Nonlinear Mechanics - Physics at Oregon State University

Nonlinear Mechanics
A. W. Stetz
January 8, 2012

Contents
1 Lagrangian Dynamics 5
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Generalized Coordinates and the Lagrangian . . . . . . . . . 6
1.3 Virtual Work and Generalized Force . . . . . . . . . . . . . . 8
1.4 Conservative Forces and the Lagrangian . . . . . . . . . . . . 10
1.4.1 The Central Force Problem in a Plane . . . . . . . . . 11
1.5 The Hamiltonian Formulation . . . . . . . . . . . . . . . . . . 13
1.5.1 The Spherical Pendulum . . . . . . . . . . . . . . . . . 15
2 Canonical Transformations 17
2.1 Contact Transformations . . . . . . . . . . . . . . . . . . . . . 17
2.1.1 The Harmonic Oscillator: Cracking a Peanut with a
Sledgehammer . . . . . . . . . . . . . . . . . . . . . . 20
2.2 The Second Generating Function . . . . . . . . . . . . . . . . 21
2.3 Hamilton’s Principle Function . . . . . . . . . . . . . . . . . . 22
2.3.1 The Harmonic Oscillator: Again . . . . . . . . . . . . 24
2.4 Hamilton’s Characteristic Function . . . . . . . . . . . . . . . 25
2.4.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.5 Action-Angle Variables . . . . . . . . . . . . . . . . . . . . . . 27
2.5.1 The harmonic oscillator (for the last time) . . . . . . . 29
3 Abstract Transformation Theory 33
3.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.1.1 Poisson Brackets . . . . . . . . . . . . . . . . . . . . . 35
3.2 Geometry in n Dimensions: The Hairy Ball . . . . . . . . . . 38
3.2.1 Example: Uncoupled Oscillators . . . . . . . . . . . . 41
3.2.2 Example: A Particle in a Box . . . . . . . . . . . . . . 43
3

4 CONTENTS
4 Canonical Perturbation Theory 45
4.1 One-Dimensional Systems . . . . . . . . . . . . . . . . . . . . 45
4.1.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.1.2 The simple pendulum . . . . . . . . . . . . . . . . . . 49
4.2 Many Degrees of Freedom . . . . . . . . . . . . . . . . . . . . 51
5 Introduction to Chaos 55
5.1 The total failure of perturbation theory . . . . . . . . . . . . 56
5.2 Fixed points and linearization . . . . . . . . . . . . . . . . . . 58
5.3 The Henon oscillator . . . . . . . . . . . . . . . . . . . . . . . 62
5.4 Discrete Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
5.5 Linearized Maps . . . . . . . . . . . . . . . . . . . . . . . . . 70
5.6 Lyapunov Exponents . . . . . . . . . . . . . . . . . . . . . . . 72
5.7 The Poincaré-Birkhoff Theorem . . . . . . . . . . . . . . . . . 74
5.8 All in a tangle . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
5.9 The KAM theorem and its consequences . . . . . . . . . . . . 80
5.9.1 Two Conditions . . . . . . . . . . . . . . . . . . . . . . 81
5.10 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

Chapter 1
Lagrangian Dynamics
1.1 Introduction
The possibility that deterministic mechanical systems could exhibit the behavior
we now call chaos was first realized by the French mathematician
Henri Poincaré sometime toward the end of the nineteenth century. His
discovery emerged from analytic or classical mechanics, which is still part
of the foundation of physics. To put it a bit facetiously, classical mechanics
deals with those problems that can be “solved,” in the sense that it is possible
to derive equations of motions that describe the positions of the various
parts of a system as functions of time using standard analytic functions.
Nonlinear dynamics treats problems that cannot be so solved, and it is only
in these problems that chaos can appear. The simple pendulum makes a
good example. The differential equation of motion is
¨θ + ω 2 sin θ = 0 (1.1)
The sin is a nonlinear function of θ. If we linearize by setting sin θ ≈ θ,
the solutions are elementary functions, sin ωt and cos ωt. If we keep the sin,
the solutions can only be expressed in terms of elliptic integrals. This is
not a chaotic system, because there is only one degree of freedom, but if we
hang one pendulum from the end of another, the equations of motion are
hopeless to find (even with elliptic integrals) and the resulting motion can
be chaotic. 1
1 I should emphasize the distinction between the differential equations of motion, which
are usually simple (though nonlinear), and the equations that describe the positions of
the elements of the system as functions of time, which are usually non-existent.
5

6 CHAPTER 1. LAGRANGIAN DYNAMICS
In order to arrive at Poincaré’s moment of discovery, we will have to
review the development of classical mechanics through the nineteenth century.
This material is found in many standard texts, but I will cover it here
in some detail. This is partly to insure uniform notation throughout these
lectures and partly to focus on those things that lead directly to chaos in
nonlinear systems. We will begin formulating mechanics in terms of generalized
coordinates and Lagrange’s equations of motion. We then study
Lagrange transformations and use them to derive Hamilton’s equations of
motions. These equations are particularly suited to conservative systems
in which the Hamiltonian is constant in time, and it is such systems that
will be our primary concern. It turns out that Lagrange transformations
can be used to transform Hamiltonians in a myriad of ways. One particularly
elegant form uses action-angle variables to transform a certain class of
problems into a set of uncoupled harmonic oscillators. Systems that can be
so transformed are said to be integrable, which is to say that they can be
“solved,” at least in principle. What happens, Poincaré asked, to a system
that is almost but not quite integrable? The answer entails perturbation
theory and leads to the disastrous problem of small divisors. This is the
path that led originally to the discovery of chaos, and it is the one we will
pursue here.
1.2 Generalized Coordinates and the Lagrangian
Vector equations, like F = ma, seem to imply a coordinate system. Beginning
students learn to use cartesian coordinates and then learn that this
is not always the best choice. If the system has cylindrical symmetry, for
example, it is best to use cylindrical coordinates: it makes the problem easier.
By “symmetry” we mean that the number of degrees of freedom of the
system is less that the dimensionality of the space in which it is imbedded.
The familiar example of the block sliding down the incline plane will make
this clear. Let’s say that it’s a two dimensional problem with an x-y coordinate
system. The block is constrained to move in a straight line, however,
so that its position can be completely specified by one variable, i.e. it has
one degree of freedom. The clever student chooses the x axis so that it lies
along the path of the block. This reduces the problem to one dimension,
since y = 0 and the x coordinate is given by one simple equation. In the
pendulum example from the previous section, it was most convenient to use
a polar coordinate system centered at the pivot. Since r is constant, the
motion can be described completely in terms of θ.

1.2. GENERALIZED COORDINATES AND THE LAGRANGIAN 7
These coordinate systems conceal a subtle point: the pendulum moves
in a circular arc and the block moves in a straight line because they are
acted on by forces of constraint. In most cases we are not interested in these
forces. Our choice of coordinates simply makes them disappear from the
problem. Most problems don’t have obvious symmetries, however. Consider
a bead sliding along a wire following some complicated snaky path in 3-d
space. There’s only one degree of freedom, since the particle’s position
is determined entirely by its distance measured along the wire from some
reference point. The forces are so complicated, however, that it is out of
the question to solve the problem by using F = ma in any straightforward
way. This is the problem that Lagrangian mechanics is designed to handle.
The basic (and quite profound) idea is that even though there may be no
coordinate system (in the usual sense) that will reduce the dimensionality of
the problem, yet there is usually a system of coordinates that will do this.
Such coordinates are called generalized coordinates.
To be more specific, suppose that a system consists of N point masses
with positions specified by ordinary three-dimensional cartesian vectors, ri,
i = 1 · · · N, subject to some constraints. The easiest constraints to deal with
are those that can be expressed as a set of l equations of the form
fj(r1, r2, . . . , t) = 0, (1.2)
where j = 1 · · · l. Such constraints are said to be holonomic. If in addition,
the equations of constraint do not involve time explicitly, they are said to be
scleronomous, otherwise they are called rheonomous. These constraints can
be used to reduce the 3N cartesian components to a set of 3N − l variables
q1, q2, . . . , q3N−l. The relationship between the two is given by a set of N
equations of the form
ri = ri(q1, q2, . . . , q3N−l, t). (1.3)
The q’s used in this way are the generalized coordinates. In the example of
the bead on a curved wire, the equations would reduce to r = r(q), where
q is a distance measured along the wire. This simply specifies the curvature
of the wire.
It should be noted that the q’s need not all have the same units. Also
note that we can use the same notation even if there are no constraints.
For example, the position of an unconstrained particle could be written r =
r(q1, q2, q3), and the q’s might represent cartesian, spherical, or cylindrical
coordinates. In order to simplify the notation, we will often pack the q’s

8 CHAPTER 1. LAGRANGIAN DYNAMICS
into an array and use vector notation,
q =
q1
q2
q3
.
(1.4)
This is not meant to imply that q is a vector in the usual sense. For one
thing, it does not necessarily posses “a magnitude and a direction” as good
vectors are supposed to have. By the same token, we cannot use the notion
of orthogonal unit vectors.
Along with the notion of generalized coordinates comes that of generalized
velocities.
˙qk ≡ dqk
(1.5)
dt
Since qi depends only on t, this is a total derivative, but when we differentiate
ri, we must remember that it depends both explicitly on time as well as
implicitly through the q’s.
˙ri = ∑ ∂ri
∂qk
k
˙qk + ∂ri
∂t
(1.6)
(In this chapter I will consistently use the index i to sum over the N point
masses and k to sum over the 3N − l degrees of freedom.) Differentiating
both sides with respect to ˙qk yields
∂ ˙ri
∂ ˙qk
= ∂ri
∂qk
which will be useful in the following derivations.
1.3 Virtual Work and Generalized Force
(1.7)
There are several routes for deriving Lagrange’s equations of motion. The
most elegant and general makes use of the principle of least action and the
calculus of variation. I will use a much more pedestrian approach based on
Newton’s second law of motion. First note that F = ma can be written in
the rather arcane form ( )
d ∂T
= Fi
(1.8)
dt ∂vi
Where Fi is i-th component of the total force acting on a particle with
kinetic energy T . The point of writing this in terms of energy rather than

1.3. VIRTUAL WORK AND GENERALIZED FORCE 9
acceleration is that we can separate out the forces of constraint, which are
always perpendicular to the direction of motion and hence do no work. The
trick is to write this in terms of generalized coordinates and velocities. This
is rather technical, but the underlying idea is simple, and the result looks
much like (1.8).
The qk’s are all independent, so we can vary one by a small amount δqk
while holding all others constant.
δri = ∑ ∂ri
δqk
∂qk
k
(1.9)
This is sometimes called a virtual displacement. The corresponding virtual
work is
δWk = ∑
(
∂ri
Fi ·
∂qk
)
δqk
(1.10)
We define a generalized force
i
ℑk = ∑
i
Fi · ∂ri
∂qk
(1.11)
The forces of constraint can be excluded from the sum for the reason explained
above. We are left with
ℑk = δWk
δqk
The kinetic energy is calculated using ordinary velocities.
T = 1
2
i
∑
mi ˙ri · ˙ri
i
∂T
=
∂qk
∑ ∂ ˙ri
mi ˙ri · =
∂qk
∑ ∂ ˙ri
pi ·
∂qk
∂T
∂ ˙qk
= ∑ ∂ ˙ri
mi ˙ri ·
∂ ˙qk
i
i
= ∑
i
pi · ∂ri
∂qk
(1.12)
(1.13)
(1.14)
(1.15)
Equation (1.7) was used to obtain the last term. A straightforward calculation
now leads to
??ℑk = d
( )
∂T
−
dt ∂ ˙qk
∂T
∂qk
(1.16)
which is the generalized form of (1.8).

10 CHAPTER 1. LAGRANGIAN DYNAMICS
1.4 Conservative Forces and the Lagrangian
So far we have made no assumptions about the nature of the forces included
in ℑ except that they are not forces of constraint. Equation (16) is therefore
quite general, although seldom used in this form. In these notes we are
primarily concerned with conservative forces, i.e. forces that can be derived
from a potential.
Fi = −∇iV (r1 · · · rN) (1.17)
Notice that V doesn’t depend on velocity. (Electromagnetic forces are velocity
dependent, of course, but they can easily be accommodated into the
Lagrangian framework. I will return to this issue later on.) Now calculate
the work done by changing some of the q’s.
∫
∑
W = Fi · dri = − ∑
∫
∇iV · dri
i
= − ∑
∫
∇iV ·
i
∑ ∂ri
dqk
∂qk k
= − ∑
∫
k
( ∑
∇iV ·
i
∂ri
)
dqk
∂qk
= − ∑
∫
∂V
dqk
∂qk
i
k
(1.18)
The integral is a multidimensional definite integral over the various q’s that
have changed. Summing over (1.12) then gives
Comparison with (1.18) yields
Finally define the Lagrangian
δW = ∑
δWk = ∑
ℑkδqk
k
W = ∑
∫
k
ℑk = − ∂V
∂qk
k
ℑkdqk
(1.19)
(1.20)
(1.21)
L = T − V (1.22)

1.4. CONSERVATIVE FORCES AND THE LAGRANGIAN 11
Equation (??) becomes
d
dt
( ∂L
∂ ˙qk
)
− ∂L
∂qk
= 0. (1.23)
Equation (1.23) represents a set of 3N −l second order partial differential
equations called Lagrange’s equations of motion. I can summarize this long
development by giving you a “cookbook” procedure for using (1.23) to solve
mechanics problems: First select a convenient set of generalized coordinates.
Then calculate T and V in the usual way using the ri’s. Use equation (1.3)
to eliminate the ri’s in favor of the qk’s. Finally substitute L into (1.23) and
solve the resulting equations.
Classical mechanics texts are full of examples in which this program is
carried to a successful conclusion. In fact, most of these problems are contrived
and of little interest except to illustrate the method. The vast majority
of systems lead to differential equations that cannot be solved in closed
form. The modern emphasis is to understand the solutions qualitatively
and then obtain numerical solutions using the computer. The Hamiltonian
formalism described in the next section is better suited to both these ends.
1.4.1 The Central Force Problem in a Plane
Consider the central force problem as an example of this technique.
V = V (r) F = −∇V (1.24)
L = T − V = 1
2 m
(
˙r 2 + r 2 ϕ˙ 2 )
− V (r) (1.25)
Let’s choose our generalized coordinates to be q1 = r and q2 = ϕ. Equation
(1.23) becomes
m¨r − mr ˙ ϕ 2 + dV
= 0
dr
d
(
mr
dt
(1.26)
2 )
ϕ˙
= 0 (1.27)
This last equation tells us that there is a quantity mr 2 ˙ ϕ that does not change
with time. Such a quantity is said to be conserved. In this case we have
rediscovered the conservation of angular momentum.
This reduces the problem to one dimension.
mr 2 ˙ ϕ ≡ lz = constant (1.28)
m¨r = l2 z dV
−
mr3 dr
(1.29)

12 CHAPTER 1. LAGRANGIAN DYNAMICS
Since there are no constraints, the generalized forces are identical with the
ordinary forces
ℑϕ = − dV
dϕ = 0 ℑr = − dV
(1.30)
dr
This equation has an elegant closed form solution in the special case of
gravitational attraction.
V = − GmM
r
≡ −k
r
(1.31)
m¨r = l2 z k
− (1.32)
mr3 r
This apparently nonlinear equation yields to a simple trick, let u = 1/r.
d2u dϕ2 + u = m2k l2 z
(1.33)
If the motion is circular u is constant. Otherwise it oscillates around the
value m2k/l2 z with simple harmonic motion. 2 The period of oscillation is
identical with the period of rotation so the corresponding orbit is an ellipse.
This problem was easy to solve because we were able to discover a nontrivial
quantity that was constant, in this case the angular momentum. The
constant enabled us to reduce the number of independent variables from
two to one. Such a conserved quantity is called an integral of the motion
or a constant of the motion. Obviously, the more such quantities one can
find, the easier the problem. This raises two practical problems. First, how
can we tell, perhaps from looking at the physics of a problem, how many
independent conserved quantities there are? Second, how are we to find
them?
In the central force problem, both of these questions answered themselves.
We know that angular momentum is conserved. This fact manifests
itself in the Lagrangian in that L depends on ˙ ϕ but not on ϕ. Such a coordinate
is said to be cyclic or ignorable. Let q be such a coordinate. Then
( )
d ∂L
= 0 (1.34)
dt ∂ ˙q
The quantity in brackets has a special significance. It is called the canonically
conjugate momentum. 3
∂L
≡ pk
(1.35)
˙qk
2
This illustrates a general principle in physics: When correctly viewed, everything is a
harmonic oscillator.
3
This notation is universally used, hence the old aphorism that mechanics is a matter
of minding your p’s and q’s.

1.5. THE HAMILTONIAN FORMULATION 13
To summarize, if q is cyclic, p is conserved.
Suppose we had tried to do the central force problem in cartesian coordinates.
Both x and y would appear in the Lagrangian, and neither px nor
py would be constant. If we insisted on this, central forces would remain an
intractable problem in two dimensions. We need to choose our generalized
coordinates so that there are as many cyclic variables as possible. The two
questions reemerge: how many are we entitled to and how do we find the
corresponding p’s and q’s?
A partial answer to the first is given by a well-known result called
Noether’s theorem: For every transformation that leaves the Lagrangian
invariant there is a constant of the motion. 4 This theorem (which underlies
all of modern particle physics) says that there is a fundamental connection
between symmetries and invariance principles on one hand and conservation
laws on the other. Momentum is conserved because the laws of physics
are invariant under translation. Angular momentum is conserved because
the laws of physics are invariant under rotation. Despite its fundamental
significance, Noether’s theorem is not much help in practical calculations.
Granted it gives a procedure for finding the conserved quantity after the
corresponding symmetry transformation has been found, but how is one to
find the transformation? The physicist must rely on his traditional tools:
inspiration, the Ouija Board, and simply pounding one’s head against a
wall. The fact remains that there are simple systems, e.g. the Henon-Heiles
problem to be discussed later, that have fascinated physicists for decades
and for which the existence of these transformations is still controversial.
I will have much more to say about the second question. As you will
see, there is a more or less “cookbook” procedure for finding the right set
of variables and some fundamental results about the sorts of problems for
which these procedures are possible.
1.5 The Hamiltonian Formulation
I will explain the Hamiltonian assuming that there is only one degree of
freedom. It’s easy to generalize once the basic ideas are clear. Lagrangians
are functions of q and ˙q. We define a new function of q and p (given by
(1.34)).
H(p, q) = p ˙q − L(q, ˙q) (1.36)
The new function is called the Hamiltonian, and the transformation L → H
is called a Lagrange transformation. The equation is much more subtle than
4 See Finch and Hand for a simple proof and further discussion.

14 CHAPTER 1. LAGRANGIAN DYNAMICS
it looks. In fact, its worth several pages of explanation.
It’s clear from elementary mechanics that q, ˙q, and p can’t all be independent
variables, since p = m ˙q. You might say that there are two ways of
formulating Newton’s second law: a (q, ˙q) formulation, F = m¨q, and a (q, p)
formulation, F = ˙p. The connection between q and its canonically conjugate
momentum is usually more complicated than this, but there is still a (q, ˙q)
formulation, the Lagrangian, and a (q, p) formulation, the Hamiltonian. The
Legendre transformation is a procedure for transforming the one formulation
into the other. The key point is that it is invertible. 5 To see what this
means, let’s first assume that q, ˙q and p are all independent.
dH =
What is the condition that H not depend on ˙q?
H(q, ˙q, p) = p ˙q − L(q, ˙q) (1.37)
(
p − ∂L
)
d ˙q + ˙q dp −
∂ ˙q
∂L
dq (1.38)
∂q
p(q, ˙q) =
∂L(q, ˙q)
∂ ˙q
OK. This is the definition of p anyhow, so we’re on the right track.
dH = ˙q dp − ∂L
∂q dq
dH = ∂H
∂p
dp +∂H
∂q dq
Adding and subtracting these two equations gives
˙q(q, p) = ∂H
∂p
− ∂L
∂q
= ∂H
∂q
Combining (1.23), (1.39), and (1.4141) gives the fourth major result.
˙p(q, p) = − ∂H
∂q
5 The following argument is taken from Finch & Hand.
(1.39)
(1.40)
(1.41)
(1.42)

1.5. THE HAMILTONIAN FORMULATION 15
Now here’s what I mean that Legendre transformations are invertible:
First follow the steps from L → H. We start with L = L(q, ˙q). Equation
(1.39) gives p = p(q, ˙q). Invert this to find ˙q = ˙q(q, p). The Hamiltonian is
now
H(q, p) = ˙q(q, p)p − L[q, ˙q(q, p)]. (1.43)
Now suppose that we start from H = H(p, q). Use (1.40) to find ˙q = ˙q(q, p).
Invert to find p = p(q, ˙q). Finally
L(q, ˙q) = ˙qp(q, ˙q) − H[q, p(q, ˙q)] (1.44)
In both cases we were able to complete the transformation without knowing
ahead of time the functional relationship among q, ˙q, and p. To summarize:
Equations (1.37), (1.39), and (1.41) enable us to transform between the (q, ˙q)
(Lagrangian) prescription and the (q, p) (Hamiltonian) prescription; while
(1.40) and (1.41) are Hamilton’s equations of motion.
1.5.1 The Spherical Pendulum
A mass m hangs from a string of length R. The string makes an angle θ
with the vertical and can rotate about the vertical with an angle ϕ.
T = 1
2 mR2 ( ˙ θ 2 + sin 2 θ ˙ ϕ 2 ) (1.45)
V = mgR(1 − cos θ) (1.46)
The mgR constant doesn’t appear in the equations of motion, so we can
forget about it. The Lagrangian is L = T − V as usual.
pθ = ∂L
∂ ˙ θ = mR2 ˙ θ (1.47)
pϕ = ∂L
∂ ˙ ϕ = mR2 sin 2 θ ˙ ϕ ≡ lϕ
(1.48)
The angle ϕ is cyclic, so pϕ = lϕ is constant. At this point we are still in the
(q, ˙q) prescription. Invert (47) and (48) to obtain ˙ θ and ˙ ϕ as functions of pθ
and lϕ.
˙θ = pθ/mR 2
(1.49)
H = p2 θ +
2mR2 ˙ϕ = lϕ/mR 2 sin 2 θ (1.50)
l2 ϕ
2mR2 sin2 − mgR cos θ (1.51)
θ

16 CHAPTER 1. LAGRANGIAN DYNAMICS
The equations of motion follow from this.
˙θ = ∂H
∂pθ
= pθ
mR 2
˙pθ = − ∂H
∂θ = l2 ϕ cos θ
˙ϕ = ∂H
∂pϕ
mR 2 sin 3 θ
=
lϕ
mR 2 sin 2 θ
(1.52)
− mgR sin θ (1.53)
(1.54)
˙pϕ = 0 (1.55)
Suppose we were to try to find an analytic solution to this system of
equations. First note that there are two constants of motion, the angular
momentum lϕ, and the total energy E = H.
1. Invert (1.49) to obtain pθ = pθ(θ, E, lϕ).
2. Substitute pθ into (1.49) and integrate
∫
mR2 dθ = t ≡ N(θ)
pθ
The integral is hopeless anyhow, so we label its output N(θ), (short
for an exceedingly nasty function).
3. Invert the nasty function to find θ as a function of t.
4. Take the sine of this even nastier function and substitute it into (1.54)
to find ˙ ϕ.
5. Integrate and invert to find ϕ as a function of t.
This makes sense in principle, but is wildly impossible in practice. Now
suppose we could change the problem so that both θ and ϕ were cyclic so
that the two constants of motion were pθ and pϕ (rather than E and pϕ).
Then
˙θ = ∂H
= ωθ
∂pθ
θ = ωθt + θ0
˙ϕ = ∂H
∂pϕ
= ωϕ ϕ = ωϕt + ϕ0
Here ωθ and ωϕ are two constant “frequencies” that we could easily extract
from the Hamiltonian. This apparently small change makes the problem
trivial! In both cases there are two constants of motion: it makes all the
difference which two constants. This is the basis of the idea we will be
pursuing in the next chapter.

Chapter 2
Canonical Transformations
We saw at the end af the last chapter that a problem in which all the
generalized coordinates are cyclic is trivial to solve. We also saw that there
is a great flixibility allowed in the choice of coordinates for any particular
problem. It turns out that there is an important class of problems for which
it is possible to choose the coordinates so that they are in fact all cyclic.
The choice is usually far from obvious, but there is a formal procedure for
finding the “magic” variables. One formulates the problem in terms of the
natural p’s and q’s and then transforms to a new set of variables, usually
called Qk and Pk, that have the right properties.
2.1 Contact Transformations
The most general transformation is called a contact transformation.
Qk = Qk(q, p, t) Pk = Pk(q, p, t) (2.1)
(In this formula and what follows, the symbols p and q when used as arguments
stand for the complete set, q1, q2, q3, · · · , etc.) There is a certain privileged
class of transformations called canonical transformations that preserve
the structure of Hamilton’s equation of motion for all dynamical systems.
This means that there is a new Hamiltonian function called K(Q, P ) for
which the new equations of motion are
˙Qk = ∂K
∂Pk
Pk
˙ = − ∂K
∂Qk
(2.2)
In a footnote in Classical Mechanics, Goldstein suggested that K be called
the Kamiltonian. The idea has caught on with several authors, and I will
use it without further apology. The trick is to find it.
17

18 CHAPTER 2. CANONICAL TRANSFORMATIONS
Theorem: Let F be any function of qk and Qk and possibly pk and Pk,
as well as time. Then the new Lagrangian defined by
¯L = L − dF
dt
(2.3)
is equivalent to L in the sense that it yields the same equations of motion.
Proof:
F ˙ = ∑ ∂F
˙qk +
∂qk k
∑ ∂F
˙Qk +
∂Qk k
∂F
∂t
(
d ∂
dt
(2.4)
˙
)
F
=
∂ ˙qk
d
( )
∂F
=
dt ∂qk
∂ ˙ F
∂qk
(
d ∂
dt
˙ F
∂ ˙ )
=
Qk
d
( )
∂F
=
dt ∂Qk
∂ ˙ F
∂Qk
These last two can be rewritten
d
dt
(
∂ ˙
F
∂ ˙qk
(
d ∂
dt
˙ F
∂ ˙ )
−
Qk
)
∂F˙ − = 0
∂qk
∂ ˙
F
∂Qk
So ˙ F satisfies Lagrange’s equation whether we regard it as a function of qk
or Qk. Obviously, if L satisfies Lagrange’s equation, then so does L − ˙ F .
(The conclusion is unchanged if F contains pk and/or Pk.) The function F
is called the generating function of the transformation.
K is obtained by a Legendre transformation just as H was.
= 0
K(Q, P ) = ∑
Pk ˙ Qk − ¯ L(Q, ˙ Q, t) (2.5)
k
This has the same form as (1.36), so the derivation of the equations of motion
(1.39) through(1.42) are unchanged as well.
Pk = ∂K
∂ ˙ Qk
˙Qk = ∂K
∂Pk
Pk
˙ = − ∂K
∂Qk
(2.6)
These simple results provide the framework for canonical transformations.
In order to use them we will need to know two more things: (1) How

2.1. CONTACT TRANSFORMATIONS 19
to find F , and given F , (2) how to find the transformation (q, p) → (Q, P ).
We deal with (2) now and postpone (1) to later sections.
Consider the variables q, Q, p, and P . Any two of these constitute a
complete set, so there are four kinds of generating functions usually called
F1(q, Q, t), F2(q, P, t), F3(p, Q, t), and F4(p, P, t). All four are discussed in
Goldstein. F1 provides a good introduction. Most of our work will make
use of F2.
Starting with F1(q, Q) (2.3) becomes
Since
we get with the help of (4)
¯L(Q, ˙ Q, t) = L(q, ˙q, t) − d
dt F1(q, Q, t) (2.7)
∂ ¯ L
∂ ˙qk
∂ ¯ L
∂ ˙qk
= ∂L
−
∂ ˙qk
= ∂L
∂ ˙ Qk
∂ ˙
F1
∂ ˙qk
= 0,
= ˙pk − ∂F1
∂qk
∂ ¯ L
∂ ˙ Qk
= Pk = ∂L
∂ ˙ Qk
∂F1 ˙
−
∂ ˙ Qk
This yields the two transformation equations
Pk = − ∂F1
∂Qk
pk = ∂F1
∂qk
= 0.
= − ∂F1
∂Qk
(2.8)
(2.9)
A straightforward set of substitutions gives our final formula for the Kamiltonian.
K = ∑
[
− ∂F
˙Qk − L +
∂Qk
∂F
˙qk +
∂qk
∂F
]
˙Qk +
Qk
∂F
∂t
To be more explicit
k
= −L + ∑
k
pk ˙qk + ∂F
∂t
K(Q, P ) = H(q(Q, P ), p(Q, P ), t) + ∂
∂t F1(q(Q, P ), Q, t) (2.10)
Summary:

20 CHAPTER 2. CANONICAL TRANSFORMATIONS
1. Here is the typical problem: We are given the Hamiltonian H =
H(q, p) for some conservative system. H = E is constant, but the
q’s and p’s change with time in a complicated way. Our goal is to find
the functions q = q(t) and p = p(t) using the technique of canonical
transformations.
2. We need to know the generating function F = F1(q, Q). This is the
hard part, and I’m postponing it as long as possible.
3. Substitute F into (2.8) and (2.9). This gives a set of coupled algebraic
equations for q, Q, p, and P . They must be combined in such a way
as to give qk = qk(Q, P ) and pk = pk(Q, P ).
4. Use (2.10) to find K. If we had the right generating function to start
with, Q will be cyclic, i.e. K = K(P ). The equations of motion are
obtained from (2.6). Pk
˙ = 0 and ˙ Qk = ωk. The ω’s are a set of
constants as are the P ’s. Qk(t) = ωkt + αk. The α’s are constants
obtained from the initial conditions.
5. Finally qk(t) = qk(Q(t), P ) and pk(t) = pk(Q(t), P ).
2.1.1 The Harmonic Oscillator: Cracking a Peanut with a
Sledgehammer
H = p2 kq2
+
2m 2
It’s useful to try a new technique on an old problem. As it turns out, the
generating function is
F = mωq2
cot Q (2.12)
2
The transformation is found from (2.8) and (2.9).
p = ∂F
∂q
= 1
2m (p2 + m 2 ω 2 q 2 ) (2.11)
= mωq cot Q
P = − ∂F mωq2
=
∂Q 2 sin2 Q
Solve for p and q in terms of P and Q and then substitute into (2.10) to
find K.
√
2P
q =
mω sin Q p = √ 2P mω cos Q

2.2. THE SECOND GENERATING FUNCTION 21
K = ωP P = E/ω
We have achieved our goal. Q is cyclic, and the equations of motion are
trivial.
˙Q = ∂K
q =
∂P = ω Q = ωt + Q0 (2.13)
√ 2E
mω 2 sin(ωt + Q0) p = √ 2mE cos(ωt + Q0) (2.14)
2.2 The Second Generating Function
There’s an old recipe for tiger stew that begins, “First catch the tiger.” In
our quest for the tiger, we now turn our attention to the second generating
function, F2 = F2(q, P, t). F2 is obtained from F1 by means of a Legendre
transformation. 1
F2(q, P ) = F1(q, Q) + ∑
(2.15)
k
QkPk
We are looking for transformation equations analogous to (refe2.8) and (2.9).
Since L = ¯ L + ˙ F1,
∑
pk ˙qk − H = ∑
Pk ˙ Qk − K + d
dt (F2 − ∑ QkPk)
k
k
k
= − ∑ Qk ˙
Pk − K + ˙
F2
Substitute
F2
˙ = ∑
[
∂F2
˙qk +
∂qk k
∂F2
]
Pk
˙ +
∂Pk
∂F2
∂t
−H = −K + ∑
[( ) ( ) ]
∂F2
∂F2
− pk ˙qk + − Qk Pk
˙ +
∂qk
∂Pk
∂F2
∂t
We are working on the assumption that ˙q and ˙
P are not independent variables.
We enforce this by requiring that
∂F2
∂qk
∂F2
∂Pk
= pk
= Qk
(2.16)
(2.17)
K(q, P ) = H(q(Q, P ), P ) + ∂
∂t F2(q(Q, P ), P ) (2.18)
1 When in doubt, do a Legendre transformation.

22 CHAPTER 2. CANONICAL TRANSFORMATIONS
2.3 Hamilton’s Principle Function
The F1 style generating functions were used to transform to a new set of
variables (q, p) → (Q, P ) such that all the Q’s were cyclic. As a consequence,
the P ’s were constants of the motion, and the Q’s were linear functions
of time. The generating function itself was hard to find, however. The
F2 generating function goes one step further; it can transform to a set of
variables in which both the Q’s and P ’s are constant and simple functions of
the initial values of the phase space variables. In essence, our transformation
is
(q(t), p(t)) ↔ (q0, p0)
This is a time dependent transformation, of course. The fact that we can
find such transformations shows that the time evolution of a system is itself
a canonical transformation.
We look for an F2 so that K in (2.18) is identically zero! Then from
(2.6), ˙ Qk = 0 and ˙
Pk = 0. The appropriate generating function will be a
solution to
H(q, p, t) + ∂F2
∂t
We eliminate pk using (2.16)
(
H q1, . . . , qn; ∂F2
, . . . ,
∂q1
∂F2
)
; t +
∂qn
∂F2
∂t
(2.19)
= 0. (2.20)
The solution to this equation is usually called S, Hamilton’s principle function.
The equation itself is the Hamilton-Jacobi equation. 2
There are two serious issues here: does it have a solution, and if it does,
can we find it? We will take a less serious approach: if we can find a solution,
then it most surely exists. Furthermore, if we can find it, it will have the
form
S = ∑
Wk(qk) − αt (2.21)
k
Partial differential equations that have solutions of the form (2.21) are said
to be separable. 3 Most of the familiar textbook problems in classical mechanics
and atomic physics can be separated in this form. The question
of separability does depend on the system of generalized coordinates used.
For example, the Kepler problem is separable in spherical coordinates but
not in cartesian coordinates. It would be nice to know whether a particular
2 See Goldstein, Classical Mechanics, Chapter 10
3 Or to be meticulous, completely separable

2.3. HAMILTON’S PRINCIPLE FUNCTION 23
Hamiltonian could be separated with some system of coordinates, but no
completely general criterion is known. 4 As a rule of thumb, Hamiltonians
with explicit time dependence are not separable.
If our Hamiltonian is separable, then when (2.21) is substituted into
(2.20), the result will look like
f1
(
q1, dW1
dq1
)
+ f2
(
q2, dW2
)
+ · · · = α (2.22)
dq2
Each function fk is a function only of qk and dWk/dqk. Since all the q’s
are independent, each function must be separately constant. This gives us a
system of n independent, first-order, ordinary differential equations for the
Wk’s.
(
fk qk, dWk
)
= αk. (2.23)
dqk
The W ’s so obtained are then substituted into (2.21). The resulting function
for S is
F2 ≡ S = S(q1, . . . , qn; α1, . . . , αn; α, t)
The final constant α is redundant for two reasons: first, ∑ αk = α, and
second, the transformations equations (2.16) and (2.17) involve derivatives
with respect to qk and Pk. When S is so differentiated, the −αt piece will
disappear. In order to make this apparent, we will write S as follows:
F2 ≡ S = S(q1, . . . , qn; α1, . . . , αn; t) (2.24)
Since the F2 generating functions have the form F2(q, P ), we are entitled to
think of the α’s as “momenta,” i.e. αk in (??) corresponds to Pk in (2.17).
In a way this makes sense. Our goal was to transform the time-dependent
q’s and p’s into a new set of constant Q’s and P ’s, and the α’s are most
certainly constant. On the other hand, they are not the initial momenta p0
that evolve into p(t). The relationship between α and p0 will be determined
later.
If we have dome our job correctly, the Q’s given by (2.17) are also constant.
They are traditionally called β, so
Qk = βk =
∂S(q, α, t)
∂αk
Again, β’s are constant, but they are not equal to q0.
We can turn this into a cookbook algorithm.
(2.25)
4 The is a very technical result, the so-called Staeckel conditions, which gives necessary
and sufficient conditions for separability in orthogonal coordinate systems.

24 CHAPTER 2. CANONICAL TRANSFORMATIONS
1. Substitute (2.21) into (2.20) and separate variables.
2. Integrate the resulting fist-order ODE’s. The result will be n independent
functions Wk = Wk(q, α). Put the Wk’s back into (2.21) to
construct S = S(q, α, t).
3. Find the constant β coordinates using
βk = ∂S
∂αk
4. Invert these equations to find qk = qk(β, α, t)
5. Find the momenta with
pk = ∂S
∂qk
2.3.1 The Harmonic Oscillator: Again
The harmonic oscillator provides an easy example of this procedure.
H = 1
[ (∂S
1
2m ∂q
[ (∂W
1
2m ∂q
2m (p2 + m 2 ω 2 q 2 )
) 2
+ m 2 ω 2 q 2
]
) 2
+ m 2 ω 2 q 2
+ ∂S
= 0
∂t
]
= α
(2.26)
(2.27)
Since there is only one q, the entire quantity on the left of the equal sign is
a constant.
W (q, α) = √ ∫
2mα
√
dq 1 − mω2q2 2α
The new transformed constant “momentum” P = α.
∂S(q, α, t)
β = =
∂α
∂W (q, α)
− t
∂α
= 1
ω sin−1
[ √ ]
mω2 q − t (2.28)
2α
Invert this equation to find q as a function of t and β.
√
2α
q = sin(ωt + βω)
mω2

2.4. HAMILTON’S CHARACTERISTIC FUNCTION 25
Evidentally, β has something to do with initial conditions: ωβ = ϕ0, the
initial phase angle.
p = ∂S
∂q = √ 2mα − m 2 ω 2 q 2
= √ 2mα cos(ωt + ϕ0)
The maximum value of p is √ 2mE, so that makes sense too.
2.4 Hamilton’s Characteristic Function
There is another way to use the F2 generating function to turn a difficult
problem into an easy one. In the previous section we chose F2 = S = W −αt,
so that K = 0. It is also possible to to take F2 = W (q) so that
(
K = H qk, ∂W
)
= E = α1
(2.29)
∂qk
The W obtained in this way is called Hamilton’s characteristic function.
W = ∑
Wk(qk, α1, . . . , αn)
k
= W (q1, . . . , qn, E, α2, . . . , αn) = W (q1, . . . , qn, α1, . . . , αn) (2.30)
It generates a contact transformation with properties quite different from
that generated by S. The equations of motion are
Pk
˙ = − ∂K
∂Qk
˙Qk = ∂K
∂Pk
= ∂K
∂αk
= 0 (2.31)
= δk1
The new feature is that ˙ Q1 = 1 so Q1 = t − t0. In general
but now β1 = t − t0.
Qk = ∂W
∂αk
pk = ∂W
∂qk
as before.
The algorithm now works like this:
= βk
(2.32)
(2.33)
(2.34)

26 CHAPTER 2. CANONICAL TRANSFORMATIONS
1. Substitute (2.30) into (2.29) and separate variables.
2. Integrate the resulting fist-order ODE’s. The result will be n independent
functions Wk = Wk(q, α). Put the Wk’s back into (2.30) to
construct S = S(q, α, t).
3. Find the constant β coordinates using
Remember that β1 = t − t0.
βk = ∂S
∂αk
4. Invert these equations to find qk = qk(β, α, t)
5. Find the momenta with
2.4.1 Examples
pk = ∂S
∂qk
(2.35)
(2.36)
Problems with one degree of freedom are virtually identical whether they
are formulated in terms of the characteristic function or the principle function.
Take for example, the harmonic oscillator from the previous section.
Equation (2.28) becomes
β =
∂W (q, α)
∂α
q =
= 1
ω sin−1
[ √
mω2 q
2α
]
= t − t0
√ 2α
mω 2 sin[ω(t − t0)] (2.37)
The following problem raises some new issues.
Consider a particle in a stable orbit in a central potential. The motion
will lie in a plane so we can do the problem in two dimensions.
H = 1
2m
(
p 2 r + p2 ψ
r 2
)
+ V (r) (2.38)
pψ = mr 2 ˙ ψ is the angular momentum. It is conserved since ψ is cyclic.
[ (∂W ) 2
1
+
2m ∂r
1
r2 ( ) ]
2
∂W
+ V (r) = α1
∂ψ
(2.39)

2.5. ACTION-ANGLE VARIABLES 27
[
r 2
( ) 2
dWr
+ 2mr
dr
2 V (r) − 2mα1r 2
]
+
( ) 2
dWψ
= 0 (2.40)
dψ
At this point we notice ∂W
∂ψ = pψ, which we know is constant. Why not
call it something like αψ? Then Wψ = αψψ. This is worth stating as a
general principle: if q is cyclic, Wq = αqq, where αq is one of the n constant
α’s appearing in (2.30).
∫
W =
√
dr 2m(α1 − V ) − α2 ψ /r2 + αψψ (2.41)
We can find r as a function of time by inverting the equation for β1, just as
we did in (2.37), but more to the point
βψ = ∂W
∫
αψdr
= − √
∂αψ r 2m(α1 − V ) − α2 ψ /r2
+ ψ (2.42)
Make the usual substitution,u = 1/r.
∫
du
ψ − βψ = − √
2m(α1 − V (r))/α2 ψ − u2
(2.43)
This is a new kind of equation of motion, which gives ψ = ψ(r) or r = r(ψ)
(assuming we can do the integral), i.e. there is no explicit time dependence.
Such equations are called orbit equations. Often it will be more useful to
have the equations in this form, when we are concerned with the geometric
properties of the trajectories.
2.5 Action-Angle Variables
We are pursuing a rout to chaos that begins with periodic or quasi-periodic
systems. A particularly elegant approach to these systems makes use of a
variant of Hamilton’s characteristic function. In this technique, the integration
constants αk appearing directly in the solution of the Hamilton-Jacobi
equation are not themselves chosen to be the new momenta. Instead, we
define a set of constants Ik, which form a set of n independent functions of
the α’s known as action variables. The coordinates conjugate to the J’s are
angles that increase linearly with time. You are familiar with a system that
behaves just like this, the harmonic oscillator!
q =
√ 2E
k sin ψ p = √ 2mE cos ψ

28 CHAPTER 2. CANONICAL TRANSFORMATIONS
Where ψ = ωt + ψ0. In the language of action-angle variables I = E/ω, so
√
2I
q =
mω sin ψ p = √ 2mIω cos ψ
I is the “momentum” conjugate to the “coordinate” ψ.
Action angle variables are only appropriate to periodic motion, and there
are other restrictions we will learn as we go along, but within these limitations,
all systems can be transformed into a set of uncoupled harmonic
oscillators. 5 To see what “periodic motion” implies, have a look at the
simple pendulum.
H = p2 θ − mgl cos θ = E = α (2.44)
2ml2 pθ = ± √ 2ml 2 (E + mgl cos θ) (2.45)
There are two kinds of motion possible. If E is small, the pendulum will
reverse at the points where pθ = 0. The motion is called libration, i.e.
bounded and periodic. If E is large enough, however, the pendulum will
swing around a complete circle. Such motion is called rotation (obviously).
There is a critical value of E = mgl for which, in principle, the pendulum
could stand straight up motionless at θ = π. An orbit in pθ - θ phase space
corresponding to this energy forms the dividing line between the two kinds
of motion. Such a trajectory is called a separatrix.
For either type of periodic motion, we can introduce a new variable I
designed to replace α as the new constant momentum.
I(α) = 1
2π
p(q, α) dq (2.46)
This is a definite integral taken over a complete period of libration or rotation.
6 I will prove (1) the angle ψ conjugate to I is cyclic, and (2) ∆ψ = 2π
corresponds to one complete cycle of the periodic motion.
1. Since I = I(α) and H = α, it follows that H is a function of I only.
H = H(I).
I ˙ = − ∂H
∂ψ
= 0
˙
ψ = ∂H
∂I
= ω(I) (2.47)
5 When correctly viewed, everything is a harmonic oscillator.
6 Textbooks are about equally divided on whether to call action I or J and whether or
not to include the factor 1/2π.

2.5. ACTION-ANGLE VARIABLES 29
2. We are using an F2 type generating function, which is a function of the
old coordinate and new momentum. Hamilton’s characteristic function
can be written as
W = W (q, I). (2.48)
The transformation equations are
Note that
so
dψ =
∂ψ
∂q
ψ = ∂W
∂I
∂ψ
∂q
∂ ∂W
dq =
∂I ∂q
p = ∂W
∂q
( )
∂ ∂W
=
∂I ∂q
∂
dq =
∂I
2.5.1 The harmonic oscillator (for the last time)
H = 1
2m (p2 + m 2 ω 2 q 2 )
p = ± √ 2mE − m2ω2q 2
I = 1
√2mE
− m2ω2q 2 dq
2π
(2.49)
p dq = ∂
(2πI) = 2π.
∂I
The integral is tricky in this form because p changes sign at the turning
points. We won’t have to worry about this if we make the substitution
q =
√
2E
sin ψ (2.50)
mω2 This substitution not only makes the integral easy and takes care of the sign
change, it also makes clear the meaning of an integral over a complete cycle,
i.e. ψ goes from 0 to 2π.
I = E
πω
cos 2 ψ dψ = E/ω
From this point of view the introduction of ψ at (50) seems nothing
more that a mathematical trick. We would have stumbled on it eventually,

30 CHAPTER 2. CANONICAL TRANSFORMATIONS
however, as the following argument shows. The Hamilton-Jacobi equation
is
[ (dW ) 2
1
+ m
2m dq
2 ω 2 q 2
]
= E
∫
√2mIω
W =
− m2ω2q 2 dq
∂W
∂I
∫
= mω
dq
√ 2mIω − m 2 ω 2 q 2
= sin −1
( √ )
mω2 q − ψ0 = ψ
2I
√
2E
q = sin(ψ − ψ0)
mω2 In the last equation ψ0 appears as an integration constant. Evidentally, ψ
is the angle variable conjugate to I.
In summary, to use action-angle variables for problems with one degree
of freedom:
1. Find p as a function of E = α and q.
2. Calculate I(E) using (2.46).
3. Solve the Hamilton-Jacobi equation to find W = W (q, I).
4. Find ψ = ψ(q, I) using (2.49).
5. Invert this equation to get q = q(I, ψ).
6. Use (2.47) to get ω(I).
7. Calculate p = p(I, q) from (2.49).
One attractive feature of this scheme is that you can find the frequency
without using the characteristic function and without finding the equations
of motion. The phase space plot is particularly important. Use polar coordinates
(what else) for (I, ψ). Every trajectory, whatever the system, is a
circle!
Our derivation was based on the following assumptions: (1) The system
had one degree of freedom. (2) Energy was conserved and the Hamiltonian
had no explicit time dependence. (3) The motion was periodic. Every such
system is at heart, a harmonic oscillator. Phase space trajectories are circles.

2.5. ACTION-ANGLE VARIABLES 31
The frequency can be found with a few deft moves. From a philosophical
point of view, (and we will be getting deeper and deeper into philosophy as
these lectures proceed) problems in this category are “as good as solved,”
nothing more needs to be said about them. The same is definitely not true
true with more than one degree of freedom. I will take a paragraph to
generalize before going on to some more abstract developments.
We must assume that the system is separable, so
W (q1, . . . , qn, α1, . . . , αn) = ∑
Wk(qk, α1, . . . , αn) (2.51)
k
pk = ∂
Wk(qk, α1, . . . , αn) (2.52)
∂qk
Ik = 1
pk(qk, α1, . . . , αn) (2.53)
2π
Next find all the q’s as function of the I’s and substitute into W .
Finally
ψk = ∂W
∂Ik
W = W (q1, . . . , qn; I1, . . . , In)
˙
Ik = 0
˙ ψk = ∂H
∂Ik
= ωk
(2.54)

32 CHAPTER 2. CANONICAL TRANSFORMATIONS

Chapter 3
Abstract Transformation
Theory
So, one-dimensional problems are simple. Given the restrictions listed in
the previous section, their phase space trajectories are circles. How does
this generalize to problems with two or more degrees of freedom? A brief
answer is that, given a number of conditions that we must discuss carefully,
the phase space trajectories of a system with n degrees of freedom, move
on the surface of an n-dimensional torus imbedded in 2n dimensional space.
The final answer is a donut! In order to prove this remarkable assertion and
understand the conditions that must be satisfied, we must slog through a
lot of technical material about transformations in general.
3.1 Notation
Our first job is to devise some compact notation for dealing with higher
dimensional spaces. I will show you the notation in one dimension. It will
then be easy to generalize. Recall Hamilton’s equations of motion.
˙p = − ∂H
∂q
We will turn this into a vector equation.
( )
q
η =
p
(
0
J =
−1
1
0
The equations of motion in vector form are
˙q = ∂H
∂p
)
∇ =
( ∂
∂q
∂
∂p
)
(3.1)
˙η = J · ∇H (3.2)
33

34 CHAPTER 3. ABSTRACT TRANSFORMATION THEORY
J is not a vector of course. Sometimes an array used in this way is called a
dyadic. At any rate this is just shorthand for matrix multiplication, i.e.
( ) ( )
˙q 0 1
=
˙p −1 0
( )
∂H
∂q
∂H
∂p
The structure of J is important. Notice that it does two things: it exchanges
p and q and it changes one sign. This is called a symplectic transformation.
I want to explore the connection between canonical transformations and
symlpectic transformations.
I’ll start with the generic canonical transformation, (q, p) → (Q, P ). How
do the velocities transform? Define
)
Using the notation
this can be written
( ˙Q
˙
P
M =
)
=
(
∂Q ∂Q
∂q
∂P
∂p
∂P
∂q ∂p
(
∂Q ∂Q
∂q
∂P
∂p
∂P
∂q ∂p
ζ =
( Q
P
)
) ( ˙q
˙p
)
(3.3)
(3.4)
(3.5)
˙ζ = M · ˙η = M · J · ∇H (3.6)
The gradient operator differentiates H with respect to q and p. These derivatives
transform e.g.
∂H ∂H ∂Q ∂H ∂P
= +
∂q ∂Q ∂q ∂P ∂q
consequently
The T stands for transpose, of course:
but
Combining (3.8) and (3.9):
∇ (q,p) = M T · ∇ (Q,P )H (3.7)
˙ζ = M · J · M T · ∇ (Q,P )H (3.8)
˙ζ = J · ∇ (Q,P )H (3.9)
J = M · J · M T
(3.10)

3.1. NOTATION 35
Those of you who have studied special relativity should find (??) congenial.
Remember the definition of a Lorentz transformation: any 4×4 matrix
Λ that satisfies
g = Λ · g · Λ T
(3.11)
is a Lorentz transformation. 1 The matrix
⎛
1 0 0
⎞
0
⎜
g = ⎜ 0
⎝ 0
−1
0
0
−1
0 ⎟
0 ⎠
0 0 0 −1
(3.12)
is called the metric or metric tensor. Forgive me for exaggerating slightly:
everything there is to know about special relativity flows out of (3.11). We
say that Lorentz transformations “preserve the metric,” i.e. leave the metric
invariant. The geometry of space and time is encapsulated in (12). By the
same token, canonical transformations preserve the metric J. The geometry
of phase space is encapsulated in the definition of J. Since J is symplectic,
canonical transformations are symplectic transformation, they preserve the
symplectic metric.
Equation (4.10) is the starting point for the modern approach to mechanics
that uses the tools of Lie group theory. I will only mention in passing
some points of contact with group theory. Both Goldstein’s and Schenk’s
texts have much more on the subject.
3.1.1 Poisson Brackets
Equation (3.10) is really shorthand for four equations, e.g.
∂Q ∂P
∂q ∂p
∂P ∂Q
− = 1 (3.13)
∂q ∂p
This combination of derivatives is called a Poisson bracket. The usual notation
is
∂X ∂Y ∂X ∂Y
− ≡ [X, Y ]q,p
∂q ∂p ∂q ∂p
(3.14)
The quantity on the left is called a Poisson bracket. Then (3.13) becomes
[Q, P ]q,p = 1 (3.15)
1 It is not a good idea to use matrix notation in relativity because of the ambiguity
inherent in covariant and contravariant indices. Normally one would write (11) using
tensor notation.

36 CHAPTER 3. ABSTRACT TRANSFORMATION THEORY
This together with the trivially true
[q, p]q,p = 1 (3.16)
are called the fundamental Poisson brackets. We conclude that canonical
transformations leave the fundamental Poisson brackets invariant. It turns
out that all Poisson brackets have the same value when evaluated with respect
to any canonical set of variables. This assertion requires some proof,
however. I will start by generalizing to n dimensions.
⎛
⎜
η = ⎜
⎝
q1
q2
.
qn
p1
p2
.
pn
⎞
⎟
⎠
J =
( 0 ℑn
−ℑn 0
)
⎛
⎜
∇ = ⎜
⎝
∂
∂q1
∂
∂q2
.
∂
∂qn
∂
∂p1
∂
∂p2
∂
∂pn
⎞
⎟
⎠
(3.17)
The symbol ℑn is the anti-diagonal n × n unit matrix. (refe3.14) becomes
[X, Y ]η ≡ ∑
(
∂X ∂Y
∂qk ∂pk
− ∂X
)
∂Y
,
∂pk ∂qk
(3.18)
or in matrix notation
The following should look familiar:
k
[X, Y ]η = (∇ηX) T · J · ∇ηY. (3.19)
[qi, qk] = [pi, pk] = 0 [qi, pk] = δik
These are, of course, the commutation relations for position and momentum
operators in quantum mechanics. The resemblance is not accidental. The
operator formulation of quantum mechanics grew out of Poisson bracket
formulation of classical mechanics. This development is reviewed in all the
standard texts. In matrix notation
[η, η]η = [ζ, ζ]η = J (3.20)
The Poisson bracket of two vectors is itself a n × n matrix. i.e.
[X, Y ]ij ≡ [Xi, Yj] (3.21)

3.1. NOTATION 37
The proof of the above assertion is straightforward.
∇ηY = M T · ∇ζY
(∇ηX) T = (M T · ∇ζX) T = (∇ζX) T · M
[X, Y ]η = (∇ζX) T · M · J · M T · ∇ζY
= (∇ζX) T · J · ∇ζY = [X, Y ]ζ
The last step makes use of (3.10). The invariance of the Poisson brackets is
a non-trivial consequence of the symplectic nature of canonical transformations.
From now on will will not bother with the subscripts on the Poisson
brackets.
Here is another similarity with quantum mechanics. Let f be any func-
tion of canonical variables.
f ˙ = ∑
(
∂f
∂qk k
= ∑
(
∂f ∂H
∂qk ∂pk
k
df
dt
˙qk + ∂f
)
˙p +
∂pk
∂f
∂t
− ∂f
∂pk
∂H
∂qk
= [f, H] + ∂f
∂t
)
+ ∂f
∂t
(3.22)
This looks like Heisenberg’s equation of motion. For our purposes it means
that if f doesn’t depend on time explicitly and if [f, H] = 0, then f is a
constant of the motion. We can use (3.22) to test if our favorite function is
in fact constant, and we can also use it to construct new constants as the
following argument shows.
Let f, g, and h be arbitrary functions of canonical variables. The following
Jacobi identity is just a matter of algebra.
[f, [g, h]] + [g, [h, f]] + [h, [f, g]] = 0 (3.23)
Now suppose h = H, the Hamiltonian, and f and g are constants of the
motion. Then
[H, [f, g]] = 0
Consequence: If f and g are constants of the motion, then so is [f, g].
This should make us uneasy. Take any two constants. Well, maybe they
commute, but if not, then we have three constants. Commute the new
constant with f and g and get two more constants, etc. How many constants
are we entitled to – anyway? This is a deep question, which has something
to do with the notion of involution. I’ll get to that later.

38 CHAPTER 3. ABSTRACT TRANSFORMATION THEORY
3.2 Geometry in n Dimensions: The Hairy Ball
Have another look at equation (3.2). Let’s call ˙η a velocity field. By this
I mean that it associates a complete set of ˙q’s and ˙p’s with each point in
phase space. In what direction does ˙η point? This is an easy question in
one dimension; ˙η evaluated at the point P points in the direction tangent
to the trajectory through P . Since trajectories can’t cross in phase space,
there is only one trajectory through P , and the direction is unambiguous.
If we use action-angle variables, the trajectory is a circle, and ˙η is what we
would call in Ph211, a tangent velocity. The same is true, no doubt, for
n > 1, but how do these circles fit together? How does one visualize this in
higher dimensions?
The answer, as I have mentioned before, is that the trajectories all lie
on the surface of an n dimensional torus imbedded in 2n dimensional space.
Your ordinary breakfast donut is a two dimensional torus imbedded in three
dimensional space. 2 This is easy to visualize, so let’s limit the discussion
to two degrees of freedom for the time being. The step from one degree of
freedom to two involves some profound new ideas. The step from two to
higher dimension is mostly a matter of mathematical generalization.
Since we are dealing with conservative systems, the trajectories are limited
by the conservation of energy, i.e. H(q1, q2; p1, p2) = E is an equation
of constraint. The trajectories move on a manifold with three independent
variables. Now the gradient of a function has a well defined geometrical
significance: at the point P , ∇f points in a direction perpendicular to the
surface or contour of constant f through P . In this case ∇H is a four component
vector perpendicular to the surface of constant energy. Unfortunately,
˙η points in the direction of J · ∇H. What direction is that? Well,
(∇H) T · J · ∇H = [H, H] = 0
Consequently, J · ∇H points in a direction perpendicular to ∇H, which is
perpendicular to the plane of constant H, i.e. ˙η lies somewhere on the three
dimensional surface of constant H.
We could have guessed that ahead of time, of course, but we can take the
argument further. H is probably not the only constant of motion. Suppose
there are others; call them F , G, etc. For each of these constants we can
2 The word “dimension” gets used in two different ways. When we talk about physical
systems, Lagrangians, Hamiltonians, etc., the dimension is equal to the number of degrees
of freedom. Here I am using dimension to mean the number of independent variables
required to describe the system.

3.2. GEOMETRY IN N DIMENSIONS: THE HAIRY BALL 39
construct a vector field using (2).
˙ηF = J · ∇F
˙ηG = J · ∇G
· · · etc. · · ·
How many such fields can we construct that are independent of one another?
To put it another way, how many independent constants of motion are there?
That’s a good question – what do you mean by “independent”? The answer
comes from differential geometry. I’m afraid I can only give a hand-waving
introduction to it. There are two related requirements:
1. Suppose F and G are independent constants of motion. Take any
trajectory from the manifold of constant F and another from constant
G. There is no continuous canonical transformation that maps the one
trajectory into another.
2. For each point P in space there must be one unique trajectory that lies
in the plane of constant F and simultaneously in the plane of constant
G.
Think about this last requirement in the case where there are two degrees
of freedom and two independent constants of motion. The trajectories must
lie on a two- dimensional surface. If we use action-angle variables, the
trajectories are circles. This sounds like a globe of the earth. Trajectories
with constant ϕ are called longitudes, lines of constant θ are latitudes. But
wait! We have a serious problem at the poles. The north and south poles
have all possible longitudes. Requirement 2 is violated. Could you rearrange
the lines so that this problem doesn’t occur? It turns out that this is not
possible. This deep result is known in mathematical circles a the Poincare-
Hopf theorem. In the sort of less exalted company we keep, it’s the Hairy
Ball Theorem. The idea is this: try to comb the hair on a hairy ball so that
there is no bald spot. It can’t be done. So long as you really use a comb, i.e.
so long as the trajectories don’t cross, you will always be left with one hair
standing straight up! This is not a proof, of course, but it is a vivid way
of visualizing the content of the theorem. It is easy to see, however, that
what is impossible on a sphere is trivially easy on the surface of a donut. It
can be done in an infinite variety of ways. The simplest is to choose your
“longitudes” so they go around the donut the long way. Latitudes go around
the short way. This also satisfies requirement 1. You can’t deform a latitude
into a longitude without cutting through the donut.

40 CHAPTER 3. ABSTRACT TRANSFORMATION THEORY
OK. Suppose you have two constants of motion F and G. How can you
tell if they are independent? The answer is surprisingly simple, [F, G] = 0
Proof: Take a point P on the surface of the donut. We should be able
to set up a local coordinate system with its origin at P to describe the
trajectories on the surface. We need two unit vectors, ˆ ξF and ˆ ξG, such that
every trajectory in the ˆ ξF - ˆ ξG plane has constant F and G. Choose
ˆξF = ϵJ · ∇F
This is guaranteed to lie in the surface of constant F ; however, G should
remain constant along ˆ ξF . This means that
0 = ( ˆ ξF ) T · ∇G = ϵ(J · ∇F ) T · ∇G
= −ϵ(∇F ) T · J · ∇G = −ϵ[F, G]
This proves the assertion. It’s worth reflecting on the fact that this construction
would be impossible on the surface of a sphere. The sphere, unlike
the donut, has only one independent constant, its radius. This theorem also
relieves our anxiety about extra constants. If F and G are independent, we
don’t get a “free” constant K = [F, G], because K = 0.
Summary and generalization:
1. A system with n degrees of freedom has at most n independent constants
of motion. Otherwise we could use the additional constants to
eliminate one or more of these degrees. For example, we could use the
Hamilton-Jacobi procedure to make all the momenta constant. The
Hamiltonian would then only be a function of the n coordinates, but
these would not be independent because of the additional constraints.
2. Let’s say there are k constants, Fi, i = 1, . . . , k. If they are independent
we must have [Fi, Fj] = 0.
3. In the best case there are exactly n independent constants. Such
constants are said to be in involution. Such a system is said to be
integrable.
4. All trajectories of integrable systems are confined to the surfaces of
n-dimensional tori imbedded in 2n-dimension space.
5. If k < n there are no general statements we can make about the
behavior of the trajectories. We will be very much concerned in the
next chapter with systems that are “almost” integrable.

3.2. GEOMETRY IN N DIMENSIONS: THE HAIRY BALL 41
6. There are no general criteria known for deciding whether or not a
system is integrable; however, if the Hamiltonian is separable, the
system is integrable.
3.2.1 Example: Uncoupled Oscillators
The Hamiltonian for two uncoupled harmonic oscillators (with m = 1) is
H = 1
2 (p2 1 + p 2 2 + ω 2 1q 2 1 + ω2q 2 2)
This is an important problem because every linear oscillating system can
be put in this form by a suitable choice of coordinates. 3 There are two
constants of motion
E1 = 1
2 (p2 1 + ω 2 1q 2 1) E2 = 1
2 (p2 2 + ω 2 2q 2 2)
In terms of action-angle variables, the constants are I1 and I2.
H = I1ω1 + I2ω2 = E1 + E2 = E
Every integrable system can be put in this form, although in general the
ω’s will be functions of the I’s. Here they are just parameters from the
Hamiltonian.
This is a simple problem, but the phase space is four dimensional. Let’s
think about all possible ways we might visualize it. In the q1 - p1 or (q2 -
p2) plane the trajectories are ellipses with
qk(max) = √ 2Ek/ωk
pk(max) = √ 2Ek,
where k = 1, 2. The area enclosed by each ellipse is significant, because
∫
area = dq dp = p dq = 2πI (3.24)
s
The first integral is a surface integral over the area of the ellipse. The second
is a line integral around the ellipse. This identity is a variant of Stokes’s
theorem. It’s useful to rescale the variables so that they both have the same
units and the trajectory is a circle. An natural choice would be
q ′ k
√
= qk ωk = √ 2Ik sin ψk
p ′ k = pk/ √ ωk = √ 2Ik cos ψk
3 This comes under the heading of “theory of small oscillations.” Most mechanics texts
devote a chapter to it.

42 CHAPTER 3. ABSTRACT TRANSFORMATION THEORY
The trajectories are now circles with radius √ 2Ik. The area enclosed is 2πIk,
as required by (3.24).
The motion in the q1 - q2 plane is more complicated. It depends on
the ratio ω1/ω2 called the winding number. If this is a rational number,
say N1/N2 then after N1 cycles of q1 and N2 cycles of q2, the trajectory
will come back to its starting point. This is called a Lissajou figure. If the
winding number is irrational, the trajectory will be confined to a limited
area but will never return to its starting point. It will eventually “color
in” all available space. In the next chapter we will be concerned with systems
that are “almost” integrable. For such systems the winding number is
all-important. Systems with irrational winding numbers tend to be stable
under perturbation. Those with rational winding numbers disintegrate at
the slightest push!
The centerpiece of this chapter is the torus. The trajectories spiral
around the donut. If the winding number is rational they “wear a path”
around the donut. If it’s irrational they cover the donut evenly. A useful
way of visualizing this was invented by Poincaré. Imagine a flat plane cutting
through the donut in such a way that every point on the plane has the
angle ψ1 = 0. Place a dot on the plane at he point where each trajectory
passes through it. If the winding number is a rational fraction, there will
be a finite number of points. Each time a trajectory passes through ψ1 = 0
it will pass through one of the dots. If the winding number is irrational the
crossings will mark out a continuous circle. The Poincaré section as it is
called (some books call it the surface of section) is a useful diagnostic tool.
Suppose you have a system of equations that are not integrable (so far as
you know) but is amenable to computer calculation. Take various Poincaré
sections. If they are circles then the system is at least approximately integrable
and can be described with action-angle variables. As we will see,
there are often regions of phase space, “islands” as it were, where motion is
simply periodic and other regions that are wildly chaotic.
Pictures of this motion appear in all the standard texts. I have yet to
see a clear explanation of the coordinates involved, however. What does it
mean really to say that the donut is a 2-d surface in a 4-d space? Your
breakfast donut, after all, is imbedded in 3-d space. If we take a Poincaré
section through the donut at the plane ψ2 = 0 and plot q ′ 1 versus p′ 1
, we
will get either a circle of dots or a continuous circle with a radius equal to
√ 2I1, or we can take a slice through ψ1 = 0 and get a circle with radius
√ 2I2. Put it this way, any point on the torus has four (polar) coordinates,
( √ 2I1, ψ1, √ 2I2, ψ2), but in 3-d space, only three of them are independent.
When the torus is in 4-d space, all four of them are independent. If we really

3.2. GEOMETRY IN N DIMENSIONS: THE HAIRY BALL 43
lived in 4-d space, we would label the axes of the donut plot (q ′ 1 , p′ 1 , q′ 2 , p′ 2 ).
This is impossible for us to imagine. The donut is easy; just remember that
there is no equation of constraint among the four variables. 4
3.2.2 Example: A Particle in a Box
Consider a particle in a two-dimensional box with elastic walls.
0 ≤ x ≤ a 0 ≤ y ≤ b
H = 1
2m (p2x + p 2 y) = π2
(
I2 1
2m a2 + I2 2
b2 )
I1 = 1
px dx =
2π
a
π |px| I2 = b
π |py|
ω1 = ∂H
∂I1
= π2
I1 ω2
ma2 = π2
I2
mb2 There are several interesting points about this apparently trivial problem.
The Hamiltonian looks linear, but in fact it contains an invisible nonlinear
potential that reverses the particle’s momentum when it hits the wall. One
symptom of this is that the frequencies depend on I. This looks odd, but
it’s just the action-angle way of saying that the particle makes a round trip
(in the x direction) in a time T = 2am/px. The loop integral in this context
is an integral over one “round trip” of the particle.
∫ a ∫ 0
pxdx = |px| dx + (−|px|) dx = 2a|px|
0
My real point in showing this example is to call your attention to the
angle variable. I will work through the calculation for the x variable. This
same thing holds for y of course.
1
2m
a
( ) 2
dWx
= E1
dx
∫
Wx = (±) √ ∫
2mE1 dx =
ψ1 = ∂Wx
∂I1
= ± π
∫
a
(±) πI1
a dx
dx = ± π
a x + ψ10 = ψ1
4 Of course, I1 and I2 are constant for any given set of initial conditions. It is this sense
in which the torus is a 2-d surface.

44 CHAPTER 3. ABSTRACT TRANSFORMATION THEORY
The term ψ10 is an integration constant. There is no reason why it must be
the same for both legs of the journey. We are free to choose it as follows:
0 → x → a: ψ1 = πx/a
0 ← x ← a: ψ1 = 2π − πx/a
While the particle is bouncing violently between the walls, the angle variables
are increasing smoothly with time, ψ1 = ω1t and ψ2 = ω2t. Even this
strange problem is equivalent to a donut! 5
5 When correctly viewed, everything is a harmonic oscillator – in this case two harmonic
oscillators.

Chapter 4
Canonical Perturbation
Theory
So far we have assumed that our systems had exact analytic solutions. One
way of stating this is that we can find a canonical transformation to action
angle variables such that the new Hamiltonian is a function of the action
variables only, H = H(I). Such problems are the exception rather than the
rule. For our purposes they are also uninteresting. All periodic integrable
systems are equivalent to a set of uncoupled harmonic oscillators. Once you
get over the thrill of this discovery, the oscillators are boring! The existence
of chaos depends on the system not being equivalent to a set of oscillators.
In order to deal with systems that are non-trivial in this sense, we need some
way of doing perturbation theory. 1
4.1 One-Dimensional Systems
I will present the theory first for systems with one degree of freedom. This
will simplify the notation, however the interesting complications only appear
in higher dimensions. Here is the basic situation: A bounded conservative
system with one degree of freedom is described by a constant Hamiltonian
H(q, p) = E. We need to obtain the equations of motion in the form q = q(t)
1 I will follow the treatment in Chaos and Integrability in Nonlinear Dynamics, Michael
Tabor, Wiley-Interscience, 1989. Another good reference is Classical Mechanics by R. A.
Metzner and L.C. Shepley, Prentice Hall, 1991. The subject is also discussed in Classical
Mechanics, Goldstein, Poole and Safko, third edition, Addison-Wesley, 2002. Goldstein
discusses time-dependent and time-independent perturbation theory. We are doing the
time-independent variety.
45

46 CHAPTER 4. CANONICAL PERTURBATION THEORY
and p = p(t), but this is impossible due to the non-linear nature of the
problem. We are able to split up the Hamiltonian
H = H0 + ϵH1
in such a way that H0 is amenable to exact solution, and H1 is in some
sense small. We indicate the smallness by multiplying it by ϵ. This is a
bookkeeping device; it will be set to one after the approximations have been
derived.
The first step is to find the canonical transformation that makes H0
cyclic, i.e. q = q(I, ψ), p = p(I, ψ), and H0 = H0(I) where I and ˙ ψ = ω0
are both constant. Unfortunately, this transformation does not render the
complete Hamiltonian cyclic, so we write
H(I, ψ) = H0(I) + ϵH1(I, ψ). (4.1)
H is still constant, and consequently I now depends on ψ. H0 is not an
explicit function of ψ, but it does depend on ψ implicitly through I.
Despite this inconvenience, I and ψ are still a perfectly good set of
canonical variables, so that the equations of motion
I ˙ = − ∂
H(I, ψ) ˙
∂ψ
∂
ψ = H(I, ψ) (4.2)
∂I
are valid without approximation, even though we are unable to solve them in
this form. The so-called time-dependent perturbation proceeds from here by
expanding the solutions of (4.2) as power series in ϵ. Our approach is to find
a second canonical transformation, i.e. (q, p) → (I, ψ) → (J, φ) such that
H(I, ψ) → K(J). This last step must be done as a series of approximations,
of course, otherwise the problem would be exactly solvable.
In order to make the transformation (I, ψ) → (J, φ) we will use a generating
function of the F2 genus, i.e. F = F (ψ, J). We need to expand
F = F0(ψ, J) + ϵF1(ψ, J) + · · · (4.3)
where F0 = Jψ. This is the identity transformation as can be seen as follows:
I = ∂
∂ψ F0 = J φ = ∂
∂J F0 = ψ
In terms of (??)the transformation equations are
I = ∂F
∂ψ
= J + ϵ∂F1 (ψ, J) + · · · (4.4)
∂ψ

4.1. ONE-DIMENSIONAL SYSTEMS 47
φ = ∂F
∂J
= ψ + ϵ∂F1 (ψ, J) + · · · (4.5)
∂J
Before going on there are some technical points about ψ and J that need
to be discussed. When ϵ = 0, ψ is the exact angle variable for the system.
This means that we can find p and q as functions of ψ such that p and q
return to their original values when ∆ψ = 2π. We can in principle invert
this transformation to find ψ as a function of p and q.
ψ = ψ(q, p) (4.6)
When p and q run through a complete cycle, ψ advances by 2π. When ϵ ̸= 0
the orbit will be different from the unperturbed case, but the functional
relationship doesn’t change, so when p and q run through a complete cycle,
we must still have ∆ψ = 2π. Of course, the exact angle variable will also
advance 2π. In summary
∆ψ = ∆φ = 2π (4.7)
for one complete cycle.
The following integrals are all equal because canonical transformations
preserve phase space volume.
J = 1
2π
p dq = 1
2π
Now integrate (4.4) around one orbit:
that is
1
2π
I dψ = 1
2π
J dφ = 1
2π
J dψ + 1
2π ϵ
∂F1
∂ψ
J = J + 1
2π ϵ
∂F1
∂ψ
dψ + · · · ;
I dψ (4.8)
dψ + · · · ;
We have just seen that ∆ψ = 2π around one cycle. Consequently
∂F1
dψ = 0 (4.9)
∂ψ
implies that the derivative of F1 is purely oscillatory with a fundamental
period of 2π in ψ. (The same is true of the higher order terms as well.)
The Hamiltonian is transformed using (??) with the new variables.
K(φ, J) = H(ψ(φ, J), I(φ, J)) + ∂
∂t F2(ψ(φ, J), J, t) (4.10)

48 CHAPTER 4. CANONICAL PERTURBATION THEORY
As explained above, we seek a transformation that makes φ cyclic so that
K = K(J). The appropriate generating function does not depend on time,
so (4.10) becomes
K(J) = H(ψ(φ, J), I(φ, J)) (4.11)
The approximation procedure consists in expanding the left and right sides
of this equation in powers of ϵ and then equating terms of zeroth and first
order. This procedure could be carried out to higher order. I’m interested
in first order corrections only.
The so-called Kamiltonian is expanded as follows:
K(J) = K0(J) + ϵK1(J) + · · ·
At first sight, this agenda looks hopeless. We need to know the exact value
of J to make use any of these terms, even the zeroth order approximation.
The exquisite point is that we can use (4.8) to calculate J exactly without
knowing the complete transformation.
The zeroth order Hamiltonian is expanded with the help of (4.4).
H0(I) = H0( ∂F
∂ψ ) = H0(J + ϵ ∂F1
∂ψ + · · · ) = H0(J) + ϵ ∂F1
∂ψ
∂H0(J)
∂J
ϵ=0
The first order term is already multiplied by ϵ.
Substitute all this into (??) gives
∂H0
∂J
ϵ=0
+ · · ·
= ∂H0(I)
∂I = ω0 (4.12)
ϵH1(ψ, I) = ϵH1(φ, J) + · · ·
K0(J) = H0(J) (4.13)
K1(J) = ∂F1
∂ψ ω0 + H1(φ, J) (4.14)
The notation H0(J) means that you take your formula for H0(I) and replace
the symbol I with the symbol J without making any change in the functional
form of H0.
Integrate (4.14) around one cycle and use (4.9); (4.14) becomes
K1(J) = H1(J) ≡ 1
∫ 2π
H1(ψ, J) dψ, (4.15)
2π 0

4.1. ONE-DIMENSIONAL SYSTEMS 49
and
∂
∂ψ F1(ψ, J) = 1 [
H1 − H1(ψ, J)
ω0(J)
] ≡ − ˜ H1(ψ, J)
ω0
(4.16)
˜H1 is the periodic part of H1. We are left with a differential equation that
is easy to integrate.
F1(ψ, J) = − 1
∫
dψ
ω0(J)
˜ H1(ψ, J) (4.17)
4.1.1 Summary
I will summarize all these technical details in the form of an algorithm for
doing first order perturbation theory. Remember that the object is to find
equations of motion in the form q = q(t) and p = p(t). We do this in three
steps: (1) Find q = q(I, ψ) and p = p(I, ψ). (2) Find I = I(J, φ) and
ψ = ψ(J, φ). (3) J and ˙φ are constant, so φ = ˙φt + φ0.
1. Identify the H0 part of the Hamiltonian. Find the transformation
equations q = q(I, ψ) and p = p(I, ψ) using the Hamiltonian-Jacobi
equation as described in the previous section. Use (??) to get ω0.
2. Equation (4.8) can be used to find J in terms of the total energy
E. The integral presents no difficulties in principle, especially if the
Hamiltonian is separable. In fact, textbooks never bother to do this.
It seems sufficient to display the results in terms of J, the assumption
being that we could find J = J(E) if we really had to.
3. The first order correction to the energy is obtained from the integral in
(4.15). Get the first order correction to the frequency by differentiating
it with respect to J.
4. The generating function F1 is calculated from (4.17). It is then substituted
into (4.4) and (4.5). These give implicit equations for ψ =
ψ(J, φ) and I = I(J, φ). Unfortunately, it is usually impossible to
invert them to obtain these formula explicitly.
4.1.2 The simple pendulum
The pendulum makes a nice example
H = l2
+ mgR(1 − cos θ)
2mR2

50 CHAPTER 4. CANONICAL PERTURBATION THEORY
The angular momentum l = mR2θ˙ is canonically conjugate to the angle θ.
H = 1
2mR2 [
l 2 + m 2 R 4 ω 2 0θ 2
(
1 − θ2
)]
+ · · ·
12
The first two terms reduce to the familiar harmonic oscillator. This is the
zeroth order problem.
l 2 =
Make the natural substitution
H0 = E0 = l2 mgRθ2
+
2mR2 2
( ) 2
dW
= 2mR
dθ
2 E0 − m 2 R 4 ω 2 0θ 2
l =
sin 2 ψ = ml2 ω 2 0
2E0
( dW
dθ
θ 2
(4.18)
)
= √ 2mR2E cos ψ (4.19)
We can look on this as a convenient change of variable, but ψ is also the
angle variable. This can be seen as follows:
I = 1
2π
l dθ =
√ 2mR 2 E0
2π
[
1 − mR2 ω 2 0
2E0
θ 2
] 1/2
Use (4.19) to get the familiar result, I = E0/ω0. The generating function is
obtained from the indefinite integral
∫ ( ) ∫
dW [2mR2 W =
dθ = ω0I − m
dθ
2 R 4 ω 2 0θ 2] 1/2
dθ (4.20)
According to the basic transformation formula we should have
ψ = ∂W
∂I
One can show by differentiating (4.20) and using (4.19) to complete the
integration, that this is indeed so.
Equations (4.18) and (4.19) can be rearranged to give
l = √ 2mR 2 Iω0 cos ψ (4.21)
dθ

4.2. MANY DEGREES OF FREEDOM 51
θ =
√ 2I
mR 2 ω0
sin ψ (4.22)
The goal of the action-angle program is to express the original coordinates
and momenta in terms of the action-angle variables. This has now been
completed to zeroth order.
The first order correction is
H1(I, ψ) = − mR2 ω 2 0 θ4
24
= − I2
6mR 2 sin4 ψ.
We are now in a position to recast our Hamiltonian à la (4.1).
(
H(I, ψ) = Iω0 + ϵ − I2
6mR2 sin4 )
ψ
We have also obtained ω0 = √ g/R “for free.” The ϵ is there for bookkeeping
purposes only. We have no further need for it.
K0(J) = H0(J) = Jω0
K1(J) = H1(J) = 1
∫ 2π
2π 0
F1(J, ψ) = − 1
˜H = H1 − H1 =
ω0
∫
J 2
H1 dψ = −
16mR2 J 2
48mR 2 (3 − 8 sin4 ψ)
dψ ˜ J
H1 =
2
192 mR2 (sin 4ψ − 8 sin 2ψ)
ω0
ω = ω0 −
J
32mR 2
4.2 Many Degrees of Freedom
For systems of two or more degrees of freedom, canonical perturbation theory
is formulated in exactly the same way as before – but now profound
difficulties arise, even to first order in ϵ. The problem centers around equation
(4.16) repeated here for reference
ω0(J) ∂F1(ψ, J)
∂ψ
= − ˜ H1(ψ, J)
We were able to solve this with a simple integration (4.17). This is not
possible for more that one degree of freedom, so we must resort to Fourier

52 CHAPTER 4. CANONICAL PERTURBATION THEORY
series. Before doing this, however, we will need to generalize our notation.
Let’s use the vectors
J = (J1, · · · , Jn) ω0 = (ω01, · · · , ω0n) ∇ψ = ( ∂
∂ψ1
, · · · , ∂
)
∂ψn
where n is the number of degrees of freedom. In this notation (4.16) becomes
where
ω0(J) · ∇ψF1(ψ, J) = − ˜ H1(J, ψ) (4.23)
¯H1(J, ψ) =
∫ 2π
0
∫ 2π
dψ1 · · · dψnH1(J, ψ) (4.24)
0
and ˜ H1 = H1 − ¯ H1. Since both sides of (4.16) are periodic, we can solve
them with Fourier series.
˜H1(J, ψ) = ∑
Ak(J)e ik·ψ
(4.25)
where k is a vector of integers
k
F1(J, ψ) = ∑
Bk(J)e ik·ψ
k
k = k1, · · · , kn
(4.26)
It seems as if we could proceed as follows: ˜ H1 is known at this point, so we
can find Ak Substitute these definitions into (4.16) we get
Bk = i Ak
ω0 · k
(4.27)
Now here’s the infamous problem. Suppose, for example, there were only
two degrees of freedom. In this case the denominator of (??) would be
ω0 · k = ω01k1 + ω02k2
(4.28)
You can see that if the winding number ω01/ω02 is a rational number, then
for some k, Bk will be infinite. It seems that the slightest perturbation
will blow this system into outer space! Even if the winding number is not
rational, there will always be values of k that will make ω0 · k arbitrarily
small.
This problem was discovered in the early twentieth century, and all the
effort of the most eminent mathematicians of the day failed to solve it.

4.2. MANY DEGREES OF FREEDOM 53
One opinion held that the slightest perturbation would cause the system to
become “ergodic,” that is to say, the trajectories would fill up all of phase
space. Numerical calculations later showed that this was often not the case.
Trajectories will often “lock in” to stable patterns. This has been the subject
of much contemporary research. When and why do trajectories lock in, and
what happens when they do not? The question of what trajectories remain
stable under small perturbations is at least partly answered by the so-called
KAM (Kolmogorov, Arnold, Moser) theorem. In the general case there is,
if not a complete theory, at least a well-developed taxonomy. We will turn
to these matters in the next chapter.

54 CHAPTER 4. CANONICAL PERTURBATION THEORY

Chapter 5
Introduction to Chaos
The canonical perturbation theory of the previous chapter is a lot of work,
and in two or more degrees of freedom it summons up the ogre of small
denominators. Many people have tried to solve this problem by pounding
their heads on it. This turns out not to be a fruitful approach. I will
illustrate the limitations of perturbation theory by considering the van der
Pol oscillator. This is a simple nonlinear, one-dimensional, second-order
differential equation closely resembling a damped harmonic oscillator. It
has stable solutions which can easily be found numerically, yet it has no
known analytic solutions, and perturbation theory, on general principles,
just can’t work! 1 We then go on to discuss linear stability theory. With
these simple techniques you can analyze most nonlinear systems (the van
der Pol oscillator is an exception) and get a qualitative picture of the phase
space dynamics. In one degree of freedom (two-dimensional phase space)
it will become immediately apparent where perturbation theory is possible
and a qualitative idea of the motion of the system where it is not.
Higher dimensional spaces are not so easy to analyze, in part because
they are hard to visualize and in part because they are often not integrable.
It is this non-intagrability that leads to chaos. Here we resort to the
Poincareé section and the notion of discrete maps. The Poincaré-Birkoff and
KAM theorems can then tell us something about the onset and structure of
chaos.
1 It should be remembered that all the major developments in elementary particle theory
over the last few decades starting with the standard model in the 1970’s are based on the
notion of spontaneous symmetry breaking. Spontaneous symmetry breaking, almost by
definition, cannot be described with perturbation theory. When perturbation theory fails
we always expect new physics. The same is true (to a lesser extent) in classical mechanics
as well.
55

56 CHAPTER 5. INTRODUCTION TO CHAOS
5.1 The total failure of perturbation theory
To get some feeling for how perturbation theory might be useless, look at
the following “toy” example.
¨x = −x + ϵ(x 2 + ˙x 2 − 1) sin( √ 2t) (5.1)
This looks like a harmonic oscillator with a resonant frequency ω = 1 and
a “small” driving term with a frequency ω = √ 2. Obvious solutions are
x(t) = sin t and x(t) = cos t, which hold for all values of ϵ. If we set ϵ = 0
then the solutions more generally are x(t) = x0 sin(t + t0). This solution
plotted on a phase space plot of x(t) versus ˙x(t) will be a circle with radius
r = x0. What would you expect for finite ϵ? There presumably are other
solutions, but don’t waste your time looking for them! You should convince
yourself however, that there are no solutions of the form
x(t) = sin t +
∞∑
ϵ n fn(t) (5.2)
Also convince yourself that the trouble comes from the non-linear terms.
The point is because of the non-linearity, it is not possible to start with
unperturbed solutions and get new solutions by adding to them.
A more interesting and oft-studied example is the van der Pol equation.
It was first introduced by van der Pol in 1926 in a study of the nonlinear
vacuum tube circuits of early radios.
n=1
¨x + ϵ(x 2 − 1) ˙x + x = 0 (5.3)
Again the ϵ = 0 equations are x(t) = x0 sin(t + t0). In phase space this is
a circle of radius x0. If we make ϵ ever so much larger than zero, however,
something remarkable happens as shown in the first of the plots in Fig.
5.1. Yes the orbit eventually becomes a circle, but regardless of the initial
conditions, the radius r ≈ 2. The same sort of behavior is shown in Fig. 5.1
for larger values of ϵ. The shape of the final orbit is determined entirely by ϵ
and is completely unaffected by the initial conditions. A curve of the sort is
called a limit cycle. It’s easy to see in vague way why the limit cycle exists.
The term proportional to ϵ in (5.3) looks like an oscillator damping term, but
its sign depends on whether x 2 is greater or less than 1. If it is greater, the
oscillation is damped; if it is smaller the oscillation is “undamped.” Indeed,
if ϵ is made negative, the orbits either collapse to zero or diverge to infinity
depending on the initial conditions. For obvious reasons the solutions with

5.1. THE TOTAL FAILURE OF PERTURBATION THEORY 57
dx/dt
dx/dt
4
2
0
−2
Epsilon=0.1
−4
−4
4
−2 0
x
Epsilon=1.5
2 4
2
0
−2
−4
−4 −2 0
x
2 4
dx/dt
dx/dt
4
2
0
−2
Epsilon=0.5
−4
−4
10
−2 0
x
Epsilon=3
2 4
5
0
−5
−10
−4 −2 0
x
2 4
Figure 5.1: The van der Pol plot for four values of ϵ and two starting values
(indicated by asterisks)

58 CHAPTER 5. INTRODUCTION TO CHAOS
positive ϵ are said to be stable and those with negative ϵ are said to be
unstable.
This simple model makes an important point. Conventional perturbation
theory starts with unperturbed, i.e. ϵ = 0 solutions, and then looks for
series solutions in powers of ϵ. This is obviously hopeless here since even
a smidgeon of ϵ is enough to completely alter the nature of the orbits. It
would be better to start with some simple function that approximated the
limit cycle and then expand in powers of some parameter that characterized
the deviation of the actual orbit from the simple function. Alas, I don’t
know how to do this. The trouble is that the limit cycle is so weird, at least
for large ϵ, that it’s hard to come up with a “lowest-order” solution. For
many systems however, this is a practical approach. The trick is to look for
the fixed points.
5.2 Fixed points and linearization
Equations of motion can always be cast in the form
˙ξ = f(ξ, t). (5.4)
With n degrees of freedom ξ and f are 2n-dimensional vectors. For example,
Hamilton’s equations with one degree of freedom are
˙q = ∂H
∂p
˙p = − ∂H
∂q
[ ]
p
ξ =
q
(5.5)
To keep the notation simple and general (and to save typing) I will keep the
notation in the form (5.5) for the time being and not type out the p’s and
q’. I will also restrict the discussion to autonomous systems, i.e. those in
which the Hamiltonian does not depend explicitly on time. 2
A fixed point (also called a stationary point, equilibrium point, or critical
point) is simply the point ξf where all the time derivatives vanish, f(ξf ) =
ξf
˙ = 0. It’s the place where nothing happens. Detailed information about
2 The notation in this section is taken from Classical Dynamics by J. V. Jose and E. J
Saletan

5.2. FIXED POINTS AND LINEARIZATION 59
the motion of a system close to a fixed point can be obtained by linearizing
the equations of motion. This is done as follows: First the origin is moved
to the fixed point by writing
ζ(t) = ξ(t; ξ0) − ξf
Second, (5.4) is written for ζ rather than for ξ.
(5.6)
˙ζ = f(ζ + ξf ) ≡ g(ζ) (5.7)
Third, g is expanded in a Taylor series about ζ = 0.
˙ζ j = dgj
dζk
ξf
+ O(ζ 2 ) ≡ A j
k ζk + O(ζ 2 ) (5.8)
I am using the Einstein summation convention in which one sums over repeated
indices. Dropping the ζ 2 terms gives the matrix equation
˙z = A · z (5.9)
A is a constant matrix called (among other things) the stability matrix. It’s
easy to solve (9) using the matrix exponential.
where
e At ≡
z(t) = e At z0
∞∑
n=0
(5.10)
A n t n /n! (5.11)
For our purposes it will be enough to take the case of one degree of freedom
in which case A is a 2 × 2 real, constant matrix. If A is diagonal
e At
=
eλ1t 0
0 eλ2t
(5.12)
where λ1 and λ2 are eigenvalues, which might be real or complex. If they
are complex they come in complex-conjugate pairs, λ∗ 1 = λ2.
Various cases can be identified. If both eigenvalues are real and positive,
all trajectories flow away from the fixed point which is then called unstable.
If they are both negative all trajectories flow toward and the fixed point is
said to be stable. If the eigenvalues have opposite signs then the trajectories
are repelled from one axis and attracted to the other. This is called a
hyperbolic fixed point or a saddle point.

60 CHAPTER 5. INTRODUCTION TO CHAOS
Figure 5.2: Unstable fixed point for real λ2 > λ1 > 0.
Figure 5.3: Unstable fixed point for real λ1 = λ2 > 0.

5.2. FIXED POINTS AND LINEARIZATION 61
Figure 5.4: Hyperbolic fixed point for real λ1 < 0, λ2 > 0.
It is possible that A cannot be diagonalized. In that case it can at least
be in upper triangular form, i.e.
then
A =
z(t) = e λt
λ 0
µ λ
1 0
µt 1
(5.13)
(5.14)
Complex eigenvalues require a bit more discussion. Let λ = α + iβ and
z = u + iv, where α and β are real numbers, and u and v are real vectors
orthogonal to one another. Separating real and imaginary parts
A · u = αu − βv A · v = βu + αv (5.15)
Evidentally A · z ∗ = λ ∗ z ∗ so z ∗ is an eigenvector with eigenvalue λ ∗ . Eigne
vectors belonging to different eigenvalues are independent. We can construct
the independent real vectors u and v as follows
u =
z + z∗
2
v =
z − z∗
2i
(5.16)

62 CHAPTER 5. INTRODUCTION TO CHAOS
Figure 5.5: Unstable fixed point for nondiagonalizable A matrix. All of the
integral curves are tangent to z2 at the fixed point.
Substituting these definitions into (5.10) gives
e At u = e αt (u cos βt − v sin βt), (5.17)
e At v = e αt (u sin βt + v cos βt). (5.18)
There are two important cases, α > 0 in which case the fixed point is
unstable and the orbits are spirals, and α = 0 when the phase portrait
consists of circles. In this case the fixed point is call a center or an elliptic
point.
It should be remembered that (5.9) is a linearized equation. It hold
in some small region of the fixed point, and of course, as is so often the
case, the theory gives us no well to tell how small that region might be. The
damped oscillator makes a good example of the method, and there are many
other examples in the textbooks. On the other hand, the the theory fails
completely for the van der Pol oscillator in the previous section.
5.3 The Henon oscillator
Although the theory from the previous section is perfectly general in the
sense that it can be applied to systems with any number of degrees of free-

5.3. THE HENON OSCILLATOR 63
Figure 5.6: Unstable fixed point for complex λ with ℜ(λ) > 0.
Figure 5.7: Stable fixed point for λ pure imaginary.

64 CHAPTER 5. INTRODUCTION TO CHAOS
dom, it is almost impossible to visualize in four or more dimensions, and the
number of cases that must be considered increases rapidly. The best tool
for visualizing higher dimensional spaces is the Poincaré section. This was
described briefly in Chapter 3, and we will make more use of it shortly. Before
doing so it will be useful to a good example of motion with two degrees
of freedom. A fascinating and oft-studied cases is the Hénon-Heiles Hamiltonian.
The Hamiltonian was originally used to model the motion of stars
in the galaxy3 . Written in terms of dimensionless variables the Hamiltonian
is
H = 1
2 ( ˙x2 + ˙y 2 + k1x 2 + k2y 2 ) + λ(x 2 y − y3
) (5.19)
3
This is the Hamiltonian of two uncoupled harmonic oscillators with a perturbation
proportional to λ. The oscillators have frequencies ω1 = √ k1 and
ω2 = √ k2. The phase space is the four-dimensional space spanned by x,
˙x, y, and ˙y. We can think of the unperturbed orbit as lying on two tori.
In this case their cross sections are circular with radii determined by the
initial conditions. If the winding number is w = r/s, x will complete r
cycles while y completes s. Let us make a Poincaré section through the y
torus at x = 0. Each time the orbit passes through from x < 0 to x > 0
we mark a point at y and ˙y on the x = 0 plane. An example is shown in
Figure (5.8) for w = 7/2. Because the winding number is rational there
are seven discrete dots on the Poincaré section. The case of an irrational
winding number is shown in Figure (5.9). The x vs. y plot is completely
filled in, and the Poincaré plot is a continuous loop. Continuous loops like
this on the Poincaré plot are a sign that the system is circulating around an
invariant torus and hence is integrable.
When we turn on the perturbation by making λ ̸= 0 something remarkable
happens. 4 Figures (5.10) through (5.13) show a progression from the
orderly motion of the uncoupled oscillators Figure (5.9) through the loop in
(5.10) suggesting motion around a single distorted torus. As the interaction
strength is increased this one torus breaks up into five separate tori. In the
next plot Figure (5.12) the points are beginning to disperse in a random
way with some structure remaining. Because of the poor resolution of the
plots one cannot see the fine details that remain. Finally in the last plot the
points are arranged in a completely random pattern. This is paradigmatic.
As the strength of the perturbation increases orderly motion disintegrates
3 See Goldstein’s Classical Dynamics for a review of the physics
4 I am following standard practice by varying the perturbation strength by changing
the total energy with λ = 1.

5.3. THE HENON OSCILLATOR 65
y
dy/dt
0.1
0.05
0
−0.05
−0.1
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
x vs y
−0.1 −0.05 0
x
0.05 0.1
dy/dt
0.2
0.15
0.1
0.05
0
−0.05
−0.1
−0.15
−0.2
Poincare Section through x=0
−0.1 −0.05 0
y
0.05 0.1
Figure 5.8: Harmonic oscillator coordinates for w = 7/2
Poincare Section through x=0
−0.3 −0.2 −0.1 0
y
0.1 0.2 0.3
Figure 5.9: Harmonic oscillator coordinates irrational winding number

66 CHAPTER 5. INTRODUCTION TO CHAOS
dy/dt
0.4
0.3
0.2
0.1
0
−0.1
−0.2
−0.3
Poincare Section through x=0
−0.4
−0.4 −0.3 −0.2 −0.1 0
y
0.1 0.2 0.3 0.4
Figure 5.10: Henon-Heiles Hamiltonian. Orbit circulates a distorted torus.
dy/dt
0.5
0.4
0.3
0.2
0.1
0
−0.1
−0.2
−0.3
−0.4
Poincare Section through x=0
−0.5
−0.4 −0.2 0 0.2 0.4 0.6
y
Figure 5.11: The orbit breaks up into smaller tori.

5.3. THE HENON OSCILLATOR 67
dy/dt
dy/dt
0.5
0.4
0.3
0.2
0.1
0
−0.1
−0.2
−0.3
−0.4
Poincare Section through x=0
−0.5
−0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8
y
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
Figure 5.12: Chaos begins to set in.
Poincare Section through x=0
−0.8
−0.5 0 0.5 1
y
Figure 5.13: Complete Chaos

68 CHAPTER 5. INTRODUCTION TO CHAOS
into chaos. One of the goals of chaos theory is to explain explain and predict
this phenomena. This will require some new formalism.
5.4 Discrete Maps
Suppose we were to number the points on the Poincaré plot in the order
they appeared as the orbit repeatedly cut through the x = 0 plane. This
would give us a series of coordinates (x1, y2), (x2, y2), · · · , (xn, yn). Think of
this in terms of a mapping operator T that maps the n’th point into the
n + 1’th point.
T(xn, yn) ≡ (xn+1, yn+1) (5.20)
In principle we could derive the exact mathematical form for this operator.
(I doubt that anyone has actually done this.) Certainly we could write
a computer program to do the mapping, and certainly we could derive a
linearized version of T that would be OK for small displacements. For my
purposes it will be enough to consider the general properties such operators
must have. The first of these (from which all others flow) is that they must
be area preserving.
Canonical transformations preserve the volume of phase space. This is
called Liouville’s theorem; it’s proved in most mechanics texts. For a onedegree
of freedom system, this is just preservation of area in the (p, q)-phase
plane. Thus for some area A, enclosed by a closed curve C, we can use
Stokes’ theorem to write
p dq = p dq (5.21)
C
where C ′ is the shape of the curve after it has been changed by some canonical
transformation including the passage of time which is itself a canonical
transformation. Another way to say the same thing is that if the (q, p) point
in phase space is transformed to (q ′ , p ′ ), then the Jacobian
∂(q ′ , p ′ )
∂(q, p)
C ′
= 1 (5.22)
These results can be extended to higher dimensions in a completely straightforward
manner.
So far, so good. There is a corollary to Liouville’s theorem that is not
so easy to prove. The transformations of the form (5.20) on the Poincaré

5.4. DISCRETE MAPS 69
J
3
2
1
0
−1
−2
−3
Standard Map, ε = 0.00
0 1 2 3 4 5 6
phi
Figure 5.14: The standard map with ϵ = 0.
section also preserve area in the sense of (5.22). 5 Discrete maps are area
preserving.
Let’s take time out for an example. The following transformation is
called the standard map, presumably because it appears in so many different
contexts. Thanks to the Jn+1 (rather than Jn) in the first of equation (5.23)
it is trivially area preserving.
ϕn+1 = (ϕn + Jn+1)mod2π (5.23)
Jn+1 = ϵ sin ϕn + Jn
This is a one-dimensional map written in terms of action-angle variables J
and ϕ. Not only ϕ, but also J is periodic with period 2π. We can imagine
all the orbits wrapped around a cylinder. In the case ϵ = 0, the (ϕn, Jn)’s lie
along parallel circles as shown in Figure 5.14. When ϵ is increased to 0.050
a new feature appears, a loop in the center of the plot. This is unusual in
the sense that it can be contracted to a point; it is topologically distinct
from all the ϵ = 0 circles. As ϵ is increased an assortment of smaller loops
appear together with a smattering of completely random points. Because of
limitations on plot resolution, computer time, and my patience you cannot
see the really significant thing about this plot: this pattern of islands of loopy
order interspersed with random dots persists at ever smaller and smaller
scales. They have a property called self similarity. In this sense they are
5 Tabor, Appendix 4.1

70 CHAPTER 5. INTRODUCTION TO CHAOS
J
3
2
1
0
−1
−2
−3
Standard Map, ε = 0.050
0 1 2 3 4 5 6
phi
Figure 5.15: The standard map with ϵ = 0.050.
similar to fractal patterns. Finally, as ϵ is increased further, all appearance
of order disappears and the dots become completely random. This is the
state of complete chaos.
5.5 Linearized Maps
Like the continuous transformations we studied in Section 5.2, discrete maps
have fixed points about which one can analyze the local topology. Consider
a generic mapping of the form 6
[ xi+1
yi+1
]
[
xi
= T
yi
]
(5.24)
A fixed point of the mapping would be a point where xi+1 = xi and yi+1 = yi.
I will argue later on that in a plot like Figure 5.16 there are an infinite
number of fixed points, but to keep the algebra simple here I will assume
that the fixed point is at the origin (0, 0). Linearizing T about this point
gives [ ] [ ] [ ]
δxi+1 T11 T12 δxi
=
(5.25)
δyi+1 T21 T22 δyi
where of course
Tij = ∂Ti
∂xj
xi,xj=0
6 I am using Tabor’s notation from section 4.3.4.
(5.26)

5.5. LINEARIZED MAPS 71
J
3
2
1
0
−1
−2
−3
Standard Map, ε = 0.750
0 1 2 3 4 5 6
phi
Figure 5.16: The standard map with ϵ = 0.750.
The eigenvalues λi of the Tij matrix must satisfy
λ 2 − λ(trace(T )) + det(T ) = 0 (5.27)
The all-important point here is that because of the area-preserving property
of T, det(T ) = 1. This greatly restricts the allowed types of fixed points.
There are only three cases to consider.
If |trace(T )| < 2, λ1 λ2 are a complex conjugate pair lying on the unit
circle, that is,
λ1 = e +iα , λ2 = e −iα
(5.28)
This is simply a rotation in the vicinity of the fixed point (0, 0). This corresponds
to a stable or elliptic point. Thus in the immediate neighborhood
of (0, 0) we expect to find invariant curves like Figure 5.7.
If |trace(T )| > 2, λ1 λ2 are real numbers satisfying
λ1 = 1/λ2
(5.29)
There are two subcases to consider here depending on whether λ is positive
or negative. If it is positive we have a regular hyperbolic fixed point in
which successive iterate stay on the same branch of the hyperbola as in
Figure 5.17 (a). If λ < 0 we have a hyperbolic-with-reflection fixed point
in which successive iterates jump backwards and forwards between opposite
branches of the hyperbola. (See Figure 5.17 (b).)

72 CHAPTER 5. INTRODUCTION TO CHAOS
Figure 5.17: (a) Hyperbolic fixed point. (b) Hyperbolic-with-reflection fixed
point.
5.6 Lyapunov Exponents
Loosely speaking, systems are chaotic because adjacent trajectories diverge
exponentially from one another. If this were literally true we could parameterize
this divergence with the function e λx , where λ is some constant and x
is the independent variable, which might be continuous or discrete depending
on the application. This is the basic idea behind Lyapunov exponentials,
a formalism with many alternate definitions (and spellings).
Let’s apply this idea first to a one-dimensional iterative map of the form
xi+1 = f(xi) (5.30)
We can characterize the divergence of two trajectories separated by ϵ upon
the n-th iteration as
|f(xn + ϵ) − f(xn)|
lim
=
df(xn)
ϵ→0 ϵ
dxn
(5.31)
A small but finite deviation at the n-th iteration, say δxn, should grow to
δxn+1 ≈
df(xn)
dxn
δxn
(5.32)
Continuing this reasoning
δxn+1
=
df(xn) df(xn−1)
δx0
dxn dxn−1
× · · · × df(x0)
dx0
(5.33)

5.6. LYAPUNOV EXPONENTS 73
=
n∏
i=0
|f ′ (xi)| = e λn
The last equality is just a hypothesis. λ will certainly depend on the point
n where we stop iterating. We should write instead
λ(n) = 1
n ln
n∏
|f ′ (xi)| (5.34)
with the understanding that the definition only makes sense if there is some
range of n over which λ(n) is more or less constant. λ defined in this way is
a Lyapunov exponent.
In the case of multidimensional mappings
i=0
xi+1 = F (xi) (5.35)
where x and F are n-dimensional vectors, there will be a set of n characteristic
exponents corresponding to the n eigenvalues of the linearized map
(??). Introducing the eigenvalues λi(N), i = 1, . . . , n, of the matrix
(LM)N = (T (xN)T (xN−1) · · · T (x1)) 1/N
(5.36)
where T (xi) is the linearization of F at the point xi, the exponents are
defined as
σi(N) = ln |λi(N)| (5.37)
Since the T ’s have unit determinant for area-preserving maps, it is clear that
the sum of the exponents must be zero.
For the final example, suppose the equation of motion is
˙x = f(x) (5.38)
Let s(t) = x(t) − x0(t) be the difference between two near-by trajectories.
If this does indeed diverge exponentially with time, then ˙s = λs. Then we
can argue that
˙s = ˙x − ˙x0 = f(x) − f(x0) = λs = λ(x − x0) (5.39)
λ =
f(x) − f(x0)
x − x0
≈ df
dx
x0
(5.40)

74 CHAPTER 5. INTRODUCTION TO CHAOS
5.7 The Poincaré-Birkhoff Theorem
The phase-space trajectories of integrable systems move on smooth tori. The
appearance of the Poincaré section depends on whether the winding number
is rational or irrational. If it is rational the section shows discrete points. If
irrational, the points are ‘ergodic’ and form a continuous loop. Under the
influence of nonlinear perturbations the tori become distorted, then break
up into smaller tori, and finally disintegrate into chaos. It turns out that
the way this happens depends on whether the winding number is rational or
irrational. If it is irrational the tori are preserved, distorted but preserved,
under small perturbations. This is a gross oversimplification of the KAM
theorem, which I will discuss in the next Section 5.8. If the winding number
is rational, the tori break up in a way that is governed by the so-called
Poincaré-Birkoff theorem, the subject of this section. This may seem like
a swindle, since every irrational number can be approximated to arbitrary
accuracy by a rational number. But, as it turns out, some numbers are more
irrational than others!
I will prove the PB theorem for the standard map equation (5.23) but it
is true under quite general assumptions. I will use the symbol Tϵ for (5.23),
i.e.
Tϵ(ϕn, Jn) = (ϕn+1, Jn+1).
Now imagine the points in Figure 5.14 (ϵ = 0) plotted in polar coordinates
(for positive J) with ϕ the angular and J the radial coordinate. The points
now lie on concentric circles of constant J. Choose J ≡ Jr = 2πj/k, with
k and j integers, i.e. Jr has a rational winding number. If we iterate T0 k
times, J remains unchanged and ϕ is incremented by k factors of 2π, which
is to say, ϕ is not changed at all. Symbolically
T k 0(ϕ, Jr) = (ϕ, Jr)
Now take a J+ slightly larger than Jr. Tk 0 will increment ϕ by slightly more
than 2πk so ϕ will increase. In the same way if J− < Jr, Tk 0 will cause ϕ to
decrease. We can imagine the values of ϕ lying on three circles J+, Jr, and
J− as shown in Figure 5.18(a).
Now turn on a small perturbation ϵ > 0. Tk ϵ will map some ϕ’s to larger
values and some to smaller, but there will be some locus of points, called C
in Figure 5.19 which are not changed at all. In other words, the curve C is
mapped purely radially.
T k ϵ (Jr, ϕ) = (Jc, ϕ)

5.7. THE POINCARÉ-BIRKHOFF THEOREM 75
Figure 5.18: (a) Three orbits of the unperturbed standard map Tk 0 . (b) The
ϕ coordinate is left invariant on C by the perturbed map Tk ϵ .
Curve C is mapped into a new curve called D in Figure 5.18(b).
T k ϵ (Jc, ϕ) = (Jd, ϕ)
The curves C and D must have the same area (remember these are areapreserving
transformations) so they must cross one another an even number
of times. This situation is shown in Figure 5.19. The crossings represent
points that are invariant under T k ϵ – they are fixed points.
This is our first result. A torus with rational winding number j/k is
invariant under T k 0 , i.e. every point on the torus is a fixed point of Tk 0 .
When ϵ is even slightly larger than zero, only a discrete (even) number of
fixed points of T k ϵ survive. You can ascertain the type of fixed points by
seeing how other points in their immediate vicinity are mapped. Compare
this flow as it’s called with the arrows in Figures 5.4 and 5.17. You should be
able to convince yourself that the points along the curve C are alternatively
hyperbolic and elliptic. Figure 5.20 should help you visualize this. Since
there are an even number of fixed points, half of them will be elliptic and
half hyperbolic. How many are there? Suppose (ϕ0, J0) is a fixed point of
T k ϵ . We can create more fixed points by multiplying by Tϵ as the following

76 CHAPTER 5. INTRODUCTION TO CHAOS
Figure 5.19: The curves C and D. Crossings, like a and b, are fixed points.
Figure 5.20: A closer look at the fixed points a and b.

5.8. ALL IN A TANGLE 77
simple argument shows.
T k ϵ [Tϵ(ϕ0, J0)] = TϵT k ϵ (ϕ0, J0) = Tϵ(ϕ0, J0)
Starting with (ϕ0, J0) we can create k − 1 additional fixed points by multiplying
repeatedly with Tϵ. To put it another way, every fixed point of Tk ϵ is a
member of a family of k fixed points obtained by multiplying by various powers
of Tϵ Because each mapping is a continuous function of ϕ and J, all the
members of an elliptic family are elliptic and all the members of a hyperbolic
family are hyperbolic. I claim that all the members of a family are distinct.
Proof: Let (ϕs, Js) be the fixed point obtained by Ts ϵ(ϕ0, J0) = (ϕs, Js) with
s < k. Then of course all such points are fixed points of Tk ϵ . The claim is
that there is no m < k such that Tm ϵ (ϕs, Js) = (ϕs, Js). Multiply both sides
of this equation with T−s ϵ . The result is Tm ϵ (ϕ0, J0) = (ϕ0, J0) It is just this
equation with m replaced by k that defines (ϕ0, J0). Hence m = k. Finally
note that none of these newly created fixed points can lie along the original
curve C. If there were there would be instances in which two hyperbolic or
two elliptic points appeared side by side. This we know to be impossible.
Consequently each torus breaks up into k fixed points for every fixed point
on the curve C. This is the Poincaré-Birkoff theorem.
5.8 All in a tangle
Have another look at the hyperbolic fixed points in Figures 5.4 and 5.17.
There are always two loci of points leading directly toward the fixed point
and two loci leading away from it. These are called the stable and unstable
manifolds respectively. Following the notation of Hand and Finch I will call
them H+ and H−. Call the fixed point pf . Any point along H+ will be
mapped asymptotically back to pf under repeated applications of Tϵ, and
any point on H− will be mapped asymptotically back to pf under repeated
applications of T −1
ϵ . Can these manifolds cross one another? I claim the
following.
• H+ and H− cannot intersect themselves, but they can and do intersect
one another.
• Stable manifolds of different fixed points cannot intersect one another.
• Unstable manifolds of different fixed points cannot intersect one another.

78 CHAPTER 5. INTRODUCTION TO CHAOS
• Stable manifolds can intersect with unstable ones. The stable and
unstable manifolds of a single fixed point intersect in what are called
homoclinic points and those of two different fixed points, in heteroclinic
points.
• Neither H+ nor H− can cross the tori surrounding elliptic fixed points.
• There are, depending on the size of ϵ, narrow bands surrounding tori
with irrational winding number that are not broken up into isolated
fixed points. This is the content of the KAM theorem to be discussed
in the next section. Neither H+ nor H− can cross these bands.
The proofs of these assertions are easy and are given in Finch and Hand.
Referring to Figure 5.21(a), x0 is a heteroclinic point that lies on the unstable
manifold H− of pf1 and the stable manifold H+ of pf2. Since both manifolds
are invariant under Tϵ, the Tk ϵ x0 are a set of discrete points that lie on both
manifolds, so the two manifolds must therefore intersect again. For instance,
because x1 = Tϵx0 is on both manifolds, H− must loop around to meet H+.
Similarly the xk = Tk ϵ must lie on both manifolds, so H− must loop around
over and over again as illustrated in Figure 5.21(b). The inverse map also
leaves H+ and H+ invariant and hence the x−k = T−k ϵ x0 are intersections
that force H+ to loop around to meet H−. As k increases and xk approaches
one of the fixed points, the spacing between the intersections gets smaller,
so the loops they create get narrower. But because Tϵ is area-preserving,
the loop areas are the same, so the loops get longer, which leads to many
intersections among them, as shown in Figure 5.21(c) and (d).
Try explaining all this to an intimate friend on a date. The more you
explain the more you will see that this mechanism produces a tangle of
fathomless complexity. 7 Nonetheless the mess is contained, at least for small
ϵ. Since stable manifolds cannot cross, the stable manifold emanating from
pf1 acts as a barrier to the stable manifold emanating from pf2. The same is
true of the unstable manifolds. The tangle also cannot cross the stable torri
surrounding the elliptic fixed points nor can they cross the KAM tori. As
a consequence we expect to see islands of chaos developing between stable
ellipses. This is clear in Figures 5.15 and 5.16. As ϵ increases, the KAM tori
also break down and chaos engulfs the entire plot.
7 Don’t try to explain this for higher-dimensional spaces. That way lies madness.

5.8. ALL IN A TANGLE 79
Figure 5.21: A hereroclinic intersection. (a) Two hyperbolic fixed points pf1
and pf2, and an intersection x0 of the unstable manifold of pf1 with the stable
manifold of pf2. (b) Adding the forward maps T k x0 of the intersection.
(c) Adding the backward maps T −1 x0 of the intersection. (d) Adding another
intersection x ′ and some of its backward maps. U – unstable manifold;
S – stable manifold.

80 CHAPTER 5. INTRODUCTION TO CHAOS
5.9 The KAM theorem and its consequences
For a system with n independent degrees of freedom to be integable, it is
a necessary and sufficient condition that n independent constants of the
motion exist. In this case the system can be transformed into a set of
action-angle variables
In this notation
ω0 ≡ (ω01, ω02, · · · , ω0n) I0 ≡ (I01, I02, · · · , I0n) (5.41)
ω0 = ∂H0
∂I0
Now suppose the system is perturbed slightly
(5.42)
H(ω0, I0, ϵ) = H0(I0) + ϵH1(ω0, I0) (5.43)
where I0 and ω0 are the AA variables of H0. According to our perturbation
formalism from Chapter 4, there are two series that must converge (4.25)
and (4.26) repeated here for convenience.
where k is a vector of integers 8
and
˜H1(I, ψ) = ∑
Ak(I)e ik·ψ
k
F1(I, ψ) = ∑
Bk(I)e ik·ψ
k
k = k1, · · · , kn
Bk = i Ak
ω0 · k
(5.44)
(5.45)
(5.46)
The rate of decrease of the |Bk| depends both on the |Ak| and the denominators
|ω0·k|, so even if the |Ak| decrease fast enough for (5.44) to converge,
(5.45) will not converge if the |ω0 · k| decrease too rapidly.
The situation seems hopeless. If any of the ω0’s yields a rational winding
number, the series will blow up immediately, and if one is working with finite
precision – on a computer for example – every number is a rational number.
And yet we have seem from our computer models that some stable periodic
8 The sum over k means the sum over all possible combinations of the n integers
k1, · · · , kn.

5.9. THE KAM THEOREM AND ITS CONSEQUENCES 81
trajectories persist even under the influence of small perturbations. The circumstances
under which this happens is spelled out in a remarkable theorem
first outlined by Kolmogorov and later proved independently by Arnold and
Moser. The theorem is extremely difficult and sophisticated although Tabor
has a nice explanation of the basic ideas, and an understandable outline of
the proof is given in Classical Dynamics by José and Saletan. I will explain
the theorem as carefully as I can and let it go at that.
5.9.1 Two Conditions
The KAM theorem claims that in regions of phase space where certain conditions
hold, the perturbation series converges to all order in ϵ. The first
condition involves the Hessian matrix.
∂ω0α
det
≡ det
∂
2
H0
̸= 0. (5.47)
∂I0β
∂I0α∂I0β
The content of this statement is as follows: We assume that each torus has
a unique frequency associated with it. Thus if we knew all the ω0’s we could
calculate all the I0’s and vice versa. Equation (5.47) ensures that this is
true. A simple (albeit artificial) example is provided by José and Saletan.
Consider the one degree of freedom Hamiltonian
H = I 3 /3 (5.48)
in which I takes on values in the interval −1 < I < +1. The above condition
requires that
d2H = 2aI ̸= 0 (5.49)
dI2 Why is this significant? Note that ω(I) = dH/dI = aJ 2 . Inverting this gives
I(ω) = ± √ ω/a. The ± is a sign that the inversion is not unique. There
are two regions separated by I = 0. In the region 0 < I ≤ 1 I = √ ω/a. In
the region −1 ≤ I < 0, I = − √ ω/a. Thus there are two “good” regions
separated by a barrier.
There is a second condition restricting the frequencies. Of course we
are only considering frequencies with irrational winding numbers. Even if
the frequencies are incommensurate, |ω0 · k| could be arbitrarily small. The
KAM theorem requires that it be bounded from below by the so-called “weak
diophantine condition”
|ω0 · k| ≥ γ|k| −κ for all integer k (5.50)

82 CHAPTER 5. INTRODUCTION TO CHAOS
where k = √ k · k and γ and κ > n are positive constants.
What is the significance of this strange inequality? The best way to
understand it, I think, is to face up to the paradox I mentioned earlier that
the series can only converge for irrational winding numbers, and yet it seems
that every irrational number is “arbitrarily close” to a rational number. It is
this last statement that needs to be examined more carefully. This requires
a brief excursion into number theory. Consider the unit interval [0, 1]. The
rationals have measure zero in the interval. That means roughly that they
don’t take up any space. This can be proved as follows. First put the
rationals in a one-to-one correspondence with the integers. Construct a
small open interval of length ϵ < 1 about the first rational, and one of
length ϵ 2 about the second and so forth. The sum of all these little intervals
(this is a geometric series) is σ = ϵ/(1 − ϵ), which can be made arbitrarily
small by choosing ϵ small enough. Thus the space occupied by the rationals
is less than any positive number. This requires taking the limit ϵ → 0. The
paradoxical thing is that it is possible to remove a finite interval around each
rational without deleting all of [0, 1]. This can be seen as follows. Write each
rational in [0, 1] in its lowest form as p/q, and about each one construct an
interval of length 1/q 3 . For each q there are at most q − 1 rationals. Thus
for a given q no more than (q − 1)/q 3 is covered by the intervals, and the
total length Q that is covered is less (because of overlaps) than the sum of
these intervals over all q.
Q <
∞∑
q=2
q − 1
<
q3 ∞∑
q=2
1
q 3
(5.51)
This sum is related to the Riemann zeta function. At any rate Q < 0.645.
We can make this number as small as we like by replacing 1/q 3 with Γ/q 3
where Γ < 1. Even if we leave Γ = 1, the fraction of [0, 1] covered by the
finite intervals is less than 1.
Now we can divide the irrationals into two sets, those covered by the
intervals around the rationals and those outside the intervals. Those uncov-
ered satisfy the condition
ω − p
q
Several comments are in order regarding this inequality.
Γ
≥ . (5.52)
q3 • Equation (5.52) makes irrationality a quantitative concept. 9 Those
irrationals that satisfy (5.52) are “more irrational” than those that
9 This can also be quantified in terms of continued fraction expansions.

5.10. CONCLUSION 83
don’t, and the extent of their irrationality can be quantified by that
value of Γ for which they just do or do not satisfy the inequality.
• Equation (5.50) is just an n-dimensional version of (5.52). The constants
γ and κ characterize the degree of irrationality of ω in the same
way that Γ and the exponent 3 characterize the irrationality of ω in
(5.52).
• The uncovered irrationals occupy isolated “islands” between the covered
intervals. We expect that as the perturbation parameter ϵ is
increased, those tori with less irrational winding numbers will be destroyed
first, but islands of stability will remain between them. Eventually
as the perturbation is increased, all will be swept away in chaos.
• The KAM theorem gives us no clue how to calculate the appropriate
values of γ and κ or the values of ϵ for which chaos will set in. Some
estimates placed the critical value of ϵ to be something around 10 −50 !
If this were true, of course, the theorem would be quite pointless.
Numerical test with specific models have found critical values of ϵ as
large as ϵc ≈ 1. I will close with a quote from José and Saletan, “To
our knowledge a rigorous formal estimate of a realistic critical value
for ϵ remains an open question.”
5.10 Conclusion
This is the end of our story about chaos. Remember that we have only
dealt with bounded, conservative systems with time-independent Hamiltonians.
(Classical mechanics is a big subject.) Systems with one degree of
freedom are trivial (in principle) to solve using the method of quadratures.
Systems with n degrees of freedom are trivial (again in principle) if they have
n constants of motion. Such a system can be reduced by using action-angle
variables to an ensemble of uncoupled oscillators. These systems are said to
be integrable and they do not display chaos. The trouble comes when we
introduce some non-integrability as a perturbation. Perturbation theory is
straightforward with one degree of freedom, but with two or more degrees
of freedom comes the notorious problem of small denominators. 10 Perturbation
theory fails immediately for all periodic trajectories with rational
10 There are other ways of doing perturbation theory in addition to the one described
here. They all suffer the same problem.

84 CHAPTER 5. INTRODUCTION TO CHAOS
winding number. According to the Poincaré-Birkoff theorem, these trajectories
on the Poincaré section break up into complicated whorls and tangles
surrounded be regions of stability corresponding to irrational winding numbers.
According to the KAM theorem these regions break down with those
with “more irrational” winding numbers surviving those with less. At last
“Universal darkness covers all,” and the trajectories though deterministic
show no order or pattern.