1.1 PRINCIPLE OF LEAST ACTION 640-213 MelatosChapter 1.docx

1.1 PRINCIPLE OF LEAST ACTION 640-213 Melatos
Chapter 1
A New Perspective on F = ma
Reference: Feynman, Lectures on Physics, Volume 2, Chapter
19
1.1 Principle of Least Action
Throw a tennis ball (mass m) straight upwards. How does it
“know” where to move?
Traditional perspective (local):
at every instant t, the ball accelerates in response to the net
force F(t) acting at
time t only, not at some earlier or later time; we update its
velocity according
to v(t + Δt) = v(t) + a(t)Δt and its position according to x(t +
Δt) =
x(t) + v(t)Δt + a(t)(Δt)2/2, with a(t) = F(t)/m.
New perspective (global):
“try” lots of different paths and choose the one which
extremises some “virtue”
What is this “virtue”? It is different for different natural laws.
For example:

• propagation of light: minimise (usually) the travel time
between two points
• Maxwell’s laws: minimise the difference between the electric
and magnetic energies
• Einstein’s theory of gravity: minimise a messy function of the
space-time curvature
1
1.1 PRINCIPLE OF LEAST ACTION 640-213 Melatos
For our tennis ball:
compute the kinetic energy at each point on the path, subtract
the potential
energy at each point, and integrate over time along the whole
path; the result
is smallest for the true path, smaller than for any other trial path
between
the same two endpoints
S =
∫ t2
t1
dt
[
1
2
m

∣ ∣ ∣ ∣ dx(t)dt
∣ ∣ ∣ ∣
2
− mgx(t)
]
(1.1)
S is called the action. (Units: J s.)
Exercise. What other famous physical constant has units of J s?
The integrand [...] in (1.1) is called the Lagrangian L. The law
is called the Principle
of Least Action (POLA). Because the potential energy increases
with altitude, the ball
“wants” to reach a high altitude as quickly as possible, in order
to minimise S in the fixed
time available (t2 − t1). But if the ball goes too fast (which it
must do if it is to rise very
high in the fixed time available), then the kinetic energy gets
too big and outweighs the
benefits of the potential energy.
The true path is the best compromise which minimises S.
Exercise. Explain why a free particle (i.e., zero potential
energy) travels at
uniform speed.
If this new perspective is to prove useful, it should provide a
neat way to solve mathe-
matically for the path x(t), as a substitute for integrating F = mẍ
directly. And it does!
We will see, in the first half of the course, how certain

quantities derived easily from L
2
1.2 OPTIMAL PATH BY THE CALCULUS OF VARIATIONS
640-213 Melatos
are conserved, greatly simplifying the task of solving for x(t).
The conserved quantities
correspond to symmetries in the physics of the problem, e.g.,
rotational symmetry ⇔
conservation of angular momentum. Sometimes the symmetries
are “hidden” and would
have been hard to guess without the new, Lagrangian
perspective. In these situations,
the beauty and utility of Lagrangian mechanics are
simultaneously on full display.
1.2 Optimal path by the calculus of variations
Before learning how to solve for the path, we should check that
our new principle gives
the same answers at Newton’s Second Law! Luckily, it does.
Return to our tennis ball. Suppose that we denote its true path
by x0(t). Now pretend
that we have another path x0(t)+x1(t), which is very close to the
true path, with |x1(t)| �
|x0(t)| at all t. We make sure that the two paths touch at t1 and
t2, ie., x1(t1) = 0 = x1(t2).
Otherwise, the POLA would have no meaning; you could get
any S you wanted by picking
different endpoints, and the Principle would be useless.

Since Strue is a minimum for x0(t), then it must remain almost
unchanged for x0(t)+x1(t),
viz.
Strial[x0(t) + x1(t)] = Strue[x0(t)] + terms of order |x1(t)|2
(1.2)
This is exactly analogous with any ordinary, garden variety
function of a single variable,
f(x), which looks “bowl-shaped” near a minimum x0, where f
′(x0) = 0; viz.
f(x0 + x1) = f(x0) +
1
2
f ′′(x0)x1
2 + ... (1.3)
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1.2 OPTIMAL PATH BY THE CALCULUS OF VARIATIONS
640-213 Melatos
Substituting the trial path in (1.1), we find
Strial =
∫ t2
t1
dt

{
1
2
m
[
dx0(t)
dt
]2
+ m
dx0(t)
dt
dx1(t)
dt
+
1
2
m
[
dx1(t)
dt
]2
− mgx0(t) − mgx1(t)

}
(1.4)
= Strue +
∫ t2
t1
dt
{
m
dx0(t)
dt
dx1(t)
dt
− mgx1(t)
}
(1.5)
using the fact that the (dx1/dt)
2 term is much smaller than all the others because it is
two small quantities multiplied together, and can therefore be
discarded.
Now we play a little trick which is the centrepiece of all
problems in the calculus of
variations: we integrate the first term in (1.5) by parts.
Strial = Strue +

[
m
dx0(t)
dt
x1(t)
]t=t2
t=t1
+
∫ t2
t1
dt
{
−md
2x0(t)
dt2
x1(t) − mgx1(t)
}
(1.6)
The “surface term” [...] evaluated at t1 and t2 is zero because
we choose x1(t1) = 0 = x1(t2)
at the outset.
The integral in (1.6), which is proportional to x1(t), must also
be zero because we already
said that Strial differs from Strue by terms of order |x1(t)|2 or

higher. Moreover, the integral
must vanish for any x1(t) you might choose (as long as it is
small). This is possible only
if
−md
2x0(t)
dt2
− mg = 0 (1.7)
i.e., the true path satisfies Newton’s Second Law!
Important philosophical point: the POLA is equivalent to
Newton’s Second Law. It is
no more or less “fundamental”; one law can be derived from the
other, even though the
former is global whereas the latter is local. (Think about what
happens if you apply the
4
1.4 [EXT] QUANTUM MECHANICS 640-213 Melatos
POLA to infinitesimal subsections of whole path.) Even more
important, neither principle
can be derived from something “deeper”, as far as we know.
They are laws of nature,
deduced from experimental enquiry. In another universe, if
there is such a thing, different
mechanical laws may apply.
1.3 [Ext] Notes

1. Can generalise to relativistic motion.
S = −mc2
∫ t2
t1
dt
γ
(1.8)
Here S does not equal
∫ t2
t1
dt (KE − PE). (Rather, it equals the integrated proper
time along the path.)
2. Can generalise to many particles with positions qi and
velocities q̇ i (i = 1, ...,N).
(We do this in Chapter 2.)
3. Can generalise to infinite degrees of freedom (N → ∞),
whereupon the qi’s and q̇ i’s
become fields and their gradients.
e.g., electromagnetism: Lagrangian replaced by Lagrangian
density
L = −ρΦ + J · A + 1
2
�0
(

−∇ Φ − ∂A
∂t
)2
− 1
2μ0
(curlA)2 (1.9)
Action is integral of L over all space and time.
S =
∫
dtd3x L(x, t) (1.10)
Minimising S with respect to Φ and A gives Maxwell’s laws!
(Try to prove this if
you are feeling brave! It is not easy; you may need to wait until
next year when you
know some more advanced calculus.)
1.4 [Ext] Quantum Mechanics
We know that particles don’t really follow classical trajectories.
There is some mutual
uncertainty in their positions and momenta, according to the
Heisenberg Uncertainty
Principle, because really particles are matter waves.
In quantum mechanics, there is some probability that a particle
starting at point P at
time t1 arrives at point Q at times t2. You can find this
probability by solving what is
known as the time-dependent Schrödinger equation for the

particle wave function ψ(x, t).
5
1.4 [EXT] QUANTUM MECHANICS 640-213 Melatos
The probability is then |ψ(x, t)|2. (You will learn about this in
your introductory QM
course. It is the standard way to do things.)
Alternatively, Feynman showed that you can get the probability
by summing up the
“arrival phases” corresponding to all possible classical paths P
→ Q:
probability ∝
∣ ∣ ∣ ∣ ∣
∑
paths
eiS/�
∣ ∣ ∣ ∣ ∣
2
(1.11)
S is different for every path. As � is so tiny, even small
differences ΔS in the action
between paths very close together produce huge differences
ΔS/� in the phase angle. So
eiS/� is effectively a “random” vector on a circle and sums up
(“cancels”) to zero. This is
familiar from optics: destructive interference of waves.

Exercise. Estimate S/� for the true path of a tennis ball thrown
vertically
upwards at 5 m s−1 on Earth’s surface.
The only paths for which cancellation does not occur are those
near the true classical
path, which minimises S. As we have already seen, Strial =
Strue + (2nd order quantities)
near the true path, i.e., S is very flat in the vicinity of Strue.
Hence ΔS is very very small
near the true path and there is much less cancellation.
The above picture is called the “path integral formulation” of
QM. It is a powerful tool in
quantum field theory. It is also a promising approach to the
quest to put the fundamentals
of QM, like the Schrödinger’s cat paradox, entanglement, the
Einstein-Podolsky-Rosen
paradox, etc., on a rigorous footing.
6
2.2 GENERALISED COORDINATES 640-213 Melatos
Chapter 2
Lagrange’s Equations
2.1 Phase space
Newton’s Second Law for a single particle, F = mẍ, is a
package of three second-order
ordinary differential equations. The resulting motion is

therefore described by six func-
tions: x(t) and ẋ(t). [In Cartesian coordinates, these are the
vector components x(t),
y(t), z(t), ẋ(t), ẏ(t), ż(t). An overdot denotes a total time
derivative d/dt.] To solve for
the motion, six initial conditions are needed: x(0) and ẋ(0).
Another way to say this: the motion is a curve in a six
dimensional phase space.
dx
dt
= v (2.1)
dv
dt
=
F
m
(2.2)
Extension note: the same is true in special relativity, except it is
easier to work in terms
of p(t) (momentum) rather than ẋ(t) (velocity):
dx
dt
=
p

m
(
1 +
p2
m2c2
)− 1
2
(2.3)
dp
dt
= F (2.4)
2.2 Generalised coordinates
Consider now N particles, whose motion is described by 3N
Cartesian coordinates xs(t)
and 3N velocities ẋs(t), 1 ≤ s ≤ N.
7
In many systems, some or all of the particles are constrained in
some way. For example:
• motion on a plane ⇒ one less coordinate needed per particle
• motion on a curve ⇒ two less coordinates per particle

• motion inside a volume ⇒ inequality
• rolling ⇒ velocity constraint at the point of contact
In the first two examples, the constraint can be expressed as an
equation relating the
xs(t)’s. Such constraints are holonomic.
e.g., a ball restricted to moving on a flat table in three
dimensions has
ax(t) + by(t) + cz(t) = d (2.5)
where the vector (a,b,c) is normal to the table. (Remember your
first year maths classes,
where you learnt about the equation of a plane.)
In the second two examples, the constraint cannot be expressed
as a closed-form equation.
Such constraints are anholonomic. They are tougher to handle.
Exercise. [Ext] Work through the example of the rolling disk in
Goldstein
et. al., Classical Mechanics, Third Edition, Section 1.3
Clearly, if there are K constraints, the number of truly
independent coordinates (which
cannot be eliminated in terms of the others) is n = 3N −K. Each
independent coordinate
has an associated velocity.
e.g., simple pendulum swinging in a plane
N = 1 (one particle)
K = 2 (confined to plane, attached to string)

⇒ one independent coordinate (call it x)
8
2.3 GENERALISED VELOCITIES AND MOMENTA 640-213
Melatos
Now let us transform from the original 3N Cartesian coordinates
xs(t) to a new set of
n = 3N −K independent generalised coordinates, q1(t), ...,qn(t).
By the above reasoning,
the former are expressible in terms of the latter.
xs(t) = xs(q1(t), ...,qn(t), t) (2.6)
Each generalised coordinate qi(t) is associated with a
(“conjugate”) generalised velocity
q̇ i(t) = dqi(t)/dt. Obviously, you can choose the qi’s in an
infinite number of perfectly
valid ways, although some may be more useful than others.
Vital point: qi(t) and q̇ i(t) are explicit functions of t. In what
follows, however, they
are also treated as independent variables in certain contexts.
E.g., the Lagrangian L
(which you met briefly in Lecture 1) is a function of the 2n + 1
independent variables
t,q1, ...,qn, q̇ 1, ..., q̇ n. One consequence is that expressions like
∂qi/∂qj and ∂q̇ i/∂qj actually
make mathematical sense; the former equals δij and the latter
equals zero.
1 Likewise, it

would be totally wrong to write ∂qi/∂qj =
dqi
dt
/
dqj
dt
= q̇ i/q̇ j . You will see how this works
below.
[If you are curious about the geometry underlying this
mathematics, called a jet bundle
space, please consult V. I. Arnold, Mathematical Methods of
Classical Mechanics, Springer
(1978). Jet bundles lie outside the scope of this subject.]
2.3 Generalised velocities and momenta
Velocities: use the chain rule:
ẋs(t) =
∂xs
∂t
+
n∑
j=1
q̇ j
∂xs
∂qj
(2.7)

From here onwards, we adopt the Einstein summation
convention: repeated indices denote
a sum over all values of that index. For example, it is neater to
drop the summation sign
and write q̇ j∂xs/∂qj as shorthand for q̇ 1∂xs/∂q1 + ... +
q̇ n∂xs/∂qn in (2.7) above.
Please note: ∂xs/∂t and ∂xs/∂qi are functions of t and the qi’s
because of how we defined
xs, but not functions of the q̇ i’s. On the other hand, q̇ i is a
function of t only, not of the
qi’s (see “vital point” made earlier).
2
1The symbol δij is called the Kronecker delta; it equals one if i
= j and zero if i = j.
2Please note also: there is a huge difference between d/dt and
∂/∂t. To illustrate why, consider the
following (meaningless for now) function of time:
F (t) = cos ωt +
1
2
mẋ(t)2 − 1
2
kx(t)2
d/dt is the total time derivative. It tells you how F (t) changes
with t due to all appearances of t, i.e.,
ωt, ẋ(t), and x(t).
dF

dt
= −ω sin ωt + mẋ(t)ẍ(t) − kx(t)ẋ(t)
9
2.4 DERIVATION OF LAGRANGE’S EQUATIONS 640-213
Melatos
Kinetic energy: function of t, qi, q̇ i
T(t) =
N∑
s=1
1
2
msẋs(t) · ẋs(t) (2.8)
Generalised momentum:
pi(t) =
∂T(t)
∂q̇ i
(2.9)
=
N∑
s=1
msẋs ·

∂ẋs
∂q̇ i
(2.10)
=
N∑
s=1
msẋs ·
∂xs
∂qi
(2.11)
where the last line comes from differentiating (2.7) with respect
to q̇ i. As a mnemonic
for the formula, think T = 1
2
mv2 ⇒ ∂T/∂v = mv = p; but beware, it is not always that
simple. Indeed, pi(t) may not be an actual physical momentum
at all!
2.4 Derivation of Lagrange’s equations
With these definitions, we can now derive Lagrange’s
equations, which tell us how the
generalised momentum changes with time (just like Newton’s
laws tell us how the physical
momentum changes with time).
dpi
dt

=
N∑
s=1
msẋs ·
d
dt
(
∂xs
∂qi
)
+ msẍs ·
∂xs
∂qi
(2.12)
=
N∑
s=1
msẋs ·
(
∂
∂t
+ q̇ j
∂

∂qj
)
∂xs
∂qi
+ msẍs ·
∂xs
∂qi
(2.13)
=
N∑
s=1
msẋs ·
∂
∂qi
(
∂
∂t
+ q̇ j
∂
∂qj
)
xs + msẍs ·

∂xs
∂qi
(2.14)
=
N∑
s=1
msẋs ·
∂ẋs
∂qi
+ msẍs ·
∂xs
∂qi
(2.15)
=
∂
∂qi
(
N∑
s=1
1
2
msẋs · ẋs
)

+ msẍs ·
∂xs
∂qi
(2.16)
∂/∂t is the partial time derivative. It tells you how F (t) changes
with t due to explicit t appearances.
∂F
∂t
= −ω sin ωt
10
2.5 CONSERVATIVE FORCES 640-213 Melatos
To get from (2.12) to (2.13), and from (2.14) to (2.15), note that
d
dt
= ∂
∂t
+ q̇ j
∂
∂qj
by the
chain rule when it operates on ∂xs/∂qi and xs, both of which
depend only on t and the

qi’s (from the definition of xs). To get from (2.13) to (2.14),
note that ∂/∂t commutes
with ∂/∂qi (independent variables) and ∂
2/∂qi∂qj = ∂
2/∂qj∂qi (mixed partial derivatives
swap).
Letting Fs = msẍs denote the net force on the sth particle (the
first and only place
where Newton is introduced), we arrive at Lagrange’s equations
(there are n of them,
i = 1, ...,n):
d
dt
(
∂T
∂q̇ i
)
− ∂T
∂qi
=
N∑
s=1
Fs ·
∂xs
∂qi

(2.17)
RHS = generalised force Qi. Even for a single particle, Qi
usually isn’t the same as the
physical Fs. Rather, it is Fs weighted by the dependence of the
particle’s position on the
ith generalised coordinate.
e.g., if xs is independent of qk, Qk will not contain any
information about Fs.
e.g., if qi is an angular coordinate, Qi doesn’t even have the
dimensions of a force! (What
dimensions does it have instead?)
Please don’t worry if you find it hard to visualise Qi physically;
we rarely work with it
directly.
Exercise. (a) Show that normal reaction forces don’t contribute
to Qi. (b)
Show that tension forces in inextensible linear connectors
between particles
do not contribute to Qi.
2.5 Conservative forces
A force like gravity, or the tension in a spring, which can be
written as the gradient of a
potential V (x, t), is called conservative.
F(x, t) = −∇ V (x, t) (2.18)
In Cartesian coordinates, ∇ = (∂/∂x,∂/∂y,∂/∂z). If a surface of
constant V looks like a
hill, F points downhill (towards lower V ) at every point.

V = −
∫
dl · F (2.19)
Independent of integral end points. [Necessary and sufficient
condition: curl F = 0]
Legrange’s equations take a simple form for a conservative
force:
Fs = F(xs, t) = −
∂V
∂xs
(2.20)
Qi = −
∂V
∂xs
· ∂xs
∂qi
= −∂V
∂qi
(chain rule) (2.21)
11
2.5 CONSERVATIVE FORCES 640-213 Melatos
As long as V is independent of the velocities ẋs (or equivalently

q̇ i)
3 we get
d
dt
(
∂L
∂q̇ i
)
− ∂L
∂qi
= 0 (2.22)
with
L = T − V. (2.23)
You will recognise L from Chapter 1! It is the Lagrangian,
whose time integral over the
path (action) is always an extremum.
Note: Our derivations made no special demands on the
generalised coordinates; they can
be anything (as long as they do not involve the velocities). So
Lagrange’s equations hold
true in the same form in any coordinate system; you can write
them straight down by
inspection without transforming from Cartesian or anything
else.
3Not always true. For example, the electromagnetic force qE +
qv × B does depend on velocity.

Luckily, Lagrangian methods can be tweaked to apply to it; you
will learn about this in later years.
12
3.2 FIRST INTEGRALS 640-213 Melatos
Chapter 3
Solving Lagrange’s Equations
3.1 Example: simple harmonic oscillator
Consider a body of mass m attached to a spring with spring
constant k, with one end
attached at the origin. Let x be the displacement of the body
from the origin.
L =
1
2
mẋ2 − 1
2
kx2 (3.1)
∂L
∂ẋ
= mẋ (3.2)
∂L

∂x
= −kx (3.3)
Lagrange’s equations:
d
dt
(mẋ) + kx = 0 (3.4)
i.e., mẍ = −kx (3.5)
This is Newton’s Second Law (F = −kx is Hooke’s law for a
spring).
Now suppose the body can move on the x-y plane, but the
spring only pulls in the x
direction (an artificial situation, but never mind).
L =
1
2
mẋ2 +
1
2
mẏ2 − 1
2
kx2 (3.6)
∂L
∂ẋ
= mẋ (3.7)

∂L
∂x
= −kx (3.8)
∂L
∂ẏ
= mẏ (3.9)
∂L
∂y
= 0 (3.10)
Two Lagrange equations. First is the same as (3.4). The second
reads d
dt
(mẏ) = 0, i.e., y
component of momentum is conserved.
3.2 First integrals
We look for constants of the motion, i.e., functions F [q1(t),
...,qn(t), q̇ 1(t), ..., q̇ n(t), t] of
the generalised coordinates and velocities which are
independent of time when Lagrange’s
13
3.2 FIRST INTEGRALS 640-213 Melatos
equations

d
dt
(
∂L
∂q̇ i
)
− ∂L
∂qi
= 0 (3.11)
are satisfied. Here, independent of time means
0 =
dF
dt
(3.12)
=
∂F
∂t
+ q̇ i
∂F
∂qi
+ q̈ i
∂F

∂q̇ i
(3.13)
(Remember: repeated indices denote summation over 1 ≤ i ≤ n)
There are two situations where constants of the motion (also
called “first integrals”) are
easy to find.
1. If qk is an ignorable coordinate for some k (there may be
more than one), such
that L is independent of qk, then the associated (conjugate)
generalised momentum
pk = ∂L/∂q̇ k is a constant of the motion.
dpk
dt
=
d
dt
(
∂L
∂q̇ k
)
(3.14)
=
d
dt
(

∂L
∂q̇ k
)
− ∂L
∂qk
(3.15)
= 0 (3.16)
We go from (3.14) to (3.15) by noting ∂L/∂qk = 0, and from
(3.15) to (3.16) using
Lagrange’s equation (3.11).
2. If L does not depend explicitly on time (i.e., ∂L/∂t = 0; note:
this is not the same
as dL/dt = 0) then the function
H = q̇ i
∂L
∂q̇ i
− L (3.17)
is a constant of the motion (sometimes called the Jacobi
integral).
dH
dt
= q̈ i
∂L

∂q̇ i
+ q̇ i
d
dt
(
∂L
∂q̇ i
)
− ∂L
∂t
− q̇ i
∂L
∂qi
− q̈ i
∂L
∂q̇ i
(3.18)
= −∂L
∂t
+ q̇ i
[
d
dt

(
∂L
∂q̇ i
)
− ∂L
∂qi
]
(3.19)
= −∂L
∂t
(3.20)
= 0 (3.21)
We use the chain and product rules to get (3.18), and use
Lagrange’s equation (3.11)
to get from (3.19) to (3.20).
14
3.3 SYMMETRIES OF NATURE 640-213 Melatos
The function H(t) is called the Hamiltonian. It is sometimes (but
not always!) the total
energy of the system.4 So it should not be surprising that it is
conserved if ∂L/∂t = 0.
To appreciate why, consider the converse situation. E.g., if
∂L/∂t = 0 in the tennis ball

example, it would mean that the Earth’s gravitational potential
varies explicitly with
time, e.g., mg(t)x. But if there is nothing else in the problem,
the tennis ball itself would
have to (somehow...) be causing the variation. Hence it would
be doing work on the
Earth. So the total energy of the tennis ball itself cannot be
conserved.
3.3 Symmetries of Nature
In general, a mechanical system has 2n−1 constants of the
motion. Initial conditions give
2n constants c1, ...,c2n (n for the coordinates, n for the
velocities). In a closed system, the
equations of motion do not depend on time explicitly, so the
origin of time is arbitrary
and can be chosen to eliminate one of the constants, leaving c1,
...,c2n−1. Subsequently,
at time t, each qi and q̇ i can be written as a function of t and c1,
...,c2n−1. We can then
eliminate t from these 2n equations to express the 2n− 1
constants as functions of qi and
q̇ i. For example, c1 = c1(q1, ...,qn, q̇ 1, ..., q̇ n,c2, ...,c2n−1) is a
constant of the motion.
Some constants are especially important because they are
additive. Examples are mo-
mentum, energy, and angular momentum. For example, if a
system is divided into two
parts, which interact weakly, the energy of the whole system is
the sum of the energies of
the two subsystems.
Some constants are also intimately related to fundamental
symmetries of the system.

Again, momentum, energy, and angular momentum are
important examples. There is a
one-to-one relation between the symmetries and conservation
laws in a mechanical system,
as shown in the table below.
symmetry L transformation conserved quantity
homogeneous t temporal translation energy
homogeneous x spatial translation momentum
isotropic x rotation angular momentum
For example, in line 1 of the table, a system that is
homogeneous in time has a Lagrangian
that is invariant under time translation (i.e. the transformation t
�→ t + t0). The total
energy of such a system is conserved.
The correspondences in the table are stated without proof; they
are special cases of a
profound result called Noether’s theorem, which we do not
attempt to prove. If you wish
to know more, please consult the extension material in Chapter
4, where we go through
the special cases in the table one by one and then talk some
more about the central role
played by Noether’s theorem in fundamental physics.
4In fact, H is the total energy only if the transformation from
Cartesian xs to generalised qi is
independent of time.
15
3.4 APPLICATION: THE KEPLER PROBLEM 640-213 Melatos

3.4 Application: the Kepler problem
Planet of mass m orbiting an immobile star, of mass M � m,
surrounded by a gravi-
tational potential V (r). The gravitational force is centripetal
(∇ V points radially). Let
the orbit be described by [r(t),θ(t),φ(t)] in spherical polar
coordinates, the generalised
coordinates most suited to this problem.
Orthogonal velocity components: radial ṙ, latitudinal rθ̇,
longitudinal r sin θφ̇. Sum their
squares to get v2 (Pythagoras).
KE =
1
2
mv2 (3.22)
=
1
2
m(ṙ2 + r2θ̇2 + r2 sin2 θφ̇2) (3.23)
PE = mV (r) (3.24)
Lagrangian:
L = KE − PE (3.25)
As pointed out in Lecture 2, the Lagrange equations are already
written in generalised
coordinates, so there is no work involved in transforming them;
you write them in exactly

the same form in spherical polars coordinates as in any other
system.
e.g., θ equation
0 =
d
dt
(
∂L
∂θ̇
)
− ∂L
∂θ
(3.26)
=
d
dt
(
mr2θ̇
)
− mr2 sin θ cos θφ̇2 (3.27)
Exercise. Show directly, starting from Cartesian coordinates,
that the right-
hand side of this equation is simply the θ component of the
acceleration (which

is zero by Newton’s Second Law for a centripetal force).
16
Three constants of the motion (actually 2 × 3 − 1 = 5 in general,
but two turn out to be
trivial; this happens often):
1. L independent of φ
⇒ pφ =
∂L
∂φ̇
= constant (call it mJ) (3.28)
r2 sin2 θφ̇ = J (3.29)
J is the angular momentum about the z axis per unit mass.
(Verify!)
2. L independent of t
⇒ H = ṙ∂L
∂ṙ
+ θ̇
∂L
∂θ̇
+ φ̇

∂L
∂φ̇
− L = constant (call it mE) (3.30)
1
2
(
ṙ2 + r2θ̇2 + r2 sin2 θφ̇2
)
+ V (r) = E (3.31)
E is the total energy per unit mass. (Verify!)
3. One which is hard to guess by Lagrange methods:
pθ
2 +
pφ
2
sin2 θ
= m2α2 (3.32)
(Hamilton-Jacobi methods, which we will mention in a later
lecture, give it to
you easily.) The physical interpretation is straightforward: α is
the total angular
momentum (not just around the z axis) per unit mass, and it is
constant because a
centripetal force exerts zero torque.

Exercise. Show that d
dt
(p2θ + p
2
φ/ sin
2 θ) = 0 using Legrange’s equations.
The third constant tells us the motion lies in a plane. Here is
why. We have
r2θ̇ = ±
(
α2 − J
2
sin2 θ
)1/2
(3.33)
and
dφ
dθ
=
φ̇
θ̇
= ± J
sin2 θ
(

α2 − J
2
sin2 θ
)−1/2
(3.34)
which can be integrated (verify!) to give
φ − φ0 = ± cos−1
(
J cot θ√
α2 − J2
)
(3.35)
Expanding using cos(A + B) = cos A cos B − sin A sin B, we
get
sin θ cos φ cos φ0 + sin θ sin φ sin φ0 −
J cos θ√
α2 − J2
= 0 (3.36)
This is of the form ax + by + cz = 0, i.e., a plane passing
through the origin with normal
(cos φ0, sin φ0,−J cos θ/
√
α2 − J2).

17
Exercise. For a simpler, more physical proof, which eschews
Lagrangian
methods, shows that the angular momentum per unit mass, x×ẋ,
is a constant
vector for a central force. [Hint: calculate d(x × ẋ)/dt.] Hence
argue that the
orbit lies in the plane perpendicular to this constant vector.
Now that we know the orbit lies in a plane, we can set θ = π/2
without loss of generality
and look at the r-φ motion.
r2φ̇ = J (3.37)
1
2
(
ṙ2 + r2φ̇2
)
+ V (r) = E (3.38)
Eliminating φ̇:
1
2
ṙ2 +
J2

2r2
+ V (r) = E (3.39)
Equivalent to a one-dimensional problem with an effective
potential Veff(r) = J
2/2r2 +
V (r), comprising the physical (e.g., gravitational) potential plus
a centrifugal term J2/2r2.
Newtonian gravity (before Einstein):
Veff =
J2
2r2
− GM
r
(3.40)
Black hole gravity (after Einstein):
Veff =
J2
2r2
− GM
r
− GMJ
2
c2r3

(3.41)
18
Exercise. [Ext] Can you show that circular orbits (radius r0)
near a black
hole are stable only if r0 ≥ 6GM/c2? [This is a tough one!
Hints: V ′eff(r0) = 0.
Then consider a perturbed orbit r(t) = r0 + ξ(t), plug into the
equation of
motion obtained by differentiating (3.39) with respect to time,
and show that
ξ(t) satisfies a differential equation
ξ̈(t) +
[
V ′′(r0) +
3J2
r04
]
ξ(t) = 0
with runaway (i.e., exponentially growing) solutions if r0 <
6GM/c
2.]
19

4.1 ENERGY 640-213 Melatos
Chapter 4
[Ext] Symmetries and Conservation
Laws
This chapter contains extension material only, for those who
want to know more about
the profound link between symmetries and conservation laws in
nature (introduced briefly
in Chapter 3).
symmetry L transformation conserved quantity
homogeneous t temporal translation energy
homogeneous x spatial translation momentum
isotropic x rotation angular momentum
4.1 Energy
Time is homogeneous. It ticks away at the same rate yesterday,
today, and tomorrow.
(Let us ignore relativity for the moment.) As Newton wrote in
the Principia,
“Absolute, true and mathematical time, of itself, and from its
own nature
flows equably without regard to anything external, and by
another name is
called duration”
Consequently, L does not depend explicitly on t in a closed
system. In other words, it is
invariant under the translation t′ = t+t0. (In an open system,

some human or other agent
can intervene to muck up this state of affairs, e.g., by
periodically jolting the system.)
From Chapter 3,
H = q̇ i
∂L
∂q̇ i
− L (4.1)
= q̇ i
∂T
∂q̇ i
− T + V (4.2)
is a constant of the motion if ∂L/∂t = 0. Equation (4.2) follows
from (4.1) assuming that
V is independent of the generalised velocities.
20
4.2 MOMENTUM 640-213 Melatos
If, in addition, the transformation to generalised coordinates is
independent of t, i.e,
xs = xs(q1, ...,qn), then we have
ẋs = q̇ j
∂xs
∂qj

(4.3)
T =
N∑
s=1
1
2
msq̇ jq̇ k
∂xs
∂qj
· ∂xs
∂qk
(4.4)
∂T
∂q̇ i
=
N∑
s=1
1
2
ms
(
q̇ k
∂xs

∂qi
· ∂xs
∂qk
+ q̇ j
∂xs
∂qj
· ∂xs
∂qi
)
(4.5)
=
N∑
s=1
msq̇ k
∂xs
∂qi
· ∂xs
∂qk
(4.6)
q̇ i
∂T
∂q̇ i
=
N∑

s=1
msq̇ iq̇ k
∂xs
∂qi
· ∂xs
∂qk
(4.7)
= 2T (4.8)
and hence
H = T + V (4.9)
i.e., H equals the total energy under these special conditions.
(Note: in the above algebra,
the repeated indices are dummies which we sum over, so we can
use any letter we like.
Hence q̇ j∂xs/∂qj = q̇ k∂xs/∂qk for example.)
4.2 Momentum
Space is homogeneous. It looks exactly the same at every point,
here or in a distant galaxy,
although its contents generally differ from point to point.
(Again we ignore relativity.)
Consequently, L is invariant under any translation x′ = x+a (i.e.,
a parallel displacement
in space from x to x′). In particular, it is invariant under any
infinitesimal translation,
where a is infinitesimally small (|a| → 0).
Let the generalised coordinates be the Cartesian coordinates in
this application. (This

is allowed because there cannot be any constraints like inclined
planes, which break the
symmetry, if L is translationally invariant.) Then, holding the
velocities fixed, L =
L(x1, ..., xN ), L
′ = L(x′1, ..., x
′
N ) = L(x1 + a, ..., xN + a), and
L′ − L =
N∑
s=1
∂L
∂xs
· a + terms of order O(|a|2). (4.10)
But invariance requires L′ = L for all possible a, which implies
N∑
s=1
∂L
∂xs
= 0. (4.11)
21
4.3 ANGULAR MOMENTUM 640-213 Melatos

From Lagrange’s equations,
0 =
N∑
s=1
d
dt
∂L
∂ẋs
− ∂L
∂xs
(4.12)
=
N∑
s=1
d
dt
∂L
∂ẋs
(4.13)
from (4.11). But if the potential is independent of the velocities,
we have ∂L/∂ẋs =
∂T/∂ẋs = msẋs and hence

N∑
s=1
msẋs = constant (4.14)
i.e., the total momentum is conserved if L is translationally
invariant. The total momen-
tum is also manifestly additive.
Exercise. (a) Show that the Lagrangian transforms as
L = L′ + V · P′ + 1
2
∑
s
msV
2 (4.15)
between two inertial frames K and K′, where K moves with
velocity V relative
to K′, and we define P′
∑
s
msẋ
′
s.
(b) If R′ is the centre-of-mass position vector in K′, show that
the action
transforms as

S = S′ +
∑
s
msV · R′ +
1
2
∑
s
msV
2t. (4.16)
4.3 Angular momentum
Space is isotropic. It looks the same in all directions when
observed from any vantage
point. (Ignore relativity.) Hence the Lagrangian of a closed
system is invariant under any
rotation about a point, including infinitesimal rotations.
Let a be a vector, whose magnitude |a| → 0 is the angle of
rotation, and which points
along the axis of rotation. Clearly position vectors and velocity
vectors rotate by the
same amount.
x′ = x + a × x (4.17)
ẋ′ = ẋ + a × ẋ (4.18)
22

4.3 ANGULAR MOMENTUM 640-213 Melatos
With L′ = L(x′1, ..., x
′
N ) = L(x1 + a × x1, ..., xN + a × xN, ẋ1 + a × ẋ1, ..., ẋN + a ×
ẋN ),
we find
L′ − L =
N∑
s=1
∂L
∂xs
· a × xs +
∂L
∂ẋs
· a × ẋs (4.19)
= a ·
(
N∑
s=1
xs ×
∂L
∂xs
+ ẋs ×

∂L
∂ẋs
)
(4.20)
using the vector identity a · b × c = a × b · c. From Lagrange’s
equations and the product
rule of differentiation,
L′ − L = a ·
(
N∑
s=1
xs ×
d
dt
(
∂L
∂ẋs
)
+ ẋs ×
∂L
∂ẋs
)

(4.21)
= a ·
N∑
s=1
d
dt
(
xs ×
∂L
∂ẋs
)
(4.22)
Again, a is arbitrary, implying
N∑
s=1
xs ×
∂L
∂ẋs
= constant (4.23)
If the potential is independent of the velocities, we have ∂L/∂ẋs
= ∂T/∂ẋs = msẋs and
hence

N∑
s=1
xs × msẋs = constant (4.24)
i.e., the total angular momentum is conserved if L is invariant
under rotations. The total
angular momentum is manifestly additive. Its value depends on
the choice of origin.
Interestingly, energy, momentum, and angular momentum are
the only additive integrals.
(Can you guess why?)
The angular momentum is not conserved in general if the system
is open, e.g., if it sits
in an external field. However, if the field is rotationally
symmetric about an axis passing
23
4.4 NOETHER’S THEOREM: SYMMETRIES OF THE
PARTICLES AND FORCES
OF NATURE 640-213 Melatos
through the origin, then the angular momentum component
along that axis is conserved
(as long as the angular momentum is measured relative to the
same origin). A special
case is a central field, i.e., one which points radially out from or
into the origin. In this
case, the field is rotationally symmetric about every axis
through the origin, so all three
independent components of the angular momentum are

conserved. This explains why
p2θ + p
2
φ/ sin
2 θ is a constant in the Kepler problem; see equation (3.32) in
Chapter 3.
4.4 Noether’s Theorem: Symmetries of the particles
and forces of Nature
Emmy Noether (1882-1935), one of the 20th Century’s leading
mathematicians5, discov-
ered a profound generalisation of the results in the preceding
sections:
any symmetry of the action is associated with a conserved
“current”, and vice
versa.
Noether’s theorem reduces to the results in the previous section
for space-time symmetries
like the homogeneity of time.
However, it also applies to the fields which describe Nature’s
particles and forces. In this
general case, the generalised coordinates qi(t) and velocities
q̇ i(t) (n degrees of freedom) are
replaces by fields ψ(x, t) and field gradients ∂ψ/∂t and ∇ ψ
(infinite degrees of freedom).
Without going into the mathematical proofs, you can still get a
good sense of how this
works:

1. a system of (usually coupled) particle and forces has a
Lagrangian density L =
L(ψ(x, t),∂ψ(x, t)/∂t,∇ ψ(x, t)) and an action S =
∫
dtd3x L
2. minimising S with respect to all possible choices of ψ gives
you the field equations
(“equations of motion”) describing the force or particle
3. if S is unchanged when you make some transformation of ψ,
then there is an asso-
ciated conservation law.
Examples (general idea; don’t worry about the maths until next
year):
1. Schrödinger’s equation for the wave function of an electron,
ψ(x, t), in a potential
V (x) can be derived by minimising the action
S =
∫
dtd3x
(
iψ∗
∂ψ
∂t
+
�

2
2m
ψ∗ ∇ 2ψ − ψ∗ V ψ
)
(4.25)
5She also discovered important results regarding the invariants
of mathematical structures like rings,
fields, and noncommutative algebras.
24
with respect to all possible choices of ψ.
But you know from your quantum mechanics lectures that we
can only ever measure
the probability |ψ(x, t)|2, not the wave function itself. So the
physics of the electron
(i.e., Schrödinger’s equation) must remain unchanged if we
replace ψ by ψ′ = eiθψ,
because |ψ′|2 = |ψ|2. Furthermore, θ can be as small as we like
so we can rewrite
ψ′ = (1 + iθ)ψ. [Note: ex ≈ 1 + x for small x.]
By Noether’s theorem, you can show that this type of
transformation guarantees a
conservation law
∂

∂t
(eψ∗ ψ) + div
[
ie�
2m
(ψ∇ ψ∗ − ψ∗ ∇ ψ)
]
= 0 (4.26)
Do you recognise this from your quantum mechanics lectures? It
is the conservation
of electric charge! eψ∗ ψ is the charge per unit volume (ρ from
electromagnetism;
when you integrate over volume you get e), while [...] is e times
the probability
current, i.e., the electric current density (J from
electromagnetism).
2. Maxwell’s equations for the electromagnetic potentials Φ(x,
t) and A(x, t) can be
derived by minimising the action from (1.9) and (1.10) with
respect to all possible
choices of Φ and A.
It turns out (see next year’s Electrodynamics course) that S is
invariant under gauge
transformations
Φ′ = Φ − 1
c
∂χ

∂t
(4.27)
A′ = A + ∇ χ (4.28)
where χ is any function. (Again, don’t worry about proving this
for now. Just
believe it.) By Noether’s theorem, you can show that this type
of transformation
guarantees
∂ρ
∂t
+ divJ = 0 (4.29)
i.e., the conservation of electric charge again!
We have just seen something profound: invariance under ψ′ =
eiθψ of Schrödinger’s
equation (which has no electromagnetism in it), and invariance
under a gauge trans-
formation of Maxwell’s equations (which have no quantum
mechanics in them), both
guarantee the same conservation law (of electric charge).
3. Strong nuclear force (quantum chromodynamics).
If you look in a table of physical data, you will notice that the
masses of the
proton and neutron are almost the same: mp = 1.6726 ×
10−27kg versus mn =
1.6749 × 10−27kg. Surely this is not a coincidence? It isn’t;
experiments have shown
that p and n are “identical” as far as the strong nuclear force is

concerned (unlike
electromagnetism, which affects p but not n).
25
Mathematically, this means that the equations describing the
strong nuclear force,
which involve the proton and neutron wave functions ψp(x, t)
and ψn(x, t), should
remain invariant if we swap ψp and ψn.
6 And indeed they do! In fact, they remain
invariant under the transformation(
ψ′p
ψ′n
)
=
(
cos α sin α
− sin α cos α
)(
ψp
ψn
)
(4.30)

which mixes up ψn and ψp in linear combinations (with α = π/2
corresponding to
a straight swap).7
Does (4.30) look familiar? It is the matrix for a rotation by an
angle α! But be
careful: it is not a rotation in the physical space, it is a rotation
in “particle space”
(i.e., neutron-proton space).
By Noether’s theorem, (4.30) implies a conservation law. It is
the conservation of
isospin. The proton is “isospin up”, the neutron is “isospin
down”, and the total
isospin of all the reactants and products in a strong nuclear
interaction must be
equal.
Symmetries, conservation laws, and Noether’s theorem are a big
deal in particle
physics. They are deeply connected to the mathematics of
groups. [Equation (4.30)
belongs to the group SU(2), for example.] Sometimes, Noether’s
theorem throws
up startling new results. For example, you might think that the
forces of nature
behave unchanged when viewed in a mirror (the parity
transformation x �→ −x).
Incredibly, they are not! The weak nuclear force (β decay)
violates this. For closely
related reasons which we cannot cover here, nature must also
violate time reversal
symmetry (corollary of the so-called CPT theorem). No one
knows why Nature
finds it necessary to behave in this way.

If you want to learn more, Chapter 4 of Introduction to
Elementary Particles, by D.
Griffiths, is a good read.
6The QCD equations are horrible to write down and even harder
to solve. They were discovered in
the late 1960’s.
7In reality, the matrix entries are a bit more complicated than in
(4.30) and involve submatrices called
the Pauli spin matrices, but we don’t worry about that here.
26
640-213 Melatos
Chapter 5
Welcome to my centrifuge, Mr Bond
Cartoon from http://xkcd.com/123/
In this lecture, we work through an example which illustrates
the abstract ideas introduced
in Chapters 1–4. The example is inspired by the above cartoon:
what is the motion of a
frictionless body sliding along the inside surface of a spherical
shell?
27

5.1 Generalised coordinates
N = 1 particle, 3N = 3 degrees of freedom, but one of them is
constrained (body cannot
move radially), leaving 2 degrees of freedom.
Choose latitude θ and longitude φ as the generalised coordinates
q1 and q2 respectively.
x1 = x1(q1,q2) (5.1)
= (R sin θ cos φ,R sin θ sin φ,R cos θ) (5.2)
= (R sin q1 cos q2,R sin q1 sin q2,R cos q1) (5.3)
R, the radius of the spherical shell, is constant; it is not a
coordinate.
Let us calculate the velocity two ways and show that they give
the same answer.
Directly, by the chain and product rules:
ẋ1 = (R cos q1q̇ 1 cos q2 − R sin q1 sin q2q̇ 2,
R cos q1q̇ 1 sin q2 + R sin q1 cos q2q̇ 2,−R sin q1q̇ 1) (5.4)
More abstractly, as in Chapters 1–4,
x1 = x1(q1(t),q2(t)) (5.5)
ẋ1 =
∂x1
∂q1

dq1
dt
+
∂x1
∂q2
dq2
dt
(chain rule) (5.6)
= q̇ 1(R cos q1 cos q2,R cos q1 sin q2,−R sin q1)
+ q̇ 2(−R sin q1 sin q2,R sin q1 cos q2, 0) (5.7)
You can see by inspection that (5.7) agreeds with (5.4). This is
why, in Chapters 1–4, we
write
ẋs =
∂xs
∂t
+ q̇ i
∂xs
∂qi
(5.8)
28
5.3 LAGRANGE’S EQUATIONS 640-213 Melatos
for the sth particle. (Remember: the repeated index i is just

shorthand for
∑n
i=1; in our
example n = 2.) The ∂xs/∂t term does not appear in (5.7)
because xs in our example
depends on t only through q1(t) and q2(t), not explicitly.
N.B. In practical problems, we don’t write out xs explicitly in
terms of qi; it is built into
the Lagrangian formalism. But it is important to do it once or
twice at this stage to get
a feel for how it all works.
5.2 Lagrangian
Kinetic energy:
T =
1
2
m
(
vr
2 + vθ
2 + vφ
2
)
(5.9)
vr = 0: can’t move in radial direction.
vθ = (lever arm) × (angular velocity in latitude) = Rθ̇

vφ = (lever arm) × (angular velocity in longitude) = R sin θφ̇
Potential energy (take midplane to be reference level):
V = mg × (altitude) (5.10)
= mgR cos θ (5.11)
Lagrangian:
L = T − V (5.12)
=
1
2
mR2
(
θ̇2 + sin2 θφ̇2
)
− mgR cos θ (5.13)
5.3 Lagrange’s equations
Remember that θ, φ, θ̇, and φ̇ are independent variables.
Latitude coordinate:
0 =
d
dt
(
∂L

∂q̇ 1
)
− ∂L
∂q1
(5.14)
=
d
dt
(
∂L
∂θ̇
)
− ∂L
∂θ
(5.15)
=
d
dt
(
mR2θ̇
)
−
(

mR2 sin θ cos θφ̇2 + mgR sin θ
)
(5.16)
29
5.4 SOLVING FOR THE MOTION 640-213 Melatos
Longitude coordinate:
0 =
d
dt
(
∂L
∂q̇ 2
)
− ∂L
∂q2
(5.17)
=
d
dt
(
∂L

∂φ̇
)
− ∂L
∂φ
(5.18)
=
d
dt
(
mR2 sin2 θφ̇
)
(5.19)
You would get the same equations starting from Newton’s
Second Law plus the con-
straints, but your life would be harder.
5.4 Solving for the motion
We do not try to solve Lagrange’s equations directly (otherwise
we might as well start
from Newton’s laws). Instead, we look for constants of the
motion.
L is independent of φ. Hence
∂L
∂φ̇

= mR2 sin2 θφ̇ = constant (call it mJ) (5.20)
L is independent of t. Hence
θ̇
∂L
∂θ̇
+ φ̇
∂L
∂φ̇
− L = 1
2
mR2
(
θ̇2 + sin2 θφ̇2
)
+ mgR cos θ (5.21)
= constant (call it mE) (5.22)
Substituting (5.20) in (5.22):
E =
1
2
R2θ̇2 +
1

2
R2 sin2 θ × J
2
R4 sin4 θ
+ gR cos θ (5.23)
=
1
2
R2θ̇2 +
J2
2R2 sin2 θ
+ gR cos θ (5.24)
This is equivalent to a one-dimensional problem (in θ) with
effective potential Veff(θ) =
J2/(2R2 sin2 θ) + gR cos θ; see Chapter 3.
30
5.5 [EXT] NOETHER’S THEOREM 640-213 Melatos
So the body bounces between turning points (where θ̇ = 0
momentarily, but φ̇ = 0) at
latitudes θ1 and θ2 where E = Veff(θ1) and E = Veff(θ2).
Note that Veff(θ) is not symmetric about the equator, because in
the northern hemisphere
both gravity and the normal force point downwards, whereas in

the southern hemisphere
the normal force points upwards. Circular orbits alwas occur in
the southern hemisphere
at latitude θ3 > π/2, where Veff(θ3) is a minimum.
5.5 [Ext] Noether’s theorem
In Section 5.3, we see that the angular momentum ∂L/∂φ̇ = mR2
sin2 θφ̇ is a constant from
Lagrange’s equations. In this section, we derive the same result
by Noether’s theorem.
We start by noting that our system is invariant under rotation
about the z axis. This is
true by common sense; alternatively, it follows because L is
independent of φ. Suppose
we rotate the system by an infinitesimal angle δφ, so that φ is
replaced by φ + δφ.
31
Coordinates (see Section 5.1):
x1(θ,φ + δφ) = [R sin θ cos(φ + δφ),R sin θ sin(φ + δφ),R cos θ]
(5.25)
= [R sin θ(cos φ cos δφ − sin φ sin δφ),
R sin θ(sin φ cos δφ + cos φ sin δφ),R cos θ] (5.26)
≈ (R sin θ cos φ − R sin θ sin φδφ,
R sin θ sin φ + R sin θ cos φδφ,R cos θ) (5.27)

= x1(θ,φ) + δφẑ × x1(θ,φ) (5.28)
Equation (5.26) follows from (5.25) by the sine and cosine sum
and product rules. Equa-
tion (5.27) follows from (5.26) because δφ is chosen to be
infinitesimally small (cos δφ ≈ 1,
sin δφ ≈ δφ). Equation (5.28) follows from (5.27) by easy
algebra; check the Cartesian
components of the cross product to make sure, noting that ẑ is a
unit vector in the z di-
rection. Note that we can just as easily reach the above result
using a Taylor expansion:
x1(θ,φ + δφ) ≈ x1(θ,φ) + δφ
∂x1(θ,φ)
∂φ
+ ... (5.29)
= x1(θ,φ) + δφ(−R sin θ sin φ,R sin θ cos φ, 0) (5.30)
= x1(θ,φ) + δφẑ × x1(θ,φ) (5.31)
Velocities work out exactly the same way. From (5.4) in Section
5.1, and following similar
steps as from (5.25) to (5.28), we find
ẋ1(θ,φ + δφ) = [R cos θθ̇ cos(φ + δφ) − R sin θ sin(φ + δφ)φ̇,
R cos θθ̇ sin(φ + δφ) + R sin θ cos(φ + δφ)φ̇,−R sin θθ̇] (5.32)
≈ [R cos θθ̇ cos φ − R cos θθ̇ sin φδφ
−R sin θ sin φφ̇ − R sin θ cos φδφφ̇,
R cos θθ̇ sin φ + R cos θθ̇ cos φδφ
+R sin θ cos φφ̇ − R sin θ sin φδφφ̇,−R sin θθ̇] (5.33)
= ẋ1(θ,φ)δφẑ × ẋ1(θ,φ) (5.34)

Now how does L transform under the infinitesimal rotation
above?
L(θ,φ + δφ) =
1
2
m|ẋ1(θ,φ + δφ)|2 − mgx1(θ,φ + δφ) · ẑ (definition of L)(5.35)
=
1
2
m
[
|ẋ1(θ,φ)|2 + 2δφẋ1(θ,φ) · ẑ × ẋ1(θ,φ)
+(δφ)2|ẑ × ẋ1(θ,φ)|2
]
−mgx1(θ,φ) · ẑ − mgδφẑ × x1(θ,φ) · ẑ (5.36)
Ignore terms of order (δφ)2 because they are doubly small.
L(θ,φ + δφ) − L(θ,φ) = mδφẋ1 · ẑ × ẋ1 − mgδφẑ × x1 · ẑ (5.37)
32
Clearly the RHS vanishes (a · b × a = 0 for any two vectors),
and so it should; we already
know that L is invariant under rotations, so L(θ,φ + δφ) =
L(θ,φ):

0 = ẋ1 · ẑ × ẋ1 − gẑ × x1 · ẑ (5.38)
Writing the equation another way, we get
0 = ẑ · (mẋ1 × ẋ1 − mgẑ × x1) (5.39)
= ẑ ·
(
∂L
∂ẋ1
× ẋ1 +
∂L
∂x1
× x1
)
(5.40)
using L = 1
2
mẋ1 · ẋ1 − mgx1 · ẑ. From Lagrange’s equations,
0 = ẑ ·
[
∂L
∂ẋ1
× ẋ1 +
d
dt

(
∂L
∂ẋ1
)
× x1
]
(5.41)
= ẑ · d
dt
(
∂L
∂ẋ1
× x1
)
(5.42)
But x1×∂L/∂ẋ1 equals the angular momentum vector. So the z-
component of the angular
momentum is constant with time, i.e.,
constant = ẑ · (x1 × mẋ1) (5.43)
= R sin θ cos φ · m(R cos θθ̇ sin φ + R sin θ cos φφ̇)
−R sin θ sin φ · m(R cos θθ̇ cos φ − R sin θ sin φφ̇) (5.44)
= mR2 sin2 θφ̇, (5.45)
exactly as predicted (much more simply!) in Section 5.3.

N.B. You do not need to go through this trouble every time you
use Noether’s Theorem.
Just quote it and apply it without proof. We follow the details
here just to illustrate the
abstract proof in lectures with a concrete example.
33
6.1 DEGREES OF FREEDOM 640-213 Melatos
Chapter 6
Rigid body rotation
6.1 Degrees of freedom
A rigid body is a collection of particles whose relative positions
remain fixed at all times.
Six generalised coordinates are needed to describe how three
orthogonal Cartesian axes
(e1, e2, e3), which are fixed in the body, move with respect to a
set of Cartesian axes in
the external space (i, j, k), i.e., in the “laboratory frame”. The
six coordinates are:
• X(t), the origin O of the body axes (e1, e2, e3). Usually we
take this to coincide with
the centre of mass, so that the kinetic energy assumes a simple
form (see below).
• Three (Euler) angles θ(t), φ(t), ψ(t) define the orientation of
(e1, e2, e3) according
to the figure below. To see this, note that the correct orientation

of e3 (say) with
respect to (i, j, k) is specified by two angles (a latitude and a
longitude). Once
e3 is correctly specified, e2 (say) can be rotated around e3 until
its orientation is
also correct (one extra angle, making three in total). e1 comes
along for the ride
(orthogonal) and automatically ends up pointing in the correct
direction.
Once (e1, e2, e3) are completely specified at any instant t, so
are the 3N coordinates of
the N particles constituting the body, because the body is rigid.
Let P be a point in the body, whose position vector is x(t) with
respect to (e1, e2, e3) and
x′(t) = X(t) + x(t) (6.1)
with respect to (i, j, k). An infinitesimal rotation by an angle dφ
about the axis n changes
any position vector from x to x + dφn × x. Defining the angular
velocity Ω(t) to be the
instantaneous rate of change of dφn, we have dφn = Ωdt and
hence
ẋ′(t) = Ẋ(t) + Ω × x(t) (6.2)
The last term on the RHS follows from
ẋ =
x(t + dt) − x(t)
dt
(6.3)
=

dφn × x(t)
dt
(6.4)
= Ω × x(t) (6.5)
Exercise. Show that if O is displaced by δX to O′, the angular
velocity
around the new origin is unchanged, while the translational
velocity of the
origin is Ẋ + Ω × δX.
34
6.1 DEGREES OF FREEDOM 640-213 Melatos
Euler angles. From E. D. Fackerell and C. J. Durrant, 1993,
Lagrangian Dynamics,
University of Sydney
35
6.2 KINETIC ENERGY 640-213 Melatos
6.2 Kinetic energy
Sum over particles s = 1, ...,N in a rigid body, each with mass
ms and position vector
x′s. Let the body have total mass M. Note the vector identities a
· b × c = a × b · c and
(a × b) · (c × d) = a · cb · d − a · db · c.

Kinetic energy:
T =
∑
s
1
2
ms|ẋ′s|2 (6.6)
=
∑
s
[
1
2
ms|Ẋ|2 + msẊ · Ω × xs +
1
2
ms|Ω × xs|2
]
(6.7)
=
1
2
|Ẋ|2

∑
s
ms + Ẋ × Ω ·
∑
s
msxs +
∑
s
1
2
ms
[
Ω2|xs|2 − (Ω · xs)2
]
(6.8)
=
1
2
M|Ẋ|2 +
∑
s
1
2

ms
[
Ω2|xs|2 − (Ω · xs)2
]
(6.9)
because
∑
s msxs = 0 is just the centre of mass, which lies at the origin of
the (e1, e2, e3)
plane (to which the xs vectors are referred). 1st term: KE of
translation of centre of
mass. 2nd term: KE of rotation about centre of mass.
If we write xs = xs1e1 + xs2e2 + xs3e3, in terms of its
components along (e1, e2, e3), a bit
of algebra confirms that we can write (verify!)
∑
s
1
2
ms
[
Ω2|xs|2 − (Ω · xs)2
]
=
1
2

(
Ω1 Ω2 Ω3
)⎛⎝ I11 I12 I13I21 I22 I23
I31 I32 I33
⎞
⎠
⎛
⎝ Ω1Ω2
Ω3
⎞
⎠ (6.10)
The matrix Iij is called the moment-of-inertia tensor. It’s (i,j)-th
entry is given by
Iij =
∑
s
ms
[
(x2s1 + x
2
s2 + x
3
s3)δij − xsixsj
]

(6.11)
e.g., I22 =
∑
s ms(x
2
s1 + x
2
s3) and I31 = I13 = −
∑
s msxs1xs3.
8
For a continuous (i.e. smooth) mass distribution, with density
ρ(x), we replace
∑
s ms by∫
d3x ρ(x) (and xs by x, of course).
Hence the Lagrangian for a rigid body in a potential V
(X,θ,φ,ψ) is
L =
1
2
M|Ẋ|2 + 1
2
ΩT IΩ − V (6.12)

where Ω is the column vector (Ω1, Ω2, Ω3) and Ω
T is its transpose, as in (6.10).
8Kronecker delta notation: δij = 0 if i = j and δij = 1 if i = j.
36
6.4 EXAMPLE: NONAXISYMMETRIC RIGID PENDULUM
640-213 Melatos
6.3 Principal moments of inertia
If you choose (e1, e2, e3) wisely, you can always ensure that
the matrix Iij comes out to be
diagonal, i.e., off-axis elements vanish. From the definition
(6.11), it is clear that Iij is real
and symmetric. Your linear algebra course last year taught you
that such matrices are
diagonalisable when pre- and post-multiplied by a matrix S
containing the eigenvectors
of I as columns; i.e., S−1IS is diagonal. The eigenvectors are
called the principal axes;
the eigenvalues I1, I2, I3 are called the principal moments of
inertia. Often, the wisest
choice of (e1, e2, e3) is obvious, e.g., dictated by symmetry.
For example, an axisymmetric top has a moment of inertia
tensor of the following form if
e3 is chosen along the symmetry axis:
I =
⎛
⎝ I1 0 00 I1 0

0 0 I3
⎞
⎠ (6.13)
I3 < I1 (left) and I3 > I1 (right)
If an axis of symmetry exists, the centre of mass lies on that
axis of symmetry. If the body
is not axisymmetric, we have I1 = I2 = I3 and the axes (e1, e2,
e3) must be constructed
by solving the eigenvalue problem Ie = λe.
Exercise. Show that (a) I3 = I1 + I2 for a two dimensional (i.e.,
flat) rigid
body, and (b) I3 =
1
2
MR2 for a uniform disk of mass M and radius R.
6.4 Example: Nonaxisymmetric rigid pendulum
Consider a strangely shaped block threaded on a spindle
(direction s) and allowed to
swing freely around it.
End-on and side-on views are shown in the figure below
37
6.4 EXAMPLE: NONAXISYMMETRIC RIGID PENDULUM
640-213 Melatos

Let e1, e2, e3 (in general out of plane of page) make angles α,
β, γ with the rotation axes
in both figures.
Ω = φ̇s (6.14)
Ω1 = φ̇s · e1 = φ̇ cos α (6.15)
Ω2 = φ̇ cos β (6.16)
Ω3 = φ̇ cos γ (6.17)
Also the centre of mass moves with speed |Ẋ| = (lever arm) ×
(angular velocity) = lθ̇.
Lagrangian:
L =
1
2
M|Ẋ|2 + 1
2
ΩT IΩ − V (6.18)
=
1
2
Ml2φ̇2 +
1
2
φ̇2
(

I1 cos
2 α + I2 cos
2 β + I3 cos
2 γ
)
− Mgl(1 − cos φ) (6.19)
(Potential energy is measured relative to vertical hang.)
For small angles we have 1 − cos φ ≈ φ2/2. The frequency of
oscillation is therefore
ω2 =
Mgl
Ml2 + I1 cos2 α + I2 cos2 β + I3 cos2 γ
(6.20)
(A simple harmonic oscillator with Lagrangian L = mv2/2 −
kx2/2 has oscillation fre-
quency ω2 = k/m.)
38
6.6 EULER’S EQUATIONS OF MOTION 640-213 Melatos
6.5 Angular momentum
Evaluated like the kinetic energy. Refer to the (i, j, k) origin.
M =
∑

s
x′s × msẋ′s (6.21)
=
∑
s
ms(X + xs) × (Ẋ + Ω × xs) (6.22)
= X × Ẋ
∑
s
ms + X × (Ω ×
∑
s
msxs)
−Ẋ ×
∑
s
msxs +
∑
s
msxs × (Ω × xs) (6.23)

= MX × Ẋ +
∑
s
ms(Ω|xs|2 − xsxs · Ω) (6.24)
using
∑
s msxs = 0 (centre of mass located at origin in body frame) and
the vector identity
a × (b × c) = b(a · c) − c(c · b).
Writing xs = xs1e1 + xs2e2 + xs3e3, as in the previous section,
we confirm after a bit of
algebra (try it!) that
∑
s
ms(Ω|xs|2 − xsxs · Ω) =
⎛
⎝ I11 I12 I13I21 I22 I23
I31 I32 I33
⎞
⎠
⎛
⎝ Ω1Ω2
Ω3
⎞
⎠ = IΩ (6.25)

where the matrix Iij is defined according to (6.11) as before.
About the centre of mass, i.e., (e1, e2, e3) origin, the MX × Ẋ
term goes away, leaving
M = IΩ (6.26)
6.6 Euler’s equations of motion
Rate of change of the total angular momentum equals the net
external torque applied, N.
N =
dM
dt
(6.27)
=
d
dt
(I1Ω1e1 + I2Ω2e2 + I3Ω3e3) (6.28)
Both Ω and the body axes change with time. The unit vectors
are vectors like any other,
so they obey [see, for example, equation (6.5)]
de1
dt
= Ω × e1 etc. (6.29)
39

6.6 EULER’S EQUATIONS OF MOTION 640-213 Melatos
Therefore
N = I1Ω̇1e1 + I1Ω1ė1 + I2Ω̇2e2 + I2Ω2ė2 + I3Ω̇3e3 + I3Ω3ė3
(6.30)
= I1Ω̇1e1 + I1Ω1(Ω × e1) + I2Ω̇2e2 + I2Ω2(Ω × e2)
+I3Ω̇3e3 + I3Ω3(Ω × e3) (6.31)
Project along the orthogonal body axes to get Euler’s equations.
Along e1:
N1 = I1Ω̇1e1 · e1 + I2Ω2(Ω × e2) · e1 + I3Ω3(Ω × e3) · e1
(6.32)
= I1Ω̇1 + I2Ω2Ω · e2 × e1 + I3Ω3Ω · e3 × e1 (6.33)
= I1Ω̇1 − I2Ω2Ω3 + I3Ω3Ω2 (since e1 × e2 = e3) (6.34)
= I1Ω̇1 − (I2 − I3)Ω2Ω3 (6.35)
Similarly along e2 and e3:
N2 = I2Ω̇2 − (I3 − I1)Ω3Ω1 (6.36)
N3 = I3Ω̇3 − (I1 − I2)Ω1Ω2 (6.37)
Euler equations (6.35), (6.36) and (6.37) can be solved for
Ω1(t), Ω2(t), Ω3(t). These
three angular velocity components are referred to the body axes,
which are themselves
tumbling through space in the lab frame.
40
7.1 WORKED EXAMPLE: FREE PRECESSION 640-213

Melatos
Chapter 7
Tops
7.1 Worked example: free precession
Consider an axisymmetric body rotating freely, i.e.,
experiencing no net external torque.
If the body rotates around an axis other than one of its three
principal axes, then its
motion is composed of two parts: a spin, and a wobble.
(Imagine what happens if you
spin a rugby ball about an axis tilted at 45◦ with respect to its
long axis.)
In body frame (Euler’s equations):
Ω̇1 + �Ω2Ω3 = 0 (7.1)
Ω̇2 − �Ω1Ω3 = 0 (7.2)
Ω̇3 = 0 (7.3)
where we have I1 = I2 = I3 and � = (I3 − I1)/I1.
Solution
is Ω3 = constant, Ω1 ∝ cos(�Ω3t), Ω2 ∝ sin(�Ω2t), with the
constants of pro-
portionality determined by the initial conditions. In other
words, in the body frame, Ω

describes a cone around e3, with period
Tpr =
2π
�Ω3
(7.4)
This motion is called free precession.
The angle χ can be large or small and is determined by initial
conditions (Ω3 = Ω cos χ).
What does this motion look like in the lab frame? We can
answer this question in two
ways. Firstly, by brute force, we can solve for the Euler angles
θ(t), φ(t), ψ(t) given Ω1(t),
Ω2(t), Ω3(t) from the equations
Ω1 = sin ψθ̇ − sin θ cos ψφ̇ (7.5)
Ω2 = cos ψθ̇ + sin θ sin ψφ̇ (7.6)
Ω3 = ψ̇ + cos θφ̇ (7.7)
41

7.1 WORKED EXAMPLE: FREE PRECESSION 640-213
Melatos
We do not prove these equations in this subject, because the
algebra is messy. Nor do
you need to remember them for the exam! But you can easily
solve them on a computer
for θ(t), φ(t), ψ(t) given Ω1(t), Ω2(t), Ω3(t), which then tells
you how the (e1, e2, e3) axes
(and hence the body) tumble in space.
Alternatively, and more elegantly, we note that the total angular
momentum M is constant
(i.e., fixed) in the lab frame, as the net external torque is zero.
Moreover, the top is
axisymmetric, so we can always arrange e2(t) to lie
perpendicular to the plane containing
M and the instantaneous axis of symmetry e3(t).
Then Ω2 = M2/I2 (see Chapter 6) must vanish (since M · e2 =
0). In other words, Ω has
zero projection along e2 and therefore lies in the plane

containing M and e3 at all times.
Note that this plane has unit normal ∝ M × e3 (or equivalently
∝ Ω × e3).
Now consider the motion of a point at some arbitrary distance
x3 along the axis of the top.
Its position vector is x′ = x3e3 and its velocity in the lab frame
is v′ = Ω×x′ = Ω×x3e3.
Clearly, v′ is parallel to the normal of the above plane. If the
velocity of the e3 axis is
always directed perpendicular to the plane containing M and e3,
it means that e3 rotates
uniformly about M (and so does Ω, since it is coplanar with M
and e3). (Experiment
with a piece of paper if you need convincing.) Simultaneously,
the top spins uniformly
about its own axis (the body-frame rotation as seen in the lab
frame).
42
7.2 [EXT] NUTATION 640-213 Melatos

The angular velocities of the two rotations can be expressed in
terms of M and the wobble
angle θ (between M and e3), which are constant. The top spins
about its own axis with
angular velocity Ω · e3 (project Ω along e3). This is just Ω3 =
M3/I3 (from Chapter 6)
= M cos θ/I3 (from diagram). The top also spins with angular
velocity Ωpr about M.
Writing Ω two ways, i.e., Ω1e1 + Ω3e3 = Ω = ΩprM
̂ + Ω
′
3e3, and dotting with e1 and e3,
we find Ω1 = Ωpr sin θ and Ω3 = Ωpr cos θ+ Ω
′
3. This gives Ωpr = Ω1/ sin θ = M1/I1 sin θ =
M/I1 (since M1 = M sin θ). Part of this comes from components
along e1 and e3. The
remainder (Ω′3e3) is already included in the above motion
(rotation about e3).
7.2 [Ext] Nutation
Now consider an axisymmetric top in a gravitational field. The

external torque is not zero.
Suppose the vertex O of the top is held fixed. The centre of
mass C obviously lies along
the axis of symmetry e3. Let |OC| = l. We need three
generalised coordinates to describe
the motion: the colatitude θ(t) and longitude φ(t) of C, and the
angular displacement
ψ(t) of the top as it spins about its symmetry axis.
43
Lagrangian:
L =
1
2
Mv2c +
1

2
I1Ω
2
1 +
1
2
I2Ω
2
2 +
1
2
I3Ω
2
3 − MgzC (7.8)
Potential energy is given by Mg × (altitude of C), as if the
whole mass is concentrated
at that single point.

MgzC = Mgl cos θ (7.9)
Translational kinetic energy of C:
1
2
Mv2C =
1
2
M(vCθ
2 + vCφ
2) can’t move radially (7.10)
=
1
2
Ml2(θ̇2 + sin2 θφ̇2) (7.11)
Ω1, Ω2, Ω3 can be written in terms of φ, θ, ψ as in previous

section. With I1 = I2, we get
1
2
I1Ω
2
1 +
1
2
I2Ω
2
2 +
1
2
I3Ω
2
3

=
1
2
I1
[
(sin ψθ̇ − sin θ cos ψφ̇)2 + (cos ψθ̇ + sin θ sin ψφ̇)2
]
+
1
2
I3(ψ̇ + cos θφ̇)
2 (7.12)
=
1
2
I1(θ̇

2 + sin2 θφ̇2) +
1
2
2 (7.13)
i.e.,
L =
1
2
(Ml2 + I1)(θ̇
2 + sin2 θφ̇2) +
1
2
2 − Mgl cos θ (7.14)

Now we see why Lagrangian methods are so powerful. Without
writing down Lagrange’s
equations, we can identify by inspection three constants of the
motion:
1. L is independent of ψ ⇒ ∂L/∂ψ̇ = constant
i.e.,
ψ̇ + cos θφ̇ = constant (call it n) (7.15)
where n absorbs a factor I−13 after dividing through both sides
2. L is independent of φ ⇒ ∂L/∂φ̇ = constant
sin2 θφ̇ + 2p cos θ = constant (call it 2λ) (7.16)
where 2p = nI3/(Ml
2 + I1) and 2λ absorbs a factor (Ml
2 + I1)
−1 after dividing
through both sides
3. L is independent of t

⇒ θ̇∂L
∂θ̇
+ ψ̇
∂L
∂ψ̇
+ φ̇
∂L
∂φ̇
− L = constant (7.17)
44
i.e.,
θ̇2 + sin2 θφ̇2 + 2q cos θ = constant (call it 2μ) (7.18)
where 2q = 2Mgl/(Ml2 + I1) and 2μ absorbs a term I3n
2/(Ml2 + I1) after dividing

through by Ml2 + I1 on both sides
Upon substituting (7.16) into (7.18), we get an equation
involving just θ:
1
2
θ̇2 +
2(λ − p cos θ)2
sin2 θ
+ q cos θ = μ (7.19)
In principle, we can always solve this equation for θ(t), e.g.,
numerically, on the computer.
This is what most working physicists would do. (Try it if you
have access to a package
like Mathematica.)
However, we can learn a good deal about the motion by
inspection, drawing on the ideas
regarding the effective potential in Chapter 3. We simplify the

task by looking at z = cos θ
(altitude l cos θ of the centre of mass about the midplane, in
units of l) instead of θ. (N.B.
−1 ≤ z ≤ 1.)
From ż = − sin θθ̇, we find
1
2
ż2 = −2(λ − pz)2 + (μ − qz)(1 − z2) (7.20)
Cubic in z, with 1
2
ż = −2(λ + p)2 at θ = π (south pole), μ − 2λ2 at θ = π/2
(equator),
and −2(λ − p)2 at θ = 0 (north pole).
Cases:
45

Motion is impossible for 1
2
ż2 < 0, i.e., cases A, B, and E in the figure.
In cases D and F, z (and hence θ) bounces between two turning
points θ1 and θ2 (see
Chapter 3), where the top is momentarily at rest in the θ (but
not the φ) direction
( 1
2
ż2 = 1
2
sin2 θθ̇2 = 0). In other words, the axis of the top wobbles in a
band of colatitude.
This is called nutation.
To see what the wobble looks like, combine (7.15) and (7.16) to
obtain
φ̇ =
2(λ − pz)

1 − z2 (7.21)
Clearly φ̇ changes sign at cos θs = λ/p. If θs lies between θ1 and
θ2, then the top travels in
longitude in opposite directions at the two turning points;
otherwise, it travels in longitude
in the same direction always. If θs coincides with θ2, the top
momentarily comes to rest
in longitude as well as colatitude. Here are pictures of the three
cases.
46
47
7.3 CHANDLER WOBBLE 640-213 Melatos
Case C is a special case of the above motion where θ1 = θ2
coincide, i.e., there is only

one colatitude where stable rotation is possible ( 1
2
ż = 0). This corresponds to steady
precession, similar qualitatively to Section 7.1 (but not torque
free).
Exercise. [Ext] Can you predict what happens in Case G? (Hint:
This is
sometimes called a “sleeping top”.)
7.3 Chandler wobble
References: Munk, W. H. & MacDonald, G. J. F., 1960, The
Rotation of the Earth,
CUP, and Lambeck, K., 1980, The Earth’s Variable Rotation:
Geophysical Causes and
Consequences, CUP.
Discovered in 1891, the Chandler wobble is a small nutation of
the Earth’s rotation axis
about its symmetry axis:
• amplitude ≈ 0.7” (≈ 22 m at pole)

• period ≈ 433 days
(Not the same as the precession of the equinoxes, whose period
is 25,700 years, which is
caused by the tidal gravitational forces exerted by the Sun and
Moon, and which causes
the constellations to shift over millennia, invalidating
astrology!)
Centrifugal bulge:
I3 − I1
I1
≈ 1
355
(Euler) (7.22)
Expect Chandler wobble period to be 355 days. Discrepancy
explained by the elasticity
of the Earth’s mantle and sloshing of the oceans.
Why isn’t the Chandler wobble damped? Friction inside the
fluid Earth should kill it
in ∼ 102 years! This major mystery was resolved in 2000, when

scientists from NASA’s
Jet Propulsion Laboratory tied down precisely the sources of
excitation: mainly pressure
fluctuations on the ocean floor, and to a lesser extent
fluctuations in atmospheric pressure.
(By contrast, atmospheric winds and ocean currents are of minor
importance.)
If you want to learn more about this cool topic, please consult
the reference books men-
tioned above.
48
8.1 WORKED EXAMPLE: TRIATOMIC MOLECULE 640-213
Melatos
Chapter 8
Coupled Linear Oscillators
In this lecture, we study the oscillation modes of coupled,
small-amplitude (i.e., linear)

oscillators. A familiar example is a pair of strings on a violin,
which vibrate individually
but are also coupled (weakly) through the bridge. If you pluck
one string, the other
vibrates in sympathy. Another example is a pair of pendula,
joined by a weak spring. If
you set one pendulum in motion, it gradually transfers its
energy to the other, and vice
versa, creating a sloshing effect.
8.1 Worked example: triatomic molecule
Consider the triatomic molecule CO2, whose bond structure is
O=C=O. In the simplest
case, the three atoms are collinear. The force between two
neutral atoms is given by
minus the gradient of the Lennard-Jones potential, VLJ ,
graphed below:
At equilibrium, the atoms are separated by a distance req (where
VLJ is a minimum). For
small displacements around req , the atoms feel a Hooke’s law
restoring force ∝ −(r−req)
which brings them back towards the equilibrium point.

e.g.,
r < req, −
dVLJ
dr
> 0 (from graph), pushes to right (8.1)
r > req, −
dVLJ
dr
< 0 pushes to left (8.2)
Hooke’s law holds almost always near an energy minimum.
[Occasionally the restoring
force scales ∝ −(r − req)3, if the minimum is quartic.]
49
Melatos

Our first job is to figure out what the equilibrium state looks
like. Let b denote the equi-
librium length of each bond (before the molecule is plucked).
The bonds are chemically
identical, so they have the same effective spring constant
(which is proportional to the
curvature of VLJ in the vicinity of req). Let’s call it k.
Exercise. (a) Guesstimate k for a C=O bond without looking up
any
books. [Hint: roughly how much energy does it cost to break the
bond, and
roughly how long is it?] (b) Hence guesstimate the oscillation
frequency of the
molecule.
Lagrangian: L = T − V
T =
1
2
mẋ1
2 +

1
2
Mẋ2
2 +
1
2
mẋ3
2 (8.3)
V =
1
2
k(x2 − x1 − b)2 +
1
2
k(x3 − x2 − b)2 (8.4)
Clearly an equilibrium (d/dt = 0) occurs when V is a minimum,

i.e., when x1 = x2 − b
and x3 = x2 + b. By symmetry, the molecule looks the same no
matter how we pick x2.
So we choose x1 = −b, x2 = 0, x3 = b for simplicity.
Now change to coordinates (ξ1,ξ2,ξ3) which measure the
displacement of each atom from
its equilibrium position. These displacements are usually small.
That is, we nudge the
50
Melatos
system slightly away from its equilibrium and watch it vibrate.
x1 = ξ1 − b ẋ1 = ξ̇1 (8.5)
x2 = ξ2 ẋ2 = ξ̇2 (8.6)
x3 = ξ3 + b ẋ3 = ξ̇3 (8.7)
L =
1

2
mξ̇21 +
1
2
Mξ̇22 +
1
2
mξ̇23
−1
2
k(ξ2 − ξ1)2 −
1
2
k(ξ3 − ξ2)2 (8.8)
Lagrange’s equations:

0 = mξ̈1 − k(ξ2 − ξ1) (8.9)
0 = Mξ̈2 + k(ξ2 − ξ1) − k(ξ3 − ξ2) (8.10)
0 = mξ̈3 + k(ξ3 − ξ2) (8.11)
Look for harmonic solutions ξi ∝ e−iωt. (NB. ξ1, ξ2, ξ3 all
have the same ω otherwise they
would fall out of sync and clash catastrophically, disrupting the
motion.)
0 =
⎛
⎝ −ω
2m + k −k 0
−k −ω2M + 2k −k
0 −k −ω2m + k
⎞
⎠
⎛
⎝ ξ1ξ2
ξ3

⎞
⎠ (8.12)
The matrix equation has the trivial solution (ξ1,ξ2,ξ3) = (0, 0,
0) unless the determinant
vanishes, which only happens at certain special frequencies
satisfying
0 = (k − mω2)[(k − mω2)(2k − Mω2) − k2] + k(−k)(k − mω2)
(8.13)
= (k − mω2)(mMω2 − kM − 2km)ω2 (8.14)
i.e.,
ω2 = 0, k/m, k(M + 2m)/Mm (8.15)
Real ω implies pure oscillations. If ω has an imaginary part,
then e−iωt ∝ e−i[Re(ω)+iIm(ω)]t ∝
e[Im(ω)]te−i[Re(ω)]t represents a growing or decaying
oscillation depending on the sign of
Im(ω). This situation does not arise unless L depends explicitly
on time t (energy not
conserved; see Chapter 3).
What kind of motions correspond to the above frequencies? To
answer this question, we

solve for (ξ1,ξ2,ξ3) in each case.
Exercise. Show that the special case ω2 = 0 reduces to ξ1 = ξ2 =
ξ3 ∝ t, i.e.,
translation of the whole molecule at uniform speed.
51
Melatos
For ω2 = k/m:
0 =
⎛
⎝ 0 −k 0−k −kM
m
+ 2k −k
0 −k 0

⎞
⎠
⎛
⎝ ξ1ξ2
ξ3
⎞
⎠ (8.16)
i.e., ξ2 = 0, ξ3 = −ξ1.
For ω2 = k(M + 2m)/Mm:
0 =
⎛
⎝ −2km/M −k 0−k kM/m −k
0 −k −2km/M
⎞
⎠
⎛
⎝ ξ1ξ2

ξ3
⎞
⎠ (8.17)
i.e., ξ1 = ξ3 = −(M/2m)ξ2.
Note that the overall amplitude is arbitrary in all cases because
we do not specify how hard
we pluck the molecule. Just like in a small-angle (i.e., linear)
pendulum, the amplitude
is independent of the frequency; it depends only on initial
conditions.
52
9.1 NONLINEAR RESPONSES 640-213 Melatos
Chapter 9
Nonlinear Oscillators

Reference: Feynman Lectures on Physics, Volume 1, Section
50.6
9.1 Nonlinear responses
Consider an electronic amplifier, e.g., in a stereo system. The
amplifier converts a voltage
Vin at its input terminals into an output voltage Vout.
• linear: Vout(t) = kVin(t)
• nonlinear: Vout(t) = k [Vin(t) + �Vin(t)2]
Suppose the input Vin(t) = A cos ωt is a pure tone.
Vout(t) = kA cos ωt + k�A
2 cos2 ωt (9.1)
= kA cos ωt +
1
2
k�A2(1 + cos 2ωt) (9.2)
Output distorted:

• anharmonic waveform
• rectification (DC component)
• harmonic generation (2ω component)
53
9.2 DAMPED, DRIVEN, SIMPLE HARMONIC OSCILLATOR
640-213 Melatos
Now suppose the input Vin(t) = A cos ω1t + B cos ω2t is a
combination of two pure tones.
Vout(t) = kA cos ω1t + kB cos ω2t
+k�
(
A2 cos2 ω1t + 2AB cos ω1t cos ω2t + B
2 cos2 ω2t
)
(9.3)

= kA cos ω1t +
1
2
k�A2(1 + cos 2ω1t)
+kB cos ω2t +
1
2
k�B2(1 + cos 2ω2t)
+k�AB [cos(ω1 + ω2)t + cos(ω1 − ω2)t] (9.4)
Output modulated:
• principle of superposition (which you know and love from
your studies of wave
diffraction and interference) does not apply in nonlinear
systems
• generate sum and difference frequencies ⇒ beat notes (e.g. at
frequencies 2ω1, 2ω2,
and ω1 ± ω2)

• if ω1 ≈ ω2 then you get a woo-woo-woo effect, e.g., wolf note
on a violin
Nonlinear effects (∝ V 2in, V 3in, etc.) are stronger at larger
amplitudes, e.g., human ear;
loudspeaker driven to saturation; laser fusion experiments.
9.2 Damped, driven, simple harmonic oscillator
An SHO is linear by definition:
ẍ + γẋ + ω20x = F0 cos ωt (9.5)
Think of it as a linear amplifier: if you drive it with a pure tone
F0 cos ωt (input), you
get back a pure tone (output) at the same frequency (but with
different amplitude and
phase, in general).
54
9.3 PERIOD-AMPLITUDE RELATION 640-213 Melatos

e.g., if we write the output as x(t) = 1
2
Ae−iωt + c.c., then the complex amplitude A of the
output satisfies
(−ω2 − iωγ + ω20)
1
2
Ae−iωt + c.c. =
1
2
F0e
−iωt + c.c. (9.6)
i.e.,
A =
F0
ω20 − ω2 − iωγ

(9.7)
=
F0√
(ω20 − ω2)2 + ω2γ2
eiθ (9.8)
with tan θ = ωγ/(ω20 − ω2). Damping causes a phase shift; the
output leads or lags the
driving force. If we sweep ω sweeps through ω0, θ flips by π
radians.
9.3 Period-amplitude relation
Now consider an undriven, undamped oscillator with (say) a
cubic nonlinearity. An
example is a pendulum with a moderate swing angle θ, which
obeys the equation 0 =
θ̈ + (g/l) sin θ ≈ θ̈ + (g/l)(θ − 1
6
θ3 + ...). In general:
ẍ + ω20x = −�x3 (9.9)

We take � to be small, so that the oscillator is approximately an
SHO. 9
Clearly, if we start with a pure tone x(t) = 1
2
Aeiωt + c.c., the nonlinear term will generate
new harmonics ∝ (Aeiωt + A∗ e−iωt)3 = A3e3iωt + 3A2A∗ eiωt
+ 3AA∗ 2e−iωt + A∗ 3e−3iωt (see
Section 9.1). If ω = ω0 (the only possibility, as the system is
undriven), the e
±iωt terms on
the right-hand side of (9.9) cause big trouble; they act as
internal drivers of the undamped
linear oscillation ẍ + ω20x = 0 and hence produce an infinite
resonant response. This is
9� is sometimes large, e.g. in relaxation oscillators like the van
der Pol oscillator or the human heart,
as discussed below.
55

9.4 [EXT] BISTABILITY IN A DRIVEN, DAMPED,
ANHARMONIC
OSCILLATION 640-213 Melatos
impossible, as it violates conservation of energy, so the
amplitude A must take a special
value that kills the e±iωt terms. This special value satisfies the
condition
(−ω2 + ω20)
1
2
Aeiωt + c.c. = −3�
8
A2A∗ eiωt + c.c. , (9.10)
obtained by susbtituting the pure tone into (9.9) and equating
the coefficients of the eiωt
terms to make them disappear. Upon rearranging and noting that
ω20 − ω2 = (ω0 −

ω)(ω0 + ω) ≈ 2ω0(ω0 −ω) for ω very close to ω0 (again, the
only possibility, as the system
is undriven), we obtain
ω = ω0 +
3�
8ω0
|A|2 . (9.11)
This result tells us that, when a nonlinear spring oscillates
stably, its oscillation frequency
ω is shifted slightly away from its natural frequency ω0. For a
cubic nonlinearity, the shift
(also called the detuning) is proportional to the square of the
oscillation amplitude |A|2.
This is an important general result with many practical
applications. It is called the
period-amplitude relation. E.g., for a pendulum with amplitude
θmax, we have � = −ω20/6
and hence ω ≈ ω0(1 − 116θ2max).
The e±3iωt terms remain and beat together with the natural
oscillation at ω0 to generate
responses at 3ω ± ω0, which are then fed back into the
nonlinear term to generate even

more harmonics. And so on.
9.4 [Ext] Bistability in a driven, damped, anharmonic
oscillation
Now suppose that the driven, damped SHO is also generalised
to include a small (� � 1)
nonlinear (cubic, say) term in the restoring force, in the spirit of
the previous section.
Equation of motion:
ẍ + γẋ + ω20x = −�x3 + F0 cos ωt (9.12)
Amplitude of response (see Section 9.2):
|A|2 = F
2
0
(ω20 − ω2)2 + ω2γ2
(9.13)
Period-amplitude relation (see Section 9.3):

ω = ω0 +
3�
8ω0
|A|2 (9.14)
Combining, and keeping terms up to order �2 in the
denominator, we find
|A|2 = F
2
0
γ2ω20 +
3�γ2
4
|A|2 + 9�2
16
(
1 + γ
2

4ω20
)
|A|4
(9.15)
Equation (9.15) is a cubic in |A|2. Plot |A| versus ω − ω0 (c.f.
Section 9.2).
56
9.5 [EXT] PARAMETRIC RESONANCE 640-213 Melatos
The system is bistable! There are two stable states QR and PS
available for a range of
detunings ω−ω0, provided that F0 is large enough. (The central
state PR is unstable.) The
history of the system determines which state is occupied. For
example, if you approach
the bistable region by sweeping ω starting from ω � ω0 (i.e. ω
− ω0 very negative), you

end up in the state QR. In contrast, if you start from ω � ω0,
you end up in PS.
9.5 [Ext] Parametric resonance
Open system whose Lagrangian depends on time through
variations in the parameters.
Energy can be added to, or subtracted from, the system in this
way.
e.g., pendulum whose length changes with time
e.g., inverted pendulum whose base is raised and lowered
harmonically
Generally speaking, such a system exhibits an infinite response
(if undamped) for a range
of parameters, usually when the parameters oscillate at a simple
fraction multiple of the
natural frequency.
Important special case: oscillating natural frequency
ẍ + ω0(1 + � cos ωt)x = 0 (9.16)

Strongest response near 2ω0. Set ω = 2ω0 + σ, σ small. Look
for solutions of the form
x(t) = A(σt) cos(ω0 +
1
2
σ)t + B(σt) sin(ω0 +
1
2
σ)t (9.17)
where A(σt) and B(σt) are slowly varying amplitudes [ d
dt
A(σt) ∼ σ � ω0 ∼ ddt cos(ω0 +
1
2
σ)t.] Substituting into (9.16) and ignoring terms at 3(ω0 +
1
2
σ), which do not resonate

with the natural oscillation at ω0, we find (from coefficients of
cos and sin)
2A′ + Bσ +
1
2
�ω0B = 0 (9.18)
2B′ − Aσ + 1
2
�ω0A = 0 (9.19)
57
9.6 [EXT] RELAXATION OSCILLATOR 640-213 Melatos
where primes denote derivatives with respect to σt. Nontrivial
solutions ∝ eλt arise for
det

(
2λ σ + 1
2
�ω0
−σ + 1
2
�ω0 2λ
)
= 0 (9.20)
Resonant solutions (i.e., explicitly growing) arise for
0 ≤ λ2 = 1
4
[
(
1
2
�ω0)

2 − δ2
]
(9.21)
i.e., 1
2
�ω0 ≤ δ ≤ 12�ω0.
9.6 [Ext] Relaxation oscillator
Studied first by van der Pol in the 1920’s in connection with
electronic circuits involving
(triode) valves (which act as amplifiers). Nonlinear damping:
ẍ + ω20x + γ(x
2 − 1)ẋ = 0 (9.22)
Oscillation sustains itself even in the absence of a driver. When
the amplitude drops below
a threshold (here, |x| < 1), the damping supports the oscillation
instead of opposing it
[i.e., γ(x2 − 1) < 0]. When the amplitude exceeds the threshold,
the damping changes
sign and opposes the oscillation.

58
9.6 [EXT] RELAXATION OSCILLATOR 640-213 Melatos
For γ � 1, amplitude ≈ 2/
√
3 and period ≈ 1.6γ. A fixed amplitude (or, more generally,
an amplitude-period relation) is one of the main points of
difference between linear and
nonlinear oscillators; a linear oscillator like a small-amplitude
pendulum has arbitrary
(small) amplitude, independent of its period. C.f. human heart,
which is constantly
subjected to electrical stimuli but maintains its amplitude and
period in a tight band.
59
10.1 HAMILTON’S EQUATIONS 640-213 Melatos

Chapter 10
Hamiltonian dynamics
10.1 Hamilton’s equations
The Lagrangian is a function of time t, generalised coordinates
q1, ...,qn, and generalised
velocities q̇ 1, ..., q̇ n. Likewise, the generalised momenta
defined in Chapter 3 are functions
of the same 2n + 1 things.
pi =
∂L(t,q1, ...,qn, q̇ 1, ..., q̇ n)
∂q̇ i
i = 1, ...,n (10.1)
What happens if we rearrange the n equations in (10.1) to write
each generalised velocity
q̇ i in terms of t,q1, ...,qn, and p1, ...,pn? (This is always doable
in principle but often very
messy. Luckily, you never need to actually do it; it is important
simply to know that it can

be done.) Then we have a new description of the motion in
terms of 2n + 1 independent
things: time t, the “old” generalised coordinates q1, ...,qn, and
the generalised momenta
p1, ...,pn. This procedure is called a Legendre transformation.
We can’t keep using the Lagrangian L in this new description,
because it depends on
the q̇ i’s as independent variables, not the pi’s. Instead we use a
new function, called the
Hamiltonian:
H = q̇ j
∂L
∂q̇ j
− L (10.2)
= q̇ j(t,q1, ...,qn,p1, ...,pn)pj
−L(t,q1, ...,qn, q̇ 1(t,q1, ...,qn,p1, ...,pn), ..., q̇ n(t,q1, ...,qn,p1,
...,pn))(10.3)
(Remember Einstein’s shorthand: repeated index j means
∑n

j=1.)
You already met H in Chapter 3 when learning about constants
of the motion. Back
then, it was a function of t, the qi’s, and the q̇ i’s. Now, we
reexpress it in terms of t, the
qi’s and the pi’s: H = H(t,q1, ...,qn,p1, ...pn).
Differentiate H with respect to the qi’s:
∂H
∂qi
=
∂q̇ j
∂qi
pj −
(
∂L
∂qi
+
∂L

∂q̇ j
∂q̇ j
∂qi
)
(chain rule) (10.4)
=
∂q̇ j
∂qi
pj −
∂L
∂qi
− pj
∂q̇ j
∂qi
(definition of pj) (10.5)
=

∂L
∂qi
(10.6)
= −dpi
dt
(from Lagrange’s equations) (10.7)
60
10.2 WORKED EXAMPLE: 2D SHO 640-213 Melatos
Differentiate H wth respect to the pi’s:
∂H
∂pi
=
∂q̇ j
∂pi

pj + q̇ i −
∂L
∂q̇ j
∂q̇ j
∂pi
(chain rule) (10.8)
=
∂q̇ j
∂pi
pj + q̇ i − pj
∂q̇ j
∂pi
(definition of pj) (10.9)
= −dqi
dt
(10.10)

Equations (10.7) and (10.10) are Hamilton’s equations. Given H
as a function of t,
the qi’s, and the pi’s, these equations are a set of 2n first-order
ODEs, whose solution
q1(t), ...,qn(t),p1(t), ...,pn(t) completely describes the motion.
This motion traces out a
path in phase space.
e.g., pendulum
Exercise. Prove ∂H/∂t = −∂L/∂t in a similar way.
10.2 Worked example: 2D SHO
Consider a block of mass m sliding on a plane and attached to
the origin by a spring with
spring constant k.
Cartesian coordinates (x1,x2):
61

10.2 WORKED EXAMPLE: 2D SHO 640-213 Melatos
L = KE − PE (10.11)
=
1
2
mẋ21 +
1
2
mẋ22 −
1
2
k(x21 + x
2
2) (10.12)
By definition, the generalised momenta are

p1 =
∂L
∂ẋ1
= mẋ1 (10.13)
p2 =
∂L
∂ẋ2
= mẋ2 (10.14)
Obviously we can now write the velocities ẋ1 and ẋ2 in terms of
the coordinates x1, x2
and momenta p1, p2:
ẋ1 = ẋ1(t,x1,x2,p1,p2) =
p1
m
(10.15)
ẋ2 = ẋ2(t,x1,x2,p1,p2) =
p2
m

(10.16)
In fact, in this simple situation, the coordinates don’t even
appear, just the momenta.
Normally things are not so clean. E.g., for the same problem in
(r,θ) coordinates, we have
L = 1
2
m(ṙ2 + r2θ̇2) − 1
2
kr2, pr = ∂L/∂ṙ = mṙ, pθ = ∂L/∂θ̇ = mr
2θ̇, and hence ṙ = pr/m,
θ̇ = pθ/mr
2, so that θ̇ is a function of both the momentum pθ and the
coordinate r.
We can now write out the Hamiltonian, which is a function of t,
the qi’s, and the pi’s.
H = ẋ1(t,x1,x2,p1,p2)p1 + ẋ2(t,x1,x2,p1,p2)p2

−L(t,x1,x2, ẋ1(t,x1,x2,p1,p2), ẋ2(t,x1,x2,p1,p2)) (10.17)
=
(p1
m
)
p1 +
(p2
m
)
p2
−
[
1
2
m
(p1
m
)2

+
1
2
m
(p2
m
)2
− 1
2
kx21 −
1
2
kx22
]
(10.18)
=
p21

2m
+
p22
2m
+
1
2
kx1
2 +
1
2
kx2
2 (10.19)
The final step is to convert Lagrange’s equations to Hamilton’s
equations (which are
equivalent, of course).
dp1

dt
=
∂L
∂x1
(Lagrange) (10.20)
= −kx1 (10.21)
= −∂H
∂x1
(by inspection from (10.19)) (10.22)
dx1
dt
= ẋ1 (by definition) (10.23)
=
p1
m
(from equation (10.15)) (10.24)

=
∂H
∂p1
(by inspection from (10.19)) (10.25)
62
10.3 CONSTANTS OF THE MOTION 640-213 Melatos
Similarly for dp2/dt = −∂H/∂x2 and dx2/dt = ∂H/∂p2.
10.3 Constants of the motion
Suppose you have some function F(t,q1(t), ...,qn(t),p1(t),
...,pn(t)) and you are curious to
know how it changes with time as the system evolves. F will
usually be some quantity
of physical importance, e.g., total energy, distance between two
particles, etc. Clearly,
F can depend on t explicitly, but it also depends on t implicitly
through the qi(t)’s and
pi(t)’s:

dF
dt
=
∂F
∂t
+
∂F
∂qj
dqj
dt
+
∂F
∂pj
dpj
dt

(chain rule) (10.26)
=
∂F
∂t
+
∂F
∂qj
∂H
∂pj
− ∂F
∂pj
∂H
∂qj
(Hamilton’s equations) (10.27)
The last two terms on the RHS (which contain sums over
repeated index j = 1, ...,n) crop
up again and again, so we give them a special name, the Poisson

bracket. The symbol for
the Poisson bracket is curly braces, viz.
{f,g} = ∂f
∂qj
∂g
∂pj
− ∂f
∂pj
∂g
∂qj
(10.28)
for any two functions f and g. In this notation, equation (10.27)
can be written as
dF
dt
=
∂F

∂t
+ {F,H} (10.29)
Exercise. From this formula and the previous exercise, prove
that H itself is
a constant of the motion in a closed system (where the
Lagrangian does not
depend on t explicitly).
Exercise. Prove from (10.28) the following useful properties of
the Poisson
bracket.
1. Antisymmetry: {f,g} = −{g,f}
2. Bilinearity: {c1f + c2g,h} = c1{f,h} + c2{g,h} for constants
c1,c2
3. Product rule: {fg,h} = f{g,h} + g{f,h}
4. Jacobi identity: {f,{g,h}} + {g,{h,f}} + {h,{f,g}} = 0
5. Explicit time dependence: ∂{f,g}/∂t = {∂f/∂t,g} + {f,∂g/∂t}
63

10.3 CONSTANTS OF THE MOTION 640-213 Melatos
From these properties and (10.29), we can show that {F,G} is a
constant of the motion
if F and G are.
d
dt
{F,G} = ∂
∂t
{F,G} + {{F,G},H} (10.30)
= {∂F
∂t
,G} + {F, ∂G
∂t
} + {{F,G},H} (10.31)
= −{{F,H},G} − {F,{G,H}} + {{F,G},H} (10.32)
= {{H,F},G} + {{G,H},F} + {{F,G},H} (10.33)
= 0 (10.34)

Sometimes, when you actually write out {F,G} for your system,
you will get something
which looks nothing like F and G. This is great! It means you
have discovered a new,
independent constant of the motion with minimal effort. Chapter
3 shows that the con-
stants of the motion are a useful solution tool. On the other
hand, {F,G} is often trivial
(e.g., zero, unity) or just a simple combination of F and G. In
this case, you have not
discovered a new constant of the motion and are obliged to keep
looking.
Exercise. (a) Argue that the hard-to-guess constant p2θ + p
2
φ/ sin
2 θ in the
central force problem (Chapter 3) cannot be constructed from
Poisson brackets
involving pφ and H. (b) Separately, prove that {p2θ + p2φ/ sin2
θ,pφ} is trivial.
64

11.1 DETERMINISM AND CHAOS 640-213 Melatos
Chapter 11
[Ext] Chaos and turbulence
11.1 Determinism and chaos
Systems obeying the laws of classical mechanics, whether in the
Newtonian, Lagrangian,
or Hamiltonian formulation, are deterministic. If the state of the
system is known at time
t, e.g., qi(t) and pi(t) for i = 1, ...,n, then the equations of
motion predict its state at
t + Δt exactly. It is therefore tempting to presume that classical
mechanical systems are
predictable, unlike their quantum mechanical counterparts,
which are inherently proba-
bilistic. The quality of prediction depends on how well you
know the initial state, which
may be “not very” if the system has a huge number of degrees
of freedom. But, until
recently, this was viewed as a practical limitation, not a

fundamental one.
‘Consider an intelligence which, at any instant, could have a
knowledge of all
forces controlling nature together with the momentary
conditions of all the
entities of which nature consists. If this intelligence we
powerful enough to
submit all this data to analysis it would be able to embrace in a
single formula
the movements of the largest bodies in the universe and those of
the lightest
atoms; for it nothing would be uncertain; the future and the past
would be
equally present to its eyes.’ (Pierre Simon Laplace, 1749-1827)
The comforting presumption of predictability was overthrown in
the 1960’s and 1970’s by
researchers studying nonlinear systems, such as the
mathematical biologist Robert May10
and the meteorologist Edward Lorenz, building on pioneering
work by Henri Poincaré
in the 1890’s. These researchers showed that many classical
deterministic systems are

chaotic. That is, they are sensitively dependent on initial
conditions: a small difference
in initial conditions, say in the kth decimal place, grows
geometrically rather than alge-
braically with time, reaching order unity after ∼ k time steps. In
the absence of perfect
initial measurements, such systems are unpredictable.
Systems are chaotic under a variety of conditions, but three
prerequisites are:
• nonlinearity
• three or more degrees of freedom
10Robert McCredie May (now Baron May of Oxford!) received
his PhD from the School of Physics at
the University of Sydney, where he developed the modern
theory of superconductivity at the same time
as Bardeen, Cooper and Schrieffer. He then moved to Harvard,
Princeton, and Oxford, where he helped
found the field of mathematical ecology, studying the evolution
of animal populations using tools like the
logistic map.
65

11.2 PLANETARY RINGS 640-213 Melatos
• driving and damping (i.e., open system)
Examples:
• gravitational 3-body problem and long term fate of our Solar
System
• double pendulum
• electronic voltages in neurons (Hodgkin-Huxley model; 1963
Nobel Prize in Physi-
ology or Medicine)
• weather and climate
• eye tracking disorder in schizophrenics
• electronic circuits like self-exciting oscillators and coupled,
superconducting Joseph-
son junctions
• crickets that chirp in unison

• lasers
• quantum chaos? (The question mark appears because, at the
time of writing, re-
searchers are engaged in a bun fight over whether quantum
chaos can exist at all!
Schrödinger’s equation for a single particle is linear in the wave
function, so it does
not exhibit chaotic dynamics. But we know that the classical
limit does! Weird...)
11.2 Planetary rings
Kirkwood gaps: depleted rings in the asteroid belt at radii where
the orbital period is a
simple fraction of Jupiter’s orbital period, e.g., 3:1 at r ≈ 2.5
AU, 5:2 at r ≈ 2.82 AU.
Saturn’s rings, e.g., Cassini’s division: large gap between the A
and B rings, near the 1:2
orbital resonance of the moon Mimas.
Orbital resonance: if the two orbital periods are commensurable
(i.e., ratio is a simple
function), the bodies occupy a limited set of relative positions
over time. Gravitational

kicks (e.g., when bodies are closest) add up over time, rather
than cancelling out randomly.
Full explanation of the Kirkwood gaps requires collisions, i.e.,
damping.
Full explanation of Cassini’s division requires self-gravity of
the rings and excitation of
spiral density waves. Cassini’s division is wider than a simple
resonance predicts, and the
1:2 resonance is at inner edge of gap, not at the centre.
66
11.3 PERIODICALLY KICKED ROTOR 640-213 Melatos
11.3 Periodically kicked rotor
Impulsive torque N(φ) applied at times 0,τ, 2τ, ....
φ̇ = ω (11.1)
ω̇ = −γω + KN(φ)

∞∑
n=0
δ(t − nτ) (11.2)
Integrate equations of motion:
ωn+1 = [ωn + KN(φn)]e
−γτ (11.3)
φn+1 = φn + γ
−1(1 − e−γτ )[ωn + KN(φn)] (11.4)
where ωn+1 and ωn correspond to times (n + 1)τ and nτ.
Take the limit where kicks and damping are both large, with K/γ
= constant (call it c).
Then e−γτ is tiny, and K/γ is much larger than ωn/γ.
φn+1 = φn + cN(φn) (11.5)
Let N(φ) have a quadratic nonlinearity, with cN(φ) = (λ − 1)φ −
λφ2.

φn+1 = λφn(1 − φn) (11.6)
67
11.3 PERIODICALLY KICKED ROTOR 640-213 Melatos
This equation is called the logistic map. You may remember it
as the discretised form of
the continuous population evolution equation
dφ
dt
= λφ(1 − φ) , (11.7)
whose solution is a simple curve, called a sigmoid.
By contrast, the iterated map has a wild solution! What is its
behaviour as n → ∞?
0 ≤ λ ≤ 1 φn → 0 (independent of φ0)
1 ≤ λ ≤ 3 φn → 1 −
1

λ
3 ≤ λ ≤ 1 +
√
6 φn →
1
2
[
1 +
1
λ
±
√
(1 − 1
λ
)2 − 4
λ2

]
hops between two fixed points
1 +
√
6 ≤ λ ≤ 3.54... φn → hops between 4 fixed points
...
3.569... ≤ λ ≤ 3.624... no limit at all, very sensitive to φ0
(chaos)
3.624... ≤ λ ≤ 6.640... φn → hops between 3 fixed points
Bifurcations! Period-doubling route to chaos.
Exercise. A fixed point of period two satisfies φn+2 = φn. Show
that two
such points exist for 3 ≤ λ ≤ 1 +
√
6, with the values in the table above.
68

11.5 BÉNARD CONVECTION 640-213 Melatos
Exercise. [Ext] If λm is the value of λ where the mth bifurcation
occurs,
show that
λm − λm−1
λm+1 − λm
→ 4.669... as m → ∞
The magic number 4.669... is called the Feigenbaum number. It
appears in
“all” chaotic maps; it is universal. [Hint: This is an extremely
tough problem!
Look it up in a book if you would like more information, e.g.
Strogatz, S.,
2001, Nonlinear Dynamics and Chaos, Perseus.]
11.4 Mandelbrot set
Iterated map for complex numbers zn:
zn+1 = z
2

n + c , z0 = 0 (11.8)
We say that c ∈ Mandelbrot set (M) if |zn| remains bounded for
all n.
e.g.,
c = 1 zn = 0, 1, 2, 5, 26, ... c /∈ M
c = i zn = 0, i,−1 + i,−i,−1 + i, ... c ∈ M
Fractal. Copies on all scales (self similar).
11.5 Bénard convection
Viscous fluid heated from below.
Top view of circulation cells: hexagons.
69
11.5 BÉNARD CONVECTION 640-213 Melatos
Look at time evolution of modes.

Flow velocity: v(x, t) = x(t) cos kz
Temperature: T(x, t) = y(t) cos kz + z(t) cos 2kz
Substitute into fluid equations (Navier-Stokes and heat
equations) and get the Lorenz
equations:
ẋ = σ(y − x) (11.9)
ẏ = rx − y − xz (11.10)
ż = −bz + xy (11.11)
Here, σ denotes the ratio of the kinematic viscosity to the
thermal diffusivity, b is the
horizontal length-scale of the convection cells, and r ∝ ΔT is
the Rayleigh number (scaled
to its critical value for convection).
Equations (11.9) – (11.11) are not a Hamiltonian system (odd
number of variables; open;
driving and dissipation).
State as t → ∞:
r ≤ 1 no motion
1 ≤ r ≤ rc two convection rolls (clockwise, anticlockwise)

r ≥ rc chaos
with
rc =
σ(σ + b + 3)
σ − b − 1 (11.12)
Exercise. (a) Find the steady states of the Lorenz equations.
[Answer:
(0, 0, 0) and (±
√
b(r − 1),
√
b(r − 1),r − 1).] (b) Show that small perturba-
tions (ξ1,ξ2,ξ3) produce coupled linear oscillations obeying
ξ̇1 = −σξ1 − σξ2 (11.13)
ξ̇2 = ξ1 − ξ2 − [b(r − 1)]1/2ξ3 (11.14)
ξ̇3 = [b(r − 1)]1/2ξ1 + [b(r − 1)]1/2ξ2 − bξ3 (11.15)

(c) Show that the perturbations grow exponentially (i.e.,
unstable) for r > rc.
70
11.6 HYDRODYNAMIC TURBULENCE 640-213 Melatos
11.6 Hydrodynamic turbulence
Still unsolved after centuries of study. Is there an efficient and
mathematically compact
way to describe the beautiful and complicated (but not random)
transient patterns seen?
E.g., flow past a cylinder of radius R, speed V , kinematic
viscosity ν.
Dimensionless Reynolds number :
Re =
RV
ν

(11.16)
71
PHYC20014 Physical Systems
Wave Theory and Fourier Analysis: Tutorial 1
Tutorial problems
1. Basic Fourier series. We start with some “get to know you”
exercises. Sketch the periodic
extension of these functions and calculate their Fourier series.
(a) The square wave ⇧ : (�⇡ , ⇡ ] ! R:
⇧ (✓) =
8
><
>:

+1 ✓ > 0
0 ✓ = 0
�1 ✓ < 0.
(b) The triangle wave � : (�⇡ , ⇡ ] ! R:
�(✓) = ⇡ � |✓|.
2. Derivatives. Check that (where di↵ erentiable) �0 = �⇧ .
Di↵ erentiating term-by-term,
verify this relation also holds for the associated Fourier series.
3. Sine, cosine and half-range. An odd function has the property
that f(�✓) = �f(✓), while
an even function satisfies f(�✓) = f(✓).
(a) Show that for odd (even) functions, the Fourier coe�cients a
n
(b
n

) vanish. Since the
cosine terms vanish, an odd function has a sine series, and
similarly, an even function
has a cosine series.
(b) Prove that you can uniquely split an arbitrary function f into
odd and even parts:
f(✓) = f+(✓) + f�(✓), f±(�✓) = ±f±(✓).
Thus, the Fourier series for f splits into a cosine series for f+
and a sine series for f�.
(c) Recall from lectures that the half-range expansion of a
function is a Fourier series valid
over [0, L]. We can use either a cosine series (a
n
terms) or sine series (b

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