n-Body Simple Harmonic Oscillator Choreographies

As show in a previous presentation, n-body simple harmonic oscillators are always soluble. This presentation will consider analogs of gravitational choreographies, solutions in which all bodies follow the same orbit at equally spaced intervals.

As developed in the presentation indicated, the general solution for an arbitrary number of equal masses interacting pairwise simple harmonically with equal frequencies is

$[\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \end{array}] = [\frac{J}{n} - \frac{M}{n} cos (\sqrt{n} ω t)] [\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \end{array}]_{0} - \frac{M}{n^{3 / 2} ω} sin (\sqrt{n} ω t) [\begin{array}{c} {\overset{\cdot}{x}}_{1} \\ {\overset{\cdot}{x}}_{2} \\ ⋮ \end{array}]_{0}$

with matrices defined by

$M = J - n I J = [\begin{array}{c} 1 & 1 & \dots \\ 1 & 1 & \dots \\ ⋮ & ⋮ & ⋱ \end{array}]$

This solution holds for all Cartesian coordinates, and so represents a general solution in an arbitrary number of dimensions. The behavior of the coordinates differs only in initial conditions.

For comparison with Lagrange choreographies, it is instructive to set $ω^{2} = \frac{2 k}{m}$ and then write the solution as

$[\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \end{array}] = [\frac{J}{n} - \frac{M}{n} cos (\sqrt{\frac{2 n k}{m}} t)] [\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \end{array}]_{0} - \frac{M}{n \sqrt{2 n k / m}} sin (\sqrt{\frac{2 n k}{m}} t) [\begin{array}{c} {\overset{\cdot}{x}}_{1} \\ {\overset{\cdot}{x}}_{2} \\ ⋮ \end{array}]_{0}$

The factor of $\sqrt{n}$ in the argument of the circular functions corresponds to a summation over angular variables when $p = 2$ in the presentation just referenced. In the notation of that presentation, with $i = n$ one has

$\begin{array}{l} ω^{2} = \frac{k}{m a} \sum_{j = 1}^{n - 1} r_{i j}^{2} = \frac{2 k}{m} \sum_{j = 1}^{n - 1} (1 - cos φ_{j}) \\ = \frac{2 k}{m} [n - 1 - \sum_{j = 1}^{n - 1} cos \frac{2 π j}{n}] \\ ω^{2} = \frac{2 k}{m} [n - \sum_{j = 1}^{n} cos \frac{2 π j}{n}] = \frac{2 n k}{m} \end{array}$

The sum in the last line is zero due to the symmetry of sines and cosines considered at equally spaced points on a circle. This is equivalent to summing over the roots of unity using the geometric series:

$\sum_{j = 1}^{n} e^{\pm 2 π i j / n} = \sum_{j = 0}^{n - 1} e^{\pm 2 π i j / n} = \frac{1 - e^{\pm 2 π i n / n}}{1 - e^{\pm 2 π i / n}} = 0$

Since sines and cosines are linearly dependent upon exponentials, their sums over roots of unity are also zero. The same applies for sums over any points equally spaced on a circle because a constant offset in phase translates to a linear combination of sines and cosines.

As also pointed out in the referenced presentation, the interaction coefficient k as well as the mass are free parameters that can be used to set the rate of evolution of choreographies to come. They can both be set to unity for convenience.

The solution as given requires a fixed center of mass, since there is no term linear in time. In the context of choreographies it is natural to put the center of mass at the origin. This choice makes any term proportional to J irrelevant since it then produces zero when multiplying initial positions and velocities. The solution for this choice simplifies to

$[\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \end{array}] = cos (\sqrt{\frac{2 n k}{m}} t) [\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \end{array}]_{0} + \frac{1}{\sqrt{2 n k / m}} sin (\sqrt{\frac{2 n k}{m}} t) [\begin{array}{c} {\overset{\cdot}{x}}_{1} \\ {\overset{\cdot}{x}}_{2} \\ ⋮ \end{array}]_{0}$

One can now see how Lagrange choreographies arise: if one seeks solutions of the form $cos (Ω t + φ_{i})$ then simply set

$x_{i} = cos φ_{i} {\overset{\cdot}{x}}_{i} = - Ω sin φ_{i} = - \sqrt{\frac{2 n k}{m}} sin φ_{i}$

where the phases $φ_{i}$ are those of the roots of unity. Then solution again simplifies as expected:

$[\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \end{array}] = [\begin{array}{c} cos (\sqrt{\frac{2 n k}{m}} t + φ_{1}) \\ cos (\sqrt{\frac{2 n k}{m}} t + φ_{2}) \\ ⋮ \end{array}]$

For solutions in between cosines and sines, include a constant offset in the phases, so for example

$cos (Ω t + φ_{i} - \frac{π}{2}) = sin (Ω t + φ_{i})$

Sums over initial conditions are automatically zero as required to keep the center of mass unmoving at the origin.

Now for a little fun! Since the base solution holds for each Cartesian coordinate in turn, one can mathematically construct a one-dimensional choreography. With $k = m = 1$ for convenience, it looks like this:

The offset φ_x determines where the red body starts in the choreography. Since this is an idealization, one should imagine the bodies as points that do not collide.

A more realistic system evolves in two dimensions, with individual phase offsets in either coordinate. The math is naturally the same for each coordinate direction. The shape of the resulting choreography is determined by the relative values of the offsets:

These choreographies interpolate between straight lines for offsets of the same absolute value, to the circular Lagrange choreographies with differences of π/2 mod π in the offsets.

The process can be continued for any number of dimensions. For three dimensions it looks like this:

And again, imagine these bodies as points that do not collide. One gets the same interpolation between straight lines and circular Lagrange choreographies as in two dimensions. This implies the same behavior in higher dimensions.

Unfortunately, this appears to be the limit of choreographies for the simple harmonic oscillator. With a single frequency in the system, there does not appear to be a way to generate more complicated choreographies, such as a figure eight. That would require anisotropies in the potential, which would imply nonphysical anisotropies in the underlying space.