# Spatial Central Configurations with  Two Twisted Regular 4-Gons

Authors: Sen Zhang; Furong Zhao
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We study the configuration formed by two squares in two parallel layers separated by a distance. We picture the two layers horizontally with thez-axis passing through the centers of the two squares. The masses located on the vertices of each square are equal, but we do not assume that the masses of the top square are equal to the masses of the bottom square. We prove that the above configuration of two squares forms a central configuration if and only if the twist angle is equal to kπ/2 or (π/4+kπ/2)  (k=1,2,3,4).


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## Body

## 1. Introduction and Main Results

This paper uses the same notations as [1]. The Newtonian N-body problems [2, 3] concern the motions of N particles with masses mj∈R+ and positions qj∈R3(j=1,2,…,N). The motion is governed by Newton’s second law and the Universal law:
(1)mjq¨j=∂U(q)∂qj,
where q=(q1,q2,…,qN) with the Newtonian potential:
(2)U(q)=∑1⩽j<k⩽N‍mjmk|qj-qk|.
Consider the space
(3)X={q=(q1,q2,…,qN)∈R3N:∑j=1N‍mjqj=0};
that is, suppose that the center of mass is fixed at the origin of the origin of the coordinate axis, because the potential is singular when two particles have the same position. It is natural to assume that the configuration avoids the collision set Δ={q:qj=qk for some k≠j}. The set X∖Δ is called the configuration space.Definition 1 (see [2, 3]).
A configurationq=(q1,q2,…,qN)∈X∖Δ is called a central configuration if there exists a constant λ such that
(4)∑j=1,j≠kN‍mjmk|qj-qk|3(qj-qk)=-λmkqk,1⩽k⩽N.
The value of constantλ in (4) is uniquely determined by
(5)λ=UI,
where
(6)I=∑k=1N‍mk|qk|2.
Consider the configuration inR3 consisting of two layer regular N-gons (N⩾2) with distance h⩾0. It is assumed that the lower layer regular N-gons lie in horizontal plane, and the upper regular N-gons parallel the lower one, and z-axis passes through both centers of two regular N-gons. Suppose that the lower layer particles have masses m1,m2,…,mN and the upper layer particles have masses m~1,m~2,…,m~N, respectively; then these assumptions can be interpreted more precisely by the following. Let ρk denote for all N complex roots of unity, that is,
(7)ρk=exp(2πkNi).
Let
(8)ρ~k=aρkeiφ,
where a>0,  i=-1,  0⩽φ⩽2π, and φ is called twist angle. It is assumed that mk(1⩽k⩽N) locates at the vertex qk of the lower layer regular N-gons; m~k(1⩽k⩽N) locates on the vertex q-k of the upper layer regular N-gons:
(9)qk=(ρk,0),q~k=(ρ~k,h),
where h≥0 is the distance between the two layers. Then the center of masses is
(10)z0=∑j=1N‍(mjqj+m~jq~j)M,
where
(11)M=∑j=1N‍(mj+m~j).
Let
(12)PK=qk-z0,P~K=q~k-z0.
If m1,m2,…,mN and m~1,m~2,…,m~N form a central configuration, then there is λ∈R+ such that
(13)∑j≠kN‍mjPj-Pk|Pj-Pk|3+∑j=1N‍m~jP~j-Pk|P~j-Pk|3=-λPk,1⩽k⩽N,∑j=1N‍mjPj-P~k|Pj-P~k|3+∑j≠kN‍m~jP~j-P~k|P~j-P~k|3=-λP~k,1⩽k⩽N.
Under the case that twist angle φ=0, Moeckel-Simo proved.Theorem 2 (see [4]).
IfN<473, there is a unique pair of spatial central configurations of parallel regular N-gons. If N≥473, there is no such central configuration for b<b0(N), where b is the mass ratio. At a unique pair bifurcates from the planar central configuration with the smaller masses on the inner polygon. This remains the unique pair of spatial central configurations until b=1/b0, where a similar bifurcation occurs in reverse, so that b>1/b0, and only the planar central configurations remain.Xie et al. [5] studied the necessary conditions for the masses of two layer regular polygon central configurations in R3, and they proved the following theorems.Theorem 3.
Under the assumptions of (9), if m1,m2,…,mN and m~1,m~2,…,m~N form a central configuration, then(i)
(14)λ=MN∑j=1N‍1|qj-q~N|3,(ii)
forN⩾2 (φ≠π/2,  when N=2),
(15)m1=m2=⋯=mN,m~1=m~2=⋯=m~N.Without loss of generality, suppose that m~j=bmj and N⩾2 (φ≠π/2, when N=2), under the above assumptions, there are four parameters, and ratio b of masses ratio, a of radius of two regular N-gons, the distance h between two layers, and the phase difference φ. As for these parameters, Xie et al. [5] proved.Theorem 4.
Under the assumptions of (9) and m~j=bmj, then m1,m2,…,mN and m~1,m~2,…,m~N form a central configuration if and only if the parameters b, a, h, and φ satisfy the following relationships:
(16)λM=1N(1+b)(∑j≠N‍1-ρj|qj-qN|3+∑j‍b(1-aρjeiφ)|q~j-qN|3),λM=1N∑j=1N‍1|qj-q~N|3,λM=e-iφNa(1+b)(∑j‍aeiφ-ρj|qj-q~N|3+∑j≠N‍b(a-aρjeiφ)|q~j-q~N|3).
Zhang and Zhu [1] proved the sufficient conditions for special cases a=1, b=1, and φ=π/N.Theorem 5.
Under the assumptions of (9) and (15) and m~j=bmj if a=1, b=1, and φ=π/N, then for every N there exists a unique central configuration.
Whenh=0, Bang and Elmabsout [6] study the twist angle.Theorem 6.
LetΠ1  (resp., Π2)  be a regular N-gon centred around a mass m0 at O, m1 being at each of its vertices (resp., m2). Then m0, Π1, and Π2 are relative equilibriums if and only if they are homothetic or cursed with an angle equal to (π/N) (and suitable ratio of radii).
A natural interesting problem is that whether (9) form a central configuration for h>0 if and only if twist angle φ=0 or π/N. In this paper, one will prove the necessary condition of twisted angle φ for a special case N=4.Theorem 7.
Under the assumptions of (9) and (15) and m~j=bmj if N=4, then φ=π/4+(π/2)k or φ=(π/2)k, (k=0,1,2,3).


## 2. The Proof of Theorem7

### 2.1. Some Lemmas

We need three lemmas.Lemma 1 (see [7]).
ForN≥3 and m~1=m~2=⋯=m~N, if m~1,m~2,…,m~N locate on vertices of a regular polygon, then they form a central configuration.
One hasm1*=m2*=m3*=m4*. Also m1*, m2*, m3*, and m4* are located on vertices of a square S; qi* is the position mi*(i=1,2,3,4) and q* is the position of m*. Plane P contains the square S. One has F*=∑i=14‍mi*m*(qi*-q*)/|qi*-q*|3.Lemma 2.
F
*’s projection on P is directed toward the center of the square if and only if the point m* is on a vertical plane of symmetry.


### 2.2. The Proof of Lemma2

Without loss of generality, suppose thatm1*=m2*=m3*=m4*=1, q1*=(1,0,0), q2*=(0,1,0), q3*=(-1,0,0), q4*=(0,-1,0), and q*=(ρcosθ,ρsinθ,h), where ρ>0. Consider the following:
(17)F*=m*(1-ρcosθ,-ρsinθ,-h)|1+ρ2+h2-2ρcosθ|3/2+m*(-ρcosθ,1-ρsinθ,-h)|1+ρ2+h2-2ρsinθ|3/2+m*(-1-ρcosθ,-ρsinθ,-h)|1+ρ2+h2+2ρcosθ|3/2+m*(-ρcosθ,-1-ρsinθ,-h)|1+ρ2+h2+2ρsinθ|3/2.IfF*’s projection on P is directed toward the center of the square, there exists λ*>0 such that
(18)m*(1-ρcosθ)|1+ρ2+h2-2ρcosθ|3/2+m*(-ρcosθ)|1+ρ2+h2-2ρsinθ|3/2+m*(-1-ρcosθ)|1+ρ2+h2+2ρcosθ|3/2+m*(-ρcosθ)|1+ρ2+h2+2ρsinθ|3/2=-λ*ρcosθ,(19)=m*(-ρsinθ)|1+ρ2+h2-2ρcosθ|3/2+m*(1-ρsinθ)|1+ρ2+h2-2ρsinθ|3/2+m*(-ρsinθ)|1+ρ2+h2+2ρcosθ|3/2+m*(-1-ρsinθ)|1+ρ2+h2+2ρsinθ|3/2=-λ*ρsinθ.Ifθ=(π/2)k, m* is on a vertical plane of symmetry.Whenθ≠(π/2)k, multiplying both sides of (18) and (19), respectively, by sinθ and cosθ, we have
(20)sinθ-ρcosθsinθ|1+ρ2+h2-2ρcosθ|3/2+-ρcosθsinθ|1+ρ2+h2-2ρsinθ|3/2+-sinθ-ρcosθsinθ|1+ρ2+h2+2ρcosθ|3/2+-ρcosθsinθ|1+ρ2+h2+2ρsinθ|3/2=-λ*ρcosθsinθ,-ρcosθsinθ|1+ρ2+h2-2ρcosθ|3/2+cosθ-ρcosθsinθ|1+ρ2+h2-2ρsinθ|3/2+-ρcosθsinθ|1+ρ2+h2+2ρcosθ|3/2+-cosθ-ρcosθsinθ|1+ρ2+h2+2ρsinθ|3/2=-λ*ρcosθsinθ.By (20) we have
(21)sinθ|1+ρ2+h2-2ρcosθ|3/2+-cosθ|1+ρ2+h2-2ρsinθ|3/2+-sinθ|1+ρ2+h2+2ρcosθ|3/2+cosθ|1+ρ2+h2+2ρsinθ|3/2=0.Then(22)1sinθ(1|1+ρ2+h2-2ρsinθ|3/2-1|1+ρ2+h2+2ρsinθ|3/2)=1cosθ(1|1+ρ2+h2-2ρcosθ|3/2-1|1+ρ2+h2+2ρcosθ|3/2).Let(23)f(x)=1x(1|1+ρ2+h2-2ρx|3/2-1|1+ρ2+h2+2ρx|3/2);
then the system (22) is equivalent to f(sinθ)=f(cosθ). It is obvious that f(x) is even and f(x)>0 when x>0. We will prove that f(x) is a strictly increasing function when x>0.When0<x≤1 we compute
(24)df(x)dx=-1x2(1|1+ρ2+h2-2ρx|3/2-1|1+ρ2+h2+2ρx|3/2)+1x(3ρ|1+ρ2+h2-2ρx|5/2+3ρ|1+ρ2+h2+2ρx|5/2)=(1+ρ2+h2+2ρx)(1+ρ2+h2-2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2-(1+ρ2+h2-2ρx)(1+ρ2+h2+2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2+3ρx(1+ρ2+h2+2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2+3ρx(1+ρ2+h2-2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2.Let(25)g(x)=(1+ρ2+h2+2ρx)×(1+ρ2+h2-2ρx)5/2-(1+ρ2+h2-2ρx)×(1+ρ2+h2+2ρx)5/2+3ρx(1+ρ2+h2+2ρx)5/2+3ρx(1+ρ2+h2-2ρx)5/2.Then(26)df(x)dx=g(x)x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2,(27)g(0)=(1+ρ2+h2)7/2-(1+ρ2+h2)7/2=0,(28)dg(x)dx=2ρ(1+ρ2+h2-2ρx)5/2-5ρ(1+ρ2+h2+2ρx)×(1+ρ2+h2-2ρx)3/2+2ρ(1+ρ2+h2+2ρx)5/2-5ρ(1+ρ2+h2-2ρx)×(1+ρ2+h2+2ρx)3/2+3ρ(1+ρ2+h2+2ρx)5/2+15ρ2x(1+ρ2+h2+2ρx)3/2+3ρ(1+ρ2+h2-2ρx)5/2-15ρ2x(1+ρ2+h2-2ρx)3/2,(29)dg(x)dx|x=0=10ρ(1+ρ2+h2)5/2-10ρ(1+ρ2+h2)5/2=0,(30)d2g(x)dx2=-10ρ2(1+ρ2+h2-2ρx)3/2-10ρ2(1+ρ2+h2-2ρx)3/2+15ρ2(1+ρ2+h2+2ρx)×(1+ρ2+h2-2ρx)1/2+10ρ2(1+ρ2+h2+2ρx)3/2+10ρ2(1+ρ2+h2+2ρx)3/2-15ρ2(1+ρ2+h2-2ρx)×(1+ρ2+h2+2ρx)1/2+15ρ2(1+ρ2+h2+2ρx)3/2+15ρ2(1+ρ2+h2+2ρx)3/2+45ρ3x(1+ρ2+h2+2ρx)1/2-15ρ2(1+ρ2+h2-2ρx)3/2-15ρ2(1+ρ2+h2-2ρx)3/2+45ρ3x(1+ρ2+h2-2ρx)1/2≥45ρ3x(1+ρ2+h2+2ρx)1/2+45ρ3x(1+ρ2+h2-2ρx)1/2>0,∀0<x≤1.Hence by (26), (27), (29), and (30) we have
(31)g(x)>0,∀0<x≤1.Hencedf(x)/dx>0, when 0<x≤1. From f(sinθ)=f(cosθ) we obtain |sinθ|=|cosθ|; that is, θ=(π/4)+(π/2)k, so m* is on a vertical plane of symmetry.It is obvious that if the pointm* is on a vertical plane of symmetry, F*’s projection on P is directed toward the center of the square.The proof of Lemma2 is completed.


### 2.3. The Proof of Theorem7

Without loss of generality, suppose thatm~1=m~2=m~3=m~4=m, m1=m2=m3=m4=1. We can let
(32)q1=(1,0,0),q2=(0,1,0),q3=(-1,0,0),q4=(0,-1,0),q~1=(ρcosφ,ρsinφ,h),q~2=(ρcos(2π4+φ),ρsin(2π4+φ),h),q~3=(ρcos(2π4×2+φ),ρsin(2π4×2+φ),h),q~4=(ρcos(2π4×3+φ),ρsin(2π4×3+φ),h),
where ρh>0. It is obvious that z0=∑j=14‍(mjqj+m~j+q~j)/M= (0,0,4mh/(4+4m))=(0,0,mh/(1+m)). By Lemma 1, since m~1, m~2, m~3,  and m~4 locate on vertices of a regular polygon, so they form a central configuration; then there exists a constant λ~ (notice that it can be different from λ in(16)), such that
(33)∑j≠k4‍m~jq~j-q~k|q~j-q~k|=-λ~(q~k-z~0),
where z~0=(∑j=14‍q~jm~j)/4m=(0,0,h).By (1) and (33) we have
(34)-λmP~k=∑j=14‍mqj-q~k|qj-q~k|3+∑j≠k4‍mmq~j-q~k|q~j-q~k|3=∑j=14‍mPj-P~k|Pj-P~k|3+∑j≠k4‍mmP~j-P~k|P~j-P~k|3=∑j=14‍mqj-q~k|qj-q~k|3-λ~m(q~k-z~0)(k=1,2,3,4).Then(35)∑j=14‍m(qj-q~k)|qj-q~k|3=-λm(q~k-z0)+λ~m(q~k-z~0)=m(λ~-λ)q~k+m(0,0,λmh1+m)-m(0,0,λ~h).
Letting F1~=∑j=14‍m(qj-q~1)/|qj-q~1|3, F1~ is the force generated by the m1, m2, m3, and m4.By (19), F1~’s projection on the plane P1 is (λ~-λ)(ρcosφ,ρsinφ,0), where P1 is the plane containing m1, m2, m3, and m4. It is obvious that the projection is directed toward the center of the lower layer regular 4-gons. By Lemma 2, we have φ=π/4+(π/2)k or φ=(π/2)k(k=0,1,2,3).The proof of Theorem7 is completed.


## 2.1. Some Lemmas

We need three lemmas.Lemma 1 (see [7]).
ForN≥3 and m~1=m~2=⋯=m~N, if m~1,m~2,…,m~N locate on vertices of a regular polygon, then they form a central configuration.
One hasm1*=m2*=m3*=m4*. Also m1*, m2*, m3*, and m4* are located on vertices of a square S; qi* is the position mi*(i=1,2,3,4) and q* is the position of m*. Plane P contains the square S. One has F*=∑i=14‍mi*m*(qi*-q*)/|qi*-q*|3.Lemma 2.
F
*’s projection on P is directed toward the center of the square if and only if the point m* is on a vertical plane of symmetry.


## 2.2. The Proof of Lemma2

Without loss of generality, suppose thatm1*=m2*=m3*=m4*=1, q1*=(1,0,0), q2*=(0,1,0), q3*=(-1,0,0), q4*=(0,-1,0), and q*=(ρcosθ,ρsinθ,h), where ρ>0. Consider the following:
(17)F*=m*(1-ρcosθ,-ρsinθ,-h)|1+ρ2+h2-2ρcosθ|3/2+m*(-ρcosθ,1-ρsinθ,-h)|1+ρ2+h2-2ρsinθ|3/2+m*(-1-ρcosθ,-ρsinθ,-h)|1+ρ2+h2+2ρcosθ|3/2+m*(-ρcosθ,-1-ρsinθ,-h)|1+ρ2+h2+2ρsinθ|3/2.IfF*’s projection on P is directed toward the center of the square, there exists λ*>0 such that
(18)m*(1-ρcosθ)|1+ρ2+h2-2ρcosθ|3/2+m*(-ρcosθ)|1+ρ2+h2-2ρsinθ|3/2+m*(-1-ρcosθ)|1+ρ2+h2+2ρcosθ|3/2+m*(-ρcosθ)|1+ρ2+h2+2ρsinθ|3/2=-λ*ρcosθ,(19)=m*(-ρsinθ)|1+ρ2+h2-2ρcosθ|3/2+m*(1-ρsinθ)|1+ρ2+h2-2ρsinθ|3/2+m*(-ρsinθ)|1+ρ2+h2+2ρcosθ|3/2+m*(-1-ρsinθ)|1+ρ2+h2+2ρsinθ|3/2=-λ*ρsinθ.Ifθ=(π/2)k, m* is on a vertical plane of symmetry.Whenθ≠(π/2)k, multiplying both sides of (18) and (19), respectively, by sinθ and cosθ, we have
(20)sinθ-ρcosθsinθ|1+ρ2+h2-2ρcosθ|3/2+-ρcosθsinθ|1+ρ2+h2-2ρsinθ|3/2+-sinθ-ρcosθsinθ|1+ρ2+h2+2ρcosθ|3/2+-ρcosθsinθ|1+ρ2+h2+2ρsinθ|3/2=-λ*ρcosθsinθ,-ρcosθsinθ|1+ρ2+h2-2ρcosθ|3/2+cosθ-ρcosθsinθ|1+ρ2+h2-2ρsinθ|3/2+-ρcosθsinθ|1+ρ2+h2+2ρcosθ|3/2+-cosθ-ρcosθsinθ|1+ρ2+h2+2ρsinθ|3/2=-λ*ρcosθsinθ.By (20) we have
(21)sinθ|1+ρ2+h2-2ρcosθ|3/2+-cosθ|1+ρ2+h2-2ρsinθ|3/2+-sinθ|1+ρ2+h2+2ρcosθ|3/2+cosθ|1+ρ2+h2+2ρsinθ|3/2=0.Then(22)1sinθ(1|1+ρ2+h2-2ρsinθ|3/2-1|1+ρ2+h2+2ρsinθ|3/2)=1cosθ(1|1+ρ2+h2-2ρcosθ|3/2-1|1+ρ2+h2+2ρcosθ|3/2).Let(23)f(x)=1x(1|1+ρ2+h2-2ρx|3/2-1|1+ρ2+h2+2ρx|3/2);
then the system (22) is equivalent to f(sinθ)=f(cosθ). It is obvious that f(x) is even and f(x)>0 when x>0. We will prove that f(x) is a strictly increasing function when x>0.When0<x≤1 we compute
(24)df(x)dx=-1x2(1|1+ρ2+h2-2ρx|3/2-1|1+ρ2+h2+2ρx|3/2)+1x(3ρ|1+ρ2+h2-2ρx|5/2+3ρ|1+ρ2+h2+2ρx|5/2)=(1+ρ2+h2+2ρx)(1+ρ2+h2-2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2-(1+ρ2+h2-2ρx)(1+ρ2+h2+2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2+3ρx(1+ρ2+h2+2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2+3ρx(1+ρ2+h2-2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2.Let(25)g(x)=(1+ρ2+h2+2ρx)×(1+ρ2+h2-2ρx)5/2-(1+ρ2+h2-2ρx)×(1+ρ2+h2+2ρx)5/2+3ρx(1+ρ2+h2+2ρx)5/2+3ρx(1+ρ2+h2-2ρx)5/2.Then(26)df(x)dx=g(x)x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2,(27)g(0)=(1+ρ2+h2)7/2-(1+ρ2+h2)7/2=0,(28)dg(x)dx=2ρ(1+ρ2+h2-2ρx)5/2-5ρ(1+ρ2+h2+2ρx)×(1+ρ2+h2-2ρx)3/2+2ρ(1+ρ2+h2+2ρx)5/2-5ρ(1+ρ2+h2-2ρx)×(1+ρ2+h2+2ρx)3/2+3ρ(1+ρ2+h2+2ρx)5/2+15ρ2x(1+ρ2+h2+2ρx)3/2+3ρ(1+ρ2+h2-2ρx)5/2-15ρ2x(1+ρ2+h2-2ρx)3/2,(29)dg(x)dx|x=0=10ρ(1+ρ2+h2)5/2-10ρ(1+ρ2+h2)5/2=0,(30)d2g(x)dx2=-10ρ2(1+ρ2+h2-2ρx)3/2-10ρ2(1+ρ2+h2-2ρx)3/2+15ρ2(1+ρ2+h2+2ρx)×(1+ρ2+h2-2ρx)1/2+10ρ2(1+ρ2+h2+2ρx)3/2+10ρ2(1+ρ2+h2+2ρx)3/2-15ρ2(1+ρ2+h2-2ρx)×(1+ρ2+h2+2ρx)1/2+15ρ2(1+ρ2+h2+2ρx)3/2+15ρ2(1+ρ2+h2+2ρx)3/2+45ρ3x(1+ρ2+h2+2ρx)1/2-15ρ2(1+ρ2+h2-2ρx)3/2-15ρ2(1+ρ2+h2-2ρx)3/2+45ρ3x(1+ρ2+h2-2ρx)1/2≥45ρ3x(1+ρ2+h2+2ρx)1/2+45ρ3x(1+ρ2+h2-2ρx)1/2>0,∀0<x≤1.Hence by (26), (27), (29), and (30) we have
(31)g(x)>0,∀0<x≤1.Hencedf(x)/dx>0, when 0<x≤1. From f(sinθ)=f(cosθ) we obtain |sinθ|=|cosθ|; that is, θ=(π/4)+(π/2)k, so m* is on a vertical plane of symmetry.It is obvious that if the pointm* is on a vertical plane of symmetry, F*’s projection on P is directed toward the center of the square.The proof of Lemma2 is completed.


## 2.3. The Proof of Theorem7

Without loss of generality, suppose thatm~1=m~2=m~3=m~4=m, m1=m2=m3=m4=1. We can let
(32)q1=(1,0,0),q2=(0,1,0),q3=(-1,0,0),q4=(0,-1,0),q~1=(ρcosφ,ρsinφ,h),q~2=(ρcos(2π4+φ),ρsin(2π4+φ),h),q~3=(ρcos(2π4×2+φ),ρsin(2π4×2+φ),h),q~4=(ρcos(2π4×3+φ),ρsin(2π4×3+φ),h),
where ρh>0. It is obvious that z0=∑j=14‍(mjqj+m~j+q~j)/M= (0,0,4mh/(4+4m))=(0,0,mh/(1+m)). By Lemma 1, since m~1, m~2, m~3,  and m~4 locate on vertices of a regular polygon, so they form a central configuration; then there exists a constant λ~ (notice that it can be different from λ in(16)), such that
(33)∑j≠k4‍m~jq~j-q~k|q~j-q~k|=-λ~(q~k-z~0),
where z~0=(∑j=14‍q~jm~j)/4m=(0,0,h).By (1) and (33) we have
(34)-λmP~k=∑j=14‍mqj-q~k|qj-q~k|3+∑j≠k4‍mmq~j-q~k|q~j-q~k|3=∑j=14‍mPj-P~k|Pj-P~k|3+∑j≠k4‍mmP~j-P~k|P~j-P~k|3=∑j=14‍mqj-q~k|qj-q~k|3-λ~m(q~k-z~0)(k=1,2,3,4).Then(35)∑j=14‍m(qj-q~k)|qj-q~k|3=-λm(q~k-z0)+λ~m(q~k-z~0)=m(λ~-λ)q~k+m(0,0,λmh1+m)-m(0,0,λ~h).
Letting F1~=∑j=14‍m(qj-q~1)/|qj-q~1|3, F1~ is the force generated by the m1, m2, m3, and m4.By (19), F1~’s projection on the plane P1 is (λ~-λ)(ρcosφ,ρsinφ,0), where P1 is the plane containing m1, m2, m3, and m4. It is obvious that the projection is directed toward the center of the lower layer regular 4-gons. By Lemma 2, we have φ=π/4+(π/2)k or φ=(π/2)k(k=0,1,2,3).The proof of Theorem7 is completed.



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*Source: 315976-2013-11-03.xml*