Central Configurations in the Trapezoidal Four-Body Problems
aa r X i v : . [ m a t h . D S ] M a r Central Configurations in the TrapezoidalFour-Body Problems
Muhammad ShoaibFaculty of Sciences, Department of Mathematics,University of Hail, Kingdom of Saudi Arabia,email: [email protected] 9, 2018
Abstract
In this paper we discuss the central configurations of the Trape-zoidal four-body Problem. We consider four point masses on thevertices of an isosceles trapezoid with two equal masses m = m at positions ( ∓ . , r B ) and m = m at positions ( ∓ α/ , r A ).We derive, both analytically and numerically, regions of centralconfigurations in the phase space where it is possible to choosepositive masses. It is also shown that in the compliment of theseregions no central configurations are possible. AMS Subject Classification : 37N05, 70F07, 70F10, 70F15,70F17
Key Words and Phrases:
Dynamical systems, Central con-figuration, four-body problem, n-body problem, inverse problem
The classical equation of motion for the n-body problem has the form([5]- [11]) m i d ~r i dt = ∂U∂~r i = X j = i m i m j ( ~r i − ~r j ) | ~r i − ~r j | i = 1 , , ..., n, (1)where the units are chosen so that the gravitational constant is equal toone, r i is the location vector of the i th body, U = X (cid:22) i
2. Therefore a number of restric-tion techniques have been used to find special solutions of the few bodyproblem. See for example [11], [12] and [13]. The two most commontechniques used to reduce the dimension of the phase space are the con-sideration of symmetries are taking one of the masses to be infinitesimal.We consider four point masses on the vertices of an isosceles trapezoidwith two equal masses m = m at positions ( ∓ . , r B ) and m = m at positions ( ∓ α/ , r A ). We derive, both analytically and numerically,regions of central configurations in the phase space where it is possibleto choose positive masses.The CC equations for a general 5-body problem derived from (3) areas under. m ~r | ~r − ~r | + m ~r | ~r − ~r | + m ~r | ~r − ~r | = − λ ( ~r − ~c ) (6) m ~r | ~r − ~r | + m ~r | ~r − ~r | + m ~r | ~r − ~r | = − λ ( ~r − ~c ) (7) m ~r | ~r − ~r | + m ~r | ~r − ~r | + m ~r | ~r − ~r | = − λ ( ~r − ~c ) (8) m ~r | ~r − ~r | + m ~r | ~r − ~r | + m ~r | ~r − ~r | = − λ ( ~r − ~c ) (9) Theorem 1:
Consider four bodies of masses m = M, m = m = m and m = M . The four bodies are placed at the vertices of a trapezoid r = ( − . , − r B ) , r = ( − α , r A ) , r = ( α , r A ) and r = (0 . , − r B ) , (10)2 (m ) (−0.5, −r B ) (0.5, −r B )P (m ) P (m ) P (m )AB C (0,0) r A r B r r r r r r r r (− α /2,r A ) ( α /2,r A ) Figure 1: Trapezoidal Model shown in figure 1. Where r A is the distance from the centre of mass ofthe system to the centre of mass of m and m and r B is the distancefrom the centre of mass of the system to the centre of mass of m and m . Then m = ab ( a + b − ab )( a + b )( a + b − ab + α ( a − b )) (11) M = abα ( a − b )( a + b )( a + b − ab + α ( a − b )) (12) where a = (cid:16)(cid:16) . − α (cid:17) + β (cid:17) / , b = (cid:16)(cid:16) . α (cid:17) + β (cid:17) / (13) make the configuration r = ( ~r , ~r , ~r , ~r ) central. Theorem 2:
Let r = ( ~r , ~r , ~r , ~r ) be a central configuration asdefined in theorem 1. Then there exist a region R = R f c ∩ R f c ) inthe αβ -plane such that for any α, β ∈ R there exists positive masses ( m, M, m, M ) making r a central configuration. The regions R f and R f are R f = { ( α, β ) | α < g ( β ) and < β < } R f = { ( α, β ) | α < g ( β ) and < β < } where g ( β ) = q − .β − . β + 2 ( β + 0 . / − . β − . r . β + − . β − . √ β +0 . − . . (14) g ( β ) = s − p h1 − h0h2 h2 − h1 h2Numerically, region R is given by the colored part of figure (3b). Lemma 1:
Let r = r A − r B (ref: figure 1) and then using the ge-ometry of our proposed problem we arrive at the following relationshipsbetween r i , i = 1 , , , , r and r . r = − mM + m r + 12 r , r = Mm + M r + α r , r = Mm + M r − α r , r = − mM + m r − r . Without loss of generality, it is assumed that the centre of mass of thesystem ~c = 0, ~r = − α~r , r BA = | ~r A − ~r B | = βr . This gives us ~r B = − mM ~r A and r A = M βm + M (15)Using these assumptions with equation (1) we obtain the followingequations of motion.¨ r = m r (cid:16)(cid:0) . − α (cid:1) + β (cid:17) / + m r (cid:16)(cid:0) . α (cid:1) + β (cid:17) / + M r , (16)¨ r = M r (cid:16)(cid:0) . − α (cid:1) + β (cid:17) / + m r α + M r (cid:16)(cid:0) . α (cid:1) + β (cid:17) / , (17)¨ r = M r (cid:16)(cid:0) . α (cid:1) + β (cid:17) / + m r α + M r (cid:16)(cid:0) . − α (cid:1) + β (cid:17) / , (18)¨ r = m r (cid:16)(cid:0) . α (cid:1) + β (cid:17) / + M r + m r (cid:16)(cid:0) . − α (cid:1) + β (cid:17) / . (19)It is clear from lemma 1 that it is enough to study the equations for r = m + M M ( r + r ) and r as r i , i = 1 , , , , are linear combination of r and r . ¨ r = [ − M + ma ( α − − mb ( α + 1)] r . (20)4 - - - Α Β - - - - Α Β Figure 2: a ) R f (white) where f >
0. b) R f (white) where f > r p = − [ m + Ma + m + Mb ] ( r + r ) . (21)Using equation (3) in conjunction with equations (20) and (21) we obtainthe following equations of central configurations for the trapezoidal four-body problem. 2 M − m ( α − a + m ( α + 1) b = λ, (22) m + Ma + m + Mb = λ. (23)It is a straightforward exercise to solve (22) and (23) to obtain (11) and(12). This completes the proof of theorem 1. Let f = a + b − ab, f = a − b, f a + b − ab + α ( a − b )To find the region where the mass function m is positive we need to findregions where1. f > f > f < f < M to be positive5 - - - Α Β - - - - Α Β Figure 3: a) Region (colored), R m = ( R f ∩ R f ) ∪ ( R f c ∩ R f c ) b)The central configuration region ( R ) where both m and M are positive( R = R f c ∩ R f c )).1. f > f > f < f < f , we will need to solve f = 0. Its nearlyimpossible to solve f > f aprox = α − . β + 0 . β p β + 0 .
25 + 0 . p β + 0 .
25 + 0 . ! − β − . β + 2 p β + 0 . β − . β + 0 . p β + 0 . − . . (24)Now it is a straightforward exercise to show that f > R f . R f = { ( α, β ) | α < g ( β ) and 0 < β < } where g ( β ) = q − .β − . β + 2 ( β + 0 . / − . β − . r . β + − . β − . √ β +0 . − . . (25)Numerically, R f is given in figure (2a). The common denominator of m ,and M can be analyzed in a similar way. f aprox = h α + h α + h (26)6here h = − β − . β + 2 (cid:0) β + 0 . (cid:1) / − . β − . h = − . β − . β p β + 0 .
25 + 0 . h = ( − . β − . β + (cid:16) − . p β + 0 . − . (cid:17) β − . p β + 0 .
25+ 0 . β p β + 0 . − . .β + 0 . (27)Using approximate techniques with help from symbolic computation inMathematica it can be shown that f attains positive values in the fol-lowing region. R f = { ( α, β ) | α < g ( β ) and 0 < β < } where g ( β ) = s − p h1 − − h12h2Numerically, R f is given in figure (2b).Therefore the central configuration region where m > R m = ( R f ∩ R f ) ∪ ( R f c ∩ R f c )Numerically, R m is given in figure (3). As f < α and β therefore M > f <
0. This region is numericallyrepresented by the colored part of figure (2b) and analytically by thecompliment of R f . Hence the central configuration region ( R ) whereboth m and M are positive is given by R = ( R f c ∩ R f c ). Numericallythis region is given in figure (3b). In this paper we model non-collinear trapezoidal four-body problem wherethe masses are placed at the vertices of an isosceles trapezoid. Expres-sions for m and M are formed as functions of α , and β which gives centralconfigurations in the trapezoidal four-body problems. We show that inthe αβ -plane, m is positive when ( α, β ) ∈ R m . Similarly M is positivewhen ( α, β ) ∈ R cf . We have identified regions in the αβ -plane whereno central configurations are possible. A central configuration region R = ( R f c ∩ R f c ) for the isosceles trapezoidal 4-body problem is iden-tified in the αβ − plane where m and M are both positive. No central7onfigurations are possible outside this region unless we allow one of themasses to become negative. Acknowledgement:
The author thanks the Deanship of research atthe University of Hail, Saudi Arabia for funding this work under grantnumber SM14014.
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