CH 26 – Special Relativity
Special relativity is a theory that describes the properties of matter at speeds near the
speed of light. It was developed by Albert Einstein and published in 1905. At speeds
small compared with the speed of light ordinary Newtonian mechanics is extremely
accurate in describing motion, but it fails at very high speeds. Special relatively is valid
for all speeds.
Frames of Reference
When making physical observations, the frame of reference of the observer is important.
A basic assumption, called the principle of Galilean relativity, is that the laws of
mechanics are the same in all frames of reference that are moving with respect to one
another at a constant velocity.
A frame of reference in which Newton’s laws are valid is called an inertial system. All
frames of reference moving at constant velocity with respect to an inertial system are also
inertial systems.
As an example, consider two coordinate systems that are moving at constant velocity
with respect to each other. If system A is an inertial system, then to an observer at rest in
A, an object at rest which experiences no net force will forever remain at rest (Newton’s
first law). To an observer in B the object would be moving at constant velocity. It would
also experience no net force and would forever remain in its state of constant motion
(again, Newton’s first law).
But then let’s consider a third coordinate system, C, which is accelerating with respect to
coordinate systems A and B. An observer at rest in coordinate system C would see the
object accelerating (in a direction opposite to the direction of acceleration of C with
respect to A and B). Thus, Newton’s first law would not appear to apply.
In Galilean relativity, velocities transform by simple vector addition. For example,
suppose a man standing in a train car throws a ball with a speed of 30 m/s as measured in
the train. If the train is moving in the direction of the ball’s motion with a speed of 50
m/s, then an observer at rest beside the train track observes the ball to be moving at a
speed of 80 m/s. This simple velocity addition rule works fine until the speeds begin to
approach the speed of light. For example, if the ball were thrown at a speed of 0.5c and
the train were moving at a speed of 0.8c, this would suggest that the stationary observer
would see the ball moving at a speed of 1.3c, which is greater than the speed of light. In
special relativity, no objects can travel faster than the speed of light.
Galilean relativity would also mean that the speed of light would depend on the observer.
For example, if the man in the moving train car were to shine a light in the direction of
motion of the train, then if he were to measure the speed of the light as c, the stationary
observer would measure the speed to be c + v, where v is the speed of the train. This is in
1
conflict with the theory of special relativity which says that the speed of light is the same
for all observers, regardless of their relative motion.
Michelson-Morley Experiment
Mechanical waves require a medium for propagation. Sound, for example, can’t
propagate in a vacuum. According to Galilean relativity, the speed of sound that one
measures depends on the motion of the medium (e.g., air) relative to the observer. For
example, in still air the speed of sound is about 340 m/s. If the air is moving at a speed of
30 m/s, then the speed of sound measured by a stationary observer would be between 310
m/s and 370 m/s, depending on the direction of the sound with respect to the moving air.
In the 19th century it was believed that light also
required a medium for propagation. Since it was
known that light could propagate in a vacuum
(e.g., from the sun to the earth), this medium was
thought to have unusual properties and to be
difficult to detect. It was referred to as a
luminiferous ether. It was believed that the earth
moved though this ether as the earth orbited the
sun. So, if the speed of light was c if measured by
an observer at rest with respect to the ether, then it
would be different if measured by an observer at
rest on the earth and would depend on the
direction of the light with respect to the motion of
the ether (or the motion of the earth in the ether).
In the late 1800’s two scientists, Albert Michelson
and Edward Morley, performed a series of
experiments to try to detect the presence of this
ether. Their apparatus shown here is now referred
to as the Michelson-Morley interferometer.
2
The apparatus essentially measures the difference between the speed of light parallel and
perpendicular to the ether. A monochromatic light source is split by a half-silvered
mirror into two beams moving in mutually perpendicular directions. The beams are
reflected back through the half-silvered mirror, recombined, and observed. If there is an
ether wind, then the speeds of these two mutually perpendicular beams should differ,
giving a slight phase shift between them and a slight change in intensity when
recombined.
These measurements are analogous to measuring the time for a boat to travel directly
across a river and back, versus the time for the boat to move the same distance down and
up the river. The crossing speed would be c 2  v 2 , whereas the speed down and up
the river would be c  v , where c is the speed of the boat relative to the water (or light
relative to the ether) and v is the speed of the water (or ether) relative to earth.
The results of the Michelson and Morley experiments were null. That is, no difference
could be measured for the speed of light between any two mutually perpendicular
directions. This provided evidence that an ether did not exist
Postulates of Relativity
Einstein made two postulates on which he based his theory of special relativity:
1. The principle of relativity: The laws of physics are the same in all frames of
reference moving at a constant velocity with respect to one another.
2. The constancy of the speed of light: The speed of light is the same to all
observers.
The first postulate is basically the same as the Galilean principle of relativity. The
second postulate is at odds with the Galilean transformation of velocities between two
inertial frames. For example, in the case of a person shining a flashlight on a moving
train, it would mean that if the train were to travel at 0.9c, then instead of the groundbased observer measuring the speed of the light as c + 0.9c = 1.9c, he would still measure
the speed to be c.
Simultaneity
The concept of simultaneity – whether two events occur at the same time – is important
when comparing measurements made in two different inertial systems. If the speed of
light is the same in all moving coordinate systems, this means that events that occur
simultaneously in one system may not be observed as being simultaneous in another
coordinate system.
3
An example is illustrated in the figure below.
An observer O’ stands in the middle of a moving boxcar and an observer O stands at rest
beside the track. When the positions of the observers coincide, a lightning bolt strikes
each end of the boxcar, leaving marks on the ground and marks on each end of the boxcar.
The light from the lightning strikes at A and B reach observer O at the same time, so
observer O concludes that the lightning strikes occurred simultaneously. But to observer
O’ in the moving boxcar, the lightning strikes do not appear to occur at the same time.
The light traveling from A’ to O’ travels further than the light from B’ to O’. Because of
the motion, O’ moves toward the incoming beam from B’ and away from the incoming
beam from A’. So to observer O’ the strike at B’ appeared to occur before the strike at
A’.
Time Dilation
One consequence of the constancy of the speed of light is that moving clocks run slower.
In the figure above a light clock consists of a laser beam being reflected from the roof of
a boxcar of height d. The time interval of the clock is the time for the beam to go up and
back down. To an observer at rest in the boxcar, this time interval is
4
t p 
2d
.
c
This time is referred to as the ‘proper’ time since the clock is at rest with respect to the
observer and the time interval is measured between two events, sending and receiving the
laser pulse, that occur at the same position to the observer.
To an observer at rest beside the track, the light must travel at an angle to reflect from the
moving roof and back to the moving observer. Thus, it travels a greater distance and the
time interval is longer. During the time t/2 that it takes for the light to reach the roof,
the boxcar has traveled a distance vt/2. The distance the light travels is given by
2
ct
 vt 
2
 
 d
2
2


If we square both sides of this equation and solve for t, we get
t 
2d / c
1 v2 / c2
Comparing this expression with that for tp, we see that
t 
t p
  t p ,
1 v2 / c2
where
 
1
2
1 v / c
2
.
Example:
An astronaut travels to a distant planet with a speed of 0.5c. According to his clock, the
trip takes one year. How long does the trip appear to take to an observer on earth?
Solution:
The time measured in the spacecraft is the proper time since the clocks in the spacecraft
are at rest with respect to the astronaut. So,
5
t 
t p
2
1 v / c
2

1 yr
1  ( 0.5 )
2
 1.15 yr
How fast would the astronaut have to travel so that the travel time was two years to the
earth observer?
t  2t p 
t p
1 v2 / c2
1 v2 / c2  1/ 2
1 v2 / c2  1/ 4
v  0.866c
Length Contraction
Time dilation leads to the concept of space contraction. Consider an observer in a rocket
ship traveling with speed v from earth to a distant planet. To an observer at rest on earth,
this distance is Lp = vt. The distance the observer measures is considered the proper
length, Lp, since it is the length of something that is at rest with respect to the observer.
(The planet is at rest with respect to the earth.) The time that the earth observer measures
for the trip, t, is not the proper time since the two events that are timed, leaving earth
and arriving at the planet, do not occur at the same position. Now the guy in the rocket
ship measures the time for the trip as tp. This would be considered the proper time since
two events occur at the same position with respect to him. He sees the planet as moving
toward him with speed v, and he is measuring a moving length, L. So,
L p  vt and L  vt p
Using the time dilation formula relating t and tp, we obtain
L  Lp 1 v2 / c2 
Lp

.
Thus, an object moving with respect to an observer appears have its dimension in the
direction of the motion shortened.
6
Example:
A meter stick zips by you with a speed of 0.9c. The length of the stick is along its
direction of motion. How long does it appear to be?
Solution:
L  L p 1  v 2 / c 2  1m 1  ( 0.9 )2  0.44 m
Muon decay in the earth’s atmosphere
Evidence of time dilation and space contraction has been obtained by measuring the
number of muons in the earth’s atmosphere as a function of elevation. Muons are
unstable particles that are produced in the earth’s atmosphere by the collision of highenergy cosmic rays with atmospheric molecules. They are created in large numbers at an
altitude of about 4,800 m with enough energy to travel close to the speed of light. When
at rest they have a half-life of 2.2 s and decay into an electron (or positron) and two
neutrinos. A particle with a speed c would travel in 2.2 s a distance
d = (3 x 108 m/s)(2.2 x 10-6 s) = 660 m.
This would mean that the muons should never reach the surface of the earth. However, a
considerable number of muons do indeed reach earth. So, how is this possible?
The result can be explained by time dilation. To an observer at rest on earth, the half-life
of the moving muons is larger by an amount that depends on their speed. If their speed
was v = 0.99c, then the half-life observed by someone on earth would be
t 
t p
2
1 v / c
2

2.2s
1  ( 0.99 )
2
 16 s
So, the person on earth would calculate the distance traveled by the dying muon to be
d’  ct = (3 x 108 m/s)(16 x 10-6 s) = 4,800 m,
so they would reach the surface of earth before decaying.
An alternative view is that the muon sees the distance to earth contracted because of the
motion of the earth respect to the muon. The contracted distance would be
d  d' 1  v 2 / c 2  ( 4800 m ) 1  ( 0.99 )2  660 m
7
Twin Paradox
The concept of motion is relative to the observer. So when thinking about time dilation, a
person at rest on earth would say that the clocks in a moving spacecraft run more slowly.
But on the other hand a person in the spacecraft could say that he was at rest and the
person on earth was moving and say that the earth clocks were running more slowly.
Actually, both observers are correct. A conflict would appear to arise if the space
traveler were to go on a long trip and return back to earth so they could compare their
clocks. This situation occurs in the so-called twin paradox.
In the twin paradox, at age 20 twin Speedo leaves earth in a spacecraft for a round trip to
Planet X while twin Goslo stays behind on earth. The planet is 20 light-years from earth
and the spacecraft travels with a speed of 0.95c. What are the ages of the twins when
Speedo returns?
To Goslo the time of travel is
t  2d / v  2( 20 c  yr ) /( 0.95c )  42 yr
(Note: A light year is the distance that light travels in one year. So 1 light-yr = speed x
time = 1 c x yr.)
Goslo perceives Speedo’s time of travel to be shorter, so he determines that Goslo has
aged by
t p  t /   t 1  v 2 / c 2  ( 42 yr ) 1  ( 0.95 )2  ( 42 yr )( 0.31 )  13 yr
However, one might think that Speedo could consider that he was as rest and Goslo was
moving, so that Goslo aged by 13 yrs and Speedo aged by 44 yrs. However, when
applying the concepts of relativity, we must compare inertial frames of reference. Goslo
is in a fixed inertial frame of reference, whereas Speedo is in one inertial frame going and
a different inertial frame returning. Thus, Goslo’s calculations are correct. Upon
Speedo’s return, Speedo is 33 yrs old and Goslo is 62 yrs old.
Another way to envision this is that from Speedo’s point of view the distance that he has
to travel to Planet X is less than 20 light-years because of space contraction. The
distance he travels is
d'  d 1  v 2 / c 2  ( 20 c  yr )( 0.31 )  6.2 c  yr
So the time for him to travel to Planet X and back is
t p  2d' / v  2( 6.2 ) / 0.95  13 yr
8
Velocity Addition
In Galilean relativity it is possible for velocities to add such that objects can exceed the
speed of light. In special relativity, the addition rule for velocities makes this impossible.
Consider a spaceship (system 2) moving at a speed v21 with respect to an observer at rest
on earth (system 1). If an observer at rest in the spaceship fires a projectile (object 3)
with speed v32 relative to the spaceship, then the speed of the projectile as measured by
the earth observer is
v31 
v32  v21
1  v32 v21 / c 2
Example:
A space ship is traveling at 0.9 times the speed of light relative to earth. A person in the
space ship fires a projectile in the direction of travel with a speed of 0.8c. What speed
does a person on earth measure for the speed of the projectile?
Solution:
We have v21 = 0.9c and v32 = 0.8c. Using the velocity addition formula, we have
v31 
0.9c  0.8c
1  ( 0.9c )( 0.8c ) / c
2

1.7c
 0.99c
1.72
If we used the Galilean transformation rule, we would have v31 = 0.9c + 0.8c = 1.7c,
which would exceed the speed of light.
Example:
A space ship is traveling at 0.9 times the speed of light relative to earth. A person in the
space ship fires shines a flashlight in the direction of travel. What speed does a person on
earth measure for the speed of the light from the flashlight?
Solution:
v
0.9c  c
1  ( 0.9c )( c ) / c
2

1.9c
c
1.9
This is in agreement with the second principle of relativity that says that all observers
measure c for the speed of light.
9
Relativistic Momentum
In the special theory of relativity, linear momentum is given by
p
mv
2
1 v / c
2
  mv
If we use the classical definition of momentum, which is p = mv, then momentum is not
conserved in high speed collisions. The relativistic momentum approaches infinity as the
speed of a particle approaches the speed of light.
Example:
A proton travels at a speed of 0.9c. Compare its relativistic and classical momenta.
Solution:
pcl  mv  ( 1.67 x10 27 kg )( 0.9  3x108 m / s )  4.51x10 19 kg  m / s
prel 
4.51x10 19 kg  m / s
1  ( 0.9 )2
 1.03  10 18 kg  m / s
So the relativistic momentum is more than four times what one would calculate using the
incorrect classical definition.
Note that if the speed is small compared with c, then   1 and both formulas give
approximately the same result.
Relativistic Energy
Classically, kinetic energy is given by KE = ½ mv2. The relativistic formula for kinetic
energy is
KE 
mc 2
1 v2 / c2
 mc 2   mc 2  mc 2
The quantity mc2 is the rest mass energy,
E0  mc 2 .
So we can think of  mc2 as the total energy,
10
E   mc 2 
mc 2
1 v2 / c2
The kinetic energy and the total energy of a particle approach infinity as its speed
approaches the speed of light. This means that it is impossible to accelerate a particle
with finite mass to the speed of light.
Example:
An electron has a speed of 0.8c. What is its kinetic energy?
Solution:
mc 2
KE 

1 v2 / c2
 mc 2
( 9.11x10  31 kg )( 3x108 m / s )2
1  ( 0.8 )
2
 ( 9.11x10  31 kg )( 3x108 m / s )2
 1.37 x10 13 J  8.2 x10 14 J  5.5 x10 14 J
For comparison, the classical expression for KE would give
KE  1 mv 2  1 ( 9.11x10  31 kg )( 0.8 x3x108 m / s )2  2.62 x10 14 J
2
2
Example:
Repeat the above calculations for v = 0.1c.
Solution:
KErel 
( 9.11x10 31 kg )( 3 x108 m / s ) 2
1  ( 0.1 )
2
 ( 9.11x10  31 kg )( 3 x108 m / s ) 2
 4.13 x10 16 J
KEcl  1 ( 9.11x10  31 kg )( 0.1x3 x108 m / s )2  4.10 x10 16 J
2
In this case the classical result and the relativistic results are nearly the same.
11
Example:
Calculate the rest mass energy of the electron in electron volts.
Solution:
An electron volt is the kinetic energy that an electron would acquire if accelerated
through a potential difference of 1 volt. This energy is eV, where e = 1.6 x 10-19 C and V
= 1 volt. That is,
1 eV = (1.6x10-19C)(1V) = 1.6x10-19 J
Then
E0  ( 9.11x10 31 kg )( 3 x108 m / s )2  8.20 x10 14 J
 ( 8.20 x10 14 J )( 1ev / 1.6 x10 19 J )  5.11x105 ev  511 kev
Example:
Find the kinetic energy released in the fusion reaction given below:
3
2
4
1
2 He  1D2 He 1H
The rest mass energies of the nuclei are
3
2 He : 2 ,809.4 Mev
2
1 D : 1,876.1 Mev
4
2 He : 3,728.4 Mev
1
938.8 Mev
1H :
Solution:
E  mHe 3c 2  mD c 2  mHe 4c 2  mH c 2
 2809.4  1876.1  3728.4  938.8  18.3 Mev
12
Relativistic Energy and Momentum Relationship
From the expressions for the relativistic energy and momentum, an expression relating
the two can be obtained. The result is
E 2  p 2 c 2  ( mc 2 )2
Pair Production
For a photon, m = 0, so
E  pc
Since for a photon E = hf, we have
p
E hf
h


c
c

Pair Production and Pair Annihilation
A photon with sufficient energy can be converted
into a particle-antiparticle pair if the photon energy
exceeds the rest mass energy of the pair. In order
to conserve momentum, the conversion must take
place near a massive particle, such as a nucleus,
that can absorb the excess momentum of the
photon. The antiparticle of an electron is a
positron, which has the mass of an electron and a
positive charge. A proton antiparticle has the mass
of a proton and a negative charge. Thus, in pair
production charge is conserved.
Example:
What photon wavelength is required to produce a proton-antiproton pair?
Solution:
The rest mass of a proton is 938 Mev. Thus
E  2mc2  2( 938 )  1876 Mev  1876 Mev  1.6 x10 13 J / Mev  3.0  10 10 J
hc ( 6.6 x10 34 J  s )( 3x108 m / s )


 6.6 x10 16 m

10
E
3.0 x10 J
13
In pair annihilation, a particle combines with its antiparticle and two or more photons are
produced. (If only one photon were produced, then momentum would not be conserved.)
Pair Annihilation
Example:
An electron and a positron with nearly zero kinetic
energies annihilate and two photons are produced.
What is the energy and wavelength of each photon.
Solution:
In order to conserve momentum, each photon must
have the same momentum and thus the same energy.
The energy of each photon would be
E  mc 2  (9.11x10 31 kg)(3x108 m / s) 2
 8.20 x10 14 J  511 kev
and their wavelength would be

hc ( 6.6 x10 34 J  s )( 3x108 m / s )

 2.4 x10 12 m

14
E
( 8.2 x10
J)
14