Exercises — Action-angle variables — integrable systems

WHAT: $\oint pdq$ literally loop se enclosed geometric area hai — symbol ka poora content yehi hai (exactly Figure s01 mein shaded region). WHY: Green's theorem se $\oint pdq=\iint dqdp=\text{area}$. Radius $R$ ke circle ka area $\pi R^2$ hai, toh $\oint pdq=\pi R^2,J=\frac{1}{2\pi }\oint pdq=\frac{R^2}{2}.$ Koi calculus nahi chahiye — "action = area / $2\pi$" ki pehchaan hi yahan poori skill hai.

(i) Critical energy se neeche → libration. Angle $\theta$, $\pm {\theta }_{\max }$ ke beech jhoolti hai aur reverse karti hai — round-trip contour ${\theta }_{\min }\to {\theta }_{\max }\to {\theta }_{\min }$ hai (usual 2 ka factor aata hai). (ii) Critical energy se upar → rotation. Yahan $\theta$ badhta rehta hai aur kabhi reverse nahi karta, toh ek period $\theta :0\to 2\pi$ hai (pendulum ek poora chakkar lagaane ke baad same physical state par wapas aata hai, ek baar traverse hoke). (iii) Exactly critical energy par → separatrix, ek infinite-period edge case. Knife-edge energy par pendulum finite time mein $2\pi$ ka chakkar pura nahi karta: woh unstable top ($\theta \to \pi$) ki taraf asymptotically creep karta hai aur wahan pahunchne mein infinitely lamba waqt laata hai, isliye $T\to \infty$ aur $\omega \to 0$. Phase space mein yeh curve separatrix kehlaati hai — yeh libration region (andar closed ovals) aur rotation region (bahar open waves) ke beech ki boundary hai, aur yeh genuinely periodic orbit bilkul nahi hai. Action-angle variables theek isi par break down karte hain (frequency zero ho jaati hai), aur yehi reason hai ki perturbation theory separatrices ke paas sabse mushkil hoti hai.

Step A — $p(q,E)$ nikalo. $H=E$ solve karo $p$ ke liye: $p=\pm \sqrt{2mE-m^2{\omega }_0^2{q}^2}$. Step B — ellipse pehchano. $H=E$ ko $E$ se divide karo: $\frac{q^2}{2E/(m{\omega }_0^2)}+\frac{p^2}{2mE}=1,$ yeh ek ellipse hai jiske semi-axes $q$ ke along $a=\sqrt{2E/(m{\omega }_0^2)}$ aur $p$ ke along $b=\sqrt{2mE}$ hain. Step C — area / $2\pi$. Ellipse area $=\pi ab=\pi \sqrt{\frac{2E}{m{\omega }_0^2}}\sqrt{2mE}=\frac{2\pi E}{{\omega }_0}$. Isliye $J=\frac{1}{2\pi }\cdot \frac{2\pi E}{{\omega }_0}=\frac{E}{{\omega }_0}\Rightarrow E={\omega }_{0}J.$ Step D — differentiate. $\omega =\frac{dE}{dJ}={\omega }_0.$ ✔ Frequency bina koi trajectory solve kiye nikal aayi.

Pehle definition ka connection: L2.1 ke $H=\frac{1}{2}m{\omega }_0^2{q}^2$ se compare karne par $H=\frac{1}{2}k{q}^2$ dikhata hai ki potential coefficient hai hi $k=m{\omega }_0^2$; yeh $k$ spring ki fixed property hai, jabki ${\omega }_0=\sqrt{k/m}$ spring par latak rahe mass par depend karta hai. L2.1 se, $J=E/{\omega }_0$. $m$ double karne par ${\omega }_0^{\prime }=\sqrt{k/(2m)}={\omega }_0/\sqrt{2}$. (a) $J^{\prime }=E/{\omega }_0^{\prime }=\sqrt{2}E/{\omega }_0=\sqrt{2}J$ — loop same energy par $\sqrt{2}$ guna zyada area enclose karta hai (bhaari ⇒ har $q$ par zyada momentum). (b) $\omega ={\omega }_0^{\prime }={\omega }_0/\sqrt{2}$ — slower, as expected for a heavier bob.

Step A: $p=\sqrt{2mE}$, magnitude mein constant hai (walls ke beech koi force nahi). Step B — the loop. Phase space mein orbit ek rectangle hai (Figure s02 dekho): top edge $p=+\sqrt{2mE}$ (daayein chalti, $q:0\to L$), bottom edge $p=-\sqrt{2mE}$ (baayein chalti, $q:L\to 0$). Dono legs $+pL$ contribute karti hain: $\oint pdq=pL+(-p)(-L)=2pL.$ Step C: $J=\frac{1}{2\pi }2pL=\frac{pL}{\pi }=\frac{L\sqrt{2mE}}{\pi }$. Invert karo: $E=\frac{{\pi }^2{J}^2}{2m{L}^2}$. Step D: $\omega =\frac{dE}{dJ}=\frac{{\pi }^{2}J}{m{L}^2}=\frac{\pi v}{L}$ jahan $v=p/m$; period $T=\frac{2\pi }{\omega }=\frac{2L}{v}$. ✔ Bilkul round-trip bounce time.

WHAT/WHY — brute force ki jagah scaling. Hume $E$-dependence chahiye, toh ek substitution use karo jo saare $E$'s ko integral ke bahar nikal de. $J=\frac{1}{2\pi }\oint \sqrt{2m(E-\alpha q^4)}dq.$ Pehle contour fix karo. Motion libration hai: $q$ dono turning points ke beech chalti hai jahan $p=0$, yaani $E-\alpha q^4=0\Rightarrow q=\pm q_{\max }$ jahan $q_{\max }=(E/\alpha {)}^{1/4}$. Closed loop $-q_{\max }\to +q_{\max }\to -q_{\max }$ hai. Substitute karo $q=q_{\max }u={\left(\frac{E}{\alpha }\right)}^{1/4}u$, toh $q^4=\frac{E}{\alpha }{u}^4$ aur $dq={\left(\frac{E}{\alpha }\right)}^{1/4}du$. Jab $q$, $-q_{\max }$ se $+q_{\max }$ tak jaata hai, naya variable $u$, $-1$ se $+1$ tak jaata hai; round-trip (wahan aur wapas) ise double karta hai. Phir $J=\frac{1}{2\pi }\oint \sqrt{2mE}\sqrt{1-u^4}{\left(\frac{E}{\alpha }\right)}^{1/4}du=\underset{\text{ek pure number }C(\text{no }E,\text{finite, }>0)}{\underbrace{\frac{\sqrt{2m}}{2\pi {\alpha }^{1/4}}\cdot 2{\int }_{-1}^1\sqrt{1-u^4}du}}{E}^{1/2+1/4}.$ Yahan $C$ sirf ek fixed positive number hai: dimensionful constants $m,\alpha$ times definite integral ${\int }_{-1}^1\sqrt{1-u^4}du\approx 1.748$ (ek plain number, $\sqrt{1-u^4}$ ke ek hump ke neeche ka area), aur return leg ke liye extra factor 2 ke saath. Isme koi energy dependence nahi hai, aur yehi poori baat hai. Toh $J=C{E}^{3/4}$, yaani $E=(J/C{)}^{4/3}\propto J^{4/3}$. Differentiate karo: $\omega =\frac{dE}{dJ}\propto J^{1/3}\propto (E^{3/4}{)}^{1/3}=E^{1/4}$. Interpretation: SHO ke unlike (jiska $\omega$ energy-independent hai kyunki well quadratic hai), ek steeper well higher-energy orbits ko zyada tezi se squeeze karta hai ⇒ higher frequency. $\omega$ ki yeh energy-dependence exactly woh hai jo anharmonic systems ko non-isochronous banati hai.

WHY 2 ka factor nahi: rotation ka matlab hai $\varphi$ monotonically $2\pi$ se badhta hai; contour ek baar traverse hoti hai, wahan-aur-wapas nahi. $J=\frac{1}{2\pi }\oint p_{\varphi }d\varphi =\frac{1}{2\pi }{p}_{\varphi }\cdot 2\pi =p_{\varphi }.$ Yahan action equals momentum — ek special symmetric case (yeh "$J=p$" temptation ka honest version hai). Phir $E=p_{\varphi }^2/2I=J^2/2I$ aur $\omega =\frac{dE}{dJ}=\frac{J}{I}=\frac{p_{\varphi }}{I}=\dot{\varphi }.$ Angle-variable frequency actual angular velocity ke barabar hai, jaisa uniform circular motion ke liye hona chahiye.

Set up: $\oint pdq=2pL=nh$. $h=2\pi \hslash$ ke saath, yeh hai $J=\frac{2pL}{2\pi }=\frac{pL}{\pi }=n\hslash$, toh $p=\frac{n\pi \hslash }{L}$. Energy: $E_n=\frac{p^2}{2m}=\frac{n^2{\pi }^2{\hslash }^2}{2m{L}^2}$. Compare: yeh exact answer se bilkul match karta hai — box un rare cases mein se hai jahan semiclassical quantization exact hai (koi zero-point offset nahi, kyunki walls hard hain). SHO se contrast karo, jahan Bohr–Sommerfeld deta hai $E_n=\hslash {\omega }_0(n+\frac{1}{2})$: $+\frac{1}{2}$ smooth turning points se aata hai (Maslov correction), jo hard walls ke liye absent hai.

Key idea: slow ⇒ $J$ fixed rehta hai. L2.3 se, $J=\frac{L\sqrt{2mE}}{\pi }$, toh $J^2=\frac{L^2\cdot 2mE}{{\pi }^2}\propto L^{2}E$. $J$ constant rakhke: $L_0^2{E}_0=L_1^2{E}_1\Rightarrow \frac{E_1}{E_0}={\left(\frac{L_0}{L_1}\right)}^2.$ Sanity check: box squeeze karna ($L_1<L_0$) energy raise karta hai — particle speed up hoti hai, exactly ek approaching wall se bounce karne waali ball ki tarah jo speed gain karti hai. Agar $L_1=L_0/2$, toh $E_1/E_0=4$.

Torus surface ko ek unit square mein unroll karke dekho jiske left/right edges glued hain aur top/bottom edges glued hain (Figure s03). $({\theta }_1/2\pi ,{\theta }_2/2\pi )$ par ek point slope ${\omega }_2/{\omega }_1$ ki seedhi line mein chalta hai; jab ek edge chodta hai toh opposite edge se re-enter karta hai. (a) Rational ratio ⇒ closed orbit. Likho ${\omega }_1=3\Omega$, ${\omega }_2=2\Omega$ ek common base rate $\Omega$ ke saath. State tab repeat hoti hai jab dono angles $2\pi$ ke whole multiples se advance ho jaayein: ${\theta }_1$ ko ek chakkar ke liye time $2\pi /{\omega }_1$ chahiye, ${\theta }_2$ ko $2\pi /{\omega }_2$. Common period woh chota sa $T$ hai jis par ${\omega }_{1}T=2\pi p$ aur ${\omega }_{2}T=2\pi q$ integers $p,q$ ke liye. Sabse chhota: $p=3,q=2$ deta hai $T=\frac{2\pi \cdot 3}{{\omega }_1}=\frac{2\pi }{\Omega }$. Curve ${\theta }_1$ ke 3 loops aur ${\theta }_2$ ke 2 loops ke baad close hoti hai — ek (3,2) torus knot (Figure s03 ka left panel). (b) Irrational ratio ⇒ kabhi close nahi hoti. Koi integers $p,q$ satisfy nahi karte $\sqrt{2}=p/q$, toh orbit kabhi exactly repeat nahi hoti; bajaaye iske woh forever wind karti hai aur poori torus surface ko densely fill karti hai (quasi-periodic motion — Figure s03 ka right panel). (c) KAM link (ek line): KAM theorem kehta hai ki integrable system mein small perturbations ke under, sufficiently irrational frequency ratios waale tori survive karte hain (deform hote hain par bane rehte hain), jabki rational ya near-rational ratios waale pehle chaos mein toot jaate hain.

Invert karo. $S\equiv J_r+L$ set karo. Phir $k\sqrt{\frac{m}{2|E|}}=S$, toh $\frac{m{k}^2}{2|E|}=S^2$, milta hai $|E|=\frac{m{k}^2}{2S^2}\Rightarrow E=-\frac{m{k}^2}{2(J_r+L{)}^2}.$ Frequency ke liye differentiate karo. Kyunki $E$ sirf sum $J_r+L$ par depend karta hai, dono frequencies barabar hain: ${\omega }_r=\frac{\partial E}{\partial J_r}=\frac{m{k}^2}{(J_r+L{)}^3}=\frac{\partial E}{\partial L}={\omega }_{\varphi }.$ Matlab: ${\omega }_r={\omega }_{\varphi }$ matlab radial aur angular motions ek frequency share karte hain — orbit exactly ek radial oscillation per revolution ke baad close hoti hai ⇒ ek closed ellipse (koi precession nahi). Kepler problem ki yeh "accidental" degeneracy isi wajah hai ki sirf $1/r$ (aur $r^2$) potentials closed orbits dete hain — woh gehri wajah jisse planetary ellipses dheere-dheere rotate nahi karti. Quantization finish karo (Bohr–Sommerfeld). Semiclassically har action $\hslash$ ka integer multiple hai: radial quantum number $n_r=J_r/\hslash$ aur angular wala $\ell =L/\hslash$ likho, toh sum $S/\hslash =n_r+\ell \equiv n$ ek single principal quantum number hai ($n=1,2,3,\dots$). $J_r+L=n\hslash$ substitute karo: $E_n=-\frac{m{k}^2}{2(n\hslash {)}^2}=-\frac{m{k}^2}{2{\hslash }^2{n}^2}.$ Coulomb coupling $k=\frac{e^2}{4\pi {\epsilon }_0}$ ke saath yeh exactly hydrogen spectrum hai $E_n=-\frac{m{e}^4}{2(4\pi {\epsilon }_0{)}^2{\hslash }^2{n}^2}=-\frac{13.6\text{eV}}{n^2}$. Upar jo degeneracy ${\omega }_r={\omega }_{\varphi }$ humne nikali woh is fact ka classical shadow hai ki hydrogen levels sirf $n$ par depend karte hain (alag se $\ell$ par nahi) — woh celebrated $\ell$-degeneracy.

Action phase-space area divided by ==$2\pi$ hoti hai. Libration ke liye 2 ka factor aata hai kyunki loop wahan aur wapas jaata hai; rotation ke liye contour ek baar traverse hoti hai. Exactly critical pendulum energy par orbit separatrix hai, jiska period infinite== hai (frequency $\to 0$). Frequency ==differentiating $E(J)$== se milti hai (yaani $\omega =dE/dJ$). Ek quantity jo slow parameter changes ke under conserved hoti hai use adiabatic invariant kehte hain, aur conserved cheez ==$J$== hai (energy nahi). $n$-torus par, ek irrational frequency ratio orbit ko torus ko densely fill karta hai (quasi-periodic); Diophantine ratio woh hoti hai jo fractions se approximate karna mushkil ho aur isliye perturbation survive karta hai.