Physics – Einstein`s Relativity

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Physics – Einstein’s Relativity
“the essentials” lectures, 2014
author(s)
These materials represent the collective effort of many teachers across the state.
The principal author of this booklet is:
Dr. Greg Wilmoth, B. Sc (Hons)., Ph.D., Dip. Ed., Grad. Dip. Computing
(Senior VCE Teacher – Haileybury College)
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INTRODUCTION
The theory of special relativity, explains how to interpret motion between different inertial
frames of reference – that is, places that are moving at constant speeds relative to each
other.
Special relativity includes only the special case (hence the name) where the motion is
uniform. The motion it explains is only if you’re travelling in a straight line at a constant
speed. As soon as you accelerate or curve – or do anything that changes the nature of the
motion in any way – special relativity ceases to apply. That’s where Einstein’s general theory
of relativity comes in, which is not part of this unit of study.
Einstein’s Special Relativity theory is based on two key principles:
•
The principle of relativity: The laws of physics don’t change, even for objects moving in
inertial (constant speed) frames of reference.
•
The principle of the speed of light: The speed of light is the same for all observers,
regardless of their motion relative to the light source. (Physicists write this speed using
the symbol c.)
In the latter part of the 19th century, physicists were searching for the mysterious thing
called ether – the medium they believed existed for light waves to wave through. Einstein
just removed the ether entirely from his thinking and assumed that the laws of physics,
including the speed of light, worked the same regardless of how you were moving – exactly
as latter day experiments and mathematics showed them to be!
Einstein’s theory of special relativity created a fundamental link between space and time.
The universe can be viewed as having three space dimensions – up/down, left/right,
forward/backward – and one time dimension. This 4-dimensional space is referred to as the
space-time continuum.
If you move fast enough through space, the observations that you make about space
(dimensions and mass) and time, differ somewhat from the observations of other people,
who are moving at different speeds.
This is the unification of space-time.
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GALILEAN RELATIVITY
Galileo’s Principle of Relativity states that all motion is relative to some particular frame of
reference, but there can be no frame of reference that has an absolute zero velocity. Galileo
stated that you cannot tell if you are stationary or moving uniformly without looking outside
your frame of reference.
Galileo showed that there was no absolute frame of reference. This would imply that there is
somewhere in the Universe that is stationary (or absolute rest).
Inertial frame of reference: Frames of reference that are at rest or moving with constant
velocity are called inertial frames of reference.
NEWTONIAN RELATIVITY
Newton knew that motion was relative, but he believed there was a fixed frame of reference
to which all others could be compared. This frame was called the ether.
The principle of relativity states that the laws of physics do not depend on the velocity of
the frame of reference. Newton’s Laws of Motion do not contradict the principle of relativity.
For Example:
There are two motor cars. Car A is at rest relative to the ground while Car B is travelling at
25 ms-1 north relative to Car A. Car A increases its speed to 50 ms-1 north in 10 seconds.
What is the acceleration of Car A from reference points of the ground and of Car B?
The acceleration of Car A relative to the ground is given by:
a( relative to ground ) =
Δv 50 − 0
=
= 5.0 ms − 2 north
t
10
Relative to Car B, the initial speed of Car A is 25 ms-1 south (which we’ll call –25 ms-1), and
the final speed is 25 ms-1 north.
Therefore the acceleration of car A relative the car B is given by:
arelative to car B
25 − − 25
=
= 5.0 ms − 2 north
10
We can see that the acceleration of Car A, and hence the force and changes in momentum
experienced on Car A and its occupants, does not depend on the frame of reference.
When you discuss the velocity of an object, you usually assume, without saying so, that the
frame of reference is the Earth. In this section, however, we will be analysing motion from
different frames of reference and so the frame of reference needs to be stated.
A stationary frame of reference and a frame of reference with constant velocity are called
inertial frames of reference. Newton’s laws of motion are valid in these inertial frames of
reference.
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Newton also assumed that physical quantities such as mass, time, distance and so on were
absolute quantities. This means that their values don’t change whatever the frame of
reference. This would seem to make sense as you wouldn’t expect the size of an object to
change as it travels faster.
Put simply, the relative velocity of two bodies is given by the following relationship:
Velocity of B relative to A = Velocity of B − Velocity of A
VBrA = VB − VA
EXAMPLE 1
As a man stands on the road, a car due north approaches at 20 ms-1.
(a)
(b)
What is the velocity of the car relative to the man?
What is the velocity of the man relative to the car?
Solution
Velocity of the car = 20 ms-1 S
Velocity of the man = 0
(a)
VBrA = VB − VA = 20 – 0 = 20 ms-1 S
(b)
V ArB = V A − VB = 0 − 20 = −20 ms −1 S = 20 ms −1 N
(Note that the vector –20 ms-1 south is the equivalent to 20 ms-1 north.)
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QUESTION 1
You are in interstellar space and know that your velocity relative to Earth is 3.5 × 106 ms–1
away from Earth. You then notice another spacecraft with a speed, towards you, of
5 × 105 ms–1. Is the other craft heading toward or away from Earth, and at what speed?
Solution
QUESTION 2
An aeroplane can fly at 130 ms–1 through the air. The pilot wants to fly to a destination
500 km due north and then fly straight back. However, there is a west wind blowing at
50 ms–1.
(a)
In the absence of any wind, how long would the return trip take?
(b)
Given the 50 ms–1 west wind, how long will a return trip take if the pilot heads the
plane so that the actual ground velocity is due north or south?
(c)
On another occasion there is a 50 ms–1 north wind blowing. Compare the time for a
return trip to the same destination on this occasion with that in part (b).
Solution
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THE DOPPLER EFFECT
The Doppler Effect is also an example of Newtonian relativity. It is an example of how
objects moving relative to an observer may seem to be different because of their movement.
Let’s consider a stationary source ‘S’ emitting a single frequency sound. An observer at all
points around the source would detect the same frequency. For example, in figure (a) the
wavelength of the sound is the same in all directions.
If, however, the source is moving at constant speed, such as to the right as in figure (b), the
sound reaching observer 1 (O1) has a shorter wavelength than the sound arriving at
observer 2 (O2). In fact the sound reaching O1 has a longer wavelength than the original
source of sound.
This phenomenon will result in observer 1 hearing a higher frequency sound than observer
2. You may have noticed this as a noisy car passes you. The faster the source, the more
pronounced is the effect.
The wavelength of the source is modified by the amount the source moves in one period.
QUESTION 3
Which one of the following statements is the best statement about inertial frames of
reference?
A
B
C
D
Inertial frames must be stationary.
Inertial frames must be accelerating.
The laws of physics have the same form in all inertial frames.
Inertial frames cannot be moving at close to the speed of light.
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MICHELSON — MORLEY EXPERIMENT
James Maxwell proposed that light was an electromagnetic phenomenon that took time to
get from one place to another and that light behaved as a wave. It was commonly accepted
that waves, such as water waves and sound, transfer energy and require a medium through
which to travel. These waves do not exist without their media. But what is the medium for
light?
It was proposed that the medium through which light travels is the luminiferous ether that
was thought to permeate all matter and space.
The speed of light was measured to be 3.0 x 108 ms-1, but relative to what? Presumably it
was relative to the ether, but no one could prove that it existed. The dilemma was that if the
speed of light was relative to the ether, then someone in another frame of reference would
observe a different speed for light.
The most famous attempt to detect this ether was the Michelson-Morley experiment in 1887.
This experiment attempted to measure the ether wind that should be blowing against the
Earth as it moved through the ether. As the Earth revolves around the Sun it travels at about
30 kms-1, relative to the Sun. While this is considerably slower than the speed of light, in fact
about
1
th of the speed of light, a beam of light emitted into the ether wind would travel
10,000
at 299970 kms-1, whereas a beam emitted in the opposite direction would travel at 300030
kms-1. A beam emitted across the wind would travel at the speed of light.
It was not possible to directly measure the speed of light to the accuracy needed, so
Michelson and Morley compared the speed of light on two paths perpendicular to each other.
The comparison was achieved by reflecting light from a single source back and forward
between two sets of mirrors mounted at right angles and fixed rigidly to a block of granite
floating on a bath of mercury.
If the light was travelling into the wind, the velocity relative to the Earth is c − v and the time
for this leg of the journey is t1 =
L
.
c−v
On the return journey the wind is behind the light so its velocity relative to the Earth is c + v
and the time is t 2 =
L
.
c+v
The total time for the journey is:
t = t1 + t 2 =
L
L
2L
+
=
c−v c+v

v2
c1 − 2
c

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Their interferometer was capable of detecting very tiny time differences based on principles
of interference patterns, which is discussed in more detail in Unit 4 Physics. Basically, since
light is assumed to have wave properties, then two light waves can interfere with each other.
If they are in phase, i.e. crest meets crest, then superposition gives us a greater crest. This
is constructive interference and with light this is represented by a band of bright light. If
however, the waves are half a wavelength out of phase, then one crest and one trough
meeting at the same point of space will cancel themselves out, which is called destructive
interference and would be seen as a dark band. These interference patterns had been
clearly documented and explained by Young. Michelson reasoned that two beams of light
split from a single source could be sent on different paths and on return produce an
interference pattern that would enable them to see whether the waves are in phase, or have
experienced a phase shift. Presumably, if these rays were sent on the same length journey
and at the same speed then the time each ray takes on the journey should be identical, and
the rays would be ‘in phase’. If, however, the rays travel at different speeds to each other,
then the time taken for the journey would be different and on return would have a slight
phase difference. In order to bring the waves back into phase, he had a path length
adjustment, enabling him to change the path length of one beam by minute amounts.
A simplified diagram of Michelson and Morley’s apparatus. (Nelson Physics Units 3 & 4 3rd Ed)
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Their equipment had sufficient precision to be able to adjust path lengths by fractions of a
wavelength, hence make allowances for minute time differences in the journeys of the split
beam.
As mentioned before, a path difference equal to half the wavelength of the light used will
result in a bright fringe being replaced with a dark fringe. Through the telescope, Michelson
could see fringes displaced by much less than this. Consider red light of wavelength
700 nm.
Using v = f λ and f =
1
λ
λ
we get v = c =
and T = .
T
T
c
An adjustment of one full wavelength of a 700 nm beam corresponds to a time difference of:
700 × 10 −9
T= =
= 2.3 × 10 −15 s
8
c
3.0 × 10
λ
Using an 11 m path for the light, Michelson predicted a time difference for the two beams in
the order of 3.7 x 10-16 s. This would produce a clearly discernible phase shift, well within the
capability of the equipment.
Michelson and Morley rotated the apparatus fully and made observations at different times of
the day and at different times of the year. This enabled one beam of light to be projected
either in the direction the earth was travelling, or away from the direction it was travelling, or
even across the direction it was travelling. Also, any time difference should change smoothly
as the apparatus is rotated so that the beams point in different directions.
It made no difference. The interference did not alter by any measurable amount. The light
took the same time for the light to travel the two paths regardless of the motion of the Earth.
Michelson’s interferometer was, however, capable of detecting these tiny time differences if
they did indeed exist.
The inability of Michelson and Morley to detect the presence of the ether, provided an
important platform for the theory on relativity that Einstein was soon to propose.
QUESTION 4
Michelson and Morley performed an experiment to measure the speed of light with respect
to the ether. They believed that Earth moved in its orbit with speed v relative to the ether.
They believed there should be a difference in the measured speed of light depending on
whether it was measured parallel or perpendicular to the direction of Earth’s movement
through the ether.
The Michelson–Morley experiment found the ratio of the speed of light measured parallel to
the Earth’s motion through the ether the speed of light measured perpendicular to the
Earth’s motion through the ether to be:
A
B
C
D
Slightly less than one.
Equal to one.
Slightly greater than one.
Significantly greater than one.
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QUESTION 5
Which of the following statements best describes what the Michelson-Morley experiment
attempted to measure?
A
B
C
D
The speed of Earth through space.
Changes in the speed of Earth through space.
Accuracy obtainable with an optical interferometer.
Differences in the speed of light in different directions.
QUESTION 6
Which one of the following best describes what follows directly from the measurements of
the Michelson–Morley experiment?
A
B
C
D
Earth travels through a stationary ether.
The ether may exist, but it is not detectable.
The speed of light near the surface of Earth depends on the direction in which it is
measured travelling.
The speed of light near the surface of Earth is the same in all directions.
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THEORY OF SPECIAL RELATIVITY
In 1905 Albert Einstein introduced his Theory of Special Relativity. It did away with the idea
of the existence of the ether, and argued that there was no absolute frame of reference. His
theory was based around two clear propositions:
First postulate: The laws of physics are the same in all inertial frames of reference – the
Principle of Special Relativity.
Second postulate: The speed of light has the same value, c, in all inertial frames. It does
not depend on either the speed of the source or the speed of the observer.
The physics based on these postulates has become known as special relativity.
The second postulate completely contradicts Newtonian relativity, and seems to go against
common logic.
Newtonian relativity tells us that if we are seated on a train going 20 ms-1 and we throw a ball
in the direction of travel also at 20 ms-1, then the ball relative to the ground would be
travelling at 40 ms-1. If light conformed to Newtonian relativity we would expect light to go
faster or slower depending upon whether we are moving towards or away from the source,
but all experiments show that this is not the case. Compare this with the findings of
Michelson and Morley. The speed of light in a vacuum is 3.00 x 108 ms-1 regardless of the
speed of the observer or the source.
Newtonian physics works as a very good approximation only for velocities much less than c.
The faster something moves, the more obvious it is that the Newtonian world does not match
reality.
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Consider the following thought experiment:
Imagine that you’re on a spaceship (top diagram) and holding a laser so it shoots a beam of
light directly up, striking a mirror you’ve placed on the ceiling. The light beam then comes
back down and strikes a detector. You see a beam of light go up, bounce off the mirror, and
come straight down.
The bottom diagram depicts another observer, who we’ll call Lucy, who is in a stationary
craft observing the moving craft go by. In this stationary craft, Lucy sees the beam travel
along a diagonal path.
However, the spaceship is traveling at a constant speed of half the speed of light (0.5c).
According to Einstein, this makes no difference to you — you can’t even tell that you’re
moving. However, if astronaut Lucy was spying on you, as in the bottom of the figure, it
would be a different story.
Lucy would see your beam of light travel upward along a diagonal path, strike the mirror, and
then travel downward along a diagonal path before striking the detector. In other words, you
and Lucy would see different paths for the light and, more importantly, those paths aren’t
even the same length. This means that the time the beam takes to go from the laser to the
mirror to the detector must also be different for you and Amber so that you both agree on the
speed of light.
This phenomenon is known as time dilation, where the time on a ship moving very quickly
appears to pass slower than on Earth. If you apply Lorentz transformation equations (later in
this booklet), the speed of light works out exactly the same for both observers.
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QUESTION 7
According to Einstein’s theory of special relativity, which one of the following does not
depend on the motion of the observer?
A
B
C
D
The order of occurrence of two events.
The time interval between two events.
The distance between two points.
The speed of light.
QUESTION 8
Is it possible for light to travel slower than 3.0 x 108 ms-1?
Solution
Here we need to emphasise that Einstein’s 1905 theory of relativity deals only with frames of
reference in constant relative motion (that is, inertial frames of reference). For this reason it
is called the special theory of relativity. Special relativity does not deal with accelerated
situations. Ten years later, Einstein put forward the general theory of relativity, which does
deal with situations in which acceleration occurs (that is, with non-inertial frames of
reference). As part of this theory he showed that in an accelerated frame of reference, time
also slows down. The General Theory of Relativity is not part of the current VCE course.
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TIME DILATION
This equation:
t=
to
1−
v2
c2
is known as the time dilation formula. It is called time dilation because time intervals
measured by a clock in a reference frame at rest with the observer are always greater than
the time intervals measured by the clock in a reference frame in motion relative to the
observer. This is an unavoidable consequence of the result that all observers will measure
the same speed for light.
The time dilation factor appears frequently in relativity so it is convenient to give it a symbol
of its own, γ (gamma).
i.e.
γ =
1
(this is also called the Lorentz factor)
v2
1− 2
c
We can therefore simply write t = toγ.
Where:
t
to
is the time measured by an observer who sees the event as moving.
is the time measured by an observer who is at rest relative to the event being measured.
It can also be called the proper time.
v2
A quick inspection of the expression for γ will convince you that, as 2 is almost zero for
c
normal speeds we are likely to encounter on a daily basis, γ is normally very close to one.
So for ordinary speeds, time dilation is not an issue.
A useful rearrangement of this relationship is:
v = c 1−
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γ2
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As the denominator of the expression for gamma must be equal to or less than 1, gamma is
always equal to or greater than 1 (although extremely close to 1 at any normal speed). Thus
time in the moving frame, as seen from the stationary frame, will always appear to elapse
more slowly (t ≥ to). Because it was H. A. Lorentz who first introduced the factor γ in an
attempt to explain the results of the Michelson–Morley experiment, it is often known as the
Lorentz factor.
Gamma, for a speed of 0.5 c, is about 1.155 s. In other words, there is a 15.5% difference in
the time intervals measured by two observers – one at rest relative to the event and the
other with a relative motion of 0.5 c. Unlike in Newtonian physics, in special relativity we find
that time intervals are not invariant but relative to the observer.
This graph demonstrates how the Lorentz factor varies according to the speed of the moving
frame.
There are two three minute clock timers, one on a spacecraft travelling at 0.5 c, and the
other kept on Earth. At the expiration of 3.0 minutes in the spacecraft it is found that 3 min
28 sec has elapsed on earth.
Those on Earth think it is the astronaut’s clock that is running slow because he is in motion
relative to them. To summarise, it is often said ‘moving clocks run slow’.
Note that to is used to time the event in the reference frame in which it is at rest; t is the time
for the same event from a reference frame in relative motion.
In general, to is referred to as proper time, not because it is any more correct than t but
because it belongs to the event (as in ‘property’). Proper time intervals are always measured
to be shorter than any other time intervals.
Remember that time dilation is about measurements in different inertial frames; it is a result
of relative movement between the frames.
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Motion Makes Clocks Run Slowly!!
EXAMPLE 2
One of the many subatomic particles discovered in the last 50 years is the mu-meson. In
laboratories, they have been found to have an average life of 2.3 × 10–6 s. They have also
been discovered arising from interactions between cosmic rays and air molecules high in the
atmosphere, at an elevation about 10 000 m. These mu-mesons have very high velocities of
above 0.999 c. Will any of these mu-mesons reach the ground?
Solution
The average life of the mu-mesons will appear longer relative to the stationary observer on
the ground. to is the proper time recorded in the frame of reference of the mu-mesons, and t
is the time recorded in the frame of reference of Earth.
to
t=
1−
and since
t=
v2
c2
v
= 0.999
c
2.3 × 10−6
1 − 0.9992
t = 5.1 × 10−5 s
From an Earth reference point of view, the mu-mesons would appear to exist for 5.1 x 10-5
seconds, but to the mu-meson itself, it only exists for 2.3 x 10–6 seconds.
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Using a simple speed relationship, we can find the time it would take a mu-meson to travel
the 10,000 m at 0.999 c.
speed =
distance
time
therefore,
time =
distance
10000
=
= 3.3 × 10−5 seconds
speed
0.999 × 3.00 × 108
This time is less than the measured lifetime of the mu-meson, therefore the mu-mesons will
be detected on the ground.
Global positioning systems depend on satellites moving in orbit. Nanosecond accuracy is
required for the GPS but the timing would be out by more than 30 microseconds if
Newtonian physics was used which would give inaccuracies of up to a kilometre. To achieve
a high degree of accuracy, it has to compensate for relativistic effects including time dilation.
Time dilation has been experimentally demonstrated. Using two identical and extremely
accurate clocks, one was placed in an orbiting satellite and the other remained on earth.
Comparison some time later revealed the clock on the rapidly moving satellite was slow
compared to the clock remaining on Earth.
QUESTION 9
A deep space traveller moving at 0.25 c sends a ten second radio transmission to a space
station that he is approaching. How long is the radio signal that is received at the space
station?
A
B
C
D
12.0s
10.3s
10.0s
9.7s
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A spacecraft takes off on a trip to a distant star system 25 light years away at a speed,
relative to earth of 0.995 c. Upon arrival it returns to Earth at a similar speed.
QUESTION 10
What is the value of gamma? (the Lorentz factor)
A
B
C
D
γ = 1.0
γ = 5.0
γ = 10.0
γ = 20.0
QUESTION 11
How long in Earth’s frame did it take the craft to reach the star system?
A
B
C
D
24.5 yrs
25.0 yrs
25.1 yrs
50 yrs
QUESTION 12
How long does it take the traveller in his reference frame to reach the star system?
A
B
C
D
2.5 yrs
24.5 yrs
25.0 yrs
25.5 yrs
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QUESTION 13
How long did the trip take the traveller?
A
B
C
D
5.0 yrs
10 yrs
50.0 yrs
51 yrs
QUESTION 14
If the traveller left a twin back on earth, what would be the age difference upon return of the
craft to earth?
A
B
C
D
5.0 yrs
45.2 yrs
50 yrs
no difference
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SIMULTANEOUS EVENTS
Measurement of time in different reference frames produces interesting effects on what may
normally be considered to be simultaneous events. What may be seen as simultaneous
events in one inertial frame may be seen to be not simultaneous when viewed from another
frame of reference. This became known as the relativity of simultaneity.
In the following example, a light is positioned midway between the ends of a train carriage.
Two sensors are positioned at each end, each triggering a green light which is used to
indicate when the light reaches the end of the carriage.
velocity
(from Nelson Physics VCE Units 3 and 4 3rd Ed)
A thought experiment consisting of one observer midway inside a speeding train carriage
and another observer standing on a platform as the train moves past. A flash of light is given
off at the centre of the carriage just as the two observers pass each other. The observer
onboard the train sees the front and back of the carriage at fixed distances from the source
of light and as such, according to this observer, the light will reach the front and back of the
carriage at the same time.
The train-and-platform experiment from the reference frame of an observer onboard the train.
Each sensor would be triggered at the same timesensor and the green indicator lights switch
on together.
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The observer standing on the platform, on the other hand, sees the rear of the carriage
moving (catching up) toward the point at which the flash was given off and the front of the
carriage moving away from it. As the speed of light is finite and the same in all directions for
all observers, the light headed for the back of the train will have less distance to cover than
the light headed for the front. Thus, the flashes of light will strike the ends of the carriage at
different times. In this reference frame sensor A is rushing towards the light coming from the
central light and will therefore be triggered first. Sensor B is moving away from the light
waves and takes longer to be triggered.
Reference frame of an observer standing on the platform (length contraction not depicted).
The simultaneous events in the reference frame of the train are not simultaneous outside the
train!
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 20
LENGTH CONTRACTION
If time measurements are different relative to different inertial frames, what happens to
measurements of space or length?
The length of objects moving at relativistic speeds undergoes a contraction along the
dimension of motion. An observer at rest (relative to the moving object) would observe the
moving object to be shorter in length. That is to say, that an object at rest might be
measured to be 20 metres long; yet the same object when moving at relativistic speeds
relative to the observer/measurer would have a measured length which is less than 20 m.
The object is actually contracted in length as seen from the stationary reference frame. The
amount of contraction of the object is dependent upon the object's speed relative to the
observer.
There is no contraction in length at right angles to the motion; therefore other dimensions
like width and height are not affected by this contraction.
If the fast moving object was a spacecraft, the dimensions of the craft to an observer inside it
would appear to be exactly the same as before the journey commenced.
L=
Lo
γ
= Lo 1 −
v2
c2
This equation is known as the Lorentz contraction.
The term proper length of an object, is the length measured by an observer at rest relative
to the object.
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The Essentials – Physics – Einstein’s Relativity
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EXAMPLE 3
The space shuttle travels at close to 8000 ms–1. Imagine that as it travels east–west it is to
take a photograph of Australia, which is close to 4000 km wide. Because of its speed, the
space camera will see everything on Earth slightly contracted.
(a)
(b)
About how much less than 4000 km will Australia appear to be in this photograph?
Will the north–south dimension of Australia be smaller as well?
Solution
(a)
γ =
γ =
1
v2
1− 2
c
1
1−
L=
(b)
= 1.00000000036
8000
2
(3 × 10 )
8 2
4 × 106
≈ 3 mm less than the 4000 km
1.00000000036
Only measurements in the direction of motion are affected. Motion is perpendicular to
the north–south direction, so this dimension is not affected.
A moving object will appear shorter, or appear to travel less distance, by the factor γ.
Einstein’s length contraction equation: L =
Lo
γ
.
QUESTION 15
Which one of the following is the best description of the proper length of an object travelling
with constant velocity?
A
B
C
D
The length when measured by any observer at the same location.
The length when measured by an observer at rest relative to the object.
The length when both ends of the object are measured at the same time.
The length when measured with a proper standard measuring stick.
QUESTION 16
Which of the following statements about proper length is the most accurate?
A
B
C
D
The proper length of an object can only be measured by an observer who is moving
relative to the object.
The proper length of an object is sometimes less than another measure of the length of
the object, and sometimes greater than or equal to another measure of the length of the
object.
The proper length of an object is always less than another measure of the length of the
object.
The proper length of an object is always greater than or equal to another measure of
the length of the object.
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 22
QUESTION 17
Two spaceships, Discovery and Enterprise, are both travelling relative to an inertial frame
of reference at 0.8 c in the same direction. Spaceship Discovery shines a light beam
forward towards Enterprise as shown.
What is the speed of the light beam according to the captain on spaceship Enterprise?
A
B
C
D
0.2 c
0.8 c
c
1.8 c
QUESTION 18
On a stationary outpost in deep space Jim observes a spacecraft heading straight towards
him at high speed (0.9c).
When the spacecraft is 1000 km away (as measured by Jim), the pilot sends a flash of light.
Which of the following is closest to the time that the flash takes to reach the post as
measured by Jim?
A
B
C
D
1.0 milliseconds
1.8 milliseconds
3.3 milliseconds
3.8 milliseconds
QUESTION 19
The pilot of the spacecraft, Allen, measures the distance between himself and the outpost at
exactly the same time that he sends the flash of light.
Which of the following is closest to the distance that he measures?
A
B
C
D
440 km
900 km
1000 km
2300 km
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 23
QUESTION 20
If we observe a speeding rocket ship we find that (one or more):
A
B
C
D
Its clocks seem to be going fast.
Its clocks seem to be going slow.
Its length appears to be shorter than normal.
Its length appears to be longer than normal.
Solution
QUESTION 21
In deep space, two spacecrafts are heading directly towards their home base. Spacecraft X
passes spacecraft Y at a relative speed of 0.6c. At the instant both craft are alongside each
other, both send a radio message to their home base.
Which of the following statements about the arrival time of the message received at the
home base is correct?
A
B
C
D
The signal from craft X will arrive before the signal from craft Y.
The signal from craft Y will arrive before the signal from craft X.
Craft X has a value of γ = 1.45 and this will mean that its signal will arrive about 45%
sooner than the signal from craft Y.
Both signals will arrive at the same time.
QUESTION 22
Two cars are approaching an intersection from opposite directions. Car A is coming from the
north, and car B approaches from the south. A strong wind is blowing from north to south.
At a particular instant they are both 100 m from the intersection and travelling at the same
speed, which is only slightly slower than the wind speed.
Which of the following best describes which signal arrives first at the intersection.
A
B
C
D
The sound from car A arrives first.
The sound from car B arrives first.
The sound from both cars arrives at the intersection at the same time.
The sound signal from car B never reaches the intersection.
QUESTION 23
If we observe a speeding rocket ship we find that (one or more):
A
B
C
D
All its dimensions appear smaller.
All its dimensions appear normal.
Its width appears smaller.
Its width appears normal.
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The Essentials – Physics – Einstein’s Relativity
Page 24
QUESTION 24
George speeds along in his rocket ship and passes Henry at 0.9 c who happens to be
looking out the window of his space station. George holds a one metre ruler parallel to the
direction of his velocity.
(a)
(b)
What is the length of the ruler as seen by George?
What is the length of the ruler as measured by Henry?
Solution
QUESTION 25
An alien spacecraft travels through the atmosphere at very high speed and perpendicular to
the ground. Just before it crashes on Earth it goes past the top of a 150 m tall tower. It
measures the height of the tower to be only 100 m. Which of the following options is the best
estimate for the speed of the spacecraft:
A
B
C
D
0.57 c
0.75 c
0.82 c
0.90 c
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 25
A space mission is planned to a distant star 12 light years away. The spacecraft will average
a velocity of 0.5 c to get there.
QUESTION 26
What distance away from Earth will the star system be from the perspective of the spacecraft
crew?
A
B
C
D
6 light yrs
10.4 light yrs
12 light yrs
24 light yrs
QUESTION 27
How long will the journey take from the perspective of the Earth-based command unit?
A
B
C
D
10.4 years
12 years
20.8 years
24 years
QUESTION 28
How long will the journey take according to the crew of the craft?
A
B
C
D
10.4 years
12 years
20.8 years
24 years
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 26
RELATIVE MASS
We have already discussed length and time as relative values. The third relative quantity to
consider is mass.
Einstein argued that the mass of an object will be measured to be greater as its speed
increases.
The relationship that demonstrates this is:
m=
mo
1−
v2
c2
or could be considered as: m = γ mo .
The term mo is known as the rest mass as measured in a stationary inertial reference frame;
m is the measurement of the mass of the object in a reference frame that is moving at
velocity v relative to the stationary frame.
This relationship demonstrates that as an object’s velocity approaches the speed of light, its
mass will greatly increase.
v2
As v approaches c, the term 1 − 2 approaches zero.
c
What happens if the speed of the moving object exceeds c? If this is indeed possible, the
term
1−
v2
would be the square root of a negative number which is invalid. Could it be
c2
that the speed of a moving object cannot exceed the speed of light? Another way of looking
at this is to consider that as it gets faster it gets heavier, therefore giving it more inertia and
making it get harder to move. As the object approaches the speed of light, the mass of the
object approaches infinity.
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 27
EXAMPLE 4
What is the relativistic mass of an electron whose speed is measured to be 2.0 x 108 ms-1?
(Note the rest mass of an electron is 9.109 x 10-31 kg.)
Solution
m=
mo
1−
=
v2
c2
9.109 × 10 −31
(2.0 × 10 )
1−
(3 × 10 )
8 2
8 2
=
9.109 × 10 −31
1 − 0.67 2
= 1.22 x 10-30 kg
As demonstrated the mass of a rapidly moving body increases. This has a number of
implications in devices involving high speed electrons such as cathode ray tubes, particle
accelerators and the synchrotron.
With the synchrotron (which is a Detailed Study topic in Unit 4) electrons are accelerated in a
circular path. We know from our studies of motion that the turning force is given by
Fc =
mv 2
. In order to maintain a fixed radius path at very high speeds, relativistic changes
r
to mass need to be taken into account, in other words, mass is also a variable. As mass
increases, so too must the turning force.
The relativistic increase in mass of the electrons in a TV tube needs to be taken into account
by the engineers who design TVs or the picture would be small and distorted.
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
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QUESTION 29
According to Einstein’s theory of relativity, the relativistic mass, m, of a body of rest mass mo,
depends on its speed. Which one of the curves below best shows how the relativistic mass
varies with speed?
QUESTION 30
What is the relativistic mass of an electron travelling at 0.9 c? The mass of a stationary
electron is 9.1 x 10-31 kg.
A
B
C
D
2.1 x 10-30kg
9.1 x 10-30kg
2.1 x 10-31kg
9.1 x 10-31kg
QUESTION 31
What is the relativistic mass of an electron travelling at 0.99 c?
A
B
C
D
2.1 x 10-31kg
2.1 x 10-30kg
6.45 x 10-30kg
9.1 x 10-29kg
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
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QUESTION 32
How many times heavier is an electron travelling at 0.999 c compared to an electron at rest?
A
B
C
D
10
22.4
25.6
1000
QUESTION 33
If a spaceship is travelling at 99% of the speed of light, why can’t it simply turn on its engine
and accelerate through and beyond the speed of light?
Solution
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 30
MASS — ENERGY
Mass-energy equivalence is a consequence of special relativity.
Einstein expressed the kinetic energy of an object as:
KE = (γ − 1) m o c 2
which can be rearranged as:
This equation shows that mass has energy and that energy has mass, and that energy is
released when mass is lost.
Conservation of energy implies that any situation where there is a decrease in mass of
particles, there must be an accompanied increase in kinetic energies of the particles, such
as a nuclear reaction. Similarly, the mass of an object can be increased by taking in kinetic
energies.
EXAMPLE 5
The rest mass-energy of an electron is 8.2 x 10-14 J. An electron moves fast enough so its
total mass-energy increases by 30%.
(a)
(b)
What is the kinetic energy of the moving electron?
What is the speed of the moving electron?
Solution
(a)
Total mass-energy = kinetic energy + rest mass-energy
KE = 30% of 8.2 x 10-14
= 0.3 x 8.2 x 10-14
= 2.5 x 10-14J
(b)
v = c 1−
v = c 1−
1
γ2
and γ = 1.3
1
1.32
v = 0.64 c
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
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A little rearranging leads to the equation: F t = γ mo v
That is, the impulse as we see it in our frame is equal to γ times the Newtonian momentum.
We can interpret this as meaning that the momentum can be seen as:
P = γ mo v = γ Po
where Po is the momentum (mv) as we would see it in classical mechanics.
This provides further evidence that speed cannot exceed the speed of light.
EXAMPLE 6
What impulse is needed to bring a rocket ship of mass 2000 kg to 0.99 c?
Solution
1
γ =
γ =
v2
1− 2
c
1
1 − 0.99 2
= 7.09
Since P = γ m v = 7.09 × 2000 × 0.99 × 3.0 × 108 = 4.2 × 1012 kgms −1
In a nuclear bomb a few grams of mass are lost with the energy as the uranium undergoes
fission, releasing the equivalent of hundreds of gigajoules (1012 J) of energy. Millions of
tonnes of TNT (chemical) explosive would be required to produce this much energy.
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
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QUESTION 34
In a particle accelerator, an alpha particle of mass 6.64 × 10–27 kg is accelerated to high
speed from rest. The total amount of work done on the alpha particle is equal to
7.71 × 10–10 J.
Which one of the following is closest to its final speed?
A
B
C
D
0.80c
0.85c
0.90c
0.95c
QUESTION 35
A stationary proton has a rest mass energy of 1.50 x 10-10J. When a proton is accelerated
from a speed of γ = 1.05 to a speed of γ = 1.10, which of the following is closest to the
amount of work done on the proton during its acceleration?
A
B
C
D
4.2 x 102 J
4.2 x 10-9 J
7.5 x 10-12 J
8.6 x 10-15 J
QUESTION 36
Melbourne uses around 200 000 GJ of energy each day. Most of this energy is provided by
burning about 100 000 tonnes of coal each day.
(a)
What loss in chemical mass would there be each day with this amount of coal being
burnt?
(b)
What daily loss would there be in the mass of the uranium fuel if this amount of energy
was produced by a nuclear reactor?
Solution
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 33
QUESTION 37
Which of the following statements best explains why it is impossible to accelerate particles to
the speed of light?
A
B
C
D
The kinetic energy of particles tends towards an infinite value as they approach light
speed.
It is directly forbidden by one of Einstein’s postulates.
As particles increase in speed their rest mass tends towards an infinite value.
The length of the path taken increases to infinity.
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
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ADDITIONAL QUESTIONS
QUESTION 38
A car is moving at high speed (0.9c) along a straight track and is heading straight for a pole
which Jason is standing next to. When the car is 1.00 km from the post (as measured by
Jason), the driver sends a flash of light from the car.
Which of the following is closest to the time that the flash of light takes to reach the pole (as
measured by Jason)?
A
B
C
D
0.75 microseconds
0.9 microseconds
1.65 microseconds
3.33 microseconds
QUESTION 39
The driver of the car measures the distance between himself and the post at exactly the
same time that he sends the flash of light.
Which one of the following is closest to the distance that he measures?
A
B
C
D
0.22 km
0.44 km
0.50 km
1.15 km
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
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QUESTION 40
On another occasion, Susan observes the racing car. She is standing exactly midway
between two posts, A and B. At the instant the car passes her, the driver sends
simultaneous flashes of light forwards and backwards towards the posts.
The car is travelling at 0.9c towards post B as shown below.
Which one of the following best describes when the flashes of light reach the posts, as
observed by Susan?
A
B
C
D
Post A receives the flash of light first.
Post B receives the flash of light first.
Post A and Post B receive a flash of light at the same time.
It is not possible to predict which receives a flash of light first.
A meteor is heading directly towards the surface of a planet at a constant speed of 0.85c.
Observers on the surface of the planet observe it at a time when it is a distance x above the
surface in their reference frame. The observers calculate the time that the meteor will take to
reach the surface of the planet as 784 microseconds.
QUESTION 41
Which one of the following is closest to the distance x?
A
B
C
D
105 km
200 km
235 km
380 km
QUESTION 42
Which one of the following is the best estimate of the time, as measured by the meteor, for it
to reach the surface of the planet?
A
B
C
D
1488 microseconds
784 microseconds
666 microseconds
413 microseconds
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 36
QUESTION 43
Which one of the following best describes the time of the meteor’s descent to the planet
surface as measured by the meteor, and the time as measured by the observers on the
surface of the planet?
A
B
C
D
Only the meteor measures the proper time.
They are both measurements of proper time in their own reference frames.
Neither are measures of proper time.
Only the observers measure the proper time.
QUESTION 44
Muons and antimuons are anti-particles of each other and they have the same mass. When
a muon meets an antimuon both are destroyed and two photons (gamma rays) are formed. If
the two particles are effectively stationary, then the two photons have a total energy of
3.38 × 10–11 J.
Using this data, which one of the following is closest to the mass of a single muon?
A
B
C
D
1.13 × 10–28 kg
1.88 × 10–28 kg
3.76 × 10–19 kg
5.64 × 10–19 kg
QUESTION 45
Scientists observe the path of a particle created in a detector that exists for only a short
period, leaving a path of only 5.4 mm. Its speed was measured at 2.5 x 108 ms-1, giving
γ = 1.81. What is the proper lifetime of the particle?
A
B
C
D
1.2 x
2.4 x
3.3 x
5.4 x
10-11 s
10-11 s
10-11 s
10-11 s
QUESTION 46
A spaceship is travelling through space with a value of γ = 1.04 relative to a stationary
observer. Which of the following values of ‘γ’ is closest to that of a spaceship travelling at
twice the speed, relative to the stationary observer.
A
B
C
D
1.08
1.12
1.16
1.20
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 37
SOLUTIONS
FOR ERRORS AND UPDATES, PLEASE VISIT
WWW.TSFX.COM.AU/VCE-UPDATES
QUESTION 1
Away from Earth at 3.0 × 106 m s–1.
QUESTION 2
(a)
(b)
(c)
128 minutes
139 minutes
150 minutes
QUESTION 3
Answer is C
QUESTION 4
Answer is B
QUESTION 5
Answer is D
QUESTION 6
Answer is D
QUESTION 7
Answer is D
QUESTION 8
Light will travel slower than 3.0 x 108 ms-1 in media with refractive indices greater than 1.00.
QUESTION 9
t=
to
t=
10.0
Answer is B
v2
1− 2
c
v
and since
= 0.25
c
1 − 0.25 2
t = 10.32 s
QUESTION 10
Answer is C
QUESTION 11
Answer is C
time =
distance
25
=
= 25.1 years
speed
0.995
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 1
Answer is A
QUESTION 12
Time for the traveller will seem to go at one-tenth (1/γ) the rate of Earth time, so when he
gets there his clocks (or calendar) will say that it took only 2.5 years. This is the time as seen
in the traveller’s frame from the Earth’s frame, as well as in his own frame.
Answer is A
QUESTION 13
The traveller perceived that it took just 5 years.
Answer Is B
QUESTION 14
From an Earth reference point the total journey would have taken 50.2 years, whereas for
the traveller it would have only been 5.0 years. The Earth-bound twin would be 45.2 years
older than the traveller twin.
QUESTION 15
Answer is B
QUESTION 16
Answer is D
QUESTION 17
Answer is C
QUESTION 18
Answer is C
QUESTION 19
Answer is A
QUESTION 20
Answer is B and C
QUESTION 21
Answer is D
QUESTION 22
Answer is A
QUESTION 23
Answer is D
QUESTION 24
(a)
(b)
1.0 m
0.44 m
QUESTION 25
Answer is B
QUESTION 26
Answer is B
Because the craft is moving relative to the star there will be a length contraction.
L=
Lo
γ
= Lo 1 −
L = 12 1 −
v2
c2
0.52 c 2
= 12 1 − 0.25 = 12 × 0.866 = 10.4 light years
c2
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 2
Answer is D
QUESTION 27
t EARTH =
distance 12
=
= 24 years
speed
0.5c
Answer is C
QUESTION 28
s o 10.4
=
= 20.8 years
v 0.5c
t MISSION =
QUESTION 29
Answer is D
Answer is A
QUESTION 30
mo
m=
=
2
v
c2
1−
9.109 × 10 −31
1 − 0.90
Answer is C
QUESTION 31
mo
m=
1−
=
2
v
c2
= 2.1 x 10-30 kg
2
9.109 × 10 −31
1 − 0.99
= 6.45 x 10-30 kg
2
Answer is B
QUESTION 32
m
=γ
m0
γ =
1
1 − 0.999 2
=
22.4
QUESTION 33
p = γmv, so as v approaches c, γ approaches infinity.
In other words, as it goes faster its mass increases and it becomes more difficult to
accelerate.
QUESTION 34
Answer is C
QUESTION 35
Answer is C
QUESTION 36
(a)
(b)
2g
2g
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 3
QUESTION 37
Answer is A
QUESTION 38
Answer is D
QUESTION 39
Answer is B
QUESTION 40
Answer is C
QUESTION 41
Answer is B
QUESTION 42
Answer is D
QUESTION 43
Answer is A
QUESTION 44
Answer is B
QUESTION 45
Answer is A
QUESTION 46
Answer is D
 The School For Excellence 2014
The Essentials – Physics – Einstein’s Relativity
Page 4
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