Spatial Central Configurations with Two Twisted Regular 4-Gons

. We study the configuration formed by two squares in two parallel layers separated by a distance. We picture the two layers horizontally with the 𝑧 -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) .


Introduction and Main Results
This paper uses the same notations as [1].The Newtonian body problems [2,3] concern the motions of  particles with masses   ∈  + and positions   ∈  3 ( = 1, 2, . . ., ).The motion is governed by Newton's second law and the Universal law: where  = ( 1 ,  2 , . . .,   ) with the Newtonian potential: Consider the space 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 = { :   =   for some  ̸ = }.The set  \ Δ is called the configuration space.
Definition 1 (see [2,3]).A configuration  = ( 1 ,  2 , . . .,   ) ∈  \ Δ is called a central configuration if there exists a constant  such that The value of constant  in ( 4) is uniquely determined by where Consider the configuration in  3 consisting of two layer regular -gons ( ⩾ 2) with distance ℎ ⩾ 0. It is assumed that the lower layer regular -gons lie in horizontal plane, and the upper regular -gons parallel the lower one, and -axis passes through both centers of two regular -gons.Suppose that the lower layer particles have masses  1 ,  2 , . . .,   and the upper layer particles have masses m1 , m2 , . . ., m , respectively; then these assumptions can be interpreted more precisely by the following.Let   denote for all  complex roots of unity, that is, Let where  > 0,  = √ − where ℎ ≥ 0 is the distance between the two layers.Then the center of masses is where Let If  1 ,  2 , . . .,   and m1 , m2 , . . ., m form a central configuration, then there is  ∈  + such that Under the case that twist angle  = 0, Moeckel-Simo proved.
Theorem 2 (see [4]).If  < 473, there is a unique pair of spatial central configurations of parallel regular -gons.If  ≥ 473, there is no such central configuration for  <  0 (), 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  = 1/ 0 , where a similar bifurcation occurs in reverse, so that  > 1/ 0 , 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  3 , and they proved the following theorems.Theorem 3.Under the assumptions of (9), if  1 ,  2 , . . .,   and m1 , m2 , . . ., m form a central configuration, then Without loss of generality, suppose that m =   and  ⩾ 2 ( ̸ = /2, when  = 2), under the above assumptions, there are four parameters, and ratio  of masses ratio,  of radius of two regular -gons, the distance ℎ between two layers, and the phase difference .As for these parameters, Xie et al. [5] proved.Theorem 4.Under the assumptions of (9) and m =   , then  1 ,  2 , . . .,   and m1 , m2 , . . ., m form a central configuration if and only if the parameters , , ℎ, and  satisfy the following relationships: Zhang and Zhu [1] proved the sufficient conditions for special cases  = 1,  = 1, and  = /.
Theorem 6.Let Π 1 (resp., Π 2 ) be a regular -gon centred around a mass  0 at O,  1 being at each of its vertices (resp.,  2 ).Then  0 , Π 1 , and Π 2 are relative equilibriums if and only if they are homothetic or cursed with an angle equal to (/) (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  = 0 or /.In this paper, one will prove the necessary condition of twisted angle  for a special case  = 4.

The Proof of Theorem 7
2.1.Some Lemmas.We need three lemmas.