Exercises — Phase space — trajectories, phase portraits

TASVEER KYA DIKHATI HAI. Ek loop jo apne aap par close hoti hai matlab state $(q,p)$ bilkul wahin wapas aati hai jahan se shuru hua tha aur repeat hoti hai — yeh periodic (oscillatory) motion hai. CLOSED ⇒ ENERGY FIXED KYUN HAI. Ek conservative system ke liye har trajectory ek hi contour $H(q,p)=E$ par baithi hoti hai. Ek curve jo kabhi ek contour nahi chordi matlab energy constant hai. Agar energy leak ho rahi hoti (friction), toh dot neeche ke contours ki taraf drift karta aur loop close nahi ho sakti — woh andar spiral karti. Answer: constant total energy ke saath ek stable oscillation.

Yeh ek fixed point (equilibrium point) hai. Isse is baat se define kiya jata hai ki state kabhi change nahi hoti: $\dot{q}=0\text{and}\dot{p}=0.$ YEH DO KYUN. Kisi bhi point par flow vector $(\dot{q},\dot{p})$ hota hai. "Kabhi move nahi karta" ka matlab hai us arrow ki length zero hai — dono components vanish ho jate hain. Yeh akeela aisa point hai jahan trajectories touch kar sakti hain, kyunki baaki jagah single-valued flow crossings forbid karta hai.

• (a) ellipse → undamped oscillator (energy fixed, closed loop). Dekho Harmonic Oscillator.
• (b) inward spiral → damped oscillator (energy bleed ho rahi hai, dot origin par sink karta hai).
• (c) wavy open line → whirling pendulum — yeh kabhi wapas nahi mudta, $q$ (angle) badhta rehta hai, isliye curve hamesha side mein chalti jaati hai. Dekho Pendulum.

KYA/KYUN — Hamilton's equations use karo (woh energy ko state point ki velocity mein convert karti hain): $\dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m},\dot{p}=-\frac{\partial H}{\partial q}=-kq.$ $q=1,p=0,m=2,k=8$ plug karo: $\dot{q}=\frac{0}{2}=0,\dot{p}=-(8)(1)=-8.$ KAISA DIKHTA HAI — Figure s01. Blue curve woh energy ellipse hai jo hamare point se guzarti hai. Yellow dot state $(q,p)=(1,0)$ ko mark karta hai — us ellipse ki sabse dayi tip, jahan $q$ sabse bada hai aur $p=0$. Wahan pink arrow seedha neeche ki taraf $8$ length ke saath point karta hai: restoring force se momentum destroy ho raha hai, isliye dot clockwise slide karta hai. Yeh bilkul hamara computed flow $(0,-8)$ hai.

Answer: $(\dot{q},\dot{p})=(0,-8)$.

$\omega =\sqrt{k/m}=\sqrt{8/2}=2$. Yahan $\omega$ (angular frequency) oscillation ke radians per second ki sankhya hai; note karo $k=m{\omega }^2$, isliye hum $H$ ko dono taraf likh sakte hain.

Ellipse derive karo (KYUN yeh ellipse hai). Energy conservation $H=E$ se shuru karo: $\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{q}^2=E.$ Hum standard ellipse shape $\frac{q^2}{a^2}+\frac{p^2}{b^2}=1$ chahte hain, kyunki us form ki equation by definition half-widths $a$ ($q$ ke along) aur $b$ ($p$ ke along) wali ellipse hai. Iske liye right-hand side ko $1$ ke barabar banane ke liye: har term ko $E$ se divide karo. $\frac{p^2}{2mE}+\frac{m{\omega }^2{q}^2}{2E}=1.$ Ab har denominator ko ek squared half-axis ke roop mein padho. $q$-term ke liye, $\frac{m{\omega }^2{q}^2}{2E}=\frac{q^2}{2E/(m{\omega }^2)}$, isliye $\frac{q^2}{2E/(m{\omega }^2)}+\frac{p^2}{2mE}=1.$ YEH READING KYUN KAAM KARTI HAI: $q^2/a^2$ mein $q^2$ ke neeche wala number hi $a^2$ hai; jo bhi wahan baitha hai, uska square root half-width hai. Isliye $a=\sqrt{\frac{2E}{m{\omega }^2}},b=\sqrt{2mE}.$ Meaning ka sanity check: $p=0$ par saari energy potential hai, $\frac{1}{2}m{\omega }^2{q}^2=E$, jo sabse door ka reach $q=a$ deta hai — max displacement. $q=0$ par saari energy kinetic hai, $p^2/2m=E$, jo $p=b$ deta hai — max momentum. Ab numbers plug karo ($m=2,\omega =2,E=16$): $a=\sqrt{\frac{2\cdot 16}{2\cdot 2^2}}=\sqrt{\frac{32}{8}}=\sqrt{4}=2,b=\sqrt{2\cdot 2\cdot 16}=\sqrt{64}=8.$ Answer: $\omega =2,a=2,b=8$. Dekho Harmonic Oscillator.

$\dot{q}=\frac{\partial H}{\partial p}=p,\dot{p}=-\frac{\partial H}{\partial q}=-1.$ Energy contours: $E=\frac{p^2}{2}+q\Rightarrow q=E-\frac{p^2}{2}$ — ek sideways parabola jo $-q$ direction mein open hoti hai. PARABOLA KYUN: constant force ka matlab hai $p$ ek steady rate se girta hai ($\dot{p}=-1$), jabki $q$ $p$ ki tarah speed karta hai; woh quadratic relation parabola hai. Answer: trajectories parabolas $q=E-p^2/2$ hain.

Fixed points ke liye $\dot{p}=-V^{\prime }(q)=0$ aur $p=0$ chahiye. Yahan $V^{\prime }(q)=q=0\Rightarrow q=0$. Isliye ekeela fixed point $(0,0)$ hai. $V$ ki shape se classify karo: $V^{\prime \prime }(0)=1>0$, isliye $q=0$, $V$ ka ek minimum hai → ek center (closed loops se ghira, stable). Dekho Stability and Fixed Points.

$V^{\prime }(q)=-q=0\Rightarrow q=0$, $p=0$ ke saath. $V^{\prime \prime }(0)=-1<0$ → $V$ ka ek maximum → ek saddle (unstable). KAISA DIKHTA HAI — Figure s02. Do dashed yellow diagonals saddle ke through special "in" aur "out" lines hain (origin par pink dot). Blue aur pink curves aas-paas ki trajectories hain: woh saddle ki taraf ek diagonal ke along toward swoosh karti hain, phir doosri diagonal ke along away peel karti hain — ek pahadi par balance kiya hua ball jise koi bhi nudge daudta bhej deta hai.

$V^{\prime }(q)=4q^3-2q=2q(2q^2-1)=0\Rightarrow q=0,q=\pm \frac{1}{\sqrt{2}}$. $V^{\prime \prime }(q)=12q^2-2$.

• $q=0:V^{\prime \prime }=-2<0$ → maximum → saddle.
• $q=\pm 1/\sqrt{2}:V^{\prime \prime }=12\cdot \frac{1}{2}-2=4>0$ → minima → centers.

Teen energy regimes (saare cases) — Figure s03. Saddle height $V(0)=0$ par hai, isliye $E=0$ separatrix energy hai. Do wells ki bottom energy $V(\pm 1/\sqrt{2})=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$ hai.

1. $-\frac{1}{4}<E<0$ (separatrix ke neeche): particle ke paas central bump $q=0$ par chadhne ki energy nahi hai. Woh ek well mein trapp hai aur ek chhoti closed loop (blue) trace karta hai. Do families hain, ek per well.
2. $E=0$ (separatrix par): yellow figure-eight. Har lobe saddle tak zero speed ke saath barely pahunchti hai; dono lobes origin par kiss karti hain.
3. $E>0$ (separatrix ke upar): particle ke paas ab central hump par ride karne ki enough energy hai, isliye woh left well se right well tak aur wapas swing karta hai. Uski trajectory ek badi loop (pink) hai jo dono wells ko encircle karti hai — ek global oscillation jo $q=0$ ko nonzero speed se cross karta hai. Yeh important edge case hai: separatrix ke upar dono wells ek single bade orbit mein merge ho jaate hain.

Dekho Energy Conservation.

$m{l}^2$ KYUN AATA HAI. Mass $m$ aur length $l$ ke rod par swing karte bob ke liye, pivot se uski distance $l$ hai, isliye uska moment of inertia (mass ka rotational analogue) $I=m{l}^2$ hai. Angular momentum $p$ ke saath rotation ki kinetic energy $p^2/(2I)=p^2/(2m{l}^2)$ hai — seedha-line motion ke $p^2/2m$ ka exact rotational twin. Isliye $H=\frac{p^2}{2m{l}^2}+V(\theta )=E.$ $p$ ke liye solve karo: $\frac{p^2}{2m{l}^2}=E+mgl\cos \theta \Rightarrow p(\theta )=\pm \sqrt{2m{l}^2(E+mgl\cos \theta )}.$ Separatrix condition: woh orbit jo top $\theta =\pi$ tak $p=0$ ke saath just pahunchti hai. $p=0,\theta =\pi$ set karo: $0=E+mgl\cos \pi =E-mgl\Rightarrow E_{\text{sep}}=mgl.$ YEH MOTIONS KO KYUN SPLIT KARTA HAI: $E_{\text{sep}}$ se neeche square root $\theta =\pi$ se pehle imaginary ho jaata hai — pendulum wapas mudta hai (libration, closed loop). Iske upar, $p$ saare $\theta$ ke liye real rehta hai — bob top ke upar whirl karta hai (rotation, open wavy line). Answer: $p(\theta )=\pm \sqrt{2m{l}^2(E+mgl\cos \theta )},E_{\text{sep}}=mgl.$ Dekho Pendulum.

Energy leak karti hai. $\dot{E}=\dot{x}\ddot{x}+{\omega }_0^{2}x\dot{x}=\dot{x}(\ddot{x}+{\omega }_0^{2}x).$ Equation se, $\ddot{x}+{\omega }_0^{2}x=-2\gamma \dot{x}$, isliye $\dot{E}=\dot{x}(-2\gamma \dot{x})=-2\gamma {\dot{x}}^2\le 0.$ Energy girti hai jab bhi object move karta hai (kyunki $\gamma >0$ aur ${\dot{x}}^2\ge 0$), isliye dot ek contour par nahi reh sakta — woh chhoti-energy ellipses par drop karta hai.

"Focus" ka MATLAB KYA HAI. Ek focus ek fixed point hai jahan trajectories spiralling karke pahunchti hain (ghumte ghumte jab ki approach karti hain), ek node ke opposite jahan woh bina circle kiye seedha andar aati hain. "Stable" ka matlab sirf yeh hai ki woh andar aati hain, bahar nahi.

FLOW SPIRAL KYUN KARTA HAI (linearised picture). Solutions $x(t)=e^{\lambda t}$ dhundho, jahan $\lambda$ ek (possibly complex) growth rate hai jo hume dhundhna hai. $\ddot{x}+2\gamma \dot{x}+{\omega }_0^{2}x=0$ mein substitute karne par characteristic equation milti hai ${\lambda }^2+2\gamma \lambda +{\omega }_0^2=0\Rightarrow \lambda =-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}.$ Light damping ($\gamma <{\omega }_0$) ke liye root ke neeche ka quantity negative hai, isliye uska square root imaginary hai aur $\lambda =-\gamma \pm i{\omega }_d,{\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}.$ Yahan ${\omega }_d$ damped frequency hai — friction hone par actual (thoda slow) wiggling rate, aur $i=\sqrt{-1}$ imaginary unit hai. Yeh roots complex numbers hain. Imaginary part $\pm i{\omega }_d$ rotation produce karta hai (sines aur cosines, yaani loop ke around jaana); real part $-\gamma$ shrinking produce karta hai (factor $e^{-\gamma t}$ jo radius ko zero ki taraf khinchta hai). Ek rotation jo continuously shrink kare wahi inward spiral hai — yeh exactly ek stable focus hai. Shrink ka characteristic timescale $1/\gamma$ hai: origin se distance $e^{-\gamma t}$ jaisi girti hai.

Conservative case se contrast — KYUN undamped loops close hote hain. Ek conservative Hamiltonian flow ke liye, phase-space area time mein preserved rehta hai (yeh Liouville's Theorem hai). Intuitively: Hamilton's equations flow ko "incompressible" banati hain — flow vector ka divergence hai $\frac{\partial \dot{q}}{\partial q}+\frac{\partial \dot{p}}{\partial p}=\frac{\partial }{\partial q}\left(\frac{\partial H}{\partial p}\right)+\frac{\partial }{\partial p}\left(-\frac{\partial H}{\partial q}\right)=\frac{{\partial }^{2}H}{\partial q\partial p}-\frac{{\partial }^{2}H}{\partial p\partial q}=0.$ Zero divergence ka matlab hai states ka ek chhota blob na expand hota hai na contract, isliye trajectories ek point par converge nahi kar sakti — loops close honi hi chahiye. Damping ise todta hai: yeh ek term add karta hai jiska negative divergence hota hai, blob shrink hota hai, aur flow origin mein spiral karne ki permission paata hai. Dekho Liouville's Theorem.

Ellipse area $=\pi ab$ jahan $a=\sqrt{2E/(m{\omega }^2)},b=\sqrt{2mE}$ hain (L2.2 mein derive kiye gaye semi-axes). $m=\omega =1$ ke saath: $a=\sqrt{2E}$ aur $b=\sqrt{2E}$, isliye $\text{area}=\pi ab=\pi \sqrt{2E}\sqrt{2E}=2\pi E.$ YAHAN DO AXES EQUAL KYUN HAIN: jab $m=\omega =1$ hota hai toh ellipse actually radius $\sqrt{2E}$ ka circle hai, isliye uska area $\pi (\sqrt{2E}{)}^2=2\pi E$ hai. Do areas ka ratio hai $\frac{{\text{area}}_2}{{\text{area}}_1}=\frac{2\pi E_2}{2\pi E_1}=\frac{E_2}{E_1}=\frac{8}{2}=4.$ Interpretation: harmonic oscillator ke liye phase-space area directly energy ke proportional hai ($2\pi E$), isliye energy chaar guna karne par enclosed area bhi chaar guna ho jaata hai. (Yeh enclosed area action ke proportional hai, woh quantity jo slow, adiabatic changes ke under conserved rehti hai.) Answer: doosri ellipse ka area $4\times$ bada hai.

(a) $V^{\prime }(q)=3q^2-3=0\Rightarrow q=\pm 1$. $V^{\prime \prime }(q)=6q$.

• $q=+1:V^{\prime \prime }=6>0$ → minimum → center at $(1,0)$.
• $q=-1:V^{\prime \prime }=-6<0$ → maximum → saddle at $(-1,0)$.

(b) Separatrix saddle se $p=0$ ke saath guzarti hai: $E_{\star }=V(-1)=(-1{)}^3-3(-1)=-1+3=2$. (c) $q=0$ se rest se release: uski energy $E=V(0)=0$ hai, aur $0<E_{\star }=2$, isliye woh $q=1$ ke around well ke andar hai. Woh us well mein aage-peechhe oscillate karega (ek closed loop), kabhi saddle tak nahi pahunchega. Answer: fixed points center $(1,0)$, saddle $(-1,0)$; $E_{\star }=2$; particle $q=1$ ke around oscillate karta hai.

Hamilton's equations $\dot{q}=p/m,\dot{p}=-m{\omega }^{2}q$ se. Pehli ko differentiate karo aur doosri substitute karo: $\ddot{q}=\frac{\dot{p}}{m}=-{\omega }^{2}q.$ YEH KEY STEP KYUN HAI: yeh do first-order equations ko ek single equation mein collapse karta hai jiska solution $q(t)=a\cos (\omega t+\varphi )$ hai — ek pure cosine jiska frequency $\omega$ equation se set hoti hai, amplitude $a$ se nahi. Kyunki $\omega$ mein koi $E$ nahi hai, period $T=\frac{2\pi }{\omega }$ har ellipse ke liye same hai. Badi loops sirf proportionally tezi se chalti hain. Yeh "amplitude-independent period" exactly wahi hai jo oscillator ko special banata hai — pendulum se compare karo, jiska period separatrix ke paas badhta hai.

Fixed points par flow speed vanish ho jaati hai. Fixed point ke paas flow vector $(\dot{q},\dot{p})\to (0,0)$, isliye dot dhheere dhheere move karta hai.

• Stable focus (damped oscillator) ke liye, L4.2 ka linearised solution $x(t)=A{e}^{-\gamma t}\cos ({\omega }_{d}t+\varphi )$ deta hai, jahan $\gamma$ damping coefficient hai aur ${\omega }_d$ L4.2 ka damped frequency hai. Envelope $e^{-\gamma t}$ hi shrink karta hua radius hai, isliye origin se distance $r(t)\sim r_0{e}^{-\gamma t}$ jaisa behave karta hai. YEH KAHAN SE AATA HAI: eigenvalue $\lambda =-\gamma \pm i{\omega }_d$ ka real part $-\gamma$ hai, aur koi bhi mode $\propto e^{\lambda t}$ us rate se decay karta hai. Kyunki sabhi finite $t$ ke liye $e^{-\gamma t}>0$, radius $0$ sirf $t\to \infty$ mein pahunchta hai — infinite time, jo dot ke kabhi khud se cross na karne ke saath consistent hai (har loop ek alag, chhota loop hai).
• Saddle mein separatrix ke liye, wahi $\dot{}\to 0$ slowdown ka matlab hai incoming curve bhi arrive karne mein infinite time leti hai. Resolution: trajectories fixed points ko sirf $t\to \pm \infty$ ki limit mein touch karti hain, kabhi finite time mein nahi, isliye "one state → one future" rule kabhi violate nahi hota. Saddle par curves ka apparent milna ek asymptotic milna hai, finite-time crossing nahi.

Center ::: $V$ ka minimum, closed loops, stable oscillation. Saddle ::: $V$ ka maximum, unstable, iske through separatrix. Separatrix energy (pendulum) ::: $E=mgl$ (zero speed ke saath top tak pahunchta hai). Oscillator period ::: $T=2\pi /\omega$, energy se independent. Why damped loops don't close ::: $\dot{E}=-2\gamma {\dot{x}}^2<0$, trajectory contours ko andar ki taraf cross karti hai. Why conservative loops close ::: Hamiltonian flow ka zero divergence hota hai (Liouville), area preserved rehta hai.