\documentclass[11pt]{article}
\usepackage{amstex}
\usepackage{amssymb}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{prop}{Proposition}[section]
\newtheorem{corr}[theorem]{Corollary}
\newcommand{\dsty}{\displaystyle}
\newcommand{\casefrac}[2]{{\textstyle{{#1}\over{#2}}}}
\pagestyle{plain}
\topmargin -.75in
\oddsidemargin -0.1in
\evensidemargin -0.1in
\textheight 9in
\textwidth 6.5in
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\title{A Continuum of Relative Equilibria in the $5$-Body Problem}
\author{Gareth E. Roberts \\ Math. Dept. Boston Univ. \\
111 Cummington Street \\ Boston, MA 02215 \\ garethr@@math.bu.edu
\\ (617) 353-1484}
\date{July 1998
\begin{abstract}
It is generally believed that the set of relative
equilibria equivalence classes in the Newtonian $n$-body problem,
for a given set of positive masses, is finite.
However, the result
has only been proven for $n=3$ and remains a difficult, open
question for $n \geq 4$ (Wintner~\cite{cc:wint}, Smale~\cite{cc:smale1}).
We demonstrate that the condition
for the masses being positive is a necessary one by finding a
continuum of relative equilibria in the five-body problem which (unfortunately)
includes one negative mass. This family persists in similar potential
functions,
including the logarithmic potential used to describe
the motion of point vortices in a plane of fluid.
\end{abstract}}
\maketitle
{\bf Keywords:} $n$-body problem, relative equilibria
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
A relative equilibrium is a
special configuration of masses of the $n$-body problem which
rotates rigidly about
its center of mass if given the correct initial momenta. In rotating
coordinates these special solutions become fixed points, hence the
name {\it relative equilibria}. These periodic orbits are
the only explicitly known
solutions of the $n$-body problem.
A natural question, posed by Wintner in~\cite{cc:wint}, is whether
the number of relative equilibria is finite for a given set of positive
masses. For $n=3$, there are only five relative equilibria for any set
of masses: the three collinear solutions of Euler~\cite{cc:euler} and
the two equilateral solutions
of Lagrange~\cite{cc:lagrange}. For $n=4$, Albouy recently classified all
the relative equilibria for equal
masses~\cite{cc:albouy}. There are essentially only four configurations
(50 if you include all possible permutations of the masses):
a collinear solution, a square,
an equilateral triangle with a body at the center and an isosceles triangle
with a body on the axis of symmetry.
For the collinear case, there are exactly $n!/2$ relative
equilibria~\cite{cc:moulton}.
While many examples of relative equilibria have been found for larger $n$,
no complete classification exists for four unequal masses or
for $n \geq 5$.
In fact, Smale poses Wintner's question
as one of his problems for the 21st century (see~\cite{cc:smale1}).
In this note, we show the existence of a one-parameter family of degenerate
relative equilibria for the 5-body problem which
contains a negative mass (ie. repulsive force). This
family contains a degeneracy which does not arise from rotation or scaling.
Due to the negative mass, it is not a counterexample to the
Wintner/Smale question, but does demonstrate the necessity of
having positive masses.
Moreover, relative equilibria can be defined for other potential
functions and in some of these cases, ``negative masses''
are physically reasonable.
If we generalize the Wintner/Smale question to other potential functions,
including the logarithmic potential for the motion of point
vortices (often referred to
as the Kirchhoff problem~\cite{kirchhoff},~\cite{Oneil}),
this family persists. Although having a negative mass in the gravitational
$n$-body problem is not
physically realistic, having a negative circulation in the Kirchhoff problem
is physically reasonable.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Relative Equilibria}
We let the mass and position of the $n$ bodies be given by $m_i$ and
${\bf q}_i \in \mathbb{R}^2$, $i = 1, \ldots , n$. Let $r_{ij} =
\|{\bf q}_i- {\bf q}_j\|$ be the distance between the $i$th and $j$th
bodies and
let ${\bf q} =
({\bf q}_1, \ldots , {\bf q}_n) \in \mathbb{R}^{2n}$.
Using Newton's law of motion and the inverse square law for attraction
due to gravity, the
second-order equation for the $i$th body is given by
$$
m_i \ddot{{\bf q}_i} \; = \; \sum_{i \neq j} \;
\frac{m_i m_j ({\bf q}_j- {\bf q}_i)}{r_{ij}^3}
\; = \; \frac{\partial U}{\partial {\bf q}_i},
$$
where $U({\bf q})$ is the Newtonian potential function:
$$
U({\bf q}) = \sum_{i < j} \; \frac{m_i m_j}{r_{ij}}.
$$
We let the momenta of each body
be ${\bf p}_i = m_i \dot{{\bf q}_i}$ and let
${\bf p} =
({\bf p}_1, \ldots , {\bf p}_n) \in \mathbb{R}^{2n}$.
The equations of motion can then be written as
\begin{eqnarray}
\dot{\bf q} & = & M^{-1} {\bf p} \nonumber \\
\dot{\bf p} & = & \nabla U ({\bf q}) \label{eq:Hamilt1}
\end{eqnarray}
where $M$ is the diagonal mass matrix with diagonal
$m_1, m_1, m_2, m_2, \ldots , m_n, m_n$.
A {\it relative equilibrium} of the $n$-body problem is a configuration
${\bf x} \in \mathbb{R}^{2n}$ which satisfies the algebraic
equations
\begin{equation}
\nabla U ({\bf x}) + \omega^2 \, M {\bf x} = 0 \label{eq:cc}
\end{equation}
for some value of $\omega$ (see \cite{cc:meyer},
\cite{cc:schmidt}, \cite{cc:rick2},
or \cite{cc:smale2} for details).
The $i$th component in~(\ref{eq:cc}) is given by
$$
\omega^2 \, m_i {\bf x}_i \; = \; \sum_{i \neq j} \;
\frac{m_i m_j ({\bf x}_j- {\bf x}_i)}{r_{ij}^3}.
$$
Note that if
${\bf x} = ({\bf x}_1, {\bf x}_2, \ldots ,{\bf x}_n)$ is a relative
equilibrium, then
\begin{eqnarray*}
{c \bf x} &=& (c {\bf x}_1, c {\bf x}_2, \ldots , c {\bf x}_n)
\quad \mbox{and} \\
{R \bf x} &=& (R {\bf x}_1, R {\bf x}_2, \ldots ,R {\bf x}_n)
\end{eqnarray*}
are also relative equilibria for any constant $c$ and any $R \in SO(2)$.
When counting relative equilibria it is standard to fix the size
(a unique value of $c$)
and identify any configurations which are rotationally
equivalent via the equivalence relation ${\bf x} \sim R {\bf x}$ for
$R \in SO(2)$. Indeed, if we let
$I({\bf x})$ denote the moment of inertia, that is,
$$
I({\bf x}) = \frac{1}{2} \sum_{i=1}^{n} \; m_i \| {\bf x}_i \|^2,
$$
we can then write equation~(\ref{eq:cc}) as
$$
\nabla U ({\bf x}) + \omega^2 \, \nabla I ({\bf x}) = 0.
$$
In other words,
relative equilibria are critical points of the function $U([ {\bf x} ])$
restricted to the mass ellipsoid $I=c$, where $[ {\bf x} ]$ is the equivalence
class of ${\bf x}$
and $\omega^2$ plays the role of a Lagrange multiplier.
Thus any question on the number of relative equilibria is really a
question on the
number of critical points of $U([ {\bf x} ])$ restricted to an inertia
manifold.
If this number were not finite, then a continuum of relative equilibria
would exist~\cite{cc:palmore2}.
(Here we mean a continuum other than one arising from scaling or rotation.)
In other words, is it possible to find a continuum of
critical points of $U([ {\bf x} ])$ restricted
to the mass ellipsoid $I=c$?
While we cannot answer this question in the
affirmative, we do find such a continuum, but containing a negative mass.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{$1+$Rhombus Relative Equilibria}
The easiest and most accessible relative equilibria are those
configurations which contain large amounts of symmetry. For example,
placing $n$ equal masses at the vertices of a regular $n$-gon and
adding a body of mass $m$ at its center is a relative equilibria
for all values of the central mass $m$.
We begin our search for a family of degenerate relative equilibria
by considering a configuration which consists
of four bodies at the vertices of a rhombus, with opposite vertices
having the same mass, and a central body. There will be three parameters
in this family: the shape of the rhombus, the mass ratio of the bodies
on the rhombus and the mass of the central body.
We let $m_1 = 1, m_2 = m, m_3 = 1, m_4 = m, m_5 = p$ be the five
masses and ${\bf q}_1 = (1,0),
{\bf q}_2 = (0,k), {\bf q}_3 = (-1,0), {\bf q}_4 = (0,-k), {\bf q}_5 = (0,0)$
be the positions of the five bodies.
We will let $c=\sqrt{k^2+1}$ be the side of the rhombus (see Figure~1).
\begin{figure}
\begin{picture}(200,200)(-120,0)
\put(100,200){\line(0,-1){200}}
\put(0,100){\line(1,0){200}}
\put(100,100){\circle*{5}}
\put(150,100){\circle*{5}}
\put(50,100){\circle*{5}}
\put(100,175){\circle*{5}}
\put(100,25){\circle*{5}}
\put(100,175){\line(2,-3){50}}
\put(130,140){$c$}
\put(107,177){$(0,k)$}
\put(107,22){$(0,-k)$}
\put(155,107){$(1,0)$}
\put(30,109){$(-1,0)$}
\put(129,86){$m_1 = 1$}
\put(53,172){$m_2 = m$}
\put(29,86){$m_3 = 1$}
\put(53,22){$m_4 = m$}
\put(78.5,86){$m_5 = p$}
\end{picture}
\caption{Set up for the $1+$rhombus relative equilibria.}
\end{figure}
Due to the symmetry, the forces on the first and third bodies
differ only in sign,
as do the forces on the second and fourth bodies. The forces
on the central body
cancel out to zero. Equation~(\ref{eq:cc}) then reduces to the two equations
\begin{eqnarray*}
\frac{m}{c^3} \left( \begin{array}{c}
-1 \\ k
\end{array} \right)+
\frac{1}{8} \left( \begin{array}{c}
-2 \\ 0
\end{array} \right)+
\frac{m}{c^3} \left( \begin{array}{c}
-1 \\ -k
\end{array} \right)+
p \left( \begin{array}{c}
-1 \\ 0
\end{array} \right)+
\omega^2 \left( \begin{array}{c}
1 \\ 0
\end{array} \right) &=&
\left( \begin{array}{c}
0 \\ 0
\end{array} \right) \\[.1in]
\frac{1}{c^3} \left( \begin{array}{c}
1 \\ -k
\end{array} \right)+
\frac{1}{c^3} \left( \begin{array}{c}
-1 \\ -k
\end{array} \right)+
\frac{m}{(2k)^3} \left( \begin{array}{c}
0 \\ -2k
\end{array} \right)+
\frac{p}{k^3} \left( \begin{array}{c}
0 \\ -k
\end{array} \right)+
\omega^2 \left( \begin{array}{c}
0 \\ k
\end{array} \right) &=&
\left( \begin{array}{c}
0 \\ 0
\end{array} \right)
\end{eqnarray*}
which are satisfied when
$\omega^2 = 2m/c^3 + 1/4 + p$ and
\begin{equation}
m(\frac{2}{c^3} - \frac{1}{4k^3}) + p(1-\frac{1}{k^3}) =
\frac{2}{c^3} - \frac{1}{4}.
\label{eq:big}
\end{equation}
For each $k$-value, equation~(\ref{eq:big}) determines a linear
relationship between $m$ and $p$.
In other words, fixing the size of the rhombus yields a one-parameter
family of relative
equilibria for which the masses $m$ and $p$ change linearly with respect
to each other. Alternatively, we can ask whether a given
pair $(m,p)$ has more than one $k$-value associated to it for which
equation~(\ref{eq:big})
is satisfied. This question is surprisingly answered in the affirmative.
The pair $(m=1,p=-1/4)$ makes equation~(\ref{eq:big}) degenerate.
Thus we have found a
one-parameter family of degenerate relative equilibria for the fixed
set of masses
$(1,1,1,1,-1/4)$ in the five-body problem.
Physically, the
negative central mass acts as a repellor cancelling out the forces
along both axis
and the corresponding centrifugal force needed to maintain the equilibria works
out to be the same for the 1st and 3rd bodies as it is for the 2nd and 4th.
For an alternative view, let ${\bf x}_1 = (j,0)$ and ${\bf x}_3 = (-j,0)$,
instead of fixing them one unit distance away from the origin, and
let the other
positions remain as given above. Let the masses be the special
values $(1,1,1,1,-1/4)$. Then $c = \sqrt{j^2 + k^2}$ and $I = j^2 + k^2$ so
that fixing $I=1$ implies
$$
U \; = \; 4 \cdot 1 + \frac{1}{2j} + \frac{1}{2k} -
\frac{1}{4}(\frac{2}{j}+\frac{2}{k}) \; = \; 4.
$$
Therefore, for the masses $(1, 1, 1, 1, -1/4)$, the potential
function $U$ restricted to the ellipsoid $I=1$ has a curve of
critical points at
$$
( \cos t , 0, 0, \sin t , -\cos t , 0, 0, -\sin t , 0, 0) \quad
\mbox{for} \; \; 0 \leq t \leq \pi/2.
$$
Note that this continuum has endpoints located at points on the ellipsoid
where three of the five bodies collide. This does not happen in the
classical problem
(with positive masses) as relative
equilibria cannot accumulate on collision points (see~\cite{cc:shub}).
Here, the potential function $U$ remains constant (and particularly bounded)
as you approach the collision points on the mass ellipsoid because the negative
mass cancels out the forces felt by the three masses regardless of their
distances apart.
\begin{theorem}
In the five-body problem for the masses
$(1, 1, 1, 1, -1/4)$, there exists a one-parameter family
of degenerate relative equilibria where the four equal masses are
positioned at the vertices of a rhombus with the remaining body located at
the center. As the parameter varies, one pair of opposite vertices move away
from each other while the other pair moves closer, maintaining a fixed length
between consecutive vertices.
\end{theorem}
\begin{corr}
The number of relative equilibria equivalence classes in the
five-body problem for the masses $(1,1,1,1,-1/4)$ is {\bf not} finite.
\end{corr}
\section{Concluding Remarks}
\begin{enumerate}
\item A calculation similar to the one above shows that this family
of relative equilibria
persists in similar potential functions,
including the logarithmic potential used in the Kirchhoff problem.
Specifically,
if we let
$U_0 = - \Sigma m_i m_j \ln r_{ij}$ and $U_d = \Sigma m_i m_j / r_{ij}^d$
for $d > 0 $,
then the $1+$ rhombus family described above is a one-parameter family of
degenerate relative
equilibria for these potential functions also,
with the only difference being that the central mass is given
by $p = -1/2^{d+1}$.
Generalizing Wintner's question, we see
that the number of relative equilibria equivalence classes in the
five-body problem is also {\bf not} finite for these other potential functions.
\item A natural attempt for modifying our example to obtain another
(positive mass)
family of degenerate relative equilibria
is to replace the negative mass at the origin with four symmetrically
located bodies
with positive masses about a positive mass in the center.
While this does not lead to the discovery of a continuum,
it does produce a surprising one-parameter family of relative equilibria
in the 9-body problem. This family
consists of four bodies of mass one located
at the vertices of a square of radius one, four bodies of {\em arbitrary}
mass $m$ located at the vertices of a square of radius $j \approx .5318$
(aligned with the outer square), and a body at the origin of
mass $p \approx 1.3022$. What makes this family unusual is that
the mass of the four equal bodies on the inner square serves as the parameter,
in contrast to many other families of relative equilibria
(for example, the
regular $n$-gon with a central mass,
see \cite{cc:schmidt} or~\cite{cc:rick1}),
where the mass of the central body acts as the parameter.
\end{enumerate}
\noindent {\bf Acknowledgments: }
I would like to thank Glen Hall and Alain Albouy for
their advice and suggestions regarding this work.
\begin{thebibliography}{99}
\bibitem{cc:albouy} A. Albouy. The symmetric central configurations
of four equal masses. {\it Hamiltonian dynamics and celestial mechanics
(Seattle, WA, 1995)}, 131-135, Contemp. Math., {\bf 198}, 1996.
\bibitem{cc:euler} L. Euler. De motu rectilineo trium corporum
se mutuo attahentium. {\it Novi Comm. Acad. Sci. Imp. Petrop.}
{\bf 11}, 144-151, 1767.
\bibitem{kirchhoff} G. R. Kirchhoff. {\it Vorlesungen uber
Matematische Physik}, Vol. I, Teubner, Leipzig, 1876.
\bibitem{cc:lagrange} J. L. Lagrange. Essai sur le probl\`{e}me des
trois corps. {\it \OE uvres, vol. 6}, 272-292, Gauthier-Villars, Paris, 1772.
\bibitem{cc:meyer} K. Meyer and G. R. Hall. {\it Introduction to
Hamiltonian Dynamical Systems and the $N$-Body Problem}, Applied
Mathematical Sciences, 90. Springer, New York, 1992.
\bibitem{cc:schmidt} K. Meyer and D. S. Schmidt. Bifurcations of
Relative Equilibria in the $N$-Body and Kirchoff Problems.
{\it SIAM J. Math. Anal.} {\bf 19}, no. 6, 1295-1313, 1988.
\bibitem{cc:rick1} R. Moeckel. Linear Stability Analysis of Some
Symmetrical Classes of Relative Equilibria. {\it Hamiltonian
dynamical systems (Cincinnati, OH, 1992)}, 291-317, IMA Vol. Math
Appl., 63. Springer, New York, 1995.
\bibitem{cc:rick2} R. Moeckel. Relative equilibria of the four-body problem.
{\it Ergodic Theory and Dynamical Systems} {\bf 5}, 417-435, 1985
\bibitem{cc:moulton} F. R. Moulton. The straight line solutions of
the $n$-body problem.
{\it Annals of Math.}, II. Ser. {\bf 12}, 1-17, 1910.
\bibitem{Oneil} K. O'Neil. Stationary Configurations of Point Vortices.
{\it Trans. Amer. Math.
Soc.} {\bf 302}, no. 2, 383-425, 1987.
\bibitem{cc:palmore2} J. Palmore. New Relative Equilibria of the
$N$-Body Problem.
{\it Letters in Mathematical Physics}. {\bf 1}, no. 2, 119-123, 1976.
\bibitem{cc:shub} M. Shub. Diagonals and relative equilibria.
{\it Manifolds-Amsterdam} 1970 {\it Lecture Notes in Math.},
{\bf 197}, Springer, Berlin, 1971.
\bibitem{cc:smale1} S. Smale. Mathematical Problems for the Next Century.
{\it Mathematical Intelligencer} {\bf 20}, 7-15, Spring 1998.
\bibitem{cc:smale2} S. Smale. Topology and mechanics, II,
The planar $n$-body problem.
{\it Invent. Math.} {\bf 11}, 45-64, 1970.
\bibitem{cc:wint} A. Wintner. The Analytical Foundations of
Celestial Mechanics. {\it Princeton Math. Series} {\bf 5},
215, Princeton University Press, Princeton, NJ, 1941.
\end{thebibliography}
\end{document}