Abstract
A stable point is taken to be the bottom of a valley. This paper shows that $L_{4}$ is a maximum of the effective potential and stable nonetheless—the fence is not a shape but a mass ratio of $\mathbf{24.9599357944}$. No new theorem or law is claimed.
First, $L_{4}$ is a stationary point—its gradient is $(0.000\times10^{0}, 1.110\times10^{-10})$, becoming $-0.040634$ when displaced by $0.05$ (Section 2).
Second, this is the core of the paper. That stationary point is a maximum—the effective potential is $-1.499998500005$ while the largest of the eight surrounding values is $-1.499998503754$, all lower (Section 2).
Third, it is stable all the same, the condition being $27\mu(1-\mu)<1$—a threshold mass ratio of $\mathbf{24.9599357944}$ (Section 3).
Fourth, the Sun-Earth ratio of $333029.7043$ lies on the stable side and a ratio of $20$ on the unstable one (Section 3).
Fifth, $L_{1}$ solves numerically to $\mathbf{1491426526}$ m against the approximation’s $1496433206$ m, a relative difference of $0.003357$ (Section 4).
Sixth, the separator is whether what holds it is a potential or a velocity-dependent force (Section 5).
Keywords: Lagrange points, restricted three-body problem, L4 and L5, linear stability, celestial mechanics.
Scope (what is not claimed)
No new theorem or law is claimed—the restricted three-body problem, the Lagrange points and the linear stability condition for $L_{4}$ and $L_{5}$ are all standard. No celestial mechanics is built—perturbations, eccentricity, radiation pressure and the actual distribution of Trojans are not the subject; only the circular restricted problem. Nonlinear stability is excluded—what is treated is linear stability; long-term behaviour and Arnold diffusion are not entered. Not all five points are located—numbers are given for $L_{1}$ and $L_{4}$ only; $L_{2}$, $L_{3}$ and $L_{5}$ are named and left. The Hill radius is not derived—Paper 337’s scope note declines to enter the three-body problem or the derivation of the Hill radius, and this paper is placed beside it. Relation to earlier papers: Paper 195 showed that “stable” is six separate words, separated by what is perturbed—this paper gives a case where two of those six part company. Paper 487 showed that a stationary point comes in three kinds—$L_{4}$ is the maximum, and stable besides. Paper 337 showed that the edge of the solar system is five edges—in the same system, “a stable place” is likewise not one. Paper 481 showed that without separators a refutation has no destination—this paper writes one of them. What is added is confirming the maximality of $L_{4}$ in eight directions, giving its gradient as $1.110\times10^{-10}$, giving the threshold as $24.9599357944$ and checking it by inversion, bracketing the boundary with $20$ and $25$, and comparing the numerical $L_{1}$ with the approximation at $0.003357$.
1. Notation and the Question
Setting: Two bodies of masses $m_{1}$ and $m_{2}$ orbit their common centre in a circle, with mass fraction $\mu=m_{2}/(m_{1}+m_{2})$. In the co-rotating frame a third body of negligible mass feels the effective potential
$$\Omega(x,y)=-\left[\frac{1}{2}(x^{2}+y^{2}) +\frac{1-\mu}{r_{1}}+\frac{\mu}{r_{2}}\right]$$in units where $G(m_{1}+m_{2})=1$, the separation is $1$ and the angular velocity is $1$. It has five stationary points, of which $L_{4}$ and $L_{5}$ form equilateral triangles with the two bodies.
Observation 1.1 A stable point is taken to be the bottom of a valley. Displace it and it returns—the picture of falling towards lower potential.
Observation 1.2 But a rotating frame carries a velocity-dependent force. The Coriolis force—which cannot be written as a potential.
Question: Is $L_{4}$ a valley or a hill.
2. $L_{4}$ Is the Summit of a Hill
Computation 2.1: At $\mu=3.0\times10^{-6}$, close to the Sun-Earth value, evaluate the gradient at $L_{4}$.
$$\left(\frac{\partial\Omega}{\partial x},\ \frac{\partial\Omega}{\partial y}\right) =\left(0.000\times10^{0},\ 1.110\times10^{-10}\right)$$Observation 2.2 The gradient vanishes. $1.110\times10^{-10}$—the size of the finite-difference step; a stationary point.
Observation 2.3 Displaced by $0.05$, it ceases to be stationary. $-0.040634$—eight orders larger, confirming that the test responds to position.
Computation 2.4: Compare $\Omega$ at $L_{4}$ with the eight surrounding directions.
| Position | Effective potential $\Omega$ |
|---|---|
| $L_{4}$ | $\mathbf{-1.499998500005}$ |
| Largest of the eight around it | $\mathbf{-1.499998503754}$ |
| Smallest of the eight around it | $-1.499998513721$ |
Observation 2.5 Everything around it is lower. Even the largest, $-1.499998503754<-1.499998500005$—the potential falls in every direction: $L_{4}$ is a maximum.
Observation 2.6 It is not a minimum. The smallest around it is $-1.499998513721$—the directions differ, so it is not flat either.
Observation 2.7 In the three kinds of Paper 487, this is the maximum. A stationary point that is a hill—there the optical path was longest; here an object rests.
3. Stable All the Same, Fenced by a Mass Ratio
Computation 3.1: Solve the linear stability condition $27\mu(1-\mu)<1$ for $L_{4}$ and $L_{5}$.
| Quantity | Value |
|---|---|
| Critical mass fraction $\mu_{c}$ | $0.038520896505$ |
| $27\mu_{c}(1-\mu_{c})$ (inverted) | $0.999999999999999$ |
| Threshold mass ratio $(1-\mu_{c})/\mu_{c}$ | $\mathbf{24.9599357944}$ |
Observation 3.2 The inversion returns $1$. $0.999999999999999$—an independent check on both the algebra and the arithmetic.
Observation 3.3 The fence is $24.9599357944$. Stable if the heavier exceeds the lighter by more than $24.96$ times—a question of ratio, not of shape.
Computation 3.4: Check both sides of the boundary.
| System | Mass ratio | Verdict |
|---|---|---|
| Sun and Earth | $\mathbf{333029.7043}$ | stable |
| A binary at $25$ | $25$ | stable (just above) |
| A binary at $20$ | $20$ | unstable |
Observation 3.5 Sun and Earth stand four orders from the fence. $333029.7043$ against $24.9599357944$—comfortably on the stable side.
Observation 3.6 $25$ and $20$ bracket the boundary. One stable, one not—confirming the threshold lies between them.
Observation 3.7 What holds it is the Coriolis force. Displaced, the body is turned sideways and enters a circulation before it can descend—being velocity-dependent, it never appears in the picture of the potential.
4. $L_{1}$ Is No Valley, but a Saddle
Computation 4.1: Solve for $L_{1}$ in the Sun-Earth system and compare with the approximation.
| Route | Distance from Earth |
|---|---|
| Numerical | $\mathbf{1491426526}$ m ($1{,}491{,}427$ km) |
| Approximation $R(m_{2}/3m_{1})^{1/3}$ | $1496433206$ m |
| Difference | $5006679$ m (relative $0.003357$) |
Observation 4.2 It matches the familiar $1.5$ million km. $1{,}491{,}427$ km—where observing spacecraft are placed, an independent check.
Observation 4.3 The approximation is $0.34$ per cent out. A difference of $5006679$ m—not agreement, but the right order: the range where the approximation works can be read.
Observation 4.4 $L_{1}$ is a saddle and is not stable. It falls in one direction—a spacecraft placed there must keep station.
Observation 4.5 So the five stationary points are not of one kind. Three saddles, two maxima—and it is the maxima that are stable.
5. The Separator
The separator: is what holds it a potential, or a velocity-dependent force.
Observation 5.1 A potential, and nothing but a minimum is stable. The floor of a valley—which $L_{4}$ is not.
Observation 5.2 A velocity-dependent force, and a maximum may be stable. The Coriolis force—and $L_{4}$ is exactly that.
Observation 5.3 Two of Paper 195’s six senses of “stable” part company here. “A minimum of energy” and “perturbations do not grow”—$L_{4}$ satisfies the second and not the first: distinct roots by the criterion of Paper 300.
Observation 5.4 So “stable, therefore in a valley” cannot be said. A counterexample exists—the Trojan asteroids rest on that summit.
Observation 5.5 That the fence is a ratio rather than a shape follows from the same place. $24.9599357944$—the balance between the Coriolis force and the slope of the hill is fixed by the mass ratio alone.
Observation 5.6 This is one of the separators of Paper 481. When something drifts off a point that was supposed to be stable—which sense of stability failed sends the repair elsewhere.
Observation 5.7 To be honest—only linear stability is treated. That perturbations do not grow to first order—what happens over long times, and how nonlinear effects enter, is another question this paper does not touch. Not all of why the Trojans have lasted billions of years is written here.
6. Roll Call and Closing
- Gradient $1.110\times10^{-10}$ at $L_{4}$ (Computation 2.1): a stationary point.
- $-0.040634$ when displaced by $0.05$ (Observation 2.3): the test responds.
- $\Omega(L_{4})=-1.499998500005$ (Computation 2.4): the core.
- Surrounding maximum $-1.499998503754$ (Observation 2.5): all lower; a maximum.
- Threshold $24.9599357944$ (Computation 3.1): inverting gives $0.999999999999999$.
- Sun and Earth at $333029.7043$ (Computation 3.4): four orders of margin.
- Bracketed by $25$ and $20$ (Observation 3.6): both sides of the boundary.
- $L_{1}$ at $1491426526$ m (Computation 4.1): $0.003357$ from the approximation.
- The separator is potential or velocity (Observation 5.3): two of Paper 195’s six part.
- Linear stability only (Observation 5.7): not the long-term question.
Closing: A stable point is taken to be the bottom of a valley—displace it and it returns, falling towards lower potential. But a rotating frame carries a velocity-dependent force, the Coriolis force, which cannot be written as a potential. $L_{4}$ is the summit of a hill—at $\mu=3.0\times10^{-6}$ the gradient at $L_{4}$ is $(0.000\times10^{0}, 1.110\times10^{-10})$, the size of the finite-difference step, and displacing by $0.05$ gives $-0.040634$, eight orders larger. The effective potential there is $\mathbf{-1.499998500005}$ while the largest of the eight surrounding values is $\mathbf{-1.499998503754}$ and the smallest $-1.499998513721$. The potential falls in every direction, so $L_{4}$ is a maximum, not a minimum, and the directions differ so it is not flat either. In the three kinds of Paper 487 this is the maximum: there the optical path was longest, here an object rests. Stable all the same, fenced by a mass ratio—solving $27\mu(1-\mu)<1$ gives $\mu_{c}=0.038520896505$, whose inversion returns $0.999999999999999$, and a threshold mass ratio of $\mathbf{24.9599357944}$. Stability requires the heavier to exceed the lighter by more than $24.96$ times: a question of ratio, not of shape. The Sun-Earth ratio $\mathbf{333029.7043}$ stands four orders from the fence, while $25$ is stable and $20$ is not, bracketing the boundary. What holds it is the Coriolis force: displaced, the body is turned sideways and enters a circulation before it can descend, and being velocity-dependent it never appears in the picture of the potential. $L_{1}$ is no valley, but a saddle—solving for the Sun-Earth system gives $\mathbf{1491426526}$ m from Earth, or $1{,}491{,}427$ km, matching where observing spacecraft are placed. The approximation $R(m_{2}/3m_{1})^{1/3}$ gives $1496433206$ m, a difference of $5006679$ m or $0.003357$: not agreement, but the right order. $L_{1}$ is a saddle and is not stable, so a spacecraft there must keep station. The five stationary points are not of one kind—three saddles, two maxima—and it is the maxima that are stable. One thing separates them—is what holds it a potential, or a velocity-dependent force. A potential, and nothing but a minimum is stable, which $L_{4}$ is not. A velocity-dependent force, and a maximum may be stable, which $L_{4}$ is. Two of Paper 195’s six senses of “stable” part company here—“a minimum of energy” and “perturbations do not grow”—and $L_{4}$ satisfies the second and not the first, distinct roots by the criterion of Paper 300. So “stable, therefore in a valley” cannot be said: the Trojan asteroids rest on that summit. That the fence is a ratio rather than a shape follows from the same place, the balance between the Coriolis force and the slope being fixed by the mass ratio alone. This is one of the separators of Paper 481: when something drifts off a point supposed to be stable, which sense of stability failed sends the repair elsewhere. To be honest—only linear stability is treated, that perturbations do not grow to first order. What happens over long times, and how nonlinear effects enter, is another question this paper does not touch, and not all of why the Trojans have lasted billions of years is written here.
References
- Lagrange, J.-L. (1772). Essai sur le problème des trois corps. Prix de l’Académie Royale des Sciences de Paris.
- Szebehely, V. (1967). Theory of Orbits: The Restricted Problem of Three Bodies. Academic Press.
- Murray, C. D., & Dermott, S. F. (1999). Solar System Dynamics. Cambridge University Press.
- Yamagishi, Y. (2026). “Stable” Is Six Separate Words, and What Separates Them Is What You Perturb [Paper 195]. Zenodo. 10.5281/zenodo.22107170
- Yamagishi, Y. (2026). The Three Hundredth Census: Whether Two Things Share a Root Is Decidable [Paper 300]. Zenodo. 10.5281/zenodo.22171914
- Yamagishi, Y. (2026). The Edge of the Solar System Is Five Edges, and the Innermost and Outermost Differ by a Factor of $5106$ [Paper 337]. Zenodo. 10.5281/zenodo.22681940
- Yamagishi, Y. (2026). A Broken Law Can Be Blamed on Any of the Assumptions [Paper 481]. Zenodo. 10.5281/zenodo.22733927
- Yamagishi, Y. (2026). A Stationary Point Comes in Three Kinds: Minimum, Maximum, and Everywhere Equal [Paper 487]. Zenodo. 10.5281/zenodo.22759965
Reference
Yamagishi, Y. (2026). Something Is Resting on the Summit of a Hill [Paper 490]. YOI: https://www.yamagishiyuuki.com/yoi/490
DOI: 10.5281/zenodo.22760033
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 490. This paper is a preprint and has not been peer reviewed.