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

We assume nothing but this: a vector is an arrow with an $x$-part and a $y$-part, and a derivative measures steepness (how fast something changes). Everything else we earn.

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Step 0 — The map: coordinate system, the effective potential $U$, and the mass ratio $\mu$

Before we can nudge anything, we need to agree on where everything sits and what the landscape $U$ actually is. This is the setup the whole page rests on.

Every symbol from the parent's boxed result now has a home: $U_{xx},U_{yy},U_{xy}$ will be second derivatives of this landscape, evaluated at this point $(x_0,y_0)$, and $\mu$ is the one knob that sets the whole picture.

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Step 1 — The stage: a frozen landscape with one marble

WHAT. In this frozen frame the two heavy primaries sit still. A tiny third body — the marble — is free to roam and feels the landscape $U$ but is far too light to push back. We place it exactly at a Lagrange point $L=(x_0,y_0)$ where $\nabla U=0$, so all forces cancel.

WHY. If forces cancel at $L$, the marble can sit there forever. But "can sit" is not "will stay". A gust — a tiny nudge — decides everything. So the whole question of stability is: after a nudge, does the marble come home or run away?

PICTURE. Below, the two primaries are cyan dots on the $x$-axis (big one at $x=-\mu$, small one at $x=1-\mu$); the amber dot is the marble parked at $L$. The faint arrows are the two gravity pulls plus the outward centrifugal push — drawn tail-to-tip, they close into zero. That closure is equilibrium.

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Step 2 — Two rules of motion, written on the offset

WHAT. The parent's boxed equations of motion, in the rotating frame, are $\ddot{x}-2\dot{y}=\frac{\partial U}{\partial x},\ddot{y}+2\dot{x}=\frac{\partial U}{\partial y}.$ Since $x_0,y_0$ are constants, $\ddot{x}=\ddot{\xi }$ and $\dot{x}=\dot{\xi }$ (adding a constant changes nothing about how fast something moves). So the same equations hold with $\xi ,\eta$ in place of $x,y$.

WHY. We want the law obeyed by the nudge itself, not the raw position — because "does the nudge grow?" is our question. Let us read each symbol:

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

The lone term to stare at is $2\dot{\eta }$: the marble's northward speed produces an eastward twist. That is the Coriolis force — it appears only because we are on a merry-go-round, and it will be the hero that saves the triangular points.

PICTURE. The red curved arrow shows Coriolis: a body moving "up" gets steered "sideways". It never speeds the marble up or down (it is always sideways to motion) — it only bends the path.

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Step 3 — The landscape near $L$ is a tilted bowl (or saddle): Taylor to the rescue

WHAT. The push $\partial U/\partial x$ depends on where the marble is. Right at $L$ it is zero (that's equilibrium). A hair away, it grows roughly in proportion to the offset. That proportion is what the second derivative measures.

WHY this tool — the Taylor expansion. We ask "how does a slope change as I step sideways?" The tool that answers exactly that is the second derivative. Near a balance point the first derivative (the force) is zero, so the first thing that survives is the second-derivative term: $\frac{\partial U}{\partial x}{|}_{\text{after nudge}}\approx \underset{\begin{array}{c}\text{how E–W push}\\ \text{changes as we go E}\end{array}}{\underbrace{U_{xx}}}\xi +\underset{\begin{array}{c}\text{how E–W push}\\ \text{changes as we go N}\end{array}}{\underbrace{U_{xy}}}\eta .$ Here $U_{xx}={\partial }^{2}U/\partial x^2$ and $U_{xy}={\partial }^{2}U/\partial x\partial y$, both frozen at their values at $L=(x_0,y_0)$. In words: $U_{xx}$ is the curvature of the landscape $U$ in the E–W direction; positive = valley (push back toward home), negative = hill (push further away).

PICTURE. Left: a valley — nudge right, the slope pushes you left, back home ($U_{xx}>0$). Right: a hilltop — nudge right, the slope pushes you further right ($U_{xx}<0$). The sign of $U_{xx}$ is the shape.

Putting Steps 2 and 3 together, the two motion rules become linear in the tiny offset: $\ddot{\xi }-2\dot{\eta }=U_{xx}\xi +U_{xy}\eta ,\ddot{\eta }+2\dot{\xi }=U_{xy}\xi +U_{yy}\eta .$

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Step 4 — Guess the shape of the answer: $e^{\lambda t}$

WHAT. We guess that both offsets grow or wiggle like the same exponential clock: $\xi (t)={\xi }_0{e}^{\lambda t},\eta (t)={\eta }_0{e}^{\lambda t}.$

WHY this tool — the exponential. These equations are linear with constant coefficients: the only function whose derivative is a multiple of itself is $e^{\lambda t}$. That is the exact property we need, because differentiating just multiplies by $\lambda$ — the calculus collapses into algebra. The number $\lambda$ (lambda) is the verdict:

what $\lambda$ looks like	what the marble does	picture
real & positive	$e^{+\lambda t}$ blows up	runs away — unstable
pure imaginary $\pm i\omega$	$e^{i\omega t}$ = pure wobble	orbits home — stable

PICTURE. Three sample $\lambda$ verdicts plotted: growing exponential (amber, unstable), a decaying one, and a clean oscillation (cyan, stable). Our whole job is to find which case $L$ falls into.

Because each time-derivative just pulls down a $\lambda$: $\dot{\xi }=\lambda \xi$, $\ddot{\xi }={\lambda }^2\xi$. Substituting kills every $e^{\lambda t}$ (it is never zero, so we divide it out) and leaves pure numbers: $({\lambda }^2-U_{xx}){\xi }_0-(2\lambda +U_{xy}){\eta }_0=0,$ $(2\lambda -U_{xy}){\xi }_0+({\lambda }^2-U_{yy}){\eta }_0=0.$

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Step 5 — Two equations, one condition: the determinant

WHAT. We now have two equations in the two unknown amplitudes ${\xi }_0,{\eta }_0$. The boring solution is ${\xi }_0={\eta }_0=0$ (no nudge at all). We want a real nudge, so we need a non-zero solution.

WHY a determinant. Two straight-line equations through the origin share a non-zero solution only when the two lines are the same line — i.e. when they are not independent. The single number that detects "these two are secretly one" is the determinant. Setting it to zero is exactly the demand "a real nudge exists".

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

Multiply out ($\text{main diagonal}-\text{off-diagonal}$): $({\lambda }^2-U_{xx})({\lambda }^2-U_{yy})-[-(2\lambda +U_{xy})(2\lambda -U_{xy})]=0.$ The off-diagonal product is $-(2\lambda +U_{xy})(2\lambda -U_{xy})=-(4{\lambda }^2-U_{xy}^2)$, so subtracting it adds $4{\lambda }^2-U_{xy}^2$:

PICTURE. The $2\times 2$ grid with the diagonal and anti-diagonal multiplications drawn as coloured sweeps, showing exactly where the $4{\lambda }^2$ is born.

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Step 6 — Read the verdict with $\Lambda ={\lambda }^2$

WHAT. The equation only ever uses ${\lambda }^2$ and ${\lambda }^4$. Rename $\Lambda ={\lambda }^2$ to shrink a quartic into a quadratic: ${\Lambda }^2+(4-U_{xx}-U_{yy})\Lambda +(U_{xx}{U}_{yy}-U_{xy}^2)=0.$

WHY. For $\lambda$ to be pure imaginary (the stable wobble), we need ${\lambda }^2=\Lambda$ to be real and negative (since $(i\omega {)}^2=-{\omega }^2<0$). So stability becomes a checklist on the roots $\Lambda$:

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

Stable demands both roots real and negative ⇒ product $>0$, sum $<0$, and the roots must not be complex.

PICTURE. A number line for $\Lambda$: both roots in the negative zone → two imaginary $\lambda$ pairs → stable; one root strays positive → a real $\lambda$ escapes → unstable.

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Step 7 — Collinear points $L_1,L_2,L_3$: the saddle that always leaks

WHAT (edge case #1). On the line joining the primaries ($y_0=0$, so the marble sits on the $x$-axis), symmetry forces $U_{xy}=0$. One finds $U_{xx}>0$ (valley E–W) but $U_{yy}<0$ (hill N–S). So the constant term is $U_{xx}{U}_{yy}-U_{xy}^2=(+)(-)-0=\text{negative}.$

WHY it dooms them. Product of roots $=$ constant term $<0$ means the two $\Lambda$ have opposite signs — one is positive. A positive $\Lambda$ gives a real positive $\lambda$: $e^{+\lambda t}$ blows up. $L_1,L_2,L_3$ are always unstable, for every $\mu$. This is why halo orbits around $L_1/L_2$ need constant fuel.

PICTURE. A saddle surface: valley one way, hill the other. A marble left on a saddle always finds the downhill escape.

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Step 8 — Triangular points $L_4,L_5$: a maximum that Coriolis rescues

WHAT (edge case #2). At $L_4,L_5$ the marble, both primaries, form an equilateral triangle: this is exactly where $r_1=r_2=1$, so the marble sits at $(x_0,y_0)=(\frac{1}{2}-\mu ,\pm \frac{\sqrt{3}}{2})$ (each primary is distance $1$ away, matching the primary–primary separation $1$).

Where the curvature numbers come from. Plug $r_1=r_2=1$ into $U(x,y)=\frac{1}{2}(x^2+y^2)+\frac{1-\mu }{r_1}+\frac{\mu }{r_2}$, take the second partials $U_{xx},U_{yy},U_{xy}$, and evaluate at that triangular point. Because $r_1=r_2=1$ collapses all the messy denominators, the answer is pure numbers (the only survivor of $\mu$ is in the cross term): $U_{xx}=\frac{3}{4},U_{yy}=\frac{9}{4},U_{xy}=\pm \frac{3\sqrt{3}}{4}(1-2\mu ).$ The $\pm$ tracks whether we picked $L_4$ (above the axis) or $L_5$ (below); the $(1-2\mu )$ factor vanishes when the primaries are equal twins ($\mu =\frac{1}{2}$), by symmetry.

Feeding these into the discriminant $D=(4-U_{xx}-U_{yy}{)}^2-4(U_{xx}{U}_{yy}-U_{xy}^2)$ from Step 6 gives, after simplification, $D=1-27\mu (1-\mu ),$ so stability ($D\ge 0$, both roots real and negative) holds iff $1-27\mu (1-\mu )\ge 0⟺\mu \le {\mu }_{\text{crit}}=\frac{1}{2}\left(1-\sqrt{\frac{23}{27}}\right)\approx \mathrm{0.03852.}$

WHY a hilltop can be stable. At $L_4,L_5$ the potential $U$ is actually at a maximum — a hilltop, seemingly hopeless. Without Coriolis it would leak, full stop. But the $4$ (Coriolis) in the coefficient bends any runaway sideways into a closed loop provided the lighter primary is under ~3.85 % of the total mass. Rotation converts "roll off the hill" into "orbit around the hill". This is precisely why the Trojan asteroids survive at Jupiter's $L_4,L_5$ ($\mu \approx 0.00095\ll {\mu }_{\text{crit}}$).

PICTURE. Left: the hilltop with a naive escape arrow. Right: the same hilltop, but the Coriolis twist curls the escape into a stable orbit — with the $\mu =0.0385$ threshold marked on a little dial.

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The one-picture summary

Everything above, compressed: set the frozen frame and the landscape $U$ (Step 0) → nudge $(\xi ,\eta )$ → linearise the landscape ($U_{xx},U_{yy},U_{xy}$) → add Coriolis ($\to$ the $4$) → determinant → quadratic in $\Lambda ={\lambda }^2$ → sign of the roots (via the discriminant) is the verdict, split by geometry into the always-unstable saddles and the conditionally-stable triangles.

Recall Feynman retelling — the whole walkthrough in plain words

First we lay a map. We ride a turntable spinning with two heavy bodies so they hold still; we put the balance point of their masses at the centre and lay them on a line, one unit apart. On this map we draw a single landscape $U$ that already contains both gravity and the outward centrifugal fling of the spin — its slope is the force, and a Lagrange point is any flat spot. The knob $\mu$ says how lopsided the two bodies are: near zero is "one giant, one crumb". Now balance a marble at a flat spot and flick it a tiny bit — that flick is $\xi$ (sideways) and $\eta$ (up-down). Near the spot the ground pushes back in proportion to the flick; the stiffness of the push is the curvature numbers $U_{xx},U_{yy},U_{xy}$. Positive stiffness = a valley that shoves you home; negative = a hill that shoves you out. The turntable adds a twist — Coriolis — that bends any moving marble sideways instead of straight off. We guess the motion is a single clock $e^{\lambda t}$: real positive $\lambda$ means explosion (unstable); imaginary $\lambda$ means pure wobble (stable). Demanding a real flick exists turns the two equations into one determinant, which becomes a neat quadratic once we call ${\lambda }^2=\Lambda$; the Coriolis twist appears as the number $4$ inside it. Whether the roots are real is decided by the discriminant, the thing under the square root. Straight-line points $L_1,L_2,L_3$ are saddles — valley one way, hill the other — so one root is always positive; they always leak. Triangle points $L_4,L_5$ sit on a hilltop, which should be hopeless, but the Coriolis $4$ curls the escape into an orbit — stable as long as the small body is under about 3.85 % of the total mass. That last number is why Jupiter's Trojans exist.

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