Existence and stability of Lagrangian points in the relativistic restricted three body problem

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by  Oscar M Perdomo

EXISTENCE AND STABILITY OF LAGRANGIAN POINTS IN
THE RELATIVISTIC RESTRICTED THREE BODY PROBLEM
OSCAR PERDOMO
6
1
0
Abstract. In this paper we reinvestigate the stability and existence of La-
2
grangianpointsinthecircularrestricted2+1bodyproblemtreatedintheframe-

n work of the post-Newtonian approximation of the general relativity. It is well
a
knownthatthestabilityoftheLagrangian pointsintheNewtoniancasedepends
J
on showing that the real parts of the eigenvalues of a matrix are zero. The rea-
5
son we are reinvestigating this topic is due to the fact that most of the papers

written so far on the stability and existence of the relativistic restricted three
]
P bodyproblemarenotmathematicallycorrect,theyhaveoneofthefollowingwell
E known mathematical errors: 1. Showing that an expression is close to a small
. positive number does not show that this expression is positive 2. Showing that
h
the approximation of an expression is zero does not show that the expression is
p zero and 3. Showing one solution of a system of equations that is obtained by
-
o doing a small perturbation of another system of equations, does not show the
r existence of a solution of thelatter system of equations.
ast allItnhethpeoNsseibwlteonreiastnriccatseedtchiercpualararm2+et1erbµod=ymp1rm+o2bml2emd.esHcreibreesm,u1patnodsmym2maeretritehse,
[
masses of the primaries. In the relativistic case, we need two parameters to
1 describe all the possible systems, one is µ and the other one is the number c
v that represents the speed of light in the units where the period of the primaries
4 is 2π and the distance between the primaries is 1. We point out that c is not
2 necessarily a big number, it is about 10065 for the Sun-Earth system, 6262 for
9 the Sun-Mercury system and it is about 683 for the Pulsar Binary star system,
0 undertheassumption that thePulsar binary star system is circular.
0
EventhoughitseemstobealmostimpossibletofindaclosedformfortheLa-
.
1 grangianpointsintherelativisticthreebodyproblem,weshowinthispaperhow
0 the Poincare-Miranda theorem can be used to prove the existence of Lagrangian
6 points. One of the main results in this paper provides the exact expression for
1 the characteristic polynomial of the matrix that determines the stability of the
:
v Lagrange points. We point out that even without having a closed form for the
i Lagrangian points, we can show that the characteristic polynomial has the form
X λ4 +a1λ2 +a2, with the expression for a1 and a2 depending on µ, c and the
r coordinates of theLagrangian point. The form of this polynomial shows that we
a
indeedhavestabilityresultssimilartothoseshownintheNewtoniancase. Atthe
endofthepaperwefindthecoordinatesofLagrangian pointsforsomeparticular
systemswithaprecisionsmallerthan10−30 andwecomparethenewresultswith
those already found in theliterature. Weconcludethat theerror in theprevious
results is big.
Date: January 6, 2016.
1
2
1. Introduction
Lagrangianpointscouldbeseenasparticularperiodicsolutionsoftherestrictedthree
body problem and they are extensively used in space missions. Even thought for
practical reasons the effect of relativity theory in the computation of the Lagrangian
points may be irrelevant to a mission, it is a good idea to know the exact position of
theLagrangianpointswhenrelativitytheoryisconsideredandtheirstabilityinorder
to make decisions whether or not working with Newtonian physics is good enough.
With the intension to set up notation for the change of units and to get familiar
with the terminology, section 2 explains the Newtonian case. Section 3 displays
the ODE for the relativistic case. We point out that it is not known that there
are also exactly five equilibrium solutions of the ODE in the relativistic case, it is
natural to conjecture that we also have 5 of these points and it is natural to also call
them Lagrangian points. We point out that the existence is far from being obvious
-especially when µ is small- due to the fact that the case µ = 0 degenerates to the
case where there is only one massive body and two massless bodies going around.
If we were to assign equilibrium points in this case, we would have to say that the
ODE has infinitely many equilibrium points, all ofthem forming acircle. Inthis way
the Newtonian restricted three body problem moves from an ODE having infinitely
many equilibrium points when µ = 0 to an ODE having only five equilibrium points
when µ > 0. In section 4 we will prove the existence of the point L for the Earth-
4
SunsystemusingthePoincare-Miranda Theorem. Wepointoutthattheexistenceof
otherequilibriumpointscanbedoneinasimilarway foranyothersystemassociated
with other parameters µ and c. The existence of the equilibrium points L , L and
1 2
L is somehow easier because it does not require the Poincare-Miranda theorem, in
3
this case the problem reduces to solve an equation with one variable and therefore
the intermediate value theorem can be applied. Regardless of how many equilibrium
points we have, or where they are located, section 5 shows that the characteristic
polynomial at any of the equilibrium solutions -Lagrangian points- has the form
λ4 + a λ2 + a . As a consequence we obtain that whenever the two roots of the
1 2
quadratic equation σ2+a σ+a = 0 are negative, then, the roots of λ4+a λ2+a
1 2 1 2
have zero real part and therefore the equilibrium point is linearly stable. At this
point we would like to point out that proving that a quantity is zero cannot be done
byconsideringapproximations. Forthisreason, forthecaseofL ,itisnotsurprising
4
tohavepaperslikeBhatnagarandHallan[2]wheretheyconcludethatL isunstable
4
becausethe realpart of someeigenvalues are closetoa positivevery smallnumber in
contrast with papers like Douskos et al [4] and Ahmen et al [1] where they conclude
thatL isstableforsomevaluesoftheparameter µbecauseafterasetofroundingof
4
order 1 and 1 they obtain that the real part of some eigenvalues is zero. So, which
c2 c3
one of these results is true having in mind that all of them have used rounding? We
RELATIVISTIC LAGRANGE POINTS 3
will show, even without having a closed form for the equilibrium points, that L is
4
stable for some open region on the set of parameters µ and c.
On section 6 we consider the relativistic restricted three body problem coming from
the choice of parameters µ = 0.0384 and c = 4, 10, 50, 100, 400, 800, 1600, 3200,
6400, 12800. Weusethis10ODEsystemstocomparethepreviousresultswiththose
obtained in this paper. We conclude that in all of these systems, the rounding error
in the papers [4] and [1] is big. We would like to point out that according to some
authors, for example [5], papers [1] and [4] are considered to be among the latest
results regarding the stability of the Lagrangian points for the relativistic restricted
three body problem.
The author would like to thank David R. Skillman and Andrés Mauricio Rivera for
his valuable comments.
2. Circular solutions of the two body problem and changing units
Let us consider two bodies (the primaries) with masses m Kg and m Kg which
1 2
moves in the space with positions x and y. Let us take the gravitational constant to
be equal to G = 6.67384 10 11m3Kg 1s 2. It is easy to check that for a given
− − −
∗
positive number a the functions
m a m a
2 1
x(t)= − (cos(ωt),sin(ωt)), y(t) = (cos(ωt),sin(ωt))
m +m m +m
1 2 1 2
with ω = G(m1+m2) s 1 satisfy the two body problem ODE
a3 −
q
m G m G
2 1
x¨ = (y x) y¨= (x y)
x y 3 − y x 3 −
| − | | − |
Thissolutionsatisfiesthatthedistancebetweenthetwobodiesisalwaysametersand
moreover, bothmotions areperiodicsincetheycomplete arevolution afterT = 2π =
ω
2π a3 s. Let us change the units of mass, distance and time in the following
G(m1+m2)
wayq: Let us denote by um the unit of mass such that 1um = (m1+m2)Kg, let us
denote by ud the unit of distance such that 1ud =am and finally let us denote by
ut, the unit of time ut such that 1ut = a3 s. Notice that using the units
G(m1+m2)
utandudwehavethatthedistancebetwqeen thetwobodiesis1udandtheperiodof
the motion is 2πut. We also have that the gravitation constant is 1ud3um 1ut 2.
− −
We point out that the speed of light is
4 OSCARPERDOMO
a ud
(2.1) c = 299792458
∗ G(m +m ) ut
1 2
r
If we denote by µ = m2 and we work in the new units ut, um, ud, then the
m1+m2
mass of the first and second body are 1 µ and µ and the motion of the primaries
−
are given by
x(t) = µ(cos(t),sin(t)) y(t)= (1 µ) (cos(t),sin(t))
− −
Moreover, if a third body with position z(t) and neglecting mass compare with m
1
and m moves under the influence of the gravitational force of the primaries, then z
2
satisfies
(1 µ) µ
(2.2) z¨= − (x z)+ (y z)
x z 3 − y z 3 −
| − | | − |
A direct computation shows that if we take
(2.3) z = (ξ(t)cos(t) η(t)sin(t),η(t)cos(t)+ξ(t)sin(t)) ,
−
then, (2.2) reduces to
∂w ∂w
(2.4) ξ¨ 2η˙ = 0 and η¨+2ξ˙= 0
− ∂ξ ∂η
where,
1 1 µ µ
(2.5) w = (ξ2+η2)+ − +
0
2 (ξ+µ)2+η2 (ξ+µ 1)2+η2
−
p p
A direct verification shows that ξ(t)= 1−2µ andη(t) = √3 is a solutions of the (2.4).
2 2
This equilibrium point (1−22µ,√23) is known as the Lagrangian point L4. In order to
analyze the stability of L we consider the function
4
∂w ∂w
F = (ξ˙,2η˙ + 0,η˙, 0 2ξ˙)
0
∂ξ ∂η −
as afunction ofthe variables φ= (ξ,ξ˙,η,η˙). Itiseasytocheck that theODE(2.4)is
equivalenttotheODEφ˙ = F (φ). Inordertoanalyzethestability oftheequilibrium
0
RELATIVISTIC LAGRANGE POINTS 5
solution φ0 = (1−22µ,0, √23,0), we compute the 4 by 4 matrix A0 = DF0 evaluated
at ξ = 1−2µ, η = √3, ξ˙ = 0, η˙ = 0. Since we can check that the characteristic
2 2
polynomial of the matrix A is equal to
0
27
λ4+λ2 (µ 1)µ
− 4 −
Then, weconclude that, wheneither 0 < µ < 1 9 √69 or 1 9+√69 < µ < 1,
18 − 18
thentherealpartofalltheeigenvaluesofA iszeroandthereforeL islinearlystable.
0 4
(cid:0) (cid:1) (cid:0) (cid:1)
It is known that there are 5 equilibrium solutions for the ODE (2.4); we have L ,
4
given above, L5 = (1−22µ,−√23) and three more of the form (ξ1,0), (ξ2,0) and (ξ3,0)
usually label as the Lagrangian points L , L and L . A similar analysis to the one
1 2 3
that we just did for L can be done for the other Lagrange points to conclude that
4
L is also linear stable for the same range of the parameter µ and, L , L and L
5 1 2 3
are linearly unstable.
3. The ODE in the relativistic case:
The equation of the motion for the restricted three body problem are very similar
to the one given by Equation (2.4), it takes the form (see Brumberg, 1972, [3] and
Bhatnagar [2])
∂w d ∂w ∂w d ∂w
(3.1) ξ¨ 2nη˙ = ( ) and η¨+2nξ˙= ( )
− ∂ξ − dt ∂ξ˙ ∂η − dt ∂η˙
where w = w + 1 w with
0 c2 1
3 1 1 2
w = 1 µ(1 µ) ρ2+ η˙2+2(η˙ξ ξ˙η)+ξ˙2+ρ2 +
1
−2 − 3 − 8 −
(cid:18) (cid:19)
(cid:16) (cid:17)
3 1 µ µ 1 (1 µ)2 µ2
− + η˙2+2(η˙ξ ξ˙η)+ξ˙2+ρ2 − + +
2 ρ ρ − − 2 ρ 2 ρ 2
(cid:18) 1 2(cid:19) (cid:18) 1 2 (cid:19)
(cid:16) (cid:17)
7ξ 1 1 1 µ 1 µ 3µ 2 1 1 3µ
µ(1 µ) 4η˙ + η2 + − + − + −
− 2 ρ − ρ − 2 ρ 3 ρ 3 2ρ − ρ ρ 2ρ
(cid:18)(cid:18) (cid:19)(cid:18) 1 2(cid:19) (cid:18) 1 2 (cid:19) (cid:18) 1 1 2 2 (cid:19)
3 1
n = 1 1 µ(1 µ)
− 2c2 − 3 −
(cid:18) (cid:19)
ρ = ξ2+η2, ρ = (ξ +η)2+η2 and ρ = (ξ+η 1)2+η2
1 2
−
p p p
6 OSCARPERDOMO
3.1. The system of equations. Theequilibriumpointsofthesystemofdifferential
equationsgivenby(3.1)arethesolutionsofthesystemf = 0andg = 0wheref = ∂w
∂ξ
and g = ∂w evaluate at ξ˙= η˙ = 0.
∂η
4. Existence of L in the relativistic case using the
4
Poincare-Miranda theorem
4.1. Existence ofL fortheEarth-Sunsystem. Forcomputationinthissection,
4
we will take earth and the sun moving with a constant distance between them of
a = 149597870700 m with the mass of the sun equal to M = 1.988544 1030Kg
0 0
∗
and the mass of the Earth equal to 5.9729 1024Kg. We will also will be taking the
m ∗
speed of light to be c = 299792458 . Using this data we have that the values for
0 s
µ and c are given by
59729
µ = 3.00365 10 6
−
19885499729 ≈ ∗
and
a0 149896229 33171841965098778778047734
c = c = 10065.3
0∗ G(m +M ) q10 ≈
0 0
r
WewillprovetheexistenceoftherelativisticpointL forthesystemEarth-Sun-mass
4
zero body. This is, we will prove the existence of a point (ξ ,η ) that is within a
0 0
distance of 10−15 of the point (1−22µ,√23) that satisfies the equation
(4.1) f(ξ ,η ) = 0 and g(ξ ,η ) = 0
0 0 0 0
In order to prove the existence of (ξ ,η ), let us consider the following five points
0 0
2499985012616009587660193140271 1082531750278361975463116188557
Z = ,
0
5000000000000000000000000000000 1250000000000000000000000000000
(cid:18) (cid:19)
2499985512616009587660193140271 4330127145451017576343432330693
P = ,
1
5000000000000000000000000000000 5000000000000000000000000000000
(cid:18) (cid:19)
2499984762616009587660193140271 4330127145451017576343432330693
P = ,
2
5000000000000000000000000000000 5000000000000000000000000000000
(cid:18) (cid:19)
2499985512616009587660193140271 2165063356219154276435264800649
P = ,
3
5000000000000000000000000000000 2500000000000000000000000000000
(cid:18) (cid:19)
2499984762616009587660193140271 2165063356219154276435264800649
P = ,
4
5000000000000000000000000000000 2500000000000000000000000000000
(cid:18) (cid:19)
RELATIVISTIC LAGRANGE POINTS 7
And let β be the line that connects P with P , β be the line that connect P with
1 2 1 2 4
P , β be the line that connect P with P and β be the line that connect P with
3 3 3 1 4 4
P . More precisely,
2
β (t) = tP +(1 t)P
1 1 2
−
β (t) = tP +(1 t)P
2 3 4
−
β (t) = tP +(1 t)P
3 1 3
−
β (t) = tP +(1 t)P
4 2 4
−
Figure 4.1. ThePoincareMirandatheoremguaranteesthatthesys-
tem of equations f = 0 and g = 0 has a solution inside the region
above
Theorem 4.1. There is solution of the system of equations f = 0 and g = 0 inside
the region delimited by the union of the curves β , β , β and β . Moreover, we have
1 2 3 4
that the first six significant digits of the functions evaluated at the points Z , P , P ,
0 1 2
P and P are given by
3 4
f(P ) = 1.124997 10 7 g(P ) = 1.94854 10 7
1 − 1 −
···∗ ···∗
f(P ) = 2.22772 10 13 g(P ) = 3.94516 10 13
2 − 2 −
− ···∗ ···∗
f(P ) = 4.60545 10 13 g(P ) = 7.63052 10 13
3 − 3 −
···∗ − ···∗
f(P ) = 1.12499 10 7 g(P ) = 1.94855 10 7
4 − 4 −
− ···∗ − ···∗
f(Z ) = 1.14508 10 32 g(Z ) = 4.25190 10 32
0 − 0 −
− ···∗ − ···∗
8 OSCARPERDOMO
We also have that the function g > 0 on β , g < 0 on β , f > 0 on β and f < 0
1 2 3
on β . As a consequence of the Poincare-Miranda Theorem we conclude that there
4
exists a point P = (ξ ,η ) inside the region bounded by the four curves β such that
0 0 0 i
f(P )= g(P ) = 0.
0 0
Proof. The proof to the theorem relies on the fact that we have an exact expres-
sion (an analytic expression) for f and g and the fact that programs like Mathe-
matica allow us to precisely compute a certain amount of real digits of an exact
expression. In order to obtain these digits, it is required that we work with exact
numbers, that is the reason we decided to use rational numbers and not decimals.
The computation for the values of the functions f and g were obtained by using
the command RealDigits[f(Z ),10,6] and RealDigits[g(Z ),10,6] form the program
0 0
Wolfram Mathematica 10. Likewise for the points f(P ) and g(P ). It is not difficult
i i
to check that the directional derivative of g along the velocity of the curves β and
1
β does not change sign, they both are approximately 1.29903, and also, the direc-
2
tional derivative of f along the curves β and β does not change sign, they both are
3 4
approximately 1.29903. Then, we conclude that the function g is monotonic along
β and β , this fact along with the values of g at the endpoints allows to prove that
1 2
g is positive on β and negative on β . A similar arguments holds for the function
1 2
f. This concludes the proof of the theorem. (cid:3)
Remark 4.2. The value Z was initially computed to have a good approximation of
0
the equations f = 0 and g = 0 near L . This good approximation is needed to find
4
the curves β that satisfy the conditions of the Poincare-Miranda Theorem.
i
5. Exact expression for the characteristic polynomial at the
equilibrium points
The following theorem provides an expression for the characteristic polynomial of a
system of the type given by the relativistic three body problem.
Theorem 5.1. Let us consider the potential U = U(x,x˙,y,y˙) and let us consider
the following system of ODE
∂U d ∂U ∂U d ∂U
x¨ 2ky˙ = ( ), y¨+2kx˙ = ( )
− ∂x − dt ∂x˙ ∂y − dt ∂y˙
where k is a constant. If L = (x ,y ) is an equilibrium point of the system above
0 0 0
2
andd = 1+∂2U +∂2U +∂2U ∂2U ∂2U isnotzero atL˜ = (x ,0,y ,0), then, the
∂x˙2 ∂y˙2 ∂x˙2 ∂y˙2 − ∂x˙∂y˙ 0 0 0
characteristic polynomial of the m(cid:16)atrix(cid:17)that describes the linearization of the ODE
at L is given by
0
RELATIVISTIC LAGRANGE POINTS 9
λ4+a λ2+a
1 2
where,
∂2U ∂2U ∂2U 2
a d =
2 ∂y2 ∂x2 − ∂x∂y
(cid:18) (cid:19)
and
∂2U ∂2U ∂2U 2 ∂2U 2 ∂2U ∂2U ∂2U
a d = 4k +4k + +
1 − ∂y∂x˙ ∂x∂y˙ ∂y∂x˙ ∂x∂y˙ − ∂y2 − ∂y2 ∂x˙2 −
(cid:18) (cid:19) (cid:18) (cid:19)
∂2U ∂2U ∂2U ∂2U ∂2U ∂2U ∂2U
2 +2 +4k2
∂y∂x˙ ∂x∂y˙ ∂x˙∂y˙ ∂x∂y − ∂y˙2 ∂x2 − ∂x2
Proof. The ODE in this theorem can be reduce to the first order ODE φ˙ = F(φ)
with φ = (x,x˙,y,y˙) and
F(φ) = (x˙,F (φ),y˙,F (φ))
2 4
and the functions F and F are given as the solution, near L˜ = (x ,0,y ,0), of the
2 4 0 0 0
system of equations
∂U ∂2U ∂2U ∂2U ∂2U
F 2ky˙ = x˙ F y˙ F
2− ∂x − ∂x˙∂x − 2∂x˙2 − ∂x˙∂y − 4∂x˙∂y˙
∂U ∂2U ∂2U ∂2U ∂2U
F +2kx˙ = x˙ F y˙ F
4 ∂y − ∂y˙∂x − 2∂x˙∂y˙ − ∂y˙∂y − 4 ∂y˙2
Recall that we have that F (L˜ ) = F (L˜ ) = 0. If we compute the partial derivative
2 0 4 0
with respect to x to the system of equations above and we evaluate at L˜ , we get
0
the following system of equation
∂2U ∂2U ∂2U
F = F F
2x ∂x2 − 2x∂x˙2 − 4x∂x˙∂y˙
∂2U ∂2U ∂2U
F = F F
4x ∂x∂y − 2x∂x˙∂y˙ − 4x ∂y˙2
This is a linear system on F and F with solution solution satisfying,
2x 4x
10 OSCARPERDOMO
∂2U ∂2U ∂2U ∂2U ∂2U
F d = +
2x ∂y˙2 ∂x2 ∂x2 − ∂x˙∂y˙ ∂x∂y
∂2U ∂2U ∂2U ∂2U ∂2U
F d = +
4x ∂x˙2 ∂x∂y ∂x∂y − ∂x˙∂y˙ ∂x2
LikewisewecanobtainexpressionforF andF andforF ,g andfinallyforF ,
2y 4y 2x˙ x˙ 2y˙
F evaluated at the point L˜ . The theorem follow after replacing these expression
4y˙ 0
forthepartialderivativeofthefunctionsF andF intothecharacteristicpolynomial
2 4
of the matrix
0 1 0 0
F F F F
2x 2x˙ 2y 2y˙
 
0 0 0 1
 F4x F4x˙ F4y F4y˙ 
 
 
(cid:3)
6. Comparing the results in this paper with those found before
As we pointed out before, the value c that represents the speed of light with respect
to the units ut and ud, varies according to the formula c = c a where
0 ∗ GM
c = 299792458 and M = m + m is the mass of the system. Since the value c
0 1 2
p
varies, it is part of the parameters that describe the relativistic restricted three body
problem.
If we want to talk about error, we would like to go back to the units meters and
seconds. Recall that a represent the distance between the primaries in meters and
M is the mass of the system in Kilograms. In this section we will be comparing
our results with those obtained in 2002 by Douskos and Perdios, [4], and the results
obtained in 2006 by Ahmed, El-Salam and El-Bar [1].
Notice that when the equilibrium point L is stable, then the roots of the charac-
4
teristic polynomial are of the form ω i and ω i. In this case, we expect to have
1 2
± ±
periodic solutions (in the synodic frame of reference (ξ,η)) with periods close to
2π 2π
(6.1) T = and T =
1 2
ω ω
1 2
Since in [1] the authors show the stability of L for all positive values of c as long
4
as 0 < µ < 0.0384 and in [4] the authors show the stability of L for all values of
4