Lecture Presentation - Mr Herman's Webpage...Slide 27-25 Reference Frames •Suppose you’re...

Chapter 27

Lecture Presentation

Relativity

© 2015 Pearson Education, Inc.

Slide 27-2

Suggested Videos for Chapter 27 • Prelecture Videos

• Space and Time

• Mass and Energy

• Class Videos

• Mass-Energy Equivalence

• Video Tutor Solutions

• Relativity


Slide 27-3

Suggested Simulations for Chapter 27 • ActivPhysics

• 17.1, 17.2

• PhETs

• Nuclear Fission


Slide 27-4

Chapter 27 Relativity

Chapter Goal: To understand how Einstein’s theory of

relativity changes our concepts of time and space.


Slide 27-5

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.


Slide 27-6

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.


Slide 27-7

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. © 2015 Pearson Education, Inc.

Slide 27-8

Chapter 27 Preview Looking Ahead


Text: p. 874

Slide 27-9

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.


Slide 27-10

Chapter 27 Preview Stop to Think

The car is moving at 10 m/s relative to Bill. How fast does

Amy see the car as moving?

A. 5 m/s

B. 10 m/s

C. 15 m/s

D. 20 m/s


Slide 27-11

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.


Slide 27-12


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.


Slide 27-13


A clock on a moving train runs _____ an identical clock at

rest.

A. Faster than

B. Slower than

C. At the same speed as

D. It depends on which direction the train is moving.


Slide 27-14


A clock on a moving train runs _____ an identical clock at

rest.

A. Faster than

B. Slower than

C. At the same speed as

D. It depends on which direction the train is moving.


Slide 27-15


Which of these topics was not discussed in this chapter?

A. Teleportation

B. Simultaneity

C. Time dilation

D. Length contraction


Slide 27-16


Which of these topics was not discussed in this chapter?

A. Teleportation

B. Simultaneity

C. Time dilation

D. Length contraction


Slide 27-17


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.


Slide 27-18


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.


Slide 27-19


The kinetic energy of a thrown baseball is _____ its rest

energy.

A. Greater than

B. Less than

C. Equal to


Slide 27-20


The kinetic energy of a thrown baseball is _____ its rest

energy.

A. Greater than

B. Less than

C. Equal to


Section 27.1 Relativity: What’s It All About?


Slide 27-22

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 fast-

moving rocket to be shorter and a clock on the rocket to

run slower compared to when the rocket is at rest.


Slide 27-23

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.


Section 27.2 Galilean Relativity


Slide 27-25

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.


Slide 27-26

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.


Slide 27-27

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.


Slide 27-28

Reference Frames

• Two reference frames, S and S, are in relative motion.


Slide 27-29

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.


Slide 27-30


• 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.


Slide 27-31


• 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.


Slide 27-32


• 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.


Slide 27-33


• 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.


Slide 27-34

QuickCheck 27.1

Which is an inertial reference frame (or at least a very good

approximation of one)?

A. A jet plane during takeoff

B. A jet plane flying straight and level at constant speed

C. A jet plane turning at constant speed

D. B and C

E. A, B, and C


Slide 27-35

QuickCheck 27.1

Which is an inertial reference frame (or at least a very good

approximation of one)?

A. A jet plane during takeoff

B. A jet plane flying straight and level at constant speed

C. A jet plane turning at constant speed

D. B and C

E. A, B, and C


Slide 27-36

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.


Slide 27-37


• 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.


Slide 27-38


• 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.


Slide 27-39


• 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.


Slide 27-40

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?


Slide 27-41

Example 27.1 Finding the speed of sound (cont.)

PREPARE Assume that the earth (frame S) and the airplane

(frame S′) are inertial reference frames. Frame S′, in which

the airplane is at rest, moves with velocity v = 200 m/s

relative to frame S. FIGURE 27.4 shows the airplane and

the sound waves.


Slide 27-42


SOLVE The speed of a mechanical wave, such as a sound

wave or a wave on a string, is its speed relative to its

medium. Thus the speed of sound is the speed of a sound

wave through a reference frame in which the air is at rest.

This is reference frame S, where wave 1 travels with

velocity u1 = 340 m/s and wave 2 travels with velocity

u2 = +340 m/s. Notice that the Galilean transformations use

velocities, with appropriate signs, not just speeds.


Slide 27-43


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.


Slide 27-44


ASSESS This isn’t surprising. If you’re driving at 50 mph, a

car coming the other way at 55 mph is approaching you at

105 mph. A car coming up behind you at 55 mph seems to

be gaining on you at the rate of only 5 mph. Wave speeds

behave the same. Notice that a mechanical wave would

appear to be stationary to a person moving at the wave

speed. To a surfer, the crest of the ocean wave remains at

rest under his or her feet.


Slide 27-45

QuickCheck 27.2

Balls 1 and 2, about to collide, have their velocities shown

in reference frame S. What is the velocity u1i of ball 1 in

frame S?

A. –6.0 m/s

B. –2.0 m/s

C. 2.0 m/s

D. 6.0 m/s


Slide 27-46

QuickCheck 27.2



frame S?

A. –6.0 m/s

B. –2.0 m/s

C. 2.0 m/s

D. 6.0 m/s


Slide 27-47

QuickCheck 27.3



frame S?

A. –9.0 m/s

B. –1.0 m/s

C. 0.0 m/s

D. 1.0 m/s


Slide 27-48

QuickCheck 27.3



frame S?

A. –9.0 m/s

B. –1.0 m/s

C. 0.0 m/s

D. 1.0 m/s


1 m/s to the left

Slide 27-49

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. 100 m/s

B. 250 m/s

C. 350 m/s

D. 450 m/s

E. Sound waves don’t travel in Sam’s reference frame.


Slide 27-50

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. 100 m/s

B. 250 m/s

C. 350 m/s

D. 450 m/s

E. Sound waves don’t travel in Sam’s reference frame.


Slide 27-51

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?


Section 27.3 Einstein’s Principle of Relativity


Slide 27-53

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.


Slide 27-54


• 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.


Slide 27-55


It seems as if the speed of light should differ from c in

a reference frame moving through the ether.


Slide 27-56


• 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:


Slide 27-57

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.


Slide 27-58


Light travels at speed c in all inertial reference frames,

regardless of how the reference frames are moving with

respect to the light source.


Slide 27-59


• 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. © 2015 Pearson Education, Inc.

Slide 27-60


Experiments find that the photons travel through

the laboratory with speed c, not the speed 1.99975c

that you might expect. © 2015 Pearson Education, Inc.

Slide 27-61

How Can This Be?

• Suppose reference frame S is moving relative to frame S.


Slide 27-62

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.


Slide 27-63

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.


Slide 27-64

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. 1.0 108 m/s

B. 2.0 108 m/s

C. 3.0 108 m/s

D. 4.0 108 m/s

E. Light waves don’t travel in Suzzy’s reference frame.


Slide 27-65

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. 1.0 108 m/s

B. 2.0 108 m/s

C. 3.0 108 m/s

D. 4.0 108 m/s

E. Light waves don’t travel in Suzzy’s reference frame.


Section 27.4 Events and Measurements


Slide 27-67

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.


Slide 27-68

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.


Slide 27-69

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.


Slide 27-70

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.


Slide 27-71

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. At the instant you hear the boom

B. At the instant you see the flash

C. Very slightly before you see the flash

D. Very slightly after you see the flash

E. There’s no unique answer because it depends on the

observer.


Slide 27-72

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. At the instant you hear the boom

B. At the instant you see the flash

C. Very slightly before you see the flash

D. Very slightly after you see the flash

E. There’s no unique answer because it depends on the

observer.


Slide 27-73

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.


Slide 27-74


• 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.


Slide 27-75



Slide 27-76

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.


Slide 27-77

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


Slide 27-78

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


C. After


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 27-79

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?


Slide 27-80

Example 27.2 Finding the time of an event (cont.)

PREPARE Experimenters A and B are in the same reference

frame and have synchronized clocks. FIGURE 27.12 shows

the two experimenters and the explosion at unknown

position x.


Slide 27-81


SOLVE 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,

(x – 0 m) = (900 m – x) + 300 m


Slide 27-82


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).


Slide 27-83


ASSESS Although the experimenters see the explosion at

different times, they agree that the explosion actually

happened at t = 2.00 s.


Slide 27-84

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.


Slide 27-85

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?


Slide 27-86

Example 27.3 Are the explosions simultaneous? (cont.)

PREPARE Light from both explosions travels toward the

experimenter at 300 m/s.

SOLVE 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.


Slide 27-87

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


C. After


Slide 27-88

QuickCheck 27.8




Peggy at the same instant. In Peggy’s reference firecracker

A explodes ___ firecracker B.

A. Before


C. After


Slide 27-89

QuickCheck 27.9




Peggy at the same instant. In Ryan’s reference firecracker A

explodes ___ firecracker B.

A. Before


C. After


Slide 27-90

QuickCheck 27.9




Peggy at the same instant. In Ryan’s reference firecracker A

explodes ___ firecracker B.

A. Before


C. After


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.

Section 27.5 The Relativity of Simultaneity


Slide 27-92

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.


Slide 27-93


• 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.


Slide 27-94


• 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.


Slide 27-95


• 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.


Slide 27-96


• 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 27-97


• 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.


Slide 27-98



Slide 27-99


• 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.


Slide 27-100


• 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.


Slide 27-101


• 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.


Slide 27-102

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?


Slide 27-103

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?


Slide 27-104

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.


Section 27.6 Time Dilation


Slide 27-106

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.


Slide 27-107

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.


Slide 27-108

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 = t2 – t1.

• In the rest frame, the light goes straight up and straight

down, so the total distance is 2h so one tick is


Slide 27-109

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.


Slide 27-110

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:


Slide 27-111

Time Dilation

• We solve for Δt:

• The time interval between the two ticks in frame S is not

the same as in frame S.


Slide 27-112

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.


Slide 27-113

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.


Slide 27-114

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.


Slide 27-115

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.


Slide 27-116

Proper Time

Event 1: Moving clock passes stationary clock A; all clocks

read 0.


Slide 27-117

Proper Time

Event 2: Moving clock passes stationary clock B.


Slide 27-118

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


Slide 27-119

QuickCheck 27.10


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.

Slide 27-120

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?


Slide 27-121

Example 27.4 Journey time from the sun to Saturn (cont.)

PREPARE 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.” © 2015 Pearson Education, Inc.

Slide 27-122


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.


Slide 27-123


SOLVE The time interval measured in the solar system

reference frame, which includes the earth, is simply

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.


Slide 27-124


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


Slide 27-125


ASSESS The time interval measured between these two

events by the astronauts is less than half the time interval

measured by experimenters on earth. The difference has

nothing to do with when earthbound astronomers see the

rocket pass the sun and Saturn.


Slide 27-126


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.


Slide 27-127

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.


Slide 27-128


• 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.


Slide 27-129


• 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.


Slide 27-130



Slide 27-131

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


Slide 27-132

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.


Slide 27-133

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.


Slide 27-134

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.


Section 27.7 Length Contraction


Slide 27-136

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


Slide 27-137

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


Slide 27-138

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.


Slide 27-139

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.


:

Slide 27-140

QuickCheck 27.11


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


Slide 27-141

QuickCheck 27.11


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.

Slide 27-142

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?


Slide 27-143

Example 27.5 Length contraction of a ladder (cont.)

PREPARE Let reference frame S′ be attached to Dan. The

ladder is at rest in this reference frame, so Dan measures the

proper length of the ladder: = 5.0 m. Dan’s frame S′

moves relative to Carmen’s frame S with velocity v = 0.98c.


Slide 27-144

Example 27.5 Length contraction of a ladder (cont.)

SOLVE We can find the length of the ladder in Carmen’s

frame from Equation 27.12. We have

ASSESS 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.


Slide 27-145

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:


Slide 27-146

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?

PREPARE 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′.


Slide 27-147

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


Slide 27-148


The amount of the length contraction is

where 1 fm = 1 femtometer = 10–15 m.


Slide 27-149


ASSESS The amount the bus “shrinks” is only slightly larger

than the diameter of the nucleus of an atom. It’s no wonder

that we’re not aware of length contraction in our everyday

lives. If you had tried to calculate this number exactly, your

calculator would have shown – L = 0.


Slide 27-150


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.


Section 27.8 Velocities of Objects in Special Relativity


Slide 27-152

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.


Slide 27-153

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?

PREPARE 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.


Slide 27-154

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.

NOTE ▶ Many relativistic calculations are much easier

when velocities are specified as a fraction of c. ◀


Slide 27-155

Example 27.7 A speeding bullet (cont.)

ASSESS The Galilean transformation of velocity would give

u = 1.85c. Now, despite the very high speed of the rocket

and of the bullet with respect to the rocket, the bullet’s

speed with respect to the earth remains less than c. This is

yet more evidence that objects cannot travel faster than the

speed of light.


Slide 27-156

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. 1.5c

B. c

C. Between 0.75c and c

D. 0.75c

E. Less than 0.75c


Slide 27-157

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. 1.5c

B. c

C. Between 0.75c and c

D. 0.75c

E. Less than 0.75c


Slide 27-158

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?


Section 27.9 Relativistic Momentum


Slide 27-160

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.


Slide 27-161

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


Slide 27-162

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?


Slide 27-163

Example 27.8 Momentum of a subatomic particle (cont.)

PREPARE Let the laboratory be reference frame S. The

reference frame S′ of the electron beam (i.e., a reference

frame in which the electrons are at rest) moves in the

direction of the electrons at v = 0.999c. The muon velocity

in frame S is u = 0.95c.


Slide 27-164


SOLVE for the muon in the laboratory reference frame is


Slide 27-165


Thus the muon’s momentum in the laboratory is


Slide 27-166


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′:


Slide 27-167


In the laboratory frame, the faster electrons are overtaking

the slower muon. Hence the muon’s velocity in the electron-

beam frame is negative. ′ for the muon in frame S′ is


Slide 27-168


The muon’s momentum in the electron-beam reference

frame is


Slide 27-169


ASSESS From the laboratory perspective, the muon moves

only slightly slower than the electron beam. But it turns out

that the muon moves faster with respect to the electrons,

although in the opposite direction, than it does with respect

to the laboratory.


Slide 27-170

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.


Slide 27-171


• 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 ∞.


Slide 27-172


• 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.


Slide 27-173


• 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.


Slide 27-174


• 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.


Section 27.10 Relativistic Energy


Slide 27-176

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


Slide 27-177

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


Slide 27-178

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

ASSESS This is an enormous energy, enough to power a

medium-sized city for about a year.


Slide 27-179

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


Slide 27-180

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.

PREPARE 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.


Slide 27-181

Example 27.10 Comparing energies of a ball and an electron (cont.)

SOLVE

a. For the ball, with m = 0.100 kg,


Slide 27-182


b. For the electron, we start by calculating

Then, using me = 9.11 10–31 kg,


Slide 27-183


ASSESS The ball’s kinetic energy is a typical kinetic energy.

Its rest energy, by contrast, is a staggeringly large number.

For a relativistic electron, on the other hand, the kinetic

energy is more important than the rest energy.


Slide 27-184

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.


Slide 27-185


• 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.


Slide 27-186


• 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.


Slide 27-187


• 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.


Slide 27-188

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.


Slide 27-189


• 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


Slide 27-190


• 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.


Slide 27-191



Slide 27-192

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. 1.0 MeV

B. 1.5 MeV

C. 2.0 MeV

D. 4.0 MeV

E. I would need my calculator and several minutes to figure

it out.


Slide 27-193

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. 1.0 MeV

B. 1.5 MeV

C. 2.0 MeV

D. 4.0 MeV

E. I would need my calculator and several minutes to figure

it out.


K = (p – 1)E0

Slide 27-194

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.

D. A or C

E. All are possible outcomes


A. B. C.

Slide 27-195

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.

D. A or C

E. All are possible outcomes


A. B. C.

The outcome must conserve

both energy (1876 MeV) and

momentum (0).

Slide 27-196

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


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Summary: Important Concepts


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Summary: Applications


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Summary


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Summary


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