IA Dynamics and Relativity (Full)

Part IA — Dynamics and Relativity

Based on lectures by G. I. Ogilvie

Notes taken by Dexter Chua

Lent 2015

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Familiarity with the topics covered in the non-examinable Mechanics course is assumed.

Basic concepts

Space and time, frames of reference, Galilean transformations. Newton’s laws. Dimen-

sional analysis. Examples of forces, including gravity, friction and Lorentz. [4]

Newtonian dynamics of a single particle

Equation of motion in Cartesian and plane polar coordinates. Work, conservative forces

and potential energy, motion and the shape of the potential energy function; stable

equilibria and small oscillations; effect of damping.

Angular velocity, angular momentum, torque.

Orbits: the

(

) equation; escape velocity; Kepler’s laws; stability of orbits; motion in

a repulsive potential (Rutherford scattering). Rotating frames: centrifugal and coriolis

forces. *Brief discussion of Foucault pendulum.* [8]

Newtonian dynamics of systems of particles

Momentum, angular momentum, energy. Motion relative to the centre of mass; the

two body problem. Variable mass problems; the rocket equation. [2]

Rigid bodies

Moments of inertia, angular momentum and energy of a rigid body. Parallel axes

theorem. Simple examples of motion involving both rotation and translation (e.g.

rolling). [3]

Special relativity

The principle of relativity. Relativity and simultaneity. The invariant interval. Lorentz

transformations in (1 + 1)-dimensional spacetime. Time dilation and length contraction.

The Minkowski metric for (1 + 1)-dimensional spacetime. Lorentz transformations

in (3 + 1) dimensions. 4-vectors and Lorentz invariants. Prop er time. 4-velocity

and 4-momentum. Conservation of 4-momentum in particle decay. Collisions. The

Newtonian limit. [7]

Contents

0 Introduction

1 Newtonian dynamics of particles

1.1 Newton’s laws of motion

1.2 Galilean transformations

1.3 Newton’s Second Law

2 Dimensional analysis

2.1 Units

2.2 Scaling

3 Forces

3.1 Force and potential energy in one dimension

3.2 Motion in a potential

3.3 Central forces

3.4 Gravity

3.5 Electromagnetism

3.6 Friction

4 Orbits

4.1 Polar coordinates in the plane

4.2 Motion in a central force field

4.3 Equation of the shape of the orbit

4.4 The Kepler problem

4.5 Rutherford scattering

5 Rotating frames

5.1 Motion in rotating frames

5.2 The centrifugal force

5.3 The Coriolis force

6 Systems of particles

6.1 Motion of the center of mass

6.2 Motion relative to the center of mass

6.3 The two-body problem

6.4 Variable-mass problem

7 Rigid bodies

7.1 Angular velocity

7.2 Moment of inertia

7.3 Calculating the moment of inertia

7.4 Motion of a rigid body

8 Special relativity

8.1 The Lorentz transformation

8.2 Spacetime diagrams

8.3 Relativistic physics

8.4 Geometry of spacetime

8.5 Relativistic kinematics

8.6 Particle physics

0 Introduction

You’ve been lied to. You thought you applied for mathematics. And here you

have a course on physics. No, this course is not created for students taking the

“Maths with Physics” option. They don’t have to take this course (don’t ask

why).

Ever since Newton invented calculus, mathematics is becoming more and more

important in physics. Physicists seek to describe the universe in a few equations,

and derive everyday (physical) phenomena as mathematical consequences of

these equations.

In this course, we will start with Newton’s laws of motion and use it to derive

a lot of physical phenomena, including planetary orbits, centrifugal forces

and

the motion of rotating bodies.

The important thing to note is that we can “prove” all these phenomena just

under the assumption that Newton’s laws are correct (plus the formulas for, say,

the strength of the gravitational force). We are just doing mathematics here.

We don’t need to do any experiments to obtain the results (of course, we need

experiments to verify that Newton’s laws are indeed the equations that describe

this universe).

However, it turns out that Newton was wrong. While his theories were

accurate for most everyday phenomena, they weren’t able to adequately describe

electromagnetism. This lead to Einstein discovering special relativity. Special

relativity is also required to describe motion that is very fast. We will have a

brief introduction to special relativity at the end of the course.

Yes, they exist.

1 Newtonian dynamics of particles

Newton’s equations describe the motion of a (point) particle.

Definition

(Particle)

A particle is an object of insignificant size, hence it can

be regarded as a point. It has a mass m > 0, and an electric charge q.

Its position at time

is described by its position vector,

(

) or

(

) with

respect to an origin O.

Depending on context, different things can be considered as particles. We

could consider an electron to be a point particle, even though it is more accurately

described by the laws of quantum mechanics than those of Newtonian mechanics.

If we are studying the orbit of planets, we can consider the Sun and the Earth

to be particles.

An important property of a particle is that it has no internal structure. It can

be completely described by its position, momentum, mass and electric charge.

For example, if we model the Earth as a particle, we will have to ignore its own

rotation, temperature etc.

If we want to actually describe a rotating object, we usually consider it as a

collection of point particles connected together, and apply Newton’s law to the

individual particles.

As mentioned above, the position of a particle is described by a position

vector. This requires us to pick an origin of the coordinate system, as well as an

orientation of the axes. Each choice is known as a frame of reference.

Definition

(Frame of reference)

A frame of reference is a choice of coordinate

axes for r.

We don’t impose many restrictions on the choice of coordinate axes. They

can be fixed, moving, rotating, or even accelerating.

Using the position vector

, we can define various interesting quantities which

describe the particle.

Definition (Velocity). The velocity of the particle is

v =

r =

Definition (Acceleration). The acceleration of the particle is

a =

v =

r =

Definition (Momentum). The momentum of a particle is

p = mv = m

is the inertial mass of the particle, and measures its reluctance to accelerate,

as described by Newton’s second law.

1.1 Newton’s laws of motion

We will first state Newton’s three laws of motion, and then discuss them individ-

ually.

Law

(Newton’s First Law of Motion)

A body remains at rest, or moves uniformly

in a straight line, unless acted on by a force. (This is in fact Galileo’s Law of

Inertia)

Law

(Newton’s Second Law of Motion)

The rate of change of momentum of a

body is equal to the force acting on it (in both magnitude and direction).

Law

(Newton’s Third Law of Motion)

To every action there is an equal and

opposite reaction: the forces of two bodies on each other are equal and in opposite

directions.

The first law might seem redundant given the second if interpreted literally.

According to the second law, if there is no force, then the momentum doesn’t

change. Hence the body remains at rest or moves uniformly in a straight line.

So why do we have the first law? Historically, it might be there to explicitly

counter Aristotle’s idea that objects naturally slow down to rest. However, some

(modern) physicists give it an alternative interpretation:

Note that the first law isn’t always true. Take yourself as a frame of reference.

When you move around your room, things will seem like they are moving around

(relative to you). When you sit down, they stop moving. However, in reality,

they’ve always been sitting there still. On second thought, this is because you,

the frame of reference, is accelerating, not the objects. The first law only holds

in frames that are themselves not accelerating. We call these inertial frames.

Definition

(Inertial frames)

Inertial frames are frames of references in which

the frames themselves are not accelerating. Newton’s Laws only hold in inertial

frames.

Then we can take the first law to assert that inertial frames exists. Even

though the Earth itself is rotating and orbiting the sun, for most purposes, any

fixed place on the Earth counts as an inertial frame.

1.2 Galilean transformations

The goal of this section is to investigate inertial frames. We know that inertial

frames are not unique. Given an inertial frame, what other inertial frames can

we obtain?

First of all, we can rotate our axes or move our origin. In particular, we can

perform the following operations:

– Translations of space:

= r − r

– Translations of time:

= t − t

– Rotation (and reflection):

= Rr

with R ∈ O(3).

These are not very interesting. They are simply symmetries of space itself.

The last possible transformation is uniform motion. Suppose that

is an

inertial frame. Then any other frame

in uniform motion relative to

is also

inertial:

Assuming the frames coincide at t = 0, then

= x − vt

= y

= z

= t

Generally, the position vector transforms as

= r − vt,

where

is the (constant) velocity of

relative to

. The velocity and acceleration

transform as follows:

r − v

Definition

(Galilean boost)

A Galilean boost is a change in frame of reference

= r − vt

= t

for a fixed, constant v.

All these transformations together generate the Galilean group, which de-

scribes the symmetry of Newtonian equations of motion.

Law

(Galilean relativity)

The principle of relativity asserts that the laws of

physics are the same in inertial frames.

This implies that physical processes work the same

– at every point of space

– at all times

– in whichever direction we face

– at whatever constant velocity we travel.

In other words, the equations of Newtonian physics must have Galilean

invariance.

Since the laws of physics are the same regardless of your velocity, velocity

must be a relative concept, and there is no such thing as an “absolute velocity”

that all inertial frames agree on.

However, all inertial frames must agree on whether you are accelerating or

not (even though they need not agree on the direction of acceleration since you

can rotate your frame). So acceleration is an absolute concept.

1.3 Newton’s Second Law

Newton’s second law is often written in the form of an equation.

Law. The equation of motion for a particle subject to a force F is

= F,

where

is the (linear) momentum of the particle. We say

is the

(inertial) mass of the particle, which is a measure of its reluctance to accelerate.

Usually, m is constant. Then

F = ma = m

Usually,

is specified as a function of

and

. Then we have a second-order

ordinary differential equation for r.

To determine the solution, we need to specify the initial values of

and

, i.e.

the initial position and velocity. The trajectory of the particle is then uniquely

determined for all future (and past) times.

2 Dimensional analysis

When considering physical theories, it is important to be aware that physical

quantities are not pure numbers. Each physical quantity has a dimension.

Roughly speaking, dimensions are what units represent, such as length, mass

and time. In any equation relating physical quantities, the dimensions must be

consistent, i.e. the dimensions on both sides of the equation must be equal.

For many problems in dynamics, the three basic dimensions are sufficient:

– length, L

– mass, M

– time, T

The dimensions of a physical quantity

, denoted by [

] are expressible

uniquely in terms of L, M and T . For example,

– [area] = L

– [density] = L

−3

– [velocity] = LT

−1

– [acceleration] = LT

−2

–

[

force

] =

LMT

−2

since the dimensions must satisfy the equation

– [energy] = L

−2

, e.g. consider E = mv

/2.

Physical constants also have dimensions, e.g. [

] =

−1

−2

GMm/r

The only allowed operations on quantities with dimensions are sums and

products (and subtraction and division), and if we sum two terms, they must

have the same dimension. For example, it does not make sense to add a length

with an area. More complicated functions of dimensional quantities are not

allowed, e.g. e

again makes no sense if x has a dimension, since

= 1 + x +

+ ···

and if

had a dimension, we would be summing up terms of different dimensions.

2.1 Units

People use units to represent dimensional quantities. A unit is a convenient

standard physical quantity, e.g. a fixed amount of mass. In the SI system, there

are base units corresponding to the basics dimensions. The three we need are

– meter (m) for length

– kilogram (kg) for mass

– second (s) for time

A physical quantity can be expressed as a pure number times a unit with the

correct dimensions, e.g.

G = 6.67 × 10

−11

−1

−2

It is important to realize that SI units are chosen arbitrarily for historical reasons

only. The equation of physics must work in any consistent system of units. This

is captured by the fact that physical equations must be dimensionally consistent.

2.2 Scaling

We’ve had so many restrictions on dimensional quantities — equations must

be dimensionally consistent, and we cannot sum terms of different dimensions.

However, this is not a hindrance when developing new theories. In fact, it is a

very useful tool. First of all, it allows us to immediately rule out equations that

do not make sense dimensionally. More importantly, it allows us to guess the

form of the equation we want.

Suppose we believe that a physical quantity

depends on 3 other physical

quantities

, X

, i.e.

(

, X

). Let their dimensions be as

follows:

– [Y ] = L

– [X

] = L

Suppose further that we know that the relationship is a power law, i.e.

Y = CX

where

is a dimensionless constant (i.e. a pure number). Since the dimensions

must work out, we know that

a = p

+ p

b = p

+ p

c = p

+ p

for which there is a unique solution provided that the dimensions of

, X

and

are independent. So just by using dimensional analysis, we can figure

out the relation between the quantities up to a constant. The constant can

then be found experimentally, which is much easier than finding the form of the

expression experimentally.

However, in reality, the dimensions are not always independent. For example,

we might have two length quantities. More importantly, the situation might

involve more than 3 variables, and we do not have a unique solution.

First consider a simple case — if

is dimensionless, then the relation

between

and

can involve more complicated terms such as

exp

(

), since

the argument of exp is now dimensionless.

In general, suppose we have many terms, and the dimensions of

are not in-

dependent. We order the quantities so that the independent terms [

]

[

]

[

]

are at the front. For each of the remaining variables, form a dimensionless quan-

tity λ

= X

. Then the relationship must be of the form

Y = f(λ

, λ

, ···)X

where f is an arbitrary function of the dimensionless variables.

Formally, this results is described by the Buckingham’s Pi theorem, but we

will not go into details.

Example (Simple pendulum).

We want to find the period P . We know that P could depend on

– mass m: [m] = M

– length `: [`] = L

– gravity g: [g] = LT

−2

– initial displacement d: [d] = L

and of course [P ] = T .

We observe that

m, `, g

have independent dimensions, and with the fourth,

we can form the dimensionless group

d/`

. So the relationship must be of the

form

P = f





where f is a dimensionless function. For the dimensions to balance,

T = M

−2p

So p

= 0, p

= −p

= 1/2. Then

P = f





While we cannot find the exact formula, using dimensional analysis, we know

that if both ` and d are quadrupled, then P will double.

3 Forces

Force is a central concept in Newtonian mechanics. As described by Newton’s

laws of motion, forces are what causes objects to accelerate, according to the

formula

. To completely specify the dynamics of a system, one only

needs to know what forces act on what bodies.

However, unlike in Star Wars, the force is given a secondary importance in

modern treatments of mechanics. Instead, the potential is what is considered to

be fundamental, with force being a derived concept. In quantum mechanics, we

cannot even meaningfully define forces.

Yet, there are certain systems that must be described with forces instead

of potentials, the most important of which is a system that involves friction of

some sort.

3.1 Force and potential energy in one dimension

To define the potential, consider a particle of mass

moving in a straight line

with position

(

). Suppose

(

), i.e. it depends on position only. We

define the potential energy as follows:

Definition

(Potential energy)

Given a force field

(

), we define the

potential energy to be a function V (x) such that

F = −

V = −

F dx.

is defined up to a constant term. We usually pick a constant of integration

such that the potential drops to 0 at infinity.

Using the potential, we can write the equation of motion as

m¨x = −

There is a reason why we call this the potential energy. We usually consider

it to be an energy of some sort. In particular, we define the total energy of a

system to be

Definition (Total energy). The total energy of a system is given by

E = T + V,

where V is the potential energy and T =

m ˙x

is the kinetic energy.

If the force of a system is derived from a potential, we can show that energy

is conserved.

Proposition. Suppose the equation of a particle satisfies

m¨x = −

Then the total energy

E = T + V =

m ˙x

+ V (x)

is conserved, i.e.

E = 0.

Proof.

= m ˙x¨x +

˙x

= ˙x



m¨x +



= 0

Example. Consider the harmonic oscillator, whose potential is given by

V =

Then the equation of motion is

m¨x = −kx.

This is the case of, say, Hooke’s Law for a spring.

The general solution of this is

x(t) = A cos(ωt) + B sin(ωt)

with ω =

k/m.

A and B are arbitrary constants, and are related to the initial position and

velocity by x(0) = A, ˙x(0) = ωB.

For a general potential energy

(

), conservation of energy allows us to solve

the problem formally:

E =

m ˙x

+ V (x)

Since E is a constant, from this equation, we have

= ±

(E − V (x))

t − t

= ±

(E − V (x))

To find

(

), we need to do the integral and then solve for

. This is usually

not possible by analytical methods, but we can approximate the solution by

numerical methods.

3.2 Motion in a potential

Given an arbitrary potential

(

), it is often difficult to completely solve the

equations of motion. However, just by looking at the graph of the potential, we

can usually get a qualitative understanding of the dynamics.

Example.

Consider

(

) =

(

−

). Note that this can be dimensionally

consistent even though we add up

and

−

, if we declare “3” to have dimension

We plot this as follows:

1 2−1−2

Suppose we release the particle from rest at

. Then

(

). We can

consider what happens to the particle for different values of x

– x

= ±1: This is an equilibrium and the Particle stays there for all t.

– −

< x

2: The particle does not have enough energy to escape the well.

So it oscillates back and forth in potential well.

– x

< −1: The particle falls to x = −∞.

– x

2: The particle has enough energy to overshoot the well and continues

to x = −∞.

– x

= 2: This is a special case. Obviously, the particle goes towards

−

But how long does it take, and what happens next? Here we have

= 2

We noted previously

t − t

= −

(E − V (x))

Let x = −1 + ε(t). Then

(E − V (x)) = 4 − 2(−1 + ε)

+ 6(−1 + ε)

= 6ε

− 2ε

t − t

= −

dε

√

6ε

− 2ε

We reach

−

1 when

ε →

0. But for small

, the integrand is approx-

imately

∝

/ε

, which integrates to

ln ε → −∞

ε →

0. So

= 0 is

achieved after infinite time, i.e. it takes infinite time to reach

= 0, or

x = −1.

Equilibrium points

In reality, most of the time, particles are not flying around wildly doing crazy

stuff. Instead, they stay at (or near) some stable point, and only move very little

in a predictable manner. We call these points equilibrium points.

Definition

(Equilibrium point)

A particle is in equilibrium if it has no tendency

to move away. It will stay there for all time. Since

m¨x

−V

(

), the equilibrium

points are the stationary points of the potential energy, i.e.

) = 0.

Consider motion near an equilibrium point. We assume that the motion is

small and we can approximate

by a second-order Taylor expansion. Then we

can write V as

V (x) ≈ V (x

) +

)(x − x

)

Then the equation of motion is

m¨x = −V

)(x − x

(

)

0, then this is of the form of the harmonic oscillator.

has a

local minimum at

, and we say the equilibrium point is stable. The particle

oscillates with angular frequency

ω =

)

(

)

0, then

has a local maximum at

. In this case, the equilibrium

point is unstable, and the solution to the equation is

x − x

≈ Ae

γt

+ Be

−γt

for

γ =

−V

)

For almost all initial conditions,

A 6

= 0 and the particle will diverge from the

equilibrium point, leading to a breakdown of the approximation.

If V

) = 0, then further work is required to determine the outcome.

Example. Consider the simple pendulum.

Suppose that the pendulum makes an angle

with the vertical. Then the energy

E = T + V =

− mg` cos θ.

Therefore

V ∝ −cos θ

. We have a stable equilibrium at

= 0, and unstable

equilibrium at θ = π.

−π

mg`

−mg`

If E > mg`, then

θ never vanishes and the pendulum makes full circles.

If 0

< E < mg`

, then

vanishes at

±θ

for some 0

< θ

< π

i.e.

−mg` cos θ

. The pendulum oscillates back and forth. It takes a quarter

of a period to reach from

= 0 to

. Using the previous general solution,

oscillation period P is given by

dθ

cos θ

Since we know that E = −mg` cos θ

, we know that

dδ

√

2 cos θ − 2 cos θ

The integral is difficult to evaluate in general, but for small

, we can use

cos θ ≈ 1 −

. So

P ≈ 4

dθ

− θ

= 2π

and is independent of the amplitude

. This is of course the result for the

harmonic oscillator.

Force and potential energy in three dimensions

Everything looks nice so far. However, in real life, the world has (at least) three

(spatial) dimensions. To work with multiple dimensions, we will have to promote

our quantities into vectors.

Consider a particle of mass

moving in 3D. The equation of motion is now

a vector equation

r = F.

We’ll define the familiar quantities we’ve had.

Definition

(Kinetic energy)

We define the kinetic energy of the particle to be

T =

m|v|

r ·

If we want to know how it varies with time, we obtain

= m

r ·

r = F ·

r = F · v.

This is the power.

Definition

(Power)

The power is the rate at which work is done on a particle

by a force. It is given by

P = F · v.

Definition

(Work done)

The work done on a particle by a force is the change

in kinetic energy caused by the force. The work done on a particle moving from

= r(t

) to r

= r(t

) along a trajectory C is the line integral

W =

F · dr =

F ·

r dt =

P dt.

Usually, we are interested in forces that conserve energy. These are forces

which can be given a potential, and are known as conservative forces.

Definition

(Conservative force and potential energy)

A conservative force is a

force field F(r) that can be written in the form

F = −∇V.

V is the potential energy function.

Proposition. If F is conservative, then the energy

E = T + V

m|v|

+ V (r)

is conserved. For any particle moving under this force, the work done is equal to

the change in potential energy, and is independent of the path taken between

the end points. In particular, if we travel around a closed loop, no work is done.

Proof.



r ·

r + V



= m

r ·

r +

∂V

∂x

= (m

r + ∇V ) ·

= (m

r − F) ·

= 0

So the energy is conserved. In this case, the work done is

W =

F · dr = −

(∇V ) · dr = V (r

) − V (r

3.3 Central forces

While in theory the potential can take any form it likes, most of the time, our

system has spherical symmetry. In this case, the potential depends only on the

distance from the origin.

Definition

(Central force)

A central force is a force with a potential

(

) that

depends only on the distance from the origin,

|r|

. Note that a central force

can be either attractive or repulsive.

When dealing with central forces, the following formula is often helpful:

Proposition. ∇r =

Intuitively, this is because the direction in which

increases most rapidly is

r, and the rate of increase is clearly 1. This can also be proved algebraically:

Proof. We know that

= x

+ x

Then

∂r

∂x

= 2x

∂r

∂x

= (

Proposition. Let F = −∇V (r) be a central force. Then

F = −∇V = −

where

r/r

is the unit vector in the radial direction pointing away from the

origin.

Proof. Using the proof above,

(∇V )

∂V

∂x

∂r

∂x

(

Since central forces have spherical symmetry, they give rise to an additional

conserved quantity called angular momentum.

Definition (Angular momentum). The angular momentum of a particle is

L = r × p = mr ×

Proposition. Angular momentum is conserved by a central force.

Proof.

= m

r ×

r + mr ×

r = 0 + r × F = 0.

where the last equality comes from the fact that

is parallel to

for a central

force.

In general, for a non-central force, the rate of change of angular momentum

is the torque.

Definition

(Torque)

The torque

of a particle is the rate of change of angular

momentum.

G =

= r × F.

Note that

and

depends on the choice of origin. For a central force, only

the angular momentum about the center of the force is conserved.

3.4 Gravity

We’ll now study an important central force — gravity. This law was discovered

by Newton and was able to explain the orbits of various planets. However, we

will only study the force and potential aspects of it, and postpone the study of

orbits for a later time.

Law

(Newton’s law of gravitation)

If a particle of mass

is fixed at a origin,

then a second particle of mass m experiences a potential energy

V (r) = −

GMm

where G ≈ 6.67 × 10

−11

−1

−2

is the gravitational constant.

The gravitational force experienced is then

F = −∇V = −

GMm

Since the force is negative, particles are attracted to the origin.

The potential energy is a function of the masses of both the fixed mass

and the second particle

. However, it is useful what the fixed mass

does

with reference to the second particle.

Definition

(Gravitaional potential and field)

The gravitational potential is the

gravitational potential energy per unit mass. It is

(r) = −

Note that potential is confusingly different from potential energy.

If we have a second particle, the potential energy is given by V = mΦ

The gravitational field is the force per unit mass,

g = −∇Φ

= −

If we have many fixed masses

at points

, we can add up their gravitational

potential directly. Then the total gravitational potential is given by

(r) = −

|r − r

Again, V = mΦ

for a particle of mass m.

An important (mathematical) result about gravitational fields is that we can

treat spherical objects as point particles. In particular,

Proposition.

The external gravitational potential of a spherically symmetric

object of mass

is the same as that of a point particle with the same mass at

the center of the object, i.e.

(r) = −

The proof can be found in the Vector Calculus course.

Example.

If you live on a spherical planet of mass

and radius

, and can

move only a small distance z  R above the surface, then

V (r) = V (R + z)

= −

GMm

R + z

= −

GMm



1 −

+ ···



≈ const. +

GMm

= const. + mgz,

where

GM/R

≈ 9.8 m s

−2

for Earth. Usually we are lazy and just say that

the potential is mgz.

Example.

How fast do we need to jump to escape the gravitational pull of the

Earth? If we jump upwards with speed v from the surface, then

E = T + V =

−

GMm

After escape, we must have

T ≥

0 and

= 0. Since energy is conserved, we

must have E ≥ 0 from the very beginning. i.e.

v > v

esc

2GM

Inertial and gravitational mass

A careful reader would observe that “mass” appears in two unrelated equations:

F = m

and

F = −

and they play totally different roles. The first is the inertial mass, which

measures the resistance to motion, while the second is the gravitational mass,

which measures its response to gravitational forces.

Conceptually, these are quite different. There is no a priori reason why these

two should be equal. However, experiment shows that they are indeed equivalent

to each other, i.e. m

= m

, with an accuracy of 10

−12

or better.

This (philosophical) problem was only resolved when Einstein introduced his

general theory of relativity, which says that gravity is actually a fictitious force,

which means that the acceleration of the particle is independent of its mass.

We can further distinct the gravitational mass by “passive” and “active”, i.e.

the amount of gravitational field generated by a particle (

), and the amount

of gravitational force received by a particle (

), but they are still equal, and we

end up calling all of them “mass”.

3.5 Electromagnetism

Next we will study the laws of electromagnetism. We will only provide a

very rudimentary introduction to electromagnetism. Electromagnetism will be

examined more in-depth in the IB Electromagnetism and II Electrodynamics

courses.

As the name suggests, electromagnetism comprises two parts — electricity

and magnetism. Similar to gravity, we generally imagine electromagnetism

working as follows: charges generate fields, and fields cause charges to move.

A common charged particle is the electron, which is currently believed to

be a fundamental particle. It has charge

−1.6 × 10

−19

. Other particles’

charges are always integer multiples of q

(unless you are a quark).

In electromagnetism, there are two fields — the electric field

(

r, t

) and the

magnetic field

(

r, t

). Their effects on charged particles is described by the

Lorentz force law.

Law

(Lorentz force law)

The electromagnetic force experienced by a particle

with electric charge q is

F = q(E + v × B).

This is the first time where we see a force that depends on the velocity of

the particle. In all other forces we’ve seen, the force depends only on the field

which implicitly depends on the position only. This is weird, and seems to

violate Galilean relativity, since velocity is a relative concept that depends on

the frame of reference. It turns out that weird things happen to the

and

fields when you change the frame of reference. You will learn about these in the

IB Electromagnetism course (if you take it).

As a result, the magnetic force is not a conservative force, and it cannot be

given a (regular) potential. On the other hand, assuming that the fields are

time-independent, the electric field is conservative. We write the potential as

(r), and E = −∇Φ

Definition

(Electrostatic potential)

The electrostatic potential is a function

(r) such that

E = −∇Φ

While the magnetic force is not conservative in the traditional sense, it always

acts perpendicularly to the velocity. Hence it does no work. So overall, energy

is conserved under the action of the electromagnetic force.

Proposition. For time independent E(r) and B(r), the energy

E = T + V =

m|v|

+ qΦ

is conserved.

Proof.

= m

r ·

r + q(∇Φ

) ·

= (m

r − qE) ·

= (q

r × B) ·

= 0

Motion in a uniform magnetic field

Consider the particular case where there is no electric field, i.e.

, and that

the magnetic field is uniform throughout space. We choose our axes such that

B = (0, 0, B) is constant.

According to the Lorentz force law,

(

v × B

) =

qv × B

. Since

the force is always perpendicular to the velocity, we expect this to act as a

centripetal force to make the particle travel in circles.

Indeed, writing out the components of the equation of motion, we obtain

m¨x = qB ˙y (1)

m¨y = −qB ˙x (2)

m¨z = 0 (3)

From (3), we see that there is uniform motion parallel to

, which is not

interesting. We will look at the x and y components.

There are many ways to solve this system of equations. Here we solve it

using complex numbers.

Let ζ = x + iy. Then (1) + (2)i gives

ζ = −iqB

ζ.

Then the solution is

ζ = αe

−iωt

+ β,

where

qB/m

is the gyrofrequency, and

and

are complex integration

constants. We see that the particle goes in circles, with center β and radius α.

We can choose coordinates such that, at

= 0,

= 0 and

= (0

, v, w

i.e. ζ = 0 and

ζ = iv, and z = 0 and ˙z = w.

The solution is then

ζ = R(1 − e

−iωt

with R = v/ω = (mv)/(qB) is the gyroradius or Larmor radius. Alternatively,

x = R(1 − cos ωt)

y = R sin ωt

z = wt.

This is circular motion in the plane perpendicular to B:

(x − R)

+ y

= R

combined with uniform motion parallel to B, i.e. a helix.

Alternatively, we can solve this with vector operations. Start with

r = q

r × B

Let B = Bn with |n| = 1. Then

r = ω

r × n,

with our gyrofrequency

qB/m

. We integrate once, assuming

(0) =

and

r(0) = v

r = ωr × n + v

. (∗)

Now we project (∗) parallel to and perpendicular to B.

First we dot (∗) with n:

r · n = v

· n = w = const.

We integrate again to obtain

r · n = wt.

This is the part parallel to B.

To resolve perpendicularly, write r = (r · n)n + r

⊥

, with r

⊥

· n = 0.

The perpendicular component of (∗) gives

⊥

= wr

⊥

× n + v

− (v

· n)n.

We solve this by differentiating again to obtain

⊥

= ω

⊥

× n = −ω

⊥

+ ωv

× n,

which we can solve using particular integrals and complementary functions.

Point charges

So far we’ve discussed the effects of the fields on particles. But how can we

create fields in the first place? We’ll only look at the simplest case, where a

point charge generates an electric field.

Law

(Columb’s law)

A particle of charge

, fixed at the origin, produces an

electrostatic potential

4πε

where ε

≈ 8.85 × 10

−12

−3

−1

The corresponding electric field is

E = −∇Φ

4πε

The resulting force on a particle of charge q is

F = qE =

4πε

Definition

(Electric constant)

. ε

is the electric constant or vacuum permittivity

or permittivity of free space.

The form of equations here are closely analogous to those of gravity. However,

there is an important difference: charges can be positive or negative. Thus

electrostatic forces can be either attractive or repulsive, whereas gravity is always

attractive.

3.6 Friction

At an atomic level, energy is always conserved. However, in many everyday

processes, this does not appear to be the case. This is because friction tends to

take kinetic energy away from objects.

In general, we can divide friction into dry friction and fluid friction.

Dry friction

When solid objects are in contact, a normal reaction force

(perpendicular

to the contact surface) prevents them from interpenetrating, while a frictional

force

(tangential to the surface) resists relative tangential motion (sliding or

slipping).

If the tangential force is small, it is insufficient to overcome friction and no

sliding occurs. We have static friction of

|F| ≤ µ

|N|,

where µ

is the coefficient of static friction.

When the external force on the object exceeds

|N|

, sliding starts, and we

have a kinetic friction of

|F| = µ

|N|,

where µ

is the coefficient of kinetic friction.

These coefficients are measures of roughness and depend on the two surfaces

involved. For example, Teflon on Teflon has coefficient of around 0.04, while

rubber on asphalt has about 0.8, while a hypothetical perfectly smooth surface

has coefficient 0. Usually, µ

> µ

> 0.

Fluid drag

When a solid object moves through a fluid (i.e. liquid or gas), it experiences a

drag force.

There are two important regimes.

(i)

Linear drag: for small things in viscous fluids moving slowly, e.g. a single

cell organism in water, the friction is proportional to the velocity, i.e.

F = −k

where

is the velocity of the object relative to the fluid, and

0 is a

constant. This

depends on the shape of the object. For example, for a

sphere of radius R, Stoke’s Law gives

= 6πµR,

where µ is the viscosity of the fluid.

(ii)

Quadratic drag: for large objects moving rapidly in less viscous fluid, e.g.

cars or tennis balls in air, the friction is proportional to the square of the

velocity, i.e.

F = −k

|v|

In either case, the object loses energy. The power exerted by the drag force is

F · v =

(

−k

|v|

−k

|v|

Example.

Consider a projectile moving in a uniform gravitational field and

experiencing a linear drag force.

At t = 0, we throw the projectile with velocity u from x = 0.

The equation of motion is

= mg − kv.

We first solve for v, and then deduce x.

We use an integrating factor exp(

t) to obtain



kt/m



= e

kt/m

v =

kt/m

g + c

v =

g + ce

−kt/m

Since v = u at t = 0, we get c = u −

g. So

v =

x =

g +



u −



−kt/m

Integrating once gives

x =

gt −



u −



−kt/m

+ d.

Since x = 0 at t = 0. So

d =



u −



x =

gt +



u −



(1 − e

−kt/m

In component form, let x = (x, y), u = (u cos θ, u sin θ), g = (0, −g). So

x =

cos θ(1 − e

−kt/m

)

y = −

mgt



u sin θ +



(1 − e

−kt/m

We can characterize the strength of the drag force by the dimensionless constant

ku/(mg), with a larger constant corresponding to a larger drag force.

Effect of damping on small oscillations

We’ve previously seen that particles near a potential minimum oscillate indefi-

nitely. However, if there is friction in the system, the oscillation will damp out

and energy is continually lost. Eventually, the system comes to rest at the stable

equilibrium.

Example.

If a linear drag force is added to a harmonic oscillator, then the

equation of motion becomes

x = −mω

x − k

where

is the angular frequency of the oscillator in the absence of damping.

Rewrite as

x + 2γ

x + ω

x = 0,

where γ = k/2m > 0. Solutions are x = e

λt

, where

+ 2γλ + ω

= 0,

λ = −γ ±

− ω

γ > ω

, then the roots are real and negative. So we have exponential decay.

We call this an overdamped oscillator.

If 0

< γ < ω

, then the roots are complex with

(

) =

−γ

. So we have

decaying oscillations. We call this an underdamped oscillator.

For details, refer to Differential Equations.

4 Orbits

The goal of this chapter is to study the orbit of a particle in a central force,

r = −∇V (r).

While the universe is in three dimensions, the orbit is confined to a plane. This is

since the angular momentum

mr×

is a constant vector, as we’ve previously

shown. Furthermore

L · r

= 0. Therefore, the motion takes place in a plane

passing through the origin, and perpendicular to L.

4.1 Polar coordinates in the plane

We choose our axes such that the orbital plane is

= 0. To describe the orbit,

we introduce polar coordinates (r, θ):

x = r cos θ, y = r sin θ.

Our object is to separate the motion of the particle into radial and angular

components. We do so by defining unit vectors in the directions of increasing

and increasing θ:

r =



cos θ

sin θ



θ =



−sin θ

cos θ



These two unit vectors form an orthonormal basis. However, they are not basis

vectors in the normal sense. The directions of these basis vectors depend on

time. In particular, we have

Proposition.

dθ



−sin θ

cos θ



dθ



−cos θ

−sin θ



= −

Often, we want the derivative with respect to time, instead of

. By the

chain rule, we have

θ,

= −

We can now express the position, velocity and acceleration in this new polar

basis. The position is given by

r = r

Taking the derivative gives the velocity as

r = ˙r

r + r

θ.

The acceleration is then

r = ¨r

r + ˙r

θ + ˙r

θ + r

θ − r

= (¨r − r

)

r + (r

θ + 2 ˙r

θ)

θ.

Definition

(Radial and angular velocity)

. ˙r

is the radial velocity, and

is the

angular velocity.

Example

(Uniform motion in a circle)

If we are moving in a circle, then

˙r

= 0

and

θ = ω = constant. So

r = rω

θ.

The speed is given by

v = |

r| = r|ω| = const

and the acceleration is

r = −rω

Hence in order to make a particle of mass

move uniformly in a circle, we must

supply a centripetal force mv

/r towards the center.

4.2 Motion in a central force field

Now let’s put in our central force. Since V = V (r), we have

F = −∇V =

So Newton’s 2nd law in polar coordinates is

m(¨r − r

)

r + m(r

θ + 2 ˙r

θ)

θ = −

The θ component of this equation is

m(r

θ − 2 ˙r

θ) = 0.

We can rewrite it as

(mr

θ) = 0.

Let

. This is the

component (and the only component) of the

conserved angular momentum L:

L = mr ×

= mr

r × ( ˙r

r + r

θ)

= mr

r ×

= mr

So the angular component tells us that

is constant, which is the conservation

of angular momentum.

However, a more convenient quantity is the angular momentum per unit

mass:

Notation

(Angular momentum per unit mass)

The angular momentum per

unit mass is

h =

= r

θ = const.

Now the radial (r) component of the equation of motion is

m(¨r − r

) = −

We eliminate

θ using r

θ = h to obtain

m¨r = −

= −

eff

where

eff

(r) = V (r) +

We have now reduced the problem to 1D motion in an (effective) potential — as

studied previously.

The total energy of the particle is

E =

+ V (r)

m( ˙r

+ r

) + V (r)

(since

r = ˙r

r + r

θ, and

r and

θ are orthogonal)

m ˙r

+ V (r)

m ˙r

+ V

eff

(r).

Example.

Consider an attractive force following the inverse-square law (e.g.

gravity). Here

V = −

for some constant k. So

eff

= −

We have two terms of opposite signs and different dependencies on

. For small

the second term dominates and

eff

is large. For large

, the first term dominates.

Then V

eff

asymptotically approaches 0 from below.

eff

min

∗

The minimum of V

eff

is at

∗

, E

min

= −

We have a few possible types of motion:

–

min

, then

remains at

∗

and

θh/r

is constant. So we have a

uniform motion in a circle.

–

min

< E <

0, then

oscillates and

˙r

h/r

does also. This is a

non-circular, bounded orbit.

We’ll now introduce a lot of (pointless) definitions:

Definition

(Periapsis, apoapsis and apsides)

The points of minimum and

maximum

in such an orbit are called the periapsis and apoapsis. They

are collectively known as the apsides.

Definition

(Perihelion and aphelion)

For an orbit around the Sun, the

periapsis and apoapsis are known as the perihelion and aphelion.

In particular

Definition

(Perigee and apogee)

The perihelion and aphelion of the

Earth are known as the perigee and apogee.

–

E ≥

0, then

comes in from

∞

, reaches a minimum, and returns to

infinity. This is an unbounded orbit.

We will later show that in the case of motion in an inverse square force, the

trajectories are conic sections (circles, ellipses, parabolae and hyperbolae).

Stability of circular orbits

We’ll now look at circular orbits, since circles are nice. Consider a general

potential energy V (r). We have to answer two questions:

– Do circular orbits exist?

– If they do, are they stable?

The conditions for existence and stability are rather straightforward. For a

circular orbit,

∗

= const for some value of

h 6

= 0 (if

= 0, then the object

is just standing still!). Since ¨r = 0 for constant r, we require

eff

∗

) = 0.

The orbit is stable if r

∗

is a minimum of V

eff

, i.e.

eff

∗

) > 0.

In terms of V (r), circular orbit requires

∗

) =

∗

and stability further requires

∗

) +

3mh

∗

= V

∗

) +

∗

) > 0.

In terms of the radial force F (r) = −V

(r), the orbit is stable if

∗

) +

F (r

∗

) < 0.

Example. Consider a central force with

V (r) = −

for some k, p > 0. Then

(r) +

(r) =



− p(p + 1) + 3p



p+2

= p(2 − p)

p+2

So circular orbits are stable for

p <

2. This is illustrated by the graphs of

eff

(

)

for p = 1 and p = 3.

eff

p = 1

p = 3

4.3 Equation of the shape of the orbit

In general, we could determine r(t) by integrating the energy equation

E =

m ˙r

+ V

eff

(r)

t = ±

E − V

eff

(r)

However, this is usually not practical, because we can’t do the integral. Instead,

it is usually much easier to find the shape r(θ) of the orbit.

Still, solving for

(

) is also not easy. We will need a magic trick — we

introduce the new variable

Notation.

u =

Then

˙r =

dθ

θ =

dθ

= −h

dθ

and

¨r =



−h

dθ



= −h

dθ

θ = −h

dθ

= −h

dθ

This doesn’t look very linear with

, but it will help linearizing the equation

when we put in other factors.

The radial equation of motion

m¨r −

= F (r)

then becomes

Proposition (Binet’s equation).

−mh



dθ

+ u



= F





This still looks rather complicated, but observe that for an inverse square

force, F (1/u) is proportional to u

, and then the equation is linear!

In general, given an arbitrary F (r), we aim to solve this second order ODE

for u(θ). If needed, we can then work out the time-dependence via

θ = hu

4.4 The Kepler problem

The Kepler problem is the study of the orbits of two objects interacting via

a central force that obeys the inverse square law. The goal is to classify the

possible orbits and study their properties. One of the most important examples

of the Kepler problem is the orbit of celestial objects, as studied by Kepler

himself.

Shapes of orbits

For a planet orbiting the sun, the potential and force are given by

V (r) =

, F (r) = −

with

(for the Coulomb attraction of opposite charges, we have the same

equation with k = −

4πε

Binet’s equation then becomes linear, and

dθ

+ u =

We write the general solution as

u =

+ A cos(θ − θ

where A ≥ 0 and θ

are arbitrary constants.

= 0, then

is constant, and the orbit is circular. Otherwise,

reaches a

maximum at

. This is the periapsis. We now re-define polar coordinates

such that the periapsis is at θ = 0. Then

Proposition. The orbit of a planet around the sun is given by

r =

1 + e cos θ

, (∗)

with

and

. This is the polar equation of a conic, with a

focus (the sun) at the origin.

Definition

(Eccentricity)

The dimensionless parameter

e ≥

0 in the equation

of orbit is the eccentricity and determines the shape of the orbit.

We can rewrite (

∗

) in Cartesian coordinates with

r cos θ

and

r sin θ

Then we obtain

(1 − e

+ 2e`x + y

= `

. (†)

There are three different possibilities:

– Ellipse: (0 ≤ e < 1). r is bounded by

1 + e

≤ r ≤

1 − e

(†) can be put into the equation of an ellipse centered on (−ea, 0),

(x + ea)

= 1,

where a =

1 − e

and b =

√

1 − e

≤ a.

and

are the semi-major and semi-minor axis.

is the semi-latus rectum.

One focus of the ellipse is at the origin. If

= 0, then

and the

ellipse is a circle.

–

Hyperbola: (

e >

1). For

e >

r → ∞

θ → ±α

, where

cos

−1

)

∈

(

π/

, π

). Then (

†

) can be put into the equation of a hyperbola centered on

(ea, 0),

(x − ea)

−

= 1,

with a =

− 1

, b =

√

− 1

This corresponds to an unbound orbit that is deflected (scattered) by an

attractive force.

is both the semi-minor axis and the impact parameter. It is the distance

by which the planet would miss the object if there were no attractive force.

The asymptote is y =

(x − ea), or

− 1 − y = eb.

Alternatively, we can write the equation of the asymptote as

(x, y) ·

√

− 1

, −

= b

or r · n = b, the equation of a line at a distance b from the origin.

– Parabola: (e = 1). Then (∗) becomes

r =

1 + cos θ

We see that

r → ∞

θ → ±π

. (

†

) becomes the equation of a parabola,

= `(` − 2x). The trajectory is similar to that of a hyperbola.

Energy and eccentricity

We can figure out which path a planet follows by considering its energy.

E =

m( ˙r

+ r

) −



dθ



+ u

− mku

Substitute u =

(1 + e cos θ) and ` =

, and it simplifies to

E =

− 1),

which is independent of θ, as it must be.

Orbits are bounded for

e <

1. This corresponds to the case

E <

0. Un-

bounded orbits have

e >

1 and thus

E >

0. A parabolic orbit has

= 1,

= 0,

and is “marginally bound”.

Note that the condition

E >

0 is equivalent to

r| >

2GM

esc

, which

means you have enough kinetic energy to escape orbit.

Kepler’s laws of planetary motion

When Kepler first studied the laws of planetary motion, he took a telescope,

observed actual planets, and came up with his famous three laws of motion. We

are now going to derive the laws with pen and paper instead.

Law

(Kepler’s first law)

The orbit of each planet is an ellipse with the Sun at

one focus.

Law

(Kepler’s second law)

The line between the planet and the sun sweeps

out equal areas in equal times.

Law

(Kepler’s third law)

The square of the orbital period is proportional to

the cube of the semi-major axis, or

∝ a

We have already shown that Law 1 follows from Newtonian dynamics and

the inverse-square law of gravity. In the solar system, planets generally have very

low eccentricity (i.e. very close to circular motion), but asteroids and comets

can have very eccentric orbits. In other solar systems, even planets have have

highly eccentric orbits. As we’ve previously shown, it is also possible for the

object to have a parabolic or hyperbolic orbit. However, we tend not to call

these “planets”.

Law 2 follows simply from the conservation of angular momentum: The area

swept out by moving dθ is dA =

dθ (area of sector of circle). So

θ =

= const.

and is true for any central force.

Law 3 follows from this: the total area of the ellipse is

πab

(by

the second law). But b

= a

(1 − e

) and h

= k` = ka(1 − e

). So

(2π)

(1 − e

)

ka(1 − e

)

(2π)

Note that the third law is very easy to prove directly for circular orbits. Since

the radius is constant, ¨r = 0. So the equations of motion give

−r

= −

= k

Since

θ ∝ P

−1

, the result follows.

4.5 Rutherford scattering

Finally, we will consider the case where the force is repulsive instead of attractive.

An important example is the Rutherford gold foil experiment, where Rutherford

bombarded atoms with alpha particles, and the alpha particles are repelled by

the nucleus of the atom.

Under a repulsive force, the potential and force are given by

V (r) = +

, F (r) = +

For Coulomb repulsion of like charges,

k =

4πε

> 0.

The solution is now

u = −

+ A cos(θ − θ

wlog, we can take A ≥ 0, θ

= 0. Then

r =

e cos θ − 1

with

` =

, e =

We know that

and

are positive. So we must have

e ≥

1. Then

r → ∞

θ → ±α, where α = cos

−1

(1/e).

The orbit is a hyperbola, again given by

(x − ea)

−

= 1,

with

−1

and

√

−1

. However, this time, the trajectory is the other

branch of the hyperbola.

2α

It seems as if the particle is deflected by O.

We can characterize the path of the particle by the impact parameter

and

the incident speed

(i.e. the speed when far away from the origin). We know

that the angular momentum per unit mass is

(velocity

perpendicular

distance to O).

How does the scattering angle

π −

depend on the impact parameter

b and the incident speed v?

Recall that the angle α is given by α = cos

−1

(1/e). So we obtain

= cos α = cos



−



= sin





b =

√

− 1

(bv)

tan

β = 2 tan

−1





We see that if we have a small impact parameter, i.e.

b  k/v

, then we can

have a scattering angle approaching π.

5 Rotating frames

Recall that Newton’s laws hold only in inertial frames. However, sometimes, our

frames are not inertial. In this chapter, we are going to study a particular kind

of non-inertial frame — a rotating frame. An important rotating frame is the

Earth itself, but there are also other examples such as merry-go-rounds.

5.1 Motion in rotating frames

Now suppose that

is an inertial frame, and

is rotating about the

axis

with angular velocity ω =

θ with respect to S.

Definition

(Angular velocity vector)

The angular velocity vector of a rotating

frame is ω = ω

z, where

z is the axis of rotation and ω is the angular speed.

First we wish to relate the basis vectors

}

and

}

and

respectively.

Consider a particle at rest in S

. From the perspective of S, its velocity is





= ω × r,

where

is the angular velocity vector (aligned with the rotation axis).

This formula also applies to the basis vectors of S





= ω × e

Now given a general time-dependent vector

, we can express it in the

}

basis

as follows:

a =

(t)e

From the perspective of

is constant and the time derivative of

is given





, however,

is not constant. So we apply the product rule to obtain the

time derivative of a:





ω × e





+ ω × a.

This key identity applies to all vectors and can be written as an operator equation:

Proposition.

is an inertial frame, and

is rotating relative to

with

angular velocity ω, then









+ ω × .

Applied to the position vector r(t) of a particle, it gives









+ ω × r.

We can interpret this as saying that the difference in velocity measured in

the two frames is the relative velocity of the frames.

We apply this formula a second time, and allow

to depend on time. Then

we have









+ ω×





+ ω × r







+ 2ω ×





ω × r + ω × (ω ×r)

Since S is inertial, Newton’s Second Law is





= F.

Proposition.





= F − 2mω ×





− m

ω × r − mω × (ω ×r).

Definition

(Fictious forces)

The additional terms on the RHS of the equation

of motion in rotating frames are fictitious forces, and are needed to explain the

motion observed in S

. They are

– Coriolis force: −2mω ×





– Euler force: −m

ω × r

– Centrifugal force: −mω ×(ω × r).

In most cases, ω is constant and can neglect the Euler force.

5.2 The centrifugal force

What exactly does the centrifugal force do? Let ω = ω

ω, where |

ω| = 1. Then

−mω × (ω × r) = −m



(ω · r)ω − (ω · ω)r



= mω

⊥

where

⊥

r−

(

r·

)

is the projection of the position on the plane perpendicular

. So the centrifugal force is directed away from the axis of rotation, and its

magnitude is mω

times the distance form the axis.

⊥

Note that

⊥

· r

⊥

= r · r − (r ·

ω)

∇(|r

⊥

) = 2r − 2(r ·

ω)

ω = 2r

⊥

−mω × (ω × r) = −∇



−

mω

⊥



= −∇



−

m|ω × r|



Thus the centrifugal force is a conservative (fictitious) force.

On a rotating planet, the gravitational and centrifugal forces per unit mass

combine to make the effective gravity,

eff

= g + ω

⊥

This gravity will not be vertically downwards. Consider a point

at latitude

on the surface of a spherical planet of radius R.

We construct orthogonal axes:

with

x into the page. So

z is up,

y is North, and

x is East.

At P , we have

r = R

g = −g

ω = ω(cos λ

y + sin λ

eff

= g + ω

⊥

= −g

z + ω

R cos λ(cos λ

z − sin λ

= −ω

R cos λ sin λ

y − (g − ω

R cos

λ)

So the angle α between g and g

eff

is given by

tan α =

R cos λ sin λ

g − ω

R cos

This is 0 at the equator and the poles, and greatest when you are halfway

between. However, this is still tiny on Earth, and does not affect our daily life.

5.3 The Coriolis force

The Coriolis force is a more subtle force. Writing





, we can write the

force as

F = −2mω × v.

Note that this has the same form as the Lorentz force caused by a magnetic

field, and is velocity-dependent. However, unlike the effects of a magnetic field,

particles do not go around in circles in a rotating frame, since we also have the

centrifugal force in play.

Since this force is always perpendicular to the velocity, it does no work.

Consider motion parallel to the Earth’s surface. We only care about the

effect of the Coriolis force on the horizontal trajectory, and ignore the vertical

component that is tiny compared to gravity.

So we only take the vertical component of

ω sin λ

. The horizontal velocity

v = v

x + v

y generates a horizontal Coriolis force:

−2mω sin λ

z × v = −2mω sin λ(v

x − v

y).

In the Northern hemisphere (0

< λ < π/

2), this causes a deflection towards

the right. In the Southern Hemisphere, the deflection is to the left. The effect

vanishes at the equator.

Note that only the horizontal effect of horizontal motion vanishes at the

equator. The vertical effects or those caused by vertical motion still exist.

Example.

Suppose a ball is dropped from a tower of height

at the equator.

Where does it land?

In the rotating frame,

r = g − 2ω ×

r − ω × (ω × r).

We work to first order in ω. Then

r = g − 2ω ×

r + O(ω

Integrate wrt t to obtain

r = gt − 2ω × (r − r

) + O(ω

where

is the initial position. We substitute into the original equation to obtain

r = g − 2ω × gt + O(ω

(where some new

terms are thrown into

(

)). We integrate twice to obtain

r = r

−

ω × gt

+ O(ω

In components, we have g = (0, 0, −g), ω = (0, ω, 0) and r

= (0, 0, R + h). So

r =



ωgt

, 0, R + h −



+ O(ω

So the particle hits the ground at

2h/g

, and its eastward displacement is





3/2

This can be understood in terms of angular momentum conservation in the

non-rotating frame. At the beginning, the particle has the same angular velocity

with the Earth. As it falls towards the Earth, to maintain the same angular

momentum, the angular velocity has to increase to compensate for the decreased

radius. So it spins faster than the Earth and drifts towards the East, relative to

the Earth.

Example.

Consider a pendulum that is free to swing in any plane, e.g. a weight

on a string. At the North pole, it will swing in a plane that is fixed in an inertial

frame, while the Earth rotates beneath it. From the perspective of the rotating

frame, the plane of the pendulum rotates backwards. This can be explained as a

result of the Coriolis force.

In general, at latitude λ, the plane rotates rightwards with period

1 day

sin λ

6 Systems of particles

Now suppose we have

interacting particles. We adopt the following notation:

particle

has mass

, position

, and momentum

. Note that the

subscript denotes which particle it is referring to, not vector components.

Newton’s Second Law for particle i is

= F

where F

is the total force acting on particle i. We can write F

= F

ext

j=1

where

is the force on particle

by particle

, and

ext

is the external force

on i, which comes from particles outside the system.

Since a particle cannot exert a force on itself, we have

. Also, Newton’s

third law requires that

= −F

For example, if the particles interact only via gravity, then we have

= −

− r

)

− r

= −F

6.1 Motion of the center of mass

Sometimes, we are interested in the collection of particles as a whole. For

example, if we treat a cat as a collection of particles, we are more interested in

how the cat as a whole falls, instead of tracking the individual particles of the

cat.

Hence, we define some aggregate quantities of the system such as the total

mass and investigate how these quantities relate.

Definition (Total mass). The total mass of the system is M =

Definition (Center of mass). The center of mass is located at

R =

i=1

This is the mass-weighted average position.

Definition (Total linear momentum). The total linear momentum is

P =

i=1

= M

Note that this is equivalent to the momentum of a single particle of mass

the center of mass.

Definition (Total external force). The total external force is

F =

i=1

ext

We can now obtain the equation of motion of the center of mass:

Proposition.

R = F.

Proof.

R =

i=1

ext

i=1

j=1

= F +

+ F

)

= F

This means that if we don’t care about the internal structure, we can treat

the system as a point particle of mass

at the center of mass

. This is why

Newton’s Laws apply to macroscopic objects even though they are not individual

particles.

Law

(Conservation of momentum)

If there is no external force, i.e.

, then

P = 0. So the total momentum is conserved.

If there is no external force, then the center of mass moves uniformly in a

straight line. In this case, we can pick a particularly nice frame of reference,

known as the center of mass frame.

Definition

(Center of mass frame)

The center of mass frame is an inertial

frame in which R = 0 for all time.

Doing calculations in the center of mass frame is usually much more convenient

than using other frames,

After doing linear motion, we can now look at angular motion.

Definition

(Total angular momentum)

The total angular momentum of the

system about the origin is

L =

× p

How does the total angular momentum change with time? Here we have to

assume a stronger version of Newton’s Third law, saying that

= −F

and is parallel to (r

− r

This is true, at least, for gravitational and electrostatic forces.

Then we have

L =

× p





ext





+ m(

)

× F

ext

× F

ext

× F

+ r

× F

)

= G +

− r

) × F

= G,

where

Definition (Total external torque). The total external torque is

G =

× F

ext

So the total angular momentum is conserved if

, ie the total external

torque vanishes.

6.2 Motion relative to the center of mass

So far, we have shown that externally, a multi-particle system behaves as if it

were a point particle at the center of mass. But internally, what happens to the

individual particles themselves?

We write

, where

is the position of particle

relative to the

center of mass.

We first obtain two useful equalities:

−

R = MR − MR = 0.

Differentiating gives

= 0.

Using these equalities, we can express the angular momentum and kinetic energy

in terms of R and r

only:

L =

(R + r

) × (

R +

)

R ×

R + R ×

R +

= MR ×

R +

T =

(

R +

) · (

R +

)

R ·

R +

R ·

We see that each item is composed of two parts — that of the center of mass

and that of motion relative to center of mass.

If the forces are conservative in the sense that

ext

= −∇

and

= −∇

− r

where

∇

is the gradient with respect to

, then energy is conserved in the from

E = T +

) +

− r

) = const.

6.3 The two-body problem

The two-body problem is to determine the motion of two bodies interacting only

via gravitational forces.

The center of mass is at

R =

+ m

where M = m

+ m

The magic trick to solving the two-body problem is to define the separation

vector (or relative position vector)

r = r

− r

Then we write everything in terms of R and r.

= R +

r, r

= R −

Since the external force

, we have

, i.e. the center of mass moves

uniformly.

Meanwhile,

r =

−





We can write this as

r = F

(r),

where

µ =

+ m

is the reduced mass. This is the same as the equation of motion for one particle of

mass

with position vector

relative to a fixed origin — as studied previously.

For example, with gravity,

r = −

|r|

r = −

|r|

For example, give a planet orbiting the Sun, both the planet and the sun moves

in ellipses about their center of mass. The orbital period depends on the total

mass.

It can be shown that

L = MR ×

R + µr ×

T =

µ|

by expressing r

and r

in terms of r and R.

6.4 Variable-mass problem

All systems we’ve studied so far have fixed mass. However, in real life, many

objects have changing mass, such as rockets, fireworks, falling raindrops and

rolling snowballs.

Again, we will use Newton’s second law, which states that

= F, with p = m

We will consider a rocket moving in one dimension with mass

(

) and velocity

(

). The rocket propels itself forwards by burning fuel and ejecting the exhaust

at velocity −u relative to the rocket.

At time t, the rocket looks like this:

m(t)

v(t)

At time

δt

, it ejects exhaust of mass

(

)

− m

(

δt

) with velocity

(

)

−

u + O(δt).

m(t)

v(t)

v(t) − u

The change in total momentum of the system (rocket + exhaust) is

δp = m(t + δt)v(t + δt) + [m(t) − m(t + δt)][v(t) − u(t) + O(δt)] − m(t)v(t)

= (m + ˙mδt + O(δt

))(v + ˙vδt + O(δt

)) − ˙mδt(v − u) + O(δt

) − mv

= ( ˙mv + m ˙v − ˙mv + ˙mu)δt + O(δt

)

= (m ˙v + ˙mu)δt + O(δt

Newton’s second law gives

lim

δ→0

δp

δt

= F

where F is the external force on the rocket. So we obtain

Proposition (Rocket equation).

+ u

= F.

Example.

Suppose that we travel in space with

= 0. Assume also that

constant. Then we have

= −u

v = v

+ u log



m(t)



Note that we are expressing things in terms of the mass remaining

, not time

Note also that the velocity does not depend on the rate at which mass is

ejected, only the velocity at which it is ejected. Of course, if we expressed

as a

function of time, then the velocity at a specific time does depend on the rate at

which mass is ejected.

Example.

Consider a falling raindrop of mass

(

), gathering mass from a

stationary cloud. In this case, u = v. So

+ v

(mv) = mg,

with

measured downwards. To obtain a solution of this, we will need a model

to determine the rate at which the raindrop gathers mass.

7 Rigid bodies

This chapter is somewhat similar to the previous chapter. We again have a

lot of particles and we study their motion. However, instead of having forces

between the individual particles, this time the particles are constrained such

that their relative positions are fixed. This corresponds to a solid object that

cannot deform. We call these rigid bodies.

Definition

(Rigid body)

A rigid body is an extended object, consisting of

particles that are constrained such that the distance between any pair of

particles, |r

− r

|, is fixed.

The possible motions of a rigid body are the continuous isometries of Euclidean

space, i.e. translations and rotations. However, as we have previously shown,

pure translations of rigid bodies are uninteresting — they simply correspond to

the center of mass moving under an external force. Hence we will first study

rotations.

Later, we will combine rotational and translational effects and see what

happens.

7.1 Angular velocity

We’ll first consider the cases where there is just one particle, moving in a circle

of radius s about the z axis. Its position and velocity vectors are

r = (s cos θ, s sin θ, z)

r = (−s

θ sin θ, s

θ cos θ, 0).

We can write

r = ω × r,

where

ω =

is the angular velocity vector.

In general, we write

ω =

n = ω

where

n is a unit vector parallel to the rotation axis.

The kinetic energy of this particle is thus

T =

Iω

where

is the moment of inertia. This is the counterpart of “mass” in

rotational motion.

Definition (Moment of inertia). The moment of inertia of a particle is

I = ms

= m|

n × r|

where s is the distance of the particle from the axis of rotation.

7.2 Moment of inertia

In general, consider a rigid body in which all

particles rotate about the same

axis with the same angular velocity:

= ω × r

This ensures that

− r

= 2(

−

) · (r

− r

) = 2



ω × (r

− r

)



· (r

− r

) = 0,

as required for a rigid body.

Similar to what we had above, the rotational kinetic energy is

T =

i=1

Iω

where

Definition

(Moment of inertia)

The moment of inertia of a rigid body about

the rotation axis

n is

I =

i=1

n × r

Again, we define the angular momentum of the system:

Definition. The angular momentum is

L =

× (ω × r

Note that our definitions for angular motion are analogous to those for linear

motion. The moment of inertia

is defined such that

Iω

. Ideally, we

would want the momentum to be

Iω

. However, this not true. In fact,

need not be parallel to ω.

What is true, is that the component of

parallel to

is equal to

Iω

. Write

ω = ω

n. Then we have

L ·

n = ω

n · (r

× (

n × r

))

= ω

n × r

) · (

n × r

)

= Iω.

What does L itself look like? Using vector identities, we have

L =



· r

)ω − (r

· ω)r



Note that this is a linear function of ω. So we can write

L = Iω,

where we abuse notation to use

for the inertia tensor. This is represented by a

symmetric matrix with components

(|r

− (r

)

where i refers to the index of the particle, and j, k are dummy suffixes.

If the body rotates about a principal axis, i.e. one of the three orthogonal

eigenvectors of

, then

will be parallel to

. Usually, the principal axes lie on

the axes of rotational symmetry of the body.

7.3 Calculating the moment of inertia

For a solid body, we usually want to think of it as a continuous substance with

a mass density, instead of individual point particles. So we replace the sum of

particles by a volume integral weighted by the mass density ρ(r).

Definition (Mass, center of mass and moment of inertia). The mass is

M =

ρ dV.

The center of mass is

R =

ρr dV

The moment of inertia is

I =

ρs

dV =

ρ|

n × r|

dV.

In theory, we can study inhomogeneous bodies with varying

, but usually

we mainly consider homogeneous ones with constant ρ throughout.

Example

(Thin circular ring)

Suppose the ring has mass

and radius

, and

a rotation axis through the center, perpendicular to the plane of the ring.

Then the moment of inertia is

I = M a

Example

(Thin rod)

Suppose a rod has mass

and length

. It rotates

through one end, perpendicular to the rod.

The mass per unit length is M/`. So the moment of inertia is

I =

dx =

Example

(Thin disc)

Consider a disc of mass

and radius

, with a rotation

axis through the center, perpendicular to the plane of the disc.

Then

I =

2π

πa

|{z}

mass per unit length

|{z}

r dr dθ

| {z }

area element

πa

2π

dθ

πa

(2π)

Now suppose that the rotation axis is in the plane of the disc instead (also

rotating through the center). Then

I =

2π

πa

|{z}

mass per unit length

(r sin θ)

| {z }

r dr dθ

| {z }

area element

πa

2π

sin

θ dθ

πa

Example.

Consider a solid sphere with mass

, radius

, with a rotation axis

though the center.

Using spherical polar coordinates (r, θ, φ) based on the rotation axis,

I =

2π

πa

|{z}

(r sin θ)

| {z }

sin θ dr dθ dφ

| {z }

volume element

πa

(1 − cos

) sin θ dθ

2π

dφ

πa

· 2π

Usually, finding the moment of inertia involves doing complicated integrals.

We will now come up with two theorems that help us find moments of inertia.

Theorem (Perpendicular axis theorem). For a two-dimensional object (a lam-

ina), and three perpendicular axes

x, y, z

through the same spot, with

normal

to the plane,

= I

+ I

where I

is the moment of inertia about the z axis.

Note that this does not apply to 3D objects! For example, in a sphere,

= I

Proof. Let ρ be the mass per unit volume. Then

ρy

ρx

ρ(x

+ y

) dA = I

+ I

Example. For a disc, I

= I

by symmetry. So I

= 2I

Theorem

(Parallel axis theorem)

If a rigid body of mass

has moment of

inertia

about an axis passing through the center of mass, then its moment of

inertia about a parallel axis a distance d away is

I = I

+ Md

Proof.

With a convenient choice of Cartesian coordinates such that the center of

mass is at the origin and the two rotation axes are x = y = 0 and x = d, y = 0,

ρ(x

+ y

) dV,

and

ρr dV = 0.

I =

ρ((x − d)

+ y

) dV

ρ(x

+ y

) dV − 2d

ρx dV +

ρ dV

= I

+ 0 + M d

= I

+ Md

Example.

Take a disc of mass

and radius

, and rotation axis through a

point on the circumference, perpendicular to the plane of the disc. Then

I = I

+ Ma

7.4 Motion of a rigid body

The general motion of a rigid body can be described as a translation of its centre

of mass, following a trajectory

(

), together with a rotation about an axis

through the center of mass. As before, we write

= R + r

Then

R +

Using this, we can break down the velocity and kinetic energy into translational

and rotational parts.

If the body rotates with angular velocity ω about the center of mass, then

= ω × r

Since r

= r

− R, we have

R + ω × r

R + ω × (r

− R).

On the other hand, the kinetic energy, as calculated in previous lectures, is

T =

| {z }

translational KE

| {z }

rotational KE

Sometimes we do not want to use the center of mass as the center. For example,

if an item is held at the edge and spun around, we’d like to study the motion

about the point at which the item is held, and not the center of mass.

So consider any point

, with position vector

(

) that is not the center of

mass but moves with the rigid body, i.e.

Q =

R + ω × (Q − R).

Usually this is a point inside the object itself, but we do not assume that in our

calculation.

Then we can write

R + ω × (r

− R)

Q − ω × (Q − R) + ω × (r

− R)

Q + ω × (r

− Q).

Therefore the motion can be considered as a translation of

(with different

velocity than the center of mass), together with rotation about

(with the same

angular velocity ω).

Equations of motion

As shown previously, the linear and angular momenta evolve according to

P = F (total external force)

L = G (total external torque)

These two equations determine the translational and rotational motion of a rigid

body.

and

depend on the choice of origin, which could be any point that is

fixed in an inertial frame. More surprisingly, it can also be applied to the center

of mass, even if this is accelerated: If

= F

then

= F

+ m

So there is a fictitious force

in the center-of-mass frame. But the total

torque of the fictitious forces about the center of mass is



−m



= −





R = 0 ×

R = 0.

So we can still apply the above two equations.

In summary, the laws of motion apply in any inertial frame, or the center of

mass (possibly non-inertial) frame.

Motion in a uniform gravitational field

In a uniform gravitational field

, the total gravitational force and torque are

the same as those that would act on a single particle of mass

located at the

center of mass (which is also the center of gravity):

F =

ext

g = Mg,

and

G =

ext

× (m

g) =

× g = M R × g.

In particular, the gravitational torque about the center of mass vanishes:

We obtain a similar result for gravitational potential energy.

The gravitational potential in a uniform g is

= −r · g.

(since g = −∇Φ

by definition)

ext

(−r

· g)

= M(−R · g).

Example

(Thrown stick)

Suppose we throw a symmetrical stick. So the center

of mass is the actual center. Then the center of the stick moves in a parabola.

Meanwhile, the stick rotates with constant angular velocity about its center due

to the absence of torque.

Example. Swinging bar.

This is an example of a compound pendulum.

Consider the bar to be rotating about the pivot (and not translating). Its

angular momentum is

with

. The gravitational torque about

the pivot is

G = −Mg

sin θ.

The equation of motion is

L = G.

θ = −M g

sin θ,

θ = −

sin θ.

which is exactly equivalent to a simple pendulum of length 2

3, with angular

frequency

This can also be obtained from an energy argument:

E = T + V =

− Mg

cos θ.

We differentiate to obtain

θ(I

θ + Mg

sin θ) = 0.

θ = −M g

sin θ.

Sliding versus rolling

Consider a cylinder or sphere of radius

, moving along a stationary horizontal

surface.

In general, the motion consists of a translation of the center of mass (with

velocity v) plus a rotation about the center of mass (with angular velocity ω).

The horizontal velocity at the point of contact is

slip

= v − aω.

For a pure sliding motion,

v 6

= 0 and

= 0, in which case

v −aω 6

= 0: the point

of contact moves relative to the surface and kinetic friction may occur.

For a pure rolling motion,

v 6

= 0 and

ω 6

= 0 such that

v − aω

= 0: the point

of contact is stationary. This is the no-slip condition.

The rolling body can alternatively be considered to be rotating instanta-

neously about the point of contact (with angular velocity

) and not translating.

Example (Rolling down hill).

Consider a cylinder or sphere of mass

and radius

rolling down a rough plane

inclined at angle α. The no-slip (rolling) condition is

v − aω = 0.

The kinetic energy is

T =

Iω



M +



The total energy is

E =



M +



˙x

− Mgx sin α,

where

is the distance down slope. While there is a frictional force, the

instantaneous velocity is 0, and no work is done. So energy is conserved, and we

have

= ˙x



M +



¨x − M g sin α



= 0.



M +



¨x = Mg sin α.

For example, if we have a uniform solid cylinder,

I =

(as for a disc)

and so

¨x =

g sin α.

For a thin cylindrical shell,

I = M a

¨x =

g sin α.

Alternatively, we may do it in terms of forces and torques,

The equations of motion are

M ˙v = Mg sin α − F

and

I ˙ω = aF.

While rolling,

˙v − a ˙ω = 0.

M ˙v = Mg sin α −

˙v,

leading to the same result.

Note that even though there is a frictional force, it does no work, since

slip

= 0. So energy is still conserved.

Example (Snooker ball).

It is struck centrally so as to initiate translation, but not rotation. Sliding occurs

initially. Intuitively, we think it will start to roll, and we’ll see that’s the case.

The constant frictional force is

F = µ

N = µ

Mg,

which applies while v − aω > 0.

The moment of inertia about the center of mass is

I =

The equations of motion are

M ˙v = −F

I ˙ω = aF

Initially, v = v

and ω = 0. Then the solution is

v = v

− µ

ω =

as long as v − aω > 0. The slip velocity is

slip

= v − aω = v

−

gt = v



1 −

roll



where

roll

7µ

This is valid up till

roll

. Then the slip velocity is 0, rolling begins and

friction ceases.

At this point,

aω

. The energy is then

. So energy

is lost to friction.

8 Special relativity

When particles move Extremely Fast

, Newtonian Dynamics becomes inaccu-

rate and is replaced by Einstein’s Special Theory of Relativity (1905).

Its effects are noticeable only when particles approach to the speed of light,

c = 299 792 458 m s

−1

≈ 3 × 10

m s

−1

This is really fast.

The Special Theory of Relativity rests on the following postulate:

The laws of physics are the same in all inertial frames

This is the principle of relativity familiar to Galileo. Galilean relativity mentioned

in the first chapter satisfies this postulate for dynamics. People then thought

that Galilean relativity is what the world obeys. However, it turns out that

there is a whole family of solutions that satisfy the postulate (for dynamics),

and Galilean relativity is just one of them.

This is not a problem (yet), since Galilean relativity seems so intuitive, and

we might as well take it to be the true one. However, it turns out that solving

Maxwell’s equations of electromagnetism gives an explicit value of the speed of

light,

. This is independent of the frame of reference. So the speed of light must

be the same in every inertial frame.

This is not compatible with Galilean relativity.

Consider the two inertial frames

and

, moving with relative velocity

Then if light has velocity

, then Galilean relativity predicts it has velocity

c − v in S

, which is wrong.

Therefore, we need to find a different solution to the principle of relativity

that preserves the speed of light.

8.1 The Lorentz transformation

Consider again inertial frames

and

whose origins coincide at

= 0. For

now, neglect the

and

directions, and consider the relationship between (

x, t

)

and (x

, t

). The general form is

= f(x, t), t

= g(x, t),

for some functions f and g. This is not very helpful.

In any inertial frame, a free particle moves with constant velocity. So straight

lines in (

x, t

) must map into straight lines in (

, t

). Therefore the relationship

must be linear.

Given that the origins of

and

coincide at

= 0, and

moves with

velocity v relative to S, we know that the line x = vt must map into x

= 0.

Combining these two information, the transformation must be of the form

= γ(x − vt), (1)

for some factor

that may depend on

|v|

(not

itself. We can use symmetry

arguments to show that γ should take the same value for velocities v and −v).

Note that Galilean transformation is compatible with this – just take

be always 1.

Now reverse the roles of the frames. From the perspective

moves with

velocity −v. A similar argument leads to

x = γ(x

+ vt

), (2)

with the same factor

, since

only depends on

|v|

. Now consider a light ray

(or photon) passing through the origin

= 0 at

= 0. Its trajectory in

S is

x = ct.

We want a γ such that the trajectory in S

= ct

as well, so that the speed of light is the same in each frame. Substitute these

into (1) and (2)

= γ(c − v)t

ct = γ(c + v)t

Multiply the two equations together and divide by tt

to obtain

= γ

− v

γ =

− v

1 − (v/c)

Definition (Lorentz factor). The Lorentz factor is

γ =

1 − (v/c)

Note that

– γ ≥ 1 and is an increasing function of |v|.

– When v  c, then γ ≈ 1, and we recover the Galilean transformation.

– When |v| → c, then γ → ∞.

–

|v| ≥ c

, then

is imaginary, which is physically impossible (or at least

weird).

–

If we take

c → ∞

, then

= 1. So Galilean transformation is the transfor-

mation we will have if light is infinitely fast. Alternatively, in the world of

Special Relativity, the speed of light is “infinitely fast”.

For the sense of scale, we have the following values of γ at different speeds:

– γ = 2 when v = 0.866c.

– γ = 10 when v = 0.9949.

– γ = 20 when v = 0.999c.

We still have to solve for the relation between

and

. Eliminate

between

(1) and (2) to obtain

x = γ(γ(x − vt) + vt

= γt − (1 − γ

−2

)

γx

= γ



t −



So we have

Law

(Principle of Special Relativity)

Let

and

be inertial frames, moving

at the relative velocity of v. Then

= γ(x − vt)

= γ



t −



where

γ =

1 − (v/c)

This is the Lorentz transformations in the standard configuration (in one spatial

dimension).

The above is the form the Lorentz transformation is usually written, and is

convenient for actual calculations. However, this lacks symmetry between space

and time. To display the symmetry, one approach is to use units such that

= 1.

Then we have

= γ(x − vt),

= γ(t − vx).

Alternatively, if we want to keep our

’s, instead of comparing

and

, which

have different units, we can compare x and ct. Then we have

= γ



x −

(ct)



= γ



ct −



Symmetries aside, to express

x, t

in terms of

, t

, we can invert this linear

mapping to find (after some algebra)

x = γ(x

+ vt

)

t = γ





Directions perpendicular to the relative motion of the frames are unaffected:

= y

= z

Now we check that the speed of light is really invariant:

For a light ray travelling in the x direction in S:

x = ct, y = 0, z = 0.

In S

, we have

γ(x − vt)

γ(t − vx/c

)

(c − v)t

(1 − v/c)t

= c,

as required.

For a light ray travelling in the Y direction in S,

x = 0, y = ct, z = 0.

In S

γ(x − vt)

γ(t − vx/c

)

= −v,

and

γ(t − vx/c

and

= 0.

So the speed of light is

+ y

+ γ

−2

= c,

as required.

More generally, the Lorentz transformation implies

− r

= c

− x

− y

− z

= c



t −



− γ

(x − vt)

− y

− z

= γ



1 −



− x

) − y

− z

= c

− x

− y

− z

= c

− r

We say that the quantity c

− x

− y

− z

is Lorentz-invariant.

So if

= c, then

= c also.

8.2 Spacetime diagrams

It is often helpful to plot out what is happening on a diagram. We plot them on

a graph, where the position

is on the horizontal axis and the time

is on the

vertical axis. We use ct instead of t so that the dimensions make sense.

world line

Definition

(Spacetime)

The union of space and time in special relativity is

called Minkowski spacetime. Each point

represents an event, labelled by

coordinates (ct, x) (note the order!).

A particle traces out a world line in spacetime, which is straight if the particle

moves uniformly.

Light rays moving in the x direction have world lines inclined at 45

◦

light raylight ray

We can also draw the axes of

, moving in the

direction at velocity

relative to

. The

axis corresponds to

= 0, i.e.

. The

axis

corresponds to t

= 0, i.e. t = vx/c

Note that the

and

axes are not orthogonal, but are symmetrical about

the diagonal (dashed line). So they agree on where the world line of a light ray

should lie on.

8.3 Relativistic physics

Now we can look at all sorts of relativistic weirdness!

Simultaneity

The first relativistic weirdness is that different frames disagree on whether two

evens are simultaneous

Definition

(Simultaneous events)

We say two events

and

are simultane-

ous in the frame S if t

= t

They are represented in the following spacetime diagram by horizontal dashed

lines.

However, events that are simultaneous in

have equal values of

, and so

lie on lines

ct −

x = constant.

The lines of simultaneity of

and those of

are different, and events simulta-

neous in

need not be simultaneous in

. So simultaneity is relative.

thinks

and P

happened at the same time, while S

thinks P

happens first.

Note that this is genuine disagreement. It is not due to effects like, it takes

time for the light conveying the information to different observers. Our account

above already takes that into account (since the whole discussion does not involve

specific observers).

Causality

Although different people may disagree on the temporal order of events, the

consistent ordering of cause and effect can be ensured.

Since things can only travel at at most the speed of light,

cannot affect

happens a millisecond after

but is at millions of galaxies away. We can

draw a light cone that denotes the regions in which things can be influenced

. These are the regions of space-time light (or any other particle) can

possibly travel to.

can only influence events within its future light cone, and

be influenced by events within its past light cone.

All observers agree that

occurs after

. Different observers may disagree on

the temporal ordering of

and

. However, since nothing can travel faster

than light,

and

cannot influence each other. Since everyone agrees on how

fast light travels, they also agree on the light cones, and hence causality. So

philosophers are happy.

Time dilation

Suppose we have a clock that is stationary in

(which travels at constant

velocity

with respect to inertial frame

) ticks at constant intervals ∆

. What

is the interval between ticks in S?

Lorentz transformation gives

t = γ





Since x

= constant for the clock, we have

∆t = γ∆t

> ∆t

So the interval measured in S is greater! So moving clocks run slowly.

A non-mathematical explanation comes from Feynman (not lectured): Sup-

pose we have a very simple clock: We send a light beam towards a mirror, and

wait for it to reflect back. When the clock detects the reflected light, it ticks,

and then sends the next light beam.

Then the interval between two ticks is the distance 2

divided by the speed

of light.

From the point of view of an observer moving downwards, by the time light

reaches the right mirror, it would have moved down a bit. So S sees

However, the distance travelled by the light beam is now

(2d)

+ a

Since they agree on the speed of light, it must have taken longer for the clock to

receive the reflected light in S. So the interval between ticks are longer.

By the principle of relativity, all clocks must measure the same time dilation,

or else we can compare the two clocks and know if we are “moving”.

This is famously evidenced by muons. Their half-life is around 2 microseconds

(i.e. on average they decay to something else after around 2 microseconds). They

are created when cosmic rays bombard the atmosphere. However, even if they

travel at the speed of light, 2 microseconds only allows it to travel

600 m

, certainly

not sufficient to reach the surface of Earth. However, we observe lots of muons

on Earth. This is because muons are travelling so fast that their clocks run

really slowly.

The twin paradox

Consider two twins: Luke and Leia. Luke stays at home. Leia travels at a

constant speed

to a distant planet

, turns around, and returns at the same

speed.

In Luke’s frame of reference,

Luke

2cT

Leia: x = vt

A (Leia’s arrival)

Leia’s arrival (A) at P has coordinates

(ct, x) = (cT, vT ).

The time experienced by Leia on her outward journey is

= γ



T −



By Leia’s return

, Luke has aged by 2

, but Leia has aged by

. So

she is younger than Luke, because of time dilation.

The paradox is: From Leia’s perspective, Luke travelled away from her at

speed and the returned, so he should be younger than her!

Why is the problem not symmetric?

We can draw Leia’s initial frame of reference in dashed lines:

Leia

Han

In Leia’s frame, by the time she arrives at

, she has experienced a time

as shown above. This event is simultaneous with event

in Leia’s frame. Then

in Luke’s frame, the coordinates of X are

(ct, x) =



, 0





, 0



obtained through calculations similar to that above. So Leia thinks Luke has

aged less by a factor of 1

/γ

. At this stage, the problem is symmetric, and Luke

also thinks Leia has aged less by a factor of 1/γ

Things change when Leia turns around and changes frame of reference. To

understand this better, suppose Leia meets a friend, Han, who is just leaving

at speed

. On his journey back, Han also thinks Luke ages

T/γ

. But in his

frame of reference, his departure is simultaneous with Luke’s event

, not

since he has different lines of simultaneity.

So the asymmetry between Luke and Leia occurs when Leia turns around.

At this point, she sees Luke age rapidly from X to Z.

Length contraction

A rod of length L

is stationary in S

. What is its length in S?

, then length of the rod is the distance between the two ends at the

same time. So we have

In S, we have

The lines

= 0 and

map into

and

/γ

. So the length

/γ < L

. Therefore moving objects are contracted in the direction

of motion.

Definition

(Proper length)

The proper length is the length measured in an

object’s rest frame.

This is analogous to the fact that if you view a bar from an angle, it looks

shorter than if you view it from the front. In relativity, what causes the

contraction is not a spatial rotation, but a spacetime hyperbolic rotation.

Question: does a train of length 2

fit alongside a platform of length

if it

travels through the station at a speed v such that γ = 2?

For the system of observers on the platform, the train contracts to a length

2L/γ = L. So it fits.

But for the system of observers on the train, the platform contracts to length

L/γ = L/2, which is much too short!

This can be explained by the difference of lines of simultaneity, since length

is the distance between front and back at the same time.

ctback of train front of train

back of

platform

front of

platform

fits in S

doesn’t fit in S

Composition of velocities

A particle moves with constant velocity

in frame

, which moves with velocity

v relative to S. What is its velocity u in S?

The world line of the particle in S

= u

In S, using the inverse Lorentz transformation,

u =

γ(x

+ vt

)

γ(t

+ (v/c

)

+ vt

+ (v/c

+ v

1 + u

v/c

This is the formula for the relativistic composition of velocities.

The inverse transformation is found by swapping

and

, and swapping the

sign of v, i.e.

u − v

1 − uv/c

Note the following:

–

v  c

, then the transformation reduces to the standard Galilean

addition of velocities u ≈ u

+ v.

– u

is a monotonically increasing function of

for any constant

(with

|v| < c).

–

When

±c

for any

, i.e. the speed of light is constant in all

frames of reference.

–

Hence

| < c

iff

|u| < c

. This means that we cannot reach the speed of

light by composition of velocities.

8.4 Geometry of spacetime

We’ll now look at the geometry of spacetime, and study the properties of vectors

in this spacetime. While spacetime has 4 dimensions, and each point can be

represented by 4 real numbers, this is not ordinary

. This can be seen when

changing coordinate systems, instead of rotating the axes like in

, we “squash”

the axes towards the diagonal, which is a hyperbolic rotation. In particular,

we will have a different notion of a dot product. We say that this space has

dimension d = 1 + 3.

The invariant interval

In regular Euclidean space, given a vector

, all coordinate systems agree on the

length |x|. In Minkowski space, they agree on something else.

Consider events

and

with coordinates (

, x

) and (

, x

) separated

by ∆t = t

− t

and ∆x = x

− x

Definition

(Invariant interval)

The invariant interval or spacetime interval

between P and Q is defined as

∆s

= c

∆t

− ∆x

Note that this quantity ∆

can be both positive or negative — so ∆

might be

imaginary!

Proposition. All inertial observers agree on the value of ∆s

Proof.

∆t

− ∆x

= c



∆t −

∆x



− γ

(∆x − v∆t)

= γ



1 −



∆t

− ∆x

)

= c

∆t

− ∆x

In three spatial dimensions,

∆s

= c

∆t

− ∆x

− ∆y

− ∆z

We take this as the “distance” between the two points. For two infinitesimally

separated events, we have

Definition (Line element). The line element is

= c

− dx

− dy

− dz

Definition

(Timelike, spacelike and lightlike separation)

Events with ∆

are timelike separated. It is possible to find inertial frames in which the two events

occur in the same position, and are purely separated by time. Timelike-separated

events lie within each other’s light cones and can influence one another.

Events with ∆

0 are spacelike separated. It is possible to find inertial

frame in which the two events occur in the same time, and are purely separated

by space. Spacelike-separated events lie out of each other’s light cones and

cannot influence one another.

Events with ∆

= 0 are lightlike or null separated. In all inertial frames, the

events lie on the boundary of each other’s light cones. e.g. different points in

the trajectory of a photon are lightlike separated, hence the name.

Note that ∆s

= 0 does not imply that P and Q are the same event.

The Lorentz group

The coordinates of an event

in frame

can be written as a 4-vector (i.e.

4-component vector) X. We write

X =













The invariant interval between the origin and

can be written as an inner

product

X · X = X

ηX = c

− x

− y

− z

where

η =







1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1







4-vectors with

X · X >

0 are called timelike, and those

X · X <

0 are spacelike.

If X · X = 0, it is lightlike or null.

A Lorentz transformation is a linear transformation of the coordinates from

one frame S to another S

, represented by a 4 × 4 tensor (“matrix”):

= ΛX

Lorentz transformations can be defined as those that leave the inner product

invariant:

(∀X)(X

· X

= X · X),

which implies the matrix equation

ηΛ = η. (∗)

These also preserve X · Y if X and Y are both 4-vectors.

Two classes of solution to this equation are:

Λ =







1 0 0 0

0 R







where

is a 3

3 orthogonal matrix, which rotates (or reflects) space and leaves

time intact; and

Λ =







γ −γβ 0 0

−γβ γ 0 0

0 0 1 0

0 0 0 1







where

, and

= 1

1 − β

. Here we leave the

and

coordinates intact,

and apply a Lorentz boost along the x direction.

The set of all matrices satisfying equation (

∗

) form the Lorentz group

3).

It is generated by rotations and boosts, as defined above, which includes the

absurd spatial reflections and time reversal.

The subgroup with det Λ = +1 is the proper Lorentz group SO(1, 3).

The subgroup that preserves spatial orientation and the direction of time is

the restricted Lorentz group

3). Note that this is different from

3),

since if you do both spatial reflection and time reversal, the determinant of the

matrix is still positive. We want to eliminate those as well!

Rapidity

Focus on the upper left 2

2 matrix of Lorentz boosts in the

direction. Write

Λ[β] =



γ −γβ

−γβ γ



, γ =

1 − β

Combining two boosts in the x direction, we have

Λ[β

]Λ[β

] =



−γ



−γ



= Λ



+ β

1 + β



after some messy algebra. This is just the velocity composition formula as before.

This result does not look nice. This suggests that we might be writing things

in the wrong way.

We can compare this with spatial rotation. Recall that

R(θ) =



cos θ sin θ

−sin θ cos θ



with

R(θ

)R(θ

) = R(θ

+ θ

For Lorentz boosts, we can define

Definition (Rapidity). The rapidity of a Lorentz boot is φ such that

β = tanh φ, γ = cosh φ, γβ = sinh φ.

Then

Λ[β] =



cosh φ −sinh φ

−sinh φ cosh φ



= Λ(φ).

The rapidities add like rotation angles:

Λ(φ

)Λ(φ

) = Λ(φ

+ φ

This shows the close relation betweens spatial rotations and Lorentz boosts.

Lorentz boots are simply hyperbolic rotations in spacetime!

8.5 Relativistic kinematics

In Newtonian mechanics, we describe a particle by its position

(

), with its

velocity being u(t) =

In relativity, this is unsatisfactory. In special relativity, space and time can

be mixed together by Lorentz boosts, and we prefer not to single out time from

space. For example, when we write the 4-vector

, we put in both the time and

space components, and Lorentz transformations are 4

4 matrices that act on

In the definition of velocity, however, we are differentiating space with respect

to time, which is rather weird. First of all, we need something to replace time.

Recall that we defined “proper length” as the length in the item in its rest frame.

Similarly, we can define the proper time.

Definition (Proper time). The proper time τ is defined such that

∆τ =

∆s

is the time experienced by the particle, i.e. the time in the particles rest frame.

The world line of a particle can be parametrized using the proper time by

t(τ) and x(τ).

Infinitesimal changes are related by

dτ =

− |dx|

1 −

|u|

dt.

Thus

dτ

= γ

with

1 −

|u|

The total time experienced by the particle along a segment of its world line is

T =

dτ =

dt.

We can then define the position 4-vector and 4-velocity.

Definition (Position 4-vector and 4-velocty). The position 4-vect or is

X(τ) =



ct(τ)

x(τ)



Its 4-velocity is defined as

U =

dτ



dτ



dτ





= γ





where u =

Another common notation is

X = (ct, x), U = γ

(c, u).

If frames

and

are related by

= Λ

, then the 4-velocity also transforms

as U

= ΛU.

Definition

(4-vector)

A 4-vector is a 4-component vectors that transforms in

this way under a Lorentz transformation, i.e. X

= ΛX.

When using suffix notation, the indices are written above (superscript) instead

of below (subscript). The indices are written with Greek letters which range

from 0 to 3. So we have

instead of

, for

= 0

3. If we write

instead, it means a different thing. This will be explained more in-depth in the

electromagnetism course (and you’ll get more confused!).

is a 4-vector because

is a 4-vector and

is a Lorentz invariant. Note

that dX/dt is not a 4-vector.

Note that this definition of 4-vector is analogous to that of a tensor — things

that transform nicely according to our rules. Then

would be a scalar, i.e.

rank-0 tensor, while t is just a number, not a scalar.

For any 4-vector

, the inner product

U · U

· U

is Lorentz invariant,

i.e. the same in all inertial frames. In the rest frame of the particle,

= (

0).

So U · U = c

In any other frame, Y = γ

(c, u). So

Y · Y = γ

− |u|

) = c

as expected.

Transformation of velocities revisited

We have seen that velocities cannot be simply added in relativity. However, the

4-velocity does transform linearly, according to the Lorentz transform:

= ΛU.

In frame

, consider a particle moving with speed

at an angle

to the

axis

in the

plane. This is the most general case for motion not parallel to the

Lorentz boost.

Its 4-velocity is

U =







u cos θ

u sin θ







, γ

1 − u

With frames

and

in standard configuration (i.e. origin coincide at

= 0,

moving in x direction with velocity v relative to S),







cos θ

sin θ













−γ

v/c 0 0

−γ

v/c γ

0 0

0 0 1 0

0 0 0 1













u cos θ

u sin θ







Instead of evaluating the whole matrix, we can divide different rows to get useful

results.

The ratio of the first two lines gives

cos θ

u cos θ − v

1 −

cos θ

just like the composition of parallel velocities.

The ratio of the third to second line gives

tan θ

u sin θ

(u cos θ − v)

which describes aberration, a change in the direction of motion of a particle due

to the motion of the observer. Note that this isn’t just a relativistic effect! If

you walk in the rain, you have to hold your umbrella obliquely since the rain

seems to you that they are coming from an angle. The relativistic part is the

factor in the denominator.

This is also seen in the aberration of starlight (

) due to the Earth’s

orbital motion. This causes small annual changes in the apparent positions of

stars.

4-momentum

Definition (4-momentum). The 4-momentum of a particle of mass m is

P = mU = mγ





The 4-momentum of a system of particles is the sum of the 4-momentum of the

particles, and is conserved in the absence of external forces.

The spatial components of P are the relativistic 3-momentum,

p = mγ

which differs from the Newtonian expression by a factor of

. Note that

|p| → ∞

as |u| → c.

What is the interpretation of the time component

(i.e. the first time

component of the P vector)? We expand for |u|  c:

= mγc =

1 − |u|



m|u|

+ ···



We have a constant term

plus a kinetic energy term

m|u|

, plus more

tiny terms, all divided by

. So this suggests that

is indeed the energy for

a particle, and the remaining

···

terms are relativistic corrections for our old

formula

m|u|

(the mc

term will be explained later). So we interpret P as

P =



E/c



Definition

(Relativistic energy)

The relativistic energy of a particle is

E = mγc

= mc

m|u|

+ ···

Note that E → ∞ as |u| → c.

For a stationary particle, we obtain

E = mc

This implies that mass is a form of energy.

is sometimes called the rest mass.

The energy of a moving particle,

mγ

, is the sum of the rest energy

and kinetic energy m(γ

− 1)c

Since

P ·P

−|p|

is a Lorentz invariant (lengths of 4-vectors are always

Lorentz invariant) and equals

in the particle’s rest frame, we have the

general relation between energy and momentum

= |p|

+ m

In Newtonian physics, mass and energy are separately conserved. In relativity,

mass is not conserved. Instead, it is just another form of energy, and the total

energy, including mass energy, is conserved.

Mass can be converged into kinetic energy and vice versa (e.g. atomic bombs!)

Massless particles

Particles with zero mass (

= 0), e.g. photons, can have non-zero momentum

and energy because they travel at the speed of light (γ = ∞).

In this case,

P · P

= 0. So massless particles have light-like (or null)

trajectories, and no proper time can be defined for such particles.

Other massless particles in the Standard Model of particle physics include

the gluon.

For these particles, energy and momentum are related by

= |p|

E = |p|c.

Thus

P =





where n is a unit (3-)vector in the direction of propagation.

According to quantum mechanics, fundamental “particles” aren’t really

particles but have both particle-like and wave-like properties (if that sounds

confusing, yes it is!). Hence we can assign it a de Broglie wavelength, according

to the de Broglie relation:

|p| =

where h ≈ 6.63 × 10

−34

kg s

−1

is Planck’s constant.

For massless particles, this is consistent with Planck’s relation:

E =

= hν,

where ν =

is the wave frequency.

Newton’s second law in special relativity

Definition (4-force). The 4-force is

F =

dτ

This equation is the relativistic counterpart to Newton’s second law.

It is related to the 3-force F by

F = γ



F · u/c



Expanding the definition of the 4-force componentwise, we obtain

dτ

= γ

F · u ⇒

= F · u

and

dτ

= γ

F ⇒

= F

Equivalently, for a particle of mass m,

F = mA,

where

A =

dτ

is the 4-acceleration.

We have

U = γ





A = γ

= γ



˙γ

a + ˙γ



where a =

and ˙γ

= γ

a·u

In the instantaneous rest frame of a particle, u = 0 and γ

= 1. So

U =





, A =





Then

U · A

= 0. Since this is a Lorentz invariant, we have

U · A

= 0 in all

frames.

8.6 Particle physics

Many problems can be solved using the conservation of 4-momentum,

P =



E/c



for a system of particles.

Definition

(Center of momentum frame)

The center of momentum (CM) frame,

or zero momentum frame, is an inertial frame in which the total 3-momentum is

p = 0.

This exists unless the system consists of one or more massless particle moving

in a single direction.

Particle decay

A particle of mass m

decays into two particles of masses m

and m

We have

= P

+ P

i.e.

= E

+ E

= p

+ p

In the CM frame (i.e. the rest frame of the original particle),

= m

+ m

≥ m

+ m

So decay is possible only if

≥ m

+ m

(Recall that mass is not conserved in relativity!)

Example. A possible decay path of the Higgs’ particle can be written as

h → γγ

Higgs’ particle → 2 photons

This is possible by the above criterion, because m

≥ 0, while m

= 0.

The full conservation equation is





= P

+ P

= p

= E

Particle scattering

When two particles collide and retain heir identities, the total 4-momentum is

conserved:

+ P

= P

+ P

In the laboratory frame

, suppose that particle 1 travels with speed

and

collides with particle 2 (at rest).

In the CM frame S

+ p

= 0 = p

+ p

Both before and after the collision, the two particles have equal and opposite

3-momentum.

1 2

The scattering angle

is undetermined and can be thought of as being random.

However, we can derive some conclusions about the angles

and

in the

laboratory frame.

(staying in

for the moment) Suppose the particles have equal mass

They then have the same speed v in S

Choose axes such that







mγ







, P







mγ

−mγ







and after the collision,







mγ

v cos θ

mγ

v sin θ







, P







mγ

−mγ

v cos θ

−mγ

v sin θ







We then use the Lorentz transformation to return to the laboratory frame

The relative velocity of the frames is v. So the Lorentz transform is

Λ =







v/c 0 0

v/c γ

0 0

0 0 1 0

0 0 0 1







and we find







mγ







, P













where

u =

1 + v

(cf. velocity composition formula)

Considering the transformations of P

and P

, we obtain

tan θ =

sin θ

(1 + cos θ

)

tan

and

tan φ =

sin θ

(1 − cos θ

)

cot

Multiplying these expressions together, we obtain

tan θ tan φ =

So even though we do not know what

and

might be, they must be related

by this equation.

In the Newtonian limit, where |v|  c, we have γ

≈ 1. So

tan θ tan φ = 1,

i.e. the outgoing trajectories are perpendicular in S.

Particle creation

Collide two particles of mass

fast enough, and you create an extra particle of

mass M .

+ P

= P

+ P

where P

is the momentum of the new particle.

In the CM frame,

1 2

v v

+ P



2mγ



We have

+ P



+ E

)/c



2mγ

= E

+ E

≥ 2mc

+ Mc

So in order to create this new particle, we must have

≥ 1 +

Alternatively, it occurs only if the initial kinetic energy in the CM frame satisfies

2(γ

− 1)mc

≥ Mc

If we transform to a frame in which the initial speeds are

and 0 (i.e. stationary

target), then

u =

1 + v

Then

= 2γ

− 1.

So we require

≥ 2



1 +



− 1 = 1 +

This means that the initial kinetic energy in this frame must be

m(γ

− 1)c

≥



2 +



which could be much larger than

, especially if

M  m

, which usually the

case. For example, the mass of the Higgs’ boson is 130 times the mass of the

proton. So it would be much advantageous to collide two beams of protons head

on, as opposed to hitting a fixed target.