Three-tangents theorem in three-body motion in three-dimensional space

Conference Paper (PDF Available) · January 2016with 32 Reads
Conference: Symposium on Celestial Mechanics and N-Body Dynamics in Chiba, 2014, At SMBC Event Space (Chiba), Chiba, Japan
Abstract
In the general three-body problem on the plane, the conservation of the center of mass and zero angular momentum has a simple geometrical meaning: three tangent lines from the three bodies meet at a point at each instant. It is called "three-tangents theorem". Kuwabara and Tanikawa extended this theorem to the three-body motion in plane and in three-dimensional space with non-zero angular momentum. In this short note, we will investi-gate an alternative three-tangents theorem compared to Kuwabara and Tanikawa's in three-dimensional space.
Proceedings of the Symposium on Celestial Mechanics and N-body Dynamics in Chiba, 2014
Eds. M. Saito, M. Shibayama and M. Sekiguchi
Three-tangents theorem in three-body motion in
three-dimensional space
Hiroshi Ozaki1,
Tetsuya Taniguchi2, Hiroshi Fukuda2, and Toshiaki Fujiwara2
1Tokai University, Education Program Center, Shimizu Campus
2Kitasato University, College of Liberal Arts and Sciences
Abstract. In the general three-body problem on the plane, the
conservation of the center of mass and zero angular momentum has
a simple geometrical meaning: three tangent lines from the three
bodies meet at a point at each instant. It is called “three-tangents
theorem”. Kuwabara and Tanikawa extended this theorem to the
three-body motion in plane and in three-dimensional space with
non-zero angular momentum. In this short note, we will investi-
gate an alternative three-tangents theorem compared to Kuwabara
and Tanikawa’s in three-dimensional space.
1. Introduction
After the discovery of the ﬁgure-eight solution to the planar equal mass
three-body problem [1,2,3,4], equal mass N-body periodic solutions has
been paid match attention to. Recently new families of periodic solutions
for three equal masses moving under Newtonian gravity in a plane were
found by numerical simulation [5]. Both the ﬁgure-eight solution and new
solutions have zero total linear momentum and zero total angular momen-
tum. This fact leads to the following simple geometrical theorem [6]:
[Theorem 1] (three-tangents)If the total linear momentum and the
total angular momentum are zero, three tangent lines at bodies meet at a
point or three lines are parallel.
This theorem is proved without using the equation of motion of three
bodies. It means that the three-tangents theorem can be applied to wide
class of potential models. For example, it holds even for the three-body
1ozaki@tokai-u.jp
2tetsuya@kitasato-u.ac.jp, fukuda@kitasato-u.ac.jp, fujiwara@kitasato-u.ac.jp
1
2
motion under the attractive logarithmic-potential accompanied by an ar-
tiﬁcial repulsive potential [6].
Kuwabara and Tanikawa [7] extended this theorem to the planar three-
body motion with non-zero total angular momentum. They showed that
[Theorem 2] (Extended three-tangents)If the total linear momentum
is zero and the total angular momentum is Lin the planar three-body mo-
tion, the area Stof the triangle formed with three tangent lines at the bodies
is given by
St·α=L2
2,
where αis the double area of the triangle formed with three momentum
vectors p1,p2,p3:
α=p1p2=p2p3=p3p1.
The symbol is the exterior product of two vectors in two-dimensional
space.
Moreover, they applied the theorem to the three-dimensional case.
When the total angular momentum is projected to the yz-, zx-, and xy-
planes, and three tangent lines are also projected to each coordinate plane.
[Theorem 3] (three-tangents in three dimensions I )If the total linear
momentum is zero and the total angular momentum is Lin the three-
dimensional space, the area St=(St
yz , St
zx, St
xy)of the triangle formed with
projected three tangent lines to the yz-, zx-, and xy-planes is given by
St·α=L·L
2,
where L=(Lyz, Lz x, Lxy ), and α=(αyz , αzx, αxy ).
Being stimulated by Kuwabara-Tanikawa’s study, we investigated al-
ternative three-tangents theorem in three-dimensional space.
2. Alternative three-tangents theorem in three-dimensional space
We will set up a Cartesian coordinate system O-xyz. Suppose that three
bodies with masses m1, m2, m3are acting the inertial forces between any
two bodies obeying Newton’s third law in the space. According to the
equation of motions, three bodies are conﬁgured in the space keeping the
center of mass being ﬁxed in the Cartesian coordinate system. It is con-
venient to take the center of mass of three bodies as the origin Oin the
system O-xyz.
Also each body has its own momentum vector pi(i= 1,2,3) so that
they satisfy ipi=0. Three lines li(i= 1,2,3) are drawn along piso that
3
each line passes through each position for each body. If one of momentum
vectors is zero vector, put a point in the space instead of drawing a line.
We will call l1, l2, l3tangent lines. Let us introduce a vector αwhich is the
double area of the triangle formed with three momentum vectors p1,p2,p3:
α=p1×p2=p2×p3=p3×p1,
where the symbol ×is the vector product of two three-dimensional vectors.
This relation is directly derived from ipi=0. Since three momentum
vectors p1,p2,p3are perpendicular to the double area α, they are always
parallel to a plane if all three momentum vectors are not zero vectors.
If α=0, either one of three momentum vectors is zero vector and others
are parallel with each other, or three momentum vectors are parallel with
one another, or all three momentum vectors are zero vectors.
On the other hand, If α6=0, three momentum vectors form a triangle
in the three-dimensional space. In this case, the following theorem is valid:
[Theorem 4] (three-tangents in three dimensions II )If the total linear
momentum is zero and three tangent lines are projected to a plane including
the total angular momentum in the three-dimensional space, the projected
three tangent lines meet at a point or three tangent lines are parallel.
3. Proof of Theorem 4
First, we will set up a Cartesian coordinate system O-x0y0z0. Three vectors
ˆ
x0= (1,0,0), ˆ
y0= (0,1,0), and ˆ
z0= (0,0,1) are taken as unit vectors along
x0-, y0-, and z0-axis, respectively. Let us choose a piece of plane τpassing
through the origin Oand being parallel to the total angular momentum L.
The normal vector of τis denoted by n. Now we set Cartesian coordinate
system O-x0y0z0so that two orthogonal x0- and y0-axis are on τ, the rest
z0-axis is perpendicular to τ.
Second, we will project position and momentum vectors (qiand pi)
of three bodies onto τ. The projected position and momentum vectors are
denoted by q0
iand p0
i. To project position and momentum vectors in the
three-dimensional space onto τ, it is convenient to introduce the projection
matrix Pin terms of nby P=Intn, where Iis the identity matrix.
Then we obtain the relations between qiand q0
i, similarly between piand
p0
i:
q0
i=Pqi
=qi(qi·n)n,(1)
and
p0
i=Ppi
=pi(pi·n)n.(2)
4
Note that q0
iand p0
iare lying on the plane τat each instant. For the
system O-x0y0z0in which nis along z0-axis, the component representations
are written by
q0
i= (qix0, qiy0,0),p0
i= (pix0, piy0,0).
Since the center of mass Oof three bodies is projected onto itself, the
conservation laws of the center of mass and the total linear momentum are
valid for the Cartesian coordinate system O-x0y0z0. Thus we have iq0
i=0
and ip0
i=0at every instant. The total angular momentum is evaluated
in the system O-x0y0z0as follows:
L=
i
q0
i×p0
i+
i
(qiz0pix0qix0piz0)ˆ
y0+
i
(qiy0piz0qiz 0Piy0)ˆ
x0
=Lz0+Ly0+Lx0.
On the other hand the total angular momentum Lis always on the plane
τ, so only the z0component of Lvarnishes:
L0
z=
i
q0
i×p0
i
=
i
(qix0piy0qiy0pix0)ˆ
z0=0.
Now let c0
tbe the intersection point of two projected tangent lines l0
1
and l0
2. The angular momentum L0
tis also
zero: i(q0
ic0
t)×p0
i=0.Also (q0
1c0
t)×p0
1=0and (q0
2c0
t)×p0
2=0.
Thus we have (q0
3c0
t)×p0
3=0. This means that the projected tangent
three linesl0
1, l0
2and l0
3meet at the same point c0
ton τ.
Before closing our statement, we will represent the intersection point
of three tangent lines c0
ton τby the projected position and momentum
vectors. It is simply written by
c0
t=[(q0
i×p0
i)·ˆ
z0]p0
j[(q0
j×p0
j)·ˆ
z0]p0
i
(p0
i×p0
j)·ˆ
z0,(3)
where (i, j) = (1,2),(2,3),(3,1). Substituting (1) and (2) into (3), c0
tcan
also be written by qi,piand normal vector nin the three-dimensional
space.
Let us rotate τin the three-dimensional space for ﬁxed time. The
intersection point c0
tgoes to inﬁnity if the plane τis parallel to α(nis
perpendicular to α). This is easily veriﬁed because
(p0
i×p0
j)·ˆ
z0= (pi×pj)·n=α·n= 0.
5
Acknowledgment
The research of one of the authors (HO) has been supported by Grand-in
-Aid for Science Research 25400408 JSPS.
References
[1] C. Moore, Phys. Rev. Lett. 70 (1993) 3675-3679.
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201)(Basel: Birkhauser) (2001) pp 101-115.
[4] C. Simo, Dynamical properties of the ﬁgure eight solution of the three-
body problem Celestial Mechanics: Dedicated to Donald Saari for
his 60th birthday. Contemporary Mechanics 292 (Providence, RI.:
American Mathematical Society) (2002) pp 209–228.
[5] M. ˘
Suvakov and V. Dmitra˘sinovi´c, Phys. Rev. Lett. 110 (2013), 114301
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(2003) 2791–2800.
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This research hasn't been cited in any other publications.
• C Moore
C. Moore, Phys. Rev. Lett. 70 (1993) 3675-3679.
• A Chencinear
• R Montgomery
A. Chencinear and R. Montgomery, Ann. Phys. 152 (2000) 881-901.
• Dynamical properties of the figure eight solution of the threebody problem Celestial Mechanics: Dedicated to Donald Saari for his 60 th birthday
• C Simo
C. Simo, Dynamical properties of the figure eight solution of the threebody problem Celestial Mechanics: Dedicated to Donald Saari for his 60 th birthday. Contemporary Mechanics 292 (Providence, RI.: American Mathematical Society) (2002) pp 209-228.
• T Fujiwara
• H Fukuda
• H Ozaki
T. Fujiwara, H. Fukuda, and H. Ozaki, J. Phys. A: Math. Gen. 36 (2003) 2791-2800.
• K H Kuwabara
• K Tanikawa
K. H. Kuwabara and K. Tanikawa, Phys. Lett A. 354 (2006) 445-448.
• C Simo
C. Simo, Proc. 3rd Europian Cong. Math. (Progress in Mathematics 201) (Basel: Birkhauser) (2001) pp 101-115.