1

Preparation Materials
FOR
IPhO

Dinesh Kandel1

NePhO
Nepal Physical Society

1 If you find any error in this booklet, please contact me at dkandel@ualberta.ca

2
Contents

1 Preface 5

2 Problem Solving Techniques 7

2.1 Hints on Solving Physics Problems2 . . . . . . . . . . . . . . . . 7
2.2 Energy Conservation: A powerful Problem Solving Technique . . 8
2.2.1 Problem: Rolling Cylinder Inside A Fixed Tube . . . . . . 8
2.3 Motions with non-constant mass . . . . . . . . . . . . . . . . . . 10
2.4 Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 Miscelleneous Topics 15
3.1 Non-Intertial Frame of Reference . . . . . . . . . . . . . . . . . . 15
3.1.1 Elevating Pulley . . . . . . . . . . . . . . . . . . . . . . . 15
3.2 Fermat’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.3 Electrostatics: Method of Images . . . . . . . . . . . . . . . . . . 18
3.3.1 Charge outside a gounded Spherical Conductor . . . . . . 19

4 Special Theory of Relativity 21

4.1 Postulates of Special Relativity . . . . . . . . . . . . . . . . . . . 21
4.2 Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.3 Length Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . 23
4.3.1 Why is there no length in the direction perpendicular to
the motion?3 . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.4 Energy Momentum Relations . . . . . . . . . . . . . . . . . . . . 25
4.5 Lorenz Transformation . . . . . . . . . . . . . . . . . . . . . . . . 27
4.6 Spacetime Diagrams4 . . . . . . . . . . . . . . . . . . . . . . . . . 27

2 Source: Problems in General Physics, I.E. Irodov

3 Source: Griffiths, David Je↵rey, and Reed College. Introduction to electrodynamics. Vol.
3. Upper Saddle River, NJ: prentice Hall, 1999.
4 This section is taken from Schutz, Bernard. A first course in general relativity. Cambridge

university press, 2009.

3
4 CONTENTS
Chapter 1

Preface

I started this note when I was involved in training the 20 selected students
from the first round of NePhO, 2015. Being a past IPhO participant, I am
highly interested in thinking and designing of interesting problems of the IPhO
level. Problem solving requires a good deal of background knowledge, and
more importantly the ability to visualise the problem situation properly and
think critically. I have tried to present some key knowledge and skills in this
short note. This note is by no means comprehensive and complete, and I will
try to add on more to this note in the future. If you have any comments,
questions and suggestions, you could always feel free to reach me though email:
dnshkandel@gmail.com.

Dinesh Kandel
January 28, 2017

5
6 CHAPTER 1. PREFACE
Chapter 2

Problem Solving Techniques

2.1 Hints on Solving Physics Problems1

1. First of all, look through the reference datas given to you, for many prob-
lems cannot be solved without them. Besides, the reference data will make
your work easier and save your time.
2. Begin the problem by recognizing its meaning and its formulation. Make
sure that the data given are sufficient for solving the problem. Wherever
possible, draw a diagram elucidating the essence of the problem; in many
cases this simplifies both the search for a solution and the solution itself.
3. Solve each problem, as a rule, in the general form, that is in a letter
notation, so that the quantity sought will be expressed in the same terms
as the given data. A solution in the general form is particularly valuable
since it makes clear the relationship between the sought quantity and the
given data. What is more, an answer obtained in the general form allows
one to make a fairly accurate judgement on the correctness of the solution
itself (see the next item).
4. Having obtained the solution in the general form, check to see if it has
the right dimensions. The wrong dimensions are an obvious indication of
a wrong solution. If possible, investigate the behaviour of the solution in
some extreme special cases. For example, whatever the form of the expres-
sion for the gravitational force between two extended bodies, it must turn
into the well-known law of gravitational interaction of mass points as the
distance between the bodies increases. Otherwise, it can be immediately
inferred that the solution is wrong.
5. When starting calculations, remember that the numerical values of phys-
ical quantities are always known only approximately. Therefore, in cal-
culations you should employ the rules for operating with approximate
1 Source: Problems in General Physics, I.E. Irodov

7
8 CHAPTER 2. PROBLEM SOLVING TECHNIQUES

Figure 2.1: Displaced position of the rolling cylinder.

numbers. In particular, in presenting the quantitative data and answers

strict attention should be paid to the rules of approximation and numerical
accuracy.
6. Having obtained the numerical answer, evaluate its plausibility. In some
cases such an evaluation may disclose an error in the result obtained. For
example, a stone cannot be thrown by a man over the distance of the order
of 1 km, the velocity of a body cannot surpass that of light in a vacuum,
etc.

2.2 Energy Conservation: A powerful Problem

Solving Technique
2.2.1 Problem: Rolling Cylinder Inside A Fixed Tube
A cylinder of radius r and mass M is rolling perfectly inside a tube of radius
R. Find the frequency of small oscillation inside the tube.
Approach: There are two ways of attacking this problem. First is taking
into consideration forces and torques, while the second is applying energy con-
servation. We will see the that the second method is much easier to implement.

There are two constraints on the cylinder. (1) It should roll inside the tube.
(2) It should roll. Note that the angluar velocity of the cylinder is ! = ✓ẑ. ˙
There exits relationship between ✓˙ and ˙ because of the rolling condition:

R✓ = r(✓ + ), (2.1)

which upon di↵erentiation gives

˙ = (R r) ˙
✓. (2.2)
r
2.2. ENERGY CONSERVATION: A POWERFUL PROBLEM SOLVING TECHNIQUE9

Figure 2.2: Forces involved in the problem.

2.2.1.1 Force Method

Consider the angular momentum of rolling cylinder about its center of mass.
As the center of mass velocity vc is parallel to the center of mass momentum
P,
vc ⇥ P = 0, (2.3)
where bold face implies that the quantity is vector. As the torque through
center mass ⌧c is
dHc
⌧c = = rF ẑ, (2.4)
dt
where F is due to friction and Hc is the angular momentum of the cylinder
about its center of mass, given by
1 2R r ˙
Hc = Ic ! = mr ✓ẑ. (2.5)
2 r
Using this equation in equation (13) gives
1 2R r ¨
rF = mr ✓ (2.6)
2 r
To find F , we use conservation of linear momentum (ref 2.2).
d
(F mg sin ✓)êt = (mvc êt )
dt
d ˙
= (m✓(R r)êt ) (2.7)
dt
¨
= m✓(R r)êt + m✓(R ˙ r)ê˙ t
¨
= m✓(R r)êt + m✓˙2 (R r)ên ,

where in the last step we have used the fact that ê˙ t = ✓ê
˙ n . It can also be seen
from (16) that m✓˙2 (R r) = 0, as there is no normal component vector on the
left side of that equation. Finally, comparing (15) and (16) gives

2
(R r)✓¨ + g sin ✓ = 0. (2.8)
3
10 CHAPTER 2. PROBLEM SOLVING TECHNIQUES

The frequency of oscillation will be found later in next sub-section by using

small angle approximation.

2.2.1.2 Energy Consideration

This is the main highlight of the whole problem. The total energy of a system is
T + V , where T is kinetic and V is potential energy. Note that although friction
is present, as there is perfect rolling, the energy loss due to friction is essentially
negligible for the situation we are interested in. The perfect rolling condition
implies
vc = (R r)✓, ˙ (2.9)
where vc is the velocity of the center of mass. Kinetic energy is given by
1 1 2
T = mv 2 + I!
2 c 2
✓ ◆✓ ◆2
1 1 1 2 R r˙
= m(R r)2 ✓˙2 + mr ✓ (2.10)
2 2 2 r
3
= m(R r)2 ✓˙2 .
4
The potential energy of the system is

V = mg(R r) cos ✓. (2.11)

Using the fact that Ė = 0, gives

2
(R r)✓¨ + g sin ✓ = 0, (2.12)
3

which is same as the result of previous sub-section, but much more clear to
compute. Finally, using small angle approximation sin ✓ ⇡ ✓ gives us
r
2 g
!= . (2.13)
3R r

2.3 Motions with non-constant mass

Consider the following problem (which appeared in the Selection test of NePhO,
2011): Let us model the physics of falling raindrop as the following: as the rain
drop falls, its mass changes at a rate proportional to its surface area. Calculate
the acceleration of the raindrop. Clearly state the assumptions you make, and
write down how valid they are in a general atmospheric setting.
Let us think again about what information we are given in the problem.
First, the raindrop is falling under the action of gravity. Since the problem says
nothing about air-resistance, it is definitely legitimate to say that we ignore air-
resistance. One can also solve the problem by assuming air drag is proportional
to the speed of the raindrop, i.e. F~drag = b~v , where negative sign implies that
2.3. MOTIONS WITH NON-CONSTANT MASS 11

the force is resistive. But this will yield a di↵erential equation which might
not be easily solvable for high school students. Therefore, we want to keep it
simple: we ignore that the raindrop falls freely under the action of gravity. At
this point, students should ask themselves: since the raindrop is falling freely,
shouldn’t the acceleration of the raindrop just be g? The answer is (as you might
have already guessed): NO. This is because mass is changing, and therefore part
of the gravitational force is compensated by the changing mass.
Let us now do some serious calculations. We start with the assumption that
the raindrop is spherical, and has a constant density. Both the assumptions are
good as sphere minimizes surface area, and the drop is made up of homogeneous
water. We first write the force equation. Let m(t) be the mass of the drop at
any time t, and let the initial mass of the drop at t = 0 be m(0) = 0. Let
v(t) be the velocity of the drop at time t. Note that both m(t) and v(t) are
changing over time. The net force acting on the spherical raindrop at any time t
is F = mg (I will ignore the directions, as it is pretty clear that the only relevant
direction is vertically downwards). Using Newton’s second law I can write:
d
F = (mv) = mg , (2.14)
dt
which upon using product rule gives
dv dm
m +v = mg , (2.15)
dt dt
or equivalently,
dv v dm
+ =g . (2.16)
dt m dt
At this point, we use the fact that rate of change of mass is proportional to the
surface area:
dm
= c ⇥ 4⇡r2 , (2.17)
dt
where c is a proportionality constant. Using m = ⇢ ⇥ 4⇡r3 /3, where ⇢ density
of the drop and is constant, we get
dm dr
= ⇢ ⇥ 4⇡r2 . (2.18)
dt dt
Using Eqs. 2.17 and 2.18, we have
dr c
= = constant , (2.19)
dt ⇢
which gives
c
r= t (2.20)
⇢
and therefore,
1 dm 1 dr 3 dr 3
= ⇢ ⇥ 4⇡r2 = = . (2.21)
m dt ⇢ ⇥ 4⇡r3 /3 dt r dt t
12 CHAPTER 2. PROBLEM SOLVING TECHNIQUES

Using Eq.(2.16), we finally have

dv 3v
+ =g . (2.22)
dt t
First, it is always useful to check if the above equation is dimensionally correct
(and it actually is). The above di↵erential equation is not very easy to solve,
therefore, you should already get most of the points just for deriving it. But, in
order to fully solve it, youR need the concept of integration factor. In this case
3
the integration factor is e t dt = t3 . Multiplying both sides of Eq. (2.22) by
this integration factor yields:

d
(vt3 ) = gt3 , (2.23)
dt
which has the solution:
gt
v(t) = . (2.24)
4
Finally, the acceleration is just the first derivative of the velocity, and is therefore

g
a(t) = . (2.25)
4

As anticipated, the acceleration of the drop is less than g, as some part of

the gravitational force is taken in changing mass. Makes sense, right? An
important point to note is that the velocity changing linearly with time is not
very sensible in the real life, otherwise raindrops falling from high up in the sky
could potentially kill us. Usually, the gravitational force and drag force balance
each other once the rain drop reaches some speed called terminal speed.

2.4 Gravity
Some of the most difficult problems in Physics Olympiad involves problems
about gravity. Generally speaking, as a high school students you have already
encountered important pieces: the concept of gravitational potential, gravita-
tional energy, etc.
In this section, you will reinforce those concepts through a bit more chal-
lenging problems. Before that, you should keep in mind: Gravitational force
is central force, i.e., it acts only radially. Since torque is ~⌧ = ~r ⇥ F~ ,
gravitational force does not produce any torque. Therefore, angular
momentum of a gravitating body is conserved. This statement is very
powerful, and can help you solve a lot of problems involving gravity. Therefore,
always keep in mind: Angular momentum is conserved for an object which is
only influenced by gravity.
We start with a problem.
2.4. GRAVITY 13

Problem
A satellite of mass m is designed to travel from earth to Planet X in an elliptical
orbit. The perihelion of the satellite is at earth and aphelion at X. Assume that
the earth and X have circular orbits around the Sun with radii RE and RX ,
respectively, and neglect the gravitational e↵ects of the planets on the satellite,
and take that the only gravitational force acting on the satellite is due to the
Sun of mass M . (a) Find the velocity relative to earth, and the direction with
which the satellite have to leave the earth in order to reach X. (b) How long
will it take to reach X? (c) With what velocity relative to X, will it reach the
orbit of X?
Solution: It is important to note that the orbit of satellite is elliptical,
not circular. Therefore, you have to be careful with what radius r means each
time. Firstly, the energy of the satellite should be conserved, and so should the
angular momentum be. The kinetic energy of satellite is sum of circular kinetic
energy and radial kinetic energy, as both radius and angle changes during its
elliptical motion. Therefore,
1 1 1 1 1 1 L
K.E = mvr2 + I! 2 = mvr2 + mr2 ! 2 = mvr2 + ,
2 2 2 2 2 2 mr2
where vr is the radial velocity, L = I! = mr2 ! =constant is the angular
momentum of the satellite. The potential energy of the satellite is purely due
to the gravitational attraction of the Sun. Therefore, it is GM m/r. Thus, the
total energy is
1 1L GM m
E = mvr2 + . (2.26)
2 2 r2 r
At the perihelion and aphelion, vr = 0 and r = RE and r = RX respectively.
Thus, energy conservation gives
1 L2 GM m 1 L2 GM m
E= 2 = 2 , (2.27)
2 mRE RE 2 mRX RX
which gives s
2GM m2 RE RX
L= . (2.28)
RE + R X
At perihelion, the velocity (purely due to rotation motion) is
s
L 2GM RX
v = !r = = . (2.29)
mRE RE (RE + RX )
We would like to minimise the speed with which the satellite has to leave the
earth, and that would occur if the satellite is launched in a direction parallel
to the earth’s revolution around the sun, as the orbital motion it already has
would act as an additional kick. Therefore, the velocity with which the satellite
should leave the earth is
s r
2GM RX GM
vr = v vE = , (2.30)
RE (RE + RX ) RE
14 CHAPTER 2. PROBLEM SOLVING TECHNIQUES

and in the same way, the velocity it should have at the aphelion is
s r
0 0 2GM RE GM
vr = v vM = . (2.31)
RM (RE + RX ) RM

From Kepler’s third law, the period T of the revolution of the Satellite around
the sun is ✓ ◆3
2 2 3 RE + R M
T = T E RE , (2.32)
2
where TE is the period of revolution of earth which is equal to 1 year. Therefore,
the period for the Satellite to reach X is
✓ ◆3/2
T 1 R E + RM
t= = years . (2.33)
2 2 2
Chapter 3

Miscelleneous Topics

3.1 Non-Intertial Frame of Reference

Newton’s laws can only be applied to an inertial frame of reference. However,
it can also be used in an inertial frame of reference if we include pseudo forces.
This concept is illustrated in the following problem.

3.1.1 Elevating Pulley

A pulley of mass mp has two masses m1 and m2 suspended on it with an elastic
and massless rope. The pulley is itself suspended to the bottom of an elevator
of mass me that is accelerating with acceleration a. Ignoring the diameter of
the pulley, find out the accelerations of each block and tensions in each of the
strings.
Solution:

Obviously, we have the gravitational forces on each object. The pulley also
has 2T acting downward on it (due to the force exerted by the rope on the
pulley) and R acting upward (the force pulling upward on the pulley through
the rope connected to the elevator). Similarly, the elevator has tension forces
R acting downward and E upward. We include the forces on the pulley and
elevator since, a priori, its not obvious that they should be ignored. We will see
that it is not necessary to solve for the forces on the pulley and elevator to find
the accelerations of the masses, but we will be able to find these forces.

Coordinate system: Remember that Newtons second law only holds in iner-
tial reference frames. Therefore, we should reference the positions of the masses
to the fixed frame rather than to the elevator. Again, denote the z coordinates
of the two masses by z1 and z2 . Let the z coordinates of the pulley and elevator
be zp and ze .

15
16 CHAPTER 3. MISCELLENEOUS TOPICS

Equations of Motion:

m1 z̈1 = m1 g + T
m2 z̈2 = m2 g + T
(3.1)
mp z̈p = R 2T mp g
me z̈e = E R me g.

where T is the tension in the rope holding the two masses, R is the tension in
the rope holding the pulley, and E is the force being exerted on the elevator
to make it ascend or descend. Note especially the way we only consider the
forces acting directly on an object; trying to unnecessarily account for forces
is a common error. For example, even though gravity acts on m1 and m2 and
some of that force is transmitted to and acts on the pulley, we do not directly
include such forces; they are implicitly included by their e↵ect on T . Similarly
for the forces on the elevator.

Comstraints:the rope length cannot change, but the constraint is more com-
plicated because the pulley can move:

z1 + z2 = 2zp l. (3.2)

As the length of rope between the pulley and the elevator is fixed,

z̈p = z̈e = a, (3.3)

so
z̈1 + z̈2 = 2a. (3.4)
Using (3) and (4) in (1) will yield us the solution:
m1 m2 2m2
z̈1 = g+ a
m1 + m2 m1 + m2
m1 m2 2m1
z̈2 = g+ a (3.5)
m1 + m2 m1 + m2
2m1 m2
T = .
m1 + m2
Note that z̈1 6= z̈2 ! But wait! Solving 6 simultaneous equation seems cumber-
son. Is there any clever and short method? Yes there is. Let us first calculate
the accelerations of blocks in the reference frame of elevator. In this reference
frame, the e↵ective gravitational acceleration will be g + a, because this frame
is not an intertial frame and to get correct results we need to include pseudo
force ma. Let the accelerations in the frame of elevator of blocks be

z̃¨1 = z̈1 a, (3.6)

and
z̃¨2 = z̈2 a (3.7)
3.2. FERMAT’S PRINCIPLE 17

Now remember simple pulley problem’s results. The accelerations are given by:
m1 m2
z̃¨1 = (g + a)
m1 + m2
(3.8)
m1 m2
z̃¨2 = (g + a).
m1 + m2
Using (8) in (6) and (7) will give us desired result (5). Solving for R and E
using (1) gives
✓ ◆
4m1 m2
R = mp + (g + a)
m1 + m2
✓ ◆ (3.9)
4m1 m2
E = me + mp + (g + a).
m1 + m2

3.2 Fermat’s Principle

Fermats principle states that when a light ray travels between any two points, it
takes the path with shortest time interval. Since the shortest distance between
two points is a straight line, light rays travel in a straight line homogeneous
medium.
In the following few steps, we will show how Fermats principle can be used
to derive Snells law of refraction. Suppose that a light ray is to travel from
point P in medium 1 to point Q in medium 2 as shown in Fig. (3.1) , where P
and Q are at perpendicular distances a and b, respectively,from the interface.
The speed of light is v1 in medium 1 and v2 in medium 2, where
c
v1 = ,
n1
c (3.10)
v2 = .
n2
Assuming that the light ray left point P at time t = 0, we can see from the
figure that the time t it takes for light ray to arrive point Q is given by
p p
r1 r2 n1 a 2 + x 2 n2 b2 + (d x)2
t= + = + . (3.11)
v1 v2 c c
To find the trajectory such that the time of arrival is minimum, we need to
di↵erentiate the above equation (of course with respect to x, which is the only
variable upon which t is dependent), and set it to 0. It is shown is the following
steps
dt n1 x n2 (d x)
= p p = 0, (3.12)
dx c a2 + x2 c b2 + (d x)2
which implies
n1 x n2 (d x)
p = p . (3.13)
2
c a +x 2 c b2 + (d x)2
18 CHAPTER 3. MISCELLENEOUS TOPICS

Figure 3.1: Geometry for deriving Snells law of refraction using Fermats prin-
ciple.

Substituting,
x
sin ✓1 = p ,
a2 + x2
d x (3.14)
sin ✓1 = p
b + (d x)2
2

in equation (13) finally yields,

n1 sin ✓1 = n2 sin ✓2 . (3.15)

Practice Problem: Derive law of reflection using Fermat’s principle.

3.3 Electrostatics: Method of Images

In Electrostatics, some problems involving conductors and charges can be solved
with a great simplicity if the shapes of conductors are special, and has some
special symmetry. In this case, the induced charges on the conductor can be
replaced by some finite image charges. The main idea is that the potential due
to charge and its image at the conductor’s surface must give a constant. This
concept will be demonstrated in the following problem (This problem partly
appeared in IPhO 2010)
3.3. ELECTROSTATICS: METHOD OF IMAGES 19

3.3.1 Charge outside a gounded Spherical Conductor

Consider a grounded conducting sphere of radius R located at a distance d from
a point charge q (the distance is measured center of the sphere). To find the
potential of this configuration, we replace the induced surface charges on the
sphere by an image charge q 0 at a distance d0 < R inside the sphere and along
the line connecting the center of sphere and charge q. Note that there is no
obvious proof that this method works. One simply postulates an image charge
with certain value at a certain position and then test whether or not obtained
values of charge and distance can make up a constant potential at the surface
of the sphere.
Since the sphere is conducting and grounded V = 0 through out the surface
of the sphere. Using this we get

q q0
V (P ) = + = 0, (3.16)
4⇡✏0 (d R) 4⇡✏0 (R d0 )

and
q q0
V (Q) = + = 0. (3.17)
4⇡✏0 (d + R) 4⇡✏0 (R + d0 )
The above equations give
d+R R + d0
= , (3.18)
d R R d0
which upon simplification gives

R2
d0 = . (3.19)
d

Using above equation in (3.16) gives

R
q0 = q . (3.20)
d

This finally gives

2 3
1 4 1 R/d 5.
V = (3.21)
4⇡✏0 (r2 + d2 2rd cos ✓)1/2 r4 + R4 2
2 ad r cos ✓
1/2
d2

If you want to understand all sorts of interesting physics related to the image
problem, I definitely recommend you to go through the full problem from IPhO
2010. I find the last part of that problem - oscillation of charged pendulum-
fun.
20 CHAPTER 3. MISCELLENEOUS TOPICS
Chapter 4

Special Theory of Relativity

The relativity theory arose from necessity, from serious and deep contradictions
in the old theory from which there seemed no escape. The strength of the new
theory lies in the consistency and simplicity with which it solves all these diffi-
culties . . . .
The twentieth century experienced for the first time two revolutionary con-
cepts of physics. First is the Theory of Relativity (both special and general), and
the second is Quantum Physics. While Einstein solely formulated the theory
of Relativity, several people (including Einstein) have importantly contributed
in the field of Quantum Physics. Both of these concepts challenge our ordinary
way of thinking, and are oftentimes not so easy to visualise. In this section,
we will discuss the Special Theory of Relativity. This theory covers phenom-
ena such as the slowing down of moving clocks and the contraction of moving
lengths. We also discuss the relativistic forms of momentum and energy.

4.1 Postulates of Special Relativity

The two basic postulates are
1. Physics is the same in all inertial frames.
2. The speed of light is the same in all inertial frames.

The first postulate is just the principle of Galilean relativity, whose impli-
cation is that there is no absolute frame of reference; every inertial reference
frame is as good as every other one.
The second postulate is the heart of special theory of relativity, as all the
strange (or nonintuitive) physics depicted by special relativity is because of the
fact the particles cannot travel faster than the speed of light. To some extent, the
second postulate is a corollary of the first once one realizes that electromagnetic
waves do not travel in a medium. If the laws of electromagnetism, which give
rise to the speed of light, are to be the same in all frames, then the speed of
light must of necessity be the same in all frames.

21
22 CHAPTER 4. SPECIAL THEORY OF RELATIVITY

Figure 4.1: Geometry for deriving Time Dilation.

4.2 Time Dilation

The basic premise for time dilation is that two events that are simultaneous in
one reference frame are, in general, not simultaneous in a second frame moving
relative to the first. That is, simultaneity is not absolute but rather depends on
the state of motion of the observer.
Consider a pulse of light emitted from a source located at the center of a
train, which is moving at speed v in the x direction with respect to a station
(Ref. 4.1). The pulse of light is directed toward a photosensor on the floor of the
train, directly under the bulb. We are interested in the time interval between
the emission of photon and the detection of it at the floor.
The time interval measured depends on who measures it. An observer at
ground and an observer at train will have a di↵erent measurements. In the frame
of train, the light pulse travels distance h before being detected. Therefore, time
interval t0 is
h
t0 = . (4.1)
c
Let the time interval in ground frame be t. Then the ground observer claims
that photon not only travel h vertically, but also travel a horizontal distance of
v t before being detected.
p Therefore, in the ground frame, the photon travels
a distance of h0 = h2 + (v t)2 before being detected, which implies
p
h2 + (v t)2
t= . (4.2)
c
Using (16) and (17) gives

c2 ( t)2 = c2 ( t0 )2 + v 2 ( t)2 ,

which finally gives

t0
t= p ⌘ t0 , (4.3)
1 v 2 /c2

with
1
=p . (4.4)
1 v 2 /c2
4.3. LENGTH CONTRACTION 23

Since 1, it is easy to deduce from the above equation that moving clock
runs slow.
Problem: The average lifetime of a ⇡ meson in its own frame of reference
is 26.0 ns. (a) If the ⇡ meson moves with speed 0.95c with respect to the Earth,
what is its lifetime as measured by an observer at rest on Earth? (b) What is
the average distance it travels before decaying as measured by an observer at
rest on Earth?
Solution: The frame of muon is the proper rest frame, and therefore its
measured life-time is the proper life-time. The life-time measure by an observer
at rest on Earth is
t0
t= p = 83.3ns,
1 v 2 /c2
and therefore the distance it travels as measured by an observer at rest on Earth
is
L = v t = 24.0m

Problem: An atomic clock is placed in a jet airplane. The clock measures

a time interval of 3600 s when the jet moves with speed 400 m/s. How much
larger a time interval does an identical clock held by an observer at rest on the
ground measure?
Solution: The proper time is the time measured by the clock placed in the
jet. Quantity we are interested in is

t t0 = ( 1) t0
v2
⇡ (1 + 1) t0
2c2
2
v
= t0 .
2c2
Note that the approximation above was absolutely necessary as ⇡ 1, and your
calculator will probably show = 1, which will give you a result 0, which is
not what we want here. Substituting values in the above equation gives 3.2 ns
which is so much smaller than the time involved in the problem. This problem
illustrates that time dilation is negligible for low speeds.

4.3 Length Contraction

Imagine the same train as shown in figure (4.1), which flashes light from the
left end towards the right end. The right end constains a perfectly reflecting
mirror, and light reflects back to the left end after it striking the mirror at right
end. If the time interval of this event (emission of light and reception of it) is
t0 , we can infer that the length of train x0 , in the train’s frame is

c t0
x0 = . (4.5)
2
24 CHAPTER 4. SPECIAL THEORY OF RELATIVITY

In the ground frame, let the time interval from emmision of light to its arrival
at right end be t1 , and let the time interval from the reflection at right end to
its reception at right end be t1 . The the following holds:
c t1 = x 0 + v t1
(4.6)
c t2 = x0 v t1 ,
which gives
x0
t1 =
c+v
(4.7)
x0
t2 = ,
c v
which makes sense because in the first trip, relative velocity of train and light is
smaller while in the second trip, it is larger.Therfore, the total interval of this
event in the ground frame is
t= t1 + t2
x0 x0
= +
c+v c v
2 x0 (4.8)
=
c (1 v 2 /c2 )
2 2 x0
= .
c
As t= t0 , we finally get
x0
x= , (4.9)

where x is the length measured by the observer at ground level. Since 1,

moving objects get contracted.Note that the length parallel to the motion of an
object is contracted.
Problem: A rod of length L0 moves with speed v along the horizontal
direction. The rod makes an angle ✓0 with respect to the x axis. (a) Determine
the length of the rod as measured by a stationary observer. (b) Determine the
angle ✓ the rod makes with the x0 axis.
Solution The key to solve this problem is to note that only the length along
the direction of velocity is contracted. As Lx = L0 cos ✓ and Ly = L0 sin ✓, we
get
L0 cos ✓0
L0x0 =

L0y0 = L0 sin ✓0 .
which gives the length in the reference frame of stationary observer as
s✓ ◆2 s ✓ 2◆
0 L 0 cos ✓ 0 2
v
L = + (L0 sin ✓0 ) = L0 1 cos2 ✓0 .
c2
4.4. ENERGY MOMENTUM RELATIONS 25

The angle made by the rod with x0 is given by

tan ✓ = L0y0 /L0x0 = tan ✓0 .

4.3.1 Why is there no length in the direction perpendic-

ular to the motion?1
A very simple thought experiment suggested by Taylor and Wheeler2 can be
used to show that length contraction can’t happen in directions perpendicular to
the direction of motion. Imagine that we build a wall beside the railroad tracks,
and 1 m above the rails, as measured on the ground, we paint a horizontal
blue line. When the train goes by, a passenger leans out the window holding
a wet paintbrush 1 m above the rails, as measured on the train, leaving a
horizontal red line on the wall. Question: Does the passenger s red line lie
above or below our blue one? If the rule were that perpendicular directions
contract, then the person on the ground would predict that the red line is
lower, while the person on the train would say it s the blue one (to the latter,
of course, the ground is moving). The principle of relativity says that both
observers are equally justified, but they cannot both be right. No subtleties
of simultaneity or synchronization can rationalize this contradiction; either the
blue line is higher or the red one is–unless they exactly coincide which is the
inescapable conclusion. There cannot be a law of contraction (or expansion)
of perpendicular dimensions, for it would lead to irreconcilably inconsistent
predictions.

4.4 Energy Momentum Relations

The classical formulas for kinetic energies do not work well in special relativity.
Details of the energy momentum relations will not be included in this section.
Only relevant equations will be presented here. The momentum of a particle of
mass m moving at a speed v is

p = mv , (4.10)

and the energy relation is

E = mc2 . (4.11)
Using the above two equations, it can be shown that
p
E= p 2 c 2 + m2 c 4 . (4.12)

Note that p takes classical form when ⇠ 1, whereas it di↵ers significantly at

speeds v ⇠ c. Moreover, energy of a rest particle ( = 1) is mc2 . Therefore, we
1 Source: Griffiths, David Je↵rey, and Reed College. Introduction to electrodynamics. Vol.

3. Upper Saddle River, NJ: prentice Hall, 1999.

2 E. F. Taylor and J. A. Wheeler, Spacetime Physics (San Francisco: W. H. Freeman, 1966)
26 CHAPTER 4. SPECIAL THEORY OF RELATIVITY

define kinetic energy as

T =E mc2 = ( 1)mc2 . (4.13)

Using binomial expansion for v ⌧ c, i.e v/c ⌧ 1, we can write,

1 v2
=p ⇡1+ , (4.14)
1 v 2 /c2 2c2

which implies,
v2 1
T ⇡ (1 + 1)mc2 = mv 2 . (4.15)
2c2 2
The above equation shows that at low speeds, Newtonian results and results
from relativity matches. Since, we have to deal mostly with low speeds in our
everyday life, Newtonian mechanics seems to work so well.
Problem: A photon of wavelength strikes an electron, initially at rest.
After the collision the photon travels at angle relative to its original direction.
Find the new wavelegth 0 of the photon after the collision. This phenomenon
is called Compton E↵ect.
Solution We need to use relativistic equations,as the problem involves elec-
tron, which is very light and usually moves at large speed. The key to solve this
problem is to use conservation of energy and momentum.
If the momentum of photon before and after collision is Pph and P0ph and
the momentum of electron after collision is Pe , then using Pph = P0ph + Pe
gives
02 0
Pe2 = Pph
2
+ Pph 2Pph Pph cos ,

while conservation of energy gives

0
p
Pph c + mc2 = Pph c+ Pe2 c2 + m2e c4 .

Taking squuare root on one side and squaring the above equation, and finally
subtracting it from the momentum conservation equation gives
!
1 1
mc 0 . (4.16)
Pph Pph

0 0
Finally, using Pph = h/ and Pph = h/ gives

0 h
= + (1 cos ).
mc
This phenomenon is called Compton Scattering.
Self- Help Problem: Solve the same problem given above, this time using
Newtonian equations.
4.5. LORENZ TRANSFORMATION 27

Figure 4.2: A spacetime diagram in natural units (i.e, c=1).

4.5 Lorenz Transformation

Imagine a particle at position x at time t measured at a rest frame. If the axis is
moved towards right by a velocity v, then according to Gallelion transformation,
the position x0 and time t0 measured with respect to moving axis is given by

x0 = x vt, (4.17)

and
t0 = t. (4.18)
However, the second equation is already in trouble if we consider relativistic
e↵ects. The correct tranformation laws called Lorenz transformation are

x0 = (x vt) , (4.19)

and
t0 = (t vx/c2 ). (4.20)

4.6 Spacetime Diagrams3

A very important part of learning the geometrical approach to SR is mastering
the spacetime diagram. Figure (4.2) shows a two dimensional slice of spacetime,
the tx plane, in which are illustrated the basic concepts. A single point in this
space is a point of fixed x and fixed t, and is called an event. A line in the
space gives a relation x = x(t), and so can represent the position of a particle
at di↵erent times. This is called the particles world line. Its slope is related to
3 This section is taken from Schutz, Bernard. A first course in general relativity. Cambridge

university press, 2009.

28 CHAPTER 4. SPECIAL THEORY OF RELATIVITY

Figure 4.3: The time-axis of a frame whose velocity is v.

its velocity,
dt 1
slope = = . (4.21)
dx v
This concept is illustrated in figure (4.2). Notice that a light ray (photon) always
travels on a 45 line in this diagram.
Since any observer is simply a coordinate system for spacetime, and since all
observers look at the same events (the same spacetime), it should be possible
to draw the coordinate lines of one observer on the spacetime diagram drawn
by another observer. To do this we have to make use of the postulates of SR.
Suppose an observer O uses the coordinates t, x as above, and that another
observer Õ, with coordinates t̃, x̃, is moving with velocity v in the x direction
relative to O. Where do the coordinate axes for t̃ and x̃ go in the spacetime
diagram of O? t̃ axis: This is the locus of events at constant x̃ = 0 (and
ỹ = z̃ = 0, too, but we shall ignore them here), which is the locus of the origin
of Õs spatial coordinates. This is Õs world line, and it looks like that shown in
Fig. (4.3). x̃ axis: To locate this we make a construction designed to determine
the locus of events at t̃ = 0, i.e. those that Õ measures to be simultaneous with
the event t̃ = x̃ = 0. Consider the picture in Õ s spacetime diagram, shown in
Fig. (4.4).
The events on the x̃ axis all have the following property: A light ray emitted
at event E from x̃ = 0 at, say, time t̃ = a will reach the x̃ axis at t̃ = a (we
call this event P); if reflected, it will return to the point t̃ = 0 at t̃ = +a, called
event R. The x̃ axis can be defined, therefore, as the locus of events that reflect
light rays in such a manner that they return to the t̃ axis at +a if they left it
at a, for any a. Now look at this in the spacetime diagram of O, Fig. (4.5).
We know where the t̃ axis lies, since we constructed it in Fig. (4.3). The
events of emission and reception, t̃ = a and t̃ = a, are shown in Fig.(4.5).
4.6. SPACETIME DIAGRAMS4 29

Figure 4.4: Light reflected at a, as measured by Õ.

Figure 4.5: The reflection in Fig.(4.4), as measured O.

30 CHAPTER 4. SPECIAL THEORY OF RELATIVITY

Figure 4.6: Spacetime diagrams of O (left) and Õ (right).

Since a is arbitrary, it does not matter where along the negative t̃ axis we place
event E, so no assumption need yet be made about the calibration of the t̃ axis
relative to the t axis. All that matters for the moment is that the event R on
the t̃ axis must be as far from the origin as event E. Having drawn them in
Fig. (4.5), we next draw in the same light beam as before, emitted from E, and
travelling on a 45 line in this diagram. The reflected light beam must arrive at
R,so it is the 45 line with negative slope through R. The intersection of these
two light beams must be the event of reflection P . This establishes the location
of P in our diagram. The line joining it with the origin the dashed line must
be the x̃ axis: it does not coincide with the x axis. If you compare this diagram
with the previous one, you will see why: in both diagrams light moves on a 45
line, while the t and t̃ axes change slope from one diagram to the other. This is
the embodiment of the second fundamental postulate of SR: that the light beam
in question has speed c = 1 (and hence slope=1) with respect to every observer.
When we apply this to these geometrical constructions we immediately find that
the events simultaneous to Õ (the line t̃ = 0, his x axis) are not simultaneous
to O (are not parallel to the line t = 0, the x axis). This failure of simultaneity
is inescapable. The following diagrams (Fig. 4.6) represent the same physical
situation. The one on the left is the spacetime diagram O, in which Õ moves
to the right.
The one on the right is drawn from the point of view of Õ, in which O moves
to the left. The four angles are all equal to arctan |v|, where |v| is the relative
speed of O and Õ.
 Practice Problems
Dinesh Kandel
January 27, 2017

Problem 1: Levitation of Conducting Loop

A long horizontal wire carries a current I (towards right) which decreases with time. A conducting loop is
suspended over a small time interval t. In that interval the loop is in a balanced position. The coil is located
in a plane vertically at a distance D below the wire, as shown in Fig. The loop is a square of side a, mass m
and resistance R. The distance D is much greater than a. Neglect the self-inductance of the coil, and define
acceleration due to gravitiy as g.
1. Make a diagram of the system clearly indicating the currents, fields and magnetic forces involved.

[2 points]

2. Using appropriate approximations, find the current induced in the loop, i, as a function of I/ t.

[3 points]

3. Find the net magnetic force on the loop, including its magnitude and direction.

[3 points]

4. Calculate the condition that the product I It must satisfy according to electromagnetic quantities given
and constant, so that the loop is maintained levitating (or floating). Be very careful with the signs of
the amounts involved.

[2 points]

Figure 1: Set up of the Problem.

1
Useful relation:
1
⇡1 nx,
(1 + x)n
when |x| ⌧ 1.

Problem 3: Rolling Cylinder

A uniform solid cylinder of mass M and radius a is set in rotation about its axis with an angular velocity
!0 . The cylinder is then lowered with its lateral surface onto a horizontal plane and released. The coefficient
of friction between the cylinder and the plane is equal to µ.
1. Draw a free body diagram showing all forces involved in this problem after the cylider is released onto
a horizontal plane. Is there any point in the cylinder about which the torque is zero? Is there any point
in the cylinder about which the angular momentum is conserved?

[2 points]

2. Write down the conditions for sliding and pure rolling. Your conditions should involve SOME of the
following parameters: center of mass velocity of the cylinder v, angluar velocity !, a, M and µ.

[2 points]

3. Find the time t0 upto which the cylinder will move with sliding.

[3 points]

4. Find the total work performed by the sliding friction force acting on the cylinder.

[3 points]

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