Dziobek Equilibrium Configurations on a Sphere

Abstract

We investigate the n-body problem on a sphere with a general interaction potential that depends on the mutual distances. We focus on the equilibrium configurations, especially on the Dziobek equilibrium configurations, which is an analogy of Dziobek central configurations of the classical n-body problem. We obtain a criterion and then reduce it to two sets of equations. Then we apply these equations to the curved n-body problem in ${\mathbb {S}}^3$. In the end, we find the derivative of the Cayley-Menger determinant.

Appendix: The Derivative of the Cayley–Menger Determinant

For a Dziobek configuration of n-body in $\mathbb S^{n-2}$, recall the $(n-1)\times n$ matrix $X=[\mathbf {q}_1, ..., \mathbf {q}_n]$. Since $\text {rank}X =n-1$, the corresponding Gram matrix $X^TX$ has rank $n-1$. Then the determinant $F=0$. We may call the quantity F the spherical Cayley-Menger determinant, [4]. For instance, for $n=4$,

$$\begin{aligned} F=\begin{vmatrix} 1&\quad \cos d_{12}&\quad \cos d_{13}&\quad \cos d_{14} \\ \cos d_{12}&\quad 1&\quad \cos d_{23}&\quad \cos d_{24} \\ \cos d_{13}&\quad \cos d_{23}&\quad 1&\quad \cos d_{34} \\ \cos d_{14}&\quad \cos d_{24}&\quad \cos d_{34}&\quad 1 \\ \end{vmatrix}. \end{aligned}$$

A by-product of equation (6) is the following. A Dziobek configuration on $\mathbb S^{n-2}$ can be parametrized by the $C_n^2$ quantities $\{\cos d_{12}, ..., \cos d_{n-1,n}\}$ with the relation $F=0$. Then any equilibrium configuration of the system (1) is the critical point of $V+\lambda F$. Then Eq. (6) implies $\frac{\partial F}{\partial \cos d_{ij}}=\alpha \Delta _i\Delta _j$ for some $\alpha$.

Proposition 5

Let $\mathbf {q}_1, ..., \mathbf {q}_n$ be a Dziobek configuration in $\mathbb S^{n-2}$. Let $d_{12}, ..., d_{n-1, n}$ and F be the corresponding mutual distances and the spherical Cayley–Menger determinant. Then we have

$$\begin{aligned} \frac{\partial F}{\partial \cos d_{ij}}=2 \Delta _i\Delta _j \ \mathrm{for\ any}\ 1\le i<j\le n. \end{aligned}$$

where $\Delta _i$ is the signed determinant defined in (5).

Proof

By the symmetry of $X^TX$, we have $\frac{\partial F}{\partial \cos d_{ij}}=2 F_{ij}$, with $F_{ij}$ being the (i, j) cofactor of matrix $X^TX$, i.e.,

$$\begin{aligned} \frac{\partial F}{\partial \cos d_{ij}}=2 (-1)^i (-1)^j |A_{ij}|, \end{aligned}$$

where $A_{ij}$ is the (i, j) minor of matrix $X^TX$. Let $X_k$ be the square matrix of order $n-1$ obtained from X by deleting the k-th column. Then $X_i^T X_j=A_{ij}$. Thus, we have $\frac{\partial F}{\partial \cos d_{ij}}=2 \Delta _i\Delta _j$. $\square$

This derivative formula enables us to obtain Eq. (6) directly.

For a Dziobek configuration ${\mathbf{x}}=({\mathbf{x}}_1, ..., {\mathbf{x}}_n )$ in $\mathbb R^{n-2}$, the mutual distances satisfy a relation and its derivative formula is similar to the above one. Due to the translational symmetry, the appropriate Gram matrix is ${\tilde{X}}^T{\tilde{X}}$, with

$$\begin{aligned} {\tilde{X}}=[{\mathbf{x}}_2-{\mathbf{x}}_1, ..., {\mathbf{x}}_n-{\mathbf{x}}_1]. \end{aligned}$$

It is easy to see that $|{\tilde{X}}^T {\tilde{X}}|=0$. Note that the entries of $X^T{\tilde{X}}$ are not in terms of the mutual distances. By using the formula $({\mathbf{x}}_i-{\mathbf{x}}_1) \cdot ({\mathbf{x}}_j-{\mathbf{x}}_1)=\frac{1}{2}(d_{1i}^2 +d_{1j}^2 - d_{ij}^2)$ and some bordering technique, [4], we can obtain another determinant

$$\begin{aligned} \Gamma =\begin{vmatrix} 0&\quad 1&\quad 1&\quad \cdots&\quad 1\\ 1&\quad 0&\quad d_{12}^2&\quad \cdots&\quad d_{1n}^2\\ 1&\quad d_{21}^2&\quad 0&\quad \cdots&\quad d_{2n}^2\\ \vdots&\quad \vdots&\quad \vdots&\quad \ddots&\quad \vdots&\\ 1&\quad d_{n1}^2&\quad d_{n2}^2&\quad \cdots&\quad 0 \end{vmatrix}, \ \mathrm{and}\ \Gamma =(-1)^n2^{n-1}|{\tilde{X}}^T{\tilde{X}}|. \end{aligned}$$

Usually, it is $\Gamma$ instead of $|{\tilde{X}}^T{\tilde{X}}|$ that is called the Cayley–Menger determinant. Let

$$\begin{aligned} X=\begin{bmatrix} 1 &{} 1&{} \cdots &{} 1\\ {\mathbf{x}}_1&{}{\mathbf{x}}_2&{}\cdots &{}{\mathbf{x}}_n \end{bmatrix} _{(n-1)\times n}, \end{aligned}$$

and $X_k$ be the square matrix of order $n-1$ obtained from X by deleting the k-th column. Let $\Delta _k =(-1)^{k-1}|X_k|$. For $n=4$, Dziobek [9] observed a formula that is equivalent to

$$\begin{aligned} \frac{\partial \Gamma }{\partial d_{ij}^2}=-8 \Delta _i\Delta _j \ \mathrm{for\ any}\ 1\le i<j\le 4. \end{aligned}$$

With the technique used to relate $\Gamma$ and $|{\tilde{X}}^T \tilde{X}|$, we have

Proposition 6

Let ${\mathbf{x}}_1, ..., {\mathbf{x}}_n$ be a Dziobek configuration in $\mathbb R^{n-2}$. Let $d_{12}, ..., d_{n-1, n}$ be the corresponding mutual distances. Let $\Gamma$ and $\Delta _i$ be the determinants defined above. Then we have

$$\begin{aligned} \frac{\partial \Gamma }{\partial d_{ij}^2}= (-2)^{n-1} \Delta _i\Delta _j \ \mathrm{for\ any}\ 1\le i<j\le n. \end{aligned}$$

Proof

By the symmetry, we have $\frac{\partial \Gamma }{\partial d_{ij}^2}=2 (-1)^i (-1)^j |B_{ij}|$ where $B_{ij}$ is the $(i+1, j+1)$ minor of $\Gamma$. On the other hand, note that

$$\begin{aligned} | X_i|=\begin{vmatrix} 1&\quad 0&\quad 0&\quad \cdots&\quad 0\\ 0&\quad 1&\quad 1&\quad \cdots&\quad 1\\ \vec {0}&\quad {\mathbf{x}}_1&\quad {\mathbf{x}}_2&\quad \cdots&\quad {\mathbf{x}}_n \end{vmatrix} =- \begin{vmatrix} 0&\quad 1&\quad 1&\quad \cdots&\quad 1\\ 1&\quad 0&\quad 0&\quad \cdots&\quad 0\\ \vec {0}&\quad {\mathbf{x}}_1&\quad {\mathbf{x}}_2&\quad \cdots&\quad {\mathbf{x}}_n \end{vmatrix}. \end{aligned}$$

Bordering $X_j$ in the same way without exchanging the first two row, we obtain

$$\begin{aligned} |X_i^TX_j|= - \begin{vmatrix} 0&\quad 1&\quad \cdots&\quad 1&\quad \cdots&\quad 1&\quad 1&\quad \cdots&\quad 1\\ 1&\quad {\mathbf{x}}_1 \cdot {\mathbf{x}}_1&\quad \cdots&\quad {\mathbf{x}}_1 \cdot {\mathbf{x}}_i&\quad \cdots&\quad {\mathbf{x}}_1 \cdot {\mathbf{x}}_{j-1}&\quad {\mathbf{x}}_1 \cdot {\mathbf{x}}_{j+1}&\quad \cdots&\quad {\mathbf{x}}_1 \cdot {\mathbf{x}}_{n}\\ \vdots&\quad \vdots&\quad \ddots&\quad \vdots&\quad \ddots&\quad \vdots&\quad \vdots&\quad \ddots&\quad \vdots&\\ 1&\quad {\mathbf{x}}_{i-1} \cdot {\mathbf{x}}_1&\quad \cdots&\quad {\mathbf{x}}_{i-1} \cdot {\mathbf{x}}_i&\quad \cdots&\quad {\mathbf{x}}_{i-1} \cdot {\mathbf{x}}_{j-1}&\quad {\mathbf{x}}_{i-1} \cdot {\mathbf{x}}_{j+1}&\quad \cdots&\quad {\mathbf{x}}_{i-1} \cdot {\mathbf{x}}_{n}\\ 1&\quad {\mathbf{x}}_{i+1} \cdot {\mathbf{x}}_1&\quad \cdots&\quad {\mathbf{x}}_{i+1} \cdot {\mathbf{x}}_i&\quad \cdots&\quad {\mathbf{x}}_{i+1} \cdot {\mathbf{x}}_{j-1}&\quad {\mathbf{x}}_{i+1} \cdot {\mathbf{x}}_{j+1}&\quad \cdots&\quad {\mathbf{x}}_{i+1} \cdot {\mathbf{x}}_{n}\\ \vdots&\quad \vdots&\quad \ddots&\quad \vdots&\quad \ddots&\quad \vdots&\quad \vdots&\quad \ddots&\quad \vdots&\\ 1&\quad {\mathbf{x}}_j \cdot {\mathbf{x}}_1&\quad \cdots&\quad {\mathbf{x}}_j\cdot {\mathbf{x}}_i&\quad \cdots&\quad {\mathbf{x}}_j \cdot {\mathbf{x}}_{j-1}&\quad {\mathbf{x}}_j \cdot {\mathbf{x}}_{j+1}&\quad \cdots&\quad {\mathbf{x}}_j \cdot {\mathbf{x}}_{n}\\ \vdots&\quad \vdots&\quad \ddots&\quad \vdots&\quad \ddots&\quad \vdots&\quad \vdots&\quad \ddots&\quad \vdots&\\ 1&\quad {\mathbf{x}}_n \cdot {\mathbf{x}}_1&\quad \cdots&\quad {\mathbf{x}}_n \cdot {\mathbf{x}}_i&\quad \cdots&\quad {\mathbf{x}}_n \cdot {\mathbf{x}}_{j-1}&\quad {\mathbf{x}}_n \cdot {\mathbf{x}}_{j+1}&\quad \cdots&\quad {\mathbf{x}}_n \cdot {\mathbf{x}}_{n} \end{vmatrix} \end{aligned}$$

We then replace ${\mathbf{x}}_i\cdot {\mathbf{x}}_j$ be $\frac{1}{2}(||{\mathbf{x}}_i||^2 +||{\mathbf{x}}_j||^2 - d_{ij}^2)$, and eliminate all the $||{\mathbf{x}}_i||^2$ by subtracting the appropriate multiple of the first row and column from the others. We obtain

$$\begin{aligned} \Delta _i\Delta _j= (-1)^{i+j}|X_i^T X_j| =(-1)^{i+j} 2^{2-n} (-1)^{n-1}|B_{ij}|. \end{aligned}$$

Hence follows the formula $\frac{\partial \Gamma }{\partial d_{ij}^2}= (-2)^{n-1} \Delta _i\Delta _j$. $\square$

Central configuration in $\mathbb R^n$ of dimension $n-2$ are considered in [11]. The equations of them are derived by vectorial method there. Note that these equations follow easily from the above derivative formula.