Lagrange Points and the James Webb Space Telescope

Abstract

This article guides the reader through Lagrange’s famous solution to the so-called “three-body problem” and shows its application to the recently launched James Webb Space Telescope. Lagrange’s original equations are presented in his original notation, supplemented by modern vector computations. The entire work is accomplished with mathematics typically taught at the college freshman and sophomore levels.

MSC:

• 01A50
• 70F07

1 Introduction

On Christmas Day, 2021, NASA launched the James Webb Space Telescope (jwst) from the European Space Agency’s facility in French Guiana. Over the next month, jwst traveled approximately one million miles from earth, and was placed into an orbit around …absolutely nothing! More precisely, jwst is now in orbit around a Lagrange point, one of five such points in the sun-earth system [4, 5].

The mathematics that made the JWST mission possible was first developed by Joseph-Louis Lagrange in his “Essai d’une nouvelle méthode pour résoudre le problême des trois corps” [2], which won the prize of the Royal Academy of Sciences of Paris in 1772. In the nearly identical version of the paper found in his collected works [3, p. 230], Lagrange writes that his work is

…intended to examine how and in which cases the three bodies can move so that their distances can always be constant, or shall at least keep constant ratios between themselves. I find that these conditions can only be satisfied in two cases: one, when the three bodies are arranged in one straight line, and the other, when they form an equilateral triangle …

(Much of Lagrange’s paper is available in English at the website [7].)

In the three-body problem consisting of the sun, the earth, and a third object of infinitesimal mass, these two cases require that the third object must be located at one of the so-called Lagrange points $L_1,L_2,L_3,L_4$, or L₅ shown in Figure 1. jwst is located at L₂, which is unstable in the sense that if an object at L₂ is displaced even slightly, it will continue to move away from L₂. Thus jwst actually orbits L₂ in a small halo orbit rather than remaining “parked” at L₂. The orbit itself is unstable, meaning that it must be corrected occasionally by small rocket thrusts. More information, including an animation of the jwst orbit, can be found at the NASA website [5].

Fig. 1 The sun, earth, and the five Lagrange points. jwst orbits L₂.

Modern treatments of the three-body problem can be found in Roy [6], Danby [1], and other sources, but it is difficult to recognize the brilliant simplicity of Lagrange’s original method in these works. Therefore the goal of this paper is to give a detailed description showing just how Lagrange arrived at the straight-line and equilateral triangle solutions. Though we shall take great care to preserve Lagrange’s original notation and original equations, which contain no vectors whatsoever, it will be useful to introduce supplementary vector notation to keep the length of the presentation manageable. Because Lagrange’s paper contains a great deal of algebra and bit of calculus, this description of his work will involve a great deal of computation. And as a part of that computation, we will even find it necessary to correct an error in Lagrange’s work that is related to the location of jwst!

2 Notation and Preliminary Ideas

In the following work, every symbol representing a scalar (including any scalar coordinate or magnitude of a vector) is Lagrange’s original notation, with a single exception that will be noted near the end of our work. Also, all symbols representing vectors and matrices have been introduced to represent Lagrange’s work in a more compact form, but are not his notation. Lagrange, like other mathematicians of his era, used prime notation x, $x\prime ,x″$ in the same way that modern mathematicians use subscripts x₀, x₁, x₂, etc. Note that the primes do not represent derivatives! Instead, Lagrange uses the familiar Leibniz notation $\frac{d}{dt}$ for derivative; in addition we shall use $\dot{\mathbf{r}},\dot{\mathbf{r}}\prime ,\stackrel{..}{\mathbf{r}}$, etc. for vector derivatives.

The specific configuration of the sun, earth, and a third body at one of the Lagrange points, shown in Figure 1, is generalized in Figure 2. This figure shows three bodies whose masses are A, B, and C, along with the vectors representing their relative positions. The vectors ${\mathbf{R}}_A$ and ${\mathbf{R}}_B$ represent positions with respect to a fixed origin; these will appear only temporarily in our work.

Fig. 2 Masses A, B, and C and their relative positions $\mathbf{r},\mathbf{r}\prime ,\mathbf{r}″$.

From Figure 2, it is easy to see that $\mathbf{r}+\mathbf{r}″=\mathbf{r}\prime$; from this, one has $\dot{\mathbf{r}}+\dot{\mathbf{r}}″=\dot{\mathbf{r}}\prime$ by differentiating. Therefore an analogous figure describes the velocity vectors. The scalars $r,r\prime$, and $r″$ are the magnitudes of the vectors $\mathbf{r},\mathbf{r}\prime$, and $\mathbf{r}″$, respectively, and the speeds $u,u\prime$, and $u″$ are the magnitudes of the velocity vectors $\dot{\mathbf{r}},\dot{\mathbf{r}}\prime$, and $\dot{\mathbf{r}}″$.

Next, for this paper to serve effectively as a roadmap for our journey through Lagrange’s work, it will be useful to display a “map legend” of symbols introduced by Lagrange and related to Figure 2: $$p=\frac{1}{2}(r{\prime }^2+r\prime {\prime }^2-r^2)=\mathbf{r}\prime ·\mathbf{r}″\nu =\frac{1}{2}(u{\prime }^2+u\prime {\prime }^2-u^2)=\dot{\mathbf{r}}\prime ·\dot{\mathbf{r}}″$$ $$p\prime =\frac{1}{2}(r^2+r\prime {\prime }^2-r{\prime }^2)=-\mathbf{r}·\mathbf{r}″\nu \prime =\frac{1}{2}(u^2+u\prime {\prime }^2-u{\prime }^2)=-\dot{\mathbf{r}}·\dot{\mathbf{r}}″$$ $$p″=\frac{1}{2}(r^2+r{\prime }^2-r\prime {\prime }^2)=\mathbf{r}·\mathbf{r}\prime \nu ″=\frac{1}{2}(u^2+u{\prime }^2-u\prime {\prime }^2)=\dot{\mathbf{r}}·\dot{\mathbf{r}}\prime$$

$q=\frac{1}{r{\prime }^3}-\frac{1}{r\prime {\prime }^3}$, $q\prime =\frac{1}{r^3}-\frac{1}{r\prime {\prime }^3}$, $q″=\frac{1}{r{\prime }^3}-\frac{1}{r^3}$ (so $q″=q-q\prime$).

The formulas that define p, $p\prime$, and $p″$ can be found by applying the Law of Cosines to the triangle formed by r, $\mathbf{r}\prime$, and $\mathbf{r}″$ in Figure 2, with the formulas for ν, $\nu \prime$, and $\nu ″$ found analogously from the velocity vectors. Finally, we note that the lettered equation labels (A), (B), etc. referred to in this article are also Lagrange’s original notation. References in this article to these equation labels might serve to guide the interested reader through Lagrange’s paper [3], if one is so inclined.

3 The gravitational forces

Referring to Figure 2, we see that the net force on mass A is the sum of the forces exerted by B and C. The gravitational forces obey an inverse square law, so with units chosen to make the gravitational constant of proportionality equal to 1, we have $$A{\stackrel{..}{\mathbf{R}}}_A=\frac{AB}{r^2}\frac{\mathbf{r}}{r}+\frac{AC}{r{\prime }^2}\frac{\mathbf{r}\prime }{r\prime }.$$

Similarly $$B{\stackrel{..}{\mathbf{R}}}_B=-\frac{AB}{r^2}\frac{\mathbf{r}}{r}+\frac{BC}{r\prime {\prime }^2}\frac{\mathbf{r}″}{r″}.$$

By recognizing $\stackrel{..}{\mathbf{r}}={\stackrel{..}{\mathbf{R}}}_B-{\stackrel{..}{\mathbf{R}}}_A$ and $\mathbf{r}″=\mathbf{r}\prime -\mathbf{r}$, we can subtract equations to obtain Lagrange’s equation (A): (A) $$\stackrel{..}{\mathbf{r}}+\left(\frac{A+B}{r^3}+\frac{C}{r\prime {\prime }^3}\right)\mathbf{r}+C\left(\frac{1}{r{\prime }^3}-\frac{1}{r\prime {\prime }^3}\right)\mathbf{r}\prime =\mathbf{0}.$$(A)

Lagrange’s equations (B) $$\stackrel{..}{\mathbf{r}}\prime +\left(\frac{A+C}{r{\prime }^3}+\frac{B}{r\prime {\prime }^3}\right)\mathbf{r}\prime +B\left(\frac{1}{r^3}-\frac{1}{r\prime {\prime }^3}\right)\mathbf{r}=\mathbf{0}$$(B)

and (C) $$\stackrel{..}{\mathbf{r}}″+\left(\frac{B+C}{r\prime {\prime }^3}+\frac{A}{r^3}\right)\mathbf{r}″+A\left(\frac{1}{r{\prime }^3}-\frac{1}{r^3}\right)\mathbf{r}\prime =\mathbf{0}$$(C) are developed analogously. (In Lagrange’s work, each of these vector equations is replaced by three scalar coordinate equations.)

4 Algebra, more algebra, and a little calculus

To follow Lagrange’s work, significant amounts of basic algebra and calculus are necessary. We begin by observing that $$\begin{array}{c}\frac{1}{2}\frac{d^2}{d{t}^2}{r}^2-u^2=\frac{1}{2}\frac{d^2}{d{t}^2}(\mathbf{r}·\mathbf{r})-u^2=\frac{1}{2}\frac{d}{dt}(2\mathbf{r}·\dot{\mathbf{r}})-u^2\\ =\mathbf{r}·\stackrel{..}{\mathbf{r}}+\dot{\mathbf{r}}·\dot{\mathbf{r}}-u^2=\mathbf{r}·\stackrel{..}{\mathbf{r}}+u^2-u^2=\mathbf{r}·\stackrel{..}{\mathbf{r}}.\end{array}$$

This (along with a formula from our map legend) allows us to express the dot product of (A) with r in a useful form, giving us the first of three equations making up Lagrange’s (F): (F) $$\frac{1}{2}\frac{d^2}{d{t}^2}{r}^2+\left(\frac{A+B}{r^3}+\frac{C}{r\prime {\prime }^3}\right)r^2+\frac{C}{2}\left(\frac{1}{r{\prime }^3}-\frac{1}{r\prime {\prime }^3}\right)(r^2+r{\prime }^2-r\prime {\prime }^2)-u^2=0.$$(F)

Similar computations for the dot product of (B) with $\mathbf{r}\prime$ and (C) with $\mathbf{r}″$ give us the other two equations in (F), which the interested reader will, of course, find in [3]. (Note: showing only the first of a set of three equations, as we have done here, will be typical in the remainder of the article. Analogous derivations produce the second and third equations in each set, and will be omitted.)

Using $\mathbf{r}-\mathbf{r}\prime =-\mathbf{r}″$, we can rearrange equation (A) and form the dot product with $\dot{\mathbf{r}}$ to obtain (1) $$\stackrel{..}{\mathbf{r}}·\dot{\mathbf{r}}+\frac{A+B}{r^3}\mathbf{r}·\dot{\mathbf{r}}+C\left(\frac{\mathbf{r}\prime }{r{\prime }^3}-\frac{\mathbf{r}″}{r\prime {\prime }^3}\right)·\dot{\mathbf{r}}=0.$$(1)

The relationships $\mathbf{r}·\mathbf{r}=r^2$ and $\dot{\mathbf{r}}·\dot{\mathbf{r}}=u^2$ can be differentiated to obtain $\mathbf{r}·\dot{\mathbf{r}}=r\frac{dr}{dt}$ and $\dot{\mathbf{r}}·\stackrel{..}{\mathbf{r}}=u\frac{du}{dt}$. By defining R via $$\frac{dR}{dt}=2\left(\frac{\mathbf{r}\prime }{r{\prime }^3}-\frac{\mathbf{r}″}{r\prime {\prime }^3}\right)·\dot{\mathbf{r}},$$equation (1)(1) $$\stackrel{..}{\mathbf{r}}·\dot{\mathbf{r}}+\frac{A+B}{r^3}\mathbf{r}·\dot{\mathbf{r}}+C\left(\frac{\mathbf{r}\prime }{r{\prime }^3}-\frac{\mathbf{r}″}{r\prime {\prime }^3}\right)·\dot{\mathbf{r}}=0.$$(1) becomes $$u\frac{du}{dt}+\frac{A+B}{r^2}\frac{dr}{dt}+\frac{1}{2}C\frac{dR}{dt}=0,$$ which can be integrated to obtain $$u^2=2\frac{A+B}{r}-CR.$$

Setting $R=-\frac{2}{r}-Q$ gives us the first of a set of three equations that Lagrange labels (J): (J) $$u^2=2\frac{A+B+C}{r}+CQ.$$(J)

The other two equations in (J) describe $u{\prime }^2$ and $u\prime {\prime }^2$ and come from similar operations on equations (B) and (C), along with similar definitions of $R\prime ,R″,Q\prime$, and $Q″$.

The relationship between R and Q gives us $$\frac{dQ}{dt}=\frac{2}{r^2}\frac{dr}{dt}-\frac{dR}{dt}=\frac{2}{r^3}r\frac{dr}{dt}-2\left(\frac{\mathbf{r}\prime }{r{\prime }^3}-\frac{\mathbf{r}″}{r\prime {\prime }^3}\right)·\dot{\mathbf{r}}.$$

Using $r\frac{dr}{dt}=\mathbf{r}·\dot{\mathbf{r}}$ and $\mathbf{r}″=\mathbf{r}\prime -\mathbf{r}$ (and the map legend!), this can be reduced to (2) $$\frac{dQ}{dt}=2\left(\frac{1}{r^3}-\frac{1}{r\prime {\prime }^3}\right)\mathbf{r}·\dot{\mathbf{r}}-2\left(\frac{1}{r{\prime }^3}-\frac{1}{r\prime {\prime }^3}\right)\mathbf{r}\prime ·\dot{\mathbf{r}}=2(q\prime \mathbf{r}-q\mathbf{r}\prime )·\dot{\mathbf{r}}.$$(2)

By defining ρ via $\frac{d\rho }{dt}=\mathbf{r}\prime ·\dot{\mathbf{r}}-\mathbf{r}·\dot{\mathbf{r}}\prime$ and making several substitutions from the map legend, one also gets (3) $$q\prime \frac{dp\prime }{dt}-q″\frac{dp″}{dt}-q\frac{d\rho }{dt}=2(q\prime \mathbf{r}-q\mathbf{r}\prime )·\dot{\mathbf{r}},$$(3) so (2) and (3) easily combine to produce the first of three equations in Lagrange’s (I): (I) $$\frac{dQ}{dt}=q\prime \frac{dp\prime }{dt}-q″\frac{dp″}{dt}-q\frac{d\rho }{dt}.$$(I)

By now, the reader should anticipate this observation: the second and third equations that make up (I) describe $\frac{dQ\prime }{dt}$ and $\frac{dQ″}{dt}$ and are derived analogously.

Next, the first equation in (F) and the first equation in (J) can be combined (using the map legend, of course) to produce the first of three equations in Lagrange’s (K): (K) $$\frac{1}{2}\frac{d^2}{d{t}^2}{r}^2-\frac{A+B+C}{r}-C(p\prime q\prime -p″q″+Q)=0.$$(K)

And, of course, equations for $\frac{d^2}{d{t}^2}r{\prime }^2$ and $\frac{d^2}{d{t}^2}r\prime {\prime }^2$ follow from the second and third equations in (F) and (J).

The relationship $\frac{d\rho }{dt}=\mathbf{r}\prime ·\dot{\mathbf{r}}-\mathbf{r}·\dot{\mathbf{r}}\prime$ can be differentiated to obtain $$\frac{d^2\rho }{d{t}^2}=\mathbf{r}\prime ·\stackrel{..}{\mathbf{r}}-\mathbf{r}·\stackrel{..}{\mathbf{r}}\prime .$$

Using (A) and (B) to replace $\stackrel{..}{\mathbf{r}}$ and $\stackrel{..}{\mathbf{r}}\prime$, respectively, and several reductions from the map legend, we can simplify this to obtain Lagrange’s equation (H): (H) $$\frac{d^2\rho }{d{t}^2}+\mathit{Cpq}-Bp\prime q\prime -Ap″q″=0.$$(H)

Here we have broken the pattern that we have grown accustomed to, for equation (H) stands alone, rather than being the first of a set of three equations.

5 A problem in linear algebra

The last of the preliminary formulas that Lagrange assembles before his grand finish is also the most challenging in its algebraic derivation. But in a page of his work filled with complicated algebra, computations that are unmistakably Cramer’s rule for solving a linear system jump out at the modern reader, offering a suggestion as to how to proceed. Of course, Lagrange uses no matrix, determinant, or vector notation, but we can once again take advantage of such notation to represent his work.

We begin by setting $\mathbf{r}\prime ·\dot{\mathbf{r}}=\frac{dV}{dt}$ and $\dot{\mathbf{r}}\prime ·\mathbf{r}=\frac{dV\prime }{dt}$. Now the relationships $$\begin{array}{c}\mathbf{r}·\dot{\mathbf{r}}=r\frac{dr}{dt}\mathbf{r}·\mathbf{r}=r^2\\ {\mathbf{r}}^{\mathbf{\prime }}·\dot{\mathbf{r}}=\frac{dV}{dt}\hspace{1em}\text{and \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}{\mathbf{r}}^{\mathbf{\prime }}·\mathbf{r}=p^{″}\\ {\dot{\mathbf{r}}}^{\prime }·\dot{\mathbf{r}}={\nu }^{″}{\dot{\mathbf{r}}}^{\prime }·\mathbf{r}=\frac{d{V}^{\prime }}{dt}\end{array}$$can be written in matrix form as $\mathbf{A}\dot{\mathbf{r}}={\mathbf{b}}_1$ and $\mathbf{Ar}={\mathbf{b}}_2$, where $\mathbf{A}$ has rows $\mathbf{r},\mathbf{r}\prime$, and $\dot{\mathbf{r}}\prime$, and ${\mathbf{b}}_1$ and ${\mathbf{b}}_2$ are given by ${(r\frac{dr}{dt},\frac{dV}{dt},\nu ″)}^T$ and ${(r^2,p″,\frac{dV\prime }{dt})}^T$, respectively. Lagrange computes $\delta =\det \mathbf{A}$, then solves these two systems via Cramer’s rule. We shall proceed in a rather different manner.

The product $\mathbf{A}{\mathbf{A}}^T$ is given by $$\mathbf{A}{\mathbf{A}}^T=\left(\begin{array}{ccc}\mathbf{r}·\mathbf{r}& \hspace{1em}\mathbf{r}·\mathbf{r}\prime & \hspace{1em}\mathbf{r}·\dot{\mathbf{r}}\prime \\ \mathbf{r}\prime ·\mathbf{r}& \hspace{1em}\mathbf{r}\prime ·\mathbf{r}\prime & \hspace{1em}\mathbf{r}\prime ·\dot{\mathbf{r}}\prime \\ \dot{\mathbf{r}}\prime ·\mathbf{r}& \hspace{1em}\dot{\mathbf{r}}\prime ·\mathbf{r}\prime & \hspace{1em}\dot{\mathbf{r}}\prime ·\dot{\mathbf{r}}\prime \end{array}\right)=\left(\begin{array}{ccc}{r}^2& p″& \frac{dV\prime }{dt}\\ p″& r{\prime }^2& r\prime \frac{dr\prime }{dt}\\ \frac{dV\prime }{dt}& r\prime \frac{dr\prime }{dt}& u{\prime }^2\end{array}\right).$$

Lagrange’s solution incorporates the nine cofactors of our matrix $\mathbf{A}$, which he identifies as $\alpha ,\beta ,\gamma ,\alpha \prime ,\beta \prime ,\gamma \prime ,\alpha ″,\beta ″,\gamma ″$. These are the entries of the adjoint matrix $\text{adj} \mathbf{A}$, which allows us to make use of the matrix identity $$\text{adj} (\mathbf{A}{\mathbf{A}}^T)={(\text{adj} \mathbf{A})}^T\text{adj} \mathbf{A}.$$

This identity appears in Lagrange’s work as a collection of scalar equations. For example, with α, β, and γ in the first column of $\text{adj} \mathbf{A}$, equating the row 1, column 1 entry on each side of the matrix identity gives us $$r{\prime }^{2}u{\prime }^2-{\left(r\prime \frac{dr\prime }{dt}\right)}^2={\alpha }^2+{\beta }^2+{\gamma }^2,$$the first of six scalar equations in his work. (One might expect a 3 × 3 matrix equation to produce nine scalar equations, but since $\text{adj} (\mathbf{A}{\mathbf{A}}^T)$ is symmetric, three of the equations are redundant.)

From $\mathbf{A}\dot{\mathbf{r}}={\mathbf{b}}_1$ and the familiar formula relating $\text{adj} \mathbf{A}$ to the matrix inverse, we find $\delta \dot{\mathbf{r}}=(\text{adj} \mathbf{A}){\mathbf{b}}_1$, so that $$\begin{array}{c}\left|\right|\delta \dot{\mathbf{r}}|{|}^2=||(\text{adj} \mathbf{A}){\mathbf{b}}_1|{|}^2={\left((\text{adj} \mathbf{A}){\mathbf{b}}_1\right)}^T(\text{adj} \mathbf{A}){\mathbf{b}}_1\\ ={\mathbf{b}}_1^T{\left(\text{adj} \mathbf{A}\right)}^T\left(\text{adj} \mathbf{A}\right){\mathbf{b}}_1={\mathbf{b}}_1^T\text{adj} (\mathbf{A}{\mathbf{A}}^T){\mathbf{b}}_1,\end{array}$$or simply $${\delta }^2{u}^2={\mathbf{b}}_1^T\text{adj} (\mathbf{A}{\mathbf{A}}^T){\mathbf{b}}_1.$$

Likewise, from $\mathbf{Ar}={\mathbf{b}}_2$, we find $${\delta }^2{r}^2={\mathbf{b}}_2^T\text{adj} (\mathbf{A}{\mathbf{A}}^T){\mathbf{b}}_2.$$

Solving the second of these equations for ${\delta }^2$ and substituting it into the first gives us $$\frac{u^2}{r^2}{\mathbf{b}}_2^T\text{adj} (\mathbf{A}{\mathbf{A}}^T){\mathbf{b}}_2={\mathbf{b}}_1^T\text{adj} (\mathbf{A}{\mathbf{A}}^T){\mathbf{b}}_1.$$

By expressing $\text{adj} (\mathbf{A}{\mathbf{A}}^T),{\mathbf{b}}_1$, and ${\mathbf{b}}_2$ in terms of $r,r\prime ,p″,\nu ″$ etc., one finds that this matrix-vector equation is exactly Lagrange’s equation (N*) $$\begin{array}{c}(r^{2}r{\prime }^2-p\prime {\prime }^2)(u^{2}u{\prime }^2-\nu \prime {\prime }^2)+{\left(r\frac{dr}{dt}r\prime \frac{dr\prime }{dt}-\frac{dV}{dt}\frac{dV\prime }{dt}\right)}^2\\ -\left[r^2{\left(r\prime \frac{dr\prime }{dt}\right)}^2-2p″r\prime \frac{dr\prime }{dt}\frac{dV\prime }{dt}+r{\prime }^2{\left(\frac{dV\prime }{dt}\right)}^2\right]u^2\\ -\left[r{\prime }^2{\left(r\frac{dr}{dt}\right)}^2-2p″r\frac{dr}{dt}\frac{dV}{dt}+r^2{\left(\frac{dV}{dt}\right)}^2\right]u{\prime }^2\\ -2\left[p″\left(r\frac{dr}{dt}r\prime \frac{dr\prime }{dt}+\frac{dV}{dt}\frac{dV\prime }{dt}\right)-r^{2}r\prime \frac{dr\prime }{dt}\frac{dV\prime }{dt}-r{\prime }^{2}r\frac{dr}{dt}\frac{dV\prime }{dt}\right]\nu ″=0.\end{array}$$(N*)

(One might want to enlist the help of Maple or similar software to verify the final step in transforming the matrix-vector equation into the equivalent scalar equation!)

A few comments are in order here. Lagrange expresses the equation that we have labeled (N*) in differential notation instead of derivative notation, as was common in the 18th century. He does not label this equation. Also, by means of utterly heroic algebra and the introduction of still more symbols, Lagrange transforms this equation into an equivalent form that he labels (N). This equation (N) is central to the final steps in his solution, but the equivalent (N*) can be used in its place.

6 Lagrange’s solution and the Lagrange points

At last all the machinery is in place for the grand finish. Lagrange writes, “The first case that presents itself is that where the three distances r, $r\prime$, and $r″$ are constant.” Immediately this tells us that $p,p\prime ,p″,q,q\prime$, and $q″$ are constant as well, and that the derivatives of all of these are zero. Now from (K) we see that Q is constant, so it is clear from (I) that $q\frac{d\rho }{dt}=0$. In the same manner, the second and third equations in (K) together with the second and third equations in (I) (which we have not shown) combine to give us $q\prime \frac{d\rho }{dt}=0$ and $q″\frac{d\rho }{dt}=0$. Therefore we must have either $q=q\prime =q″=0$ or $\frac{d\rho }{dt}=0$.

The first case is easily addressed. If $q=q\prime =q″=0$, it follows immediately from the map legend that $r=r\prime =r″$. Lagrange concludes that “the system of the three bodies can move so that the three bodies always form an equilateral triangle.” He goes on to show that “the three bodies will always necessarily be in one given plane …” and that “the bodies B and C just rotate around body A with a constant angular velocity ….” This describes the sun-earth system, with the third body at one of the Lagrange points L₄ or L₅, as shown in Figure 1.

The second case, which will be of greater interest to us, requires $\frac{d\rho }{dt}=0$ in addition to the constant values mentioned above. (So ρ is constant as well.) It is not hard to verify from our definitions of $p″,\frac{d\rho }{dt},\frac{dV}{dt}$, and $\frac{dV\prime }{dt}$ that $$\frac{dV}{dt}=\frac{1}{2}\left(\frac{dp″}{dt}+\frac{d\rho }{dt}\right)\text{ and }\frac{dV\prime }{dt}=\frac{1}{2}\left(\frac{dp″}{dt}-\frac{d\rho }{dt}\right).$$

Therefore, in the present case, $\frac{dV}{dt}=\frac{dV\prime }{dt}=0.$ Now all of the derivatives in equation (N*) are zero, so the equation reduces to (4) $$(r^{2}r{\prime }^2-p\prime {\prime }^2)(u^{2}u{\prime }^2-\nu \prime {\prime }^2)=0.$$(4)

The area of the triangle formed by the masses in Figure 2 is given by $$\mathcal{A}=\frac{1}{2}rr\prime \hspace{0.17em}\sin \hspace{0.17em}\zeta ″.$$

(Here $\mathcal{A}$ is not Lagrange’s notation—the single exception to our notational convention!) Now $$4{\mathcal{A}}^2=r^{2}r{\prime }^2(1-{\hspace{0.17em}\cos \hspace{0.17em}}^2\zeta ″)=r^{2}r{\prime }^2-{(rr\prime \hspace{0.17em}\cos \hspace{0.17em}\zeta ″)}^2=r^{2}r{\prime }^2-{(\mathbf{r}·\mathbf{r}\prime )}^2=r^{2}r{\prime }^2-p\prime {\prime }^2.$$

Lagrange applies Heron’s formula to obtain the area of the same triangle in the form $${\mathcal{A}}^2=\frac{1}{2}(r+r\prime +r″)·\frac{1}{2}(r\prime +r″-r)·\frac{1}{2}(r+r″-r\prime )·\frac{1}{2}(r+r\prime -r″).$$

One solution to (4) occurs when $r^{2}r{\prime }^2-p\prime {\prime }^2=0$, and from the two area formulas it is clear that this forces $$r=r\prime +r″\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\text{ or }\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}r\prime =r+r″\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\text{ or }\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}r″=r+r\prime ,$$each of which describes one of the so-called straight-line solutions. These correspond to L₁, L₂, and L₃, respectively, as shown in Figure 1. (One might consider these straight-line configurations as “degenerate triangles” with zero area.)

The other solution to (4) occurs when $u^{2}u{\prime }^2-\nu \prime {\prime }^2=0$. But Lagrange shows that when this second factor is zero, the first factor must also be zero, reducing this case to the case already considered. The details, of course, can be found in [3].

To complete the straight-line solution, Lagrange assumes that the masses are arranged in the order A, B, C, so that $r\prime =r+r″$. He then sets $r\prime =mr$, so that $r″=(m-1)r$, making $p=\frac{1}{2}(r{\prime }^2+r\prime {\prime }^2-r^2)=m(m-1)r^2$. Similarly $p\prime ,p″,q,q\prime$ and $q″$ can all be expressed in terms of m and r. When these values are substituted into equation (H) and the result is multiplied through by $m^2{(m-1)}^{2}r$, the r’s cancel out, and Lagrange obtains (d) $$C(-3m^2+3m-1)+B{m}^3(m^2-3m+3)-A{(m-1)}^2(1-m^3)=0.$$(d)

Equation (d) would be just the tool to use to determine where to locate the jwst, except for one thing: Lagrange got it wrong! Figure 3 shows Lagrange’s value for $q\prime$ from both [2] and [3]. There is certainly a typesetting error here, as the blurry but correct $\frac{1}{r^3}$ at the end of the first description of $q\prime$ becomes $\frac{1}{r^2}$ in the second. Pursuing the algebra leading to equation (d) makes it clear that Lagrange uses the correct $\frac{1}{r^3}$ value when he substitutes $q\prime$ into equation (H). But there is another error in the formula for $q\prime$, in both versions shown in Figure 3, that makes Lagrange’s equation (d) wrong. Do you see it? His algebra in converting $1-\frac{1}{{(m-1)}^3}$ to common denominator form is incorrect. This is not merely a typesetting error: Lagrange’s equation (d) clearly uses the incorrectly simplified fraction!

Fig. 3 Values of $q\prime$ in the straight-line configuration from the original publication of Lagrange’s paper (top) and from his collected works (bottom).

After correcting this error, equation (d) becomes (d*) $$C(-3m^2\mathrm{}+\mathrm{}3m\mathrm{}\text{-}\mathrm{}1)\mathrm{}+\mathrm{}B{m}^2(m^3\mathrm{}\text{-}\mathrm{}3m^2\mathrm{}+\mathrm{}3m\mathrm{}\text{-}\mathrm{}2)\mathrm{}\text{-}\mathrm{}A{(m\mathrm{}\text{-}\mathrm{}1)}^2(1\mathrm{}\text{-}\mathrm{}{m}^3)=0.$$(d*)

(The label (d) is Lagrange’s notation; (d*) is not.) Fortunately, at some point in the two-and-a-half centuries between Lagrange’s work and the launch of jwst, this error has been recognized and corrected. For example, an equation equivalent to (d*) appears in Roy [6], though in slightly different notation.

With this correction, it is straightforward to determine the distance from earth to the Lagrange point L₂ where jwst is located. Now r is the distance from sun to earth, $r″$ is the distance from earth to jwst, and $r\prime =r+r″$ is the distance from the sun to jwst. The mass of the sun is $A=1.9885\times {10}^{30}$ kg, the mass of the earth is $B=5.9724\times {10}^{24}$ kg, and the mass of jwst is $C\approx 0$. With these numbers, the real root of (d*) is $m\approx 1.0100374$. Setting r = 1 au, we find $r″=(m-1)r=0.0100374$ au which is approximately 1,502,000 km, or 933,000 miles.

Lagrange addressed one last detail in his essay [3, p. 280]:

We have supposed above that the radii $r,r\prime ,r″$ were constant, and we have seen that that can only occur in two cases, namely: when the three radii are all equal, and when one of them is equal to the sum of the other two. Let us suppose now that these three radii are only in a constant ratio among themselves, and let us see how that condition can occur.

He goes on to show that this slightly weaker assumption produces exactly the same results: the only solutions are the equilateral triangle and straight line configurations, and in the latter case, equation (d) (well, actually (d*)) still holds. For the sun-earth-jwst problem, r varies from approximately 0.983 au to 1.0167 au, and the corresponding values of $r″$ vary from 917,000 to 949,000 miles. The work is complete!

7 Conclusion

The mathematical tools that Lagrange uses to arrive at a solution to this three-body problem lie entirely within the scope of modern courses in algebra, trigonometry, and first-semester calculus. But surely no ordinary person could have pursued the many extraordinarily complicated threads in his work to their ends, let alone woven them together into a magnificent solution to the problem as he has done. Lagrange noted in the introduction to his essay, “This research is really no more than for pure curiosity …” If only he could have watched on Christmas Day as the James Webb Space Telescope began its journey to the Lagrange point L₂!

Acknowledgment

The author wishes to thank J. R. Stockton, who made this article possible by providing a translation of Lagrange’s work.

Additional information

Notes on contributors

Donald Teets

Donald Teets received his Doctor of Arts (D.A.) in mathematics from Idaho State University in 1988, and has taught at the South Dakota School of Mines and Technology since then. He received the Carl B. Allendoerfer writing award from the MAA in 2000 for an article on the astronomical work of Gauss, and the Burton W. Jones Distinguished Teaching award from the Rocky Mountain Section of the MAA in 2004.