Poisson summation

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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 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

[ 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

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)

1.2 OPTIMAL PATH BY THE CALCULUS OF VARIATIONS 640-213 Melatos

Substituting the trial path in (1.1), we find

Strial =

∫ t2 t1

{ 1

2 m

[ dx0(t)

]2 + m

dx0(t)

dx1(t)

+ 1

2 m

[ dx1(t)

]2 − mgx0(t) − mgx1(t)

} (1.4)

= Strue +

∫ t2 t1

{ m dx0(t)

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

{ −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

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

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 q̇i (i = 1, ...,N). (We do this in Chapter 2.)

3. Can generalise to infinite degrees of freedom (N → ∞), whereupon the qi’s and q̇i’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).

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.

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.

dt = v (2.1)

dt =

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):

dt =

( 1 +

m2c2

)− 1 2

(2.3)

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.

2.2 GENERALISED COORDINATES 640-213 Melatos

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)

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 q̇i(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 q̇i(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, q̇1, ..., q̇n. One consequence is that expressions like ∂qi/∂qj and ∂q̇i/∂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

= q̇i/q̇j . 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

q̇j ∂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 q̇j∂xs/∂qj as shorthand for q̇1∂xs/∂q1 + ... + q̇n∂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 q̇i’s. On the other hand, q̇i is a function of t only, not of the qi’s (see “vital point” made earlier).

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).

dt = −ω sin ωt + mẋ(t)ẍ(t) − kx(t)ẋ(t)

2.4 DERIVATION OF LAGRANGE’S EQUATIONS 640-213 Melatos

Kinetic energy: function of t, qi, q̇i

T(t) =

N∑ s=1

2 msẋs(t) · ẋs(t) (2.8)

Generalised momentum:

pi(t) = ∂T(t)

∂q̇i (2.9)

= N∑

s=1

msẋs · ∂ẋs ∂q̇i

(2.10)

N∑ s=1

msẋs · ∂xs ∂qi

(2.11)

where the last line comes from differentiating (2.7) with respect to q̇i. 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

( ∂xs ∂qi

) + msẍs ·

∂xs ∂qi

(2.12)

= N∑

s=1

msẋs · ( ∂

∂t + q̇j

∂

∂qj

) ∂xs ∂qi

+ msẍs · ∂xs ∂qi

(2.13)

= N∑

s=1

msẋs · ∂

∂qi

( ∂

∂t + q̇j

∂

∂qj

) xs + msẍs ·

∂xs ∂qi

(2.14)

N∑ s=1

msẋs · ∂ẋs ∂qi

+ msẍs · ∂xs ∂qi

(2.15)

= ∂

∂qi

( N∑

s=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

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

+ q̇j ∂

∂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):

( ∂T

∂q̇i

) − ∂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)

2.5 CONSERVATIVE FORCES 640-213 Melatos

As long as V is independent of the velocities ẋs (or equivalently q̇i) 3 we get

( ∂L

∂q̇i

) − ∂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.

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:

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 +

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), q̇1(t), ..., q̇n(t), t] of the generalised coordinates and velocities which are independent of time when Lagrange’s

3.2 FIRST INTEGRALS 640-213 Melatos

equations d

( ∂L

∂q̇i

) − ∂L ∂qi

= 0 (3.11)

are satisfied. Here, independent of time means

0 = dF

dt (3.12)

= ∂F

∂t + q̇i

∂F

∂qi + q̈i

∂F

∂q̇i (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/∂q̇k is a constant of the motion.

dpk dt

= d

( ∂L

∂q̇k

) (3.14)

= d

( ∂L

∂q̇k

) − ∂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 = q̇i ∂L

∂q̇i − L (3.17)

is a constant of the motion (sometimes called the Jacobi integral).

dt = q̈i

∂L

∂q̇i + q̇i

( ∂L

∂q̇i

) − ∂L ∂t

− q̇i ∂L

∂qi − q̈i

∂L

∂q̇i (3.18)

= −∂L ∂t

+ q̇i

[ d

( ∂L

∂q̇i

) − ∂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).

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 q̇i 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 q̇i. For example, c1 = c1(q1, ...,qn, q̇1, ..., q̇n,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.

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

( ∂L

∂θ̇

) − ∂L ∂θ

(3.26)

= d

( 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).

3.4 APPLICATION: THE KEPLER PROBLEM 640-213 Melatos

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)

( 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

sin2 θ

)1/2 (3.33)

and dφ

dθ = φ̇

θ̇ = ± J

sin2 θ

( α2 − J

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).

3.4 APPLICATION: THE KEPLER PROBLEM 640-213 Melatos

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

( ṙ2 + r2φ̇2

) + V (r) = E (3.38)

Eliminating φ̇: 1

2 ṙ2 +

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

c2r3 (3.41)

3.4 APPLICATION: THE KEPLER PROBLEM 640-213 Melatos

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.]

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 = q̇i ∂L

∂q̇i − L (4.1)

= q̇i ∂T

∂q̇i − 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.

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 = q̇j ∂xs ∂qj

(4.3)

T =

N∑ s=1

2 msq̇jq̇k

∂xs ∂qj

· ∂xs ∂qk

(4.4)

∂T

∂q̇i =

N∑ s=1

2 ms

( q̇k ∂xs ∂qi

· ∂xs ∂qk

+ q̇j ∂xs ∂qj

· ∂xs ∂qi

) (4.5)

= N∑

s=1

msq̇k ∂xs ∂qi

· ∂xs ∂qk

(4.6)

q̇i ∂T

∂q̇i =

N∑ s=1

msq̇iq̇k ∂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 q̇j∂xs/∂qj = q̇k∂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)

4.3 ANGULAR MOMENTUM 640-213 Melatos

From Lagrange’s equations,

0 =

N∑ s=1

∂L

∂ẋs − ∂L ∂xs

(4.12)

N∑ s=1

∂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′ + ∑

msV · R′ + 1

∑ 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)

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

( ∂L

∂ẋs

) + ẋs ×

∂L

∂ẋs

) (4.21)

= a · N∑

s=1

( 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

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 q̇i(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.

4.4 NOETHER’S THEOREM: SYMMETRIES OF THE PARTICLES AND FORCES OF NATURE 640-213 Melatos

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).

4.4 NOETHER’S THEOREM: SYMMETRIES OF THE PARTICLES AND FORCES OF NATURE 640-213 Melatos

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.

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?

5.1 GENERALISED COORDINATES 640-213 Melatos

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 q1q̇1 cos q2 − R sin q1 sin q2q̇2, R cos q1q̇1 sin q2 + R sin q1 cos q2q̇2,−R sin q1q̇1) (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)

= q̇1(R cos q1 cos q2,R cos q1 sin q2,−R sin q1) + q̇2(−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

+ q̇i ∂xs ∂qi

(5.8)

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) =

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

( ∂L

∂q̇1

) − ∂L ∂q1

(5.14)

= d

( ∂L

∂θ̇

) − ∂L ∂θ

(5.15)

= d

( mR2θ̇

) − ( mR2 sin θ cos θφ̇2 + mgR sin θ

) (5.16)

5.4 SOLVING FOR THE MOTION 640-213 Melatos

Longitude coordinate:

0 = d

( ∂L

∂q̇2

) − ∂L ∂q2

(5.17)

= d

( ∂L

∂φ̇

) − ∂L ∂φ

(5.18)

= d

( 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 +

2 R2 sin2 θ × J

R4 sin4 θ + gR cos θ (5.23)

= 1

2 R2θ̇2 +

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.

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.

5.5 [EXT] NOETHER’S THEOREM 640-213 Melatos

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)

5.5 [EXT] NOETHER’S THEOREM 640-213 Melatos

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 +

( ∂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.

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.

6.1 DEGREES OF FREEDOM 640-213 Melatos

Euler angles. From E. D. Fackerell and C. J. Durrant, 1993, Lagrangian Dynamics, University of Sydney

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 = ∑

2 ms|ẋ′s|2 (6.6)

= ∑

[ 1

2 ms|Ẋ|2 + msẊ · Ω × xs +

2 ms|Ω × xs|2

] (6.7)

= 1

2 |Ẋ|2 ∑

ms + Ẋ × Ω · ∑

msxs + ∑

2 ms [ Ω2|xs|2 − (Ω · xs)2

] (6.8)

= 1

2 M|Ẋ|2 +

∑ s

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

2 ms [ Ω2|xs|2 − (Ω · xs)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 = ∑

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.

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.

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

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 +

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.)

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 = ∑

x′s × msẋ′s (6.21)

= ∑

ms(X + xs) × (Ẋ + Ω × xs) (6.22)

= X × Ẋ ∑

ms + X × (Ω × ∑

msxs)

−Ẋ × ∑

msxs + ∑

msxs × (Ω × xs) (6.23)

= MX × Ẋ + ∑

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)

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.

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)

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).

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 = ΩprM̂ + 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.

7.2 [EXT] NUTATION 640-213 Melatos

Lagrangian:

L = 1

2 Mv2c +

2 I1Ω

2 1 +

2 I2Ω

2 2 +

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:

2 Mv2C =

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

2 I1Ω

2 1 +

2 I2Ω

2 2 +

2 I3Ω

2 3

= 1

2 I1

[ (sin ψθ̇ − sin θ cos ψφ̇)2 + (cos ψθ̇ + sin θ sin ψφ̇)2

] +

2 I3(ψ̇ + cos θφ̇)

2 (7.12)

= 1

2 I1(θ̇

2 + sin2 θφ̇2) + 1

2 I3(ψ̇ + cos θφ̇)

2 (7.13)

i.e.,

L = 1

2 (Ml2 + I1)(θ̇

2 + sin2 θφ̇2) + 1

2 I3(ψ̇ + cos θφ̇)

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)

7.2 [EXT] NUTATION 640-213 Melatos

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:

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

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:

7.2 [EXT] NUTATION 640-213 Melatos

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.

7.2 [EXT] NUTATION 640-213 Melatos

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.

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.]

8.1 WORKED EXAMPLE: TRIATOMIC MOLECULE 640-213 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 +

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

8.1 WORKED EXAMPLE: TRIATOMIC MOLECULE 640-213 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 +

2 Mξ̇22 +

2 mξ̇23

−1 2 k(ξ2 − ξ1)2 −

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.

8.1 WORKED EXAMPLE: TRIATOMIC MOLECULE 640-213 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.

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)

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).

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. =

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.

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

( 1 + γ

4ω20

) |A|4

(9.15)

Equation (9.15) is a cubic in |A|2. Plot |A| versus ω − ω0 (c.f. Section 9.2).

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 +

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)

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.

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.

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 q̇1, ..., q̇n. Likewise, the generalised momenta defined in Chapter 3 are functions of the same 2n + 1 things.

pi = ∂L(t,q1, ...,qn, q̇1, ..., q̇n)

∂q̇i i = 1, ...,n (10.1)

What happens if we rearrange the n equations in (10.1) to write each generalised velocity q̇i 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 q̇i’s as independent variables, not the pi’s. Instead we use a new function, called the Hamiltonian:

H = q̇j ∂L

∂q̇j − L (10.2)

= q̇j(t,q1, ...,qn,p1, ...,pn)pj

−L(t,q1, ...,qn, q̇1(t,q1, ...,qn,p1, ...,pn), ..., q̇n(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 q̇i’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 =

∂q̇j ∂qi

pj − ( ∂L

∂qi + ∂L

∂q̇j

∂q̇j ∂qi

) (chain rule) (10.4)

= ∂q̇j ∂qi

pj − ∂L

∂qi − pj

∂q̇j ∂qi

(definition of pj) (10.5)

= ∂L

∂qi (10.6)

= −dpi dt

(from Lagrange’s equations) (10.7)

10.2 WORKED EXAMPLE: 2D SHO 640-213 Melatos

Differentiate H wth respect to the pi’s:

∂H

∂pi =

∂q̇j ∂pi

pj + q̇i − ∂L

∂q̇j

∂q̇j ∂pi

(chain rule) (10.8)

= ∂q̇j ∂pi

pj + q̇i − pj ∂q̇j ∂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):

10.2 WORKED EXAMPLE: 2D SHO 640-213 Melatos

L = KE − PE (10.11) =

2 mẋ21 +

2 mẋ22 −

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 +

2 m (p2 m

)2 − 1

2 kx21 −

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)

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:

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

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}

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.

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.

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.

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.

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)

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 −

3 ≤ λ ≤ 1 + √

6 φn → 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.

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.

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)

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)