The Pennsylvania State University The Graduate School
EFFECT OF TIDAL DISSIPATION ON THE MOTION OF
CELESTIAL BODIES
A Dissertation in Mathematics by Chong Ai
c 2012 Chong Ai
Submitted in Partial Fulfillment of the Requirements for the Degree of
Doctor of Philosophy
December 2012 The dissertation of Chong Ai was reviewed and approved∗ by the following:
Mark Levi Professor of Mathematics Department Dissertation Advisor, Chair of Committee
Sergei Tabachnikov Professor of Mathematics Department
Diane Henderson Professor of Mathematics Department
Milton Cole Distinguished Professor of Physics Department
Yuxi Zheng The Head of Mathematics Department
∗Signatures are on file in the Graduate School. Abstract
Tidal effects in celestial bodies manifest themselves in many ways. Tides cause periodic changes in sea and ground levels, they affect the length of day, and even volcanic activity. Tides cause effects on the scale larger than that of an individual body, affecting entire orbits of planets and moons.
In this thesis we focus on the effect of tides on the dynamics of orbits, leav- ing aside internal effects of tides on planets. This thesis addresses a gap in the literature. On the one hand, the mathematical theory of celestial mechanics is a classical subject going back to Newton, and it reached a high level of develop- ment by people like Legendre, Lagrange, Laplace, Jacobi, Poincar´e,Moser, Arnold and others. Without exception (to our knowledge) this theory treats planets as point masses subject to Newtonian gravitational attraction, and without account for tidal effects. On the other hand, astronomers take more realistic models of the planets, but get few if any rigorous results. In this thesis we study problems which fall in the gap between these two approaches: they do include tidal dissipation on the one hand, making them more realistic than the classical system which com- pletely ignores them, but we make this dissipation simple enough to be tractable mathematically.
To build dissipation into the equations of motion, we use the Routh method of introducing dissipation into Lagrangian equations of motion. According to this method, to write the equations of motion one only needs, in addition to the La- grangian of the system, also the so–called Routh dissipation function: the power dissipated as a function of generalized coordinates and generalized velocities of the system. We choose a simple class of dissipation functions, leaving more general questions for future work.
iii In this thesis we study tidal dissipation in two problems: the Kepler problem, and the restricted three–body problem, and ask the question of the long–term be- havior of these problems with dissipation. There are two main results. First, we show that all the negative energy solutions of Kepler’s problem approach circular motion, and do so with an additional interesting feature.
The second main result of this thesis deals with the restricted three body prob- lem with dissipation. We show that the Lagrangian equilateral configurations become unstable due to tidal dissipation. This is a rather surprising result of dis- sipation causing instability. In addition, we show that almost all (in the Lebesgue sense) motions end either in a collision or an escape to infinity.
The restricted three body problem, which is infinitely delicate in the classical conservative case, thus admits an essentially complete analysis if one introduces an arbitrarily small dissipation.
iv Table of Contents
List of Figures vii
List of Tables viii
Chapter 1 Tidal Effect 1 1.1 Introduction ...... 1 1.2 The properties of the tidal effect ...... 6
Chapter 2 Tidal Dissipation in General 11 2.1 General dissipation formulation ...... 11 2.2 Hamiltonian equations with dissipation under canonical transfor- mations ...... 13 2.3 LaSalle’s invariance principle ...... 14
Chapter 3 Tidal Dissipation for Kepler Problem 16 3.1 Assumptions on the tidal dissipation ...... 16 3.2 Effect of dissipation on the evolution of the dynamics of the 2-body problem ...... 17 3.3 One choice of dissipation function ...... 20
Chapter 4 Tidal dissipation in the restricted planar 3-body problem 26 4.1 Introduction ...... 26 4.2 A model in rotating frame ...... 26 4.3 Assumptions about the tidal effects ...... 29
v 4.4 The asymptotic behavior of the dissipative system ...... 30 4.5 Effect of dissipation on stability ...... 33 4.5.1 Stability of the dissipative problem at relative equilibria . . 33 4.5.2 A particular dissipation function ...... 36 4.5.3 Motion of the spectrum ...... 37
Bibliography 40
vi List of Figures
1.1 the bulge of Earth ...... 7 1.2 the dumbbell model ...... 8 1.3 the spin lag ...... 9
3.1 the graph of h(r) ...... 19 3.2 Pictures of Kepler Problem with dissipation ε = 10−5 ...... 21 3.3 Symmetry of the system ...... 24
4.1 the neighborhood, U ...... 32 4.2 Hill’s regions and a level set of E ...... 32 4.3 How dissipations causes instability µ = 0.01 and ε = 0.01. Both figures are in configuration space...... 37
vii List of Tables
1.1 Data from wikipedia about Jupiter’s moons ...... 2 1.2 Data from wikipedia about Roche Limits ...... 4 1.3 Estimation of tidal forces ...... 8
viii Chapter 1
Tidal Effect
1.1 Introduction
The word ”tides” is a generic term used to define the alternating rise and fall in sea level with respect to the land, produced by the gradient of the gravitational attraction of the moon and the sun. Tidal forces affect not only the shape of the celestial bodies, but also their orbits.
Tides on Earth
Tides on Earth are due to both the Moon and the Sun. The Moon pull away the closest water body of oceans from the solid shell of the earth and also the solid shell of the earth from the farthest water body of the ocean. So two symmetric high tides always simultaneously appear on opposite positions on the earth. And on certain position there are two high tides every day.
Because the tidal effect of the Sun is the smaller than that of the Moon, we usually consider that the tidal effect of the Sun enforce or diminish the tides by the Moon. The twice-monthly highest tides, called spring tides, reach at syzygy(full Moon and new Moon), when the three bodies staying in a straight line. And the lowest tides, called neap tides, occur when the Sun and the Moon form a right 2 angle with the Earth.
Tidal Locking and Orbital Resonances
In a time scale of thousands and millions of years, tidal forces tend to synchro- nize in integer ratios, the self-spin of satellites and orbital rotation of them. It is not an accident that the same side of the Moon always faces the Earth. The phe- nomenon, called synchronous rotation causes by the tidal force. When the Moon was spinning at some other speed, Earth had exert torque on the Moon, slowing or speeding its spin until this spin is locked with orbital period, at which point the bulge faced the Earth. Particularly when the masses of the primary and the satellite differs not much, both bodies will be tidally locked to each other, as Pluto and Charon do.
Tidal locking is so widespread that, as Veverka in 1987 indicates, almost all satellites close enough to the primary conduct synchronous rotation. A good exam- ple is the moons of Jupiter, Callisto, which itself is, separates tidally locked moons from non-tidally locked moons. Also, the first three Galilean, Io, and Europa and Ganymede, in a 4:2:1 orbital resonance (some data from wiki about these moons in table (1.1)), give an example of higher-order resonances.
Mass Semi-major Orbit period Eccen- Incli- (kg) axis(km) (day) (relative) tricity nation(◦) Io 8.93 × 1022 421,800 1.769 (1) 0.0041 0.050 Europa 4.80 × 1022 671,100 3.551 (2) 0.0094 0.471 Ganymede 1.48 × 1023 1,070,400 7.155 (4) 0.0011 0.204 Callisto 1.08 × 1023 1,882,700 16.69 (9.4) 0.0074 0.205
Table 1.1: Data from wikipedia about Jupiter’s moons
The satellites may reach an orbital resonance rather than tidally locked when the orbit of it is eccentric. It was thought that Mercury was tidally locked with the Sun [6]. But, by observations of Doppler radar, people proved in 1965 that Mercury has, in fact, a 3:2 spin-orbit resonance, spinning three times for every two 3 revolutions around the Sun. More recently Correia and Laskar [7] explained the 3:2 orbital resonance by using the equations for an oscillating damped pendulum.
Tidal Heating
Besides changing the orbits and the spins, tidal forces can also internally heat a satellite. A satellite spins on its center at a constant speed but, in an eccentric orbit, its orbital speed varies with the distance from the primary, according to Kepler’s 2nd law. Thus the tidal bulge of a tidally locked satellite cannot always point precisely at its primary: most of time it will fail to do so and the primary will exert a small pull on the bulge, causing the deformation. The resulting internal friction heats the interior of the body.
With over 400 active volcanoes, Io’s internal heat mainly come from tidal heat- ing resulting from the pulls by Jupiter and other Galilean moons. The height of Io’s tidal bulge varies as much as 100 meters over its course. Peal et al [8] explained the tidal heat as follows. While all Galilean moons’ orbits are nearly circular and the orbits of Ganymede, Europa, and Io are in a 1:2:4 resonance, Io frequently encounters Europa and Ganymede, causing frequent changes of tidal deformations on Io, in turn generating significant friction and in extreme case volcanic activities. They predicted volcanic activities on Io, and one week after the prediction, analy- sis of an image from Voyager-1, a probe passed by Io on March 5, 1979, proved it.
A similar process is probably the reason of keeping the interior warm, melting ice crust into liquid water, and powering water vapor geysers on Enceladus, which is a moon of Saturn and experiences orbital resonance with Dione, another moon of Saturn.
The Roche Limit
The Roche limit named after French astronomy Edouard Roche, is the distance outside that a large satellites, hold together mainly by its self-gravity, can exist, while within the limit the tidal force exceeds the gravitational self-attraction and 4 hence tears apart the satellite. In 1848, Roche published the first theoretical es- timate of the limit, about 2.5 times of the radius of the primary for fluid satellite and equal densities between the satellite and the primary.
The Roche Limit [11] for two bodies is roughly approximated by a function of their densities: For rigid satellites: ρ 1/3 L = R 2.44 planet ρsatellite where R is the radius of primary, and the ρ values are the densities.
For liquid satellite: ρ 1/3 L = 2.44R planet ρsatellite where R is the radius of primary, and the ρ values are the densities. Using datas from wikipedia and taking the average density of comets to be around 500kg/m3, the table (1.2) gives some Roche limits.
Roche Limit (rigid) Roche Limit (fluid) Body Satellite Distance (km) R Distance (km) R Moon 9,496 1.49 18,261 2.86 Earth average Comet 17,880 2.80 34,390 5.39 Earth 554,400 0.80 1,066,300 1.53 Jupiter 890,700 1.28 1,713,000 2.46 Sun Moon 655,300 0.94 1,260,300 1.81 average Comet 1,234,000 1.78 2,374,000 3.42
Table 1.2: Data from wikipedia about Roche Limits
As mentioned before, both natural and artificial satellites hold together by forces other than self-gravity do not necessarily disintegrate inside their primary’s Roche Limit, such as the Jupiter’s moon Metis and Saturn’s moon Pan bonded by tensile strength, and some small satellites self-attracted by their electrochemical bonds. 5
Rings and Satellites
Tidal force plays a very important factor in the process of rings-forming. All gas giants of the solar system, Jupiter, Saturn, Uranus and Neptune, possess ring systems, composed of swarms of orbiting cosmic dust and small particles in shape of flat disc.
Historically Maxwell’s theoretical conclusion that Saturn’s rings must be com- posed of small particles confirmed in 1895 by spectroscope analysis that inner and edge and outer edge of the ring are rotating in different speed. Further studies sug- gests three possible sources for these small particles of the ring system of a primary.
The first is original material in the disk. The second source is the debris of satellites that are disrupted by large impact and then captured by primary’s grav- ity, moved into Roche limit and possibly destroyed by the tidal force as happened to comet Shoemaker-Levy 9 at Jupiter. Within Roche limit the tidal force pre- vents both original material and new captured fragments to coalesce a satellite. The last source is debris disrupted from satellites by tidal force when they passing within Roche limit. The photographs from 1996 to 1997 show debris disintegrate from Amalthea and Thebe and into the rings of Jupiter, supporting this possibility.
Based on these suggestions, the existence of sharp boundaries of some rings can be explained by shepherd satellites, like Cordelia and Ophelia, confining as inner and outer shepherds of Uranus’ ε ring respectively.
Catastrophic Events
Tidal forces is also responsible for some extraordinary events. Neptune’s moon Triton, a unique large moon in solar system with retrograde orbit, is predicted to pass within Roche limit in 3.6 billion years from now, then will be captured by Neptune or torn up by tidal force, forming a ring similar to that of Saturn. Mars’ moon Phobos is also predicted to break up and forming a ring in about 50 million years. 6
More recently, collision between the comet Shoemaker-Levy 9 and Jupiter was predicted and oberserved in July 1994. Calculations shows that it was broken up into more than twenty pieces when it passed within Jupiter’s Roche limit in 1992. These fragments were captured and orbiting Jupiter on March 24, 1993 and finally colliding in 1994.
In summary we already know that the tidal force raised on planets and satellites by satellites, planets and the Sun can affect rotations of planets and satellites. Also, the tidal forces raised on planets by satellites or on satellites by planets can affect the orbits of the satellites. While tidal locking of the spin of planets and satellites is a popular topic in astronomy, the tidal effect on the orbit of the planets and satellites is much less understood. This question is one of dynamical systems. In this paper we will study the effect of tidal forces on the orbits of the satellites in the 2-body and 3-body problems. As tidal forces distort planets and satellites, the energy of the satellite and the system is constantly converted into heat—–except in some special motions. Because of this energy loss, the tidal force causes the orbit the moon to approach a circle under some appropriate assumptions, as we show in chapter 3. we discuss the effect on the orbit of 3-body problem in chapter 4. In the next section we give some additional background on tidal forces.
1.2 The properties of the tidal effect
Let the radius of Earth be re. A point mass m on the Earth closest to the Moon feels a gravitational force of
mmoonm Fclostest = G 2 (r − re) while the point of the same mass farthest from the Moon feels a gravitational force of mmoonm Ffarthest = G 2 (r + re) where G is the gravitational constant and r is distance from the center of the earth to the center of the Moon. 7
The Earth is therefore stretched along an axis pointing toward the Moon, with the graviational force gradually decreasing to the far side.
Earth Moon
Figure 1.1: the bulge of Earth
Using some simplifying assumptions (described in the next paragraph), we will estimate that the tidal force pulling the Earth apart along the centers-line as follows: 2GMmr F = , (1.2.1) R3 where M and m are the masses of the Moon and of the Earth, R is the center-center distance, and r is the radius of the Earth. This inverse cube function decreases rapidly with distance.
This formula is obtained from the dumbbell model, the Earth is imagined as a m dumbbell with two equal point masses . The mass closest to the moon feels the 2 GmM force as F = , while the mass on the far side feels a force from moon as 1 2(R − r)2 GmM F = . We estimate the deference to the first order: 2 2(R + r)2 8
GmM 1 1 ∆F = F − F = − 1 2 2 (R − r)2 (R + r)2 GmM 2 2GmMr 2r . ≈ 2 (R − r)3 ≈ R3
the dumbbell model
m/2 r R M r m/2
Figure 1.2: the dumbbell model
Using the expression (1.2.1), we can estimate tidal forces for familiar bodies in table (1.3).
Apogees (farthest) Perigees (closest) Moon feels from Earth 1.51 × 1018 N 2.25 × 1018 N Earth feels from Moon 5.55 × 1018 N 8.25 × 1018 N Earth feels from Sun 2.87 × 1018 N 3.18 × 1018 N
Table 1.3: Estimation of tidal forces
Now let us account for the Moon’s rotation. If Earth’s self-spin and the rota- tion of the Moon around Earth were at the same rate, the long axis of the bulge (a prolate spheroid ) would always be along with the center-center line (the line from the center of Earth to the center of the Moon). 9
If the Earth were spinning a little less slowly than the Moon’s orbital period, and the nearest point of the bulge to the Moon was displaced from the center-center axis, slight behind the center-center axis. The Moon’s gravity applied therefore a small but significant torque to the Earth to speed up the self-spin of it and the radius of the Moon’s orbit was gradually shortened to conserve the angular mo- mentum.
Since the Earth is spinning faster than the Moon’s rotation, the bulge is located ahead of the center-center axis, and the Moon is steadily slowing down the Earth’s spin, and moving away from Earth.
Tidal bulge created by moon
Earth Moon
True position
Imaginary position
The spin of the Earth is slower than the rotation of the moon.
Figure 1.3: the spin lag
This has been a rough description of the tidal effect. Two principal formalisms [10] has been established by astronomers to discussing the effect of the tidal dis- sipation on an orbit: the equilibrium tide model and the dynamical tide model. However both of these models are not perfect and paid too much attention on the inner structures of celestial bodies so that they are difficult to apply on compli- cated context. 10
In this paper we treat a simplified model of tidal effect by ignoring the spin. Namely our models account for dissipation caused by the changes of the mutual distances only, but not the spin. On the other hand, our results are mathematically rigorous and hopefully they point a way to a rigorous treatment of more realistic models of tides. Generally speaking, discussions in the paper are based on the two assumptions. i. the rate of energy dissipation is a non-negative definite function of the posi- tion and velocity of the satellite.
ii. the rate of energy dissipation is zero only when the distance between the planet and the satellite is fixed. Chapter 2
Tidal Dissipation in General
2.1 General dissipation formulation
In real world the orbits of planets and satellites are not exactly planar due to the presence of out-of-plane perturbations. Here we consider only planar motions, however.
Let D(q, q˙) be the so-called Routh dissipation function [12] defined as follows:
∂D q˙ = rate of dissipation of energy per second. ∂q˙ ∂D One can therefore think of as the generalized dissipation force; sinceq ˙ = ∂q˙ generalized velocity, the above simply says: velocity · force = power.
It is convenient to introduce dissipation when the system is written in La- grangian (rather than Newtonian ) form. Let L(q, q˙) be the Lagrangian of the system. The equations governing the perturbative system are
d ∂L ∂L ∂D − + ε = 0 (2.1.1) dt ∂q˙ ∂q ∂q˙ where 0 < ε 1.
Our assumptions on the tidal effect can be restated in the following natural 12 way.
∂D i. the energy is always non-increasing, i.e. −q˙ ≤ 0. ∂q˙ ii. the energy does not dissipate only when the relative distance does not change, that is when |q|· = 0.
To understand the effect of tidal dissipation, like how tidal dissipation causes the Kepler ellipses to deform, it is more convenient to use Delaunay Variables. Since these are the action-angle variables, we must first write the equation (2.1.1) as a (dissipatively perturbed) Hamiltonian system. This is done in the next section in more detail.
Here we note only that the standard way to rewrite a Lagrangian system in the Hamiltonian form is to take the Hamiltonian H(q, p) = pq˙ − L(q, q˙) where q˙ =q ˙(q, p) is defined by Lq˙(q, q˙) = p implicitly. The Hamiltonian form is
∂H q˙ = ∂p (2.1.2) ∂H ∂D p˙ = − − ε ∂q ∂q˙
Under the assumption (i) only, the Hamiltonian equation above suggests the energy of the system, i.e. the Hamiltonian H of unperturbed Hamiltonian system, is always non-increasing.
Proposition 2.1.1. If the assumption (i) holds, then the Hamiltonian H of the corresponding unperturbed system of (2.1.2) is always non-increasing, and, more- dH over, if = 0, then |q|· = 0. dt Proof. We have
dH ∂H ∂H = p˙ + q˙ dt ∂p ∂q ∂D =q ˙p˙ + (−ε − p˙)q ˙ (2.1.3) ∂q˙ ∂D = −εq˙ ≤ 0 ∂q˙ 13 by assumption (i). The second part of this property follows from assumption (ii) and the remark #2.
2.2 Hamiltonian equations with dissipation un- der canonical transformations
Now we will rewrite the Hamiltonian equations with dissipative perturbation under new coordinates (especially the Delaunay coordinates). Our system (2.1.2) is of the form ! 0 z˙ = J∇H + (2.2.1) f(p, q) ! 0 I where z = (q, p) and J = . −I 0
Theorem 2.2.1. Under the substitution Z = φ(z), φ : R4 −→ R4 is symplec- tic, the perturbed Hamiltonian system (2.2.1) is equivalent to the new perturbed Hamiltonian system ∂φ Z˙ = J∇K + f φ−1(Z), (2.2.2) ∂p where K = H ◦ φ−1.
Proof. Let ϕ = φ−1, so that z = ϕ(Z). We have
∇Z K(Z) = ∇Z H(ϕ(Z)) (2.2.3) 0 T = (ϕ (Z)) ∇zH(ϕ(Z))
Because φ is a canonical transformation, ϕ is also a canonical transformation, i.e. both φ0(z) and ϕ0(Z) are symplectic matrices. Then (ϕ0(Z))−1J((ϕ0(Z))−1)T = J, 14 we then have
Z˙ = (ϕ0(Z))−1z˙ ! 0 −1 0 = (ϕ (Z)) (J∇zH(ϕ(Z)) + ) f(z) ! 0 −1 0 T −1 0 −1 0 = (ϕ (Z)) J((ϕ (Z)) ) ∇Z K(Z) + (ϕ (Z)) f(z) ! (2.2.4) 0 −1 0 −1 T 0 −1 0 = (ϕ (Z)) J((ϕ (Z)) ) ∇Z K(Z) + (ϕ (Z)) f(z) ! 0 −1 0 = J∇Z K(Z) + (ϕ (Z)) f(z) ∂φ = J∇K + f φ−1(Z). ∂p
2.3 LaSalle’s invariance principle
To find the limit orbits in a dissipatively perturbed system, we need a theorem similar to LaSalle’s Invariance principle. We extend this principle to problems with singularities, such as the Kepler problem, or the 3-body problem.
Theorem 2.3.1. Let ϕt(x) be the flow of the autonomous system x˙ = f(x) on a subset U of Rn, where f(x) is a smooth function. Let V be a real valued function on Rn, possibly infinite, and differentiable where finite. Assume that V (ϕt(x)) is non-increasing with respect to t for every given x ∈ U when it is well defined and finite. Define A = {x|∇V · f = 0} to be the zero dissipation set and let t S = {x|V (x) = −∞}. Let x0 be a point such that ϕ (x0) is defined for all t ≥ 0. t Then either ω(x0) ⊂ A or ω(x0) ⊂ S. In other words, either ϕ (x0) approaches a t connected component of A or ϕ (x0) approaches to a connected component of S.
t Proof. V (ϕ (x0)) is non-increasing real function and well defined for any t ≥ 0. So t lim V (ϕ (x0)) is either −∞ or a finite number. There is nothing to discuss when t→∞ ω(x0) is empty. Since in that case the orbit leaves the domain U or approaches 15
infinity, we assume ω(x0) 6= ∅. Choose y0 ∈ ω(x0). There exists a subsequence tn tn {tn} with tn → ∞ such that ϕ (x0) → y0, while V (ϕ (x0)) → V (y0)(V (y0) may be −∞). Choose another y1 ∈ ω(x0) if exists. It gives us another limit value t0 t0 V (y1) of another subsequence of the orbit ϕ m (x0), i.e. V (ϕ m (x0)) → V (y1). By t monotonicity of V (ϕ (x)) with respect to t, V (y0) = V (y1). This shows that V restricted to ω(x0) is constant, call it c(x0).
Case I: c(x0) = −∞. That implies that ω(x0) ⊂ S by the definition of S. By continuity ω(x0) is a connected component of the set S.
t Case II: c(x0) > −∞. Since ω(x0) is invariant, V (ϕ (y)) = c, for every y ∈ d ω(x ) and any t ∈ +. So for every y ∈ ω(x ), 0 = V (ϕt(y))| = ∇V (y) · f(y), 0 R 0 dt t=0 i.e. ω(x0) ⊂ A. Chapter 3
Tidal Dissipation for Kepler Problem
3.1 Assumptions on the tidal dissipation
The orbit of the two body problem, i.e. that of the Kepler problem
q q¨ = − , |q|3 are ellipses in the case of negative energy. Now we introduce a model of tidal dissipation into this classical problem as follows. We use the same model to study restricted planar 3-body problem in the next chapter as well.
The Lagrangian equation of the dissipative Kepler problem is also in the form of d ∂L ∂L ∂D − + ε = 0, (3.1.1) dt ∂q˙ ∂q ∂q˙ where ε > 0 and the Lagrangian is
q˙2 +q ˙2 1 L = 1 2 + . (3.1.2) 2 p 2 2 q1 + q2
For this problem we add one more assumption about the symmetry of the dissipa- tion force and restate all assumptions as following.
∂D i. the energy is always non-increasing, i.e. −q˙ ≤ 0. ∂q˙ 17
ii. the energy does not dissipate only when the relative distance does not change, that is when |q| = constant.
iii. D(q, q˙) = D(Rq, Rq˙) for any 2 × 2 rotation matrix R.
Remark #1: Note the assumption (iii) is equivalent to saying that D(q, q˙) is a function of |q| and (q, q˙), and hence a function of q and |q|·,
D(q, q˙) = f(|q|, |q|·).
We can see it by choosing R in assumption (iii) as a rotation of −θ, where θ is the angle of q.
Remark #2: The assumption (ii) is equivalent to stating that f2 6= 0, where f2 is the partial derivative of f with respect to the second variable. Indeed, note first d √ q · q˙ q · q˙ that |q|· = q · q = √ = . Then dt q · q |q|
∂D ∂ q · q˙ ∂ q · q˙ q = f |q|, = f = f . (3.1.3) ∂q˙ ∂q˙ |q| 2 ∂q˙ |q| 2 |q|
Hence ∂D q f q˙ =q ˙ · f = 2 (q · q˙) = f |q|·. ∂q˙ 2 |q| |q| 2 ∂D So that ifq ˙ = 0, then either f = 0, or |q|· = 0. ∂q˙ 2
3.2 Effect of dissipation on the evolution of the dynamics of the 2-body problem
In this section we prove that, under the three assumptions stated in section 3.1, angular momentum of the system determined by the dissipative Kepler problem with negative energy is conserved and every orbit of the system approaches a circle. 18
The Lagrangian equation of the dissipative Kepler problem is in the form of (3.1.1), and the Hamiltonian equations are
∂H q˙ = ∂p (3.2.1) ∂H ∂D p˙ = − − ε ∂q ∂q˙ where 0 < ε 1 and the Hamiltonian is
q˙2 +q ˙2 1 p2 + p2 1 H = 1 2 − = 1 2 − , 2 p 2 2 2 p 2 2 q1 + q2 q1 + q2 where q = (q1, q2) and p = (p1, p2). Elliptic orbits correspond to H < 0.
The Angular momentum is
G = q1q˙2 − q˙1q2 = q1p2 − p1q2
Proposition 3.2.1. If the dissipation function D satisfies the assumption (iii), then the angular momentum, G is conserved.
Proof.
dG = q q¨ − q¨ q dt 1 2 1 2 q2 ∂D q1 ∂D = q1(− 3 − ε ) − q2(− 3 − ε ) |q| ∂q˙2 |q| ∂q˙1
∂D ∂D ∂D ∂D T = ( q2 − q1) = ε( , ) · (q2, −q1) ∂q˙1 ∂q˙2 ∂q˙1 ∂q˙2 = 0, using (3.1.3) in the last step. The angular momentum, G is preserved.
Now combining the proposition (3.2.1) and theorem (2.3.1), we show that the limit orbit must be a circle.
Theorem 3.2.1. Under the three assumptions, the orbit of dissipative Kelper’s problem, i.e. the system (3.2.1), in any initial condition, approaches a circle with nonzero angular momentum. 19
Proof. First of all observe that the nonzero angular momentum precludes collisions.
The angular momentum G in polar coordinates, q1 = r cos θ and q2 = r sin θ, takes form 2 ˙ G = q1q˙2 − q˙1q2 = r θ.
Since G(t) = G0 6= 0, we show below that collisions never happen. Since the q˙ · q˙ 1 energy H is non-increasing, H = − ≤ H where H is the energy of the 2 r 0 0 system at time 0. Using polar coordinates, we have
r˙2 + (rθ˙)2 1 H = − ≤ H 2 r 0 r˙2 G2 1 + 0 − ≤ H 2 2r2 r 0 G2 2 0 − ≤ 2H r2 r 0
G2 2 Define h(r) := 0 − . The graph of h(r) is illustrated in figure (3.1). So r2 r r ≥ rmin > 0, where rmin is the minimum value of r satisfying the inequality, h(r) ≤ 2H0, and as shown in the figure.
6 5 4 3 2 1 0 3 6 9 12 15 18 -1 rmin -2 y=2H0 -3 -4 -5 -6
Figure 3.1: the graph of h(r)
Since H is non-increasing, let Hmin = limt→∞ H(t). From the form of the energy H we see H is infinity only when q is 0, i.e. a collision happens. So Hmin 20 is a finite number. Now we can apply theorem (2.3.1) on the function H since dH it is bounded from above and below. Enegy dissipation = 0 if and only if dt q · q˙ = rr˙ = 0 which implies a circular orbit.
3.3 One choice of dissipation function
In this section we will plot numerical solutions of the dissipation equation (3.2.1) by choosing a specific dissipation function to confirm the result on the limit orbit, theorem (3.2.1). And we will also explain an interesting phenomena that the or- bit of the dissipative Kepler problem repeatedly passes through two points on the vertical axis (as explained below and illustrated in figure(3.2) ).
Now we start by choosing D to be a quadratic function of both q andq ˙, i.e.
(q · q˙)2 D = , (3.3.1) 2 then the power dissipation is given by
∂D −εq˙ = −ε(q · q˙)2. ∂q˙
A graph of the numerical solution of the equation (3.1.1) is illustrated in figure (3.2). We can see the orbit approach a circle, which is determined by the initial condition. This orbit has a surprising feature: it seems to go repeatedly through two points on the vertical axis. This axis passes through the primary and is perpendicular to the major axis of the Kepler ellipse.
To explain this phenomenon, we consider the perturbed Kepler problem (3.1.1) under a canonical transformation to the two dimensional Delaunay variables φ : (q, p) 7−→ (k, K, g, G). Apply theorem (2.2.1) here. In these variables the per- 21
10 10
5 5
-25 -20 -15 -10 -5 5 -25 -20 -15 -10 -5 5
- 5 -5
-10 -10
(a) t=500 (b) t=3000 Figure 3.2: Pictures of Kepler Problem with dissipation ε = 10−5 turbed Kepler system takes form:
∂H ∂K ∂D K˙ = − − ε · ∂k ∂p ∂q˙ ∂H ∂G ∂D G˙ = − − ε · ∂g ∂p ∂q˙ (3.3.2) ˙ ∂H ∂k ∂D k = − ε · ∂K ∂p ∂q˙ ∂H ∂g ∂D g˙ = − ε · ∂G ∂p ∂q˙ where 1 H = − (3.3.3) 2K2 ∂ ∂ is the Hamiltonian, and , must be expressed in term of partial derivatives ∂p ∂q˙ ∂H ∂H ∂H with respect to (k, K, g, G). Note that all , and are actually zero by ∂k ∂g ∂G the form of the Hamiltonian in K, due to (3.3.3). Thus (3.3.2) simplifies to
∂K ∂D K˙ = −ε · ∂p ∂q˙ ∂G ∂D G˙ = −ε · ∂p ∂q˙ (3.3.4) 1 ∂k ∂D k˙ = − ε · 3 K ∂p ∂q˙ ∂g ∂D g˙ = −ε · . ∂p ∂q˙ 22
The variable (q, p) defines the Kepler ellipse, as initial data. Then
K = pmajor axis of Kepler ellipse G = angular momentum of the system
k = 2πt/T g = the angle from the major axis to a coordinate axis where T is the period of the Kepler ellipse with (q(t), p(t)) as initial data.
Equation (3.3.5) below explains figure (3.2) as follows. Since K,G and g are almost constant compared with k, the first equation of (3.3.4) can be averaged over one period of the unperturbed ellipse, i.e. over 0 ≤ l ≤ 2π.
Z 2π ˙ 1 ∂K ∂D Kav = − ε · dk (3.3.5) 2π ∂p ∂q˙ 0 substitute K by Kav
1 We estimate the integrand. Since K = √ , −2H
∂K ∂H 1 = (−2H)−3/2 = − q.˙ ∂p ∂p K3
So we have from (3.3.5)
Z 2π ˙ ε 1 ∂D Kav = − q˙ · dk ≤ 0 (3.3.6) 2π K3 ∂q˙ 0 substitute K by Kav by proposition (2.1.1). This calculation shows that the major axis of the Kepler ellipse shortens due to the presence of dissipation and suggests that the orbit ap- proaches a circle. We proved it rigorously in theorem 3.2.1.
Another fact about the averaged system that will be in the next theorem is that 1 for sufficiently small ε, |K(t) − K (t)| < cε, for all t, 0 < t < , see Arnold[1], av ε page 292.
It is not clear a priori why the direction of the major axis should ”remember” 23 as figure (3.2) suggested. We now give the reason in the following theorem.
Theorem 3.3.1. The major axis of the orbit, of the system (3.3.4) with dissipation function D defined in (3.3.1), rotates at the average speed of O(ε2) over the time 1 interval 0 ≤ t ≤ . ε 1 Proof. Since |K(t) − K (t)| = O(ε) for all 0 < t < , it is safe to use the or- av ε bit(ellipse) of the Kepler Problem to calculate the variation of g, the angle from the major axis to the reference axis.
Set (q, p) = (q(t), p(t)) and (q0, p0) = (q(−t), p(−t)). So q and q0 are symmetric w.r.t. the x-axis (the major axis), and p and p0 are symmetric w.r.t. the y-axis. Introduce the notations ! ! 1 0 −1 0 Rx = and Ry = , 0 −1 0 1 representing reflection over the x-axis and the y-axis respectively, as shown in figure (3.3). Then the variation of g in one period is
Z ∂g ∂D g(T ) − g(0) = ε · dt ∂p ∂q˙ −T/2 ∂g ∂D ∂g ∂D We now show that · = − · . Let ∆g denote the ∂p ∂q˙ (q,p) ∂p ∂q˙ (q0,p0) 24 p (q,p)=(q(t),p(t)) q g p' q' (q',p')=(q(-t),p(-t)) Figure 3.3: Symmetry of the system 0 0 change due to ∆p. Then a change −Rx∆p at point (q , p ) induces a change of −∆g. So ∂g ∂g = , ∂p1 (q0,p0) ∂p1 (q,p) ∂g ∂g = − . ∂p2 (q0,p0) ∂p2 (q,p) We use this symmetry result to show that the O(ε) term in (3.3.7) vanishes. ∂g ∂D ∂g 0 0 0 ∂g 0 0 · = · q (q · p ) = Rx · (Rxq)(q · p ) ∂p ∂q˙ (q0,p0) ∂p (q0,p0) ∂p (q,p) ∂g ∂g = · q (−Rxq) · (Ryp) = − · qλ(q, p) ∂p (q,p) ∂p (q,p) ∂g ∂D = − · . ∂p ∂q˙ (q,p) (3.3.8) 25 Plugging the equation (3.3.8) into the equation (3.3.7) we proved this theorem. Consider the Kepler ellipse with the initial condition (q(t), p(t)). Since K = O(ε) and G = const., the y-intercept of the Kepler ellipse is G2, a constant. Chapter 4 Tidal dissipation in the restricted planar 3-body problem 4.1 Introduction In the study of the 3-body problem, Euler discovered 3 linear central configura- tions while Lagrange discovered another two equilateral configurations. The Sun, Jupiter and the cluster of the Trojan asteroids form an approximately equilateral triangle. It is therefore of interest to study the effect of tidal dissipation on these configurations. In this chapter we will build a model of the restricted (the third mass is negligible compared with the first two masses) planar 3-body problem and a similar model of the corresponding problem with dissipation. Based on the model we will study the asymptotical behavior of the orbits of this problem. And we will also study the stability in the relative equilibria. 4.2 A model in rotating frame In this section we will build a model of the restricted circular planar 3-body prob- lem in rotating frame and write the equations in different forms of this problem for the use in the following sections. 27 We consider the motion of the planar 3-body problem in inertial frame first. The three masses m1, m2 and m3 move under Newtonian mutual attraction; as- sume m3 to be negligible compared with m1 and m2. The masses m1 and m2 then travel along conic sections. In fact, we assume that they describe circles about their common center of mass and let ω be the angular velocity of their revolution. Let Qi (i = 1, 2, 3) denote the positions of the three masses in an inertial frame with the origin at the center of mass. Without loss of generality we can normalize the total mass of the system to be 1, taking m1 = 1 − µ, m2 = µ and m3 = 0. We also normalize the distance between positive masses to 1; this implies ω = 1, ! ! −µ 1 − µ and Q1(0) = and Q2(0) = . Then 0 0 ! ! −µ 1 − µ Q1(t) = R(t) and Q2(t) = R(t) , 0 0 where ! cos t − sin t R(t) = . sin t cos t The position vector q = Q3 of the third mass satisfies Newton’s Law q¨ = −Uq (4.2.1) where (1 − µ) µ U(q, t) = − − |q − Q1| |q − Q2| is the time-dependent potential energy of the third particle, and where Uq denotes the gradient of U: (1 − µ)(q − Q1) µ(q − Q2) Uq = 3 + 3 . |q − Q1| |q − Q2| In the next section we will need a Hamiltonian from of these equations in the 28 rotation frame. Equation (4.2.1) can be written as a Hamiltonian system: q˙ = p (4.2.2) p˙ = −Uq By choosing the rotating frame of the heavier masses, we turn equation (4.2.1) ! x into an autonomous system, as follows. The postion x = 1 and the momentum x2 ! y1 y = of the third mass m3 in the rotating frame satisfies y2 q = R(t)x and p = R(t)y. (4.2.3) The new system is Hamiltonian with the new Hamiltonian function 1 E = y2 + x y − x y + U(x), (4.2.4) 2 2 1 1 2 where (1 − µ) µ U(x) = − 2 2 1/2 − 2 2 1/2 (4.2.5) ((x1 + µ) + x2) ((x1 + µ − 1) + x2) is now time-independent. ! 0 1 The equation (4.2.2) in the rotating frame takes the form with J = : −1 0 x˙ = y + Jx y˙ = −Ux + Jy, or more explicitly x˙ 1 = y1 + x2 x˙ 2 = y2 − x1 (1 − µ)(x + µ) µ(x + µ − 1) y˙ = y − 1 − 1 (4.2.6) 1 2 3 3 r1 r2 (1 − µ)x2 µx2 y˙2 = −y1 − 3 − 3 , r1 r2 29 2 2 1/2 where r1 = ((x1 + µ) + x2) , the distance from the mass m3 to the mass m1, and 2 2 1/2 r2 = ((x1 + µ − 1) + x2) , the distance from the mass m3 to the mass m2. An alternative way to transform the Newtonian form (4.2.1) into the rotating frame is via complex notation. Write z = x1 +ix2 andz ˙ =x ˙ 1 +ix˙ 2 and q = q1 +iq2. Equation q = R(t)x amounts to q = eitz. Then the Newton’s equation (4.2.1) becomes (1 − µ)eit(z + µ) µeit(z + µ − 1) eit(¨z + 2iz˙ − z) = − − . |z + µ|3 |z + µ − 1|3 (1 − µ) µ Letting V (z) = − − , we get the equations of motion in the |z + µ| |z + µ − 1| rotating frame: z¨ = 2iz˙ + z − ∇zV (z). (4.2.7) Here 2iz˙ is the Coriolis force and z is the centrifugal force. Note that the energy |z˙|2 |z|2 − + V (z) (4.2.8) 2 2 is conserved. The corresponding Hamiltonian equation (4.2.6) in complex notation is z˙ = w − iz (4.2.9) w˙ = −iw − ∇zV (z) with the Hamiltonian 1 ww + V (z) + Im zw. (4.2.10) 2 4.3 Assumptions about the tidal effects Now we discuss the effect of the tides on the motion of the third mass in the model of restricted planar 3-body problem. Write z = x1 + ix2, dissipation can be included in the Lagrangian equation according to Routh-Lagrange, as mentioned 30 before: d ∂L ∂L ∂D − + ε = 0 (4.3.1) dt ∂z˙ ∂z ∂z˙ 1 where L(z, z˙) = ((x ˙ − x )2 + (x ˙ + x )2) − U(z) and U(z) is defined in (4.2.5). 2 1 2 2 1 Equation (4.3.1) can also be written in a form of first order z˙ = w ∂D (4.3.2) w˙ = 2iw + z − ∇ V (z) − ε . z ∂z˙ ∂D ∂D(z, z˙) where = , and D is the same function as in equation (4.3.1). ∂z˙ ∂z˙ z˙=w Note that the energy |w|2 |z|2 E = − + V (z) (4.3.3) 2 2 dE ∂D changes according to = −εw . dt ∂z˙ We assume that the energy dissipation takes place due to changes of the dis- tances, Q1Q3 and Q2Q3, as in the dissipative Kepler problem. That is, we take D(z, z˙) = Da(z + µ, z˙) + Db(z + µ − 1, z˙), where Da and Db satisfy the three as- sumptions stated in section 3.1. ∂D Remark #3: The three assumptions imply thatz ˙ a ≥ 0 with equality if ∂z˙ ∂D and only if |z + (µ, 0)| = constant, andz ˙ b ≥ 0 with equality if and only if ∂z˙ ∂D |z + (1 − µ, 0)| = constant. Soz ˙ = 0 impliesz ˙ = 0. ∂z˙ 4.4 The asymptotic behavior of the dissipative system In this section we show that every solution of the dissipative 3-body problem approaches one of five central configurations, escapes to infinity, or approaches a collision with one of the primaries. 31 Theorem 4.4.1. Any orbit of (4.3.2) with D satisfying the three assumptions approaches infinity, a singularity, or an equilibrium in the configuration space. Proof. The energy function E defined in (4.2.8) satisfies all the conditions for the function in the theorem (2.3.1). Under the first two assumptions on the dissipation force, the zero-dissipation set is ∂D(z, w) A = {x = (z, w): w · = 0}\S ∂w = {x = (z, w): w = 0}\S (4.4.1) = {x = (z, 0) : z 6= −µ and z 6= 1 − µ}, where the collision set S := S1 ∪ S2, S1 = {x = (z, w): z = −µ}, and S2 = {x = (z, w): z = 1 − µ}. Here treat z, w, µ, 1 − µ and 0 as complex numbers. Now let ϕt(x) be the flow defined by the equation (4.3.2), and consider the t orbit of ϕ (x0). For the limit set ω(x0) = ∅, the orbit approaches either infinity t or one of the singularities. Otherwise, there exist two subsequences of ϕ (x0) re- spectively approaching two different singularities or one singularity and infinity. t However, since such an orbit ϕ (x0) must cross the level sets of the energy set, it is a contradiction with the fact that the energy is always non-increasing. For ω(x0) 6= ∅, the orbit is connected and invariant under the flow of the equation (4.3.2). The orbit must cross the level set of the energy too, as long as there are more than one element in ω(x0). So ω(x0) must be one equilibrium in the w = 0 plane. This proves the theorem. We give an alternative proof of the theorem (4.4.1), which does not rely on our extension of the LaSalle’s principle and is more direct and intuitive. 4 Proof #2. In system (4.3.2), the 2-planes S1 and S2 in R are the sets of singu- larities. All five equilibria — three Eulerian and two Lagrangian — lie in the zero-dissipation set A, which is defined in (4.4.1). 32 ε ε ε ε Now choose a compact neighborhood U = B1/ε\(S1 ∪ S2 ∪ A ), where Br is a ε ε ε closed ball of radius r, S1 , S2 and A are the ε neighborhood of corresponding set. For convenience, we write U ε = U. As illustrated in figure (4.1), U is in R4 with grey areas excluded. z w=z! u S1 S2 (-!,0) (1-!,0) A z Figure 4.1: the neighborhood, U The energy E(x), defined in (4.3.3), restricted to U is bounded from above and below. Moreover, there exists δ = δ(ε), such that D(z, w) < −δ < 0 for any x = (z, w) ∈ U. Hence E(x(t)) is strictly decreasing in U and thus U contains no limit points of any solution. Hence limit points lie outside of U. According to the ε ε ε arbitrariness of U, we conclude that all limit points lie in the set S1 ∪S2 ∪A . If the t t limit set ω(x0) = , then ϕ (x0) → ∞. If ω(x0) 6= and lim E(ϕ (x0)) = ∞, then ∅ ∅ t→∞ t t ϕ (x0) → S = {x = (z, w)|z = µ or z = 1 − µ}. If ω(x0) 6= and lim E(ϕ (x0)) ∅ t→∞ is finite, then ω(x0) ⊂ A = {x = (z, w)|w = 0 and z 6= µ and z 6= 1 − µ} and, because ω(x0) is invariant under the flow of the differential equation, ω(x0) consists of equilibria. Since the set of equilibria consists of disconnected points, ω(x0) could only be one of them. Figure 4.2: Hill’s regions and a level set of E 33 4.5 Effect of dissipation on stability In this section we will study the stability of equilibria of the system in the rotating frame, i.e. the relative equilibria of the original system. We will show that a large class of dissipation functions always cause these equilibria to become unstable. And then we illustrate this stability loss numerically by choosing some specific dissipation function. We also give a formula for the eigenvalue changes caused by dissipation ε. 4.5.1 Stability of the dissipative problem at relative equi- libria Definition :A relative equilibrium of the Newtonian n-body problem is a planar configuration of the n point masses such that the acceleration vector of each mass produced by the gravitational attraction of the other n − 1 masses is directed toward the center of mass with magnitude proportional to the distance from the center of mass. The acceleration can be exactly balanced by an outward centrifugal accelera- tion if the configuration is uniformly rotated at the appropriate angular velocity. In our setting relative equilibria are just equilibria in the rotational frame. The N-body problem has no stationary equilibria for N ≥ 2. There are five relative equilibria of the 3-body problem, particularly of the restricted planar 3- body problem. Richard Moeckel [4] proved the linear instability of collinear relative equilibria as well as linear stability at equilateral triangles (solutions of Lagrange) under some condition in the system without dissipation. Let us now consider the dissipative problem in Hamiltonian form z˙ = w − iz ∂D (4.5.1) w˙ = −iw − ∇ V (z) − ε , z ∂z˙ 34 (1 − µ) µ where V (z) = − − and D is a dissipation function to be de- |z + µ| |z + µ − 1| scribed later. Since we assume D(z, 0) = 0, the equilibria of this system are the same as those of the classical conservative one. We find these by settingz ˙ = 0 and w˙ = 0 in the system (4.5.1), getting w = iz and concluding that z is the solution of the algebraic equation (1 − µ)(z + µ) µ(z + µ − 1) z = + . (4.5.2) |z + µ|3 |z + µ − 1|3 This equation expresses the balance between the centrifugal force and gravitational pull. Two solutions satisfying |z + µ|√= |z + µ − 1|√= 1 give the Lagrange configu- 1 3 3 rations, z = x + ix , x = − µ ± and x = . 1 2 1 2 2 2 2 For ε = 0, Moeckel’s result [4] gives stability of the Lagrange’s equilibria, pro- vided 27µ(1 − µ) < 4. Our goal is to understand the effect of the dissipation on this stability. Equation (4.2.7) written as the first order system is x˙ = u ∂D (4.5.3) u˙ = 2Ju + x − U − ε . x ∂x˙ The linearization at an equilibrium is ! ! ξ˙ ξ = (M + εN) (4.5.4) η˙ η where ! ! ! 0 I 1 0 0 1 M = , I = , J = , (4.5.5) I − Uxx 2J 0 1 −1 0 ! 0 0 and N = , Dxx˙ Dx˙x˙ and Uxx is the Hessian Matrix of the function U. Its entries are 35 2 2 1 − µ µ 3(1 − µ)(x1 + µ) 3µ(x1 + µ − 1) Ux1x1 = 3 + 3 − 5 − 5 r1 r2 r1 r2 3(1 − µ)(x1 + µ)x2 3µ(x1 + µ − 1)x2 Ux1x2 = Ux2x1 = − 5 − 5 r1 r2 2 2 1 − µ µ 3(1 − µ)x2 3µx2 Ux2x2 = 3 + 3 − 5 − 5 r1 r2 r1 r2 where r1, r2 are the distances to the primaries. To find the characteristic polynomial for M, note that √ ! 1 1 −3 3(1 − 2µ) Uxx = √ , 4 −3 3(1 − 2µ) −5 where we have substituted the Lagrange equilibrium into it. In particular, for ε = 0, ! λI −I P0(λ) = det(λI − M) = det −I + Uxx λI − 2J 27 = λ4 + λ2 + µ(1 − µ). 4 The equilibrium is spectrally stable if and only if 27µ(1−µ) < 4, which reproduces Moeckel’s result [4]. We now prove that a rather general class destroys stability. Theorem 4.5.1. Let the dissipation function D in (4.5.3) be a quadratic form in x˙ as D(x, x˙) = (Ax,˙ x˙), where (·, ·) is the dot product and A = A(x) is a 2 × 2 positive definite matrix. The Lagrangian equilibrium of the dissipative problem (4.5.3) is always spectrally unstable if 0 < ε 1 and small and 0 < µ < 1. ! 0 0 Proof. For convenience write N as the block matrix N = , where N1 N2 36 ! a b N1 = Dxx˙ (x, x˙)|x˙=0 = 0, and N2 = −Dx˙x˙ = −2A = . Then b c ! λI −I Pε(λ) = det(λI − (M + εN)) = det −I + Uxx − εN1 λI − 2J − εN2 = λ4 −ε(a + c) λ3 + (1 + (ac − b2)ε2) λ2+ | {z } | {z } a a 3 √ 2 9 3 3 3 √ 27 ε( a + c − b + 3 3µb) λ + µ(1 − µ) . 4 4 2 4 | {z } | {z } a1 a0 According to the Routh-Hurwitz criterion, this polynomial is stable, i.e. its roots λi satisfy Re λi < 0 if and only if (here a4 = 1) ai > 0, i = 0, 1, 2, 3, 4. a3a2 > a4a1 2 2 and a3a2a1 > a4a1 + a3a0. Assume that our polynomial is stable, i.e. the above inequalities hold, in par- 2 ticular, a1 > 0. Since N2 is negative definite,√ac − b > 0√ and a + c < 0. Hence 9 3 3 3√ 3 3 both a < 0 and c < 0. Observe a + c ≤ − ac < |b|. Then 4 4 2 2 √ √ 9 3 3 3 √ 3 3 √ a = ε( a + c − b + 3 3µb) < ε(− (|b| + b) + 3 3µb). 1 4 4 2 2 √ If b ≤ 0, then a = 3 3µb ≤ 0, a contradiction with a > 0. If b > 0 and 1√ 1 0 < µ < 1, a1 = 3 3b(µ − 1) < 0, it is a contradiction with a1 > 0 again. Corollary 4.5.1. In the settings of the last theorem every solution of (4.5.3) not on the stable manifold of one of the 5 equilibria approaches ∞ or a collision. Proof. It directly follows from theorem (2.3.1) and the previous theorem. 4.5.2 A particular dissipation function Theorem (4.5.1) applies to a large class of quadratic dissipation functions. To sim- ulate the result numerically, we must choose a specific such function. A reasonable 37 choice that includes the tidal effect of both of the primaries on the third mass is 1 1 D = ((x + µ)x ˙ + x x˙ )2 + ((x + µ − 1)x ˙ + x x˙ )2 (4.5.6) 2 1 1 2 2 2 1 1 2 2 This function is in the class covered by theorem (4.5.1). We show the result intuitively by graphing the numerical solution in the figures (4.3). As corollary (4.5.1) proved, we see after the dissipation breaks the stability of the Lagrangian equilibria, almost all orbits approach singularities or infinity except those orbits consisting of a low-dimensional stable manifolds of equilibria. The figure (4.3a) illustrates one orbit starting around a Lagrangian equilib- rium going to infinity, while figure (4.3b) illustrates another orbit also close to the Lagrangian equilibrium which goes to a singularity. 6 1.0 4 2 0.5 -6 -4 -2 2 4 6 -1.0 -0.5 0.5 1.0 -2 -4 -0.5 -6 √ √ 3 3 (a) To the infinity with z = 0.4899 + i (b) To a singularity with z = 0.489 + i 0 2 0 2 Figure 4.3: How dissipations causes instability µ = 0.01 and ε = 0.01. Both figures are in configuration space. 4.5.3 Motion of the spectrum Our result on the dissipation-induced instability, proven in theorem (4.5.1) and il- lustrated numerically can be confirmed in an alternative way, by adapting an idea 38 of Krein, which are invented to study Hamiltonian systems, to dissipative systems. Our result in this section is a dissipative counterpart of Krein’s theorem on the Hamiltonian system that we state here. In [5] Krein gave an interesting lemma which showed how the eigenvalues of a linear Hamiltonian system change under ! 0 I perturbation of the Hamiltonian. Below, J = , where I is a 2n×2n identity I 0 matrix. Theorem 4.5.2 (Krein). Consider a linear Hamiltonian system Jx˙ = (H(t) + εQ(t))x depending on the real parameter ε with H(t) and Q(t) 1-periodic, and with Q(t) positive definite for all t, or, more generally, Jx˙ = H(t, ε)x ∂H with positive definite. Assume that the system is strongly stable at ε = 0. ∂ε Then the eigenvalues λk(ε) of the monodromy matrix of positive(negative) type are moving clockwise (counterclockwise) on the unit circle as ε increases in the small neighborhood of 0. In a similar spirit we describe how the eigenvalues λ change with ε in the dissipative problem. Theorem 4.5.3. Assume that λ is a simple purely imaginary eigenvalue of M, where M and N are defined in equations (4.5.4) and (4.5.5). Then there exists a simple eigenvalue λε of M + εN for 0 < ε < ε0 for some ε0. Let vε be the eigenvector of M + εN and in particular, let v = v0 be the eigenvector of λ. Then dλ [Nv, v] | = , dε ε=0 [v, v] 1 1 where [x, y] = (Jx, y) = (Jx)T y. 2i 2i 39 Proof. Differentiating the equation (M +εN)vε = λεvε with respect to ε at ε = 0, we have Nv + Mv0 = λ0v + λv0. acting on this by the skew-product with v, we get [Nv, v] + [Mv0, v] = [λ0v, v] + [λv0, v]. (4.5.7) According to our definition of the skew product, [Mv0, v] = [λv0, v], and using H = H = HT , we get 1 1 [Mv0, v] = (JMv0, v) = (JJHv0, v) 2i 2i 1 1 T = − (Hv0, v) = − (v0, H v) 2i 2i 1 1 = − (v0, Hv) = − (Jv0, JHv) 2i 2i = −[v0, Mv] = −[v0, λv] = λ[v0, v], where we used λ = −λ in the last step. Then the 2nd and the 4th terms in (4.5.7) cancel, and that completes the proof. dλ [Nv, v] A direct numerical calculation of both | and Re for the example of dε ε=0 [v, v] dλ section (4.5.2) gives | > 0 for one of the eigenvalues of M + εN and confirms dε ε=0 the instability observed by displaying transformations. Corollary 4.5.2. In the setting of the last theorem, λ0 loses stability for ε > 0 if [Nv, v] Re > 0 [v, v] Bibliography [1] Arnold, V.I. Mathematical Methods of Classical Mechanics, Graduate Texts in Mathematics 60, Springer, New York-Heidelberg, 1978. [2] Katok, A. and Hasselblatt, B. Introduction to the Modern Theory of Dynam- ical Systems, Cambridge University Press, New York, 1995. [3] Moser, J. and Zehnder, E.J. Notes on Dynamical Systems, American Mathe- matical Society, CLN 12, Rhode Island, 2005. [4] Moeckel, R. Linear Stability Analysis of Some Symmetrical Classes of Relative Equilibria, preprint. [5] Krein, S. W. A Basic Attitude Instability of Spacecraft with Imperfect Mo- mentum Wheels, Automatica, Vol. 16, 1980 pp. 415–418. [6] Chaisson, E. and Mcmillan, S. Astronomy Today, volume 1, the solar system, 5th edition, Prentice-Hall, New Jersey, 2005. [7] Correia, A., and Laskar, J. Mercurys Capture into the 3/2 Spin-orbit Reso- nance as a Result of its Chaotic Dynamics, Nature, 429, 24 June 2004. [8] Peal, S.J., Cassen, P. and Reynolds, R. Melting of Io by Tidal Dissipation, Science, v. 203, pp 892-894, 1979. [9] McDonald, R. Tidal Forces and their Effects in the Solar System. [10] Eggletion, P.P. and Kiseleva, L.G. and Hut P. The Equilibrium-tide Model for Tidal Friction, Astrophysical Journal, v.499, p.853, 1998. [11] Shu, Frank H. The Physical Universe: an Introduction to Astronomy, p. 431, University Science Books, 1982. [12] Goldstein, H and Poole, C.P. and Safko, J.L. Classical Mechanics, Addison Wesley, 2002. 41 [13] Meyer, K.R. and Hall, G.R. and Offin, D Introduction to Hamiltonian Dy- namical Systems and the N-Body Problem, Springer, New York, 2009. [14] Ott, E Chaos in Dynamical Systems, Cambridge University Press, New York, 2002. Vita Chong Ai I like challenges. The form of this personality was owing to my mother. ”You can choose not to do it. But as long as you decide to do, try to be the best,” my mother told me when I was 4-year old. As she wished, I did get a five-year long straight rank-one in my elementary school. She also encouraged me to take a lot of contests from writing to speaking, science to art, music to sports, do-it-yourself to inventing, and even a race on brushing teeth and washing face. Thanks to this strategy, I was stimulated to continuously improve myself, to face a failure or a success, to make teamwork, and more important, to seek my own interests, which later enriched and directed my life. My interest in math is a critical one. At middle school I once suspected if ”to be the best” was my life and later then accepted Descartes’ philosophy, ”I think, therefore I am”. Ignoring other things, I believed my life was meaningful by reflection questions in my own interest. One question was on how to trisect an arbitrary angle. Unaware of the unsolvability of it and after a three-month attempt, I could only conclude that we could approxi- mate the trisection points as close enough. However this experience helped me to understand the meanings of limit and Galois theory later in college. The illegitimate exercise of the philosophy might be an good excuse of my lazi- ness. It did hurt myself so that my parents had to pay extra tuitions to send me into my ideal high school. With a chip on my shoulder, I rapidly grew up and learned on methodology of study. My teacher, Yichun Yang, urged me to think about who I was and I wanted to be. I also made a lot of friends, who influenced me and are great treasures of my life. I finally built up my own philosophy about the world and myself and decided to study math in college after graduate. At Shandong University from 1999 to 2003, I took about 50 courses in pure math, applied math, statistics, even economics and management. An succeed in a project of my real analysis course about Riemann integral stired me to do research in math. I enrolled in graduate school there then. During three years study in the theory of uniqueness of meromorphic function, my research interest gradually transferred to dynamical system, which is an important and flourishing field in math itself with applications in differential geometry, number theory, operator al- gebras, celestial and statistical mechanics, biology and etc. From 2006 to 2012, I studied in dynamical systems at Penn State University and finished my dissertation named effect of tidal dissipation on the motion of celestial bodies. Guiding by my interest and logical thinking, I am looking forward to new challenges in new life.