Central configurations

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Curator and Contributors

1.00 - Richard Moeckel

A central configuration is a special arrangement of point masses interacting by Newton's law of gravitation with the following property: the gravitational acceleration vector produced on each mass by all the others should point toward the center of mass and be proportional to the distance to the center of mass. Central configurations (or CC's) play an important role in the study of the Newtonian N-body problem. For example, they lead to the only explicit solutions of the equations of motion, they govern the behavior of solutions near collisions and near infinity, and they influence the topology of the integral manifolds.

Figure 1: A central configuration of five equal masses with gravitational acceleration vectors.

Equations for Central Configurations

Let $$m_i>0, i=1\dots N$$ be the masses and $$q_i\in R^d$$ the position vectors of N point particles in d-dimensional space. The configuration of the N particles is represented by $$q = (q_1,\ldots,q_N)\in R^{Nd}\ .$$

The Basic CC Equations

Newton's law of motion for the gravitational N-body problem is

$\tag{1} m_i \ddot q_i = F_i = \sum_{j\ne i}\frac{m_im_j(q_j-q_i)}{r^3_{ij}}$

where $$r_{ij} = |q_i-q_j|$$ is the Euclidean distance between particles i and j (the gravitational constant is normalized to G= 1). The force vector $$F_i\in R^d$$ on the right-hand side can also be written as a partial gradient vector $$F_i = \nabla_i U$$ where

$\tag{2} U = \sum_{j\ne i}\frac{m_im_j}{r_{ij}}$

is the Newtonian potential function and $$\nabla_i$$ denotes the vector of partial derivatives with respect to the d components of $$q_i\ .$$

The acceleration of the i-th body is $$F_i\,/m_i$$ so the condition for $$q$$ to be a central configuration is

$\tag{3} \nabla_i U = \lambda m_i (q_i-c)$

where

$\tag{4} c=(m_1q_1+\ldots+m_Nq_N)/(m_1+\ldots+m_N)\in R^d$

is the center of mass and $$\lambda\in R$$ is a constant. By definition, $$q\in R^{Nd}$$ is a central configuration for the masses $$m_i$$ if and only if (3) and (4) hold for some constant $$\lambda\ .$$

It turns out, however, that the values of $$\lambda$$ and $$c$$ are uniquely determined by (3). To see this, note that translation invariance and degree -1 homogeneity of the Newtonian potential give

\begin{align} \sum_i \nabla_i U &= 0\\ \sum_i q_i^T \nabla_i U &= -U. \end{align}

Together these give $$\sum_i (q_i-c)^T \nabla_i U = -U$$ and then (3) shows that

$\lambda = -U/I$

where

$\tag{5} I = \sum m_i|q_i-c|^2$

is the moment of inertia with respect to c. Thus $$\lambda <0$$ is uniquely determined. Finally, summing (3) shows that c must be the center of mass.

Thus $$q$$ is a central configuration for the given masses if and only if (3) holds for some $$\lambda\in R, c\in R^d\ .$$

Equivalent Central Configurations and Normalized Equations

The central configuration equation (3) is invariant under the Euclidean similarities of $$R^d$$ -- translations, rotations, reflections and dilations. Call two configurations $$q, q' \in R^{Nd}$$ equivalent if there are constants $$k\in R, b\in R^d$$ and an dxd orthogonal matrix $$Q$$ such that $$q'_i = k A q_i + b, i=1,\ldots,N\ .$$ If $$q$$ satisfies (3) with constant $$\lambda, c$$ then $$q'$$ satisfies (3) with constant $$\lambda'= k^3\lambda, c' = c+b\ .$$ So one can speak of an equivalence class of central configurations.

Translation invariance can be used to eliminate the center of mass. Setting $$c=0$$ and canceling a factor of $$m_i$$ the central configuration equation becomes

$\tag{6} \lambda \, q_i = \sum_{j\ne i}\frac{m_j(q_j-q_i)}{r^3_{ij}}.$

Alternatively, substituting (4) into (3) leads, after some simplification, to

$\tag{7} \sum_{j\ne i}m_jS_{ij}(q_j-q_i)\qquad S_{ij} = \frac{1}{r_{ij}^3}+\lambda'$

where $$\lambda' = \lambda/(m_1+\ldots+m_N)\ .$$

Dilation invariance can be used to normalize the size of a central configuration. The moment of inertia (5) is a natural measure of the size and setting $$I = 1$$ is a popular normalization. Alternatively, one can fix the size by normalizing $$\lambda = -1$$ in (6) or $$\lambda = -1$$ in (7). .

CC's as Critical Points

Central configurations can be viewed as constrained critical points of the Newtonian potential. To see this note that the basic equation (3) can be written

$\nabla_i U = \lambda\; \nabla_i I$

where I is the moment of inertia. Thus a central configuration is a critical point of the restriction of $$U$$ subject to the constraint $$I=const\ .$$ From this point of view, $$\lambda$$ is a Lagrange multiplier.

Alternatively, with the normalization $$\lambda = 1\ ,$$ central configurations are unconstrained critical point of $$U/\sqrt{I}\ .$$

Examples

The Two-Body Problem

Any two configurations of N=2 particles in $$R^d$$ are equivalent. Moreover, (7) reduces to just one equation

$S_{12}(q_1-q_2) = 0\qquad S_{12} = \frac{1}{r_{12}^3}+\lambda$

which holds for $$\lambda = -r_{12}^{-3}\ .$$ Thus every configuration of two bodies is central.

In this case, the possible motions are well-known -- each mass moves on a conic section according to Kepler's laws. In particular, one has a family of elliptical periodic motions ranging from circular (eccentricity $$\epsilon=0$$) to collinear (limit as $$\epsilon\rightarrow 1\,$$). The latter is an example of a total collision solution, that is, one for which all N bodies collide at the center of mass.

Symmetrical Configurations of Equal Masses

Figure 2: Seven equal masses forming a centered hexagon.
Figure 3: Asymmetric central configuration of eight equal masses.

When all of the masses are equal, it is obvious from symmetry that certain configurations are central. In the plane (d=2), one can place the masses at the vertices of a regular N-gon or at the vertices of a regular (N-1)-gon with the last mass at the center (see Figure 2 ). Similarly, in space (d=3), a regular polyhedron or centered regular polyhedron are central configurations but these are possible only for special values of N. For d>3, regular polytopes are central configurations for certain values of N. Note however, that equal masses may admit other central configurations with less symmetry as in Figure 1 or with no symmetry at all as in Figure 3 .

When some but not all of the masses are equal it is again possible to look for central configurations which are symmetric under similarities of $$R^d$$ which permute equal masses. For example, it is clear that the centered regular polygon is still be a central configuration when the mass at the center is different from the others.

Euler

The first nontrivial examples of central configurations were discovered by Euler in 1767, who studied the case N=3, d=1, that is, three bodies on a line. When two masses are equal, one can get a central configuration by putting an arbitrary mass at their midpoint (a centered 2-gon). For three unequal masses it is not obvious that any central configurations exist. But Euler showed that, in fact, there will be exactly one equivalence class of collinear central configurations for each possible ordering of the masses along the line.

For example, if the order is $$q_1<q_2<q_3\ ,$$ one can reduce to the case $$q_1=0, q_2=1, q_3 = 1+r$$ where $$r>0\ .$$ (7) give three equations for $$r, \lambda$$ of which only two are independent. Eliminating $$\lambda$$ leads to a fifth degree polynomial equation

\tag{8} \begin{align} (m_1+m_2)r^5 &+ (3m_1+2m_2)r^4 + (3m_1+m_2)r^3 \\ &-(m_2+3m_3)r^2-(2m_2+3m_3)r-(m_2+m_3) = 0. \end{align}

Descartes' rule of signs shows that there is exactly one positive real root, $$r(m_1,m_2,m_3)>0\ .$$

(8) illustrates the nature of the problem of finding central configurations. In the general case, one has a complicated set of algebraic equations for the configuration, $$q\ ,$$ depending on the masses as parameters. The goal is to find or at least count the equivalence classes of real solutions.

Lagrange

Lagrange found next example in the planar three-body problem $$N=3, d=2$$. Remarkably, an equilateral triangle is a central configuration, not only for equal masses, but for any three masses $$m_1, m_2, m_3$$. Moreover, it is the only non-collinear central configuration for the three-body problem.

When the masses are not equal, the center of mass will not be the center of the triangle and it is not at all obvious that the configuration is central. But it is easy to see it using mutual distance coordinates. The three mutual distances $$r_{12}, r_{31}, r_{23}$$ can be used as local coordinates on the space of non-collinear configurations of three bodies in the plane up to symmetry. The potential and the moment of inertia can be expressed in these coordinates as

$U = \frac{m_1 m_2}{r_{12}}+\frac{m_3 m_1}{r_{31}}+\frac{m_2 m_3}{r_{23}}\qquad\qquad\qquad I = (m_1m_2 r_{12}^2+m_3m_1 r_{31}^2+m_2m_3 r_{23}^2)/(m_1+m_2+m_3).$

Now use the characterization of CC's as critical points of $$U$$ with fixed $$I$$. Setting

$\frac{\partial U}{r_{ij}} = \lambda \frac{\partial I}{r_{ij}}$

gives

$r_{ij}^3 = \frac{m_1+m_2+m_3}{-2\lambda}$

showing that the three distances are equal.