# Stacked Central Configurations for the Spatial Nine-Body Problem

Authors: Su Xia; Deng Chunhua
Journal: The Scientific World Journal
 (2013)

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## Abstract

We show the existence of the twisted stacked central configurations for the 9-body problem. More precisely, the position vectorsx1, x2, x3, x4, and x5 are at the vertices of a square pyramid Σ; the position vectors x6, x7, x8, and x9 are at the vertices of a square Π.


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## Body

## 1. Introduction and Main Results

The classicaln-body problem [1, 2] concerns the motion of n mass points moving in space according to Newton’s law:
(1)mix¨i=-∑j=1,j≠in‍mimj(xi-xj)rij3,i=1,2,…,n.
Here, xi∈ℝd is the position of mass mi>0, the gravitational constant is taken equal to 1, and rij=|xi-xj| is the Euclidean distance between xi and xj.The space of configuration is defined by(2)X={(x1,…,xn)∈(ℝd)n:xi≠xj∀i≠j},
while the center of mass is given by
(3)c=m1x1+⋯+mnxnM,
where M=m1+⋯+mn is the total mass.A configurationx=(x1,…,xn)∈X is called a  central configuration [2, 3] if there exists a constant λ, called the multiplier, such that
(4)-λ(xi-c)=∑j=1,j≠in‍mj(xj-xi)rij3,i=1,2,…,n.
It is easy to see that a central configuration remains a central configuration after a rotation in ℝd and a scalar multiplication. More precisely, let A∈SO(d) and a>0, if x=(x1,…,xn) is a central configuration, so are Ax=(Ax1,…,Axn) and ax=(ax1,…,axn).Two central configurations are said to be equivalent if one can be transformed to the other by a scalar multiplication and a rotation. In this paper, when we say a central configuration, we mean a class of central configurations as defined by the above equivalent relation.Central configurations of then-body problem are important because they allow the computation of homographic solutions; if the n bodies are heading for a simultaneous collision, then the bodies tend to a central configuration (see [3, 4]); there is a relation between central configurations and the bifurcations of the hypersurfaces of constant energy and angular momentum (see [5]).In this paper, we are interested in spatial central configurations, that is,d=3. In 2005, Hampton [6] provides a new family of planar central configurations for the 5-body problem with an interesting property: the central configuration has a subset of three bodies forming a central configuration of the 3-body problem. The authors [7] find new classes of central configurations of the 5-body problem which are the ones studied by Hampton [6] having three bodies in the vertices of an equilateral triangle, but the other two, instead of being located symmetrically with respect to a perpendicular bisector, are on the perpendicular bisector. The stacked central configurations studied by Hampton [6] were completed by Llibre et al. [8] (see also [9]).Zhang and Zhou [10] showed the existence of double pyramidal central configurations of N+2-body problem. The authors [11–13] provided new examples of stacked central configurations for the spatial 7-body problem where four bodies are at the vertices of a regular tetrahedron and the other three bodies are located at the vertices of an equilateral triangle.In this paper, we find new classes of stacked spatial central configurations for the 9-body problem which have five bodies at the vertices of a square pyramid, and the other four bodies are located at the vertices of a square. More precisely, the spatial central configurations considered here satisfy the following (see Figure1): the position vectors x1,x2,x3,x4, and x5 are at the vertices of a square pyramid Σ; the position vectors x6,x7,x8, and x9 are at the vertices of a square Π.Figure 1
The configuration for the 9-body problem.Without loss of generality, we can assume that(5)x1=(1,0,0),x2=(0,1,0),x3=(-1,0,0),x4=(0,-1,0),x5=(0,0,h),x6=(x,0,y),x7=(0,x,y),x8=(-x,0,y),x9=(0,-x,y),
where x>0, y∈ℝ, and y≠0; the positive constant h satisfies the equation
(6)2r153=1r123+1r133,
(see [10] and the references therein); that is, h=1.26276522.The main results of this paper are the following.Theorem 1.
Consider the spatial configurations according to Figure1, in order that the nine mass points are in a central configuration, the following statements are necessary:(1)
the massesm1,m2,m3, and m4 must be equal;(2)
the massesm6,m7,m8, and m9 must be equal.Theorem 2.
There exist points(x0,y0)∈T-1(0)∩D (see Figure 2) such that the nine bodies take the coordinates
(7)x1=(1,0,0),x2=(0,1,0),x3=(-1,0,0),x4=(0,-1,0),x5=(0,0,h),x6=(x0,0,y0),x7=(0,x0,y0),x8=(-x0,0,y0),x9=(0,-x0,y0).
Then, there are positive solutions of m1,m5,m6 such that these bodies form a spatial central configuration according to Figure 1.Figure 2
The regionD.The proofs of the theorems are given in the next sections.


## 2. Proof of Theorem1

For the spatial central configurations, instead of working with (4), we consider the Dziobek-Laura-Andoyer equations (see [9, 11–13] and the references therein):
(8)fijk=∑l=1,l≠i,j,kn‍ml(dil-djl)Δijkl=0
for 1≤i<j≤n, k=1,…,n,k≠i,j. Here, dij=1/rij3 and Δijkl=(xi-xj)∧(xi-xk)·(xi-xl). Thus, Δijkl gives six times the signed volume of the tetrahedron formed by the bodies with positions xi,xj,xk, and xl; (8) is a system of n(n-1)(n-2)/2 equations.For the 9-body problem, (8) is a system of 252 equations. According to Figure 1, our class of configurations with nine bodies must satisfy
(9)r12=r23=r34=r14=2,r13=r24=2,r67=r78=r89=r69=2x,r68=r79=2x,r16=r27=r38=r49=(x-1)2+y2,r17=r19=r26=r28=r37=r39=r4600=r48=x2+1+y2,r18=r29=r36=r47=(x+1)2+y2,r15=r25=r35=r45=1+h2,r56=r57=r58=r59=x2+(y-h)2.
Due to assumption (5) and the definition of Δijkl, we have several symmetries in the signed volumes.By using the symmetries and the properties ofΔijkl, we obtain the following results.Lemma 3.
In order to have a spatial central configuration according to Figure1, a necessary condition is that the masses m1, m2, m3, and m4 must be equal.Proof.
It is sufficient to consider the equationsf687=0 and f796=0:
(10)f687=(m1-m3)(d16-d18)Δ6871=0,f796=(m2-m4)(d16-d18)Δ7962=0.
For our class of central configurations, we have d16-d18≠0, Δ6871≠0, and Δ7962≠0. So the above equations hold if and only if m1=m3, m2=m4. Consider the expression of f678=0:
(11)f678=(m1-m2)(d16-d17)Δ6781+(m3-m4)(d18-d17)Δ6783=0.
Substituting m1=m3, m2=m4 into the above equation, we have
(12)f678=(m1-m2)(d16+d18-2d17)Δ6781=0.
For our class of central configurations, we have d16+d18-2d17≠0, since the function g(x)=x-3/2 is convex for all x>0, and Δ6781≠0. So the above equation holds if and only if m1=m2. So statement 1 of Theorem 1 is proved.Lemma 4.
If the configuration, according to Figure1, is a central configuration, a necessary condition is that the masses m6, m7, m8, and m9 must be equal.Proof.
It is sufficient to consider the equationsf132=0 and f241=0:
(13)f132=(m6-m8)(d16-d18)Δ1326=0,f241=(m7-m9)(d16-d18)Δ2417=0.
For our class of central configurations, we have d16-d18≠0, Δ1326≠0, and Δ2417≠0. So the above equations hold if and only if m6=m8, m7=m9. Consider the expression of f123=0:
(14)f123=(m6-m7)(d16-d17)Δ1236+(m8-m9)(d18-d17)Δ1238=0.
Substituting m6=m8, m7=m9 into the above equation, we have
(15)f123=(m6-m7)(d16+d18-2d17)Δ1236=0.
For our class of central configurations, we have d16+d18-2d17≠0, and Δ1236≠0. So the above equation holds if and only if m6=m7. Hence, statement 2 of Theorem 1 is proved.
The proof Theorem1 is completed.We restrict the set of admissible masses tom1=m2=m3=m4=α and m6=m7=m8=m9=β. Substituting m1=m2=m3=m4=α and m6=m7=m8=m9=β into (8), they reduce to the following 4 equations:
(16)f152=β((d16+d17-2d56)Δ152600+(d17+d18-2d56)Δ1528)=0,(17)f162=α(d12+d13-d17-d18)Δ1623+m5(d15-d56)Δ1625+β(d17+d18-d67-d68)Δ1628=0,(18)f175=α((d12-d16)Δ1752+(d13-d17)Δ175300+(d12-d18)Δ1754)+β((d16-d67)Δ1756+(d18-d67)Δ17580000+(d17-d68)Δ1759)=0,(19)f562=α((d15-d16)Δ5621+(d15-d18)Δ562300+(d15-d17)Δ5624)+β((d56-d67)Δ5627+(d56-d68)Δ56280000+(d56-d67)Δ5629)=0.If we writef152=βT=β((d16+d17-2d56)Δ1526+(d17+d18-2d56)Δ1528)=0, it follows that T=0 in order to have central configurations. So in the following, we restrict our central configurations to the set T-1(0).Lemma 5.
According to one’s assumptions and the setT-1(0), (8) is satisfied if (17) and (18) are satisfied.Proof.
Under the assumptions (5), we have
(20)T=(d16+2d17+d18-4d56)(y-h)+hx(d16-d18)=0;
that is,
(21)4(y-h)d56=(y-h)(d16+2d17+d18)+hx(d16-d18).
Substituting (21) into (19), we obtain the equation f175=0.
Hence in the setT-1(0), f175=0 implies f562=0. This completes the proof.From Lemma5, in order to study central configurations according to Figure 1 in the set T-1(0), it is sufficient to study the following 2 equations:
(22)f162=0,f175=0.
Denote by A=(aij) the matrix of the coefficients of the homogeneous linear system in the variables α,m4,β defined by (22). Thus,
(23)a11=(d12+d13-d17-d18)Δ1623=-2y(1((x+1)2+y2)3/2122+18-1(x2+1+y2)3/2000000-1((x+1)2+y2)3/2),a12=(d15-d56)Δ1625=(-y-hx+h)×(1(1+h2)3/2-1(x2+(y-h)2)3/2),a13=(d17+d18-d67-d68)Δ1628=-2xy(1(x2+1+y2)3/2+1((x+1)2+y2)3/200000000-18x3-122x31((x+1)2+y2)3/2),a21=(d12-d16)Δ1752+(d13-d17)Δ1753+(d12-d18)Δ1754=(y-h)(1((x+1)2+y2)3/2000000000-1((x-1)2+y2)3/2)+hx(14+12-1((x-1)2+y2)3/20000000-1((x+1)2+y2)3/2-2(x2+1+y2)3/2),a22=0,a23=(d16-d67)Δ1756+(d18-d67)Δ1758+(d17-d68)Δ1759=hx2(1((x+1)2+y2)3/2-1((x+1)2+y2)3/2)-x(y-h)×(1((x+1)2+y2)3/2+1((x+1)2+y2)3/2000+2(x2+1+y2)3/2-14x3-12x31((x+1)2+y2)3/2).
Let x=(αm4β). Then in order to get the spatial central configuration as Figure 1, we need to find a positive solution α,m4,β of the following system:
(24)Ax=0,
where T=0.


## 3. The Existence of Spatial Central Configurations

In order to prove the existence of positive solutions of (24) in the set T-1(0), it is sufficient to prove that the entries in each row of A change the signs. So if the entries of some row of A have the same signs, there are no admissible masses such that the bodies are in a central configuration according to Figure 1.Proof of Theorem2.
Since the rank of matrixA is two in the set T-1(0), there are nontrivial solutions of (24) in the set T-1(0).
Now we prove the existence of spatial central configurations according to Figure1 for some points in the set D (see Figure 2). In order to prove the existence of positive solutions of (24) in the set T-1(0), the entries a21,a23 of the second line in the matrix A should have opposite signs. Thus, we consider the following set D, where D is surrounded by curves x=0, y=0, a21=0, and a23=0.
In the setD, the entries of matrix A have the following signs: a21>0, a23<0 (see Figures 3 and 4); a11>0, a12<0, a13>0 because the set D is included in the set E, where E is surrounded by curves x=0, y=0, and y=h(1-x) (see Figures 5, 6, 7, and 8). In short, the signs of the entries of the matrix A restricted to the set D are the following:
(25)A=(+-++0-).
In the rest of the proof, we show that the setT-1(0) has intersection with the set D. We consider the subset of D:
(26)L={(x,y):x=x1,0<y<y1},
where x1∈(0,1). Obviously L is a segment with endpoints
(27)P1=(x1,0),P2=(x1,y1),
(see Figure 9), and the point (x1,y1) satisfies the equation a21=0. Evaluating the function T at these points, we have
(28)T(P1)<0,T(P2)>0.
Thus, there exists a point P0=(x0,y0)∈L, such that T(P0)=0. So at the point P0 we have nontrivial positive solutions of (24), since the signs of the entries of the matrix A at this point are the following:
(29)A(P0)=(+-++0-).
Thus, the proof of Theorem 2 is completed.Figure 3
The curvea21=0.Figure 4
The curvea23=0.Figure 5
The regionE.Figure 6
The curvea11=0.Figure 7
The curvea12=0.Figure 8
The curvea13=0.Figure 9
The existence of central configurations for the 9-body problem.



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