While the classical gravitational *n*-body problem is in general far from soluble, there is a family of particular solutions in the plane that is unique in this regard. If the masses are placed at the vertices of a regular *n*-gon, there is a frequency of rotation that always leads to a solution to the equations of motion. The three-body version was first solved by Lagrange, and the appellation “choreography” indicates that the masses follow one another equally spaced along the orbit.

The Lagrangian for a planar gravitational *n*-body problem is

$L=\frac{m}{2}\sum _{i}({\stackrel{\xb7}{x}}_{i}^{2}+{\stackrel{\xb7}{y}}_{i}^{2})+\sum _{\mathrm{pairs}}\frac{k}{{r}_{ij}}\phantom{\rule{3em}{0ex}}{r}_{ij}=\sqrt{({x}_{i}-{x}_{j}{)}^{2}+({y}_{i}-{y}_{j}{)}^{2}}$

with the corresponding equations of motion

$m{\stackrel{\xb7\xb7}{x}}_{i}=k\sum _{j\ne i}\frac{{x}_{j}-{x}_{i}}{{r}_{ij}^{3}}\phantom{\rule{5em}{0ex}}m{\stackrel{\xb7\xb7}{y}}_{i}=k\sum _{j\ne i}\frac{{y}_{j}-{y}_{i}}{{r}_{ij}^{3}}$

A rotating regular *n*-gon of unit radius has vertices at the coordinates

$$ \begin{array}{l}{x}_{i}=cos(\omega t+{\phi}_{i})\\ {y}_{i}=sin(\omega t+{\phi}_{i})\end{array}\phantom{\rule{4em}{0ex}}{\phi}_{i}=\frac{2\pi}{n}i\phantom{\rule{1em}{0ex}},\phantom{\rule{1em}{0ex}}i=1,\dots ,n$$

and sides of length $l=2sin\left(\frac{\pi}{n}\right)$ . The squares of the separations be between vertices are

${r}_{ij}^{2}=2-2cos({\phi}_{i}-{\phi}_{j})$

Inserting these quantities into the equations of motion and equating coefficients leads to the two apparently different conditions

$\begin{array}{l}-m{\omega}^{2}cos{\phi}_{i}=k\sum _{j\ne i}\frac{cos{\phi}_{j}-cos{\phi}_{i}}{{r}_{ij}^{3}}\\ -m{\omega}^{2}sin{\phi}_{i}=k\sum _{j\ne i}\frac{sin{\phi}_{j}-sin{\phi}_{i}}{{r}_{ij}^{3}}\end{array}$

which are seen to be equivalent when all angles are shifted by $\frac{\pi}{2}$ . Taking $i=n$ in the first form for convenience, the square of the frequency of rotation that produces a solution to the equations of motion is

$\begin{array}{l}{\omega}^{2}=\frac{k}{2m}\sum _{j=1}^{n-1}\frac{1}{\sqrt{2-2cos{\phi}_{j}}}\\ \phantom{{\omega}^{2}}=\frac{k}{m}[\sum _{j=1}^{int\left(\frac{n-1}{2}\right)}\frac{1}{\sqrt{2-2cos{\phi}_{j}}}+\frac{1}{4}{\delta}_{\left(n\phantom{\rule{.1em}{0ex}}\mathrm{even}\right)}]\end{array}$

Plotting the first few values of the frequency produces a rather pleasantly continuous-looking behavior:

The points are nicely interpolated for small *n* by an expression proportional to
$(n-1{)}^{0.63}$ ,
but this does not hold for large values of the argument. One implication of this graphic is that if a continuous ring of matter were held together solely by gravitational attraction, it would have to rotate at an infinite speed.

Another way to exhibit the values of frequency is with a visualization of the action as a function of the number of bodies and frequency. Since extrema of the action are only physically significant when the endpoints of the motion are fixed, here in terms of the temporal variable, periods of all solutions need to 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}\phantom{\rule{4em}{0ex}}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:

${L}_{\mathrm{normalized}}=\frac{n}{2}{w}^{4/3}+{w}^{-2/3}\sum _{\mathrm{pairs}}\frac{1}{{r}_{ij}}$

The distances between vertices in the sum over pairs are evaluated as for the sides of the *n*-gon, with an additional multiplier in the argument. There are *n* such distances for each increasing step over half of the *n*-gon, along with separations across the figure for *n* even:

$\sum _{\mathrm{pairs}}\frac{1}{{r}_{ij}}=\sum _{j=1}^{int\left(\frac{n-1}{2}\right)}\frac{n}{2sin\left(\frac{\pi j}{n}\right)}+\frac{n}{4}{\delta}_{\left(n\phantom{\rule{.1em}{0ex}}\mathrm{even}\right)}$

The action is then simply

$S=2\pi {L}_{\mathrm{normalized}}$

and a two-dimensional plot of the action, omitting the overall factor of *n*, for a given number of bodies is

Since the action is surprisingly flat over much of its domain, the locations of frequency that minimize the action stand out more clearly in plot of its numerical derivative:

The points where the derivative crosses the *x*-axis match as expected with the frequencies evaluated as a solution to the equations of motion.

Since the distances between bodies are constant, the potential for this system is identically constant. This appears at first to be a bit strange, since a constant potential normally implies at most rectilinear motion. In this system the centrifugal tendencies of the bodies balance the gravitational attractions for uniform circular motion in a constant potential. This is akin to the circular orbits that exist for any central potential.

The energy of the system is easily evaluated,

$\begin{array}{l}E=\frac{m}{2}\sum _{i}({\stackrel{\xb7}{x}}_{i}^{2}+{\stackrel{\xb7}{y}}_{i}^{2})-\sum _{\mathrm{pairs}}\frac{k}{{r}_{ij}}\\ \phantom{E}=\frac{m}{2}n{\omega}^{2}-nk[\sum _{j=1}^{int\left(\frac{n-1}{2}\right)}\frac{1}{2sin\left(\frac{\pi j}{n}\right)}+\frac{1}{4}{\delta}_{\left(n\phantom{\rule{.1em}{0ex}}\mathrm{even}\right)}]\\ E=\frac{nk}{2}\left[\sum _{j=1}^{int\left(\frac{n-1}{2}\right)}\right(\frac{1}{\sqrt{2-2cos\left(\frac{2\pi j}{n}\right)}}-\frac{1}{sin\left(\frac{\pi j}{n}\right)})-\frac{1}{4}{\delta}_{\left(n\phantom{\rule{.1em}{0ex}}\mathrm{even}\right)}]\end{array}$

and the energy per body has this behavior:

An implication of this graphic is that the continuous ring of matter rotating at infinite speed would have infinite binding energy.

*Uploaded 2018.06.24 — Updated 2018.07.21*
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