Lecture
Presentation
Chapter 27
Relativity
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Chapter 27 Relativity
Chapter Goal: To understand how Einstein’s theory of
relativity changes our concepts of time and space.
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Slide 27-2
Chapter 27 Preview
Looking Ahead: Simultaneity
• The lightning strikes are simultaneous to you, but to
someone who is moving relative to you they occur at
different times.
• You’ll learn how to compute the order in which two events
occur according to observers moving relative to each other.
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Slide 27-3
Chapter 27 Preview
Looking Ahead: Time and Space
• Time itself runs faster on the surface of the earth than on
GPS satellites that are in rapid motion relative to the earth.
• You’ll learn how moving clocks run slower and moving
objects are shorter than when they are at rest.
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Slide 27-4
Chapter 27 Preview
Looking Ahead: Mass and Energy
• The sun’s energy comes from converting 4 billion
kilograms of matter into energy every second.
• You’ll learn how Einstein’s famous equation E = mc2
shows that mass and energy are essentially equivalent.
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Slide 27-5
Chapter 27 Preview
Looking Ahead
Text: p. 874
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Slide 27-6
Chapter 27 Preview
Looking Back: Relative Motion
• In Section 3.5 you learned how to find the velocity of a
ball relative to Ana given its velocity relative to Carlos.
• In this chapter, we’ll see how our commonsense ideas
about relative motion break down when one or more of the
velocities approach the speed of light.
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Slide 27-7
Reading Question 27.1
The principle of relativity states that
A. No object can travel faster than light.
B. All motion is relative.
C. All the laws of physics are the same in all inertial
reference frames.
D. The speed of light is constant.
E. Energy is given by E = mc2.
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Slide 27-8
Reading Question 27.1
The principle of relativity states that
A. No object can travel faster than light.
B. All motion is relative.
C. All the laws of physics are the same in all inertial
reference frames.
D. The speed of light is constant.
E. Energy is given by E = mc2.
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Slide 27-9
Reading Question 27.2
A clock on a moving train runs _____ an identical clock at
rest.
A.
B.
C.
D.
Faster than
Slower than
At the same speed as
It depends on which direction the train is moving.
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Slide 27-10
Reading Question 27.2
A clock on a moving train runs _____ an identical clock at
rest.
A.
B.
C.
D.
Faster than
Slower than
At the same speed as
It depends on which direction the train is moving.
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Slide 27-11
Reading Question 27.4
Proper time is
A. The time calculated with the correct relativistic
expression.
B. The longest possible time interval between two events.
C. The time interval between two events that occur at the
same position.
D. The time measured by a light clock.
E. Not discussed in Chapter 27.
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Slide 27-12
Reading Question 27.4
Proper time is
A. The time calculated with the correct relativistic
expression.
B. The longest possible time interval between two events.
C. The time interval between two events that occur at the
same position.
D. The time measured by a light clock.
E. Not discussed in Chapter 27.
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Slide 27-13
Section 27.1 Relativity: What’s It All About?
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Relativity: What’s It All About?
• In Newtonian mechanics, space and time are absolute
quantities; the length of a meter stick and the time between
ticks on a clock are the same to any observer, whether
moving or not.
• Einstein’s special theory of relativity challenges these
commonsense notions.
• Ground-based observers measure the length of a fastmoving rocket to be shorter and a clock on the rocket to
run slower compared to when the rocket is at rest.
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Slide 27-15
What’s Special About Special Relativity?
• Special relativity deals exclusively with inertial reference
frames. Inertial reference frames are reference frames that
move relative to each other with constant velocity.
• General relativity is a more encompassing theory that
considers accelerated motion and its connection to gravity.
• Special relativity is a “special case” of general relativity
where the acceleration of the reference frames is zero.
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Slide 27-16
Section 27.2 Galilean Relativity
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Reference Frames
• Suppose you’re driving along a freeway at 60 mph. A car
passes you going 65 mph. Is 65 mph that car’s “true”
speed?
• The car will appear to travel at 65 mph to someone
standing on the side of the road. But relative to you, that
car’s speed is 5 mph.
• Your speed is 120 mph relative to a driver approaching
from the other direction at 60 mph.
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Slide 27-18
Reference Frames
• An object does not have a “true” speed or velocity.
• The definition of velocity, Δv/Δt, assumes the existence of
a coordinate system.
• We must specify an object’s velocity relative to, or with
respect to, the coordinate system in which it is measured.
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Slide 27-19
Reference Frames
• We define a reference frame to be a coordinate system in
which experimenters equipped with meter sticks,
stopwatches, or any other needed equipment make
position and time measurements on moving objects.
• Three ideas are implicit:
• A reference frame extends infinitely far in all directions.
• The experiments are at rest in the reference frame.
• The number of experimenters and the quality of their
equipment are sufficient to measure positions and velocities
to any level of accuracy needed.
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Slide 27-20
Reference Frames
• Two reference frames, S and S, are in relative motion.
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Slide 27-21
Inertial Reference Frames
• A student cruising at constant velocity on an airplane
places a ball on the floor. The ball does not move.
•
in the airplane’s coordinate system when
satisfying Newton’s first law.
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Slide 27-22
Inertial Reference Frames
• We define an inertial reference frame as one in which
Newton’s first law is valid.
• An inertial reference frame is one in which an isolated
particle, on which there are no forces, either remains at
rest or moves in a straight line at a constant speed, as
measured by experimenters at rest in the frame.
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Slide 27-23
Inertial Reference Frames
• If a student places a ball on the floor of an airplane as it
accelerates during takeoff, the ball will roll to the back of
the plane.
• The ball is accelerating in the plane’s reference frame. Yet
there is no identifiable force that causes the acceleration.
• This violates Newton’s first
law, so the plane is not an
inertial reference frame
during takeoff.
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Slide 27-24
Inertial Reference Frames
• In general, accelerating reference frames are not
inertial reference frames.
• When you’re in a jet flying smoothly at 600 mph—an
inertial reference frame—Newton’s laws are valid. You
can pour drinks or toss and catch a ball.
• When the jet is diving or shaking from turbulence, simple
“experiments” like these would fail. A ball thrown straight
up would land far from your hand.
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Slide 27-25
Inertial Reference Frames
• The simple observations of the differences in reference
frames can be stated as the Galilean principle of relativity:
• Any reference frame that moves at a constant velocity with
respect to an inertial reference frame is itself an inertial
reference frame.
• A reference frame that accelerates with respect to an
inertial reference frame is not an inertial reference frame.
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Slide 27-26
QuickCheck 27.1
Which is an inertial reference frame (or at least a very good
approximation of one)?
A.
B.
C.
D.
E.
A jet plane during takeoff
A jet plane flying straight and level at constant speed
A jet plane turning at constant speed
B and C
A, B, and C
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Slide 27-27
QuickCheck 27.1
Which is an inertial reference frame (or at least a very good
approximation of one)?
A.
B.
C.
D.
E.
A jet plane during takeoff
A jet plane flying straight and level at constant speed
A jet plane turning at constant speed
B and C
A, B, and C
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Slide 27-28
The Galilean Velocity Transformation
• Special relativity is concerned with how physical
quantities such as position and time are measured by
experimenters in different reference frames.
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Slide 27-29
The Galilean Velocity Transformation
• Suppose Sue is standing beside a highway as Jim drives by
at 50 mph.
• Sue’s reference frame, S, is attached to the ground. Jim’s
reference frame, S, is attached to
Jim’s car. The velocity of reference
frame S relative to S
is v = 50 mph.
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Slide 27-30
The Galilean Velocity Transformation
• Sue measures a motorcyclist’s velocity u = 75 mph. The
motorcycle’s velocity u relative to Jim is therefore 25
mph.
• This is the difference between his
speed relative to the ground and
Jim’s speed relative to
the ground.
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Slide 27-31
The Galilean Velocity Transformation
• An object’s velocity measured in a frame S is related to its
velocity measured in frame S by
• These equations are the Galilean velocity
transformations.
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Slide 27-32
Example 27.1 Finding the speed of sound
An airplane is flying at speed 200 m/s with respect to the
ground. Sound wave 1 is approaching the plane from the
front, while sound wave 2 is catching up from behind. Both
waves travel at 340 m/s relative to the ground. What is the
velocity of each wave relative to the plane?
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Slide 27-33
Example 27.1 Finding the speed of sound
(cont.)
The airplane travels to the right with reference frame S′ at
velocity v. We can use the Galilean transformations of
velocity to find the velocities of the two sound waves in
frame S′:
u′1 = u1  v = 340 m/s  200 m/s = 540 m/s
u′2 = u2  v = 340 m/s  200 m/s = 140 m/s
Thus wave 1 approaches the plane with a speed of 540 m/s,
while wave 2 approaches with a speed of 140 m/s.
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Slide 27-34
QuickCheck 27.2
Balls 1 and 2, about to collide, have their velocities shown
in reference frame S. What is the velocity u1i of ball 1 in
frame S?
A.
B.
C.
D.
–6.0 m/s
–2.0 m/s
2.0 m/s
6.0 m/s
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Slide 27-35
QuickCheck 27.2
Balls 1 and 2, about to collide, have their velocities shown
in reference frame S. What is the velocity u1i of ball 1 in
frame S?
A.
B.
C.
D.
–6.0 m/s
–2.0 m/s
2.0 m/s
6.0 m/s
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Slide 27-36
QuickCheck 27.3
Balls 1 and 2, about to collide, have their velocities shown
in reference frame S. What is the velocity u2i of ball 2 in
frame S?
A.
B.
C.
D.
–9.0 m/s
–1.0 m/s
0.0 m/s
1.0 m/s
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Slide 27-37
QuickCheck 27.3
Balls 1 and 2, about to collide, have their velocities shown
in reference frame S. What is the velocity u2i of ball 2 in
frame S?
A.
B.
C.
D.
–9.0 m/s
–1.0 m/s
0.0 m/s
1.0 m/s
1 m/s to the left
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Slide 27-38
QuickCheck 27.4
Race car driver Sam is heading down the final straightaway,
approaching the finish line at 100 m/s. His fans, at the finish
line straight ahead, are shouting. The speed of sound on this
very hot day is 350 m/s. How fast are the sound waves of
the shouts approaching in Sam’s reference frame?
A.
B.
C.
D.
E.
100 m/s
250 m/s
350 m/s
450 m/s
Sound waves don’t travel in Sam’s reference frame.
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Slide 27-39
QuickCheck 27.4
Race car driver Sam is heading down the final straightaway,
approaching the finish line at 100 m/s. His fans, at the finish
line straight ahead, are shouting. The speed of sound on this
very hot day is 350 m/s. How fast are the sound waves of
the shouts approaching in Sam’s reference frame?
A.
B.
C.
D.
E.
100 m/s
250 m/s
350 m/s
450 m/s
Sound waves don’t travel in Sam’s reference frame.
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Slide 27-40
Example Problem
You are running at 10 m/s relative to the ground. A person
standing still behind you throws ball A toward you at 30
m/s. A person standing still in front of you throws ball B
toward you at 30 m/s. And another person standing still in
front of you throws ball C straight up into the air at 30 m/s.
What are the speeds of balls A, B, and C relative to you?
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Slide 27-41
Section 27.3 Einstein’s Principle of Relativity
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Einstein’s Principle of Relativity
• If light is a wave, what is the medium in which it travels?
• The medium in which it was thought to travel was called
ether.
• Experiments measuring the speed of light were thought to
be measuring the speed of light through ether.
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Slide 27-43
Einstein’s Principle of Relativity
• Maxwell’s theory of electromagnetism predicted that light
waves travel with the speed
• This prediction seemed only valid in the reference frame
of the ether.
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Slide 27-44
Einstein’s Principle of Relativity
It seems as if the speed of light should differ from c in
a reference frame moving through the ether.
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Slide 27-45
Einstein’s Principle of Relativity
• Einstein considered how a light wave would look to
someone traveling alongside the wave at the wave speed.
• An electromagnetic wave sustains itself with changing
electric and magnetic fields, but to someone moving with
the wave, the fields would not change.
• After many years of thinking about the connection
between electromagnetic waves and reference frames,
Einstein concluded that all the laws of physics, not just
Newton’s laws of mechanics, hold in any inertial reference
frame:
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Slide 27-46
The Constancy of the Speed of Light
• According to the principle of relativity, Maxwell’s
equations of electromagnetism must be true in every
inertial reference frame.
• Maxwell’s equations predict that electromagnetic waves
travel at a speed c = 3.00 × 108 m/s.
• Therefore, light travels at speed c in all inertial
reference frames.
• This implies that all experimenters, regardless of how they
move with respect to each other, find that all light waves,
regardless of their source, travel in their reference frame
with the same speed c.
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Slide 27-47
The Constancy of the Speed of Light
Light travels at speed c in all inertial reference frames,
regardless of how the reference frames are moving with
respect to the light source.
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Slide 27-48
The Constancy of the Speed of Light
• Recent experiments to measure the speed of light in
different reference frames use the unstable elementary
particles called π mesons that decay into high-energy
photons, or particles of light.
• The π mesons are created in a particle accelerator and move
at the velocity .99975c. They emit photons at speed c in
their reference frame.
• You’d expect the photon to travel in the laboratory’s
reference frame at c + .99975c = 1.99975c. Instead, the
photon is measured to travel at 3.00 × 108 m/s.
• In every experiment, we have found that light travels at
speed c, regardless of how the reference frames are moving.
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Slide 27-49
The Constancy of the Speed of Light
Experiments find that the photons travel through
the laboratory with speed c, not the speed 1.99975c
that you might expect.
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Slide 27-50
How Can This Be?
• Suppose reference frame S is moving relative to frame S.
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Slide 27-51
How Can This Be?
• As the ray of light moves from Dan to Eric, they measure
it having traveled a distance Δx. Laura will measure a
longer distance Δx, simply because Eric is moving to the
right, so as seen by Laura, the ray has to travel farther to
reach him.
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Slide 27-52
How Can This Be?
• The definition of velocity is v = Δx/Δt. In order for the
speed of light to be measured as c in both frames, the time
Δt as measured by Laura cannot be the same amount of
time as measured by Dan and Eric Δt.
• This means that our assumptions for the nature of time
must be reevaluated.
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Slide 27-53
QuickCheck 27.5
Race rocket driver Suzzy is heading down the final
straightaway, approaching the finish line at 1.0  108 m/s.
Her fans, at the finish line straight ahead, are shooting laser
beams toward her. The speed of light is 3.0  108 m/s. How
fast are the light waves approaching in Suzzy’s reference
frame?
A.
B.
C.
D.
E.
1.0  108 m/s
2.0  108 m/s
3.0  108 m/s
4.0  108 m/s
Light waves don’t travel in Suzzy’s reference frame.
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Slide 27-54
QuickCheck 27.5
Race rocket driver Suzzy is heading down the final
straightaway, approaching the finish line at 1.0  108 m/s.
Her fans, at the finish line straight ahead, are shooting laser
beams toward her. The speed of light is 3.0  108 m/s. How
fast are the light waves approaching in Suzzy’s reference
frame?
A.
B.
C.
D.
E.
1.0  108 m/s
2.0  108 m/s
3.0  108 m/s
4.0  108 m/s
Light waves don’t travel in Suzzy’s reference frame.
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Slide 27-55
Section 27.4 Events and Measurements
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Events
• The fundamental
element of relativity is
called an event. An
event is a physical
activity that takes place
at a definite point in
space and at a definite
time.
• Events can be observed
and measured by
experimenters in different reference frames.
• Spacetime coordinates are defined by four letters: x, y, z,
and t.
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Slide 27-57
Measurements
• We can measure an event using a two-part measurement
scheme:
• The (x, y, z) coordinates of an event are determined by the
intersection of the meter sticks closest to the event.
• The event’s time t is the time displayed on the clock nearest
to the event.
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Slide 27-58
Measurements
• Several important issues need to be noted:
1. The clocks and meter sticks in each reference frame are
imaginary, so they have no difficulty passing through
each other.
2. Measurements of position and time made in one
reference frame must use only the clocks and meter sticks
in that reference frame.
3. There’s nothing special about the sticks being 1 m long
and the clocks 1 m apart. The lattice spacing can be
altered to achieve whatever level of measurement
accuracy is desired.
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Slide 27-59
Measurements (cont.)
4. We’ll assume that the experimenters in each reference
frame have assistants sitting beside every clock to record
the position and time of nearby events.
5. Perhaps most important, t is the time at which the event
actually happens, not the time at which an experimenter
sees the event or at which information about the event
reaches an experimenter.
6. All experimenters in one reference frame agree on the
spacetime coordinates of an event. In other words, an
event has a unique set of spacetime coordinates in each
reference frame.
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Slide 27-60
QuickCheck 27.6
A firecracker explodes high overhead. You notice a slight
delay between seeing the flash and hearing the boom. At
what time does the event “firecracker explodes” occur?
A.
B.
C.
D.
E.
At the instant you hear the boom
At the instant you see the flash
Very slightly before you see the flash
Very slightly after you see the flash
There’s no unique answer because it depends on the
observer.
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Slide 27-61
QuickCheck 27.6
A firecracker explodes high overhead. You notice a slight
delay between seeing the flash and hearing the boom. At
what time does the event “firecracker explodes” occur?
A.
B.
C.
D.
E.
At the instant you hear the boom
At the instant you see the flash
Very slightly before you see the flash
Very slightly after you see the flash
There’s no unique answer because it depends on the
observer.
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Slide 27-62
Clock Synchronization
• It is important that all clocks in a reference frame are
synchronized, meaning that all the clocks in the reference
frame have the same reading at any one instant of time.
• We could use a master clock that would be used in every
lattice of the different reference frames, but because the
master clock would move in some reference frames, we
cannot assume it will keep track of time the same way as
stationary clocks.
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Slide 27-63
Clock Synchronization
• In order to synchronize clocks in moving reference frames,
we can utilize the clock’s distance from the origin.
Because the speed of light is known, we can calculate
exactly how long light will take to travel the distance from
the origin to the location of each clock.
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Slide 27-64
Clock Synchronization
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Slide 27-65
Events and Observations
• During an event, t is the time when the event actually
happens.
• Light takes time to travel, so the event is observed by an
experimenter at a later time when the light waves reaches
the observer’s eyes.
• Our interest is in the time of the event itself, not when the
observer sees it.
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Slide 27-66
QuickCheck 27.7
Firecrackers A and B are 600 m apart. Sam is standing
halfway between them. Suzzy is standing 300 m on the
other side of firecracker A (and thus 900 m from firecracker
B). Sam sees two flashes, from the two explosions, at
exactly the same instant. According to Suzzy, firecracker A
explodes ______ firecracker B.
A. Before
B. At the same time as
C. After
© 2015 Pearson Education, Inc.
Slide 27-67
QuickCheck 27.7
Firecrackers A and B are 600 m apart. Sam is standing
halfway between them. Suzzy is standing 300 m on the
other side of firecracker A (and thus 900 m from firecracker
B). Sam sees two flashes, from the two explosions, at
exactly the same instant. According to Suzzy, firecracker A
explodes ______ firecracker B.
A. Before
B. At the same time as
C. After
© 2015 Pearson Education, Inc.
She won’t see the flashes at the same
time, but when you see an event is
not the same as when the event
occurred.
Slide 27-68
Example 27.2 Finding the time of an event
Experimenter A in reference frame S stands at the origin
looking in the positive x-direction. Experimenter B stands at
x = 900 m looking in the negative x-direction. A firecracker
explodes somewhere between them. Experimenter B sees
the light flash at t = 3.00 s. Experimenter A sees the light
flash at t = 4.00 s. What are the spacetime coordinates of
the explosion?
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Slide 27-69
Example 27.2 Finding the time of an event
(cont.)
The two experimenters observe light flashes at two
different instants, but there’s only one event. Light travels at
300 m/s, so the additional 1.00 s needed for the light to
reach experimenter A implies that distance (x – 0 m) from x
to A is 300 m longer than distance (900 m – x) from B to x;
that is,
SOLVE
(x – 0 m) = (900 m – x) + 300 m
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Slide 27-70
Example 27.2 Finding the time of an event
(cont.)
This is easily solved to give x = 600 m as the position
coordinate of the explosion. The light takes 1.00 s to travel
300 m to experimenter B and 2.00 s to travel 600 m to
experimenter A. The light is received at 3.00 s and 4.00 s,
respectively; hence it was emitted by the explosion at
t = 2.00 s. The spacetime coordinates of the explosion are
(600 m, 0 m, 0 m, 2.00 s).
© 2015 Pearson Education, Inc.
Slide 27-71
Simultaneity
• Events are said to be simultaneous if they take place at
different positions x1 and x2, but at the same time t1 = t2.
• In general, simultaneous events are not seen at the same
time because of the difference in light travel times from
the event to an experimenter.
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Slide 27-72
Example 27.3 Are the explosions
simultaneous?
An experimenter in reference frame S stands at the origin
looking in the positive x-direction. At t = 3.0 s she sees
firecracker 1 explode at x = 600 m. A short time later, at
t = 5.0 s, she sees firecracker 2 explode at x = 1200 m. Are
the two explosions simultaneous? If not, which firecracker
exploded first?
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Slide 27-73
Example 27.3 Are the explosions
simultaneous? (cont.)
Light from both explosions travels toward the
experimenter at 300 m/s.
PREPARE
The experimenter sees two different explosions, but
perceptions of the events are not the events themselves.
When did the explosions actually occur? Using the fact that
light travels at 300 m/s, it’s easy to see that firecracker 1
exploded at t1 = 1.0 s and firecracker 2 also exploded at
t2 = 1.0 s. The event are simultaneous.
SOLVE
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Slide 27-74
QuickCheck 27.8
Peggy is standing at the center of a railroad car as it passes
Ryan. Firecrackers A and B at the ends of the car explode. A
short time later, flashes from the two explosions reach
Peggy at the same instant. In Peggy’s reference firecracker
A explodes ___ firecracker B.
A. Before
B. At the same time as
C. After
© 2015 Pearson Education, Inc.
Slide 27-75
QuickCheck 27.8
Peggy is standing at the center of a railroad car as it passes
Ryan. Firecrackers A and B at the ends of the car explode. A
short time later, flashes from the two explosions reach
Peggy at the same instant. In Peggy’s reference firecracker
A explodes ___ firecracker B.
A. Before
B. At the same time as
C. After
© 2015 Pearson Education, Inc.
Slide 27-76
QuickCheck 27.9
Peggy is standing at the center of a railroad car as it passes
Ryan. Firecrackers A and B at the ends of the car explode. A
short time later, flashes from the two explosions reach
Peggy at the same instant. In Ryan’s reference firecracker A
explodes ___ firecracker B.
A. Before
B. At the same time as
C. After
© 2015 Pearson Education, Inc.
Slide 27-77
QuickCheck 27.9
Peggy is standing at the center of a railroad car as it passes
Ryan. Firecrackers A and B at the ends of the car explode. A
short time later, flashes from the two explosions reach
Peggy at the same instant. In Ryan’s reference firecracker A
explodes ___ firecracker B.
A. Before
B. At the same time as
C. After
© 2015 Pearson Education, Inc.
Ryan has to agree that the flashes reach Peggy
simultaneously because their arrivals could be
measured with detectors. In Ryan’s frame,
Peggy is moving away from the point in space
where A exploded, and toward B. Firecracker
A had to explode first if the light waves from
A are to reach Peggy at the same instant as
light waves from B.
Slide 27-78
Section 27.5 The Relativity of Simultaneity
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The Relativity of Simultaneity
• We begin our investigation of the nature of time with a
thought experiment similar to one suggested by Einstein.
• Imagine Peggy is standing in the center of a long railroad
car traveling at a velocity that is
an appreciable fraction of the
speed of light.
• A firecracker is attached to
each end of the car and will
leave a mark on the ground
when it explodes.
© 2015 Pearson Education, Inc.
Slide 27-80
The Relativity of Simultaneity
• In the thought experiment, Peggy has a box at her feet
with two light detectors and a signal on top. Each light
detector is pointed to one of the firecrackers.
• Ryan is standing on the ground
as the railroad car passes by.
© 2015 Pearson Education, Inc.
Slide 27-81
The Relativity of Simultaneity
• When the firecrackers go off, if a flash of light is received
by the detector facing the right firecracker (as seen by
Ryan) before a flash is received by the left detector, then
the light on the top of the box turns green.
• If the left detector receives
the signal from the flash
before or at the same time
as the right detector, then
the light turns red.
© 2015 Pearson Education, Inc.
Slide 27-82
The Relativity of Simultaneity
• The fireworks explode as the railroad car passes Ryan. He
sees the two light flashes simultaneously.
• He measures his distance to the burn marks and finds he is
equidistant from where the
explosions occurred.
• Because light travels equal
distances in equal times, he
concludes that the explosions
were simultaneous.
© 2015 Pearson Education, Inc.
Slide 27-83
The Relativity of Simultaneity
• From Ryan’s perspective, the light from the right
firecracker is detected first, so the signal on the box turns
green.
© 2015 Pearson Education, Inc.
Slide 27-84
The Relativity of Simultaneity
• In Peggy’s reference frame, Ryan is moving to the left
with velocity v.
• If the explosions were simultaneous, then the light waves
should travel at a velocity c and reach Peggy at the same
time, since she is directly between the fireworks.
• In Peggy’s reference frame, then, the detectors would
receive signals at the same time and the light on top of the
box would be red.
© 2015 Pearson Education, Inc.
Slide 27-85
The Relativity of Simultaneity
© 2015 Pearson Education, Inc.
Slide 27-86
The Relativity of Simultaneity
• According to Peggy, the light on the box turns red, but
according to Ryan it turns green. It cannot be both!
• We know with certainty:
• Ryan detected the flashes simultaneously.
• Ryan was halfway between the firecrackers when they
exploded.
• The light from the two explosions traveled toward Ryan at
equal speeds.
• Peggy made the assumption that the explosions were
simultaneous. Peggy has a different set of clocks than
Ryan that correspond with her reference frame.
© 2015 Pearson Education, Inc.
Slide 27-87
The Relativity of Simultaneity
• The fact that tR = tL in Ryan’s
reference frame S does not
mean that they are equal in
Peggy’s frame S.
• The right firecracker must
explode before the left
firecracker in frame S.
© 2015 Pearson Education, Inc.
Slide 27-88
The Relativity of Simultaneity
• One of the most disconcerting conclusions of relativity is
that two events occurring simultaneously in reference
frame S are not simultaneous in any reference frame S
that is moving relative to S.
• This is called the relativity of simultaneity.
© 2015 Pearson Education, Inc.
Slide 27-89
Example Problem
Ann and Bill are standing 1200 m apart. A firecracker
explodes 900 m from Ann, and she sees the light flash at
t = 5.0 μs.
A. At what time did the explosion occur? (Use c = 300 m/μs.)
B. Are “sees flash” and “firecracker explodes” the same
event? If not, which is more significant?
C. At what time does Bill see the flash?
© 2015 Pearson Education, Inc.
Slide 27-90
Example Problem
Ann and Bill are still standing 1200 m apart, and
firecrackers explode 300 m on either side of Bill (900 m and
1500 m from Ann). Ann sees the two flashes at the same
time.
A. According to Ann, were the two explosions
simultaneous?
B. According to Bill, were the two explosions simultaneous?
© 2015 Pearson Education, Inc.
Slide 27-91
Example Problem
Two volcanoes, Mt. Newton and Mt. Einstein, are 600 km
apart. You are at rest exactly halfway between the volcanoes
and your friend is at rest at the base of Mt. Newton. Both
volcanoes erupt. Your friend, based on measurements she
makes, determines that the two eruptions are simultaneous.
Do you see Mt. Newton erupt first, Mt. Einstein erupt first,
or both erupt at the same instant of time? Explain.
© 2015 Pearson Education, Inc.
Slide 27-92
Section 27.6 Time Dilation
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Time Dilation
• A light clock is a box with height h, a light source at the
bottom, and a mirror at the top.
• The light source emits a short pulse of light that travels to
the mirror and reflects back to a detector.
• The clock
advances one
“tick” each time
the detector receives
a light pulse.
© 2015 Pearson Education, Inc.
Slide 27-94
Time Dilation
• A light clock is at
rest in reference
frame S. It is called
the rest frame.
• S moves relative
to S.
• We define event 1 to be the emission of a light pulse and
event 2 to be the detection of a pulse.
• Experimenters are able to measure when and where these
events occur in their frame.
© 2015 Pearson Education, Inc.
Slide 27-95
Time Dilation
• In frame S, the time interval Δt = t2 – t1 is one tick of the
clock.
• In frame S, the time interval is Δt = t2 – t1.
• In the rest frame, the light goes straight up and straight
down, so the total distance is 2h so one tick is
© 2015 Pearson Education, Inc.
Slide 27-96
Time Dilation
• As seen in frame S, the light clock is moving to the right at
speed v.
• Thus the mirror has moved a distance ½ v(Δt) during the
time ½(Δt) in which the pulse moves to the mirror.
• The light must travel
farther from the source to
the mirror (the diagonal
path) than in the rest frame
for the clock.
© 2015 Pearson Education, Inc.
Slide 27-97
Time Dilation
• The length of the diagonal is easy to calculate because the
speed of light is equal in all inertial frames. The length of the
diagonal is
distance = speed × time = c(½ Δt) = ½ cΔt.
• We can apply the
Pythagorean Theorem to
the right triangle:
© 2015 Pearson Education, Inc.
Slide 27-98
Time Dilation
• We solve for Δt:
• The time interval between the two ticks in frame S is not
the same as in frame S.
© 2015 Pearson Education, Inc.
Slide 27-99
Time Dilation
• It is useful to define β = v/c, the speed as a fraction of the
speed of light.
• Now we relate the time intervals between events in two
reference frames as:
• If reference frame S´ is at rest relative to frame S, then
β = 0 and Δt = Δt. When the frames are moving, they will
measure different time intervals.
© 2015 Pearson Education, Inc.
Slide 27-100
Time Dilation
• We are unaware of the differences in time intervals in our
everyday lives because our typical speeds are much less
than c.
• The differences do affect precise timekeeping and are
important for accurate location measurements with a GPS
receiver.
© 2015 Pearson Education, Inc.
Slide 27-101
Proper Time
• The time interval between two events that occur at the
same position is called the proper time Δτ.
• Only one inertial frame measures the proper time, and it
can do so with a single clock that is present at both events.
• Experimenters in an inertial frame moving with a speed
v = βc relative to the proper-time frame must use two
clocks to measure the time interval: one at the position of
the first event and the other at the position of the second
event.
© 2015 Pearson Education, Inc.
Slide 27-102
Proper Time
• The time interval between two ticks is the shortest in
the reference frame in which the clock is at rest.
• Another way to view this equation is to say that a moving
clock runs slowly compared to an identical clock at
rest.
• The “stretching out” of a time interval is called time
dilation.
© 2015 Pearson Education, Inc.
Slide 27-103
Proper Time
Event 1: Moving clock passes stationary clock A; all clocks
read 0.
© 2015 Pearson Education, Inc.
Slide 27-104
Proper Time
Event 2: Moving clock passes stationary clock B.
© 2015 Pearson Education, Inc.
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QuickCheck 27.10
Peggy passes Ryan at velocity . Peggy and Ryan both
measure the time it takes the railroad car, from one end to
the other, to pass Ryan. The time interval Peggy measures is
____ the time interval Ryan measures.
A. Longer than
B. At the same as
C. Shorter than
© 2015 Pearson Education, Inc.
Slide 27-106
QuickCheck 27.10
Peggy passes Ryan at velocity . Peggy and Ryan both
measure the time it takes the railroad car, from one end to
the other, to pass Ryan. The time interval Peggy measures is
____ the time interval Ryan measures.
A. Longer than
B. At the same as
C. Shorter than
Ryan measures the proper time because both events occur at
the same position in his frame. Time intervals measured in any
other reference frame are longer than the proper time.
© 2015 Pearson Education, Inc.
Slide 27-107
Example 27.4 Journey time from the sun to
Saturn
Saturn is 1.43  1012 m from the sun. A rocket travels along
a line from the sun to Saturn at a constant speed of exactly
0.9c relative to the solar system. How long does the journey
take as measured by an experimenter on earth? As measured
by an astronaut on the rocket?
© 2015 Pearson Education, Inc.
Slide 27-108
Example 27.4 Journey time from the sun to
Saturn (cont.)
Let the solar system be in reference frame S and
the rocket be in reference frame S′ that travels with velocity
v = 0.9c relative to S. Relativity problems must be stated in
terms of events. Let event 1
be “the rocket and the sun
coincide” (the experimenter
on earth says that the rocket
passes the sun; the astronaut
on the rocket says that the
sun passes the rocket) and
event 2 be “the rocket and
Saturn coincide.”
PREPARE
© 2015 Pearson Education, Inc.
Slide 27-109
Example 27.4 Journey time from the sun to
Saturn (cont.)
FIGURE 27.19 shows the two events as seen from the two
reference frames. Notice that the two events occur at the
same position in S′, the position of the rocket.
© 2015 Pearson Education, Inc.
Slide 27-110
Example 27.4 Journey time from the sun to
Saturn (cont.)
The time interval measured in the solar system
reference frame, which includes the earth, is simply
SOLVE
Relativity hasn’t abandoned the basic definition v = Δx/Δt,
although we do have to be sure that Δx and Δt are measured
in just one reference frame and refer to the same two events.
© 2015 Pearson Education, Inc.
Slide 27-111
Example 27.4 Journey time from the sun to
Saturn (cont.)
How are things in the rocket’s reference frame? The two
events occur at the same position in S′. Thus the time
measured by the astronauts is the proper time Δτ between
the two events. We can then use Equation 27.6 with
 = 0.9 to find
© 2015 Pearson Education, Inc.
Slide 27-112
Example 27.4 Journey time from the sun to
Saturn (cont.)
Δt is the time interval from when the rocket actually passes
the sun, as measured by a clock at the sun, until it actually
passes Saturn, as measured by a synchronized clock at
Saturn. The interval between seeing the events from earth,
which would have to allow for light travel times, would be
something other than 5300 s. Δt and Δτ are different because
time is different in two reference frames moving relative to
each other.
© 2015 Pearson Education, Inc.
Slide 27-113
Experimental Evidence
• What evidence is there that clocks moving relative to each
other tell time differently?
• In 1971 an atomic clock was sent around the world on a jet
plane while an identical clock remained in the laboratory.
After the flight, the clock on the plane was 60 ns behind
the laboratory clock, exactly as predicted by general
relativity.
© 2015 Pearson Education, Inc.
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Experimental Evidence
• More evidence of time dilation comes from the number of
unstable particles called muons that are detected at ground
level on the earth. Muons are created at the top of the
atmosphere at a height of 60 km.
• Muons decay with a half-life of 1.5 μs. The decays can be
used as a clock.
• The time for muons to travel to the surface is
approximately 200 μs, so only 1 out of every 1040 muons
should make it to the surface. Instead we detect 1 out of
10.
• The discrepancy is due to time dilation.
© 2015 Pearson Education, Inc.
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Experimental Evidence
• Two events occur as a muon travels to the surface: The
“muon is created”, and the “muon hits the ground.” The
events take place at two different places in the earth’s
reference frame.
• The events occur at the same position in the muon’s
reference frame.
• The time dilated interval Δt = 200 μs in the earth’s
reference frame corresponds to Δt= 5 μs in the muon’s
reference frame.
• This is only 3.3 half lives for the muons, so the fraction of
muons reaching the ground is much higher.
© 2015 Pearson Education, Inc.
Slide 27-116
Experimental Evidence
© 2015 Pearson Education, Inc.
Slide 27-117
The Twin Paradox
• George and Helen are twins. On their 25th birthday, Helen
departs on a starship voyage at a speed of 0.95c to a star
9.5 light years away from Earth and then returns.
• A light year (ly) is the distance light travels in one year.
c = 1 ly/year.
• According to George, the time Helen is away for is
© 2015 Pearson Education, Inc.
Slide 27-118
The Twin Paradox
• Time dilation will cause Helen to age more slowly than
George.
• Helen’s clock is always with her. The clock measures the
proper time:
• So 20 years will have passed for George, while only 6.25
years will have passed for Helen.
• This means George will be 45 years old and Helen will be
31 and 3 months.
© 2015 Pearson Education, Inc.
Slide 27-119
The Twin Paradox
• The paradox occurs when you consider how George
should age relative to Helen’s reference frame.
• Relative to Helen, George and the earth move away from
her and then towards her at 0.95c. Helen should expect the
earth’s clock to run slowly and that when she returns she
will be older than her twin brother.
• Who is right? We assumed that George and Helen’s
experiences were symmetrical, but they are not. Helen
accelerates from the earth and is therefore not in an inertial
time frame. The situation is not symmetrical.
© 2015 Pearson Education, Inc.
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The Twin Paradox
• There is no paradox.
• The calculation done on earth is correct because it is in an
inertial reference frame throughout Helen’s journey, while
Helen was not always in an inertial reference frame.
• The principle of relativity applies only to inertial
reference frames.
© 2015 Pearson Education, Inc.
Slide 27-121
Section 27.7 Length Contraction
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Length Contraction
• Peggy and Ryan want to measure the length of the train
car.
• Ryan measures the length L as the train moves past him by
measuring the time Δt that it
takes for the car to move
past a fixed cone:
L = vΔt
© 2015 Pearson Education, Inc.
Slide 27-123
Length Contraction
• Peggy is in reference frame S. She measures the length of
the car L by measuring the time Δt it takes for the cone to
move from one end of the car to the other:
L = vΔt
© 2015 Pearson Education, Inc.
Slide 27-124
Length Contraction
• Speed v is the relative speed between S and S and is the same for
both Peggy and Ryan:
• The two events that occur are the front end of the car passing the
cone and the back end passing the cone. In Ryan’s frame, these
events occur at the same position, the cone, so the proper time Δτ is
measured in Ryan’s frame S:
• The length of the car in Ryan’s frame is different from the
length in Peggy’s frame.
© 2015 Pearson Education, Inc.
Slide 27-125
Length Contraction
• Peggy’s frame, however, is the only inertial frame in
which the train car is at rest.
• The length of an object measured in the reference frame in
which the object is at rest is called the proper length :
• The length of an object is greatest in the reference frame in
which the object is at rest.
• The “shrinking” of the length of an object or the distance
between two objects is called length contraction.
© 2015 Pearson Education, Inc.
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QuickCheck 27.11
Peggy passes Ryan at velocity . Peggy and Ryan both
measure the length of the railroad car, from one end to the
other. The length Peggy measures is ____ the length Ryan
measures.
A. Longer than
B. The same as
C. Shorter than
© 2015 Pearson Education, Inc.
Slide 27-127
QuickCheck 27.11
Peggy passes Ryan at velocity . Peggy and Ryan both
measure the length of the railroad car, from one end to the
other. The length Peggy measures is ____ the length Ryan
measures.
A. Longer than
B. The same as
C. Shorter than
Peggy measures the proper length because the railroad car is
at rest in her frame. Lengths measured in any other reference
frame are shorter than the proper length.
© 2015 Pearson Education, Inc.
Slide 27-128
Example 27.5 Length contraction of a ladder
Dan holds a 5.0-m-long ladder parallel to the ground. He
then gets up to a good sprint, eventually reaching 98% of
the speed of light. How long is the ladder according to Dan,
once he is running, and according to Carmen, who is
standing on the ground as Dan goes by?
© 2015 Pearson Education, Inc.
Slide 27-129
Example 27.5 Length contraction of a ladder
(cont.)
We can find the length of the ladder in Carmen’s
frame from Equation 27.12. We have
SOLVE
The length of the moving ladder as measured by
Carmen is only one-fifth its length as measured by Dan.
These lengths are different because space is different in two
reference frames moving relative to each other.
ASSESS
© 2015 Pearson Education, Inc.
Slide 27-130
The Binomial Approximation
• A useful mathematical tool is the binomial
approximation.
• If x is much less than 1, we can approximate:
(1 + x)n ≈ 1 + nx
if
x << 1
• The binomial approximation is very useful when we need
to calculate relativistic expression for a speed much less
than c, so v << c. Because β = v/c, a reference frame
moving with v2/c2 << 1 has β << 1:
© 2015 Pearson Education, Inc.
Slide 27-131
Example 27.6 The shrinking school bus
An 8.0-m-long school bus drives past at 30 m/s. By how
much is its length contracted?
The school bus is at rest in an inertial reference
frame S′ moving at velocity v = 30 m/s relative to the
ground frame S. The given length, 8.0 m, is the proper
length in frame S′.
PREPARE
© 2015 Pearson Education, Inc.
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Example 27.6 The shrinking school bus (cont.)
• Solve In frame S, the school bus is length-contracted to
• The bus’s speed v is much less than c, so we can use the
binomial approximation to write
© 2015 Pearson Education, Inc.
Slide 27-133
Example 27.6 The shrinking school bus (cont.)
The amount of the length contraction is
where 1 fm = 1 femtometer = 10–15 m.
© 2015 Pearson Education, Inc.
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Example 27.6 The shrinking school bus (cont.)
The difficulty is that the difference between and L shows
up only in the 14th decimal place. A scientific calculator
determines numbers to 10 or 12 decimal places, but that
isn’t sufficient to show the difference. The binomial
approximation provides an invaluable tool for finding the
very tiny difference between two numbers that are nearly
identical.
© 2015 Pearson Education, Inc.
Slide 27-135
Section 27.8 Velocities of Objects in Special
Relativity
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Velocities of Objects in Special Relativity
• The Galilean transformation of velocity needs to be
modified for objects moving at relativistic speeds.
• An object’s velocity measured in frame S is related to its
velocity measured in frame S by the Lorentz velocity
transformation:
• When either u or v is much less than c, the denominator is
~1 and therefore agrees with the Galilean transformations.
© 2015 Pearson Education, Inc.
Slide 27-137
Example 27.7 A speeding bullet
A rocket flies past the earth at precisely 0.9c. As it goes by,
the rocket fires a bullet in the forward direction at precisely
0.95c with respect to the rocket. What is the bullet’s speed
with respect to the earth?
The rocket and the earth are inertial reference
frames. Let the earth be frame S and the rocket be frame S′.
The velocity of frame S′ relative to frame S is v = 0.9c. The
bullet’s velocity in frame S′ is u′ = 0.95c.
PREPARE
© 2015 Pearson Education, Inc.
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Example 27.7 A speeding bullet (cont.)
SOLVE
We can use the Lorentz velocity transformation to
find
The bullet’s speed with respect to the earth is 99.7% of the
speed of light.
▶ Many relativistic calculations are much easier
when velocities are specified as a fraction of c. ◀
NOTE
© 2015 Pearson Education, Inc.
Slide 27-139
QuickCheck 27.12
Sam flies past earth at 0.75c. As he goes by, he fires a bullet
forward at 0.75c. Suzzy, on the earth, measures the bullet’s
speed to be
A.
B.
C.
D.
E.
1.5c
c
Between 0.75c and c
0.75c
Less than 0.75c
© 2015 Pearson Education, Inc.
Slide 27-140
QuickCheck 27.12
Sam flies past earth at 0.75c. As he goes by, he fires a bullet
forward at 0.75c. Suzzy, on the earth, measures the bullet’s
speed to be
A.
B.
C.
D.
E.
1.5c
c
Between 0.75c and c
0.75c
Less than 0.75c
© 2015 Pearson Education, Inc.
Slide 27-141
Example Problem
The earth is 1.5 × 1011 m from the sun. An alien spaceship
crosses the distance in 4.0 minutes, as measured by the crew
on the spaceship. How long does the passage take according
to the earthly astronomers who are tracking the spaceship?
© 2015 Pearson Education, Inc.
Slide 27-142
Section 27.9 Relativistic Momentum
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Relativistic Momentum
• In Newtonian physics, the total momentum of a system is
a conserved quantity.
• If we use Lorentz transformations, we see Newtonian
momentum p = mu is not conserved in a frame moving
relative to a frame in which momentum is conserved.
• Momentum conservation is a central and important feature
of mechanics, so it seems likely it will hold in relativity as
well.
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Relativistic Momentum
• A relativistic analysis of particle collisions shows that
momentum conservation does hold, provided we redefine
the momentum of a particle as
• This reduces to the classical momentum p = mu when the
particle’s speed u << c.
• We define the quantity
© 2015 Pearson Education, Inc.
Slide 27-145
Example 27.8 Momentum of a subatomic
particle
Electrons in a particle accelerator reach a speed of 0.999c
relative to the laboratory. One collision of an electron with
a target produces a muon that moves forward with a speed
of 0.950c relative to the laboratory. The muon mass is
1.90  10–28 kg. What is the muon’s momentum in the
laboratory frame and in the frame of the electron beam?
© 2015 Pearson Education, Inc.
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Example 27.8 Momentum of a subatomic
particle (cont.)
SOLVE
 for the muon in the laboratory reference frame is
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Example 27.8 Momentum of a subatomic
particle (cont.)
Thus the muon’s momentum in the laboratory is
© 2015 Pearson Education, Inc.
Slide 27-148
Example 27.8 Momentum of a subatomic
particle (cont.)
The momentum is a factor of 3.2 larger than the Newtonian
momentum mu. To find the momentum in the electron-beam
reference frame, we must first use the velocity
transformation equation to find the muon’s velocity in
frame S′:
© 2015 Pearson Education, Inc.
Slide 27-149
Example 27.8 Momentum of a subatomic
particle (cont.)
In the laboratory frame, the faster electrons are overtaking
the slower muon. Hence the muon’s velocity in the electronbeam frame is negative. ′ for the muon in frame S′ is
© 2015 Pearson Education, Inc.
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Example 27.8 Momentum of a subatomic
particle (cont.)
The muon’s momentum in the electron-beam reference
frame is
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The Cosmic Speed Limit
• For a Newtonian particle with p = mu, the momentum is
directly proportional to the velocity.
• The relativistic expression for momentum agrees with the
Newtonian value if u << c, but p approaches ∞ as u
approaches c.
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The Cosmic Speed Limit
• From the impulse-momentum theorem we know
Δp = mu = Ft.
• If Newtonian physics were correct, the velocity of a
particle would increase without limit.
• We see from relativity
that the particle’s
velocity approaches c
as the momentum
approaches ∞.
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The Cosmic Speed Limit
• The speed c is the “cosmic speed limit” for material
particles.
• A force cannot accelerate a particle to a speed higher than
c because the particle’s momentum becomes infinitely
large as the speed approaches c.
• The amount of effort required for each additional
increment of velocity becomes larger and larger until no
amount of effort can raise the velocity any higher.
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The Cosmic Speed Limit
• At a fundamental level, c is the speed limit for any kind of
causal influence.
• A causal influence can be any kind of particle, wave, or
information that travels from A to B and allows A to be the
cause of B.
• For two unrelated events, the relativity of simultaneity
tells us that in one reference frame, A could happen before
B, but in another reference frame, B could happen before
A.
© 2015 Pearson Education, Inc.
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The Cosmic Speed Limit
• For two causally related events —A causes B—it would be
nonsense for an experimenter in any reference frame to
find that B occurs before A.
• According to relativity, a causal influence traveling faster
than the speed of light could result in B causing A, a
logical absurdity.
• Thus, no causal events of any kind—a particle, wave, or
other influence—can travel faster than c.
© 2015 Pearson Education, Inc.
Slide 27-156
Section 27.10 Relativistic Energy
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Relativistic Energy
• Space, time, velocity, and momentum are changed by
relativity, so it seems inevitable that we’ll need a new view
of energy.
• One of the most profound results of relativity is the
fundamental relationship between energy and mass.
• Einstein found that the total energy of an object of m
mass moving at speed u is
© 2015 Pearson Education, Inc.
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Relativistic Energy
• Let’s examine the behavior of objects traveling at speeds
much less than the speed of light.
• We use the binomial approximation to find
• For low speeds, u, the object’s total energy is then
• The second term is the Newtonian kinetic energy.
• The additional term is the rest energy given by
© 2015 Pearson Education, Inc.
Slide 27-159
Example 27.9 The rest energy of an apple
What is the rest energy of a 200 g apple?
SOLVE
From Equation 27.21 we have
E0 = mc2 = (0.20 kg)(3.0  108 m/s)2 = 1.8  1016 J
This is an enormous energy, enough to power a
medium-sized city for about a year.
ASSESS
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Relativistic Energy
• For high speeds, we must use the full expression for
energy.
• We can find the relativistic expression for kinetic energy K
by subtracting the rest energy E0 from the total energy:
• Thus we can write the total energy of an object of mass m
as
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Example 27.10 Comparing energies of a ball
and an electron
Calculate the rest energy and the kinetic energy of (a)
a 100 g ball moving with a speed of 100 m/s and (b) an
electron with a speed of 0.999c.
The ball, with u << c, is a classical particle. We
don’t need to use the relativistic expression for its kinetic
energy. The electron is highly relativistic.
PREPARE
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Example 27.10 Comparing energies of a ball
and an electron (cont.)
SOLVE
a. For the ball, with m = 0.100 kg,
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Example 27.10 Comparing energies of a ball
and an electron (cont.)
b. For the electron, we start by calculating
Then, using me = 9.11  10–31 kg,
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The Equivalence of Mass and Energy
• Now we are ready to explore the significance of Einstein’s
famous equation E = mc2.
• When a high-energy electron collides with an atom in the
target material, it can knock one electron out of the atom.
• Thus we would expect to see two electrons: the
incident electron and the ejected electron.
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The Equivalence of Mass and Energy
• Instead of two electrons, four particles emerge from the
target: three electrons and a positron.
• A positron is the antimatter of an electron. It is identical to
the electron in all respects other than having a charge
q = +e. The positron has the same mass me as an
electron.
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The Equivalence of Mass and Energy
• In chemical-reaction notation, the collision is
• The electron and positron appear to have been created out
of nothing.
• Although the mass increased, it was not “out of nothing”:
The new particles were created out of energy.
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The Equivalence of Mass and Energy
• Not only can particles be created out of energy, particles
can return to energy.
• When a particle and an antiparticle meet, they annihilate
each other.
• The mass disappears, and the energy equivalent of the
mass is transformed into two high-energy photons.
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Conservation of Energy
• Neither mass nor the Newtonian definition of energy is
conserved, however the total energy—the kinetic energy
and the energy equivalent of mass—remains a conserved
quantity.
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Conservation of Energy
• The most well-known application of the conservation of
total energy is nuclear fission.
• The Uranium isotope 236U, containing 236 protons and
neutrons, does not exist in nature. It can be created when a
235U nucleus absorbs a neutron, increasing its atomic mass.
• The 236U nucleus quickly fragments into two smaller
nuclei and several extra neutrons in a process called
nuclear fission. One way it fissions is
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Conservation of Energy
• The mass after the 236U fission is
0.186 u less than the mass before
the fission.
• The mass has been lost, but the
equivalent energy of the mass has
not. It has been converted to
kinetic energy:
ΔK = mlostc2
• The energy released from one
fission is small, but the energy
from all the nuclei fission is
enormous.
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Conservation of Energy
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QuickCheck 27.13
An electron has rest energy 0.5 MeV. An electron traveling
at 0.968c has p  4. The electron’s kinetic energy is
A.
B.
C.
D.
E.
1.0 MeV
1.5 MeV
2.0 MeV
4.0 MeV
I would need my calculator and several minutes to figure
it out.
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QuickCheck 27.13
An electron has rest energy 0.5 MeV. An electron traveling
at 0.968c has p  4. The electron’s kinetic energy is
A.
B.
C.
D.
E.
1.0 MeV
1.5 MeV K = (p – 1)E0
2.0 MeV
4.0 MeV
I would need my calculator and several minutes to figure
it out.
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QuickCheck 27.14
A proton has rest energy 938 MeV.
A proton and an antiproton are each
traveling at the same slow (p 1) speed in opposite
directions. They collide and annihilate. What is the
outcome? Each
is a photon.
A.
B.
C.
D. A or C
E. All are possible outcomes
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QuickCheck 27.14
A proton has rest energy 938 MeV.
A proton and an antiproton are each
traveling at the same slow (p 1) speed in opposite
directions. They collide and annihilate. What is the
outcome? Each
is a photon.
A.
B.
D. A or C
E. All are possible outcomes
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C.
The outcome must conserve
both energy (1876 MeV) and
momentum (0).
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Example Problem
Through what potential difference must an electron be
accelerated to reach a speed 99% of the speed of light? The
mass of an electron is 9.11 × 10−31 kg.
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Summary: General Principles
Text: p. 901
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Summary: Important Concepts
Text: p. 901
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Summary: Important Concepts
Text: p. 901
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Summary: Important Concepts
Text: p. 901
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Summary: Important Concepts
Text: p. 901
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Summary: Important Concepts
Text: p. 901
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Summary: Important Concepts
Text: p. 901
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Summary: Applications
Text: p. 901
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Summary: Applications
Text: p. 901
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Summary: Applications
Text: p. 901
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Summary
Text: p. 901
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Summary
Text: p. 901
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Summary
Text: p. 901
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