($\theta =0$ pe lowest). $\dot{\theta }$ substitute karne se $H$ sirf $(\theta ,p_{\theta })$ mein reh jaata hai.
3. Equations. $\dot{\theta }=\frac{\partial H}{\partial p_{\theta }}=\frac{p_{\theta }}{m{\ell }^2}$, ${\dot{p}}_{\theta }=-\frac{\partial H}{\partial \theta }=-mg\ell \sin \theta$. Yeh step kyun? Doosra, pehle ke saath combine karke $m{\ell }^2\ddot{\theta }=-mg\ell \sin \theta$ deta hai — exact nonlinear pendulum.
4. Small-angle reduction. $|\theta |\ll 1$ ke liye, $\sin \theta \approx \theta$, toh $\ddot{\theta }\approx -\frac{g}{\ell }\theta$ — SHO with $\omega =\sqrt{g/\ell }$. Yeh step kyun? Hum check karte hain ki nonlinear pendulum us harmonic oscillator pe degrade hota hai jis par hume already trust hai, yeh parent ke Example B se jodta hai. (SHO ka actual limiting behaviour — $\omega \to 0$ aur large energy — alag se Example 7 mein dissect kiya gaya hai.)

Example 5 — Trap: charge in a magnetic field, $p\ne m\dot{q}$ [C5]

1. Momentum — surprise. $p=\frac{\partial L}{\partial \dot{x}}=m\dot{x}+a$. Yeh step kyun? Velocity term $a\dot{x}$ contributes $a$. Toh $p\ne m\dot{x}$; "canonical" momentum "kinetic" momentum $m\dot{x}$ se $a$ differ karta hai.
2. Sahi invert karo. $\dot{x}=\frac{p-a}{m}$. Yeh step kyun? Hume actual $L$ ke liye $p=\partial L/\partial \dot{x}$ solve karna hai, $p/m$ reuse nahi karna. Yahi is cell ka poora point hai.
3. Hamiltonian. $H=p\dot{x}-L=p\cdot \frac{p-a}{m}-\left(\frac{1}{2}m\right(\frac{p-a}{m}{)}^2+a\cdot \frac{p-a}{m})$. Simplify karo: $H=\frac{(p-a{)}^2}{2m}$. Yeh step kyun? Sahi $\dot{x}$ substitute karke expand karne par linear pieces cancel ho jaate hain, shifted momentum $p-a$ ke saath kinetic energy bachti hai — exactly electromagnetism mein $\frac{(p-qA{)}^2}{2m}$ ka structure.
4. Equations. $\dot{x}=\frac{\partial H}{\partial p}=\frac{p-a}{m}$ ✓ (step 2 se match), aur $\dot{p}=-\frac{\partial H}{\partial x}=0$ ($H$ mein koi $x$ nahi). Yeh step kyun? $p$ conserved hai, lekin note karo ki $m\dot{x}=p-a$ actually move karta hai — physical velocity, canonical momentum nahi, jo tum measure karoge.

Example 6 — $H$ conserved lekin $H\ne E$: rotating wire pe bead [C6]

1. Momentum. $p=\frac{\partial L}{\partial \dot{r}}=m\dot{r}$. Yeh step kyun? Sirf $\dot{r}$ velocity part mein appear karta hai, toh $p$ ordinary radial momentum hai; invert karo $\dot{r}=p/m$.
2. Hamiltonian. $H=p\dot{r}-L=\frac{p^2}{m}-(\frac{1}{2}m\frac{p^2}{m^2}+\frac{1}{2}m{\Omega }^2{r}^2)=\frac{p^2}{2m}-\frac{1}{2}m{\Omega }^2{r}^2$. Yeh step kyun? Notice karo ${\Omega }^2{r}^2$ term pe minus — Legendre transform ne iski sign flip kar di. Isliye $H$ woh $T+V$ sum nahi hai jo tum expect karte.
3. Kya $H$ conserved hai? $H$ mein explicit $t$ nahi hai (motor speed $\Omega$ constant hai), toh $dH/dt={\partial }_{t}H=0$. Conserved. Yeh step kyun? Conservation sirf explicit time ki absence pe depend karta hai, jo yahan holds karta hai.
4. Kya $H=E$ hai? True kinetic energy hai $T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{\Omega }^2{r}^2$ (dono radial aur rotational parts), aur koi potential nahi toh $E=T=\frac{p^2}{2m}+\frac{1}{2}m{\Omega }^2{r}^2$. Lekin $H=\frac{p^2}{2m}-\frac{1}{2}m{\Omega }^2{r}^2$, toh $H\ne E$: yeh $H-E=-m{\Omega }^2{r}^2$ se differ karte hain. Yeh step kyun? Hume dono quantities directly compare karni hain. Yeh full rotational term $m{\Omega }^2{r}^2$ se differ karte hain. Toh $H$ ek conserved quantity hai jo energy nahi hai — yeh cell C6 ka flagship hai. Jab bhi $r\ne 0$ hoga difference nonzero hoga; sirf axis $r=0$ pe yeh momentarily agree karte hain.

Example 7 — SHO orbit ka limiting behaviour [C7]

1. Ellipse equation. Yahan $H=E$ genuinely holds karta hai (quadratic kinetic energy, time-independent — Example 6 se contrast karo). $H=E$ set karo: $\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2=E$. Yeh step kyun? $H$ conserved hai (parent §3), toh orbit $H$ ke ek level curve pe lie karta hai; aur kyunki yahan $H=E$ hai, woh curve constant physical energy ki curve hai — ek ellipse.
2. Semi-axes. $x$ mein: $p=0$ set karo $\Rightarrow x_{\max }=\sqrt{\frac{2E}{m{\omega }^2}}$. $p$ mein: $x=0$ set karo $\Rightarrow p_{\max }=\sqrt{2mE}$. Yeh step kyun? Axis intercepts yeh poochhne se milte hain ki "ellipse har axis ko kahan cross karta hai?" — $x$-axis pe momentum momentarily zero hota hai (turning point, sari energy potential hai), $p$-axis pe position zero hota hai (sari energy kinetic hai). Woh crossing points hi semi-axis lengths hain.
3. Enclosed area. Ellipse ka area $=\pi x_{\max }{p}_{\max }=\pi \sqrt{\frac{2E}{m{\omega }^2}}\sqrt{2mE}=\frac{2\pi E}{\omega }$. Yeh step kyun? Yeh parent ka $\oint pdx=2\pi E/\omega$ confirm karta hai — action, quantization ka hint.
4. Limit $\omega \to 0$ (weak spring). $x_{\max }=\sqrt{2E/(m{\omega }^2)}\to \infty$ jab $p_{\max }=\sqrt{2mE}$ fixed rehta hai; area $2\pi E/\omega \to \infty$. Ellipse horizontally stretch hota jaata hai jab tak Example 3 ki free-particle line $p=p_0$ nahi ban jaata. Yeh step kyun? Restoring force hatana free particle recover karna chahiye; yeh check C7 ko C3 se jodta hai aur confirm karta hai ki formula sensibly degrade hota hai.
5. Limit $E\to \infty$ (large energy). Dono semi-axes $\sqrt{E}$ ki tarah grow karte hain, toh ellipse same shape rakhta hai (aspect ratio $p_{\max }/x_{\max }=m\omega$ energy-independent hai) lekin size mein blow up karta hai; area linearly grow karta hai, $\propto E$. Yeh step kyun? Large-energy behaviour model ke sensible hone ke liye bounded/self-similar hona chahiye — koi shape distortion nahi aata, toh SHO har energy pe well-behaved hai.
6. Limit $E\to 0$ (collapse). Dono axes $\to 0$; ellipse origin pe fixed point tak shrink ho jaata hai — bead permanently rest pe. Yeh step kyun? Vanishing limit check karna degenerate rest state confirm karta hai, Example 2 ki unstable/stable fixed-point discussion ke saath loop close karta hai.

Example 8 — Real-world word problem: rocket sled on a spring bumper [C1] [C8]

1. Words ko $L$ mein translate karo. Kinetic $T=\frac{1}{2}m{\dot{x}}^2$, elastic potential $V=\frac{1}{2}k{x}^2$, toh $L=T-V=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2$. Yeh step kyun? Word problem pehle Lagrangian banana chahiye; machine ka wahi ek entry point hai.
2. Momentum & Hamiltonian. $p=m\dot{x}\Rightarrow \dot{x}=p/m$; $H=\frac{p^2}{2m}+\frac{1}{2}k{x}^2$. Yeh step kyun? $(x,p)$ mein $H$ build karne se equations directly padh sakte hain; yahan $H=T+V=E$ (Example 1-type, scleronomic).
3. Equations & frequency. $\dot{x}=p/m$, $\dot{p}=-kx$, toh $m\ddot{x}=-kx$, yaani $\ddot{x}=-\frac{k}{m}x$, jo $\omega =\sqrt{k/m}$ deta hai. Yeh step kyun? $p$ eliminate karne se Newton wapas aata hai aur angular frequency identify hoti hai.

Example 9 — Exam twist: Legendre transform ulta chalao [C9]

1. $H$ se $\dot{q}$ nikalo. $\dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m}\Rightarrow p=m\dot{q}$. Yeh step kyun? Inverse Legendre transform same pairing $\dot{q}={\partial }_{p}H$ use karta hai; hum ise $\dot{q}$ ke terms mein $p$ ke liye solve karte hain.
2. $L=p\dot{q}-H$ apply karo. $L=(m\dot{q})\dot{q}-(\frac{(m\dot{q}{)}^2}{2m}+V(q))=m{\dot{q}}^2-\frac{1}{2}m{\dot{q}}^2-V(q)$. Yeh step kyun? Legendre transform apna hi inverse hai (sign convention tak): $L=\sum p\dot{q}-H$ with $p$ re-expressed $\dot{q}$ ke through.
3. Simplify karo. $L=\frac{1}{2}m{\dot{q}}^2-V(q)$. Yeh step kyun? Hum exactly wahi Lagrangian recover karte hain jo expected tha — kinetic $T=\frac{1}{2}m{\dot{q}}^2$ minus potential $V(q)$, yaani $L=T-V$ — yeh round trip confirm karta hai.

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