Exercises — Hamilton's equations of motion

KYA: $p=\partial L/\partial \dot{q}$ apply karo. Sirf pehle term mein $\dot{q}$ hai. $p=\frac{\partial }{\partial \dot{q}}(\frac{1}{2}m{\dot{q}}^2-mgq)=m\dot{q}.$ Invert: $\dot{q}=p/m$. YEH KYUN MATTER KARTA HAI: yahi inversion woh step hai jo hume $(q,\dot{q})$ ki jagah $(q,p)$ mein sab kuch express karne deta hai — $H$ ka darwaza.

(a) Stiffness $k$ ki spring par mass $m$ — ek simple harmonic oscillator, kyunki $\frac{1}{2}k{q}^2$ spring potential energy hai. (b) $H$ mein koi explicit $t$ nahi hai, isliye $dH/dt=\partial H/\partial t=0$ se, $H$ conserved hai. (c) $\frac{p^2}{2m}+\frac{1}{2}k{q}^2=\text{const}$ $(q,p)$ plane mein ek ellipse ki equation hai — ek closed loop jis par state hamesha ghoomti rehti hai. ("Ellipse" word ko recall cue ke roop mein cloze kiya gaya hai, link nahi hai.) Neeche figure dekho.

Hamilton har coordinate ke liye do equations deta hai (${\dot{q}}_i$ aur ${\dot{p}}_i$), isliye $2n=6$ first-order equations. Lagrange har coordinate ke liye ek second-order equation deta hai, isliye $3$ equations — lekin har ek second-order hai. Order ko number se trade karke, "information ki total amount" ($6$ initial conditions) same rehti hai. Yahi $2n$-dimensional picture bilkul phase space hai.

Step 1 (momentum): $p=m\dot{q}\Rightarrow \dot{q}=p/m$. Kyun: $\dot{q}$ eliminate karne ke liye. Step 2 (Hamiltonian): $H=p\dot{q}-L=p\cdot \frac{p}{m}-\left(\frac{1}{2}m\right(\frac{p}{m}{)}^2-mgq)=\frac{p^2}{m}-\frac{p^2}{2m}+mgq=\frac{p^2}{2m}+mgq.$ Kyun: $\dot{q}=p/m$ har jagah substitute karo taaki $H$ sirf $(q,p)$ par depend kare. Yeh $T+V$ ke barabar hai. Step 3 (equations): $\dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m},\dot{p}=-\frac{\partial H}{\partial q}=-mg.$ Check: $\dot{p}=-mg$ ka matlab hai $m\ddot{q}=-mg$, yaani $\ddot{q}=-g$ — free fall. Newton phir se aata hai. ✓

Step 1 (momentum). Kyun: hume woh variable chahiye jo $\dot{\theta }$ ki jagah lega, isliye $L$ ko $\dot{\theta }$ ke respect mein differentiate karo — sirf kinetic term $\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$ mein yeh hai. $p_{\theta }=\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \dot{\theta }=\frac{p_{\theta }}{m{\ell }^2}.$ Invert kyun: $H$ ko $(\theta ,p_{\theta })$ par depend karna chahiye, kabhi $\dot{\theta }$ par nahi, isliye $\dot{\theta }$ ko $p_{\theta }$ ke terms mein likhna padega. Step 2 (Hamiltonian). Kyun: $\dot{\theta }=p_{\theta }/(m{\ell }^2)$ ko $H=p_{\theta }\dot{\theta }-L$ mein plug karo taaki har $\dot{\theta }$ gayab ho jaye. $H=p_{\theta }\dot{\theta }-L=\underset{p_{\theta }\dot{\theta }}{\underbrace{\frac{p_{\theta }^2}{m{\ell }^2}}}-\underset{L}{\underbrace{(\frac{1}{2}\frac{p_{\theta }^2}{m{\ell }^2}+mg\ell \cos \theta )}}=\frac{p_{\theta }^2}{2m{\ell }^2}-mg\ell \cos \theta .$ Isko padhna: pehla term rotational kinetic energy hai, doosra gravitational potential energy $-mg\ell \cos \theta$ hai (sabse neeche, $\theta =0$ par). Step 3 (equations). Kyun: dono canonical rules apply karo — $\dot{\theta }$ $H$ ka $p_{\theta }$-slope read karta hai, aur ${\dot{p}}_{\theta }$ minus $\theta$-slope read karta hai. $\dot{\theta }=\frac{\partial H}{\partial p_{\theta }}=\frac{p_{\theta }}{m{\ell }^2},{\dot{p}}_{\theta }=-\frac{\partial H}{\partial \theta }=-mg\ell \sin \theta .$ Aakhri minus kyun: potential slope $\partial H/\partial \theta =+mg\ell \sin \theta$ momentum ko hill se neeche push karta hai, bilkul "p fights q minus" rule ki tarah. Check. Kyun: known physics recover karne ki confirm karne ke liye, $p_{\theta }$ eliminate karo $\dot{\theta }$ differentiate karke: $m{\ell }^2\ddot{\theta }={\dot{p}}_{\theta }=-mg\ell \sin \theta \Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta$ — exact pendulum equation. ✓

Step 1. Kyun: time law $dH/dt=\partial H/\partial t$ kehta hai ki $H$ motion ke along sirf tab change ho sakta hai jab formula mein explicit $t$ ho — isliye hum usi ko dhundhte hain. Yahan woh $\cos (\Omega t)$ ke andar baitha hai. Step 2. Kyun: $H$ ko $t$ ke respect mein differentiate karo jab $q,p$ fixed labels maane jaate hain (iska matlab "explicit" hai): $\frac{\partial H}{\partial t}=-Fq\cdot (-\Omega \sin (\Omega t))=+Fq\Omega \sin (\Omega t),$ jo zero nahi hai. Step 3. Kyun: isko time law mein daalo taaki trajectory ke along actual rate of change mile: $\frac{dH}{dt}=\frac{\partial H}{\partial t}=Fq\Omega \sin (\Omega t)\ne 0.$ Isliye $H$ conserved nahi hai — external drive energy andar aur bahar pump karta hai. $H$ mein explicit $t$ iska tell-tale sign hai.

Pehle consistency. Kyun: charon points ek energy contour par hone chahiye, warna unka flow compare karna bekar hai. $a=\sqrt{2E/(m{\omega }^2)}$ aur $b=\sqrt{2mE}$ se, $\frac{b}{a}=\frac{\sqrt{2mE}}{\sqrt{2E/(m{\omega }^2)}}=\sqrt{m^2{\omega }^2}=m\omega \Rightarrow b=m\omega a.$ Isliye hum $a>0$ aur $b=m\omega a$ lete hain — yeh independent nahi hain. Har point par velocity vector $(\dot{q},\dot{p})$ evaluate karo:

• $(+a,0)$: $\dot{q}=0,\dot{p}=-m{\omega }^{2}a<0$ → neeche ja raha hai.
• $(0,+b)$: $\dot{q}=b/m=\omega a>0,\dot{p}=0$ → daayein ja raha hai.
• $(-a,0)$: $\dot{q}=0,\dot{p}=+m{\omega }^{2}a>0$ → upar ja raha hai.
• $(0,-b)$: $\dot{q}=-b/m=-\omega a<0,\dot{p}=0$ → baayein ja raha hai.

Upar par daayein, daayein par neeche, neeche par baayein, baayein par upar — yeh circulation clockwise hai. Arrows dekho.

Momentum. Kyun: replacement variable pane ke liye $L$ ko $\dot{r}$ se differentiate karo; sirf $\frac{1}{2}m{\dot{r}}^2$ mein $\dot{r}$ hai. $p_r=\frac{\partial L}{\partial \dot{r}}=m\dot{r}\Rightarrow \dot{r}=p_r/m.$ Hamiltonian. Kyun: $\dot{r}$ hatane ke liye $\dot{r}=p_r/m$ ko $H=p_r\dot{r}-L$ mein substitute karo. $H=p_r\dot{r}-L=\frac{p_r^2}{m}-(\frac{1}{2}\frac{p_r^2}{m}+\frac{1}{2}m{\omega }^2{r}^2)=\frac{p_r^2}{2m}-\frac{1}{2}m{\omega }^2{r}^2.$ Compare karo. $E$, $(r,p_r)$ mein likha, $E=\frac{p_r^2}{2m}-\frac{1}{2}m{\omega }^2{r}^2$ bhi hai — yahan yeh coincide hote hain, kyunki $\dot{\theta }$ term effective potential aur Lagrangian mein absorb ho gayi, aur single coordinate $r$ tak reduced Lagrangian time-independent hai isliye $H$ conserved hai. Subtle point: yahan $H$ moving bead ki lab frame mein total kinetic energy $\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{\omega }^2{r}^2$ nahi hai. Reduced $H$ ek conserved quantity (Jacobi integral) hai, lekin yeh lab-frame kinetic energy plus zero potential nahi hai. Yeh canonical example hai jahan "$H$ = energy" mein care chahiye: wire kaam karta hai, aur $H$ conserved Jacobi function measure karta hai, naive $T$ nahi.

${\dot{p}}_y=-\frac{\partial H}{\partial y}$ apply karo. Kyunki $H$ mein koi $y$ nahi hai, $\frac{\partial H}{\partial y}=0$, isliye ${\dot{p}}_y=0$ — $p_y$ conserved hai. Jabki ${\dot{p}}_x=-\frac{\partial H}{\partial x}=-V^{\prime }(x)\ne 0$ generally — $p_x$ conserved nahi hai. Insight: $H$ se absent ek coordinate (ek cyclic coordinate) tumhe ek conserved momentum free mein deta hai — Noether's Theorem ka beej.

Step 1 (momentum — cross-term dhyaan se dekho). Kyun: actual $L$ ko $\dot{x}$ se differentiate karo; yahan linear term $eA\dot{x}$ bhi contribute karta hai, isliye $p$ sirf $m\dot{x}$ nahi hai. $p=\frac{\partial L}{\partial \dot{x}}=m\dot{x}+eA(x).$ Isliye $p\ne m\dot{x}$! Invert kyun: humein abhi bhi $\dot{x}$ ko $(x,p)$ ke terms mein chahiye taaki use eliminate kar sakein — linear relation solve karo: $\dot{x}=\frac{p-eA}{m}$. Step 2 (Hamiltonian). Kyun: woh $\dot{x}$ $H=p\dot{x}-L$ mein substitute karo aur simplify karo. $H=p\dot{x}-L=p\dot{x}-\frac{1}{2}m{\dot{x}}^2-eA\dot{x}=(p-eA)\dot{x}-\frac{1}{2}m{\dot{x}}^2.$ $(p-eA)$ group kyun: yeh exactly $m\dot{x}$ hai, isliye algebra cleanly collapse hoti hai. $\dot{x}=(p-eA)/m$ rakho: $H=(p-eA)\frac{p-eA}{m}-\frac{1}{2}m(\frac{p-eA}{m}{)}^2=\frac{(p-eA{)}^2}{m}-\frac{(p-eA{)}^2}{2m}=\frac{(p-eA{)}^2}{2m}.$ Lesson: canonical momentum $p=m\dot{x}+eA$, kinetic momentum $m\dot{x}$ se alag hai; $H$ gauge-invariant combination $(p-eA)$ mein quadratic hai. Yahi quantum mechanics mein use hone wale "minimal coupling" $p\to p-eA$ ki origin hai.

Derivation. Chain rule se kyun start karein: $f$ sirf isliye change hota hai kyunki $q$ aur $p$ move karte hain, isliye $\frac{df}{dt}=\frac{\partial f}{\partial q}\dot{q}+\frac{\partial f}{\partial p}\dot{p}.$ Hamilton's equations substitute kyun karein: rate ko purely $H$ ke terms mein express karne ke liye. $\dot{q}=\partial H/\partial p$ aur $\dot{p}=-\partial H/\partial q$ daalo: $\frac{df}{dt}=\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial H}{\partial q}=\{f,H\}.$ $dH/dt$ recompute karo. $f=H$ kyun set karein: $H$ par hi bracket test karne ke liye. Tab $\{H,H\}=\frac{\partial H}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial H}{\partial p}\frac{\partial H}{\partial q}=0.$ Isliye (explicit $t$ nahi hone par) $dH/dt=0$ — parent note ke saath consistent, aur ab ek bracket mein package. Jis bhi cheez ka $H$ ke saath bracket zero ho woh conserved hai.

Momenta. Kyun: $L$ ko har velocity se alag differentiate karo; dono coordinates mix nahi hote. $p_x=m\dot{x},p_y=m\dot{y}\Rightarrow \dot{x}=p_x/m,\dot{y}=p_y/m.$ Hamiltonian. Kyun: dono inversions ko $H={\sum }_i{p}_i{\dot{q}}_i-L$ mein substitute karo. $H=\frac{p_x^2+p_y^2}{2m}+\frac{1}{2}m{\omega }^2(x^2+y^2).$ Chaar equations. Kyun: har coordinate ke liye ek $\dot{q}=\partial H/\partial p$ aur ek $\dot{p}=-\partial H/\partial q$. $\dot{x}=\frac{p_x}{m},\dot{y}=\frac{p_y}{m},{\dot{p}}_x=-m{\omega }^{2}x,{\dot{p}}_y=-m{\omega }^{2}y.$ Conservation. Kyun: $H$ mein koi explicit $t$ nahi, isliye $dH/dt=\partial H/\partial t=0$. Energy conserved. ✓ (Bonus: $p_{x}y-p_{y}x$ bhi conserved hai — angular momentum — kyunki potential rotationally symmetric hai.)

(i) Yeh check kyun: ek term jo total time derivative hai woh action mein constant-endpoint piece add karta hai aur isliye Euler–Lagrange equations untouched chhod deta hai. Indeed $\beta q\dot{q}=\frac{d}{dt}\left(\frac{1}{2}\beta q^2\right)$, isliye motion same simple harmonic oscillator hai. (ii) Momentum. Kyun: diye gaye $L$ ko $\dot{q}$ se differentiate karo; $\beta q\dot{q}$ term $\dot{q}$ mein linear hai aur $\beta q$ contribute karta hai. $p=\frac{\partial L}{\partial \dot{q}}=m\dot{q}+\beta q\Rightarrow \dot{q}=\frac{p-\beta q}{m}.$ Hamiltonian. Kyun: $\dot{q}=(p-\beta q)/m$ ko $H=p\dot{q}-L$ mein term by term substitute karo. $H=p\dot{q}-L=p\dot{q}-\frac{1}{2}m{\dot{q}}^2+\frac{1}{2}m{\omega }^2{q}^2-\beta q\dot{q}=(p-\beta q)\dot{q}-\frac{1}{2}m{\dot{q}}^2+\frac{1}{2}m{\omega }^2{q}^2.$ $(p-\beta q)$ group kyun: yeh $m\dot{q}$ ke barabar hai, isliye pehle do terms plain kinetic term ki tarah combine hote hain. $\dot{q}=(p-\beta q)/m$ rakho: $H=\frac{(p-\beta q{)}^2}{m}-\frac{(p-\beta q{)}^2}{2m}+\frac{1}{2}m{\omega }^2{q}^2=\frac{(p-\beta q{)}^2}{2m}+\frac{1}{2}m{\omega }^2{q}^2.$ Punchline: Hamiltonian naive $\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{q}^2$ nahi hai — isme shift $p\to p-\beta q$ hai. Yeh abhi bhi conserved hai (koi explicit $t$ nahi), aur yeh abhi bhi same SHO motion generate karta hai, lekin $(q,p)$ mein iska form $L$ mein ek mere total-derivative gauge se badal gaya. Yeh ek hands-on view hai jo Canonical Transformations formalize karte hain: alag $H$'s same physics describe kar sakte hain.

Ellipse kyun: orbit $\frac{p^2}{2mE}+\frac{q^2}{2E/(m{\omega }^2)}=1$ hai, semi-axes wala ek ellipse $q_{\max }=\sqrt{\frac{2E}{m{\omega }^2}},p_{\max }=\sqrt{2mE}.$ Area formula kyun: semi-axes $A,B$ wala ellipse $\pi AB$ area enclose karta hai, isliye $\oint pdq=\pi q_{\max }{p}_{\max }=\pi \sqrt{\frac{2E}{m{\omega }^2}}\sqrt{2mE}=\pi \cdot \frac{2E}{\omega }=\frac{2\pi E}{\omega }.$ Meaning: yeh enclosed area action variable $J=\oint pdq$ hai. Slow parameter changes (adiabatic invariance) ke under iska constancy aur iska quantization $J=nh$ Hamilton-Jacobi Theory aur old quantum theory ke gateways hain.

Fundamental bracket. Kyun: $\{f,g\}=\frac{\partial f}{\partial q}\frac{\partial g}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial g}{\partial q}$ mein $f=q,g=p$ plug karo; note karo $\partial q/\partial q=1$, $\partial q/\partial p=0$, etc. $\{q,p\}=1\cdot 1-0\cdot 0=1.$ Equations reproduce karo. Kyun: $f=q$ phir $f=p$ ke saath $\{f,H\}$ use karne par Hamilton's equations wapas milne chahiye agar bracket formulation equivalent hai. $\{q,H\}=\frac{\partial q}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial q}{\partial p}\frac{\partial H}{\partial q}=\frac{\partial H}{\partial p}=\dot{q}.✓$ $\{p,H\}=\frac{\partial p}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial p}{\partial p}\frac{\partial H}{\partial q}=-\frac{\partial H}{\partial q}=\dot{p}.✓$ Poori Hamiltonian mechanics ko is tarah phrase kiya ja sakta hai "time evolution = bracket with $H$," aur $\{q,p\}=1$ ki invariance exactly woh condition hai ki variables ka ek change ek canonical transformation hai.

Ek cyclic coordinate (jo $H$ mein absent ho) conserved momentum kyun deta hai? ::: Kyunki ${\dot{p}}_i=-\partial H/\partial q_i=0$ jab $H$ mein $q_i$ nahi hota. L4.1 mein, $p\ne m\dot{x}$ kyun hai? ::: Lagrangian mein $\dot{x}$ mein linear term hai, isliye $p=\partial L/\partial \dot{x}=m\dot{x}+eA$. Energy $E$ ke SHO ke liye $\oint pdq$ kya hai? ::: $2\pi E/\omega$, action variable.