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
u
(
θ
) 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
a
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.
a
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
t
is described by its position vector,
r
(
t
) or
x
(
t
) 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
r
, we can define various interesting quantities which
describe the particle.
Definition (Velocity). The velocity of the particle is
v =
˙
r =
dr
dt
.
Definition (Acceleration). The acceleration of the particle is
a =
˙
v =
¨
r =
d
2
r
dt
2
.
Definition (Momentum). The momentum of a particle is
p = mv = m
˙
r.
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
0
= r r
0
Translations of time:
t
0
= t t
0
Rotation (and reflection):
r
0
= 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
S
is an
inertial frame. Then any other frame
S
0
in uniform motion relative to
S
is also
inertial:
S
S
0
y
x
y
0
x
0
v
Assuming the frames coincide at t = 0, then
x
0
= x vt
y
0
= y
z
0
= z
t
0
= t
Generally, the position vector transforms as
r
0
= r vt,
where
v
is the (constant) velocity of
S
0
relative to
S
. The velocity and acceleration
transform as follows:
˙
r
0
=
˙
r v
¨
r
0
=
¨
r
Definition
(Galilean boost)
.
A Galilean boost is a change in frame of reference
by
r
0
= r vt
t
0
= 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
dp
dt
= F,
where
p
=
mv
=
m
˙
r
is the (linear) momentum of the particle. We say
m
is the
(inertial) mass of the particle, which is a measure of its reluctance to accelerate.
Usually, m is constant. Then
F = ma = m
¨
r.
Usually,
F
is specified as a function of
r,
˙
r
and
t
. Then we have a second-order
ordinary differential equation for r.
To determine the solution, we need to specify the initial values of
r
and
˙
r
, 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
X
, denoted by [
X
] are expressible
uniquely in terms of L, M and T . For example,
[area] = L
2
[density] = L
3
M
[velocity] = LT
1
[acceleration] = LT
2
[
force
] =
LMT
2
since the dimensions must satisfy the equation
F
=
ma
.
[energy] = L
2
MT
2
, e.g. consider E = mv
2
/2.
Physical constants also have dimensions, e.g. [
G
] =
L
3
M
1
T
2
by
F
=
GMm/r
2
.
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
x
again makes no sense if x has a dimension, since
e
x
= 1 + x +
1
2
x
2
+ ···
and if
x
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
m
3
kg
1
s
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
Y
depends on 3 other physical
quantities
X
1
, X
2
, X
3
, i.e.
Y
=
Y
(
X
1
, X
2
, X
3
). Let their dimensions be as
follows:
[Y ] = L
a
M
b
T
c
[X
i
] = L
a
i
M
b
i
T
c
i
Suppose further that we know that the relationship is a power law, i.e.
Y = CX
p
1
1
X
p
2
2
X
p
3
3
,
where
C
is a dimensionless constant (i.e. a pure number). Since the dimensions
must work out, we know that
a = p
1
a
1
+ p
2
a
2
+ p
3
a
3
b = p
1
b
1
+ p
2
b
2
+ p
3
b
3
c = p
1
c
1
+ p
2
c
2
+ p
3
c
3
for which there is a unique solution provided that the dimensions of
X
1
, X
2
and
X
3
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
X
2
1
X
2
is dimensionless, then the relation
between
Y
and
X
i
can involve more complicated terms such as
exp
(
X
2
1
X
2
), since
the argument of exp is now dimensionless.
In general, suppose we have many terms, and the dimensions of
X
i
are not in-
dependent. We order the quantities so that the independent terms [
X
1
]
,
[
X
2
]
,
[
X
3
]
are at the front. For each of the remaining variables, form a dimensionless quan-
tity λ
i
= X
i
X
q
1
1
X
q
2
2
X
q
3
3
. Then the relationship must be of the form
Y = f(λ
4
, λ
5
, ···)X
p
1
1
X
p
2
2
X
p
3
3
.
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).
m
`
d
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
d
l
m
p
1
`
p
2
g
p
3
,
where f is a dimensionless function. For the dimensions to balance,
T = M
p
1
L
p
2
L
p
3
T
2p
3
.
So p
1
= 0, p
2
= p
3
= 1/2. Then
P = f
d
`
s
`
g
.
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
F
=
ma
. 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
m
moving in a straight line
with position
x
(
t
). Suppose
F
=
F
(
x
), i.e. it depends on position only. We
define the potential energy as follows:
Definition
(Potential energy)
.
Given a force field
F
=
F
(
x
), we define the
potential energy to be a function V (x) such that
F =
dV
dx
.
or
V =
Z
F dx.
V
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 =
dV
dx
.
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 =
1
2
m ˙x
2
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 =
dV
dx
.
Then the total energy
E = T + V =
1
2
m ˙x
2
+ V (x)
is conserved, i.e.
˙
E = 0.
Proof.
dE
dt
= m ˙x¨x +
dV
dx
˙x
= ˙x
m¨x +
dV
dx
= 0
Example. Consider the harmonic oscillator, whose potential is given by
V =
1
2
kx
2
.
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 =
p
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
V
(
x
), conservation of energy allows us to solve
the problem formally:
E =
1
2
m ˙x
2
+ V (x)
Since E is a constant, from this equation, we have
dx
dt
= ±
r
2
m
(E V (x))
t t
0
= ±
Z
dx
q
2
m
(E V (x))
.
To find
x
(
t
), we need to do the integral and then solve for
x
. 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
V
(
x
), 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
V
(
x
) =
m
(
x
3
3
x
). Note that this can be dimensionally
consistent even though we add up
x
3
and
3
x
, if we declare “3” to have dimension
L
2
.
We plot this as follows:
x
V
O
1 212
Suppose we release the particle from rest at
x
=
x
0
. Then
E
=
V
(
x
0
). We can
consider what happens to the particle for different values of x
0
.
x
0
= ±1: This is an equilibrium and the Particle stays there for all t.
1
< x
0
<
2: The particle does not have enough energy to escape the well.
So it oscillates back and forth in potential well.
x
0
< 1: The particle falls to x = −∞.
x
0
>
2: The particle has enough energy to overshoot the well and continues
to x = −∞.
x
0
= 2: This is a special case. Obviously, the particle goes towards
x
=
1.
But how long does it take, and what happens next? Here we have
E
= 2
m
.
We noted previously
t t
0
=
Z
dx
q
2
m
(E V (x))
.
Let x = 1 + ε(t). Then
2
m
(E V (x)) = 4 2(1 + ε)
3
+ 6(1 + ε)
= 6ε
2
2ε
3
.
So
t t
0
=
Z
ε
3
dε
0
6ε
2
2ε
3
We reach
x
=
1 when
ε
0. But for small
ε
, the integrand is approx-
imately
1
, which integrates to
ln ε −∞
as
ε
0. So
ε
= 0 is
achieved after infinite time, i.e. it takes infinite time to reach
ε
= 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
0
(
x
), the equilibrium
points are the stationary points of the potential energy, i.e.
V
0
(x
0
) = 0.
Consider motion near an equilibrium point. We assume that the motion is
small and we can approximate
V
by a second-order Taylor expansion. Then we
can write V as
V (x) V (x
0
) +
1
2
V
00
(x
0
)(x x
0
)
2
.
Then the equation of motion is
m¨x = V
00
(x
0
)(x x
0
).
If
V
00
(
x
0
)
>
0, then this is of the form of the harmonic oscillator.
V
has a
local minimum at
x
0
, and we say the equilibrium point is stable. The particle
oscillates with angular frequency
ω =
r
V
00
(x
0
)
m
.
If
V
00
(
x
0
)
<
0, then
V
has a local maximum at
x
0
. In this case, the equilibrium
point is unstable, and the solution to the equation is
x x
0
Ae
γt
+ Be
γt
for
γ =
r
V
00
(x
0
)
m
.
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
00
(x
0
) = 0, then further work is required to determine the outcome.
Example. Consider the simple pendulum.
m
`
d
θ
Suppose that the pendulum makes an angle
θ
with the vertical. Then the energy
is
E = T + V =
1
2
m`
2
˙
θ
2
mg` cos θ.
Therefore
V cos θ
. We have a stable equilibrium at
θ
= 0, and unstable
equilibrium at θ = π.
θ
V
π
π
mg`
mg`
If E > mg`, then
˙
θ never vanishes and the pendulum makes full circles.
If 0
< E < mg`
, then
˙
θ
vanishes at
θ
=
±θ
0
for some 0
< θ
0
< π
i.e.
E
=
mg` cos θ
0
. The pendulum oscillates back and forth. It takes a quarter
of a period to reach from
θ
= 0 to
θ
=
θ
0
. Using the previous general solution,
oscillation period P is given by
P
4
=
Z
θ
0
0
=
dθ
q
2E
m`
2
+
2g
`
cos θ
.
Since we know that E = mg` cos θ
0
, we know that
P
4
=
s
`
g
Z
θ
0
0
dδ
2 cos θ 2 cos θ
0
.
The integral is difficult to evaluate in general, but for small
θ
0
, we can use
cos θ 1
1
2
θ
2
. So
P 4
s
`
g
Z
θ
0
0
dθ
p
θ
2
0
θ
2
= 2π
s
`
g
and is independent of the amplitude
θ
0
. 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
m
moving in 3D. The equation of motion is now
a vector equation
m
¨
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 =
1
2
m|v|
2
=
1
2
m
˙
r ·
˙
r.
If we want to know how it varies with time, we obtain
dT
dt
= 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
1
= r(t
1
) to r
2
= r(t
2
) along a trajectory C is the line integral
W =
Z
C
F · dr =
Z
t
2
t
1
F ·
˙
r dt =
Z
t
2
t
1
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
=
1
2
m|v|
2
+ 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.
dE
dt
=
d
dt
1
2
m
˙
r ·
˙
r + V
= m
¨
r ·
˙
r +
V
x
i
dx
i
dt
= (m
¨
r + V ) ·
˙
r
= (m
¨
r F) ·
˙
r
= 0
So the energy is conserved. In this case, the work done is
W =
Z
C
F · dr =
Z
C
(V ) · dr = V (r
1
) V (r
2
).
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
V
(
r
) that
depends only on the distance from the origin,
r
=
|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 =
ˆ
r.
Intuitively, this is because the direction in which
r
increases most rapidly is
r, and the rate of increase is clearly 1. This can also be proved algebraically:
Proof. We know that
r
2
= x
2
1
+ x
2
2
+ x
2
3
.
Then
2r
r
x
i
= 2x
i
.
So
r
x
i
=
x
i
r
= (
ˆ
r)
i
.
Proposition. Let F = −∇V (r) be a central force. Then
F = −∇V =
dV
dr
ˆ
r,
where
ˆ
r
=
r/r
is the unit vector in the radial direction pointing away from the
origin.
Proof. Using the proof above,
(V )
i
=
V
x
i
=
dV
dr
r
x
i
=
dV
dr
(
ˆ
r)
i
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 ×
˙
r.
Proposition. Angular momentum is conserved by a central force.
Proof.
dL
dt
= m
˙
r ×
˙
r + mr ×
¨
r = 0 + r × F = 0.
where the last equality comes from the fact that
F
is parallel to
r
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
G
of a particle is the rate of change of angular
momentum.
G =
dL
dt
= r × F.
Note that
L
and
G
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
M
is fixed at a origin,
then a second particle of mass m experiences a potential energy
V (r) =
GMm
r
,
where G 6.67 × 10
11
m
3
kg
1
s
2
is the gravitational constant.
The gravitational force experienced is then
F = −∇V =
GMm
r
2
ˆ
r.
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
M
and the second particle
m
. However, it is useful what the fixed mass
M
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
Φ
g
(r) =
GM
r
.
Note that potential is confusingly different from potential energy.
If we have a second particle, the potential energy is given by V = mΦ
g
.
The gravitational field is the force per unit mass,
g = −∇Φ
g
=
GM
r
2
ˆ
r.
If we have many fixed masses
M
i
at points
r
i
, we can add up their gravitational
potential directly. Then the total gravitational potential is given by
Φ
g
(r) =
X
i
GM
i
|r r
i
|
.
Again, V = mΦ
g
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
M
is the same as that of a point particle with the same mass at
the center of the object, i.e.
Φ
g
(r) =
GM
r
.
The proof can be found in the Vector Calculus course.
Example.
If you live on a spherical planet of mass
M
and radius
R
, and can
move only a small distance z R above the surface, then
V (r) = V (R + z)
=
GMm
R + z
=
GMm
R
1
z
R
+ ···
const. +
GMm
R
2
z
= const. + mgz,
where
g
=
GM/R
2
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 =
1
2
mv
2
GMm
R
.
After escape, we must have
T
0 and
V
= 0. Since energy is conserved, we
must have E 0 from the very beginning. i.e.
v > v
esc
=
r
2GM
R
.
Inertial and gravitational mass
A careful reader would observe that “mass” appears in two unrelated equations:
F = m
i
¨
r
and
F =
GM
g
m
g
r
2
ˆ
r,
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
i
= m
g
, 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 (
M
), and the amount
of gravitational force received by a particle (
m
), 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
q
e
=
1.6 × 10
19
C
. Other particles’
charges are always integer multiples of q
e
(unless you are a quark).
In electromagnetism, there are two fields the electric field
E
(
r, t
) and the
magnetic field
B
(
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
B
and
E
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
Φ
e
(r), and E = −∇Φ
e
.
Definition
(Electrostatic potential)
.
The electrostatic potential is a function
Φ
e
(r) such that
E = −∇Φ
e
.
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 =
1
2
m|v|
2
+ qΦ
e
is conserved.
Proof.
dE
dt
= m
¨
r ·
˙
r + q(Φ
e
) ·
˙
r
= (m
¨
r qE) ·
˙
r
= (q
˙
r × B) ·
˙
r
= 0
Motion in a uniform magnetic field
Consider the particular case where there is no electric field, i.e.
E
=
0
, 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,
F
=
q
(
E
+
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
B
, 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
m
¨
ζ = iqB
˙
ζ.
Then the solution is
ζ = αe
t
+ β,
where
ω
=
qB/m
is the gyrofrequency, and
α
and
β
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
t
= 0,
r
= 0 and
˙
r
= (0
, v, w
),
i.e. ζ = 0 and
˙
ζ = iv, and z = 0 and ˙z = w.
The solution is then
ζ = R(1 e
t
).
with R = v = (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)
2
+ y
2
= R
2
,
combined with uniform motion parallel to B, i.e. a helix.
Alternatively, we can solve this with vector operations. Start with
m
¨
r = q
˙
r × B
Let B = Bn with |n| = 1. Then
¨
r = ω
˙
r × n,
with our gyrofrequency
ω
=
qB/m
. We integrate once, assuming
r
(0) =
0
and
˙
r(0) = v
0
.
˙
r = ωr × n + v
0
. ()
Now we project () parallel to and perpendicular to B.
First we dot () with n:
˙
r · n = v
0
· 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
˙
r
= wr
× n + v
0
(v
0
· n)n.
We solve this by differentiating again to obtain
¨
r
= ω
˙
r
× n = ω
2
r
+ ωv
0
× 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
Q
, fixed at the origin, produces an
electrostatic potential
Φ
e
=
Q
4πε
0
r
,
where ε
0
8.85 × 10
12
m
3
kg
1
s
2
C
2
.
The corresponding electric field is
E = −∇Φ
e
=
Q
4πε
0
ˆ
r
r
2
.
The resulting force on a particle of charge q is
F = qE =
Qq
4πε
0
ˆ
r
r
2
.
Definition
(Electric constant)
. ε
0
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
N
(perpendicular
to the contact surface) prevents them from interpenetrating, while a frictional
force
F
(tangential to the surface) resists relative tangential motion (sliding or
slipping).
N
F
mg
If the tangential force is small, it is insufficient to overcome friction and no
sliding occurs. We have static friction of
|F| µ
s
|N|,
where µ
s
is the coefficient of static friction.
When the external force on the object exceeds
µ
s
|N|
, sliding starts, and we
have a kinetic friction of
|F| = µ
k
|N|,
where µ
k
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
s
> µ
k
> 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
1
v.
where
v
is the velocity of the object relative to the fluid, and
k
1
>
0 is a
constant. This
k
1
depends on the shape of the object. For example, for a
sphere of radius R, Stoke’s Law gives
k
1
= 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
2
|v|
2
ˆ
v.
In either case, the object loses energy. The power exerted by the drag force is
F · v =
(
k
1
|v|
2
k
2
|v|
3
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
m
dv
dt
= mg kv.
We first solve for v, and then deduce x.
We use an integrating factor exp(
k
m
t) to obtain
d
dt
e
kt/m
v
= e
kt/m
g
e
kt/m
v =
m
k
e
kt/m
g + c
v =
m
k
g + ce
kt/m
Since v = u at t = 0, we get c = u
m
k
g. So
v =
˙
x =
m
k
g +
u
m
k
g
e
kt/m
.
Integrating once gives
x =
m
k
gt
m
k
u
m
k
g
e
kt/m
+ d.
Since x = 0 at t = 0. So
d =
m
k
u
m
k
g
.
So
x =
m
k
gt +
m
k
u
m
k
g
(1 e
kt/m
).
In component form, let x = (x, y), u = (u cos θ, u sin θ), g = (0, g). So
x =
mu
k
cos θ(1 e
kt/m
)
y =
mgt
k
+
m
k
u sin θ +
mg
k
(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
m
¨
x =
2
x k
˙
x,
where
ω
is the angular frequency of the oscillator in the absence of damping.
Rewrite as
¨
x + 2γ
˙
x + ω
2
x = 0,
where γ = k/2m > 0. Solutions are x = e
λt
, where
λ
2
+ 2γλ + ω
2
= 0,
or
λ = γ ±
p
γ
2
ω
2
.
If
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
Re
(
λ
) =
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,
m
¨
r = −∇V (r).
While the universe is in three dimensions, the orbit is confined to a plane. This is
since the angular momentum
L
=
mr×
˙
r
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
z
= 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
r
and increasing θ:
ˆ
r =
cos θ
sin θ
,
ˆ
θ =
sin θ
cos θ
.
x
y
r
ˆ
r
ˆ
θ
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.
d
ˆ
r
dθ
=
sin θ
cos θ
=
ˆ
θ
d
ˆ
θ
dθ
=
cos θ
sin θ
=
ˆ
r.
Often, we want the derivative with respect to time, instead of
θ
. By the
chain rule, we have
d
ˆ
r
dt
=
˙
θ
ˆ
θ,
d
ˆ
θ
dt
=
˙
θ
ˆ
r.
We can now express the position, velocity and acceleration in this new polar
basis. The position is given by
r = r
ˆ
r.
Taking the derivative gives the velocity as
˙
r = ˙r
ˆ
r + r
˙
θ
ˆ
θ.
The acceleration is then
¨
r = ¨r
ˆ
r + ˙r
˙
θ
ˆ
θ + ˙r
˙
θ
ˆ
θ + r
¨
θ
ˆ
θ r
˙
θ
2
ˆ
r
= (¨r r
˙
θ
2
)
ˆ
r + (r
¨
θ + 2 ˙r
˙
θ)
ˆ
θ.
Definition
(Radial and angular velocity)
. ˙r
is the radial velocity, and
˙
θ
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 =
ˆ
θ.
The speed is given by
v = |
˙
r| = r|ω| = const
and the acceleration is
¨
r =
2
ˆ
r.
Hence in order to make a particle of mass
m
move uniformly in a circle, we must
supply a centripetal force mv
2
/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 =
dV
dr
ˆ
r.
So Newton’s 2nd law in polar coordinates is
m(¨r r
˙
θ
2
)
ˆ
r + m(r
¨
θ + 2 ˙r
˙
θ)
ˆ
θ =
dV
dr
ˆ
r.
The θ component of this equation is
m(r
¨
θ 2 ˙r
˙
θ) = 0.
We can rewrite it as
1
r
d
dt
(mr
2
˙
θ) = 0.
Let
L
=
mr
2
˙
θ
. This is the
z
component (and the only component) of the
conserved angular momentum L:
L = mr ×
˙
r
= mr
ˆ
r × ( ˙r
ˆ
r + r
˙
θ
ˆ
θ)
= mr
2
˙
θ
ˆ
r ×
ˆ
θ
= mr
2
˙
θ
ˆ
z.
So the angular component tells us that
L
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 =
L
m
= r
2
˙
θ = const.
Now the radial (r) component of the equation of motion is
m(¨r r
˙
θ
2
) =
dV
dr
.
We eliminate
˙
θ using r
2
˙
θ = h to obtain
m¨r =
dV
dr
+
mh
2
r
3
=
dV
eff
dr
,
where
V
eff
(r) = V (r) +
mh
2
2r
2
.
We have now reduced the problem to 1D motion in an (effective) potential as
studied previously.
The total energy of the particle is
E =
1
2
m|
˙
r|
2
+ V (r)
=
1
2
m( ˙r
2
+ r
2
˙
θ
2
) + V (r)
(since
˙
r = ˙r
ˆ
r + r
˙
θ
ˆ
θ, and
ˆ
r and
ˆ
θ are orthogonal)
=
1
2
m ˙r
2
+
mh
2
2r
2
+ V (r)
=
1
2
m ˙r
2
+ V
eff
(r).
Example.
Consider an attractive force following the inverse-square law (e.g.
gravity). Here
V =
mk
r
,
for some constant k. So
V
eff
=
mk
r
+
mh
2
2r
2
.
We have two terms of opposite signs and different dependencies on
r
. For small
r
,
the second term dominates and
V
eff
is large. For large
r
, the first term dominates.
Then V
eff
asymptotically approaches 0 from below.
r
V
eff
E
min
r
The minimum of V
eff
is at
r
=
h
2
k
, E
min
=
mk
2
2h
2
.
We have a few possible types of motion:
If
E
=
E
min
, then
r
remains at
r
and
˙
θh/r
2
is constant. So we have a
uniform motion in a circle.
If
E
min
< E <
0, then
r
oscillates and
˙r
=
h/r
2
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
r
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.
If
E
0, then
r
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,
r
=
r
= const for some value of
h 6
= 0 (if
h
= 0, then the object
is just standing still!). Since ¨r = 0 for constant r, we require
V
0
eff
(r
) = 0.
The orbit is stable if r
is a minimum of V
eff
, i.e.
V
00
eff
(r
) > 0.
In terms of V (r), circular orbit requires
V
0
(r
) =
mh
2
r
3
and stability further requires
V
00
(r
) +
3mh
2
r
4
= V
00
(r
) +
3
r
V
0
(r
) > 0.
In terms of the radial force F (r) = V
0
(r), the orbit is stable if
F
0
(r
) +
3
r
F (r
) < 0.
Example. Consider a central force with
V (r) =
mk
r
p
for some k, p > 0. Then
V
00
(r) +
3
r
V
0
(r) =
p(p + 1) + 3p
mk
r
p+2
= p(2 p)
mk
r
p+2
.
So circular orbits are stable for
p <
2. This is illustrated by the graphs of
V
eff
(
r
)
for p = 1 and p = 3.
V
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 =
1
2
m ˙r
2
+ V
eff
(r)
t = ±
r
m
2
Z
dr
p
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
r
(
θ
) is also not easy. We will need a magic trick we
introduce the new variable
Notation.
u =
1
r
.
Then
˙r =
dr
dθ
˙
θ =
dr
dθ
h
r
2
= h
du
dθ
,
and
¨r =
d
dt
h
du
dθ
= h
d
2
u
dθ
2
˙
θ = h
d
2
u
dθ
2
h
r
2
= h
2
u
2
d
2
u
dθ
2
.
This doesn’t look very linear with
u
2
, but it will help linearizing the equation
when we put in other factors.
The radial equation of motion
m¨r
mh
2
r
3
= F (r)
then becomes
Proposition (Binet’s equation).
mh
2
u
2
d
2
u
dθ
2
+ u
= F
1
u
.
This still looks rather complicated, but observe that for an inverse square
force, F (1/u) is proportional to u
2
, 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
2
.