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 -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 or () .
1. Introduction and Main Results
This paper uses the same notations as . The Newtonian -body problems [2, 3] concern the motions of particles with masses and positions . The motion is governed by Newton’s second law and the Universal law: where 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 for some . The set is called the configuration space.
Definition 1 (see [2, 3]). A configuration 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 consisting of two layer regular -gons with distance . 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 and the upper layer particles have masses , respectively; then these assumptions can be interpreted more precisely by the following. Let denote for all complex roots of unity, that is, Let where , , , and is called twist angle. It is assumed that locates at the vertex of the lower layer regular -gons; locates on the vertex of the upper layer regular -gons: where is the distance between the two layers. Then the center of masses is where Let If and form a central configuration, then there is such that Under the case that twist angle , Moeckel-Simo proved.
Theorem 2 (see ). If , there is a unique pair of spatial central configurations of parallel regular -gons. If , there is no such central configuration for , 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 , where a similar bifurcation occurs in reverse, so that , and only the planar central configurations remain.
Xie et al.  studied the necessary conditions for the masses of two layer regular polygon central configurations in , and they proved the following theorems.
Theorem 3. Under the assumptions of (9), if and form a central configuration, then(i)(ii) for (, when ), Without loss of generality, suppose that and (, when ), under the above assumptions, there are four parameters, and ratio of masses ratio, of radius of two regular -gons, the distance between two layers, and the phase difference . As for these parameters, Xie et al.  proved.
Theorem 4. Under the assumptions of (9) and , then and form a central configuration if and only if the parameters , , , and satisfy the following relationships: Zhang and Zhu  proved the sufficient conditions for special cases , , and .
Theorem 6. Let (resp., ) be a regular -gon centred around a mass at O, being at each of its vertices (resp., ). Then , , and 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 if and only if twist angle or . In this paper, one will prove the necessary condition of twisted angle for a special case .
2. The Proof of Theorem 7
2.1. Some Lemmas
We need three lemmas.
Lemma 1 (see ). For and , if locate on vertices of a regular polygon, then they form a central configuration.
One has . Also , , , and are located on vertices of a square ; is the position and is the position of . Plane contains the square . One has .
Lemma 2. ’s projection on is directed toward the center of the square if and only if the point is on a vertical plane of symmetry.
2.2. The Proof of Lemma 2
Without loss of generality, suppose that , , , , , and , where . Consider the following:
If ’s projection on is directed toward the center of the square, there exists such that
If , is on a vertical plane of symmetry.
By (20) we have
Let then the system (22) is equivalent to . It is obvious that is even and when . We will prove that is a strictly increasing function when .
When we compute
Hence , when . From we obtain ; that is, , so is on a vertical plane of symmetry.
It is obvious that if the point is on a vertical plane of symmetry, ’s projection on is directed toward the center of the square.
The proof of Lemma 2 is completed.
2.3. The Proof of Theorem 7
Without loss of generality, suppose that , . We can let where . It is obvious that . By Lemma 1, since , , , and locate on vertices of a regular polygon, so they form a central configuration; then there exists a constant (notice that it can be different from in(16)), such that where .
Then Letting , is the force generated by the , , , and .
By (19), ’s projection on the plane is , where is the plane containing , , , and . It is obvious that the projection is directed toward the center of the lower layer regular 4-gons. By Lemma 2, we have or .
The proof of Theorem 7 is completed.
The authors express their gratitude to Professor Zhang Shiqing for his discussions and helpful suggestions. This work is Supported by NSF of China and Youth Fund of Mianyang Normal University.
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