Lecture Presentation Chapter 27 Relativity

Chapter 27 Relativity Chapter Goal: To understand how Einstein s theory of relativity changes our concepts of time and space. 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. 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. 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 = mc 2 shows that mass and energy are essentially equivalent. Slide 27-5

Chapter 27 Preview Looking Ahead Text: p. 874 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. 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 = mc 2. Slide 27-8

Reading Question 27.2 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 27-10

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. Slide 27-12

Section 27.1 Relativity: What s It All About?

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. 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. Slide 27-16

Section 27.2 Galilean Relativity

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 27-18

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. 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. Slide 27-20

Reference Frames Two reference frames, S and S, are in relative motion. 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. 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. 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. 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. 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. Slide 27-26

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 27-27

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 27-29

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. 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. 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. 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? 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 = u 1 v = 340 m/s 200 m/s = 540 m/s u 2 = u 2 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 27-34

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. 6.0 m/s B. 2.0 m/s C. 2.0 m/s D. 6.0 m/s Slide 27-35

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. 9.0 m/s B. 1.0 m/s C. 0.0 m/s D. 1.0 m/s Slide 27-37

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 27-39

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? Slide 27-41

Section 27.3 Einstein s Principle of Relativity

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 27-43

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. 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. 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: 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 10 8 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 27-47

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. 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 10 8 m/s. In every experiment, we have found that light travels at speed c, regardless of how the reference frames are moving. 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. Slide 27-50

How Can This Be? Suppose reference frame S is moving relative to frame S. 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. 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. Slide 27-53

QuickCheck 27.5 Race rocket driver Suzzy is heading down the final straightaway, approaching the finish line at 1.0 10 8 m/s. Her fans, at the finish line straight ahead, are shooting laser beams toward her. The speed of light is 3.0 10 8 m/s. How fast are the light waves approaching in Suzzy s reference frame? A. 1.0 10 8 m/s B. 2.0 10 8 m/s C. 3.0 10 8 m/s D. 4.0 10 8 m/s E. Light waves don t travel in Suzzy s reference frame. Slide 27-54

Section 27.4 Events and Measurements

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 27-57

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. 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. 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. 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. 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 27-61

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 27-63

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. Slide 27-64

Clock Synchronization 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. 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 Slide 27-67

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 27-69

Example 27.2 Finding the time of an event (cont.) 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 27-70

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). Slide 27-71

Simultaneity Events are said to be simultaneous if they take place at different positions x 1 and x 2, but at the same time t 1 = t 2. 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 27-72

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? Slide 27-73

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 t 1 = 1.0 s and firecracker 2 also exploded at t 2 = 1.0 s. The event are simultaneous. 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 Slide 27-75

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 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

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 27-80

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. 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. 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. 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. 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. Slide 27-85

The Relativity of Simultaneity 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. Slide 27-87

The Relativity of Simultaneity The fact that t R = t L 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 27-88

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. 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? 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? 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. Slide 27-92

Section 27.6 Time Dilation

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 27-94

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. Slide 27-95

Time Dilation In frame S, the time interval Δt = t 2 t 1 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 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. 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: 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. 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. 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. 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. 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. Slide 27-103

Proper Time Event 1: Moving clock passes stationary clock A; all clocks read 0. Slide 27-104

Proper Time Event 2: Moving clock passes stationary clock B. Slide 27-105

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. Slide 27-107

Example 27.4 Journey time from the sun to Saturn Saturn is 1.43 10 12 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 27-108

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. 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. Slide 27-110

Example 27.4 Journey time from the sun to Saturn (cont.) 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 27-111

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 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. 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. Slide 27-114

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 10 40 muons should make it to the surface. Instead we detect 1 out of 10. The discrepancy is due to time dilation. Slide 27-115

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. Slide 27-116

Experimental Evidence Slide 27-117

The Twin Paradox George and Helen are twins. On their 25 th 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 27-118

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. 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. Slide 27-120

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. Slide 27-121

Section 27.7 Length Contraction

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 27-123

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 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. 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. Slide 27-126

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. 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? Slide 27-129

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 27-130

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 v 2 /c 2 << 1 has β << 1: 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? 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 27-132

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 27-133

Example 27.6 The shrinking school bus (cont.) The amount of the length contraction is where 1 fm = 1 femtometer = 10 15 m. Slide 27-134

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. Slide 27-135

Section 27.8 Velocities of Objects in Special Relativity

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 27-137

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? 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 27-138

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 27-139

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. 1.5c B. c C. Between 0.75c and c D. 0.75c E. Less than 0.75c Slide 27-140

Example Problem The earth is 1.5 10 11 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? Slide 27-142

Section 27.9 Relativistic Momentum

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 27-144

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 27-145

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? Slide 27-146

Example 27.8 Momentum of a subatomic particle (cont.) SOLVE for the muon in the laboratory reference frame is Slide 27-147

Example 27.8 Momentum of a subatomic particle (cont.) Thus the muon s momentum in the laboratory is 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 : 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 Slide 27-150

Example 27.8 Momentum of a subatomic particle (cont.) The muon s momentum in the electron-beam reference frame is Slide 27-151

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 27-152

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. Slide 27-153

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. Slide 27-154

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. Slide 27-155

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. Slide 27-156

Section 27.10 Relativistic Energy

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 27-158

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 27-159

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 E 0 = mc 2 = (0.20 kg)(3.0 108 m/s) 2 = 1.8 10 16 J ASSESS This is an enormous energy, enough to power a medium-sized city for about a year. Slide 27-160

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 E 0 from the total energy: Thus we can write the total energy of an object of mass m as Slide 27-161

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 27-162

Example 27.10 Comparing energies of a ball and an electron (cont.) SOLVE a. For the ball, with m = 0.100 kg, Slide 27-163

Example 27.10 Comparing energies of a ball and an electron (cont.) b. For the electron, we start by calculating Then, using m e = 9.11 10 31 kg, Slide 27-164

The Equivalence of Mass and Energy Now we are ready to explore the significance of Einstein s famous equation E = mc 2. 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 27-165

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 m e as an electron. Slide 27-166

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. Slide 27-167

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. Slide 27-168

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 27-169

Conservation of Energy The most well-known application of the conservation of total energy is nuclear fission. The Uranium isotope 236 U, containing 236 protons and neutrons, does not exist in nature. It can be created when a 235 U nucleus absorbs a neutron, increasing its atomic mass. The 236 U nucleus quickly fragments into two smaller nuclei and several extra neutrons in a process called nuclear fission. One way it fissions is Slide 27-170

Conservation of Energy The mass after the 236 U 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 = m lost c 2 The energy released from one fission is small, but the energy from all the nuclei fission is enormous. Slide 27-171

Conservation of Energy Slide 27-172

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 27-173

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 K = ( p 1)E 0 C. 2.0 MeV D. 4.0 MeV E. I would need my calculator and several minutes to figure it out. Slide 27-174

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 Slide 27-175

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. Slide 27-177