1
PHYC20015: Special Relativity
School of Physics, University of Melbourne, Parkville, Victoria, Australia
Notes heavily based on the book Special Relativity , by A. P. French.
Originally authored by Prof. Stuart Wyithe and adapted by Dr. Katie Auchettl.
Contents
1 Preliminaries
1.1 Short history
[French: P. 269] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Features of Special Relativity . . . . . . . . . . . . . . . . . . . . . . . .
1
2 Dynamics
2.1 Revision of Newtonian Mechanics: An ultimate speed
[French PP. 5-11] . . . . . . . . . . . . . . . . . . . . .
2.2 Photons, their energy and momentum
[French: PP. 11-16] . . . . . . . . . . . . . . . . . . .
2.3 Inertia, energy and Einstein’s box
[French: PP. 16-20] . . . . . . . . . . . . . . . . . . . .
2.4 Relations between Energy, Momentum and Mass
[French: PP. 20-24] . . . . . . . . . . . . . . . . . . .
2.5 A revised dynamics
[French: PP. 26] . . . . . . . . . . . . . . . . . . . . . .
3
3 Propagation of light and the search
3.1 Waves, particles and the Ether
[French PP. 38-39] . . . . . . . . .
3.2 Stellar aberration
[French PP. 41-44] . . . . . . . . .
3.3 Michelson-Morley Experiment
[French PP. 49-51] . . . . . . . . .
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for the Ether
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4 Lorentz Transformations
4.1 Galilean and Newtonian relativity
[French PP. 66-67] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Transforming Newton’s Laws
[French PP. 68-69] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 The constant speed of light
[French PP. 70-75] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
27
28
29
CONTENTS
4.4
4.5
2
Lorentz transformations
[French 76-81] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Minkowski diagrams and space-time
[French PP. 82-85] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 Lengths and Time
5.1 Events in different frames
[French PP. 92-95] . . . . . . .
5.2 Lorentz contraction
[French PP. 96-97] . . . . . . .
5.3 Time dilation
[French PP. 97-104] . . . . . . .
5.4 Proper time and proper length
[French PP. 106-109] . . . . . .
5.5 Causality
[PP. 117-119] . . . . . . . . . .
33
38
40
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40
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48
6 Relativistic Kinematics
6.1 Observing velocities in different frames
[French PP. 125-126] . . . . . . . . . . . . . . . .
6.2 Observing transverse motions in different frames
[French PP. 132-133] . . . . . . . . . . . . . . . .
6.3 The Doppler effect
[French PP. 134-144] . . . . . . . . . . . . . . . .
6.4 Moving clocks and objects
[French PP. 149-152] . . . . . . . . . . . . . . . .
6.5 Accelerated motions
[French PP. 152-154] . . . . . . . . . . . . . . . .
6.6 The twin paradox
[French PP. 154-158] . . . . . . . . . . . . . . . .
7 Collisions and Conservation Laws
7.1 Conservation Laws
[French PP. 167-168] . . . . . . . . . . . . . . . .
7.2 Elastic collisions viewed from different frames
[French PP. 169-171] . . . . . . . . . . . . . . . .
7.3 In-Elastic collisions viewed from different frames
[French PP. 172-175] . . . . . . . . . . . . . . . .
7.4 The Photon Rocket
[French PP. 183-184] . . . . . . . . . . . . . . . .
7.5 Emission and absorption of light by atoms
[French PP. 176-178] . . . . . . . . . . . . . . . .
7.6 Mossbauer effect and the Doppler effect
[French P. 181] . . . . . . . . . . . . . . . . . . .
52
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80
CONTENTS
7.7
7.8
7.9
3
Scattering
[French P. 191] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Compton scattering
[French PP. 194-195] . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Doppler effect (again)
[French PP. 197-199] . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 Forces in Special Relativity
8.1 The energy-momentum invariant
[French PP. 205-206] . . . . . . . . .
8.2 Transforming energy and momentum
[French PP. 208-212] . . . . . . . . .
8.3 Force and acceleration
[PP214-217] . . . . . . . . . . . . . .
8.4 Four vectors
[French P. 213] . . . . . . . . . . . .
9 List of formulae
81
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88
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88
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Chapter 1
Preliminaries
This subject will introduce Einstein’s Special Theory of Relativity, and develop the fundamental principles of electromagnetism and Maxwell’s equations in differential form.
The first part of the course, covered by these notes, will deal with Special Relativity.
The goal is to provide sufficient background for a relativistic understanding of the electrodynamics in the second part of the course. We will cover a number of topics in Special
relativity, including foundations of special relativity, spacetime invariance, simultaneity,
and Minkowski diagrams, relativistic kinematics, the Doppler effect, relativistic dynamics, and nuclear reactions.
1.1
Short history
[French: P. 269]
We will approach the course by describing the key observations and events that led to
the development of Einstein’s theory of Special Relativity and Maxwell’s equations of
electromagnetism. We will describe how the principles underpinning relativity reconciled
difficulties posed for the classical theories by these observations.
The origins of the Special Theory of Relativity is credited to Einstein’s famous paper
of 1905. The paper is a purely theoretical work, and indeed is devoid of references to
either experiment or subsequent theoretical studies. However others had previously
come close to understanding parts of what we know as Special Relativity. For example,
Poincare had recognised the fact that an absolute motion could not be measured using
light. In 1904 he wrote about topics including the increase of mass with velocity, and
that objects could not exceed the speed of light, as well as ideas of comparing the time
on clocks through exchange of light signals; all essential aspects of Special Relativity.
However he did not abandon the idea of an ether providing a fixed frame of reference.
Also in 1904, Lorentz had derived the Lorenz transformations (previously also discovered
by Voight in 1887 along with the Doppler effect). However Lorenz too considered the
problem in terms of an ether describing an absolute space. In another instance Lamour,
in 1900 recognised that the time measured by a moving body must be include a correction
if the equations describing the ether were to remain invariant.
1.2 Features of Special Relativity
2
We will work towards demonstrating that Einstein’s critical insight was with respect
to the problem of measuring time. From 100 years on we can look towards the great
experiments of the 19th century and chart a pedagocical progression through the observations towards a theory of Special Relativity (and this is what we will do). However the
lack of references in Einstein’s papers, and his comments in the decades after the discovery apparently leave some uncertainty regarding how aware of all these developments
he was.
1.2
Features of Special Relativity
Although Newtonian mechanics provides an excellent description of Nature and our
every day world, it is not universally valid. In particular, when we approach extreme
conditions such as the very small, the very heavy or the very fast, Newtonian Mechanics
breaks down. One such extreme condition is when particles travel very fast, and special
relativity is what replaces Newtonian mechanics when one considers speeds of particles
that are comparable or similar to the speed of light in the vacuum.
Special Relativity rests on 2 postulates.
• Postulate 1: All inertial frames are equivalent with respect to laws of physics
• Postulate 2: The speed of light in vacuum always has the same value, c
These are postulates in the theory, but are backed up by experimental evidence. However
from these postulates follow predictions for a very rich description of physical phenomena
which are always correct. These predictions are often counter-intuitive compared to our
view of the world. If boiled down to the basic outcomes, for two observations in frames
S and S 0 moving at relative speed v, we are able to use Special Relativity to calculate
• the relation between the measured positions
• the relation between the measured times
• the relation between the measured velocities
• the relation between the measured masses
• the relation between the measured accelerations
• the relation between the measured forces
• that nothing can go faster than light
• the famous equivalence between mass and energy.
However, before we get into this, let’s begin with a brief revision of Newtonian mechanics,
where it breaks down, and the first steps towards a Special Theory of Relativity.
Chapter 2
Dynamics
This chapter begins with a brief revision of the Newtonian mechanics, and then looks
at some of the simple evidence for how this needed revision.
Newtonian Mechanics describes the motions of particles under application of forces
• These equations are normally described in terms of positions and times, but since
we define an origin for our co-ordinate system, these equations actually describe
the change in relative position and time between systems.
• This relativity implies that Newtonian mechanics need not be applied in an absolute frame of reference, or that one must exist. However the acceleration (a)
provides a fundamental property which relates position and velocity at different
times. For a constant acceleration this is:
v2 (t2 ) = v1 (t1 ) + a × (t2 − t1 )
(2.1)
• The acceleration is seen to be related to the applied force through the mass of
an object. This inertial mass (m) is a fundamental property of the object, and
so Newtonian mechanics tells us the acceleration resulting from an applied force
through F = ma.
• Application of the force requires that a Force Law be known, as it is in the case
of gravitation or classical electrostatics. In addition, Newtonian Mechanics has
a key postulate that the total momentum p~ = m~v is constant. This has been
experimentally verified.
• By applying a force F to move an object over a distance d, we can add energy to
the object by doing work W . If the velocity increases from v1 to v2 , we have
1
F × d = W = m(v22 − v12 )
2
(2.2)
This energy could be lost to heat or to a change in potential energy, but it must
be conserved.
2.1 Revision of Newtonian Mechanics: An ultimate speed
[French PP. 5-11]
2.1
4
Revision of Newtonian Mechanics: An ultimate speed
[French PP. 5-11]
If we apply the laws from Newtonian dynamics we see that we can have bodies with
unlimited motion. In particular, velocities of systems described by Newtonian motion
can even exceed the speed of light! We can see this in three simple examples:
• If we have a body acted on by gravity, which we can take equal to the acceleration
on the Earth’s surface of 9.8m/s2 , then after 1 year, equation (2.1) tells us that
the speed would be v = 9.8m/s2 × 3.15 × 107 s∼ 3 × 108 m/s, or approximately the
speed of light. After 2 years, the speed would be twice c.
• Alternatively, consider the end of a stick with length d rotating at ω rad/s. The
end of this stick has a speed of v = ω × d. The end of a stick with length d > c/ω
moves faster than light.
• Thirdly, one can imagine a body in a vacuum to which a constant force is applied.
By the conservation of energy we have
F × d = W = 1/2m(v22 − v12 )
(2.3)
so that a body is accelerated to a speed faster than light if
F × d > mc2 .
(2.4)
These accelerations must be very large or applied for long times to exceed the speed
of light. You could argue that these cases are impractical for this reason. We should
therefore consider a case closer to our experience.
Electrons have small mass (compared to say a proton) but large charge, and so can
be accelerated to large speeds. An electron moving across one Volt of electric potential
receives 1eV of energy. We can calculate the resulting velocity of the electron in this
situation if we equate this energy to the kinetic energy K. Lets assume an every day
value of 100V, crossing a spacing of a few mm.
Ex 2.1.A (French P. 7) We find the following velocity and acceleration:
2.1 Revision of Newtonian Mechanics: An ultimate speed
[French PP. 5-11]
5
The resulting values of ∼ 6000km/s and acceleration 1015 g are far beyond our everyday experience. Its easy to see from the above that millions of Volts would accelerate
our electron beyond c.
Ex. 2.1.B (French P. 7) We can find the voltage required:
This (application of the Voltage) of course is possible. Figure 2.1 shows the setup
of an experiment due to Bertozzi in 1964 which demonstrates what actually happens
when large accelerations are applied. Short pulses of electrons are accelerated through a
linear accelerator. The time difference between the electrons at the start and end of the
2.1 Revision of Newtonian Mechanics: An ultimate speed
[French PP. 5-11]
Figure 2.1: From Figure 1.1 of French
Figure 2.2: From Figure 1.3 and Table 1.1 of French
6
2.1 Revision of Newtonian Mechanics: An ultimate speed
[French PP. 5-11]
7
acceleration are recorded. The length of the flight path is l = 8.4m and so the electron
speed is v = l/t. For example, with an acceleration of 0.5M eV the measured time was
t = 3.3 × 10−8 s, and we have
v=
8.4m
l
∼
∼ 2.5 × 108 m/s.
t
3.3 × 10−8 s
(2.5)
The results for a range of input values are shown in Figure 2.2.
The figure shows a very important result. The work done by the accelerator provides
kinetic energy K, and for Newtonian mechanics we would expect
1
1
W = K = m(v22 − v12 ) ∼ mv22
2
2
(2.6)
v2 ∝ K 1/2 .
(2.7)
so that
However the table and plot in Figure 2.2 shows that v2 asymptotes for large K, with a
dependence that is not as strong as expected.
We could still appeal to the conservation of energy, if the electron were losing energy,
or the expected energy were not provided to the electron because of deficiencies in the
experiment. However the electron energy can be measured at the end of the flight, and
is found to have the expected value.
This experiment implies that there is a maximum speed which cannot be exceeded by
an electron. We are used to hearing this but the result is startling. Consider an electron
accelerated through the first 0.5MeV, so that it has a speed of about 2.6 × 108 m/s. Now
do a thought experiment in which an apparatus is accelerated with this electron, which
can impart a further 0.5MeV of acceleration. We should have a velocity of 2.6 × 108 +
2.6 × 108 m/s= 5.2 × 108 m/s. This seems no different from applying the second 0.5MeV
in the accelerator in the lab (and it is no different in Newtonian mechanics), yet the
velocity never exceeds 3 × 108 m/s
Ex. 2.1.C Diagram to support above
2.2 Photons, their energy and momentum
[French: PP. 11-16]
2.2
8
Photons, their energy and momentum
[French: PP. 11-16]
Previously we showed that electrons have a maximum speed. This maximum speed
corresponds to the measured speed of light. The constant speed of light was known,
initially from the experiment due to Roemer. This nicely illustrates how the constant
speed of light affects our view of the world, as measured by events
Ex. 2.2.A (example based on Q 2-5 in French)
2.2 Photons, their energy and momentum
[French: PP. 11-16]
9
The speed of light is measured to be constant at all energies, with the value c =
3 × 108 m/s. The momentum and energy of photons is indirectly seen in the interactions,
or collisions between particles. We will study this in detail later in the course. However
experiments directly measure the momentum of populations of photons from continuous
beams of light using the phenomenon of radiation pressure1 .
If a photon has energy E, and speed c, then it has momentum
p = E/c
(2.8)
Suppose we have a plate onto which a beam of light with incident power dE/dt is shone.
If we think of photons as particles undergoing either elastic or inellastic collisions, then
our understanding from Newtonian mechanics is that an absorbed photon will impart
its initial momentum to the plate, and that a reflected photon will impart twice the
initial momentum.
Ex. 2.2.B Drawing of reflected beams
1
Radiation pressure is defined as the mechanical pressure exerted upon a surface due to the exchange
of momentum between the object and the electromagnetic field. This includes the momentum of light or
electromagnetic radiation of any wavelength that is absorbed, reflected, or otherwise emitted by matter.
2.3 Inertia, energy and Einstein’s box
[French: PP. 16-20]
10
If a fraction f of the radiation is reflected, then since the force F is given by the
change in momentum, and c is constant, there is a relation
F =
d
1 dE
p=
× (1 + f ).
dt
c dt
(2.9)
Since F , f and dE/dt can be measured, the value of c can be determined in this way. It is
found to be in excellent agreement with geometric measurements. Thus, the momentum
of radiation, and its relation to force behave in analogy with our Newtonian expectations.
What about energy? Since in our Newtonian view kinetic energy is K = 0.5mv 2 =
0.5vp, for photons we could therefore propose the expression
1
K = cp.
2
(2.10)
However this turns out not to be correct, because just as the speed of the electron
became significantly slower than expected for large accelerations (see discussion below
equation 2.7), so too does the momentum.
2.3
Inertia, energy and Einstein’s box
[French: PP. 16-20]
With this experimentally verified measure of the momentum of radiation, combined with
the experimentally measured constant speed of light, we can get to the most famous
equation of relativity via a simple thought experiment. Suppose a burst of photons with
energy E is emitted from one end of a stationary box with mass M and length L. The
box moves in response to the momentum carried by the photons for a short distance
before it is brought to rest again when the photons exert the opposing impulse when
absorbed on the other side. Since the center of mass cannot have moved in this isolated
system, there must be mass associated with the energy in the photons which maintains
the position of the center-of-mass.
Ex. 2.3.A (French PP. 16-17)
2.3 Inertia, energy and Einstein’s box
[French: PP. 16-20]
11
Furthermore, the absorption of the photons may result in an increase in thermal or other
form of energy. Thus we are led to the idea that there is an equivalence between mass
and any form of energy
The most notable example of this mass-energy equivalence is provided by thermonuclear reactions in the Sun, producing Helium from 4 Hydrogen atoms. One of the steps
involved combines 1 proton with 1 deuterion (a nucleus with 1 proton and 1 neutron)
to produce 3 He and a gamma ray photon. The 3 He nucleus is lighter than the combined mass of the proton and deuterion, with a mass corresponding to the energy of the
gamma ray divided by c2 .
Ex. 2.3.B (French PP. 18-19)
2.4 Relations between Energy, Momentum and Mass
[French: PP. 20-24]
12
As a consequence of Einstein’s insight into the equivalence between mass and any
form of energy, this suggests that the sun is losing mass. Since we know the luminosity
of the sun (Lsun = 3.78 × 1026 J s−1 ), we can use E=mc2 to estimate how much mass
the sun loses every second. This corresponds to mass loss rate of 4.2 × 109 kg s−1 .
2.4
Relations between Energy, Momentum and Mass
[French: PP. 20-24]
We are now in a position to utilise these two fundamental results for regarding mass
and energy for photons
E
E = cp
and
m= 2
(2.11)
c
to develop a general relation between energy and momentum.
Combining the above we have m = p/c. This should be compared with the Newtonian expression m = p/v, so that perhaps the expression m = p/c obtained from
equation (2.11) is just Newtonian momentum with v = c. However the fusion experiment implies that an equivalence between mass and energy, so that m = cE2 should apply
to a universal equivalence of mass and energy which does not apply just to photons. So
we instead combine m = cE2 with the Newtonian expression m = p/v, obtaining
E=
c2 p
.
v
(2.12)
2.4 Relations between Energy, Momentum and Mass
[French: PP. 20-24]
13
The above expression has an absolute energy associated with mass. This is in contrast to
Newtonian mechanics, where there is no absolute energy (kinetic or potential). Rather,
forces exerted over a change in distance produce a change in kinetic energy as follows
dE = F dx =
dp
dx = v dp.
dt
(2.13)
We can now combine these last two equations to obtain a relation, which can be posed
to be a general relation, between energy and momentum (noting that equation 2.10,
without accounting for mass energy equivalence led us to the wrong expression).
Ex. 2.4.A (French P. 21) Begin by multiplying the two sides of equations 2.12 and
2.13
E dE = .....
(2.14)
2.4 Relations between Energy, Momentum and Mass
[French: PP. 20-24]
14
Thus we have the relation
E(v) =
E0
,
(1 − v 2 /c2 )1/2
(2.15)
where E0 is a constant. To look at the meaning of E0 , we look at the energy at low
velocities and identify the result with our Newtonian experience.
Ex. 2.4.B (French P. 22)
2.5 A revised dynamics
[French: PP. 26]
15
Thus we are led to the idea of a velocity dependent mass
m(v) =
m0
(1 − v 2 /c2 )1/2
(2.16)
The Newtonian inertial mass is now seen to take the role of rest mass for a body with
v = 0. Correspondingly, E0 is the rest mass energy, and both energy and mass increase
with velocity.
Now that we are associating energy with the rest mass of a particle, the kinetic
energy becomes the difference between the total energy and the rest mass energy
1
2
K(v) = m0 c
−1 .
(2.17)
(1 − v 2 /c2 )1/2
Importantly, this is not the same as the revised mass times velocity squared, i.e. K 6=
1
2
2 m(v) v .
To summarise, the velocity dependent mass m(v) describes the inertial property of
a body such that the momentum is p = m(v) v and the total energy E = m(v) c2 .
Because the quantity in the denominator above appears so often we use the short hand
γ = (1 − v 2 /c2 )−1/2 . We can summarise what we have found:
γ(v) ≡
1
(1 −
v 2 /c2 )1/2
(2.18)
m = γ(v) m0
(2.19)
p = γ(v) m0 v
(2.20)
E = γ(v) m0 c2
(2.21)
So far we have not postulated a Special Theory of Relativity!
2.5
A revised dynamics
[French: PP. 26]
With our new idea of mass and energy, we can revisit the simplest dynamical problem
of motion under a constant force. Let a force F act on a body for time t. The body is
initially at rest, and gains velocity v
2.5 A revised dynamics
[French: PP. 26]
16
Ex. 2.5.A (French P. 26)
F t = m v = .........
(2.22)
Thus, if the energy provided is small compared to the rest mass energy, then the
speed is small and our problem reduces to simple Newtonian mechanics. However if the
energy provided is large compared to the rest mass energy, then the body approaches
the limiting velocity c.
Chapter 3
Propagation of light and the
search for the Ether
The first chapter discussed the measurements of the finite speed of light, and the observation that this must correspond to a maximum observed speed for all observers.
However our experience tells us that if we are speaking about a speed, then that speed
must be relative to a frame of reference, and that this speed must therefore depend on
the frame in question. So two questions arise
• how can we incorporate a unique velocity into dynamics?
• how can we speak about the velocity of light without specifying a reference frame?
In this chapter we will look at the experiments performed to understand the wave nature
of light, and the puzzles these presented.
3.1
Waves, particles and the Ether
[French PP. 38-39]
Given the role of light in Special Relativity, it is important to step back and talk briefly
about what we mean by light. From ancient times, the fact that light travelled in a
straight line, and apparently through a vacuum (unlike sound) implied it to be a kind
of particle. However by the 1600s, phenomena including diffraction at shadow edges
and interference in thin films implied a wave nature for light. Such waves should need
a medium in which to travel. Furthermore, in the mid 1800s, James Maxwell (who you
will hear more about in the electromagnetism section of this course) had shown that
light could be described as oscillations of the electric and magnetic fields. Here, Maxwell
related the speed of light to two constants the permittivity (0 )1 and permeability (µ0 )2
1
The permittivity is a measure of the electric polarisability of a dielectric material. Here a dielectric
material is a material that becomes polarised by an electric field. For a vaccum, 0 = 8.8541878128(13)×
10?12 F m−1 .
2
The permeability is a measure of the magnetisation of a material when a magnetic field is applied.
For a vaccum, µ0 = 1.25663706 × 10−6 N A−2 .
3.2 Stellar aberration
[French PP. 41-44]
18
p
of free space: c = (1/0 µ0 ).
But, as we have seen in the previous chapter, Newtonian mechanics tells us that
speeds are relative. As such, if Maxwells equation predicts a value for the speed of
light, it was thought at the time that this equation (and Maxwell’s other equations)
must have been valid in a preferred reference frame. This does not seem unreasonable
if you consider that if light is a wave then there must be something ”waving”. Similar
to surf/waves we see in the ocean need water, and the musics from your favourite
band needs air to travel through, it was thought that light waves need a material to
propagate in. This material that it was thought light travelled through was called the
the luminiferous ether and it was thought that Maxwells equations must only be valid
in the frame at rest with respect to this ether.
The great success of the wave description of light in explaining phenomena of interference, diffraction and polarisation drove the search for this medium in which the
waves must vibrate and this ether was thought to have the following properties.
• the medium must be stiff enough to support a wave many orders of magnitude
faster than any known.
• the medium must none-the-less be tenuous enough to allow the unimpeded travel
of the planets.
• although the ether had not been observed, Maxwell’s electromagnetic theory allowed prediction of the speed of light in a medium of known electromagnetic properties.
• given a medium, the speed of light for a wave must be defined relative to that
medium.
• like sound, the velocity of the light in the ether is independent of the speed of the
emitting source.
This line of argument suggested methods to search for the ether and lead to the famous
Michelson-Morley Experiment.
3.2
Stellar aberration
[French PP. 41-44]
One such experiment that can test the properties of this ether is stellar aberration.
Stellar aberration is the change in the apparent position of a star due to the (relative)
velocity of the Earth. As a result, when observing the position a star, the telescope
must be tilted slightly into the direction of motion of Earth relative to the star.
However, before deriving the effect of stellar aberration this lets look at a simple
situation of the observed trajectory of rain which is falling directly down with speed w
relative to a stationary person. If the person walks at speed v they observe the rain
falling along straight lines which are inclined to the vertical.
3.2 Stellar aberration
[French PP. 41-44]
19
Ex. 3.2. A (French P. 42)
Now imagine a star (sufficiently distant that parallax is negligible) viewed from 4
different positions around the Earth’s orbit (Figure 3.1). The true altitude relative to
the Earth’s orbital plane is θ0 . If the speed of light were infinite, then the altitude would
be largest at point 2 and smallest at point 1. However the motion of the Earth and the
finite speed of light introduce an aberration.
3.2 Stellar aberration
[French PP. 41-44]
20
Figure 3.1: From Figure 2.1 of French
We can derive the apparent position of the star at different times during the orbit
Ex. 3.2.B (French P. 42)
3.3 Michelson-Morley Experiment
[French PP. 49-51]
21
The aberration angle is proportional to v/c. Since the orbital speed of the Earth
was well known, these aberration measurements, made by the astronomer Bradley in
the 1720s, were used to improve measurements of the speed of light. None of this needs
relativity, and the explanation of the observed aberration is just as in the falling rain
example. No need for waves.
However we could also explain this aberration result in terms of wave phenomena,
but only if the motion of the earth does not affect the Ether, so that the velocity of
the light remains constant in the Ether frame. On the other hand, if the Earth passes
through the Ether without disturbing it, how could it be detected?
3.3
Michelson-Morley Experiment
[French PP. 49-51]
To measure the motion of the Earth through the Ether, James Clerk Maxwell suggested
using the moons of Jupiter to measure the speed of light, but to do this 6 years apart,
so that Jupiter would be on opposite sides of the Earth relative to its motion. The point
is that the speed of light would be measured differently in the two directions of motion
through the either. The timing difference of eclipse separations across the diameter of
the Earth’s orbit l would therefore be t1 = l/(v + c) in the direction of Earth’s motion,
and t2 = l/(v − c) in the direction against the Earths motion, where v is the velocity of
Earth relative to the Ether. So the difference in eclipse times should be
Ex. 3.3.A (French P. 50)
∆t = t2 − t1 = ............
(3.1)
3.3 Michelson-Morley Experiment
[French PP. 49-51]
22
Given ∆t = 2t0 vc , the value of t0 = l/c is ∼ 16 minutes. Detection of a motion of
150km/s, which is faster than the velocity of stars in the solar system, would require a
measurement of ∆t ∼ 1sec, which was not feasible across 6 years in the 19th century.
The attractiveness of this experiment was in the fact that the dependence of the
measurement was on the first power of v/c, (i.e. ∆t = 2t0 v/c). This is in contrast to
terrestrial experiments where the return of light to its origin renders the measurements
sensitive to the second power of v/c. Since v c, this makes a big difference. To see
why, calculate the total time for light to cross the Earth’s orbit l and then return
Ex. 3.3.B (French P. 50) Instead of subtracting, we now add
t = t2 + t1 = .......
(3.2)
Thus the dependence of the change in time measurement due to the motion of the
Earth through the Ether should be on the second power of v/c, (i.e. ∆t = 2t0 (v/c)2 ).
Despite this, Michelson developed an instrument to detect this minute second order
effect and to discover the motion of the Earth through the Ether. A schematic of the
apparatus is shown in Figure 3.2. The basic idea is to pass mono-chromatic light from
a source S, which is split along 2 arms, and then re-joined at a telescope T , where the
3.3 Michelson-Morley Experiment
[French PP. 49-51]
23
Figure 3.2: From Figures 2.10 and 2.9 of French
light from the 2 optical paths interferes producing interference fringes as in the lower
panel of Figure 3.2. The setup includes a beam splitter P , and a compensation plate
C to ensure symmetry of the optical paths. The mirrors are referred to as M1 and M2 ,
and if the optical paths of the two arms P M1 and P M2 are said to have lengths l1 and
l2 , then the condition giving rise to fringes is
2(l1 − l2 ) = mλ,
(3.3)
where m is an integer. Such a set up could be used to measure the change in length of
one of the arms. In particular, if the length of the arm is changed by λ/2 (by moving a
mirror for example), then the interference pattern moves by 1 fringe.
However this experiment was used to look for the motion of the Earth (and therefore
the experiment) relative to the ether. Suppose that the experiment is aimed into the
ether wind along the P M1 arm. From the frame of the experiment, there is an Ether
wind blowing past with velocity v towards the plate P
Ex. 3.3.C (French P. 54) We first calculate the time difference between arrival times
along the two arms. Note that the arm 2 must be pointed into the Ether wind, so
that the velocity relative to the interferometer along P M2 is (c2 − v 2 )1/2
3.3 Michelson-Morley Experiment
[French PP. 49-51]
24
Assume v c,
∆ = t2 − t1 =
(3.4)
Now suppose that the experiment is turned by 90 degrees, so that P M2 points
along the direction of motion and so into the Ether wind. [Note here that t01 still
corresponds to the time along the arm to M1 , not to the arm now pointing into the
Ether wind.]
∆0 = t02 − t01 = .......
(3.5)
3.3 Michelson-Morley Experiment
[French PP. 49-51]
25
Thus the change in time difference should result in a shift in the interference pattern
by a number of fringes
(l1 + l2 )v 2
δ=
.
(3.6)
λc2
The values of λ, l1 , l2 and c are known. For the velocity of the Earth (about 30km/s)
v/c ∼ 10−4 , light with λ ∼ 6 × 10−7 m and an arm length of 1.2m, we have an expected
fringe shift of δ ∼ 0.04. This could be detected by rotating the apparatus and looking
for oscillation of the fringes. The size of the shift can be increased by making a bigger
interferometer.
However no shift was detected! One can imagine the confusion for physicists of the
time. The interference and diffraction properties of light (including the fringes used in
the Michelson-Morley experiments) required a wave explanation with a medium for the
waves to travel through. However the precise measurements of the Michelson-Morley
experiment, and the stellar aberration experiments are more readily explained via a
ballistic approach to light as a particle.
Figure 3.3: From Table 2.1 of French
Table 3.3 summarises the situation described so far. Particle and wave theories
see some support from different experiments. The particle model is ruled out by the
observation that velocity of light is independent of source motion. On the other hand,
3.3 Michelson-Morley Experiment
[French PP. 49-51]
26
stellar aberration can only be explained by waves if the Earth travels relative to an
Ether, a situation ruled out by the Michelson-Morley experiment.
The next chapter will discuss the ideas of Einstein that led to the resolution of these
inconsistencies and to Special Relativity.
Chapter 4
Lorentz Transformations
4.1
Galilean and Newtonian relativity
[French PP. 66-67]
A relativity principle is an assertion about the laws of Physics as determined by observations from within different frames of reference. Observers in different frames might make
different measurements of the properties of an object, like its speed, but they should all
agree on the laws of physics that they derive from these observations.
Relativity of speed between two inertial frames1 moving relative to one another
with constant speed, is embedded in our thinking about Newtonian mechanics. Beginning with two frames S and S 0 having relative speed v along the x-direction, we can
write down the equations relating the positions (x, y, z), speeds (ux , uy , uz ), accelerations
(ax , ay , az ) and time t in the different frames. These are the Galilean transformations
x0 = x − vt
y
0
= y
z
0
= x
0
= t
t
u0x
u0y
u0z
a0x
a0y
a0z
(4.1)
= ux − v
= uy
= uz
= ax
= ay
= az
Normally we would assume the relation t = t0 following the idea of a universal time.
1
An inertial frame of reference is a frame of reference that is not undergoing acceleration. In an
inertial frame of reference, a physical object with zero net force acting on it moves with a constant
velocity (which might be zero)?or, equivalently, it is a frame of reference in which Newton’s first law of
motion holds.
4.2 Transforming Newton’s Laws
[French PP. 68-69]
28
This makes all of the above relations except the first redundant. In particular, note that
because the first equation completely defines the relative velocity, the acceleration is the
same in all inertial frames.
4.2
Transforming Newton’s Laws
[French PP. 68-69]
What about forces, and how these transform? In Newtonian mechanics the basic statement is F = ma. In the absence of a force law, this is a definition of force given an
acceleration. However we can use a specific example of a force F1,2 that arises from
the interaction of two bodies with position x1 and x2 as a function of their separation.
Examples would be gravitational or electrostatic forces.
Ex. 4.2.A (French P. 69)
0 under Galilean transformation?
We have F1,2 = f (x2 − x1 ). What is F1,2
4.3 The constant speed of light
[French PP. 70-75]
29
Thus, we see that the Newtonian law of force is unaltered by the Galilean transformation. This is always true where the force depends on relative positions. If an absolute
position were involved, so that the force law contained non-linear terms like x22 − x21 ,
then the force would be different in different frames. In such a case all inertial frames
would not be equivalent in the physics they observe, but no such situation has been
observed.
On the other hand, at sufficiently high speeds, Newtons laws are observed to break
down. Moreover, we have seen some of the difficulties that the assumption of a universal
frame defined by the ether lead to, as well as the failed attempts to detect it. The solution
lies in the statement that
All inertial frames are equivalent (even at very large relative velocity), but the
transformation of the laws of motion must be modified
This is the subject of the next section.
4.3
The constant speed of light
[French PP. 70-75]
The key point underpinning the development of Special Relativity as a theory to describe
the dynamics of high speed objects, and to solve the puzzles outlined above, was to
note that all analyses had been based on the concept of an absolute time. While a
natural concept that leads to great successes in Newtonian mechanics, absolute time is
an assumption, which Special Relativity shows to be untenable.
The issue lies in describing the time-dependent properties of a body as observed from
two different locations. For example, we would like to measure the velocity of a body
v=
r2 − r1
t2 − t1
(4.2)
With absolute time, this is easily understood. However, in the absence of this assumption, how do we relate the positions and times? We should use a clock to measure the
arrival t1 that is simultaneous with the arrival at r1 (i.e. a clock at r1 ), and a separate
clock to measure the arrival t2 that is simultaneous with the arrival at r2 . The question
then becomes what we mean by the same time at two different locations. This must be
defined, because otherwise the difference t2 − t1 has no meaning.
4.3 The constant speed of light
[French PP. 70-75]
30
If we have observations at places A and B, then we have a clock that can record
times at A and another than can record times at B.
Ex. 4.3.A (Diagram describing synchronisation of clocks)
To define a common time for these clocks, we state that the time for light to travel
from A to B should be the same as the time from B to A. If a photon travels from A at
t = 0, to a mirror at B where it is reflected back to A arriving at time t0 , then we are
able to define the time when the photon arrived at B as t = t0 /2. This provides a specific
procedure for synchronising clocks in different locations. In our everyday world the very
large speed of light allows us to relate measurements in different places providing an
experience that is very close to an absolute time.
For this definition of time measurements at different locations to work, we require
that the speed of light in vacuum be the same in all circumstances. So, based on the idea
that absolute time was misleading attempts to describe high velocity dynamics, together
with evidence for a constant maximum velocity, we are led to the two postulates on which
Special Relativity is founded:
Postulate 1: All inertial frames are equivalent with respect to laws of physics
Postulate 2: The speed of light in vacuum always has the same value c
4.3 The constant speed of light
[French PP. 70-75]
31
An important consequence of the definition of clock synchronisation, is that simultaneity of events is relative rather than absolute. To aid visualisation of this important
concept it is convenient to introduce the space-time diagram, which plots position on
the x-axis (only considering one dimension of space) vs time on the y-axis. A point in
the plane refers to an event (location and time pair for a particle). The World Line is
defined as the path that a particle takes in the space-time diagram.
Figure 4.1: From Figure 3.2 of French
Figure 4.1 shows an example of a space-time diagram for an experiment to define
simultaneity at stations A and C for observation of light emitted at the mid-point B at
time t = 0. The left panel shows the case where A, B and C are all at rest in frame
S. The world lines of the particles are vertical lines. The world lines of the photons
emitted from B follow the relation
x = xB ± ct
(4.3)
The observation by A and B of the emitted photons, are given by the events A1 and B1 ,
which correspond to the intersection of the world lines for the photons and observers.
The line joining A1 and C1 , which is parallel to the x-axis defines the simultaneity of
events A and B.
Now suppose A, B and C are at rest in a frame S 0 , moving with respect to S with
velocity v. The situation is shown in the right panel. Since the speed of light is constant
in all frames, the world lines for the photons in S remain x = xB + ct, but the world
lines for A and B are now inclined x = xA + vt and so on. The arrival at A and C is now
given by the events A01 and C10 , but the line A01 − C10 is no longer parallel to the x-axis
if observed from S. However viewed in the frame S 0 , there is no difference from the first
case. Each of A, B and C are at rest, and B is at the mid-point, so the events would
again be viewed as being simultaneous. Thus simultaneity is relative, and depends on
the frame of reference from which the measurements are made.
4.3 The constant speed of light
[French PP. 70-75]
32
Ex. 4.3.B Example. Two events x1 = x0 , t1 = x0 /c, and x2 = 2x0 , t1 = x0 /2c. What
is the velocity of a frame S 0 in which these particles are stationary so that the events
are observed simultaneously in frame S.
4.4 Lorentz transformations
[French 76-81]
4.4
33
Lorentz transformations
[French 76-81]
The Lorentz transformations replace our Galilean transformations to allow us to transform between frames with large relative velocity. We can derive these transformations
from the relativity of simultaneity. We therefore start with our space-time diagram
defining the observation of photons from B at A and C in frames S and S 0 . Figure 4.2
adds lines to the space-time diagram in Figure 4.1 corresponding to the co-ordinate axes
of S 0 .
First consider axis t0 . This is simply the world line of the origin in S (i.e. the position
of a stationary object at x = 0 in S). Since the relative velocity is v, this line is x = vt
(where we defined the origins of S and S 0 to coincide at t = t0 = 0). The axis x0 is the
line of points corresponding to t0 = 0, and hence is parallel to lines of constant t0 . Lines
of constant t0 pass through simultaneous events, of which A01 and C10 are an example
(since these events are simultaneous in S 0 as we saw above). Thus the x0 axis is parallel
to the line A1 − C1 , and passes through the origin.
Having defined the relation between axes in different frames, we can now use such
a diagram to identify the key features of the kinematic transformations. Figure 4.2
shows how the space-time point P is depicted in frames S and S 0 using the coordinates
(x, t) or (x0 , t0 ) respectively. The construction of this diagram implies that the relation
between (x, t) and (x0 , t0 ) and vice-versa should be linear. This linearity is fundamental.
4.4 Lorentz transformations
[French 76-81]
34
Figure 4.2: From Figures 3.3 and 3.4 of French
If the relations were not linear, constant motion in one inertial frame would map to an
acceleration in another inertial frame, violating the condition that all inertial frames
should be equivalent.
We are now ready to derive the Lorentz transformations.
Ex. 4.4.A (French P. 78) Noting that the transformations should reduce to the
Galilean transformations for low velocity (and so should have similar form), and
by symmetry we have
x = ax0 + bt0
(4.4)
and
x0 = ax − bt
(4.5)
We first find the relation between a and b by noting that the motion of the origin
in S in S 0 is found using x = 0 in the first equation. Similarly the origin in S 0 in S
is found using x0 = 0 in the second equation. Noting that the relative velocities are
equal and opposite in sign we have
4.4 Lorentz transformations
[French 76-81]
35
For a complete picture, we also need the connection between measures in the y and
z directions. If space is isotropic, then all measurements made transverse to the line of
motion (chosen to be x) are equivalent. This implies that
y0 = y
z0 = z
(4.6)
4.4 Lorentz transformations
[French 76-81]
36
Ex. 4.4.B (French P. 79)
We can also find the transformations for the time co-ordinates.
Note that these time transformations are the explicit statement that time is not
absolute in Special Relativity. For convenience we define the gamma factor
γ(v) = (1 − v 2 /c2 )−1/2 ,
(4.7)
where v is the velocity of S 0 as measured in S. Summarising the transformations we
have
x0 = γ(x − vt)
y0 = y
z0 = z
t0 = γ(t − vx/c2 )
x = γ(x0 + vt0 )
y = y0
z = z0
t = γ(t0 + vx0 /c2 )
These equations are referred to as the Lorentz-Einstein Transformations. They follow
from the two postulates of Special Relativity. However they were discovered a year earlier
4.4 Lorentz transformations
[French 76-81]
37
by Lorentz in order to reconcile the null result of the Michelson-Morley experiment with
a unique inertial frame provided by the Ether by modifying electromagnetic theory.
Ex. 4.4.C (French PP. 81-82) Example: If isotropic light is emitted from a point at
the origin of two frames S and S 0 , with relative velocity v, at time t = t0 = 0, what
is the radius in the two frames at later times?
4.5 Minkowski diagrams and space-time
[French PP. 82-85]
38
This anti-intuitive result is an example of the relativity of simultaneity.
4.5
Minkowski diagrams and space-time
[French PP. 82-85]
The space-time diagrams used to represent the relation between events in different inertial frames were very useful in the derivation of the Lorentz transformations in the
previous sections. Such diagrams were introduced by Minkowski, and are referred to as
Minkowski diagrams. It is convenient to use ct (a length) rather than t for the verticalaxis. With this definition, and with length scales equal on the x and ct axes, the world
line for a light signal originating at the origin becomes the bisector of the axes in all
inertial frames (Figure 4.3). One frame S can be shown with orthogonal axes, but there
is nothing special about this frame.
Figure 4.3: From Figure 3.6 of French
Although there is no absolute time or position or speed in Special Relativity, with
these quantities all being dependent on the frame of the observer, there is a quantity
known as the interval which is invariant. With observers in S and S 0 measuring an event
to have (x, ct) and (x0 , ct0 ) respectively, we have
x0 = γ(x − vt)
t0 = γ(t − vx/c2 )
(4.8)
4.5 Minkowski diagrams and space-time
[French PP. 82-85]
39
Ex. 4.5.A (French P. 84) We can evaluate (ct0 )2 − (x0 )2 :
This quantity, referred to as the interval s2 = (ct)2 − x2 has no dependence on the
relative velocity v and so is the same for all observers, i.e. invariant. Note that unit
distance on a Minkowski diagram is not equal on the different axes x, x0 , x00 . To see this
the hyperbola
x2 − (ct)2 = 1
(4.9)
is plotted in Figure 4.3. Noting that the interval s2 is constant in all frames, this
hyperbola therefore defines a unit distance on each axis. It is apparent that the unit
distance is not equal on all axes in this figure. On the other hand, this hyperbola does
provide a calibration for the co-ordinate axes in any frame.
Chapter 5
Lengths and Time
The Lorentz transformations provide a recipe for transforming between two different
descriptions of the same event. This transformation was the subject of the previous
chapter. We now turn to what these transformations imply for measurements of length
and time.
5.1
Events in different frames
[French PP. 92-95]
Suppose that we have two point events (x1 , ct1 ) and (x2 , ct2 ). Each of these points can
be transformed separately, allowing us to find the separation of the space-time event in
either frame
x02 − x01 = γ[(x2 − x1 ) − v(t2 − t1 )]
(5.1)
t02 − t01 = γ[(t2 − t1 ) − v(x2 − x1 )/c2 ]
(5.2)
Ex. 5.1.A (French P. 94) Frame S 0 has speed v = 0.6c relative to frame S, and we
define t = t0 = 0 at x = x0 = 0. There are 2 events, (x1 , t1 ) = (10m, 2 × 10−7 s), and
(x1 , t1 ) = (50m, 3 × 10−7 s). What is the distance measured in S 0 ?
5.1 Events in different frames
[French PP. 92-95]
41
We next look at differences in time of events measured from different frames. We saw
in the previous chapter that each inertial frame has its own criterion for simultaneity,
from which we derived the transformations of time. The key point to remember here,
is that the time at which an event is recorded to have occurred in one frame is different
to the time that the same event was recorded to have occurred in some other frame,
even when clocks are synchronised at one particular instant and location (e.g. the origin
x = x0 = 0 at t = t0 = 0.
Ex. 5.1.B (French P. 95) What is the time difference measured in S 0 for the above
example § 5.1.A?
Note that in these examples, the answer is not equal to what one would expect from
the naive calculation. Note also that we have considered two arbitrary, un-related points
in space-time. Importantly, if we had chosen different times for the measurements, then
we would have obtained different values for the difference in length. Similarly, if we
had taken different spatial points, then we would calculate a different separation in
time. This motivates the question of how we define the length of an object or the delay
between two events, and leads to two of the most famous results of Special Relativity.
5.2 Lorentz contraction
[French PP. 96-97]
42
Figure 5.1: From Figure 4.2 of French
5.2
Lorentz contraction
[French PP. 96-97]
From our previous discussion we know that a point particle at rest in an inertial frame
S is described by constant values of (x, y, z), and a changing time ct. The two world
lines corresponding to points x1 and x2 in Figure 5.1 could therefore represent the ends
of a body of length
l0 = x2 − x1
(5.3)
as measured in S. We next want to find the length of the body as measured at some
time t0 in frame S 0 . Inspecting Figure 5.1 we note that any line t0 = const will intersect
the world lines of the two ends of the object, at co-ordinates (x01 , ct0 ) and (x02 , ct0 ). The
length l measured in frame S 0 is therefore
l = x02 − x01
Ex. 5.2.A (French P. 97) What is the relation between l and l0 ?
(5.4)
5.3 Time dilation
[French PP. 97-104]
43
Thus we have length contraction, or Lorentz contraction
l = l0 (1 − v 2 /c2 )1/2 .
(5.5)
It is important to properly interpret what exactly is meant by l. It is the distance
between objects at the same instant as judged by an observer in S 0 . The criteria for
simultaneity is different for every frame as we have seen, so the measurement of length
also depends on the frame of reference.
5.3
Time dilation
[French PP. 97-104]
We can also consider the observed separation in time between the ticks of a clock which
is at rest in frame S relative to those at rest in frame S 0 moving with speed v. Consider
2 events with co-ordinate x0 and times t1 and t2 in frame S. In frame S 0 moving with
speed v, we can find the corresponding t02 − t01 .
Ex. 5.3.A (French P. 100) What is the relation between t2 − t1 and t02 − t01 ?
Defining the differences τ0 = (t2 − t1 ) and τ = (t02 − t01 ), we have
τ=
τ0
.
(1 − v 2 /c2 )1/2
(5.6)
While the moving length appears contracted, the ticks on the moving clock appear to
slow down. The key point here is that the clock in the observer’s frame S is the same for
5.3 Time dilation
[French PP. 97-104]
44
both measurements, but different clocks are observed in the moving frame S 0 at position
x0 at times t01 and t02 .
In order to define our space-time grid we would like to measure the time at which an
event occurs anywhere in space. To do this we would like a synchronised clock at each
grid point. The phenomena of time dilation illustrates why we must synchronise clocks
via exchange of light signals. Suppose instead that we took our clocks, synchronised
them all at the origin, and then transported them to their grid point. In order to get
the clocks from the origin to the grid point, they must attain a velocity v.
Ex. 5.3.B (French P. 109) We can calculate the error that this introduces
Thus the error can be made arbitrarily small by making v arbitrarily small, but then
the grid takes an arbitrarily long time to set up!
Because time dilation is such a remarkable result, it is worth diverting to look at the
classic experiment showing this behaviour. The experiment concerns the observation
of cosmic ray meson decay. Radioactive decay is a phenomena that provides a clock
through the buildup of individually random processes with a particular and known
half-life. If we take two groups of particles, one moving at large speed relative to the
observer, equation (5.6) implies that we should see fewer decays (or longer decay time)
in the moving particles. Such an experiment was performed by Rossi and Hall in 1941
using muons produced by cosmic rays that enter the atmosphere. These muons decay
to produce electrons, positrons and neutrinos.
5.3 Time dilation
[French PP. 97-104]
45
These muons are travelling at speeds very close to c. A particle detector can be
used to count the rate of arival of muons. Some of the muons that reach the detector
are stopped, allowing detection of the decay, which produces electrons, positrons and
neutrinos after a short delay. These decays allow the decay time at rest to be measured.
With this measured decay time, we can calculate the fraction of muons that should
decay during a flight time of ∼ l/c during travel through a path-length l. By comparing
the measured counts of muons in detectors on mountain tops vs at sea level, it is found
that muons arrive in far greater numbers than implied by their decay time at rest.
The measured time dilation in this case was a factor of 9! This implies
1 − v 2 /c2 = (τ0 /τ )2 ∼ 1/81,
(5.7)
or that v/c ∼ 0.994.
One can also consider this experiment from the point of view of the muons, travelling
downward at close to c. In the muon rest frame, the decay time is the same as the one
measured in the lab. This is at first sight a contradiction, because we have measured
from the frame of the Earth that the decay rate is much slower than expected. However
from the point of view of the muons the distance from the mountaintop to sea level is
very strongly length contracted, so that the muons do not travel for as long between the
mountain top and sea level in their frame, providing less time for the decays.
Ex. 5.3.C (French P. 104)
5.4 Proper time and proper length
[French PP. 106-109]
46
So observers in both the frame of the Earth and the frame of the muons measure the
same fraction of decays. The effect is attributed to time dilation in one case (where the
ruler is stationary in the observers frame, and to length contraction where the clock is
stationary in the observers frame.
5.4
Proper time and proper length
[French PP. 106-109]
In deriving our results we have chosen frames of reference fixed on a particular object or
observer, and which moves with that object or observer. This is defined as the rest frame
for the observer or object, and we define the quantities within it as proper quantities.
The length in the rest frame is the proper length, the time in the rest frame is proper
time etc. Note that this is not a retreat from relativity, as we could just as well work
with any other frame, but just a convenient notation.
Figure 5.2: From Figures 4.3 and 4.4 of French
Because of the importance of time dilation and length contraction, it is instructive
to re-derive these phenomena using specific examples of proper and non-proper measurements together with the postulate that the speed of light is constant. Consider a
pulse clock at rest in S 0 moving transverse to its length with speed v relative to another
frame S. The pulse clock consists of two mirrors separated by proper length l0 , so that
the interval between counts is 2l0 /c of proper time.
5.4 Proper time and proper length
[French PP. 106-109]
47
Ex. 5.4.A (French P. 107) Initially, align the clock perpendicular to the direction of
motion (see left panel of Figure 5.2).
Thus we retrieve the time dilation formula. A similar analysis can be used to investigate
length contraction.
Ex. 5.4.B (French PP. 108-9) Consider this time a pulse clock moving in the direction
of its length (see right panel of Figure 5.2).
5.5 Causality
[PP. 117-119]
48
Thus we retrieve the length contraction formula.
5.5
Causality
[PP. 117-119]
We earlier introduced the invariant quantity
s2 = (ct)2 − (x)2 = (ct0 )2 − (x0 )2
(5.8)
If we consider 2 events, we obtain the space-time interval
(∆s)2 = (c∆t)2 − (∆x)2
(5.9)
which is equal in all inertial frames. The value of (∆s)2 is zero for events connected by
a light signal.
Because the space-time interval (∆s)2 = 0 corresponds to two events connected by
the travel of a light signal, lines corresponding to (∆s)2 = 0 in a space-time diagram
can be used to define regions that are causally connected, and those which are not.
The left panel of Figure 5.3 shows a space-time diagram, now with an x−axis and a
ct−axis running towards negative values from the origin as well as positive values. In the
latter case, a negative value of ct can be considered as being in the past with-respectto an event along the line ct = 0. World lines are shown corresponding to photons
emitted at the origin (ct = ±x). These photon world-lines extend into both the future
(+ve ct) and the past (−ve ct), and divide space-time into 4 distinct regions. Points
within the sections of diagram labeled ”future” or ”past” can be connected to events
5.5 Causality
[PP. 117-119]
49
Figure 5.3: From Figures 4.9 and 4.10 of French
at the origin via light signals, whereas it is apparent that regions marked ”elsewhere”
cannot be reached by a light signal which goes through the origin. Below we look at
this more quantitatively.
Consider the relationship between two points in this space time. As an example, two
points P and Q are labeled in left panel of Figure 5.3. If c∆t > ∆x for these points, the
the angle with-respect-to the x−axis is greater than 45 degrees (where we assume equal
scales for x and ct). In this case it is possible to find a frame of reference S 0 whose ct0 −
axis is parallel to the line P − Q, so that P and Q are occurring at the same place in
S 0 , separated by a time ∆t0 . Such events are called time-like. Conversely, if c∆t < ∆x
for these points, the angle with-respect-to the x−axis is less than 45 degrees. In this
case it is possible to find a frame of reference S 0 whose x0 − axis is parallel to the line
P − Q, so that P and Q are occurring at the same time in S 0 , separated by a distance
∆x0 . Such events are called space-like. If we tried to connect two space-like events we
require a velocity larger than c.
To illustrate this, suppose we have an event P (x, ct) that can cause an event Q(x +
∆x, ct + c∆t) using a signal with u > c.
Ex. 5.5.A (French P. 118) The time interval in a frame S 0 with relative velocity v is
5.5 Causality
[PP. 117-119]
50
Whilst we cannot have information transmitted between one event and another at a
speed faster than c, this does not prevent us from observing velocities that are apparently
larger than c. A simple example is a beam of light (e.g. a laser on Earth), being swept
over a distant surface (e.g. the Moon).
Ex. 5.5.B (French PP. 119) What is the velocity of the spot on the surface?
In this simple example the spot moves at a speed greater than c. However there
is no sense in which the events at one point on the Moon are influencing the events
at another causally unconnected point. Rather the photons are travelling with speed c
from the laser to the point on the Moon, and so the two non-causally connected events
are receiving different information from different photons emitted at different times at
Earth.
As an application of the space-time interval and causality, consider the situation of
two stars located in different galaxies, each at a distance of 100 light-years from Earth,
but in opposite directions. One star goes supernova a year after the other as seen from
Earth.
5.5 Causality
[PP. 117-119]
51
Ex. 5.5.C Is there a frame of reference where the two stars go supernova at the same
time?
Before leaving this topic, lets consider a two-dimensional world, rather than the 1-D
world from the space-time diagrams considered thus-far. In this case the space-time
interval can be defined
(∆s)2 = (c∆t)2 − (∆x)2 − (∆y)2
(5.10)
Considering points relative to the origin, the equation
(c∆t)2 = x2 + y 2
(5.11)
defines a conic surface containing world-lines of light that pass through the origin. This
surface is shown graphically in the right hand panel of Figure 5.3. This surface is
therefore referred to as the light-cone, which separates regions of space that are causally
connected to the origin from those that are not.
Chapter 6
Relativistic Kinematics
6.1
Observing velocities in different frames
[French PP. 125-126]
Velocities are the rate of change of position with time, or more precisely the time derivative of position. As we now have descriptions for how positions ant times transform
between different frames, manifesting as phenomena including Lorentz contraction and
time dilation, it is therefore relatively straight forward to calculate the transformation
of velocity.
While an event occurs in 3 spatial dimensions, when talking about individual spacetime events, it has so far been sufficient to talk about position in one dimension since
the co-ordinate system can be defined so that y = z = 0, with the relative velocity
between frames defined along the x−direction. However once velocity is introduced,
there are two spatial co-ordinates required to describe the problem. Two dimensions are
sufficient because it is always possible to define a plane containing the vector of relative
motion between the frames S and S 0 (which is the unique direction in the system) and
the vector describing the motion in frame S, and to chose the x−direction to lie along
the direction of relative motion v, with the y−axis describing the orthogonal direction
of the plane in S. As a particle moves in S it maintains z = 0, and hence z 0 = 0 in S 0 .
Recall the transformation equations
x = γ(x0 + vt0 )
y = y0
t = γ(t0 + vx0 /c2 )
where
γ = (1 − v 2 /c2 )−1/2
(6.1)
Let an object have velocity components measured in S 0 of u0x and u0y . From the definition
of velocity,
u0x = dx0 /dt0 and u0y = dy 0 /dt0
(6.2)
6.1 Observing velocities in different frames
[French PP. 125-126]
53
To find the measured velocity in S, we differentiate the equations (6.1) with respect
to t0 to get
dx
dt0
dy
dt0
dt
dt0
= γ(u0x + v)
= u0y
= γ(1 + vu0x /c2 ).
(6.3)
Noting that dx/dt = dx/dt0 × dt0 /dt, and dy/dt = dy/dt0 × dt0 /dt, we obtain
dx
dt
dy
uy =
dt
ux =
=
=
u0x + v
1 + vu0x /c2
u0y /γ
1 + vu0x /c2
(6.4)
and
dx0
dt0
dy 0
u0y = 0
dt
u0x =
=
=
ux − v
1 − vux /c2
uy /γ
1 − vux /c2
(6.5)
The second set of equations (6.5) above for the reverse transformation are included
for completeness, and are derived as follows
Ex. 6.1.A
6.1 Observing velocities in different frames
[French PP. 125-126]
54
We note that in cases where u and v are both small compared with c, that these velocity
transformations reduce to the Galilean addition of velocities. However results depart
from Galilean addition when velocities become large. Parameterising the velocities u0x
and v in units of c we define β1 = v/c and β2 = u0x /c. We can show that ux is always
less than c:
Ex. 6.1.B (French P. 127)
Consider the case where u0x = c, so that β2 = 1. Using ux /c = (β1 + β2 )/(1 + β1 β2 )
derived as part of the above example, we find that ux = c for all values of v.
Thus the velocity transformations equation (6.4) contain the result that
the speed of light is measured to be constant in all frames, regardless of the
velocity of the body from which the light was emitted.
Discussion of measurements of the constant c are found in French PP.128-130.
6.2 Observing transverse motions in different frames
[French PP. 132-133]
6.2
55
Observing transverse motions in different frames
[French PP. 132-133]
We noted previously that co-ordinates describing positions orthogonal to the direction
of relative motion between two frames S and S 0 should not be modified, so that if the
motion is along the x−direction, then we have y = y 0 and z = z 0 . What about velocity?
It is tempting to think that velocities should also be unaffected. However velocity is
the rate of change of position with time, and although the transverse separation is not
affected, there is time dilation. This is seen explicitly in the derivation of equation (6.4),
where the time dilation formula is included to convert from time intervals in S 0 to time
intervals in S through the derivative dt0 /dt. This is also apparent through considering
the special case of u0x = 0, so that
uy = u0y /γ.
(6.6)
Thus, there is a measured slowing of transverse motion.
The phenomenon of stellar aberration provides an application of this phenomenon.
Let frame S be the rest frame of the Sun, and S 0 be the rest frame of the Earth. The
direction of the star relative to the the Earths orbital plane is θ in S and θ0 in S 0 , and
we consider the case when the orbital motion is in the direction of the star (as shown in
Figure 6.2)
Figure 6.1: From Figure 5.3 of French
Ex. 6.2.A (French P. 133) Consider the components of the velocity of light ux =
−c cos θ and uy = −c sin θ (-ve because light is travelling towards the origin of S):
6.2 Observing transverse motions in different frames
[French PP. 132-133]
56
We find a stellar aberration of α = β sin θ, in agreement with our previous discussion of
rain drops analogy of stellar aberration.
6.3 The Doppler effect
[French PP. 134-144]
6.3
57
The Doppler effect
[French PP. 134-144]
The term Doppler effect encompasses aspects of the phenomena where the wavelength
(or energy) of light is observed in frame S to be different from that emitted in the rest
frame of the emitting system or object in S 0 . Before discussing the Doppler phenomenon
in Special Relativity, it is worth considering the acoustic doppler effect in which the
frequency of sound is observed to change with the relative motion of a source and
observer. Consider a source and receiver moving along the same line (as in the left
panel of Figure 6.2), with speeds u1 and u2 relative to air. The signal has frequency ν
and period τ = 1/ν, with 1 pulse per τ . The speed of sound is w.
Ex. 6.3.A (French P. 135)
For the special cases of a stationary source, and moving receiver we get
(u1 = 0, u2 = v):
For the special cases of a stationary receiver, and moving source we get
(u1 = −v, u2 = 0):
6.3 The Doppler effect
[French PP. 134-144]
58
Figure 6.2: From Figures 5.4 and 5.5 of French
The source and receiver are moving away from each other in both these cases, and the
frequency is lower as a result.
Lets now look at the same problem in Special Relativity. Consider a pulsing light
source with period τ at the origin in S (x0 = ct0 = 0), and an observer with location
x = x0 at t = 0 in frame S 0 moving with velocity v relative to S. The nth pulse is emitted
at nτ , and the frequency is ν = 1/τ . What does the observer see? The world lines of
the pulses and the observer are illustrated in the space-time diagram in Figure 6.2 (right
panel).
Ex. 6.3.B (French PP. 136-137) The intersections occur in S at (x1 , t1 ), (x2 , t2 ) etc:
6.3 The Doppler effect
[French PP. 134-144]
59
Thus we obtain the relation between frequencies ν and ν 0 observed in frames S and S 0
0
ν =
1−β
1+β
1/2
ν.
(6.7)
If this expression is expanded to lowest order, then we find that we can reproduce either
of the special cases in § 6.3.A. However the lack of a special reference frame in which the
speed of light is constant, introduces symmetry into the analysis which is not present in
the case of sound.
Figure 6.3: From Figure 5.7 of French
We next extend the time-dilation formula to calculate the observed frequency of radiation observed from a source moving in an arbitrary direction relative to the observer.
Consider a short portion of a satellite orbit having speed v, which can be approximated
by a straight line passing with height h above an observation point at t = 0. Pulses
are emitted at (x1 , t1 ) and (x2 , t2 ) in the observers frame. The time interval between
6.3 The Doppler effect
[French PP. 134-144]
60
pulses with emitted frequency ν in the satellite rest frame of is τ = 1/ν. The geometry
is shown in Figure 6.3.
Ex. 6.3.C (French PP. 142-143)
Thus we have the more general formula for Doppler shift
λ0 = λ
1 − β cos θ
= γλ(1 − β cos θ)
(1 − β 2 )1/2
and
ν0 = ν
(1 − β 2 )1/2
ν
1
=
(6.8)
1 − β cos θ
γ (1 − β cos θ)
Note the important result here, that if the motion is transverse to the line-of-sight
(θ = π/2), that there is still a Doppler shift, which simply has the value of γ. Thus this
transverse component of doppler shift is due purely to time-dilation. One could therefore
look for time-dilation experimentally by trying to measure Doppler shift. Unfortunately
6.3 The Doppler effect
[French PP. 134-144]
61
the time-dilation contribution to the Doppler shift is second order in β, and so inspection
of the equation for frequency shift derived in the previous example shows that the error
in θ needs to be ∆θ β or else the first order term dominates.
However one can compare the measured wavelengths for radiation from a source
moving towards and away from an observer:
Ex. 6.3.D (French PP. 144-145)
In the absence of time dilation, there would be no higher order correction, providing
evidence for the relativistic interpretation of the Doppler effect.
6.4 Moving clocks and objects
[French PP. 149-152]
6.4
62
Moving clocks and objects
[French PP. 149-152]
Having derived the properties of both time dilation and the Doppler effect, it is instructive to pause and consider the relation between the two. We previously considered the
comparison between the time on a stationary clock in our rest frame S at point x, with
a grid of clocks (synchronised by light signals) moving in frame S 0 with speed v past
point x where their time is observed. We found that the time on the moving clocks is
observed to run slow by a factor of γ.
Let’s instead look at a moving clock from our stationary position at x in S. The time
we observe is now the time that the clock had at time t − r/c, where r is the distance to
the clock (in frame S) at this earlier time. This light-travel-time delayed time is often
referred to as the retarded time. The ticking of the clock can then be thought of as
light pulses with equal intervals τ in the clocks frame of reference S 0 . We see these ticks
with duration τ 0 , and in the case where the clock moves directly towards us we have a
situation that is precisely the same as the Doppler effect with
0
τ =
1+β
1−β
1/2
τ.
(6.9)
Retarded time also plays an important role in the apparent size and shape of moving
objects. We noted in our discussion of Lorentz contraction that it is important to
properly interpret what exactly is meant by a length l. The Lorentz contracted length
is the distance between objects at the same instant as judged by an observer in S 0 . This
is not the same as the length that is observed, because of the finite speed of light which
means the we observe different parts of the object at different retarded times, and hence
at different locations in S 0 .
A moving object therefore appears distorted. It is simplest to illustrate this through
the example of an object at a large distance so that it subtends a small angle (implying
that the light providing the information on the objects shape is close to parallel.) Consider the rectangle in Figure 6.4 of a rectangle with length L0 and width W0 as measured
in its rest-frame, moving as speed v parallel to side L0 . The rectangle is viewed from a
point perpendicular to the direction of motion:
Ex. 6.4.A (French PP. 151-152)
6.4 Moving clocks and objects
[French PP. 149-152]
63
Figure 6.4: From Figure 5.11 of French. a) and c) are top views. b) is the observers
view.
6.5 Accelerated motions
[French PP. 152-154]
64
Thus the object appears to have been rotated through an angle sin θ = v/c.
6.5
Accelerated motions
[French PP. 152-154]
Having discussed transformations for positions, times and velocities, it is natural to
discuss the transformations of acceleration. Acceleration is required to discuss forces
in Special Relativity, which will be the end point in this part of the course. It is
worth mentioning that general relativity is based on the equivalence of dynamics and
physics under any form of acceleration (gravitational or otherwise), and is as a result
more ”general” than Special Relativity which is built on the idea that physical laws do
not depend on the inertial frame in which they are being measured. However we have
Lorentz transformations for position and time, and so are able to discuss positions and
their time derivatives, including accelerations.
Ex. 6.5.A (French PP. 153-154) Recalling the velocity transformations (equation 6.4)
ux =
u0x + v
1 + vu0x /c2
Therefore dux /du0x =
and
uy =
u0y /γ
1 + vu0x /c2
and
t = γ(t0 + vx0 /c2 )
6.5 Accelerated motions
[French PP. 152-154]
65
6.5 Accelerated motions
[French PP. 152-154]
66
Thus we have
ax =
a0x
γ 3 (1 + vu0x /c2 )3
(6.10)
The expression for ay is complex, but in the special cases of either uy = 0 or a0x = 0
ay =
a0y
γ 2 (1 + vu0x /c2 )2
(6.11)
Importantly accelerations are not invariant, as they are in Newtonian mechanics. On
the other hand, we readily see that if v c, the acceleration is independent of inertial
frame, which is the cornerstone of Newtonian mechanics.
In Example 2.5.A we found the equation of motion for a body under a constant force
in frame S. To illustrate an example that uses the transformations of acceleration lets
instead, lets suppose that a moving body experiences a constant acceleration in it’s own
rest frame S’ (i.e. acceleration constant in S’). An example might be a rocket.
Ex. 6.5.B We can calculate the velocity of the rocket as observed in frame S as a
function of time. We begin by noting that the acceleration in S’ is constant
a0x = g,
(6.12)
where g is constant, and find the acceleration ax in S.
We next prove the useful identity
d
(vγ) = γ 3 a
dt
(6.13)
6.6 The twin paradox
[French PP. 154-158]
67
Thus we have
d
(vγ) = g,
dt
(6.14)
which we integrate
Thus we have
t2 g 2
v = tg 1 + 2
c
−1/2
(6.15)
for this case. In the limit of small velocities (i.e. tg c) we find that this reduces
to v ∼ tg as expected from a Galilean transformation. However in the limit of large
velocities (i.e. tg c), we obtain v ∼ c as expected for the maximum speed in special
relativity.
6.6
The twin paradox
[French PP. 154-158]
The Twins Paradox is one of the more famous paradoxes of Special Relativity. Suppose
one clock remains fixed in an inertial frame, while a second clock which initially is
synchronised with the stationary clock is transported along a path and returned to the
starting point. Consider the simplest version of the problem. Event 1) A traveller starts
at point O, instantaneously gaining speed v. Event 2) At some point, this traveller
instantaneously reverses the direction of travel (point D). Event 3) The traveller then
arrives back at the starting point and stop, again instantaneously. A representation of
this simple case is shown in the space-time diagram in Figure 6.5.
Suppose that the observer on Earth (O) records a total time T between Event 1)
and Event 3), and so infers that each leg took time T /2. Owing to time dilation, the
traveller records a time of T /2γ for each leg, implying that the time recorded by the
traveller was T /γ. The same point can be made considering length contraction. If the
distance to the turn around is L in the frame of reference of O, then the length contracted
distance is only L/γ for the traveller. Thus, owing to time dilation the transported clock
should have lost time relative to the stationary clock. The apparent paradox lies in the
6.6 The twin paradox
[French PP. 154-158]
68
Figure 6.5: From Figure 5.12 of French.
statement that due to the equivalence of inertial frames, we should be able to consider
either clock to be at rest with the other travelling. However once brought back together,
each clock cannot be simultaneously both slower and faster than the other depending
on whether it was considered to be the one moving or not.
In fact, once brought back together both the traveller and the stationary clock agree
that the travelling clock is slow. We can show this more rigorously by considering
the integrated space-time interval along different paths. While the interval between
two points in spacetime is invariant and seen to be constant by observers in different
inertial frames. The integral of elements of interval depends on the path taken through
space-time
Ex. 6.6.A (French P. 157) We always have
ds2 = c2 dt2 − dx2 = (c2 dt0 )2 − (dx0 )2
(6.16)
6.6 The twin paradox
[French PP. 154-158]
69
Thus we find that the total space-time interval traversed by the two clocks is different.
One final comment. We could transfer to the inertial frame of the travelling clock
on the way out, and then the the inertial frame of the travelling clock on the way back.
Because of the Doppler shift each would see slower arrival of the clock ticks on the way
out. Once A reverses, they immediately start to see clock ticks from B at an enhanced
rate (immediately, because the inertial frame has changed), and over the return journey
these cancel out the Doppler shift from the outbound journey. However owing to the
finite speed of light, B continues to receive slower ticks from A for a significant fraction
of A’s reverse journey (because the inertial frame of B does not change). Thus B receives
slower ticks for more than half of the time, and so sees an overall lower rate.
Chapter 7
Collisions and Conservation Laws
7.1
Conservation Laws
[French PP. 167-168]
Having discussed how measured lengths, speeds and ages of moving objects depend
on the observers frame of reference, we next turn to observations of the interactions
between moving objects. Specifically we will consider the relativistic discussion of elastic
and inelastic collisions. In Newtonian mechanics, elastic and inelastic collisions are
synonymous with conservation of momentum and energy. In nearly all situations from
our everyday experience we are able to describe the outcomes of events accurately using
these laws because the relative velocities involved are low (v c). However we have
already discussed the collisions occurring during a cosmic ray shower where v ∼ c.
Indeed, in particle physics experiments the goal is to collide particles with v ∼ c. Is
there an appropriate conservation law?
Consider a closed system in which the interaction completes in a finite time. We will
consider the relationship between the properties of two particles before and after a collision. Earlier in the course (Chapter 2.4), we developed expressions for the relationships
between energy, mass, and momentum for a particle moving with speed v in frame S
m(v) = γ(v)m0
E(v) = mc2 = γ(v)m0 c2
p(v) = m(v)v = γ(v)m0 v
(7.1)
These expressions were derived based on thought (and real) experiments. However it
is particularly important to note here that these results explicitly assumed conservation
of energy and momentum in the interactions in these experiments. For example, in
Einstein’s box, the momentum of photons was assumed to be transferred to the mirror.
The resulting expressions in equation (7.1) the followed from the forms required for these
assumed laws to hold.
Special Relativity does not lead to the prediction of conservation laws. As in Newtonian mechanics these must be postulated. However with Special Relativity we have
7.2 Elastic collisions viewed from different frames
[French PP. 169-171]
71
developed new prescriptions for calculating the dynamics extended the range of situations (specifically to large velocities) in which we can accurately express the kinetic
energy and momentum of particles. We are therefore now in a position to develop the
appropriate relativistic formulation of the laws of conservation of mass and energy.
7.2
Elastic collisions viewed from different frames
[French PP. 169-171]
Before discussion the relativistic calculation for the elastic collision between two bodies, we first consider the very general statement in Newtonian dynamics describing the
collision between 2 bodies of mass m1 and m2 , with initial velocity vectors u1 and u2
respectively. For any collision, we are able to find a set of scalars αi so that
α1 u1 + α2 u2 = α3 v1 + α4 v2
(7.2)
However experimentally, it is found that conservation of momentum yields the simpler
form
m1 u1 + m2 u2 = m1 v1 + m2 v2 .
(7.3)
This observation implies the existence of an inertial mass in Newtonian dynamics, which
is a constant property of a body.
What is the analog in Special Relativity? We have seen that the ideas of relativity
led us to the concept of a velocity dependent mass (or rather a departure of momentum
from the Newtonian form, since rest mass is a property of the body, and momentum is
the result of its motion). We therefore assert conservation of momentum in a particular
frame S as follows
m1 (u1 )u1 + m2 (u2 )u2 = m1 (v1 )v1 + m2 (v2 )v2 .
(7.4)
It is important to note here that we have postulated conservation of momentum where
the mass is given a velocity dependence, but we have not specified what that dependence
should be. However we can use the kinematics of Special Relativity developed so far to
derive what this dependence should be.
Consider the collision shown in Figure 7.1. The measurements of the collision withrespect-to two frames S and S 0 are shown. The frames S and S 0 have a large relative velocity v along the x−direction. Particles A and B do not have motion in the x−direction
as measured in S and S 0 respectively. They therefore move past each other with speed
±v as measured in S and S 0 respectively. However particles A and B are also given (a
small) motion in the y-direction of u0 and −u0 as measured in S and S 0 respectively,
timed such that they collide. Note the symetry of this collision.
The speeds mentioned above for particles A and B have been quoted relative to their
own frame. However our relativistic kinematics tell us how to calculate the velocity of
both particles in the same frame.
7.2 Elastic collisions viewed from different frames
[French PP. 169-171]
72
Figure 7.1: From Figure 6.1 of French.
Ex. 7.2.A (French P. 171) In the y-direction the velocity of particle B as measured
in S is given by considering the transformation equation for velocity components
transverse to the relative motion of the reference frames
u0y /γ
uy =
1 + vu0x /c2
(7.5)
7.3 In-Elastic collisions viewed from different frames
[French PP. 172-175]
73
Thus we obtain, in the limit of u0 v, the result that
m(v) = γm0 =
m0
,
(1 − v 2 /c2 )1/2
(7.6)
implying that p = γm0 v was the correct definition for momentum. However this derivation is unsatisfying as it relies on limiting the analysis to small velocities in 1-dimension.
7.3
In-Elastic collisions viewed from different frames
[French PP. 172-175]
We next investigate the case of a fully inelastic collision, from the points of view of
two different frames of reference. Two particles, each with rest mass m0 collide to
produce a new composite particle, and we assume that there are no other products
of the interaction. In frame S 0 , the two particles approach each other with speed ±u,
resulting in a stationary composite particle with rest mass M0 . In a second frame S, one
Figure 7.2: From Figure 6.2 of French. Note that the initial particle in case (b) should
be a function of U not u as labeled in the Figure.
7.3 In-Elastic collisions viewed from different frames
[French PP. 172-175]
74
particle is initially stationary with rest mass m0 , implying that the composite particle
must have speed u in frame S (this is the key point; putting frame S centered on the
right hand particle implies that the composite particle which was stationary, must now
have speed u). Hence the relative speed between the frames S and S 0 in this setup
must be u. The initial velocity of the moving particle in frame S is designated U . The
diagram in Figure 7.2 illustrates the situation.
Ex. 7.3.A (French P. 173) Begin with conservation of momentum and mass in S
To proceed, we then find the speed of the moving particle in frame S using the
velocity addition formula
u0x + v
ux =
(7.7)
1 + vu0x /c2
7.3 In-Elastic collisions viewed from different frames
[French PP. 172-175]
75
Thus we find
m(U ) = m0 γ(U ).
(7.8)
Unlike the previous example, this calculation is exact. The calculation can be extended
to naturally demonstrate the connection between mass and energy.
Ex. 7.3.B (French P. 174)
7.3 In-Elastic collisions viewed from different frames
[French PP. 172-175]
76
Thus we have total energy
E=
m0 c2
= m(U )c2 ,
(1 − U 2 /c2 )1/2
(7.9)
where E0 = m0 c2 is the rest mass energy. Finally, we can use equation (7.8) to show
the relationship between the rest masses before and after the collision
Ex. 7.3.C (French P. 174)
7.4 The Photon Rocket
[French PP. 183-184]
77
Thus we find
M0 =
2m0
.
(1 − u2 /c2 )1/2
(7.10)
Note the deep result that the rest mass of the final particle is larger than the sum of
the initial particle rest masses. This again illustrates the mass-energy equivalence.
In examples Ex. 7.3.A-C, we postulated the conservation of momentum with a velocity dependent mass, and conservation of velocity dependent mass. We found that
the appropriate expression is m(v) = γm0 . Multiplying by c2 provides the energy
E = γm0 c2 . Hence we postulate conservation of momentum and energy.
Etotal = mtotal c2 = constant
ptotal = constant
(7.11)
To summarise, we have found the appropriate description of the energy, mass, and
momentum which satisfy the corresponding conservation laws for a particle moving with
speed v in frame S
m(v) = γ(v)m0
E(v) = mc2 = γ(v)m0 c2
p(v) = m(v)v = γ(v)m0 v.
The energy E of a moving particle is E 2 = c2 p2 + E02 where E0 = m0 c2 is the rest mass
energy.
7.4
The Photon Rocket
[French PP. 183-184]
The photon rocket, which has sometimes been proposed as a method for driving spacecraft to very high velocities, provides a nice illustration of energy and momentum conservation. Start with a rocket of rest mass m0 and a payload (the part you want to get
to where it is going) of f m0 . We want this payload to reach a speed v, and by doing
this we send Er of energy in the reverse direction.
Ex. 7.4.A (French P. 183) Energy and momentum conservation yield
7.5 Emission and absorption of light by atoms
[French PP. 176-178]
78
The relationship required between f and γ is
f=
γ+
1
− 1)1/2
(γ 2
(7.12)
Thus, for γ = 10 (to result in a time dilation sufficient for getting to a star in less than
a human lifetime), we find f ∼ 0.05. Unfortunately, the rocket needs to slow down,
and then accelerate and slow down again on the way home, implying that the required
fraction for the payload is actually f 4 m0 ∼ 10−5 m0 .
7.5
Emission and absorption of light by atoms
[French PP. 176-178]
A fundamental application of the conservation laws are provided by considering the
emission and absorption of photons by atoms. Consider first absorption, for which a
stationary atom with rest mass M0 absorbs a photon of energy Q. Conservation of
energy yields the relation
E = M0 c2 + Q = M 0 c2 ,
(7.13)
and hence
M 0 = M0 + Q/c2 ,
(7.14)
where M 0 is the mass of the recoiling atom which has a velocity v. Thus the atom gains
mass. To find the velocity of the recoiling atom with momentum p(v) we look at the
conservation of momentum
p = Q/c = M 0 v,
(7.15)
yielding
β = v/c =
Q
.
M0 c2 + Q
(7.16)
7.5 Emission and absorption of light by atoms
[French PP. 176-178]
79
Note that if the photon energy is low so that Q M0 c2 , then β ∼ Q/M0 c2 or v ∼
(Q/c)/M0 corresponding to the Newtonian calculation of the velocity change of a mass
M0 given an impulse (Q/c).
Next consider the process of emission of a photon with energy Q by a stationary
atom with rest mass M0 . Let the recoiling atom have mass M 0 , momentum p0 and
velocity v.
Ex. 7.5.A (French P. 177) Consider conservation of energy and momentum
7.6 Mossbauer effect and the Doppler effect
[French P. 181]
Thus we find the relation
Q = Q0 1 −
Q0
2M0 c2
80
,
(7.17)
where Q0 is the total energy released, and so the photon is slightly redshifted relative
to the energy corresponding to a transition. Thus, if atomic transitions were perfectly
sharp, this recoil effect would make an atomic gas transparent to its own radiation.
However thermal motions and finite transition widths normally lower the impact of this
effect, particularly where the photon is in the optical band.
7.6
Mossbauer effect and the Doppler effect
[French P. 181]
Unlike optical photons, higher energy gamma rays can create recoils with energies larger
than the width of a transition between an individual atom in normal and excited states.
One example is 191 Ir, which has a transition corresponding to Q0 = 129keV (photon
energy in the absence of recoils), but with a transition width of only 5×10−6 eV, implying
a relative width of ∆Q/Q0 = 5×10−11 . This tiny width is much smaller than the value of
Q0 /2M0 c2 ∼ 3×10−7 in equation (7.17) (where Q0 = 2×10−14 J and M0 ∼ 3×10−25 kg),
implying that the energy in the recoil is sufficient to lower the energy of the emitted
γ-ray sufficiently that it will not be re-absorbed.
However crystal lattices can have the property that the recoil is born by all atoms in a
lattice, which number ∼ 1010 even in a tiny crystal. In this case the ratio Q0 /2M0 c2 becomes vanishingly small, and much smaller than the fractional width of the line ∆Q/Q0 .
Thus, if both the emitting and absorbing atom are part of crystals, the recoil effect may
not affect the re-absorption, even for very narrow lines. This is known as the Mossbauer
7.7 Scattering
[French P. 191]
81
Figure 7.3: From Figure 6.4 of French.
effect, and it can be used to make a very nice demonstration of the Doppler effect at
low velocities.
The reabsorption could be removed if the Doppler effect from a relative velocity v
between source and absorber introduces a fractional shift in frequency that is larger
than the fractional transition width. The required velocity in the above example is
v
∆Q
∼>
∼ 5 × 10−11 ,
c
Q0
(7.18)
or v ∼ 2cm/s. This small velocity can be measured using an apparatus like the one
illustrated in Figure 7.3.
7.7
Scattering
[French P. 191]
Conservation of energy and momentum can be used to discuss the initial and final states
of a collision just as in Newtonian dynamics. Consider the special case of a stationary
particle impacted by an identical particle with total energy E1 and momentum p1 , and
in which the momentum of the particles after the collision are each p2 in a direction
±θ/2 from the impact direction (Figure 7.4).
Ex 7.7.A (French PP. 192-193) Conservation of energy and momentum give
7.7 Scattering
[French P. 191]
82
Figure 7.4: From Figure 6.9 of French.
7.8 Compton scattering
[French PP. 194-195]
83
Thus
cos(θ) =
K1
,
4E0 + K1
(7.19)
where we have defined K1 as the kinetic energy of the incoming particle.
Note the important result above for how the angle of the scattering event depends
on the initial energy. Small energies (K1 E0 ) result in scattering at right angles with
θ = 90 degrees (note here that θ is the angle between the two trajectories, not the angle
to the initial momentum direction). For the non-relativistic case, this result therefore
reduces to the well known Newtonian result that a purely elastic collision always results
in momenta of particles that are separated in direction by 90 degrees. However large
energies (K1 E0 ) result in a small angle, which is the relativistic compression of
scattering angles or beaming that is a feature of particle physics experiments.
7.8
Compton scattering
[French PP. 194-195]
The Compton effect refers to the elastic collision between a free electron and a photon.
The collision is elastic because there is no transition for the electron, and so the electron
recoils and the photon loses momentum, and therefore energy. Assume a stationary
electron impacted by a photon with energy Q0 . The photon is scattered in direction θ
with energy Q and the electron in direction φ with energy E and momentum p. The
Compton effect, which is illustrated schematically in Figure 7.5, presents a very direct
and convincing case for the particle-like behaviour of light.
Figure 7.5: From Figure 6.11 of French.
7.8 Compton scattering
[French PP. 194-195]
84
Ex 7.8.A (French PP. 194-195) Defining n0 and n as unit vectors of the initial and
final photon directions, conservation of energy and momentum in this case give
7.9 Doppler effect (again)
[French PP. 197-199]
85
Thus, in terms of wavelength, the effect is described by
λ − λ0 =
h
(1 − cos θ)
m0 c
(7.20)
where λ and λ0 are the final and initial wavelengths. Equation (7.20) tells us that
photons scattered through larger angles are more redshifted than those which continue
on a small angle of deflection.
7.9
Doppler effect (again)
[French PP. 197-199]
For the last topic illustrating the special relativistic interpretation of the conservation of
energy and momentum, we look at the general case of photon emission from a moving
particle. We assume a particle of mass M and momentum p emits a photon with energy
Q0 at an angle θ to the initial direction. After emission the particle has a different mass
M 0 and momentum p0 (which need not be in the same direction as p. This generalises
the discussion in § 7.5, and is illustrated by the schematic in Figure 7.6.
Figure 7.6: From Figure 6.14 of French.
Ex. 7.9.A (French PP. 197-199) Defining n as unit vector along the final photon
directions, conservation of energy and momentum in this case give
7.9 Doppler effect (again)
[French PP. 197-199]
86
7.9 Doppler effect (again)
[French PP. 197-199]
87
Thus we have
Q0 (β, θ) = Q
(1 − β 2 )1/2
,
1 − β cos θ
(7.21)
where Q is the energy of the photon for v = 0 (equation 7.17). The above equation
describes the energy of a photon emitted in any direction, including the effects of recoil.
Note however, that it is of identical form to the kinematic result for the relation between
emitted and received frequency from a moving source (equation 6.8), and follows directly
from application of Q = hν = hc/λ.
Chapter 8
Forces in Special Relativity
8.1
The energy-momentum invariant
[French PP. 205-206]
Up until now we have calculated separate values for E and p in different frames based
on transformations of velocities. We next relate the quantities E and p as measured in
different frames directly, and show that these are related through an invariant quantity,
in analogy to how the interval s2 provides an invariant quantity in our discussion of the
transformation of position and time.
To see this, we start with the expressions for energy and momentum of a single
particle
p = γm0 v
E = γm0 c2
(8.1)
We have already shown that these lead to the relation
E 2 = (cp)2 + E02 .
(8.2)
Restated, this equation says that the difference between the energy and momentum in
any frame equals the rest energy of the particle (i.e. the total energy in a frame for
which the momentum equals zero), i.e.
E 2 − (cp)2 = E02 .
(8.3)
This expression holds in any frame, implying that E02 is an invariant quantity that
specifies the relation between energy and momentum of a particle in any frame,
E 2 − (cp)2 = (E 0 )2 − (cp0 )2 = E02 .
(8.4)
This is valid for a single particle. However we will show below in § 8.2 that it holds true
also for a collection of particles where E0 is interpreted as the energy of the system in
a frame where the vector sum of momenta is zero.
8.2 Transforming energy and momentum
[French PP. 208-212]
89
Figure 8.1: From Figure 7.2 of French.
8.2
Transforming energy and momentum
[French PP. 208-212]
Since energy and momentum are related via the relativistic invariant of the rest energy,
it is natural to find the expressions that allow us to directly transform the energy and
momentum between frames, just as we did for the Lorentz transformations of space and
time.
Consider the velocity of a particle to be u in frame S and u0 in frame S 0 , as shown
in Figure 8.1. Frame S 0 is measured to be moving at v in frame S. These velocities are
related by equation (6.5), which are reproduced below
u0x =
u0y =
ux − v
1 − vux /c2
uy /γ(v)
1 − vux /c2
(8.5)
Also recall the expressions for the energy and momentum of the particles
E = γ(u)m0 c2
px = γ(u)m0 ux
py = γ(u)m0 uy
(8.6)
8.2 Transforming energy and momentum
[French PP. 208-212]
90
in frame S, and
E 0 = γ(u0 )m0 c2
p0x = γ(u0 )m0 u0x
p0y = γ(u0 )m0 u0y
(8.7)
in frame S 0 .
We would like to relate E and E 0 , and p and p0 .
Ex. 8.2.A (French PP. 209-210) Note that γ is a measure of the velocity in the
different frames S and S 0 . The first step is therefore to calculate γ(u0 ) in terms of
properties measured in S (or vice versa).
8.2 Transforming energy and momentum
[French PP. 208-212]
91
Collecting the above transformations we have
p0x = γ(px − vE/c2 )
p0y = py
p0z = pz
E 0 = γ(E − vpx )
px = γ(p0x + vE 0 /c2 )
py = p0y
pz = p0z
E = γ(E 0 + vp0x )
(8.8)
8.2 Transforming energy and momentum
[French PP. 208-212]
92
There are two very important properties of these equations. The first is that the
equations follow precisely the same form as the Lorentz transformations for x and t, with
x → p and t → E/c2 . We will return to this below briefly in § 8.4 when we discuss 4vectors. The second is that unlike in Newtonian mechanics, the relations between energy
and momentum are linear. Thus the transformations above, and the relativistic invariant
relating the energy and momentum of a particle in different frames (equation 8.4) can
be simply extended to a system of particles through direct summation. As noted above,
care should be taken that the invariant is the total energy of the system in the frame
where the total vector sum of momentum is zero, which is not equal to the sum of the
rest energy of particles in a system.
To show the utility of these relations, we use the example already discussed in § 7.7
of the elastic scattering event that corresponds to the collision between two particles.
The schematic is shown in Figure 8.2, illustrating the observation of the scattering event
in the zero momentum frame S 0 (left), and in the frame S of one of the particles (right).
Figure 8.2: From Figure 7.3 of French.
Observed in S 0 , each particle has p0 both before and after the collision, but the
collision turns each momentum vector through 90 degrees. Note that this must be the
case simply because the momenta are measured in the frame with zero momentum. Now
v=
p0
c2 p0
p0
= 0
=
m(v)
E (v)/c2
E0
(8.9)
is the velocity relating S and S 0 , and therefore also the speed of the particle in frame S 0
which was stationary in S (and vicer versa).
8.2 Transforming energy and momentum
[French PP. 208-212]
93
Ex. 8.2.B (French P. 212) Consider the Energy and momentum of the particles after
the collision,
Thus we have
tan
and solve for γ
1
θ
=
,
2
γ(v)
(8.10)
8.3 Force and acceleration
[PP214-217]
94
Thus we have
θ
tan =
2
2m0 c2
E1 + m0 c2
1/2
(8.11)
Finally, this expression can be related to the expression previously derived for this
problem in equation (8.12) using conservation of energy and momentum
cos(θ) =
K1
,
4E0 + K1
(8.12)
where K1 is the kinetic energy of the incoming particle
8.3
Force and acceleration
[PP214-217]
Up until this point the course has been concerned with position and time, and with
energy and momentum. We did investigate transformations for velocity (§ 6.1), and
have used these to study important phenomena including the transformations of energy
and momentum (§ 8.2). We also considered the transformations of acceleration (§ 6.5),
and showed that an essential difference from Newtonian mechanics arises because the
direction of the acceleration differs in different inertial frames. However much of physics
is concerned with determining the physical laws for various forces. Thus, while force does
not play the central role in relativistic mechanics that it does in Newtonian mechanics,
it is never-the-less very useful to specify the transformation.
We start with a definition for force
F=
dp
d
= (m(v)v),
dt
dt
(8.13)
8.3 Force and acceleration
[PP214-217]
95
which is the simplest extension of the Newtonian result. With this definition, the force
is expressed in terms of positions, times and velocities, which can be transformed using
the Lorentz transformations. Thus, the force F must also transform the same way. We
will first use the definition of force to demonstrate that the direction of the force in not
necessarily parallel to the acceleration, and then derive the transformation equations.
Consider a particle at an instant where it has velocity v in a frame the lab frame S. In
the particles rest frame S 0 forces
F0,x = a0,x m0
and
F0,y = a0,y m0
(8.14)
are applied in the direction of and transverse to v, causing accelerations a0,x and a0,y
of the rest mass m0 .
Ex. 8.3.A (French PP. 215-217) We calculate the x- and y-components of the force
from the derivative of the momentum in the lab frame.
Thus we find Fx = F0,x (i.e. no change in the x-component of force!). What about
transverse force?
8.3 Force and acceleration
[PP214-217]
96
Thus we find Fy = γ −1 F0,y , which is not invariant.
Hence, in the lab frame S, we have
Fy
1 ax
= 2 .
Fx
γ ay
(8.15)
Thus the force is not found to be in the same direction as the acceleration!
We next calculate explicit transformations for the force F. This follows the previous
calculation for the special case in which one of the two frames of reference is the instantaneous rest frame of the particle. We again begin with the definition of force, but this
time in two separate frames S and S 0 .
dp
dp0
and F0 = 0
(8.16)
dt
dt
Calculation of the transformation requires the transformations for momentum, enF=
8.3 Force and acceleration
[PP214-217]
97
ergy, velocity, position and time. The minimal number of formulae needed are as follows
x0 = γ(x − vt)
y0 = y
t0 = γ(t − vx/c2 )
ux − v
u0x =
1 − vux /c2
uy /γ(v)
u0y =
1 − vux /c2
p0x = γ(px − vE/c2 )
(8.17)
p0y
0
(8.18)
E
= py
= γ(E − vpx )
Ex. 8.3.B (French PP. 222-225) We would like to calculate the components Fx0 and
Fy0 . First Fx0
The quantity F.u is the rate at which the force does work in Newtonian mechanics.
We can show that this holds in relativistic mechanics as follows.
8.4 Four vectors
[French P. 213]
98
The expression for Fy0 is similarly
Hence, we have
Fx0 =
Fx − (v/c2 )(F.u)
1 − vux /c2
and
Fy0 =
Fy
γ(1 − vux /c2 )
(8.19)
The equations tell us that the force in one frame depends on the power developed by the
force in the other frame, and provides the generalisation of the earlier transformations
for Fx and Fy in the special case of an instantaneous restframe.
8.4
Four vectors
[French P. 213]
As we noted above, the equations (9.7) have an identical form to the Lorentz transformations for x and t, with x → p and t → E/c2 . This implies that the 3 components
of a momentum vector transform in the same way as the 3 components of a position
vector, and that the total energy transforms in the same way as time. It follows then
that the quantity E 2 − (cp)2 is invariant, just as (ct)2 − x2 is invariant. However in the
momentum-energy transformations it is the quantity E0 that is invariant (rather than
the space-time-interval), as we argued above at the start of § 8.1.
The Lorentz transformations (both for space-time and momentum-energy) embody
the feature of Special Relativity that the specification of time in one frame requires the
specification of both time and position in another frame. Similarly, the specification
of energy in one frame requires the specification of both energy and momentum in
another frame. Thus it is convenient to treat space-time and momentum energy as 4dimensional quantities, rather than as 3-dimensional quantities as a function of time, or
as 3-dimensional momentum as a function of energy.
Indeed the kinematic and dynamic states of a particle can be expressed in any frame
using 4-vectors with components x = (ct, x, y, z) and p ≡ (E/c, px , py , pz ). The components of these vectors individually transform between frames according to the Lorentz
8.4 Four vectors
[French P. 213]
99
transformations for space-time and energy-momentum respectively. Hence, with this formalism, the Lorentz transformations become a means of transforming the corresponding
4-vectors from one set of axes onto another. The transformations then become a problem
in Matrix multiplication.
In addition we have found that there is an invariant length s such that
s2 = (ct)2 − x2 − y 2 − z 2 = (ct0 )2 − (x0 )2 − (y 0 )2 − (z 0 )2 = (s0 )2 ,
(8.20)
and a corresponding invariant mass E0 /c such that
E02 /c2 = (E/c)2 − p2x − p2y − p2z = (E/c)2 − (p0x )2 − (p0y )2 − (p0z )2 = (E00 /c)2 .
(8.21)
The quantities X.X ≡ (ct)2 − |~x|2 = (ct)2 − x2 − y 2 − z 2 and P.P ≡ (E/c)2 − |~
p|2 =
(E/c)2 − p2x − p2y − p2z are therefore scalars that are invariant under Lorentz transformations. We call these 4-scalars, which represent properties of the particle that are
invariant under Lorentz transformations. We are therefore led to the definition for the
scalar product of a 4-vector A = (A0 , ~a) = A = (A0 , A1 , A2 , A3 ) with itself
A.A = A0 A0 − ~a.~a = A20 − A21 − A22 − A23 .
(8.22)
We also have the definition of the scalar product of two 4-vectors A and B
A.B = A0 B0 − ~a.~b = A0 B0 − A1 B1 − A2 B2 − A3 B3
(8.23)
where ~a.~b is the usual vector dot-product.
Note that the conservation of energy and momentum are now encapsulated in a
statement of the conservation of the 4-momentum Ptot of a system
tot
Ptot
initial = Pfinal .
(8.24)
For example, if two particles undergo an elastic collision, we have the statement of
conservation of 4-momentum
P1,initial + P2,initial = P1,final + P2,final ,
(8.25)
where subscripts refer to the particle number. Use of the 4-vectors to solve problems in
special relativity is physically equivalent to explicit treatment of the components, but
can be mathematically more convenient.
Chapter 9
List of formulae
• Space-time Lorentz transformations
x0 = γ(x − vt)
y0 = y
z0 = z
t0 = γ(t − vx/c2 )
x = γ(x0 + vt0 )
y = y0
z = z0
t = γ(t0 + vx0 /c2 )
• Energy, mass, and momentum for a particle moving with speed v in frame S
m(v) = γ(v)m0
E(v) = mc2 = γ(v)m0 c2
p(v) = m(v)v = γ(v)m0 v
• Space-time interval
s2 = (ct)2 − (x)2 = (ct0 )2 − (x0 )2
(9.1)
l = l0 (1 − v 2 /c2 )1/2 .
(9.2)
• Lorentz contraction
101
• Time dilation between the differences τ0 = (t2 − t1 ) and τ = (t02 − t01 )
τ=
τ0
(1 − v 2 /c2 )1/2
(9.3)
• The Doppler effect for a photon emitted at an angle θ to the direction of particle
motion
(1 − (v/c)2 )1/2
ν 0 (v/c, θ) = ν
(9.4)
1 − v/c cos θ
• Velocity transformations
u0x =
u0y =
u0z =
ux =
uy =
uz =
ux − v
1 − vux /c2
uy /γ
1 − vux /c2
uz /γ
1 − vux /c2
u0x + v
1 + vu0x /c2
u0y /γ
1 + vu0x /c2
u0z /γ
1 + vu0x /c2
(9.5)
(9.6)
• Energy-momentum Lorentz transformations
p0x = γ(px − vE/c2 )
p0y = py
p0z = pz
E 0 = γ(E − vpx )
px = γ(p0x + vE 0 /c2 )
py = p0y
pz = p0z
E = γ(E 0 + vp0x )
(9.7)
102
• Force transformations
Fx − (v/c2 )(F.u)
1 − vux /c2
Fy
Fy0 =
γ(1 − vux /c2 )
Fz
Fz0 =
γ(1 − vux /c2 )
Fx0 =
Fx0 + (v/c2 )(F0 .u0 )
1 + vu0x /c2
Fy0
Fy =
γ(1 + vu0x /c2 )
Fz0
Fz =
γ(1 + vu0x /c2 )
(9.8)
Fx =
(9.9)