- Email: [email protected]
Twisted stacked central configurations for the spatial seven-body problem
Journal of Geometry and Physics 70 (2013) 164–171
Contents lists available at SciVerse ScienceDirect
Journal of Geometry and Physics journal homepage: www.elsevier.com/locate/jgp
Twisted stacked central configurations for the spatial seven-body problem Xia Su a,b,∗ , Tianqing An a a
College of Science, Hohai University, Nanjing 210098, China
b
Faculty of Mathematics and Physics, Huaiyin Institute of Technology, Huai’an 223003, China
article
abstract
info
Article history: Received 30 December 2012 Received in revised form 23 March 2013 Accepted 23 March 2013 Available online 1 April 2013
In this paper, we show the existence of the twisted stacked central configurations for 7-body problem. More precisely, the position vectors x1 , x2 , x3 and x4 are at the vertices of a regular tetrahedron Σ ; the position vectors x5 , x6 and x7 are at the vertices of an equilateral triangle Π ; the triangle (x1 , x2 , x3 ) and the triangle (x5 , x6 , x7 ) have twisted angle π3 . © 2013 Elsevier B.V. All rights reserved.
MSC: 34C15 34C25 Keywords: Seven-body problem Stacked central configuration Celestial mechanics
1. Introduction and main results The classical n-body problem concerns with the motion of n mass points moving in space according to Newton’s second law and the gravitational law: mi x¨ i =
m k m i ( xk − xi ) k̸=i
|xk − xi |3
,
i = 1, 2, . . . , n.
(1.1)
Here xi ∈ R3 is the position of mass mi > 0. Alternatively, system (1.1) can be rewritten as mi x¨ i =
∂ U (x) , ∂ xi
i = 1, 2, . . . , n
(1.2)
where U (x) = U (x1 , x2 , . . . , xn ) =
mk mj
|xk − xj | 1≤k
(1.3)
is the Newtonian potential of system (1.1). The position vector x = (x1 , x2 , . . . , xn ) ∈ (R3 )n is often referred to the configuration of the system; the vectors xi (i = 1, 2, . . . , n) are vertices of the configuration x.
∗
Corresponding author at: College of Science, Hohai University, Nanjing 210098, China. Tel.: +86 15152359290. E-mail address: [email protected] (X. Su).
0393-0440/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.geomphys.2013.03.026
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
165
Let M = m1 + · · · + mn be the total mass and 1 c= (m1 x1 + · · · + mn xn ) M be the center of mass of the configuration x. Because the potential is singular when two particles have the same position, it is natural to assume that the configuration avoids the set
△ = {x ∈ (R3 )n : xi = xj for some i ̸= j}. A configuration x = (x1 , . . . , xn ) ∈ (R3 )n \ △ is called a central configuration if there exists some positive constant λ, called the multiplier, such that n mj (xj − xi ) , |xj − xi |3 j=1,j̸=i
−λ(xi − c ) =
i = 1, 2, . . . , n.
(1.4)
It is easy to see that a central configuration remains a central configuration after a rotation in R3 and a scalar multiplication. More precisely, let A ∈ SO(3) and a > 0, if x = (x1 , . . . , xn ) is a central configuration, so are Ax = (Ax1 , Ax2 , . . . , Axn ) and ax = (ax1 , ax2 , . . . , axn ). Two central configurations are said to be equivalent if one can be transformed to another 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 equivalence relation. The study of central configuration goes back to Euler and Lagrange. For n = 3, it is a classical result there are three collinear, called Euler, central configurations and one equilateral triangular, called Lagrange, central configurations. For n = 4, Moulton [1] proved that there is exactly one collinear central configuration for each arrangement of the mass points on the line. There are several reasons why central configurations are of special importance in the study of the n-body problem; see [2–4] for details. A stacked central configuration is a central configuration in which a proper subset of the n bodies is already on a central configuration. This class of central configuration of 5-body problem was introduced by Hampton in [5]. The work of [5] was complemented by Llibre in [6,7]. Zhang and Zhou [8] showed the existence of double pyramidal central configurations of N + 2-body problem. Hampton and Santoprete [9] provided some examples of stacked central configurations for the spatial 7-body problem where the bodies are arranged as concentric three and two dimensional simplex. Mello and Fernandes [10] provided new examples of stacked central configuration for spatial 7-body problem where the 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 the exterior of regular tetrahedron. In this paper, we show the existence of the twisted stacked central configurations of the 7-body problem. The spatial central configuration considered here satisfies (see Fig. 1): the position vectors x1 , x2 , x3 and x4 are at the vertices of a regular tetrahedron Σ ; the position vectors x5 , x6 and x7 are at the vertices of an equilateral triangle Π ; the triangle (x1 , x2 , x3 ) and the triangle (x5 , x6 , x7 ) have twisted angle π3 . Without loss of generality, we can assume that
x1 = (1, 0, 0),
x5 =
2
x,
− ,
x2 =
√ 1
3 2
√ 1
2
3 2
,0 ,
,
x6 = (−x, 0, z ),
3
− ,− ,0 , 2 2 √
x3 =
x, z
√ 1
1
x7 =
2
x, −
3
2
x, z
x4 = (0, 0,
√
2), (1.5)
,
where x > 0 is the radius of the circle that contains the equilateral triangle Π and z ∈ R is the signed distance between the plane that contains x1 , x2 and x3 and the plane that contains Π . The main results of this paper are the following. Theorem 1.1. According to Fig. 1, in order that the seven mass points are in a central configuration, the following statements are necessary: 1. The masses m1 , m2 and m3 must be equal. 2. The masses m5 , m6 and m7 must be equal. Theorem 1.2. There exist points P0 (x0 , y0 ) ∈ T −1 (0) ∩ D1 such that the seven bodies take the coordinates
x1 = (1, 0, 0),
x5 =
x2 =
√ 1 2
x0 ,
3 2
√ 1
− , 2
3 2
,0 ,
x3 =
√ 1
− ,− 2
x0 , z0
,
x6 = (−x0 , 0, z0 ),
x7 =
3 2
,0 ,
x4 = (0, 0,
√ 1 2
x0 , −
3 2
√
2),
x0 , z0
.
Then there are positive solutions of m1 , m4 , m5 such that these bodies form a spatial central configuration according to Fig. 1.
166
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
Fig. 1. The configuration of the spatial seven-body problem.
2. Proof of Theorem 1.1 For the spatial central configurations, instead of working with Eq. (1.4), we consider the Dziobek–Laura–Andoyer equations (see [9], p. 295) gijh =
n
mk (dik − djk )∆ijhk = 0
(2.1)
k=1,k̸=i,j,h
for 1 ≤ i < j ≤ n, h = 1, 2, . . . , n, h ̸= i, j. Here dij = |x −1x |3 and ∆ijhk = (xi − xj ) ∧ (xi − xh ) · (xi − xk ). Thus ∆ijhk gives six i j times the signed volume of the tetrahedron formed by the bodies with positions xi , xj , xh and xk . For the 7-body problem, Eq. (2.1) is a system of 105 equations. According to Fig. 1, our class of configurations with seven bodies must satisfy 3 d12 = d13 = d14 = d23 = d24 = d34 = 3− 2 3
d15 = d17 = d25 = d26 = d36 = d37 = (1 + x2 − x + z 2 )− 2 2 − 32
d16 = d27 = d35 = ((1 + x)2 + z ) d45 = d46 = d47 = (x√ + (z − d56 = d57 = d67 = ( 3x)−3 . 2
√
(2.2)
2 − 32
2) )
Due to the assumption (1.5) and the definition of ∆ijhk , we have several symmetries in the signed volumes. For example,
∆1235 ∆1256 ∆1267 ∆5671
= ∆1236 = ∆1257 = ∆1356 = ∆5672
= ∆1237 , = ∆1357 = ∆1376 , = ∆2357 , = ∆5673
and much more ones. By using the symmetries and the properties of ∆ijhk , we obtain the following results. Lemma 2.1. In order to have a spatial central configuration according to Fig. 1, a necessary condition is that the masses m1 , m2 and m3 must be equal. Proof. It is sufficient to consider the equations g567 = 0 and g576 = 0. Consider the expressions of g567 and g576 : g567 = m1 (d15 − d16 )∆5671 + m2 (d25 − d26 )∆5672 + m3 (d35 − d36 )∆5673 + m4 (d45 − d46 )∆5674 = 0, g576 = m1 (d15 − d17 )∆5761 + m2 (d25 − d27 )∆5762 + m3 (d35 − d37 )∆5763 + m4 (d45 − d47 )∆5764 = 0.
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
167
By assumption (1.5) and symmetries (2.2), we have g567 = (m1 − m3 )(d15 − d16 )∆5671 = 0, g576 = (m2 − m3 )(d25 − d27 )∆5762 = 0. For our class of central configurations, we have d15 − d16 ̸= 0, d25 − d27 ̸= 0, ∆5671 ̸= 0, ∆5672 ̸= 0. So the above equations hold if and only if m1 = m2 = m3 . So the statement 1 of Theorem 1.1 is proved. Lemma 2.2. If the configuration according to Fig. 1 is a central configuration, a necessary condition is that the masses m5 , m6 and m7 must be equal. Proof. It is sufficient to consider the equations g123 = 0 and g132 = 0. Consider the expressions of g123 and g132 : g123 = m4 (d14 − d24 )∆1234 + m5 (d15 − d25 )∆1235 + m6 (d16 − d26 )∆1236 + m7 (d17 − d27 )∆1237 = 0, g132 = m4 (d14 − d34 )∆1324 + m5 (d15 − d35 )∆1325 + m6 (d16 − d36 )∆1326 + m7 (d17 − d37 )∆1327 = 0. By assumption (1.5) and symmetries (2.2), we have g123 = (m6 − m7 )(d16 − d26 )∆1236 = 0, g132 = (m5 − m6 )(d15 − d35 )∆1325 = 0. For our class of central configurations, we have d16 − d26 ̸= 0, d15 − d35 ̸= 0, ∆1236 ̸= 0, ∆1325 ̸= 0. So the above equations hold if and only if m5 = m6 = m7 . Hence the statement 2 of Theorem 1.1 is proved. We restrict the set of admissible masses to m1 = m2 = m3 = α and m5 = m6 = m7 = β . Substituting m1 = m2 = m3 = α and m5 = m6 = m7 = β into Eq. (2.1), one can easily verify that the following equations are already satisfied: g123 g136 g237 g561 g574
= g124 = g137 = g247 = g562 = g576
= g125 = g146 = g274 = g563 = g671
= g126 = g164 = g345 = g564 = g672
= g127 = g231 = g354 = g567 = g673
= g132 = g234 = g453 = g571 = g674
= g134 = g235 = g461 = g572 = g675
= g135 = g236 = g472 = g573 = 0.
= 0, = 0, = 0, = 0,
While the remaining 66 equations can be put into 6 sets of equivalent equations. Set 1. Equations equivalent to g145 = 0: g142 = 0,
g143 = 0,
g147 = 0,
g241 = 0,
g243 = 0,
g246 = 0,
g341 = 0,
g342 = 0,
g346 = 0,
g347 = 0.
g362 = 0,
g371 = 0.
g245 = 0,
Set 2. Equations equivalent to g152 = 0: g173 = 0,
g251 = 0,
g263 = 0,
Set 3. Equations equivalent to g153 = 0: g156 = 0,
g162 = 0,
g163 = 0,
g165 = 0,
g167 = 0,
g172 = 0,
g176 = 0,
g253 = 0,
g257 = 0,
g261 = 0,
g267 = 0,
g271 = 0,
g273 = 0,
g275 = 0,
g276 = 0,
g351 = 0,
g352 = 0,
g356 = 0,
g357 = 0,
g361 = 0,
g365 = 0,
g372 = 0,
g375 = 0.
g364 = 0,
g374 = 0.
g367 = 0,
g376 = 0.
g463 = 0,
g465 = 0,
g467 = 0,
Set 4. Equations equivalent to g154 = 0: g174 = 0,
g254 = 0,
g264 = 0,
Set 5. Equations equivalent to g157 = 0: g175 = 0,
g256 = 0,
g265 = 0,
Set 6. Equations equivalent to g451 = 0: g452 = 0,
g456 = 0,
g457 = 0,
g462 = 0,
g471 = 0,
g473 = 0,
g475 = 0,
g476 = 0.
If we write g145 = β T = β[(d16 − d46 )∆1456 + (d17 − d47 )∆1457 ] = 0, it follows that T = 0. So in the following, we restrict our central configurations in the set T −1 (0). Lemma 2.3. Under our assumptions and in the set T −1 (0), the remaining 54 equations in the set 2, 3, . . . , 6 are satisfied if g152 = g153 = g154 = 0 are satisfied.
168
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
Proof. Consider the expressions of g157 and g153 − g152 : g157 = α(2d12 − d25 − d35 )∆1572 + m4 (d14 − d45 )∆1574 + β(d16 − d56 )∆1576 , g153 − g152 = α(2d12 − d25 − d35 )∆1532 + m4 (d14 − d45 )(∆1534 − ∆1524 ) − β(d16 − d56 )∆1526 . From our assumptions we have
∆1574 ∆1576 ∆1572 = = = x ̸= 0. ∆1532 ∆1534 − ∆1524 −∆1526 Thus g152 = 0 and g153 = 0 imply g157 = 0. Consider the expression of g154 and g451 : g154 = α[(d12 − d25 )∆1542 + (d13 − d35 )∆1543 ] + β[(d16 − d56 )∆1546 + (d17 − d57 )∆1547 ], g451 = α[(d24 − d25 )∆4512 + (d34 − d35 )∆4513 ] + β(d46 − d56 )(∆4516 + ∆4517 ). In the set T −1 (0), we have d46 =
d16 ∆1456 + d17 ∆1457
∆1456 + ∆1457
.
(2.3)
Substituting (2.3) into g451 , it follows that g451 = α[(d24 − d25 )∆4512 + (d34 − d35 )∆4513 ] + β(d16 ∆1456 + d17 ∆1457 − d56 ∆1456 − d56 ∆1457 )
= α[(d24 − d25 )∆4512 + (d34 − d35 )∆4513 ] + β[(d16 − d56 )∆1456 + (d17 − d56 )∆1457 ]. From our assumptions we have
∆1542 ∆1543 ∆1546 ∆1547 = = = = −1 ̸= 0. ∆4512 ∆4513 ∆1456 ∆1457 Hence in the set T −1 (0), g154 = 0 implies g451 = 0. This completes the proof.
From Lemma 2.3, in order to study central configurations according to Fig. 1 in the set T −1 (0), it is sufficient to study the following 3 equations: g152 = 0,
g153 = 0,
g154 = 0.
(2.4)
Denote by A = (aij ) the matrix of the coefficients of the homogeneous linear system in the variables α, m4 , β defined by Eqs. (2.4). Thus a11 = (d13 − d35 )∆1523 ,
a12 = (d14 − d45 )∆1524 ,
a13 = (d16 + d17 − 2d56 )∆1526 ,
a21 = (d12 − d25 )∆1532 ,
a22 = (d14 − d45 )∆1534 ,
a23 = (d17 − d57 )∆1537 ,
a31 = (d12 − d25 )∆1542 + (d13 − d35 )∆1543 , a32 = 0, Let x =
a33 = (d16 − d56 )∆1546 + (d17 − d57 )∆1547 .
α m4
β
. Then in order to get the spatial central configuration like Fig. 1, we need to find positive solution α, m4 , β of
the following system: Ax = 0
(2.5)
where T = 0. By elementary computation we have det (A) = 0, since the third row of A is a linear combination of the first two rows. Thus in the set T −1 (0) Eq. (2.5) have nontrivial solutions. 3. The existence of spatial central configurations Consider the expression of T : T = (d16 − d46 )∆1456 + (d17 − d47 )∆1457 . Now we study the sign of each term of T . The term (d16 − d46 ) is negative if z > √
vanishes on the straight line z = vanishes on the straight line z =
2 4
√
(1 − 2x). The term ∆1456 is negative if z <
√ 2 4
√
(1 − 2x), positive if z <
2(1 + x), positive if z >
2(1 + x). The term (d17 − d47 ) is negative if z >
and vanishes on the straight line z =
√
2 4
(1 + x). The term ∆1457 is negative if z <
√
2 √4 2 2
√ 2 4 √
(1 − 2x) and
2(1 + x) and
(1 + x), positive if z <
(2 − x), positive if z >
√
2
√4 2 2
(1 + x) ( 2 − x)
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
169
Fig. 2. The regions D1 and D2 .
√
and vanishes on the straight line z = D1 ∪ D2 , where D1 = D11 ∪ D12 ,
2 2
(2 − x). Thus the connected components of the set T −1 (0) are restricted to the set
D2 = D21 ∪ D22
and
√
D11 =
(x, z )|0 < x ≤ 1,
D12 =
(x, z )|x > 1,
2 4
D22 =
(x, z )|x > 1,
(2 − x) < z <
(1 + x) < z < 2 4
4
2(1 + x) ,
√
2(1 + x) ,
√ (1 − 2x) < z <
√ 2
√
√
(x, z )|0 < x ≤ 1,
2
√
D21 =
2
√ (1 − 2x) < z <
2 2
2 4
( 1 + x) ,
(2 − x) .
See Fig. 2. In order to prove the existence of positive solutions of (2.5) 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 Fig. 1. Let
A1 = D2 ∩ {(x, z )|2x2 + x − z 2 < 1}, A2 = D2 ∩ {(x, z )|x2 − x + 1 + z 2 > 3} and A = A1 ∪ A2 , see Fig. 3. Then we have the following lemma for (x, z ) ∈ A.
170
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
Fig. 3. The regions A1 , A2 and B .
Lemma 3.1. In the set A, there are no spatial central configurations according to Fig. 1. Proof. Consider the entries a31 = (d12 − d25 )∆1542 + (d13 − d35 )∆1543 and a33 = (d16 − d56 )∆1546 + (d17 − d57 )∆1547 . It is simple to see ∆1542 > 0, ∆1543 > 0, ∆1546 > 0 and ∆1547 > 0 for (x, z ) ∈ A. While d12 − d25 , d13 − d35 , d16 − d56 , d17 − d57 are negative for (x, z ) ∈ A1 , and d12 − d25 , d13 − d35 , d16 − d56 , d17 − d57 are positive for (x, z ) ∈ A2 . Thus the entries a31 and a33 have the same signs. The lemma is proved. Now we prove the existence of spatial central configuration according to Fig. 1 for some points in the set D1 . Proof of Theorem 1.2. Since the matrix A is singular in the set T −1 (0), there are nontrivial solutions of (2.5) in the set T −1 (0). In order to simplify our analysis, we consider the following subset of D1 (see Fig. 3):
B = B1 ∪ B2 where
√
(x, z ) ∈ D1 : 1 < x <
B1 =
√
B2 =
(x, z ) ∈ D1 :
10 2
10 √
2
,
2
2x2
+x−1 ,
√ √ √ 2 ≤ x < 3, 2 < z < 3 − x + 2 .
In the set B , the entries of the matrix A have the following signs:
− A= + +
+ + 0
∗ − . −
Consider the set L = B ∩ {(x, z ) : 1.09419029 < x < 1.69932798, T = 0}.
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
171
Without loss of generality, let α = 1. It is easy to see that the mass
β=−
(d12 − d25 )∆1542 + (d13 − d35 )∆1543 (d16 − d56 )∆1546 + (d17 − d57 )∆1547
is positive. By direct computation, the mass m4 = −
(d12 − d25 )∆1532 + (d17 − d57 )∆1537 β (d14 − d45 )∆1534
is increasing in (1.09419029, 1.69932798) about x in the set L and m4 (1.09419029) = 0. So at every point in the set L , we have nontrivial positive solutions of (2.5). The proof of Theorem 1.2 is complete. Acknowledgments The first author is supported by Natural Science Foundation of China (NFSC11201168) and the Scientific Research Foundation of Huaiyin Institute of Technology (HGA1102). The authors sincerely thank the referees and the editor for their many valuable comments which helped us improve the paper. References [1] [2] [3] [4] [5] [6] [7]
F.R. Moulton, Straight line solutions of the problem of n bodies, 12, 1910, 1–17. D. Sarri, On the role and properties of n-body central configurations, Celestial Mech. 21 (1980) 9–20. S. Smale, Mathematical problems for the next century, Math. Intelligencer. 20 (1998) 7–15. R. Moeckel, On central configurations, Math. Z. 205 (1990) 499–517. M. Hampton, Stacked central configurations: new examples in the planar five-body problem, Nonlinearity 18 (2005) 2299–2304. J. Llibre, L.F. Mello, New central configurations for the planar 5-body problem, Celestial Mech. Dynam. Astronomy. 100 (2008) 141–149. J. Llibre, L.F. Mello, E. Perez-Chavela, New stacked central configurations for the planar 5-body problem, Celestial Mech. Dynam. Astronom. 110 (2011) 45–52. [8] S.Q. Zhang, Q. Zhou, Double pyramidal central configurations, Phys. Lett. A. 281 (2001) 240–248. [9] M. Hampton, M. Santoprete, Seven-body central configurations: a family of central configurations in the spatial seven-body problem, Celestial Mech. Dynam. Astronom. 99 (2007) 293–305. [10] L.F. Mello, A.C. Fernandes, Stacked central configurations for the spatial seven-body problem. Preprint.
Contents lists available at SciVerse ScienceDirect
Journal of Geometry and Physics journal homepage: www.elsevier.com/locate/jgp
Twisted stacked central configurations for the spatial seven-body problem Xia Su a,b,∗ , Tianqing An a a
College of Science, Hohai University, Nanjing 210098, China
b
Faculty of Mathematics and Physics, Huaiyin Institute of Technology, Huai’an 223003, China
article
abstract
info
Article history: Received 30 December 2012 Received in revised form 23 March 2013 Accepted 23 March 2013 Available online 1 April 2013
In this paper, we show the existence of the twisted stacked central configurations for 7-body problem. More precisely, the position vectors x1 , x2 , x3 and x4 are at the vertices of a regular tetrahedron Σ ; the position vectors x5 , x6 and x7 are at the vertices of an equilateral triangle Π ; the triangle (x1 , x2 , x3 ) and the triangle (x5 , x6 , x7 ) have twisted angle π3 . © 2013 Elsevier B.V. All rights reserved.
MSC: 34C15 34C25 Keywords: Seven-body problem Stacked central configuration Celestial mechanics
1. Introduction and main results The classical n-body problem concerns with the motion of n mass points moving in space according to Newton’s second law and the gravitational law: mi x¨ i =
m k m i ( xk − xi ) k̸=i
|xk − xi |3
,
i = 1, 2, . . . , n.
(1.1)
Here xi ∈ R3 is the position of mass mi > 0. Alternatively, system (1.1) can be rewritten as mi x¨ i =
∂ U (x) , ∂ xi
i = 1, 2, . . . , n
(1.2)
where U (x) = U (x1 , x2 , . . . , xn ) =
mk mj
|xk − xj | 1≤k
(1.3)
is the Newtonian potential of system (1.1). The position vector x = (x1 , x2 , . . . , xn ) ∈ (R3 )n is often referred to the configuration of the system; the vectors xi (i = 1, 2, . . . , n) are vertices of the configuration x.
∗
Corresponding author at: College of Science, Hohai University, Nanjing 210098, China. Tel.: +86 15152359290. E-mail address: [email protected] (X. Su).
0393-0440/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.geomphys.2013.03.026
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
165
Let M = m1 + · · · + mn be the total mass and 1 c= (m1 x1 + · · · + mn xn ) M be the center of mass of the configuration x. Because the potential is singular when two particles have the same position, it is natural to assume that the configuration avoids the set
△ = {x ∈ (R3 )n : xi = xj for some i ̸= j}. A configuration x = (x1 , . . . , xn ) ∈ (R3 )n \ △ is called a central configuration if there exists some positive constant λ, called the multiplier, such that n mj (xj − xi ) , |xj − xi |3 j=1,j̸=i
−λ(xi − c ) =
i = 1, 2, . . . , n.
(1.4)
It is easy to see that a central configuration remains a central configuration after a rotation in R3 and a scalar multiplication. More precisely, let A ∈ SO(3) and a > 0, if x = (x1 , . . . , xn ) is a central configuration, so are Ax = (Ax1 , Ax2 , . . . , Axn ) and ax = (ax1 , ax2 , . . . , axn ). Two central configurations are said to be equivalent if one can be transformed to another 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 equivalence relation. The study of central configuration goes back to Euler and Lagrange. For n = 3, it is a classical result there are three collinear, called Euler, central configurations and one equilateral triangular, called Lagrange, central configurations. For n = 4, Moulton [1] proved that there is exactly one collinear central configuration for each arrangement of the mass points on the line. There are several reasons why central configurations are of special importance in the study of the n-body problem; see [2–4] for details. A stacked central configuration is a central configuration in which a proper subset of the n bodies is already on a central configuration. This class of central configuration of 5-body problem was introduced by Hampton in [5]. The work of [5] was complemented by Llibre in [6,7]. Zhang and Zhou [8] showed the existence of double pyramidal central configurations of N + 2-body problem. Hampton and Santoprete [9] provided some examples of stacked central configurations for the spatial 7-body problem where the bodies are arranged as concentric three and two dimensional simplex. Mello and Fernandes [10] provided new examples of stacked central configuration for spatial 7-body problem where the 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 the exterior of regular tetrahedron. In this paper, we show the existence of the twisted stacked central configurations of the 7-body problem. The spatial central configuration considered here satisfies (see Fig. 1): the position vectors x1 , x2 , x3 and x4 are at the vertices of a regular tetrahedron Σ ; the position vectors x5 , x6 and x7 are at the vertices of an equilateral triangle Π ; the triangle (x1 , x2 , x3 ) and the triangle (x5 , x6 , x7 ) have twisted angle π3 . Without loss of generality, we can assume that
x1 = (1, 0, 0),
x5 =
2
x,
− ,
x2 =
√ 1
3 2
√ 1
2
3 2
,0 ,
,
x6 = (−x, 0, z ),
3
− ,− ,0 , 2 2 √
x3 =
x, z
√ 1
1
x7 =
2
x, −
3
2
x, z
x4 = (0, 0,
√
2), (1.5)
,
where x > 0 is the radius of the circle that contains the equilateral triangle Π and z ∈ R is the signed distance between the plane that contains x1 , x2 and x3 and the plane that contains Π . The main results of this paper are the following. Theorem 1.1. According to Fig. 1, in order that the seven mass points are in a central configuration, the following statements are necessary: 1. The masses m1 , m2 and m3 must be equal. 2. The masses m5 , m6 and m7 must be equal. Theorem 1.2. There exist points P0 (x0 , y0 ) ∈ T −1 (0) ∩ D1 such that the seven bodies take the coordinates
x1 = (1, 0, 0),
x5 =
x2 =
√ 1 2
x0 ,
3 2
√ 1
− , 2
3 2
,0 ,
x3 =
√ 1
− ,− 2
x0 , z0
,
x6 = (−x0 , 0, z0 ),
x7 =
3 2
,0 ,
x4 = (0, 0,
√ 1 2
x0 , −
3 2
√
2),
x0 , z0
.
Then there are positive solutions of m1 , m4 , m5 such that these bodies form a spatial central configuration according to Fig. 1.
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X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
Fig. 1. The configuration of the spatial seven-body problem.
2. Proof of Theorem 1.1 For the spatial central configurations, instead of working with Eq. (1.4), we consider the Dziobek–Laura–Andoyer equations (see [9], p. 295) gijh =
n
mk (dik − djk )∆ijhk = 0
(2.1)
k=1,k̸=i,j,h
for 1 ≤ i < j ≤ n, h = 1, 2, . . . , n, h ̸= i, j. Here dij = |x −1x |3 and ∆ijhk = (xi − xj ) ∧ (xi − xh ) · (xi − xk ). Thus ∆ijhk gives six i j times the signed volume of the tetrahedron formed by the bodies with positions xi , xj , xh and xk . For the 7-body problem, Eq. (2.1) is a system of 105 equations. According to Fig. 1, our class of configurations with seven bodies must satisfy 3 d12 = d13 = d14 = d23 = d24 = d34 = 3− 2 3
d15 = d17 = d25 = d26 = d36 = d37 = (1 + x2 − x + z 2 )− 2 2 − 32
d16 = d27 = d35 = ((1 + x)2 + z ) d45 = d46 = d47 = (x√ + (z − d56 = d57 = d67 = ( 3x)−3 . 2
√
(2.2)
2 − 32
2) )
Due to the assumption (1.5) and the definition of ∆ijhk , we have several symmetries in the signed volumes. For example,
∆1235 ∆1256 ∆1267 ∆5671
= ∆1236 = ∆1257 = ∆1356 = ∆5672
= ∆1237 , = ∆1357 = ∆1376 , = ∆2357 , = ∆5673
and much more ones. By using the symmetries and the properties of ∆ijhk , we obtain the following results. Lemma 2.1. In order to have a spatial central configuration according to Fig. 1, a necessary condition is that the masses m1 , m2 and m3 must be equal. Proof. It is sufficient to consider the equations g567 = 0 and g576 = 0. Consider the expressions of g567 and g576 : g567 = m1 (d15 − d16 )∆5671 + m2 (d25 − d26 )∆5672 + m3 (d35 − d36 )∆5673 + m4 (d45 − d46 )∆5674 = 0, g576 = m1 (d15 − d17 )∆5761 + m2 (d25 − d27 )∆5762 + m3 (d35 − d37 )∆5763 + m4 (d45 − d47 )∆5764 = 0.
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
167
By assumption (1.5) and symmetries (2.2), we have g567 = (m1 − m3 )(d15 − d16 )∆5671 = 0, g576 = (m2 − m3 )(d25 − d27 )∆5762 = 0. For our class of central configurations, we have d15 − d16 ̸= 0, d25 − d27 ̸= 0, ∆5671 ̸= 0, ∆5672 ̸= 0. So the above equations hold if and only if m1 = m2 = m3 . So the statement 1 of Theorem 1.1 is proved. Lemma 2.2. If the configuration according to Fig. 1 is a central configuration, a necessary condition is that the masses m5 , m6 and m7 must be equal. Proof. It is sufficient to consider the equations g123 = 0 and g132 = 0. Consider the expressions of g123 and g132 : g123 = m4 (d14 − d24 )∆1234 + m5 (d15 − d25 )∆1235 + m6 (d16 − d26 )∆1236 + m7 (d17 − d27 )∆1237 = 0, g132 = m4 (d14 − d34 )∆1324 + m5 (d15 − d35 )∆1325 + m6 (d16 − d36 )∆1326 + m7 (d17 − d37 )∆1327 = 0. By assumption (1.5) and symmetries (2.2), we have g123 = (m6 − m7 )(d16 − d26 )∆1236 = 0, g132 = (m5 − m6 )(d15 − d35 )∆1325 = 0. For our class of central configurations, we have d16 − d26 ̸= 0, d15 − d35 ̸= 0, ∆1236 ̸= 0, ∆1325 ̸= 0. So the above equations hold if and only if m5 = m6 = m7 . Hence the statement 2 of Theorem 1.1 is proved. We restrict the set of admissible masses to m1 = m2 = m3 = α and m5 = m6 = m7 = β . Substituting m1 = m2 = m3 = α and m5 = m6 = m7 = β into Eq. (2.1), one can easily verify that the following equations are already satisfied: g123 g136 g237 g561 g574
= g124 = g137 = g247 = g562 = g576
= g125 = g146 = g274 = g563 = g671
= g126 = g164 = g345 = g564 = g672
= g127 = g231 = g354 = g567 = g673
= g132 = g234 = g453 = g571 = g674
= g134 = g235 = g461 = g572 = g675
= g135 = g236 = g472 = g573 = 0.
= 0, = 0, = 0, = 0,
While the remaining 66 equations can be put into 6 sets of equivalent equations. Set 1. Equations equivalent to g145 = 0: g142 = 0,
g143 = 0,
g147 = 0,
g241 = 0,
g243 = 0,
g246 = 0,
g341 = 0,
g342 = 0,
g346 = 0,
g347 = 0.
g362 = 0,
g371 = 0.
g245 = 0,
Set 2. Equations equivalent to g152 = 0: g173 = 0,
g251 = 0,
g263 = 0,
Set 3. Equations equivalent to g153 = 0: g156 = 0,
g162 = 0,
g163 = 0,
g165 = 0,
g167 = 0,
g172 = 0,
g176 = 0,
g253 = 0,
g257 = 0,
g261 = 0,
g267 = 0,
g271 = 0,
g273 = 0,
g275 = 0,
g276 = 0,
g351 = 0,
g352 = 0,
g356 = 0,
g357 = 0,
g361 = 0,
g365 = 0,
g372 = 0,
g375 = 0.
g364 = 0,
g374 = 0.
g367 = 0,
g376 = 0.
g463 = 0,
g465 = 0,
g467 = 0,
Set 4. Equations equivalent to g154 = 0: g174 = 0,
g254 = 0,
g264 = 0,
Set 5. Equations equivalent to g157 = 0: g175 = 0,
g256 = 0,
g265 = 0,
Set 6. Equations equivalent to g451 = 0: g452 = 0,
g456 = 0,
g457 = 0,
g462 = 0,
g471 = 0,
g473 = 0,
g475 = 0,
g476 = 0.
If we write g145 = β T = β[(d16 − d46 )∆1456 + (d17 − d47 )∆1457 ] = 0, it follows that T = 0. So in the following, we restrict our central configurations in the set T −1 (0). Lemma 2.3. Under our assumptions and in the set T −1 (0), the remaining 54 equations in the set 2, 3, . . . , 6 are satisfied if g152 = g153 = g154 = 0 are satisfied.
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X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
Proof. Consider the expressions of g157 and g153 − g152 : g157 = α(2d12 − d25 − d35 )∆1572 + m4 (d14 − d45 )∆1574 + β(d16 − d56 )∆1576 , g153 − g152 = α(2d12 − d25 − d35 )∆1532 + m4 (d14 − d45 )(∆1534 − ∆1524 ) − β(d16 − d56 )∆1526 . From our assumptions we have
∆1574 ∆1576 ∆1572 = = = x ̸= 0. ∆1532 ∆1534 − ∆1524 −∆1526 Thus g152 = 0 and g153 = 0 imply g157 = 0. Consider the expression of g154 and g451 : g154 = α[(d12 − d25 )∆1542 + (d13 − d35 )∆1543 ] + β[(d16 − d56 )∆1546 + (d17 − d57 )∆1547 ], g451 = α[(d24 − d25 )∆4512 + (d34 − d35 )∆4513 ] + β(d46 − d56 )(∆4516 + ∆4517 ). In the set T −1 (0), we have d46 =
d16 ∆1456 + d17 ∆1457
∆1456 + ∆1457
.
(2.3)
Substituting (2.3) into g451 , it follows that g451 = α[(d24 − d25 )∆4512 + (d34 − d35 )∆4513 ] + β(d16 ∆1456 + d17 ∆1457 − d56 ∆1456 − d56 ∆1457 )
= α[(d24 − d25 )∆4512 + (d34 − d35 )∆4513 ] + β[(d16 − d56 )∆1456 + (d17 − d56 )∆1457 ]. From our assumptions we have
∆1542 ∆1543 ∆1546 ∆1547 = = = = −1 ̸= 0. ∆4512 ∆4513 ∆1456 ∆1457 Hence in the set T −1 (0), g154 = 0 implies g451 = 0. This completes the proof.
From Lemma 2.3, in order to study central configurations according to Fig. 1 in the set T −1 (0), it is sufficient to study the following 3 equations: g152 = 0,
g153 = 0,
g154 = 0.
(2.4)
Denote by A = (aij ) the matrix of the coefficients of the homogeneous linear system in the variables α, m4 , β defined by Eqs. (2.4). Thus a11 = (d13 − d35 )∆1523 ,
a12 = (d14 − d45 )∆1524 ,
a13 = (d16 + d17 − 2d56 )∆1526 ,
a21 = (d12 − d25 )∆1532 ,
a22 = (d14 − d45 )∆1534 ,
a23 = (d17 − d57 )∆1537 ,
a31 = (d12 − d25 )∆1542 + (d13 − d35 )∆1543 , a32 = 0, Let x =
a33 = (d16 − d56 )∆1546 + (d17 − d57 )∆1547 .
α m4
β
. Then in order to get the spatial central configuration like Fig. 1, we need to find positive solution α, m4 , β of
the following system: Ax = 0
(2.5)
where T = 0. By elementary computation we have det (A) = 0, since the third row of A is a linear combination of the first two rows. Thus in the set T −1 (0) Eq. (2.5) have nontrivial solutions. 3. The existence of spatial central configurations Consider the expression of T : T = (d16 − d46 )∆1456 + (d17 − d47 )∆1457 . Now we study the sign of each term of T . The term (d16 − d46 ) is negative if z > √
vanishes on the straight line z = vanishes on the straight line z =
2 4
√
(1 − 2x). The term ∆1456 is negative if z <
√ 2 4
√
(1 − 2x), positive if z <
2(1 + x), positive if z >
2(1 + x). The term (d17 − d47 ) is negative if z >
and vanishes on the straight line z =
√
2 4
(1 + x). The term ∆1457 is negative if z <
√
2 √4 2 2
√ 2 4 √
(1 − 2x) and
2(1 + x) and
(1 + x), positive if z <
(2 − x), positive if z >
√
2
√4 2 2
(1 + x) ( 2 − x)
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
169
Fig. 2. The regions D1 and D2 .
√
and vanishes on the straight line z = D1 ∪ D2 , where D1 = D11 ∪ D12 ,
2 2
(2 − x). Thus the connected components of the set T −1 (0) are restricted to the set
D2 = D21 ∪ D22
and
√
D11 =
(x, z )|0 < x ≤ 1,
D12 =
(x, z )|x > 1,
2 4
D22 =
(x, z )|x > 1,
(2 − x) < z <
(1 + x) < z < 2 4
4
2(1 + x) ,
√
2(1 + x) ,
√ (1 − 2x) < z <
√ 2
√
√
(x, z )|0 < x ≤ 1,
2
√
D21 =
2
√ (1 − 2x) < z <
2 2
2 4
( 1 + x) ,
(2 − x) .
See Fig. 2. In order to prove the existence of positive solutions of (2.5) 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 Fig. 1. Let
A1 = D2 ∩ {(x, z )|2x2 + x − z 2 < 1}, A2 = D2 ∩ {(x, z )|x2 − x + 1 + z 2 > 3} and A = A1 ∪ A2 , see Fig. 3. Then we have the following lemma for (x, z ) ∈ A.
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X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
Fig. 3. The regions A1 , A2 and B .
Lemma 3.1. In the set A, there are no spatial central configurations according to Fig. 1. Proof. Consider the entries a31 = (d12 − d25 )∆1542 + (d13 − d35 )∆1543 and a33 = (d16 − d56 )∆1546 + (d17 − d57 )∆1547 . It is simple to see ∆1542 > 0, ∆1543 > 0, ∆1546 > 0 and ∆1547 > 0 for (x, z ) ∈ A. While d12 − d25 , d13 − d35 , d16 − d56 , d17 − d57 are negative for (x, z ) ∈ A1 , and d12 − d25 , d13 − d35 , d16 − d56 , d17 − d57 are positive for (x, z ) ∈ A2 . Thus the entries a31 and a33 have the same signs. The lemma is proved. Now we prove the existence of spatial central configuration according to Fig. 1 for some points in the set D1 . Proof of Theorem 1.2. Since the matrix A is singular in the set T −1 (0), there are nontrivial solutions of (2.5) in the set T −1 (0). In order to simplify our analysis, we consider the following subset of D1 (see Fig. 3):
B = B1 ∪ B2 where
√
(x, z ) ∈ D1 : 1 < x <
B1 =
√
B2 =
(x, z ) ∈ D1 :
10 2
10 √
2
,
2
2x2
+x−1 ,
√ √ √ 2 ≤ x < 3, 2 < z < 3 − x + 2 .
In the set B , the entries of the matrix A have the following signs:
− A= + +
+ + 0
∗ − . −
Consider the set L = B ∩ {(x, z ) : 1.09419029 < x < 1.69932798, T = 0}.
X. Su, T. An / Journal of Geometry and Physics 70 (2013) 164–171
171
Without loss of generality, let α = 1. It is easy to see that the mass
β=−
(d12 − d25 )∆1542 + (d13 − d35 )∆1543 (d16 − d56 )∆1546 + (d17 − d57 )∆1547
is positive. By direct computation, the mass m4 = −
(d12 − d25 )∆1532 + (d17 − d57 )∆1537 β (d14 − d45 )∆1534
is increasing in (1.09419029, 1.69932798) about x in the set L and m4 (1.09419029) = 0. So at every point in the set L , we have nontrivial positive solutions of (2.5). The proof of Theorem 1.2 is complete. Acknowledgments The first author is supported by Natural Science Foundation of China (NFSC11201168) and the Scientific Research Foundation of Huaiyin Institute of Technology (HGA1102). The authors sincerely thank the referees and the editor for their many valuable comments which helped us improve the paper. References [1] [2] [3] [4] [5] [6] [7]
F.R. Moulton, Straight line solutions of the problem of n bodies, 12, 1910, 1–17. D. Sarri, On the role and properties of n-body central configurations, Celestial Mech. 21 (1980) 9–20. S. Smale, Mathematical problems for the next century, Math. Intelligencer. 20 (1998) 7–15. R. Moeckel, On central configurations, Math. Z. 205 (1990) 499–517. M. Hampton, Stacked central configurations: new examples in the planar five-body problem, Nonlinearity 18 (2005) 2299–2304. J. Llibre, L.F. Mello, New central configurations for the planar 5-body problem, Celestial Mech. Dynam. Astronomy. 100 (2008) 141–149. J. Llibre, L.F. Mello, E. Perez-Chavela, New stacked central configurations for the planar 5-body problem, Celestial Mech. Dynam. Astronom. 110 (2011) 45–52. [8] S.Q. Zhang, Q. Zhou, Double pyramidal central configurations, Phys. Lett. A. 281 (2001) 240–248. [9] M. Hampton, M. Santoprete, Seven-body central configurations: a family of central configurations in the spatial seven-body problem, Celestial Mech. Dynam. Astronom. 99 (2007) 293–305. [10] L.F. Mello, A.C. Fernandes, Stacked central configurations for the spatial seven-body problem. Preprint.