Phase space — trajectories, phase portraits: Question bank

2.1.13 · D5 · Physics › Analytical Mechanics › Phase space — trajectories, phase portraits

Definition Is page par use hone wale symbols — pehle padho

Neeche sab kuch ek chhote dictionary se bana hai. Yahan koi cheez is box se pehle use nahi ki gayi hai.

$q$ — system ki position (ya angle). Spring par mass ke liye yeh displacement $x$ hai; Pendulum ke liye yeh angle $θ$ hai. To " $q$ " aur " $x$ " ek hi tarah ki cheez matlab rakhte hain.
$p$ — momentum, "system mein kitni motion hai": $p = m \overset{q}{˙}$ , jahan $\overset{q}{˙}$ , $q$ ka rate of change (velocity) hai aur $m$ mass hai. Dot ka matlab hamesha "rate of change per unit time" hota hai.
State woh pair $(q, p)$ hai jo ek point ki tarah plot hoti hai; phase plane mein $q$ across aur $p$ upar hota hai.
$H$ — Hamiltonian, jo ek conservative 1-D system ke liye state ke terms mein likhi gayi total energy hai: $H (q, p) = kinetic \frac{p ^{2}}{2 m} + potential V (q) .$
$E$ — ek fixed number, ek orbit ki total energy; har trajectory apna $E$ banaye rakhti hai.
$\partial_{q} H, \partial_{p} H$ — $H$ kitna badalta hai jab tum akele $q$ ya $p$ ko thoda nudge karo (energy landscape ke partial slopes).
Flow rule (Hamiltonian Mechanics) jo ek state point ko move karta hai: $\overset{q}{˙} = \partial_{p} H$ aur $\overset{p}{˙} = - \partial_{q} H$ . Ye har point par velocity arrow $(\overset{q}{˙}, \overset{p}{˙})$ dete hain.

Figure — Phase space — trajectories, phase portraits

Upar di gayi figure har question ke peeche ki mental picture hai. Abhi isko dekho: horizontal axis position $q$ hai, vertical axis momentum $p$ hai. Cyan loop ek energy contour ( $H = E$ ) hai. Amber dot ek state point $(q, p)$ hai jo uss par baitha hai, aur white arrow uski flow velocity $(\overset{q}{˙}, \overset{p}{˙})$ hai — hamesha loop ke tangent, clockwise ghumti hui. White square origin par ek center (fixed point) hai. Jab bhi koi reveal "the loop," "the arrow," ya "the center" mention kare, isko wapas refer karo.

Energy kyun constant rehti hai (yeh ek baar karo, phir reuse karo)

Neeche ke kai traps iss fact par hinge karte hain ki $H$ ek conservative orbit ke saath kabhi nahi badalta. Yahan woh cancellation step by step spell out ki gayi hai, taaki tum ise khud samjho.

Sahi hai ya galat — justify karo

Phase plane ke axes position versus time hain.

Galat. Axes position

q

versus momentum

p

hain (equivalently

x

\overset{x}{˙}

); time woh hidden parameter hai jo har curve ke saath saath chalta hai, koi axis nahi — figure ko dobara dekho, jisme koi time axis nahi hai.

Sirf mass ki position

q

jaanna uski future motion fix kar deta hai.

Galat. Newton's law second order hai, isliye tumhe velocity/momentum

p

bhi chahiye — rest se choda gaya spring aur same point se ghuma hua spring alag alag move karte hain. Exactly isliye phase space ko do axes chahiye.

Phase space mein ek closed loop matlab motion periodically repeat hoti hai.

Conservative 1-D system ke liye Sahi. State

(q, p)

par wapas aati hai, aur kyunki wahan flow arrow fixed hai, isse wahi loop dobara trace karni hi padti hai — ek genuine period. (Harmonic Oscillator mein har loop ek ellipse hai.)

Ek conservative system ke liye, har trajectory constant energy

H = E

ki curve par hoti hai.

Sahi. Jaise upar dikhaya gaya,

d H / d t = \partial_{q} H \overset{q}{˙} + \partial_{p} H \overset{p}{˙} = A B - A B = 0

jab flow rule daala jaaye, to

H (q, p) = E

orbit ke saath fixed rehta hai. Trajectories hain hi energy contours (dekho Energy Conservation).

Do phase trajectories ek ordinary (non-fixed) point par cross kar sakti hain.

Galat. Flow arrow

(\overset{q}{˙}, \overset{p}{˙})

single-valued hai, isliye ek crossing ek state ko do futures degi — determinism se forbidden. Curves sirf wahan milti hain jahan arrow vanish ho jaata hai (fixed points).

Bada phase loop matlab state point loop ke har point par tez move karta hai.

Galat. Bada loop = zyada energy, lekin phase speed

\overset{q}{˙}^{2} + \overset{p}{˙}^{2}

loop ke around vary karti hai — fast jahan

∣ p ∣

bada hai, turning points ke paas almost ruk jaata hai jahan

p \to 0

Frictionless (Hamiltonian) system ke liye trajectories ke bundle ka area shrink ho sakta hai.

Galat. Liouville's Theorem kehta hai ki phase-space area (volume) Hamiltonian flow se conserved hota hai — yeh deform ho sakta hai lekin kabhi shrink nahi ho sakta. Area shrink hone ke liye dissipation chahiye, jaise damped oscillator mein.

Ek damped oscillator ki phase trajectories closed hoti hain.

Galat. Energy girती hai,

\dot{E} = - 2 γ \overset{x}{˙}^{2} \leq 0

(friction constant

γ > 0

), isliye har baar ek chhoti ellipse cross hoti hai — orbit origin ki taraf spiral karti hai (ek stable focus), kabhi close nahi hoti.

Phase space mein har fixed point stable hota hai.

Galat.

V (q)

ke minima centers dete hain (stable, closed loops); maxima saddles dete hain (unstable — states ek direction mein ud jaati hain). Dekho Stability and Fixed Points.

Separatrix iske around ke loops ki tarah hi ek trajectory hai.

Galat. Halanki yeh hai ek trajectory, yeh ek alag hai: yeh exactly critical energy carry karta hai, saddle se guzarta hai, aur qualitatively alag motions ko divide karta hai (Pendulum ke liye swinging vs whirling) — ordinary loops mein se koi bhi aisa nahi karta.

Ek fixed point par state ki velocity zero hai, isliye trajectories wahan touch kar sakti hain.

Sahi.

\overset{q}{˙} = \overset{p}{˙} = 0

matlab flow arrow zero hai, isliye "no-crossing" rule (jo ek point par do nonzero arrows ko forbid karta hai) apply nahi hota — separatrix branches saddles par genuinely milti hain.

Error dhundho

\overset{p}{˙} = + \partial_{q} H

\overset{q}{˙} = \partial_{p} H

ke saath symmetry se."

Minus sign missing hai:

\overset{p}{˙} = - \partial_{q} H

. Iske bina

d H / d t

mein dono terms add ho jaate cancel hone ki bajaye, isliye energy conserved nahi hoti aur oscillator loops close hone ki jagah spiral out kar jaate.

"Harmonic oscillator

(q, p)

plane mein ek circle trace karta hai."

Sirf tab jab axes aise scale kiye jayein ki dono semi-axes match karein. Energy

\frac{p ^{2}}{2 m} + \frac{1}{2} m ω^{2} q^{2} = E

se shuru karo aur dono sides ko

E

se divide karo:

\frac{p ^{2}}{2 m E} + \frac{q ^{2}}{2 E / ( m ω ^{2} )} = 1

. Dono denominators semi-axes-squared bante hain, isliye raw

(q, p)

units mein yeh ek ellipse hai, generally circle nahi.

"Oscillator loop counter-clockwise chalta hai kyunki time aage move karta hai."

Yeh clockwise chalta hai:

q > 0

ke liye force deta hai

\overset{p}{˙} = - m ω^{2} q < 0

, isliye momentum kam hota hai — point daayein taraf neeche sweep karta hai, jo clockwise hai (figure mein white arrow se match karta hai).

"Pendulum ke liye,

θ = π

(bob seedha upar) ek stable resting point hai."

Yeh

V = - m g l cos θ

ka ek maximum hai (sabse bada jab

cos θ = - 1

), isliye yeh ek saddle hai — unstable. Stable center

θ = 0

hai (bob neeche latka hua,

V

lowest).

"Kyunki damped oscillator ke liye

d H / d t = 0

, iske orbits energy contours hain."

Damped system conservative nahi hai:

\dot{E} = - 2 γ \overset{x}{˙}^{2} \neq = 0

, isliye koi fixed

H

exist nahi karta. Sirf time-independent conservative

H

ke liye trajectories energy contours ke saath coincide karte hain.

"Fixed point wahan hota hai jahan velocity

\overset{q}{˙}

zero ho."

Tumhe dono

\overset{q}{˙} = 0

aur

\overset{p}{˙} = 0

chahiye. Ek oscillation ka turning point

\overset{q}{˙} = 0

hai (momentum zero) phir bhi

\overset{p}{˙} \neq = 0

, isliye yeh fixed point nahi hai — state move karta rehta hai.

Why questions

Conservative system ka phase portrait ek contour map kyun ban jaata hai?

Kyunki

H

har orbit ke saath constant hai (

d H / d t = 0

, upar dikhaya gaya), har trajectory ek level set

H (q, p) = E

hai — energy

H

ke contours draw karna hi portrait draw karna hai.

Flow arrow trajectory par har jagah tangent kyun hona chahiye?

Trajectory woh path hai jo moving state trace karta hai, aur

(\overset{q}{˙}, \overset{p}{˙})

uski instantaneous velocity hai — velocity by definition uss path ke tangent hoti hai jo woh generate karta hai (figure mein loop par white arrow dekho).

Friction add karne se loop ek inward spiral kyun ban jaati hai smaller loop ki bajaye?

Friction energy continuously in time remove karta hai (

\dot{E} = - 2 γ \overset{x}{˙}^{2} < 0

), isliye state apni starting point par return nahi kar sakta — har cycle ek lower-energy curve par land karta hai, ek spiral produce karta hai jo equilibrium par converge karta hai.

Separatrix, koi random energy nahi, pendulum ke libration aur rotation ke beech boundary kyun mark karta hai?

Yeh exactly woh energy hai jo unstable top (

θ = π

) par zero speed ke saath pahunchti hai; iske neeche bob wapas murta hai (swings), iske upar yeh top ke upar se guzarta rehta hai (whirls).

Ek smart reader phase portrait mein equations solve kiye bina instability "dekh" kyun sakta hai?

Instability dikhti hai trajectories ke roop mein jo ek point (saddle) se door diverge hoti hain, iske aas paas loop karne ki bajaye (center) — flow ki geometry seedhi dynamics encode karti hai.

Edge cases

Stable equilibrium par absolute rest mein ek mass ki phase trajectory kaisi dikhti hai?

Ek single point — state kabhi move nahi karta, isliye "curve" fixed point par hi degenerate ho jaata hai (zero-energy loop of zero size), figure mein origin par white square ki tarah.

Jab energy

E \to 0

to harmonic-oscillator ellipse ka kya hota hai?

Dono semi-axes

2 E / (m ω^{2})

aur

2 m E

zero tak shrink ho jaate hain, isliye ellipse origin par center point par collapse ho jaata hai — yeh consistent hai ek mass ke saath jo neeche bilkul still baitha hai.

Ek oscillation ke turning point par phase space mein kya ho raha hai?

Trajectory

q

-axis cross karti hai jahan

p = 0

: momentum momentarily zero, lekin

\overset{p}{˙} = - \partial_{q} H \neq = 0

, isliye point maximum displacement par hai aur reverse hone wala hai — yeh rest mein nahi hai.

Top ke upar tezi se ghoomte pendulum ke liye trajectory ek open wavy line kyun hai, loop nahi?

Kyunki

θ

bina bound ke badhta rehta hai — bob circulate karta rehta hai. Flat plane par yeh right ki taraf hamesha ke liye bhaag jaata hai; wavy line reflect karta hai

p

ka rise aur fall jab bob bottom se tezi se guzarta hai aur top par slow hota hai.

Pendulum phase space flat plane ki bajaye cylinder "banana chahta hai" aisa kyun hai?

Angle

θ

periodic hai:

θ

aur

θ + 2 π

same physical position hain.

q

-axis ko roll karke loop banane se "open" whirling trajectories cylinder ke around ghumne wale closed circles ban jaati hain — woh topological edge case jo plane pictures chhupa deti hain.

Ek free particle (koi potential nahi,

V = 0

) ki phase space mein trajectory kaisi hoti hai?

Ek horizontal straight line

p = const

H = p^{2} / (2 m)

ke saath momentum conserved hai isliye kabhi nahi badalta, jabki

q = q_{0} + (p / m) t

steadily drift karta hai — energy contours sirf horizontal lines hain.

Recall One-line self-test

Sab kuch cover karo: axes ka naam bolo, flow rule bolo, no-crossing ka reason bolo, aur woh ek cheez jo friction change karta hai. Axes ::: position $q$ aur momentum $p$ (time curve ke saath hidden). Flow rule ::: $\overset{q}{˙} = + \partial_{p} H, \overset{p}{˙} = - \partial_{q} H$ (minus $p$ ke saath rehta hai). No-crossing reason ::: single-valued flow arrow ⇒ ek state, ek future (determinism). Friction kya change karta hai ::: energy girती hai, isliye closed loops inward spirals ban jaati hain.