Worked examples — Phase space — trajectories, phase portraits

2.1.13 · D3 · Physics › Analytical Mechanics › Phase space — trajectories, phase portraits

Yeh page ek drill hai har us case ke through jo phase-space problem mein aa sakta hai. Hum seedha parent topic ke upar build karte hain. Agar koi symbol nazar aaye, toh woh wahan earn kiya gaya tha ya yahan earn hoga.

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The scenario matrix

Neeche har cell mein kam se kam ek worked example hai. Label jaise (C1) us example ko tag karta hai jo uss cell ko cover karta hai.

Case class	Specific scenario	Covered by
Sign of $V^{\prime \prime }$	Minimum of $V$ → center (stable loop)	C1, C6
	Maximum of $V$ → saddle (unstable)	C2, C6
	$V^{\prime \prime }=0$ → degenerate / non-hyperbolic equilibrium (cusp)	C7
Quadrant of flow arrow	Oscillator ke liye $(q,p)$ ke saare chaar quadrants	C1
Full level-curve family	Hyperbolae ($E\ne 0$) jo ek saddle portrait fill karte hain	C2
Zero / degenerate input	Fixed point ($\dot{q}=\dot{p}=0$), zero-velocity turning point	C1, C3
	Degenerate fixed point jahan usual $V^{\prime \prime }$ test fail karta hai	C7
Energy sign / level	Below-, at-, above-separatrix energy	C3
Limiting behaviour	$E\to 0$ (tiny loop), $E\to \infty$ (huge loop), $t\to \infty$ decay	C4, C5
Non-conservative	Damping → focus / node; sign of $\dot{E}$; saare 3 damping regimes	C5
Real-world word problem	Pendulum: swing vs. whirl threshold real numbers mein	C3
Exam-style twist	Sirf $V(q)$ diya ho, poora portrait sketch karo	C6
Numeric readout	Period / max momentum ek ellipse se	C4

Reference figure — oscillator flow arrows, labelled $q$–$p$ axes aur quadrant markers ke saath:

Horizontal axis hai position $q$, vertical axis hai momentum $p$. Chaar quadrants mark hain: QI (dono positive), QII ($q<0,p>0$), QIII (dono negative), QIV ($q>0,p<0$); chaar orange dots woh points hain jo C1 mein evaluate hue hain.

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C1 — Oscillator flow ke saare chaar quadrants (cells: center, four quadrants, zero input)

Forecast: aage padhne se pehle — rightmost point $(1,0)$ par, dot aage upar jaayega ya neeche? Abhi guess karo.

1. Flow likhon. Hamilton's equations $m=\omega =1$ ke saath: $\dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m}=p,\dot{p}=-\frac{\partial H}{\partial q}=-m{\omega }^{2}q=-q.$ Yeh step kyun? Ek point par arrow IS $(\dot{q},\dot{p})$ hai; yeh do formulas IS the arrow field hain.

2. $(1,0)$ par evaluate karo (QI/QIV boundary, "east" point): $(\dot{q},\dot{p})=(0,-1)$ — seedha neeche QIV mein point kar raha hai. Yeh step kyun? Yeh loop ka "east" point hai; dot ko clockwise circle karne ke liye south jaana chahiye.

3. $(0,1)$ (top) par evaluate karo: $(\dot{q},\dot{p})=(1,0)$ — right mein QI ki taraf point kar raha hai. $(-1,0)$ (west) par: $(0,1)$ — upar QII mein point kar raha hai. $(0,-1)$ (bottom) par: $(-1,0)$ — left mein QIII ki taraf point kar raha hai. Yeh step kyun? Top par right, left par up, bottom par left — yeh exactly clockwise circulation hai (figure mein labelled arrows dekho).

4. Origin. $(0,0)$ par: $\dot{q}=0,\dot{p}=0$. Arrow ki length zero → kuch move nahi karta → yeh ek fixed point hai, aur kyunki $V=\frac{1}{2}{q}^2$ ka yahan minimum hai, yeh ek center hai.

Verify: Loop ke saath speed $=\sqrt{{\dot{q}}^2+{\dot{p}}^2}=\sqrt{p^2+q^2}$. Unit circle $q^2+p^2=1$ par yeh har jagah $1$ hai — constant, jo isotropic case $m=\omega =1$ mein uniform angular rate par traverse kiye perfect circle ke saath match karta hai.

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C2 — Inverted oscillator: saddle aur uski poori hyperbolic family (cells: maximum of $V$ → saddle, full level-curve family)

Forecast: $(1,0)$ par, kya dot spiral karega, loop karega, ya shoot away karega? Aur woh orbits jo do special lines nahi hain, unki shape kya hai? Step 5 se pehle guess karo.

1. Flow likhon. $m=\lambda =1$ ke saath, $\dot{q}=\frac{\partial H}{\partial p}=p,\dot{p}=-\frac{\partial H}{\partial q}=+m{\lambda }^{2}q=+q.$ Yeh step kyun? C1 se crucial sign flip note karo: kyunki $V$ ek maximum hai, $-\partial V/\partial q=+q$ state ko $q=0$ ki taraf wapas push karne ki jagah door push karta hai.

2. Origin ek fixed point hai: $(0,0)$ par, $\dot{q}=\dot{p}=0$. Aur $V^{\prime \prime }(0)=-1<0$ (ek maximum), toh hum instability expect karte hain. Yeh step kyun? Fixed point ke liye $p=0$ aur $V^{\prime }(q)=0$ chahiye; $V^{\prime \prime }$ ka sign uska character decide karta hai.

3. $(1,0)$ par evaluate karo: $(\dot{q},\dot{p})=(0,+1)$ — seedha upar point kar raha hai, phir badhta hua $p$, $q$ ko aur bahar push karta hai. State run away karta hai → origin ek saddle hai (pehle forecast ka answer: shoots away). Yeh step kyun? Center ke unlike (jo arrow ko loop mein wapas curve karta hai), yahan arrow escape ko feed karta hai — elliptic nahi, hyperbolic flow.

4. Do straight-line orbits (eigen-directions). $p=\pm q$ try karo. $p=+q$ par: $\dot{q}=p=q$ aur $\dot{p}=q=p$, toh state line $p=q$ ke saath outward move karta hai — unstable direction ($e^{+t}$ ki tarah badhta hai). $p=-q$ par: $\dot{q}=p=-q$ aur $\dot{p}=q=-p$, toh yeh $p=-q$ ke saath origin ki taraf inward move karta hai — stable direction ($e^{-t}$ ki tarah decay karta hai). Yeh do crossing lines saddle ka separatrix banate hain. Yeh step kyun? Saddle mein hamesha exactly ek incoming aur ek outgoing straight-line orbit hoti hai.

5. Orbits ki poori family ($E\ne 0$). Har trajectory ek contour $H=\frac{1}{2}(p^2-q^2)=E=\text{const}$ par lie karti hai. Rearrange karein, $p^2-q^2=2E$ — yeh hyperbola ki equation hai.

   • $E>0$ (energy hilltop se upar): $p^2-q^2=2E>0$, hyperbolae upar/neeche open hote hain ($p$-axis cross karte hain, $q$-axis ko touch nahi karte). Dot door se swoops in, origin ke paas bend karta hai, aur escape karta hai — kabhi top nahi pahunchta.
   • $E<0$: $p^2-q^2=2E<0$, yaani $q^2-p^2=2|E|$, hyperbolae left/right open hote hain ($q$-axis cross karte hain). Dot chhoti momentum ke saath aata hai, hill se turn back hota hai, aur retreat karta hai.
   • $E=0$: degenerate hyperbola $p^2=q^2$, yaani do straight lines $p=\pm q$ — exactly step 4 ka separatrix. Figure dekho: do red separatrix lines plane ko chaar regions mein divide karti hain, aur ek nested hyperbola har region mein baithta hai, lines ko hug karta hai jab $E\to 0$ aur bulge out karta hai jab $|E|$ badhta hai. Yeh step kyun? Yeh doosre forecast ka answer hai — poora portrait hyperbolae ki ek family hai jo separatrix ke asymptotic hai, do lines se door har point fill karte hain.

Verify: $p=q$ ke saath, $H=\frac{1}{2}(q^2-q^2)=0$ — separatrix exactly $E=0$ par baitha hai, hilltop ki energy $V(0)=0$ ✓. $E=\frac{1}{2}$ lo: contour $p^2-q^2=1$ point $(0,1)$ se guzarta hai — plug in: $1-0=1$ ✓, ek up/down hyperbola jaisa claim kiya. C1 se contrast: same origin, $V^{\prime \prime }$ ka opposite sign, loops hyperbolae se replace ho gaye.

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C3 — Pendulum: swing-vs-whirl threshold real numbers ke saath (cells: energy levels, separatrix, word problem, zero velocity)

Forecast: guess karo kya $6\text{ m/s}$ top pe jaane ke liye kaafi hai. Threshold se upar ya neeche?

1. Bottom par energy ($\theta =0,\cos 0=1$): $E=\frac{1}{2}m{l}^2{\dot{\theta }}_0^2-mgl=\frac{1}{2}m{v}_0^2-mgl.$ Yeh step kyun? $l{\dot{\theta }}_0=v_0$ toh $l^2{\dot{\theta }}_0^2=v_0^2$; launch state ke saath Energy Conservation substitute karo.

2. Top par energy chahiye ($\theta =\pi ,\cos \pi =-1$, zero speed): $E_{\text{sep}}=0-mgl(-1)=+mgl.$ Yeh step kyun? Separatrix woh orbit hai jo just barely top tak pahunchti hai — wahan kinetic energy zero hai, toh saari energy raised potential $V(\pi )=+mgl$ hai (figure se hilltop value).

3. Equal set karo, threshold speed solve karo: $\frac{1}{2}m{v}_0^2-mgl=mgl\Rightarrow v_0^2=4gl\Rightarrow v_0=2\sqrt{gl}.$ Numerically $v_0=2\sqrt{9.8\times 1}=6.261\text{ m/s}$. Yeh step kyun? $m$ cancel karo; threshold mass par depend nahi karta — light ya heavy bob same launch speed par whirl karta hai. Bottom par kinetic energy $\frac{1}{2}m{v}_0^2$ ko poora $2mgl$ climb supply karna hoga.

4. $v_0=6$ ke liye decide karo: kyunki $6<6.261$, energy separatrix se neeche hai → pendulum swing back karta hai (libration), phase space mein ek closed loop. Yeh step kyun? Below-separatrix energy $\theta =\pi$ tak nahi pahunch sakti; bob kisi ${\theta }_{\max }<\pi$ par turn around karta hai jahan $\dot{\theta }=0$ (ek zero-velocity turning point).

Verify: Units: $\sqrt{gl}=\sqrt{({\text{m/s}}^2)(\text{m})}=\text{m/s}$ ✓. Sanity: $6.261>6$, toh "just short" → swings, intuition ke saath consistent hai ki loop the loop ke liye thoda aur push chahiye.

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C4 — Ellipse readout: period aur max momentum (cells: limiting values, numeric readout)

Forecast: kya bigger energy loop ko faster ghumaati hai? Guess karo haan/nahi.

1. Parent ke ellipse form $\frac{q^2}{a^2}+\frac{p^2}{b^2}=1$ se semi-axes: $a=\sqrt{\frac{2E}{m{\omega }^2}},b=\sqrt{2mE}.$ Yeh step kyun? $H=E$ set karke aur $E$ se divide karne par exactly yeh axis lengths parent derivation mein aayi theen.

2. Numbers plug karo: $a=\sqrt{\frac{2\cdot 9}{2\cdot 9}}=\sqrt{1}=1\text{ m},b=\sqrt{2\cdot 2\cdot 9}=\sqrt{36}=6\text{kg⋅m/s}.$ Yeh step kyun? $m{\omega }^2=2\cdot 9=18$; $2E=18$; ratio $=1$. Max momentum $6$ hai.

3. Period. SHM ke liye $T=\frac{2\pi }{\omega }=\frac{2\pi }{3}\approx 2.094\text{ s}$, $E$ se independent. Yeh step kyun? Angular frequency $\omega$ ghoomne ki rate set karta hai; ellipse $E$ ke saath scale hoti hai lekin uss par circle karne ka time nahi. Toh bigger loop ≠ faster circulation — forecast ka answer hai nahi.

4. Limits. $E\to 0$: $a,b\to 0$, loop center point tak shrink ho jaata hai. $E\to \infty$: $a,b\to \infty$, ever-larger ellipse, lekin period $T=2\pi /3$ hamesha. Yeh step kyun? Family ke degenerate ($E=0$) aur unbounded ($E\to \infty$) ends confirm karta hai.

Verify: Consistency check — ellipse area $\pi ab=\pi \cdot 1\cdot 6=6\pi$ hai, aur action $\oint pdq=2\pi E/\omega =2\pi \cdot 9/3=6\pi$ ✓ (yeh match karne chahiye; yeh adiabatic invariants aur Liouville's Theorem ka link hai).

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C5 — Damped oscillator: $\dot{E}$ ka sign aur TEEN damping regimes (cells: non-conservative, $t\to \infty$ decay, focus vs. node)

Forecast: kya energy ek constant rate par leak hoti hai, ya fast move karne par faster? Aur kaunsa regime phase space mein spiral karta hai vs. kaunsa seedha origin mein slide karta hai? Guess karo.

1. $E$ differentiate karo: $\dot{E}=\dot{x}\ddot{x}+{\omega }_0^{2}x\dot{x}.$ Yeh step kyun? Do energy pieces par term-by-term chain rule.

2. $\ddot{x}=-2\gamma \dot{x}-{\omega }_0^{2}x$ substitute karo: $\dot{E}=\dot{x}(-2\gamma \dot{x}-{\omega }_0^{2}x)+{\omega }_0^{2}x\dot{x}=-2\gamma {\dot{x}}^2.$ Yeh step kyun? ${\omega }_0^{2}x\dot{x}$ terms exactly cancel ho jaate hain, sirf ek pure loss term bachta hai.

3. Sign: ${\dot{x}}^2\ge 0$ aur $\gamma >0$, toh $\dot{E}=-2\gamma {\dot{x}}^2\le 0$ — energy kabhi nahi badhti, aur wahan sabse fast drain hoti hai jahan bob sabse fast move karta hai (pehle forecast ka answer: constant nahi). Yeh step kyun? Dissipation rate ${\dot{x}}^2$ ke saath scale hoti hai, $q$-axis ke loop-crossings par sabse badi.

4. Ek period baad fraction (underdamped, $\gamma =0.1$). Envelope se, amplitude $\propto e^{-\gamma t}$, toh energy $\propto e^{-2\gamma t}$. $T=2\pi /{\omega }_0=2\pi$ par: $\frac{E(T)}{E(0)}=e^{-2\gamma T}=e^{-2(0.1)(2\pi )}=e^{-0.4\pi }\approx \mathrm{0.2846.}$ Yeh step kyun? Energy amplitude-squared hai; spiral ever-smaller ellipses cross karta hai, toh ek loop baad lagbhag $28.5\%$ bachta hai.

5. Decibels mein convert karo. Decibel ek logarithmic level unit hai: $\text{level}=10{\log }_{10}(\text{ratio})$ dB. Toh $10{\log }_{10}(0.2846)=10\times (-0.5459)\approx -5.46\text{ dB}.$ Yeh step kyun? Prompt ek decibel estimate maangta hai — log multiplicative energy fraction ko ek additive level drop mein convert karta hai, lagbhag $-5.5$ dB per swing.

6. Teeno regimes mein fixed point classify karo (roots $s=-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$, dono roots negative real part rakhte hain → hamesha attracting):

   • Underdamped $\gamma <{\omega }_0$: complex roots → phase point circulate karta hai shrinking ke saath → ek stable focus (spiral sink). Yeh hamara $\gamma =0.1$ case hai: ek inward spiral.
   • Critically damped $\gamma ={\omega }_0$: repeated real root → ek stable degenerate (improper) node; trajectories ek line ke tangent aa jaati hain, koi spiralling nahi.
   • Overdamped $\gamma >{\omega }_0$: do real negative roots → ek stable node; trajectories do eigen-directions ke saath seedha slide karti hain, koi loop nahi. Yeh step kyun? Yeh doosre forecast ka answer hai: sirf underdamped regime spiral karta hai; critical aur overdamped seedha origin mein jaate hain. General focus-vs-node criterion ke liye Stability and Fixed Points dekho.

Verify: $e^{-0.4\pi }\approx e^{-1.2566}\approx 0.2846$, $0$ aur $1$ ke beech mein ✓; $10{\log }_{10}(0.2846)\approx -5.46$ dB, negative jaisa ek loss hona chahiye ✓. Regime boundary check: discriminant ${\gamma }^2-{\omega }_0^2$ hai $-0.99<0$ for $\gamma =0.1,{\omega }_0=1$ (complex → focus ✓), $=0$ at $\gamma =1$ (critical), $>0$ for $\gamma =2$ (real → node ✓).

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C6 — Exam twist: $V(q)$ akele se poora portrait (cells: min→center, max→saddle, sketch task)

Forecast: kitne fixed points hain, aur kitne stable hain?

1. Fixed points $=$ extrema of $V$: $V^{\prime }(q)=q^3-q=q(q^2-1)=0\Rightarrow q=0,q=+1,q=-1$, sab $p=0$ ke saath. Yeh step kyun? Fixed point ke liye $\dot{p}=-V^{\prime }(q)=0$ AUR $\dot{q}=p/m=0$ chahiye, toh $p=0$ aur $V^{\prime }=0$.

2. $V^{\prime \prime }(q)=3q^2-1$ se classify karo:

   • $q=\pm 1$: $V^{\prime \prime }=3-1=2>0$ → minima → centers (har well mein stable loops).
   • $q=0$: $V^{\prime \prime }=-1<0$ → maximum → saddle (unstable). Yeh step kyun? $V$ ke minima oscillatory closed orbits dete hain; maximum hyperbolic saddle flow deta hai (exactly C2 mechanism).

3. Separatrix. Saddle $q=0$ par energy $E=V(0)=0$ par baitha hai. Curve $H(q,p)=0$ hai figure-eight separatrix: do loops, ek har well ke around, origin par kissing. Yeh step kyun? $E=0$ exactly woh energy hai jo central hilltop tak zero speed se pahunchti hai.

4. Upar aur neeche. $E<0$ ke liye (hilltop se neeche, well bottoms $V_{\min }=-\frac{1}{4}$ se upar): dot ek well mein trapped hai → ek chhota loop. $E>0$ ke liye: ek single bada loop dono wells ko encircle karta hai. Yeh step kyun? Central hill clear karne ke liye kaafi energy bob ko dono wells visit karne deti hai → ek bada orbit.

Verify: $V_{\min }=V(1)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$ aur $V(0)=0$. Barrier height $=0-(-\frac{1}{4})=\frac{1}{4}$ J·(unit) $>0$ ✓, confirm karta hai ki wells genuinely ek hill se separated hain.

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C7 — Degenerate case: jab $V^{\prime \prime }$ test fail karta hai (cells: $V^{\prime \prime }=0$, non-hyperbolic equilibrium)

Forecast: $q=0$ par potential second order tak flat hai — kya equilibrium stable hai, unstable hai, ya kuch beech mein? Step 4 se pehle guess karo.

1. Fixed point. $V^{\prime }(q)=q^2=0\Rightarrow q=0$; $p=0$ ke saath yeh akela equilibrium hai. Yeh step kyun? Same rule hamesha — $\dot{q}=p/m=0$ aur $\dot{p}=-V^{\prime }(q)=0$.

2. Second-derivative test fail hota hai. $V^{\prime \prime }(q)=2q$, toh $V^{\prime \prime }(0)=0$ — na positive na negative. C6 ka "min → center, max → saddle" dichotomy yahan koi answer nahi deta. Yeh step kyun? $V^{\prime \prime }=0$ matlab quadratic approximation flat hai; fixed point non-hyperbolic hai, aur linearization isse classify nahi kar sakta. Hume $V$ ki curvature nahi balki $V$ khud dekhna hoga.

3. Potential ki shape. $V=\frac{1}{3}{q}^3$ negative hai $q<0$ ke liye aur positive hai $q>0$ ke liye, origin par flat inflection ke saath zero se guzar raha hai. Toh origin ek taraf se min hai aur doosri taraf se max — ek one-sided hill. Yeh step kyun? Restoring force $-V^{\prime }(q)=-q^2\le 0$ ki direction asli kahani hai: yeh hamesha negative $q$ ki taraf push karta hai, left se origin ki taraf wapas kabhi nahi.

4. Character: ek cusp / degenerate saddle-node. Kyunki force $-q^2$ har jagah $\le 0$ hai, $q>0$ par nudge kiya gaya state $0$ ki taraf wapas push hota hai (right se stable lagta hai), lekin $q<0$ par state aur negative ki taraf push hota hai (left se unstable). Do behaviours origin par ek cusp mein milte hain: trajectories $q<0$ side se approach karti hain aur... wait — right se approach, left ko flee. Equilibrium isliye half-stable (degenerate) hai — phase curves ek cusp banate hain, true saddle ki clean crossing lines nahi aur na hi center ke closed loops (forecast ka answer: na stable na unstable — half-stable). Yeh step kyun? Isliye matrix $V^{\prime \prime }=0$ ko alag list karta hai: hyperbolic classification (center/saddle) apply nahi hoti, aur tumhe $V$ ki global shape se reason karna hoga.

Verify: Force $-V^{\prime }(q)=-q^2$: $q=+0.5$ par yeh $-0.25<0$ hai (left push karta hai, $0$ ki taraf — stable side); $q=-0.5$ par yeh $-0.25<0$ hai (bhi left push karta hai, yaani $0$ se door — unstable side). Dono sides par same-sign force ek half-stable degenerate point ka algebraic signature hai ✓. Energy contours $H=\frac{p^2}{2}+\frac{1}{3}{q}^3=E$ origin par ek crossing nahi balki cusp rakhte hain.

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Recall Quick self-test — har answer deciding algebra/step ka naam deta hai

Pendulum ke whirl karne ka threshold launch speed ::: $v_0=2\sqrt{gl}$ from $\frac{1}{2}m{v}_0^2=2mgl$ (poora top-to-bottom climb $2mgl$ supply karna hoga). Kya bigger oscillator energy period shorten karti hai? ::: Nahi — $T=2\pi /\omega$ mein koi $E$ nahi; ellipse scale hoti hai lekin circulation time nahi karta (C4 step 3). Damped oscillator mein energy loss ki rate ::: $\dot{E}=-2\gamma {\dot{x}}^2\le 0$ (${\omega }_0^{2}x\dot{x}$ terms cancel ho jaate hain; C5 step 2), jahan speed sabse zyada hoti hai wahan sabse badi. Ek damped period baad energy fraction aur dB ($\gamma =0.1,{\omega }_0=1$) ::: $e^{-0.4\pi }\approx 0.285$, yaani $10{\log }_{10}(0.285)\approx -5.46$ dB (C5 steps 4–5). Teen damping regimes aur unke fixed-point types ::: Underdamped $\gamma <{\omega }_0$ → stable focus (spiral); critical $\gamma ={\omega }_0$ → degenerate node; overdamped $\gamma >{\omega }_0$ → stable node (C5 step 6). Inverted oscillator ke off-axis orbits ::: Hyperbolae $p^2-q^2=2E$ — $E>0$ upar/neeche open, $E<0$ left/right open, $E=0$ do separatrix lines (C2 step 5). $V$ ke maximum par fixed-point type ::: Ek saddle: sign flip $-\partial V/\partial q=+q$ run-away flow deta hai stable line $p=-q$ aur unstable line $p=+q$ ke saath (C2 step 4). Double-well separatrix ki energy value yahan ::: $E=V(0)=0$ (hilltop value; C6 step 3). Kya hota hai jab fixed point par $V^{\prime \prime }=0$? ::: Min/max test fail hota hai — point non-hyperbolic hai; $V$ ki global shape se reason karo. $V=\frac{1}{3}{q}^3$ ke liye yeh ek half-stable cusp hai (C7).