A Generalized approach for computing the trajectories associated with the Newtonian N Body Problem
AbuBakr Mehmood, Syed Umer Abbas Shah and Ghulam Shabbir Faculty of Engineering Sciences, GIK Institute of Engineering Sciences and Technology Topi, Swabi, NWFP, Pakistan Email:
[email protected]
Abstract: The Classical Newtonian problem of describing the free motions of N gravitating bodies which form an isolated system in free space has been considered. It is well known from the Poincare’s Dictum that the problem is not exactly solvable. Sets of N body systems composed of masses having spherical symmetry, appropriate angular velocities (< 1 rad/s) and bounded position vectors are examined. A procedure has been developed which yields expressions approximately defining the trajectories executed by the masses.
Introduction: The problem has been solved in a two dimensional plane. Sets of N body systems in which the gravitating masses possess spherically symmetric mass distributions, small angular velocities (< 1 rad/s) and bounded position vectors have been considered. Other approaches can be found in [1-5]. The following figure shows our configuration for the system of N bodies
1
y
eθ1 ex1
ex2 m1
m2 eθ2
x2 x1
xk
x
o xN
mK
exk
mK
eθN
eθk
Figure 1 G G here x k is the position vector of mass mk, θ k is the rotation angle of x k , e xk is the radial unit G vector along x k and e θ k is a unit vector perpendicular to e xk ∀ 1 ≤ k ≤ N . We assume that the G kth body having mass mk and position vector x k remains approximately in two body motion N G G mn x ( n ≠ k ) N with a body of mass M k = ∑ n =1 mn (n ≠ k ) , placed at a point given by x M k = ∑ n=N1 n . The ∑ n=1 mn ( n ≠ k )
position vectors of mk and M k are assumed to remain approximately collinear for all time in accordance with this assumption. It then follows that the respective unit vectors in those directions also remain approximately collinear for all time, that is e xk .e xM k −1 ∀ t. Since we are free to scale each position vector along its direction, the boundedness of the position vectors G G ( | x k (t ) |< ∞ ∀ 1 ≤ k ≤ N ) can be modified as | x k (t ) | ∞ ∀ 1 ≤ k ≤ N . The closed form approximations that we will find for the case of N bodies will be valid for the case .
G
θ k (t ) < 1 rad/s & | x k (t ) | ∞ ∀ 1 ≤ k ≤ N . Note that these constraints include almost all practically encountered situations, in the sense that angular velocities encountered in celestial motion are normally much less than 1 rad/s and position vectors are bounded. These conditions therefore negligibly limit the application of our results.
2
Main Results: Figure 2 presents the configuration of our two body approximation of the system of N bodies
y
eθk exk
xk
o
x
xM k Mk
exM k
eθM k
Figure 2 G Here e xM k is a unit vector in the direction of x M k and e θ M k is a unit vector perpendicular to G G e xM k . The vectors x k and x M k are taken to be approximately collinear for all time, resulting from
the
two
body
motion .
analogue .
..
in
figure
2.
This
in
turn
implies
that
..
θ k (t ) = θ M (t ) + π ⇒ θ k (t ) = θ M (t ) ⇒ θ k (t ) = θ M (t ) ∀ t. Using Newton’s Laws to model the k
k
k
system in figure 2 we get G.. Gmk M k mk x k = − G G 2 e xk xMk − xk
(1)
G.. Gmk M k M k xMk = − G G 2 e xM k xMk − xk
(2)
G G where G is the universal gravitation constant. The representation of vectors x k and x M k as xk (t) e xk and xM k (t) e xM k respectively, will find use in the scalar representation of (1) and (2)
which can be shown to be 3
.. . 2 GM k x k − xk θ k = − G G xM k − xk
2
(3)
.. . . xk θ k + 2 x k θ k = 0 ..
.
x M k − xM k θ M k
2
(4)
Gmk = − G G xMk − xk
..
.
2
(5)
.
xM k θ M k + 2 x M k θ M k = 0
(6)
The system of equations (3), (4), (5) and (6) is not exactly solvable. We recall that the .
problem is being solved for the case θ k (t ) < 1 rad/s and xk (t ) ±∞ ( ∀ 1 ≤ k ≤ N ). We .
2
.
2
could find an approximation to equation (3) by noting that θ k 0 ∀ t ⇒ xk θ k 0 ∀ ..
.
.
2
..
t (since θ k (t ) < 1 rad/s and xk (t ) ±∞ ). This implies that x k − xk θ k x k and hence
the following is found to be an approximation to (3). .. GM k xk − G G xMk − xk
(7)
2
G G It can be shown that if | x k | ∞ ∀ 1 ≤ k ≤ N , then | x M k | ∞ ∀ 1 ≤ k ≤ N . Also, if .
.
θ k (t ) < 1 rad/s, then θ M (t ) < 1 rad/s. Using these arguments, just like we found (7) to be k
an approximation to (3), we can find (8) to be an approximation to (5). ..
xMk
Gmk = − G G xMk − xk
G G G We now define the vector x = x M k − x k and therefore
2
G2 G G x = xMk − xk
(8)
2
G G Since x is a vector along the direction of x M k , it follows that a unit vector along the G G direction of x is the same as a unit vector along the direction of x M k . We define e x to be G G G G a unit vector along the direction of x . Hence it follows that e x = e xM k . Also x , x k and x M k can be represented as products of scalar time functions and respective unit vectors. 4
(9)
G x = x(t )e x G x k = xk (t )e xk G x M k = xM k (t )e xM k
(10)
and e x = e xM k = − e xk
(11)
Using the first equation in relation (12) we can rewrite (11) as G2 G G 2 2 x = xe x = x M k − x k = x 2 Rewriting
(12)
(7) and (8) while making use of (12) we get .. GM xk = − 2 k x
(13)
.. Gm xMk = − 2 k x
(14)
.. .. G (mk + M k ) xk + xMk = − x2
(15)
Adding the above two result we get
Making use of (10) and (11) it can be shown that ..
..
..
x(t ) = x k (t ) + x M k (t )
(16)
Substitution of equation (15) in equation (16) then yields .. G (mk + M k ) x = − x2
(17)
We now integrate (13) and (14) twice with respect to time to get (18) and (19) respectively. .
t t xk (t ) = xko + x ko (t − to ) − GM k ∫ ∫ x −2 (t )dt dt t0 t 0
.
t t xM k (t ) = xM k 0 + x M ko (t − to ) − Gmk ∫ ∫ x −2 (t )dt dt to t o
.
.
.
(18)
(19)
.
where xko = xk (0) , xM k 0 = xM k (0) , x ko = x k (0) , x M ko = x M k (0) and t0 is the starting time. In both (18) and (19) the expression
∫ ∫ t
t
to
to
x −2 (t )dt dt is unknown. We can integrate equation
(17) twice with respect to time to find an alternative form for the above expression as
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. x −2 (t )dt dt = x o (t − to ) + x0 − x(t ) ∫to ∫to G (mk + M k ) t
.
t
(20)
.
where x0 = x(0) and x o = x(0) . Now substituting equation (20) into (18) we get . . Mk − xk (t ) = xko + x ko (t − to ) + [ x ( t ) x o (t − to ) − xo ] (mk + M k )
(21)
In order to find the analytic expression for the trajectory of mk by use of equation (21) , we must find t (θ k ) . Therefore we rewrite (21) as Mk M k x. o − x. ko to xk (θ k ) = xko − x0 + (mk + M k ) (mk + M k ) . . Mk Mk + x ko − x o t (θ k ) + x(θ k ) (mk + M k ) (mk + M k )
(22)
In order to find xk (θ k ) explicitly, we need to find t (θ k ) and x(θ k ) and substitute these expressions into equation (22) . Using equations (1), (2), (9), (10) and (11), we can show that G.. G (mk + M k ) x = − ex x2 The above equation can be represented in its scalar form as .. . 2 G (mk + M k ) x− xθ = − x2 ..
.
(23)
.
xθ + 2 x θ = 0 (24) G G where ′θ ′ is the rotation angle of vector x , the same as the rotation angle of vector x M k . The
above two equations can be solved to give x(θ ) =
1
c cosθ + c 1
2
sin θ +
G (mk + M k . 2
xo4 θ o
)
(25)
where c1 and c2 are constants of integration which can be determined by incorporation of the initial conditions. Now since θ = θ M k θ k − π , we can write x(θ ) x(θ k − π ) and show that
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1
x(θ k ) =
G (mk + M k ) . 2
x θo 4 o
.
− c1 cos θ k − c2 sin θ
.
where θ o = θ (0) . This can be simplified and written
k
as 1 k cos(θ k − φk ) + k2
x(θ k ) =
where k1 = ± c12 + c22 , k2 =
(26)
1
c2 . Having found x (θ ) , what k c 1
G (mk + M k )
and φk = tan −1
. 2
xo4 θ o
remains to be done is to solve for t (θ k ) , which can be shown to satisfy t (θ k ) =
1
θk
.
x θo 2 o
∫θ
ko
x 2 (θ k )dθ k + to
(27)
where θko = θk(0). A substitution of (26) in (27) yields 1
2
1 dθ + t t (θ k ) = k o . ∫θ ko 2 θ φ k − + k cos( ) k k 1 2 xo θ o
.
θk
.
where θ o = θ (0) . Evaluation of the integral gives −1 2 k k + 2 − 12 − k1tan (0.5φk − 0.5θ k ) + k1 1 2 t (θ k ) = . [ (k + k )(k − k ) ] x 2 θ k2 − k1 1 2 2 1 + k2 + k2 tan 2 (0.5φk − 0.5θ k ) o o
(k − k ) tan(0.5φk − 0.5θ k ) 1 k1 ((k1 + k2 )(k2 − k1 )) 2 tan(0.5φk − 0.5θ k ) + k1 tan −1 2 1 1 ((k1 + k2 )(k2 − k1 )) 2 ( k2 −k1 )tan(0.5φk −0.5θ k ) − 2 1 + k2 tan (0.5φk −0.5θk )−k1k2 tan 1 2 + − (( k k )( k k )) 1 2 2 1 ( k2 −k1 )tan(0.5φk −0.5θ k ) − 2 2 1 − k2 tan (0.5φk −0.5θk )tan 1 2 ((k1 +k2 )(k2 −k1 )) φ θ − . − . ( k k ) tan (0 5 0 5 ) k k − k22tan−1 2 1 1 ((k1 +k2 )(k2 −k1 )) 2 − 1 2 2 k +k − 1 − k1tan (0.5φk − 0.5θ ko ) + k1 − . 1 2 [ (k1 + k2 )(k2 − k1 )] 2 x 2 θ k2 − k1 + k2 + k2 tan 2 (0.5φk − 0.5θ ko ) o o
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(28)
(k − k ) tan(0.5φk − 0.5θ ko ) 1 k1 ((k1 + k2 )(k2 − k1 )) 2 tan(0.5φk − 0.5θ ko ) + k1 tan −1 2 1 1 ((k1 + k2 )(k2 − k1 )) 2 + k2 tan 2 (0.5φk −0.5θko )−k1k2 tan −1 (k2 −k1 )tan(0.5φk −0.51 θko ) ((k1 +k2 )(k2 −k1 )) 2 ( k2 −k1 )tan(0.5φk −0.5θ ko ) − 2 2 1 − k2 tan (0.5φk −0.5θko )tan 1 ((k1 +k2 )(k2 −k1 )) 2 − . − . ( k k ) tan (0 5 0 5 ) φ θ k ko − k22tan−1 2 1 1 2 ((k1 + k2 )(k2 −k1 )) + to
(29)
Substitution of (26) and (29) into equation (22) would then yield the following expression for xk (θ k ) .
Mk xk (θ k ) = xko − mk + M k
{
M k xo + mk + M k
. Mk 1 + x ko − k1 cos(θ k − φk ) + k2 mk + M k
}
. . M k xo − xko to + mk + M k
. xo *
2 − 12 − k1tan (0.5φ k − 0.5θ k ) + k1 2 k1 + k2 ( )( ) + − k k k k [ 1 2 2 1 ] . 2 + k2 + k2 tan (0.5φk − 0.5θ k ) xo2 θ ko k2 − k1
(k − k ) tan(0.5φk − 0.5θ k ) 1 k1 ((k1 + k2 )(k2 − k1 )) 2 tan(0.5φk − 0.5θ k ) + k1 tan −1 2 1 1 ((k1 + k2 )(k2 − k1 )) 2 −1 ( k 2 − k1 ) tan(0.5φk − 0.5θ k ) 2 + k2 tan (0.5φk − 0.5θ k ) − k1k2 tan 1 2 ((k1 + k2 )(k2 − k1 )) −1 ( k 2 − k1 ) tan(0.5φk − 0.5θ k ) 2 2 − k2 tan (0.5φk − 0.5θ k )tan 1 ((k1 + k2 )(k2 − k1 )) 2 −1 ( k 2 − k1 )tan(0.5φk − 0.5θ k ) 2 − k2 tan 1 2 (( k k )( k k )) + − 1 2 2 1 2 2 k1 + k2 − 12 − k1tan (0.5φk − 0.5θ ko ) + k1 − [ (k + k )(k − k ) ] x 2 θ. k2 − k1 1 2 2 1 + k2 + k2 tan 2 (0.5φk − 0.5θ ko ) ko o
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(k − k ) tan(0.5φk − 0.5θ ko ) 1 k1 ((k1 + k2 )(k2 − k1 )) 2 tan(0.5φk − 0.5θ ko ) + k1 tan −1 2 1 1 ((k1 + k2 )(k2 − k1 )) 2 −1 ( k 2 − k1 ) tan(0.5φk − 0.5θ ko ) 2 + k2 tan (0.5φk − 0.5θ ko ) − k1k2 tan 1 ((k1 + k2 )(k2 − k1 )) 2 + t 0 (30) k k − . − . ( ) tan(0 5 φ 0 5 θ ) k ko − k22tan 2 (0.5φk − 0.5θ ko )tan −1 2 1 1 ((k1 + k2 )(k2 − k1 )) 2 −1 ( k 2 − k1 )tan(0.5φk − 0.5θ ko ) 2 − k2 tan 1 2 ((k1 + k2 )(k2 − k1 )) Note that the result presented above is for the generalized case, and therefore it follows that the above result holds valid for the kth body of mass mk ∀ 1 ≤ k ≤ N .
Summary: A set of N gravitating masses was considered and a different procedure was developed to find expressions defining the executed trajectories. It was shown that approximate solutions can be found in closed form, in spite of the fact that exact solutions to such a problem do not exist. We assumed the N body system to be composed of masses having spherically symmetric distributions, small angular velocities (< 1 rad/s) and bounded position vectors. N number of two body motion analogues were then used to approximately replicate N body interaction. The solutions approximately describe the trajectories associated with N body motion in free space.
References: 1. H. Airault, H. P. Mckean and J. Moser, Rational and Elliptic solutions of the Korteweg-de Vries equation and related many body problem, Commun. Pure Appl. Math. 30 (1977) 95-148. 2. F. Calogero, Exactly solvable one dimensional many body problems, Lett. Nuovo Cimento 13 (1975) 411-416. 3. F. Calogero and A. M. Perelomov, Properties of certain matrices related to the equilibrium configuration of the one dimensional many body problems with the pair potentials V1(x) = - ln |sin x| and V2(x) = 1/sin2(x), Comm. Math. Phys. 59 (1978) 109-116. 4. H. W. Braden and V. M. Buchstaber, Integrable systems with pair wise interactions and functional equations, Rev. Math. & Math. Phys. 10 (1997) 3-120. 5. M. Bruschi and F. Calogero, Solvable and/or integrable and/or linearizable N body systems in ordinary (three dimensional) space. I, J. Nonlinear Math. Phys. 7 (2000) 303-386.
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