Visual walkthrough — Three-body problem — restricted (CR3BP), characteristic equation

3.2.32 · D2 · Physics › Orbital Mechanics & Astrodynamics › Three-body problem — restricted (CR3BP), characteristic equa

Hum sirf yeh ek cheez maante hain: ek vector ek arrow hai jisme ek $x$-part aur ek $y$-part hota hai, aur ek derivative steepness measure karta hai (koi cheez kitni tezi se change hoti hai). Baaki sab hum khud kamaenge.

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Step 0 — Naksha: coordinate system, effective potential $U$, aur mass ratio $\mu$

Kuch bhi nudge karne se pehle, humein yeh agree karna hoga ki sab cheez kahan baithe hain aur landscape $U$ asal mein kya hai. Yahi woh setup hai jis par poora page tika hua hai.

Parent ke boxed result ke har symbol ka ab ek ghar hai: $U_{xx},U_{yy},U_{xy}$ isi landscape ke second derivatives honge, isi point $(x_0,y_0)$ par evaluate kiye gaye, aur $\mu$ woh ek knob hai jo poori picture set karta hai.

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Step 1 — Stage: ek frozen landscape mein ek marble

KYA. Is frozen frame mein do heavy primaries still baithte hain. Ek tiny third body — marble — ghoomne ke liye free hai aur landscape $U$ feel karta hai lekin itna halka hai ki push back nahi kar sakta. Hum use exactly ek Lagrange point $L=(x_0,y_0)$ par rakhte hain jahan $\nabla U=0$ hai, toh saare forces cancel ho jaate hain.

KYUN. Agar $L$ par forces cancel ho jaate hain, toh marble wahan hamesha baith sakta hai. Lekin "baith sakta hai" ka matlab "wahan reh sakta hai" nahi hai. Ek gust — ek tiny nudge — sab kuch decide karta hai. Toh stability ka poora sawaal yeh hai: nudge ke baad, kya marble ghar aata hai ya bhaag jaata hai?

PICTURE. Neeche, do primaries $x$-axis par cyan dots hain (bada $x=-\mu$ par, chhota $x=1-\mu$ par); amber dot woh marble hai jo $L$ par khada hai. Faint arrows do gravity pulls plus outward centrifugal push hain — tail-to-tip draw kiye gaye hain, woh zero mein close ho jaate hain. Woh closure hi equilibrium hai.

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Step 2 — Motion ke do rules, offset par likhe hue

KYA. Parent ke boxed equations of motion, rotating frame mein, yeh hain: $\ddot{x}-2\dot{y}=\frac{\partial U}{\partial x},\ddot{y}+2\dot{x}=\frac{\partial U}{\partial y}.$ Kyunki $x_0,y_0$ constants hain, $\ddot{x}=\ddot{\xi }$ aur $\dot{x}=\dot{\xi }$ (koi constant add karne se yeh change nahi hota ki koi cheez kitni tezi se move karti hai). Toh wohi equations $x,y$ ki jagah $\xi ,\eta$ ke saath bhi hold karti hain.

KYUN. Hum woh law chahte hain jo nudge khud follow kare, raw position nahi — kyunki "kya nudge bada hota hai?" hamaara sawaal hai. Har symbol ko padhte hain:

$\underset{\begin{array}{c}\text{E–W}\\ \text{acceleration}\end{array}}{\underbrace{\ddot{\xi }}}-\underset{\begin{array}{c}\text{Coriolis:}\\ \text{N–S speed}\\ \text{E–W bend karta hai}\end{array}}{\underbrace{2\dot{\eta }}}=\underset{\begin{array}{c}\text{landscape se}\\ \text{E–W push}\end{array}}{\underbrace{\frac{\partial U}{\partial x}}}$

Akela term jise ghaur se dekhna hai woh hai $2\dot{\eta }$: marble ki northward speed ek eastward twist produce karti hai. Yahi Coriolis force hai — yeh sirf isliye aata hai kyunki hum ek merry-go-round par hain, aur yahi woh hero hoga jo triangular points ko bachayega.

PICTURE. Red curved arrow Coriolis dikhata hai: "upar" move karta body "sideways" steer hota hai. Yeh marble ko kabhi speed up ya slow down nahi karta (yeh hamesha motion ke sideways hota hai) — yeh sirf path ko bend karta hai.

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Step 3 — $L$ ke paas landscape ek tilted bowl (ya saddle) hai: Taylor to the rescue

KYA. Push $\partial U/\partial x$ is baat par depend karta hai ki marble kahan hai. Bilkul $L$ par yeh zero hai (yahi toh equilibrium hai). Ek baal bhar door, yeh roughly offset ke proportion mein badhta hai. Woh proportion hi woh cheez hai jo second derivative measure karta hai.

KYUN yeh tool — Taylor expansion. Hum poochhte hain "slope kaise change hota hai jab main sideways step leta hoon?" Woh tool jo exactly yeh jawab deta hai woh hai second derivative. Balance point ke paas first derivative (force) zero hoti hai, toh pehli cheez jo bachti hai woh second-derivative term hai: $\frac{\partial U}{\partial x}{|}_{\text{nudge ke baad}}\approx \underset{\begin{array}{c}\text{E–W push kitna}\\ \text{change hota E jaane par}\end{array}}{\underbrace{U_{xx}}}\xi +\underset{\begin{array}{c}\text{E–W push kitna}\\ \text{change hota N jaane par}\end{array}}{\underbrace{U_{xy}}}\eta .$ Yahan $U_{xx}={\partial }^{2}U/\partial x^2$ aur $U_{xy}={\partial }^{2}U/\partial x\partial y$ hain, dono apni values par frozen hain $L=(x_0,y_0)$ par. Shabdon mein: $U_{xx}$ landscape $U$ ki curvature hai E–W direction mein; positive = valley (ghar ki taraf push back karta hai), negative = hill (aur door push karta hai).

PICTURE. Left: ek valley — right nudge karo, slope tumhe left push karta hai, ghar wapas ($U_{xx}>0$). Right: ek hilltop — right nudge karo, slope tumhe aur right push karta hai ($U_{xx}<0$). $U_{xx}$ ka sign hi shape hai.

Steps 2 aur 3 ko mila kar, do motion rules tiny offset mein linear ho jaate hain: $\ddot{\xi }-2\dot{\eta }=U_{xx}\xi +U_{xy}\eta ,\ddot{\eta }+2\dot{\xi }=U_{xy}\xi +U_{yy}\eta .$

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Step 4 — Answer ki shape guess karo: $e^{\lambda t}$

KYA. Hum guess karte hain ki dono offsets usi ek exponential clock ki tarah badhte ya wiggle karte hain: $\xi (t)={\xi }_0{e}^{\lambda t},\eta (t)={\eta }_0{e}^{\lambda t}.$

KYUN yeh tool — exponential. Yeh equations linear with constant coefficients hain: ek hi function hai jiska derivative khud ka ek multiple hota hai, woh hai $e^{\lambda t}$. Yahi exact property humein chahiye, kyunki differentiate karna sirf $\lambda$ se multiply karta hai — calculus algebra mein collapse ho jaata hai. Number $\lambda$ (lambda) verdict hai:

$\lambda$ kaisa dikhta hai	marble kya karta hai	picture
real aur positive	$e^{+\lambda t}$ blow up ho jaata hai	bhaag jaata hai — unstable
pure imaginary $\pm i\omega$	$e^{i\omega t}$ = pure wobble	ghar orbit karta hai — stable

PICTURE. Teen sample $\lambda$ verdicts plot kiye gaye: growing exponential (amber, unstable), ek decaying wala, aur ek clean oscillation (cyan, stable). Hamaara poora kaam yeh find karna hai ki $L$ kis case mein aata hai.

Kyunki har time-derivative bas ek $\lambda$ kheeench laati hai: $\dot{\xi }=\lambda \xi$, $\ddot{\xi }={\lambda }^2\xi$. Substituting karne se har $e^{\lambda t}$ khatam ho jaata hai (woh kabhi zero nahi hota, toh hum ise divide kar dete hain) aur pure numbers reh jaate hain: $({\lambda }^2-U_{xx}){\xi }_0-(2\lambda +U_{xy}){\eta }_0=0,$ $(2\lambda -U_{xy}){\xi }_0+({\lambda }^2-U_{yy}){\eta }_0=0.$

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Step 5 — Do equations, ek condition: determinant

KYA. Ab hamare paas do unknown amplitudes ${\xi }_0,{\eta }_0$ mein do equations hain. Boring solution hai ${\xi }_0={\eta }_0=0$ (koi nudge hi nahi). Hum ek real nudge chahte hain, toh humein ek non-zero solution chahiye.

KYUN determinant. Origin se guzarne waali do straight-line equations ek non-zero solution share karti hain sirf tab jab do lines ek hi line hoon — matlab jab woh independent na hon. Woh single number jo "yeh dono secretly ek hain" detect karta hai woh determinant hai. Ise zero set karna exactly yeh demand hai ki "ek real nudge exist karta hai".

$\det \left(\begin{array}{cc}{\lambda }^2-U_{xx}& -(2\lambda +U_{xy})\\ 2\lambda -U_{xy}& {\lambda }^2-U_{yy}\end{array}\right)=0.$

Multiply out karo ($\text{main diagonal}-\text{off-diagonal}$): $({\lambda }^2-U_{xx})({\lambda }^2-U_{yy})-[-(2\lambda +U_{xy})(2\lambda -U_{xy})]=0.$ Off-diagonal product hai $-(2\lambda +U_{xy})(2\lambda -U_{xy})=-(4{\lambda }^2-U_{xy}^2)$, toh ise subtract karne par $4{\lambda }^2-U_{xy}^2$ add hota hai:

PICTURE. $2\times 2$ grid jisme diagonal aur anti-diagonal multiplications coloured sweeps ki tarah drawn hain, exactly dikhate hue ki $4{\lambda }^2$ kahan paida hota hai.

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Step 6 — $\Lambda ={\lambda }^2$ se verdict padho

KYA. Equation mein sirf ${\lambda }^2$ aur ${\lambda }^4$ use hote hain. $\Lambda ={\lambda }^2$ rename karo taaki ek quartic ek quadratic mein shrink ho jaaye: ${\Lambda }^2+(4-U_{xx}-U_{yy})\Lambda +(U_{xx}{U}_{yy}-U_{xy}^2)=0.$

KYUN. $\lambda$ ke pure imaginary hone ke liye (stable wobble), humein ${\lambda }^2=\Lambda$ ka real aur negative hona zaroori hai (kyunki $(i\omega {)}^2=-{\omega }^2<0$). Toh stability roots $\Lambda$ par ek checklist ban jaata hai:

• roots ka product $=U_{xx}{U}_{yy}-U_{xy}^2$ (constant term),
• roots ka sum $=-(4-U_{xx}-U_{yy})$.

Stable hone ke liye dono roots real aur negative chahiye ⇒ product $>0$, sum $<0$, aur roots complex nahi hone chahiye.

PICTURE. $\Lambda$ ke liye ek number line: dono roots negative zone mein → do imaginary $\lambda$ pairs → stable; ek root positive ho jaata hai → ek real $\lambda$ escape karta hai → unstable.

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Step 7 — Collinear points $L_1,L_2,L_3$: woh saddle jo hamesha leak karta hai

KYA (edge case #1). Primaries ko milane wali line par ($y_0=0$, matlab marble $x$-axis par baitha hai), symmetry force karti hai ki $U_{xy}=0$ ho. Pata chalta hai $U_{xx}>0$ (E–W mein valley) lekin $U_{yy}<0$ (N–S mein hill). Toh constant term hai: $U_{xx}{U}_{yy}-U_{xy}^2=(+)(-)-0=\text{negative}.$

KYUN yeh unhe doom karta hai. Roots ka product $=$ constant term $<0$ ka matlab hai ki do $\Lambda$ ke opposite signs hain — ek positive hai. Ek positive $\Lambda$ ek real positive $\lambda$ deta hai: $e^{+\lambda t}$ blow up ho jaata hai. $L_1,L_2,L_3$ hamesha unstable hote hain, har $\mu$ ke liye. Isi wajah se $L_1/L_2$ ke aaspaas halo orbits ko constant fuel chahiye.

PICTURE. Ek saddle surface: ek taraf valley, doosri taraf hill. Saddle par chhoda gaya marble hamesha downhill escape dhoondh leta hai.

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Step 8 — Triangular points $L_4,L_5$: ek maximum jise Coriolis bachata hai

KYA (edge case #2). $L_4,L_5$ par marble, dono primaries ek equilateral triangle banate hain: yah bilkul wahi jagah hai jahan $r_1=r_2=1$ hai, toh marble $(x_0,y_0)=(\frac{1}{2}-\mu ,\pm \frac{\sqrt{3}}{2})$ par baithta hai (har primary $1$ doori par hai, jo primary–primary separation $1$ se match karta hai).

Curvature numbers kahan se aate hain. $r_1=r_2=1$ ko $U(x,y)=\frac{1}{2}(x^2+y^2)+\frac{1-\mu }{r_1}+\frac{\mu }{r_2}$ mein plug karo, second partials $U_{xx},U_{yy},U_{xy}$ lo, aur us triangular point par evaluate karo. Kyunki $r_1=r_2=1$ saare messy denominators ko collapse kar deta hai, answer pure numbers mein aata hai (sirf $\mu$ ka cross term mein survivor hai): $U_{xx}=\frac{3}{4},U_{yy}=\frac{9}{4},U_{xy}=\pm \frac{3\sqrt{3}}{4}(1-2\mu ).$ $\pm$ track karta hai ki humne $L_4$ (axis ke upar) liya ya $L_5$ (neeche); $(1-2\mu )$ factor tab vanish hota hai jab primaries equal twins hon ($\mu =\frac{1}{2}$), symmetry se.

Inhe discriminant $D=(4-U_{xx}-U_{yy}{)}^2-4(U_{xx}{U}_{yy}-U_{xy}^2)$ mein Step 6 se feed karne par, simplification ke baad milta hai: $D=1-27\mu (1-\mu ),$ toh stability ($D\ge 0$, dono roots real aur negative) hold karti hai iff $1-27\mu (1-\mu )\ge 0⟺\mu \le {\mu }_{\text{crit}}=\frac{1}{2}\left(1-\sqrt{\frac{23}{27}}\right)\approx \mathrm{0.03852.}$

KYUN ek hilltop stable ho sakta hai. $L_4,L_5$ par potential $U$ actually ek maximum par hai — ek hilltop, jo apparently hopeless lagta hai. Coriolis ke bina woh leak karta, full stop. Lekin coefficient mein $4$ (Coriolis) kisi bhi runaway ko sideways bend karke ek closed loop mein le jaata hai bas tabhi jab chhota primary total mass ka ~3.85% se kam ho. Rotation "hill se gir jao" ko "hill ke aaspaas orbit karo" mein convert karta hai. Exactly isi wajah se Trojan asteroids Jupiter ke $L_4,L_5$ par survive karte hain ($\mu \approx 0.00095\ll {\mu }_{\text{crit}}$).

PICTURE. Left: hilltop ek naive escape arrow ke saath. Right: wohi hilltop, lekin Coriolis twist escape ko ek stable orbit mein curl kar deta hai — $\mu =0.0385$ threshold ek chhote dial par mark kiya gaya hai.

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Ek-picture summary

Upar sab kuch, compressed: frozen frame aur landscape $U$ set karo (Step 0) → nudge $(\xi ,\eta )$ → landscape linearise karo ($U_{xx},U_{yy},U_{xy}$) → Coriolis add karo ($\to$ woh $4$) → determinant → $\Lambda ={\lambda }^2$ mein quadratic → roots ka sign (discriminant ke zariye) verdict hai, geometry ke hisaab se always-unstable saddles aur conditionally-stable triangles mein split hota hai.

Recall Feynman retelling — poora walkthrough plain words mein

Pehle hum ek map bichchhate hain. Hum ek turntable par sawaar hote hain jo do heavy bodies ke saath ghoom raha hai taaki woh still rahen; hum unke masses ka balance point centre par rakhte hain aur unhe ek line par rakhte hain, ek unit door. Is map par hum ek single landscape $U$ draw karte hain jo pehle se dono gravity aur spin ke outward centrifugal fling ko contain karta hai — iska slope force hai, aur Lagrange point koi bhi flat spot hai. Knob $\mu$ kehta hai ki do bodies kitni unequal hain: zero ke paas hai "ek giant, ek crumb". Ab ek marble ko flat spot par balance karo aur ise thoda sa flick karo — woh flick hai $\xi$ (sideways) aur $\eta$ (up-down). Spot ke paas ground flick ke proportion mein push back karta hai; push ki stiffness curvature numbers $U_{xx},U_{yy},U_{xy}$ hain. Positive stiffness = valley jo tumhe ghar dhakelta hai; negative = hill jo tumhe bahar dhakelta hai. Turntable ek twist add karta hai — Coriolis — jo kisi bhi moving marble ko seedha off ki jagah sideways bend karta hai. Hum guess karte hain ki motion ek single clock $e^{\lambda t}$ hai: real positive $\lambda$ matlab explosion (unstable); imaginary $\lambda$ matlab pure wobble (stable). Real flick exist karne ki demand do equations ko ek determinant mein turn kar deta hai, jo ek neat quadratic ban jaata hai jab hum ${\lambda }^2=\Lambda$ kehte hain; Coriolis twist usmein number $4$ ki tarah appear hota hai. Roots real hain ya nahi yeh discriminant decide karta hai, woh cheez jo square root ke andar hai. Straight-line points $L_1,L_2,L_3$ saddles hain — ek taraf valley, doosri taraf hill — toh ek root hamesha positive hota hai; woh hamesha leak karte hain. Triangle points $L_4,L_5$ hilltop par baithe hain, jo hopeless lagta hai, lekin Coriolis $4$ escape ko ek orbit mein curl kar deta hai — stable rehta hai jab tak chhota body total mass ka roughly 3.85% se kam ho. Yahi aakhri number hai jiske wajah se Jupiter ke Trojans exist karte hain.

Related: Lagrange points · Jacobi constant & zero-velocity curves · Eigenvalues & linear stability analysis · Two-body problem & Kepler orbits