Gravitational choreographies are periodic solutions of gravitational systems in which all 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=m2 (x· 12 +y· 12 +x· 22 +y· 22 +x· 32 +y· 32) +kr12 +kr13 +kr23

where k>0 and the radial separations are

r12=( x1-x2 )2 +(y1 -y2 )2 r13=( x1-x3 )2 +(y1 -y3 )2 r23=( x2-x3 )2 +(y2 -y3 )2

The Lagrange equations

ddt L q· i =L qi

when evaluated for the given Lagrangian are

mx ··1 =kx1 -x2 r123 -kx1 -x3 r133 mx ··2 =kx1 -x2 r123 -kx2 -x3 r233 mx ··3 =kx1 -x3 r133 +kx2 -x3 r233   my ··1 =ky1 -y2 r123 -ky1 -y3 r133 my ··2 =ky1 -y2 r123 -ky2 -y3 r233 my ··3 =ky1 -y3 r133 +ky2 -y3 r233

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

d2 dt2 (x1 +x2 +x3) =0 d2 dt2 (y1 +y2 +y3) =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

x1+x2 +x3 =0 y1+y2 +y3 =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

x1=acosωt x2=cos( ωt+b) x3=cos( ωt+c) y1=asinωt y2=sin( ωt+b) y3=sin( ωt+c)

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

x1+x2 +x3 =(a+cosb +cosc) cosωt -(sinb +sinc) sinωt y1+y2 +y3 =(a+cosb +cosc) sinωt +(sinb +sinc) cosωt

which can both be satisfied with the choices

a=2cosb c=b

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

r122 =[acosωt -cos(ωt+b) ]2 +[asinωt -sin(ωt+b) ]2 r122 =a2+1 -2acosb =8cos2b+1   r132 =[acosωt -cos(ωt-b) ]2 +[asinωt -sin(ωt-b) ]2 r132 =a2+1 -2acosb =8cos2b+1   r232 =[cos(ωt+b) -cos(ωt-b) ]2 +[sin(ωt+b) -sin(ωt-b) ]2 r232 =2-2cos2b =4sin2b

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

x1-x2 =3cosbcosωt +sinbsinωt x1-x3 =3cosbcosωt -sinbsinωt x2-x3 =2sinbsinωt   y1-y2 =3cosbsinωt -sinbcosωt y1-y3 =3cosbsinωt +sinbcosωt y2-y3 =2sinbcosωt

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

mx ··1 =2mω2 cosbcosωt =k6cosb cosωt (8cos2b +1) 3/2   mx ··2 =mω2 (cosbcosωt -sinbsinωt) mx ··2 =k3cosb cosωt -sinbsinωt (8cos2b +1) 3/2 +k2sinb sinωt 8sin3b   mx ··3 =mω2 (cosbcosωt +sinbsinωt) mx ··3 =k3cosb cosωt +sinbsinωt (8cos2b +1) 3/2 -k2sinb sinωt 8sin3b   my ··1 =2mω2 cosbsinωt =k6cosb sinωt (8cos2b +1) 3/2   my ··2 =mω2 (cosbsinωt +sinbcosωt) my ··2 =k3cosb sinωt +sinbcosωt (8cos2b +1) 3/2 -k2sinb cosωt 8sin3b   my ··3 =mω2 (cosbsinωt -sinbcosωt) my ··3 =k3cosb sinωt -sinbcosωt (8cos2b +1) 3/2 +k2sinb cosωt 8sin3b

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

ω2=? 3(8cos2b +1) 3/2 km

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

ω2=? [1 (8cos2b +1) 3/2 +1 4sin3b ] km

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

2 (8cos2b +1) 3/2 =1 4sin3b cosb=±12 b=π3 , 2π3

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

ωLagrange =k 3m

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

x1 =cosωLagranget x2=cos( ωLagranget +2π3) x3=cos( ωLagranget -2π3) y1 =sinωLagranget y2=sin( ωLagranget +2π3) y3=sin( ωLagranget -2π3)

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

a=2cosb 0 b=π2

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

ωEuler =5k 4m

and the full parametrization is in this case

x1=0 x2=cos( ωEulert +π2) x3=cos( ωEulert -π2) y1=0 y2=sin( ωEulert +π2) y3=sin( ωEulert -π2)


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:

xi (ωt) w2/3 xi(t) yi (ωt) w2/3 yi(t) w=mk ω

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:

Lnormalized =w4/3 (2cos2b +1) +w2/3 (2 8cos2b+1 +12sinb)

The action for the visualization is simply

S=2πLnormalized

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