Gravitational choreographies are periodic solutions of gravitational systems in which bodies traverse the same orbit at equally spaced intervals. Since gravitational systems with more than two masses are not analytically soluble in general, choreographies are particular solutions to the equations of motion that are highly dependent on initial conditions. A system that starts off entirely in a plane, with no velocity components normal to the plane, will remain in the plane and can be treated spatially as a two-dimensional problem.

For the three-body problem with equal masses, there are two rather simple such solutions. One is due to Euler for which the three masses are colinear, with one fixed at the origin and the other two circling at a constant rate on opposite ends of a diameter. The other due to Lagrange has the three masses circling their common origin at a constant rate, located at the vertices of a rotating equilateral triangle. One interesting question that can be posed is whether there is an analytic continuation between these two solutions that can be controlled by a single parameter.

The Lagrangian for a planar three-body gravitational system with equal masses is

$L=\frac{m}{2}({\stackrel{·}{x}}_1^2+{\stackrel{·}{y}}_1^2+{\stackrel{·}{x}}_2^2+{\stackrel{·}{y}}_2^2+{\stackrel{·}{x}}_3^2+{\stackrel{·}{y}}_3^2)+\frac{k}{r_{12}}+\frac{k}{r_{13}}+\frac{k}{r_{23}}$

where $k>0$ and the radial separations are

$\begin{array}{l}{r}_{12}=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}\\ r_{13}=\sqrt{(x_1-x_3{)}^2+(y_1-y_3{)}^2}\\ r_{23}=\sqrt{(x_2-x_3{)}^2+(y_2-y_3{)}^2}\end{array}$

The Lagrange equations

$\frac{d}{dt}\frac{\partial L}{\partial {\stackrel{·}{q}}_i}=\frac{\partial L}{\partial q_i}$

when evaluated for the given Lagrangian are

$\begin{array}{l}m{\stackrel{··}{x}}_1=-k\frac{x_1-x_2}{r_{12}^3}-k\frac{x_1-x_3}{r_{13}^3}\\ m{\stackrel{··}{x}}_2=k\frac{x_1-x_2}{r_{12}^3}-k\frac{x_2-x_3}{r_{23}^3}\\ m{\stackrel{··}{x}}_3=k\frac{x_1-x_3}{r_{13}^3}+k\frac{x_2-x_3}{r_{23}^3}\\ \\ m{\stackrel{··}{y}}_1=-k\frac{y_1-y_2}{r_{12}^3}-k\frac{y_1-y_3}{r_{13}^3}\\ m{\stackrel{··}{y}}_2=k\frac{y_1-y_2}{r_{12}^3}-k\frac{y_2-y_3}{r_{23}^3}\\ m{\stackrel{··}{y}}_3=k\frac{y_1-y_3}{r_{13}^3}+k\frac{y_2-y_3}{r_{23}^3}\end{array}$

It is worth keeping in mind that the denominators here contain cubes, not squares, of the separations. Adding triples of equations together gives

$\frac{d^2}{d{t}^2}(x_1+x_2+x_3)=0\frac{d^2}{d{t}^2}(y_1+y_2+y_3)=0$

so that the two sums are linear in time. If the center of mass of the system is taken to be the origin and total linear momentum equal to zero, one has the exact constraints

$x_1+x_2+x_3=0y_1+y_2+y_3=0$

The Euler and Lagrange choreographies are both circular motions at a constant rate, which indicates a temporal parametrization using circular functions. Let the parametrization take the form

$\begin{array}{l}{x}_1=acos\omega t\\ x_2=cos(\omega t+b)\\ x_3=cos(\omega t+c)\end{array}\begin{array}{l}{y}_1=asin\omega t\\ y_2=sin(\omega t+b)\\ y_3=sin(\omega t+c)\end{array}$

where the parameter a controls movement of one mass from the Lagrange circle to the Euler origin, and the parameters b and c set the relative phases of the other two masses on the circle. With this parametrization the exact constraints become

$\begin{array}{l}{x}_1+x_2+x_3=(a+cosb+cosc)cos\omega t-(sinb+sinc)sin\omega t\\ y_1+y_2+y_3=(a+cosb+cosc)sin\omega t+(sinb+sinc)cos\omega t\end{array}$

which can both be satisfied with the choices

$a=-2cosbc=-b$

So far so good! With these choices, the squares of the radial separations are

$\begin{array}{l}{r}_{12}^2=[acos\omega t-cos(\omega t+b){]}^2+[asin\omega t-sin(\omega t+b){]}^2\\ \phantom{r_{12}^2}=a^2+1-2acosb=8{cos}^{2}b+1\\ \\ r_{13}^2=[acos\omega t-cos(\omega t-b){]}^2+[asin\omega t-sin(\omega t-b){]}^2\\ \phantom{r_{13}^2}=a^2+1-2acosb=8{cos}^{2}b+1\\ \\ r_{23}^2=[cos(\omega t+b)-cos(\omega t-b){]}^2+[sin(\omega t+b)-sin(\omega t-b){]}^2\\ \phantom{r_{23}^2}=2-2cos2b=4{sin}^{2}b\end{array}$

which are all encouragingly temporally constant as one would want for the two end cases. Individual coordinate differences are

$\begin{array}{l}{x}_1-x_2=-3cosbcos\omega t+sinbsin\omega t\\ x_1-x_3=-3cosbcos\omega t-sinbsin\omega t\\ x_2-x_3=-2sinbsin\omega t\\ \\ y_1-y_2=-3cosbsin\omega t-sinbcos\omega t\\ y_1-y_3=-3cosbsin\omega t+sinbcos\omega t\\ y_2-y_3=2sinbcos\omega t\end{array}$

Now all that remains is to collect expressions and insert them in the Lagrange equations, trying to find a relationship between the constant rate ω and the parameter b. The equations in intermediate form are

$\begin{array}{l}m{\stackrel{··}{x}}_1=2m{\omega }^{2}cosbcos\omega t=k\frac{6cosbcos\omega t}{(8{cos}^{2}b+1{)}^{3/2}}\\ \\ m{\stackrel{··}{x}}_2=-m{\omega }^2(cosbcos\omega t-sinbsin\omega t)\\ \phantom{m{\stackrel{··}{x}}_2}=-k\frac{3cosbcos\omega t-sinbsin\omega t}{(8{cos}^{2}b+1{)}^{3/2}}+k\frac{2sinbsin\omega t}{8{sin}^{3}b}\\ \\ m{\stackrel{··}{x}}_3=-m{\omega }^2(cosbcos\omega t+sinbsin\omega t)\\ \phantom{m{\stackrel{··}{x}}_3}=-k\frac{3cosbcos\omega t+sinbsin\omega t}{(8{cos}^{2}b+1{)}^{3/2}}-k\frac{2sinbsin\omega t}{8{sin}^{3}b}\\ \\ m{\stackrel{··}{y}}_1=2m{\omega }^{2}cosbsin\omega t=k\frac{6cosbsin\omega t}{(8{cos}^{2}b+1{)}^{3/2}}\\ \\ m{\stackrel{··}{y}}_2=-m{\omega }^2(cosbsin\omega t+sinbcos\omega t)\\ \phantom{m{\stackrel{··}{y}}_2}=-k\frac{3cosbsin\omega t+sinbcos\omega t}{(8{cos}^{2}b+1{)}^{3/2}}-k\frac{2sinbcos\omega t}{8{sin}^{3}b}\\ \\ m{\stackrel{··}{y}}_3=-m{\omega }^2(cosbsin\omega t-sinbcos\omega t)\\ \phantom{m{\stackrel{··}{y}}_3}=-k\frac{3cosbsin\omega t-sinbcos\omega t}{(8{cos}^{2}b+1{)}^{3/2}}+k\frac{2sinbcos\omega t}{8{sin}^{3}b}\end{array}$

Comparing expressions on each side of each equation, the first and fourth equations, along with the first term of all remaining equations, give the relationship

${\omega }^2\stackrel{?}{=}\frac{3}{(8{cos}^{2}b+1{)}^{3/2}}\frac{k}{m}$

Unfortunately the second and third terms of the remaining equations give a consistent but different relationship:

${\omega }^2\stackrel{?}{=}[\frac{1}{(8{cos}^{2}b+1{)}^{3/2}}+\frac{1}{4{sin}^{3}b}]\frac{k}{m}$

A plot of the dimensionless parts of the right-hand sides of the two relationships indicates that they are only equal for two distinct points:

The two points of intersection are determined by

$\frac{2}{(8{cos}^{2}b+1{)}^{3/2}}=\frac{1}{4{sin}^{3}b}\to \begin{array}{c}cosb=\pm \frac{1}{2}\\ b=\frac{\pi }{3},\frac{2\pi }{3}\end{array}$

and both correspond to the Lagrange choreography: the first with the equilateral triangle initially pointing towards the left and the second initially pointing towards the right. The frequency for this choreography is

${\omega }_{\mathrm{Lagrange}}=\sqrt{\frac{k}{\sqrt{3}m}}$

and taking the second choice for b, the full parametrization is

$\begin{array}{l}{x}_1=cos{\omega }_{\mathrm{Lagrange}}t\\ x_2=cos({\omega }_{\mathrm{Lagrange}}t+\frac{2\pi }{3})\\ x_3=cos({\omega }_{\mathrm{Lagrange}}t-\frac{2\pi }{3})\end{array}\begin{array}{l}{y}_1=sin{\omega }_{\mathrm{Lagrange}}t\\ y_2=sin({\omega }_{\mathrm{Lagrange}}t+\frac{2\pi }{3})\\ y_3=sin({\omega }_{\mathrm{Lagrange}}t-\frac{2\pi }{3})\end{array}$

The Euler choreography requires the first mass to remain identically at the origin:

$a=-2cosb\equiv 0\to b=\frac{\pi }{2}$

The frequency of this choreography is determined from the second relationship, since the first drops out of all equations of motion,

${\omega }_{\mathrm{Euler}}=\sqrt{\frac{5k}{4m}}$

and the full parametrization is in this case

$\begin{array}{l}{x}_1=0\\ x_2=cos({\omega }_{\mathrm{Euler}}t+\frac{\pi }{2})\\ x_3=cos({\omega }_{\mathrm{Euler}}t-\frac{\pi }{2})\end{array}\begin{array}{l}{y}_1=0\\ y_2=sin({\omega }_{\mathrm{Euler}}t+\frac{\pi }{2})\\ y_3=sin({\omega }_{\mathrm{Euler}}t-\frac{\pi }{2})\end{array}$

One must unfortunately conclude that there is no single relationship as a function of the parameter b that interpolates between the two choreographies using circular functions. The two are isolated particular solutions in parameter space and this can be seen quite clearly in a graphical visualization of the action.

In order for the visualization to be accurate, keep in mind that extrema of the action are only physically significant when the endpoints of the motion are fixed, in this case in terms of the temporal variable. The dimensioned quantities can easily be absorbed into the time, but the chaotic nature of the three-body system allows a variety of periods of motion. The periods of all solutions can be normalized to the usual 2π of circular functions along with a spatial scaling of all coordinates:

$\begin{array}{c}{x}_i\left(\omega t\right)\to w^{2/3}{x}_i\left(t\right)\\ y_i\left(\omega t\right)\to w^{2/3}{y}_i\left(t\right)\end{array}w=\sqrt{\frac{m}{k}}\omega$

The exponent in the spatial scaling is determined from the equations of motion, and the parameter w is a dimensionless frequency. The Lagrangian representing normalized periods is evaluated with the replacement functions omitting dimensioned quantities:

$L_{\mathrm{normalized}}=w^{4/3}(2{cos}^{2}b+1)+w^{-2/3}(\frac{2}{\sqrt{8{cos}^{2}b+1}}+\frac{1}{2sinb})$

The action for the visualization is simply

$S=2\pi L_{\mathrm{normalized}}$

and looks like this:

It is now immediately clear that the Lagrange choreography is a solution stable in the plane that minimizes the action in two ways. The Euler choreography is an unstable solution on the saddle point between.

Uploaded 2018.05.25 — Updated 2018.06.21 analyticphysics.com