Time Dilation. For centuries philosophers have debated the concept of time. Does time proceed at the same rate everywhere in the world or, for that ma...

46 downloads 120 Views 2MB Size

11.2

Having introduced the consequences of Einstein’s two postulates by demonstrating the relativity of simultaneity, we now draw a further, still more sweeping conclusion on time and overturn commonsense views on distance.

Time Dilation For centuries philosophers have debated the concept of time. Does time proceed at the same rate everywhere in the world or, for that matter, on the Moon? We know time goes forward. Can it go backward? Despite thoughts to the contrary, it turns out that all time is relative, relative to the observer. There is no such thing as absolute time. To illustrate, let us perform a thought experiment where two observers measure different time intervals for the same sequence of events. A spaceship contains two parallel mirrors (which we call “top” and “bottom”) and a method of sending a pulse of light from the bottom mirror to the top, at right angles to the mirrors (Figure 1(a)). Inside the bottom mirror the astronaut has placed a clock that records a “tick” at the instant the pulse leaves the mirror and a “tock” when the pulse returns.

(a) ∆ts

Figure 1 (a) The astronaut, stationary relative to the clock incorporated in the bottom mirror, measures a time interval 2ts for the light to make a round trip and so infers ts to be the duration of the upward journey. (b) The time interval as measured by an observer on Earth, who records it as tm. (c) The distance triangle. (The stopwatches are drawn to depict the actual time dilation occurring for a spaceship travelling at 0.83c.)

(b)

∆tm

0

observer on Earth

(c)

2∆ t m

∆tm

0 c ∆t m

c ∆ts

v ∆tm

NEL

Einstein’s Special Theory of Relativity 569

LEARNING

TIP

Representing Distance In general, for an object moving d at a constant speed, v t and d = vt. In particular, for d light, c and d = ct. t

To the astronaut, stationary with respect to the clock, the pulse goes up and down (causing the clock to register first a tick, then a tock) whether the spaceship is at rest relative to Earth or moving with a constant high velocity relative to Earth. We write 2ts for the interval, in the inertial frame of the spaceship, separating tick from tock (with “s” for “stationary,” as a reminder that the clock is stationary in the frame we have just chosen). The interval required for light to travel from the bottom mirror to the top is thus ts and the distance between the mirrors is cts. The spaceship is moving with a speed v relative to an observer on Earth. From the observer’s viewpoint the pulse takes a longer interval tm to travel the longer distance ctm, where tm is the time interval, in the frame of Earth, for light to travel from the bottom mirror to the top. We write “m” as a reminder that in this frame, the bottom mirror and top mirror are moving. In the same time that the pulse moves to the mirror, the spaceship moves a distance of vtm relative to the observer on Earth (Figure 1(b)). How can we determine if ts and tm are different? According to Einstein’s second postulate, light has the same speed c for both observers. Using the distance triangle in Figure 1(c), it follows from the Pythagorean theorem that (ctm)2 (vtm)2 (cts )2

We will isolate tm to see how it compares with ts. (ctm)2 (vtm)2 (cts )2 c 2(tm)2

v 2(tm)2 c 2(ts )2

Dividing both sides by c 2: v2 (tm )2 2 (tm )2 (t s )2 c

v2 (tm)2 1 2 c

(t ) s

2

Thus, rearranging to isolate tm2, and then taking the square root of both sides, we get: ts tm v2 1 c2

DID YOU

KNOW

?

Use Gamma As a Shortcut 1 The expression is so v2 1 c2 common in relativity applications that it is given the symbol γ, so the

t s expression tm v2 1 c2 becomes tm γts.

570 Chapter 11

where ts is the time interval for the observer stationary relative to the sequence of events and tm is the time interval for an observer moving with a speed v relative to the sequence of events. This equation shows that, for any v such that 0 v c, tm ts. The time interval 2tm seen by the Earthbound observer, moving relative to the mirrors, must be greater than the corresponding time interval 2ts seen by the observer inside the spaceship, stationary with respect to the mirrors. Another way of looking at this phenomenon is to consider the positions of the events between which the time interval is measured. The initial and final events in our thought experiment are the emission of the light from the bottom mirror (tick) and the detection of the light returning to the bottom mirror (tock). In the frame of reference of the spaceship, tick and tock occur at the same position. In the frame of reference of Earth, tick and tock occur at two positions that are separated by a distance of 2vtm. We can now identify the specific fact underlying the thought experiment: the time interval between events that occur at the same position in a frame of reference is less than the time interval between the same events as measured in any other frame of reference. In summary, “one position time” is less than “two position time” by a factor of NEL

Section 11.2

1 2 1 v c2

(for any speed v such that 0 v c)

The duration of a process (in this example 2ts), as measured by an observer who sees the process begin and end in the same position, is called the proper time. The observation that time on a clock that is moving with respect to an observer is seen to run slower than time on the clock that is stationary with respect to that observer is called time dilation. Intuitively, we may think of time as “dilated” or “expanded” on the spaceship. How can this be, you may ask? Did we not find ts to be less than tm? But dilation, not contraction, is the right description. To the Earthbound observer, it seems that tick and tock are widely separated in time, so it seems that a supposedly swift process has slowed down. This holds for any “clock,” whether it is your watch, a bouncing spring, a pendulum, a beating heart, a dividing cell, or a galactic event. All time is relative to the observer. There is no absolute time. Notice that in the equation above, ts can be a real number only if the expression v2 is positive; that is, 1 – c2 v2 1 0 c2 2 v 1 c2 v c

This is the famous “speed limit” proposed by Einstein. No material object can have a speed that is equal to or greater than that of light. We will address this again in the next section. Time dilation effects have been measured. In 1971, four exceedingly accurate atomic clocks were flown around the world twice, on regularly scheduled passenger flights. The experimental hypothesis was that the clocks would have times that differed from a similar clock at rest at the U.S. Naval Observatory. Because of Earth’s rotation, two trips were made, one eastward and one westward. Using the time dilation relationship, the predicted loss in time should have been 40 ± 23 ns for the eastward journey and 275 ± 21 ns for the westward journey. It was found that the clocks flying eastward lost 59 ± 10 ns, and the clocks flying westward lost 273 ± 7 ns. The theory of time dilation was consequently validated within the expected experimental error. Muons, which we will encounter again in Chapter 13, are subnuclear particles that are similar to electrons but are unstable. Muons decay into less massive particles after an average “life” of 2.2 ms. In 1976, muons were accelerated to 0.9994c in the circular accelerator at CERN, in Switzerland. With a lifetime of 2.2 ms, the muons should last only 14 to 15 trips around the ring before they decay. In fact, because of time dilation, the muons did not decay until after 400 circuits, approximately 30 times longer. The mean life of the muons had increased by 30 times relative to Earth, or we could say that the time elapsed relative to Earth was longer; it was dilated. It’s as if the muons were living life in slow motion. From this viewpoint, the moving muons will exist longer than stationary ones, but “the amount of life” the muons themselves will experience, relative to their own frame of reference, is exactly the same. From the evidence provided by the life of a muon, we could reach the same conclusion for people moving in a spaceship at nearly the speed of light. The passengers would have a life expectancy of hundreds of years, relative to Earth. But from the standpoint of life on the spaceship, it’s life as usual. From our perspective in the inertial frame of Earth, they are living life at a slower rate, and their normal life cycle takes an enormous amount of our time. A sample problem will illustrate this point. NEL

time dilation the slowing down of time in a system, as seen by an observer in motion relative to the system

LEARNING

TIP

Dilation The word “dilation” means “widening.” For example, the pupils in our eyes dilate to let more light in.

DID YOU

v2 c2

and, finally,

proper time (ts ) the time interval between two events measured by an observer who sees the events occur at one position

KNOW

?

Pretty Relative! Einstein said, “When a man sits next to a pretty girl for an hour, it seems like a minute. But let him sit on a hot stove for one minute and it’s much longer than an hour— that’s relativity!”

DID YOU

KNOW

?

General Theory Since the airliners validating time dilation were travelling above the surface of Earth, the gravitational field in flight was lower than at the U.S. Naval Observatory. Thus, the experimenters had to include general relativity as well as special relativity in their calculations.

Einstein’s Special Theory of Relativity 571

SAMPLE problem 1 An astronaut whose pulse frequency remains constant at 72 beats/min is sent on a voyage. What would her pulse beat be, relative to Earth, when the ship is moving relative to Earth at (a) 0.10c and (b) 0.90c?

Solution c 3.0 108 m/s pulse frequency 72 beats/min pulse period at relativistic speed tm ?

The pulse period measured on the spaceship (where pulses occur in the same position) is 1 t s 72 min1 t s 0.014 min

(a) At v = 0.10c, ts

tm

v2 1 c2

0.014 min

(0.10c)2 1 c2

0.014 min 0.995 tm 0.014 min, to 2 significant digists 1 f T 1 0.014 min f 72 beats/min

The pulse frequency at v 0.10c, relative to Earth, is to two significant digits unchanged, at 72 beats/min. (b) At v 0.90c, 0.014 min tm (0.90c)2 1 c2

0.014 min 0.436 tm 0.032 min 1 f T 1 0.032 min f 31 beats/min

The pulse frequency at v = 0.90c, relative to Earth, is 31 beats/min. This result is much lower than the 72 beats/min that the astronaut would measure for herself on the ship.

572 Chapter 11

NEL

Section 11.2

Practice Understanding Concepts Answers

1. Are airline pilots’ watches running slow in comparison with clocks on the

2. 1.2 108 s

ground? Why or why not? 2. A beam of unknown elementary particles travels at a speed of 2.0

108

m/s. Their average lifetime in the beam is measured to be 1.6 108 s. Calculate their average lifetime when at rest.

3. 75.0 h 4. 2

3. A Vulcan spacecraft has a speed of 0.600c with respect to Earth. The Vulcans

determine 32.0 h to be the time interval between two events on Earth. What value would they determine for this time interval if their ship had a speed of 0.940c with respect to Earth? 4. The K+ meson, a subatomic particle, has an average rest lifetime of 1.0 108 s.

If the particle travels through the laboratory at 2.6 108 m/s, by how much has its lifetime, relative to the laboratory, increased?

The Twin Paradox One of Einstein’s most famous thought experiments is the “twin paradox,” which illustrates time dilation. Suppose that one of a pair of young identical twins takes off from Earth and travels to a star and back at a speed approaching c. The other twin remains on Earth. We may possibly expect, in view of our result in Sample Problem 1 concerning an astronaut’s heartbeat, that the moving twin ages less than his Earthbound counterpart does. But wouldn’t the twin in the spaceship think the reverse, seeing his twin on Earth first receding at high speed and then returning? Would he not expect the Earthbound twin to have aged more for the same reason (Figure 2)? Is this not a paradox? No! The consequences of the special theory of relativity only apply in an inertial frame of reference, in this case Earth. The twin on Earth is in the same frame of reference for the other twin’s whole journey. Until now, the situations we have discussed have been symmetrical, with inertial observers moving relative to one another, seeing each other’s clocks run slow. The travelling twin is in a frame of reference whose velocity relative to Earth must change at the turn-around position, which means it is a noninertial frame at that stage. Since the situation is not symmetrical, the observations can be different. It is indeed the twin in the spaceship who returns younger than his twin on Earth.

Figure 2 (a) As the twins depart, they are the same age. (b) When the astronaut twin returns, he has aged less than the twin who stayed on Earth.

Length Contraction Since time, once assumed to be absolute, is perceived differently by different observers, it seems natural to ask if length, another supposedly absolute concept, also changes. Consider a spaceship making a trip from planet A to planet B, at rest with respect to each other and a distance Ls apart (relative to the planets) and at a speed v as measured by an observer on either planet (Figure 3). (Just as ts was defined as the proper time, Ls is called the proper length.) For the captain in the spaceship, the process (the departure of A, the arrival of B) occurs at one single place, the spaceship. The captain can thus assign a proper-time duration ts to the process, using a single clock on the spacecraft. Since she sees A recede and B approach at the speed v, she finds the distance separating the two events to be Lm vts. For observers on the planets, the process (the takeoff from A, the landing on B) occurs at two different places. Such observers can use a pair of synchronized clocks to assign a duration tm to the two-event process consisting of the takeoff and the landing and would find the distance separating the events as Ls vtm. We can now argue as follows: NEL

proper length (Ls ) the length, in an inertial frame, of an object stationary in that frame

LEARNING

TIP

Proper Length Note that “proper length” does not necessarily mean “correct length.”

Einstein’s Special Theory of Relativity 573

L0s

A

B v

ts

tm

Figure 3 Planets A and B are at rest with respect to each other. The chart below summarizes the rotation used in this example: Observer

Distance

Time

stationary observer

Ls

tm

spacecraft observer

Lm

ts

From the time dilation relationship ts tm v2 1 c2

If we multiply both sides by v we obtain vts vtm v2 1 c2

But Lm vts and Ls vtm. Substituting above we obtain Lm Ls v2 1 c2

or

Lm Ls

DID YOU

KNOW

?

The Lorentz Contraction This relativistic change in length is known as the Lorentz contraction, named after the Dutch mathematician H.A. Lorentz, who created the same theoretical correction for Maxwell’s electromagnetic equations before Einstein developed it independently 20 years later. As a result, the expression is also referred to as the Lorentz–Einstein contraction.

length contraction the shortening of distances in a system, as seen by an observer in motion relative to that system

where Ls is the proper distance between A and B, directly measurable in the inertial frame in which A and B are at rest, and Lm is the distance—which we did not try to measure directly but deduced from kinematics—between A and B as seen in an inertial frame in which A and B are moving.

v2 1 1, Lm Ls. In our example, occupants on-board the spaceship c2 would measure the distance between planets as less than observers on either planet would measure it to be. The observers on the planets are not moving with respect to the space between planets, and therefore, measure the distance as Ls. The passengers onboard the spaceship are moving through that same space and measure its distance to be Lm. Our analysis shows that this length contraction occurs in the direction of motion. So, for example, a cylindrical spaceship 10 m in diameter, moving past Earth at high speed, is still 10 m in diameter in the frame of Earth (though shortened from tip to tail). Since

If you were in a spaceship flying by a tall, stationary building, at say, 0.9c, the width of the front of the building will be much thinner but the height will be the same. You would notice that the building’s sides are slightly curved in toward the middle since the light has travelled different distances to you, the observer, and you will see the sides of the building as you approach and recede from it. This description in words does not give you a true picture, nor will it be the same at different relativistic speeds. The best way to see what happens with length contraction is to view computer simulations of such events. GO

574 Chapter 11

v2 1 c2

www.science.nelson.com

NEL

Section 11.2

SAMPLE problem 2 A UFO heads directly for the centre of Earth at 0.500c and is first spotted when it passes a communications satellite orbiting at 3.28 103 km above the surface of Earth. What is the altitude of the UFO at that instant as determined by its pilot?

Solution Before solving a problem in special relativity, we must work out which lengths or durations are proper. In this case, the UFO pilot is determining the length separating two events in a process moving through his frame (the first event being the arrival at the UFO of the satellite, the second the arrival at the UFO of the surface of Earth). The length that the pilot is determining is thus not a proper length so it can appropriately be called Lm. On the other hand, the given length of 3.28 103 km is the length separating two events in a process stationary in the frame to which it is referred (the first event being the arrival of the UFO at the satellite, the second the arrival of the UFO at the surface of Earth). The given length is thus a proper length so it can appropriately be called Ls. v 0.500c Ls 3.28 103 km Lm ? Lm Ls

DID YOU

KNOW

?

The Fourth Dimension Time is often thought of as the fourth dimension. An event may be specified by four quantities: three to describe where it is in space and the fourth to describe where it is in time. When objects move at speeds near that of light, space and time become intertwined. Each inertial frame of reference represents one way of setting up a coordinate system for space-time. Comparing measurements made by observers in different inertial frames requires a change of coordinates, the interwining of space and time.

v2 1 c2

(3.28 103 km)

(0.500c) 1 c 2

2

Lm 2.84 103 km

The altitude as observed from the UFO is 2.84 103 km. Since the speed of the UFO is fairly slow, as far as relativistic events are concerned, the distance contraction is relatively small.

SAMPLE problem 3 A spaceship travelling past Earth with a speed of 0.87c, relative to Earth, is measured to be 48.0 m long by observers on Earth. What is the proper length of the spaceship?

Solution Since the spaceship is moving relative to the observers on Earth, 48.0 m represents Lm. v 0.87c Lm 48.0 m Ls ? Lm Ls

v2 1 c2

Lm Ls v2 1 c2

48.0 m (0.87c)2 1 c2

Ls 97.35 m, or 97.4 m

The proper length of the spaceship is 97.4 m. NEL

Einstein’s Special Theory of Relativity 575

Practice Understanding Concepts Answers

5. A spaceship passes you at the speed of 0.90c. You measure its length to be

50.0 m. What is its length when at rest?

5. 115 m

6. You are a space traveller, moving at 0.60c with respect to Earth, on your way to a

6. 6.0 ly 7. 3.95

102

star that is stationary relative to Earth. You measure the length of your trajectory to be 8.0 light-years (ly). Your friend makes the same journey at 0.80c with respect to Earth. What does your friend measure the length of the trajectory to be?

m

8. (a) 37.7 ly (b) 113 a

7. A spacecraft travels along a space station platform at 0.65c relative to the

platform. An astronaut on the spacecraft determines the platform to be 3.00 102 m long. What is the length of the platform as measured by an observer on the platform?

9. 0.89c

8. A star is measured to be 40.0 ly from Earth, in the inertial frame in which both

LEARNING

star and Earth are at rest. (a) What would you determine this distance to be if you travelled to the star in a spaceship moving at 1.00 108 m/s relative to Earth? (b) How long would you determine the journey to take?

TIP

When to Round Off When working with calculations involving relativistic quantities, don’t round off until the final result is achieved; otherwise, your results could be erroneous.

9. The proper length of one spaceship is twice the proper length of another. You,

an observer in an inertial frame on Earth, find the two spaceships, travelling at constant speed in the same direction, to have the same length. The slower spaceship is moving with a speed of 0.40c relative to Earth. Determine the speed of the faster spaceship relative to Earth.

Relativistic Momentum Momentum is one of the most important concepts in physics. Recall that Newton’s laws can also be stated in terms of momentum and that momentum is conserved whenever there is no external, unbalanced force on a system. In subatomic physics (see Chapter 13), much of what we know involves the collision of particles travelling at relativistic speeds, so it is important to address relativistic momentum. Using a derivation beyond the scope of this book but based on the Newtonian concept of conservation of momentum, Einstein was able to show that the momentum of an object is given by mv

p

v2 1 c2

rest mass mass measured at rest, relative to the observer

DID YOU

KNOW

?

Rest Mass Einstein wrote: “It is not good to introduce the concept of a relativistic mass of a body for which no clear definition can be given. It is better to introduce no other mass than the rest mass.” 576 Chapter 11

where p is the magnitude of the relativistic momentum, m is the mass of the object, and v is the speed of the object relative to an observer at rest. The m in the equation is the mass of the object measured by an observer at rest relative to the mass. This mass is referred to as the rest mass. How do we measure rest mass? We measure it the same way we always have, using Newtonian physics. We either measure it in terms of inertia, using Newton’s second law F F m , or measure it gravitationally m = . For low-speed objects these inertial a g and gravitational masses are equivalent. When an object is accelerated to high speeds, the mass cannot be defined uniquely and physicists only use the rest mass. In the next section we will see that rest mass can itself change, but only if the total energy of the system changes. As with the relativistic equations for time and length, the nonrelativistic expression for the momentum of a particle (p mv) can be used when v is very small in comparison

NEL

with c. As a rule, anything travelling faster than about 0.1c can be called relativistic, and significant correction using the special relativity relationships is required. 2 The momentum equation predicts that as v approaches c, the quantity 1 v c2 approaches 0, and, consequently, for an object of nonzero rest mass, p approaches infinity. Relativistic momentum would thus become infinite at the speed of light (Figure 4) for any object of nonzero rest mass. The unbounded increase of relativistic momentum reflects the fact that no body of nonzero (rest) mass can be accelerated from an initial speed less than c to a speed equal to or greater than c. (In Chapter 12, you will learn about photons, which act like particles of light. Since photons travel at the speed of light, they must have zero rest mass. Our relativistic momentum formula does not apply to this situation. In fact, it has been shown experimentally that photons do have well-defined momentum, even though they have no mass.)

SAMPLE problem 4 Linear accelerators accelerate charged particles to nearly the speed of light (Figure 5). A proton is accelerated to 0.999 994c. (a) Determine the magnitude of the relativistic momentum.

1 relativistic momentum = nonrelativistic momentum √1 − v 2/c 2

Section 11.2

6.0 5.0 4.0 3.0 2.0 1.0 0

0.2c 0.4c 0.6c 0.8c c Speed

Figure 4 Relativistic momentum, here measured by taking the ratio of relativistic momentum to the nonrelativistic momentum (the product of speed and rest mass), increases without bound as the speed approaches c.

(b) Make an order-of-magnitude comparison between the relativistic and the nonrelativistic momenta.

Solution (a) v 0.999 994c m 1.67 1027 kg (from Appendix C) p ? mv p v2 1 c2

(1.67 1027 kg)(0.999 994c) (0.999 994c)2 1 c2

5.010 1018 kg m/s 3.4641 10–3 p 1.45

1015

kg m/s

Figure 5 A linear accelerator

The magnitude of the relativistic momentum of the proton is 1.45 1015 kg m/s. (This value agrees with the value that can be measured in the accelerator.) (b) The magnitude of the nonrelativistic momentum, given by the classical Newtonian equation, is p mv (1.67 1027 kg)(0.999 994c) (1.67 1027 kg)(0.999 994)(3.00 108 m/s) p 5.01 1019 kg m/s

The nonrelativistic momentum is 5.01 1019 kg m/s. The relativistic momentum is more than three orders of magnitude larger than the nonrelativistic momentum.

NEL

Einstein’s Special Theory of Relativity 577

Practice Understanding Concepts Answers

10. What is the relativistic momentum of an electron moving at 0.999c in a linear

accelerator? (me 9.11 1031 kg)

10. 6.11 1021 kg·m/s

11. A small space probe of mass 2.00 kg moves in a straight line at a speed of 0.40c

11. 2.62 108 kg·m/s 12. 3.76

1019

relative to Earth. Calculate its relativistic momentum in the Earth frame.

kg·m/s

12. A proton is moving at a speed of 0.60c with respect to some inertial system.

Determine its relativistic momentum in that system. (mp 1.67 1027 kg)

SUMMARY

Relativity of Time, Length, and Momentum

•

Proper time ts is the time interval separating two events as seen by an observer for whom the events occur at the same position.

•

Time dilation is the slowing down of time in a system, as seen by an observer in motion relative to the system.

•

ts The expression tm represents time dilation for all moving objects. v2 1 c2

•

Time is not absolute: both simultaneous and time duration events that are simultaneous to one observer may not be simultaneous to another; the time interval between two events as measured by one observer may differ from that measured by another.

•

Proper length Ls is the length of an object, as measured by an observer at rest relative to the object.

•

Length contraction occurs only in the direction of motion and is expressed as Lm Ls

•

v2 1 . c2

The magnitude p of the relativistic momentum increases as the speed increases mv according to the relationship p . v2 1 c2

578 Chapter 11

•

The rest mass m of an object is its mass in the inertial frame in which the object is at rest and is the only mass that can be uniquely defined.

•

It is impossible for an object of nonzero rest mass to be accelerated to the speed of light.

NEL

Section 11.2

Section 11.2 Questions Understanding Concepts 1. The time dilation effect is sometimes misleadingly stated in

the words “Moving clocks run slowly.” Actually, this effect has nothing to do with motion affecting the functioning of clocks. What, then, does it deal with? 2. To whom does the elapsed time for a process seem to be

longer, an observer moving relative to the process or an observer that is stationary relative to the process? Which observer measures proper time? 3. How would length and time behave if Einstein’s postulates

were true but with the speed of light infinite? (Hint: Would there be no relativistic effects, or would the relativistic effects be in some sense exceptionally severe?)

(c) According to the observer on the planet, the spaceship takes 8.0 s to reach the next planet in the solar system. How long does the astronaut consider the journey to take? 10. A cube of aluminum 1.00 m 1.00 m 1.00 m is moving at

0.90c, in the orientation shown in Figure 6. The rest density of aluminum is 2.70 103 kg/m3. (a) Which of the three dimensions, a, b, or c, is affected by the motion? (b) Calculate the relativistic volume of the cube. (c) Calculate the relativistic momentum of the cube.

4. How would our lives be affected if Einstein’s postulates

were true but the speed of light were 100 km/h? (Hint: Consider simple specific examples, such as the nature of a highway journey at 60 km/h. What difficulty arises as we try to accelerate the car to 110 km/h?)

v = 0.90c

a

5. “Matter has broken the sound barrier but will not be able to

Figure 6

c

exceed the speed of light.” Explain this statement.

b

6. Muons created in collisions between cosmic rays and

atoms high in our atmosphere are unstable and disintegrate into other particles. At rest these particles have an average lifetime of only 2.2 106 s. (a) What is the average lifetime in the laboratory inertial frame of muons travelling at 0.99c in that frame? (b) If muons did not experience a relativistic increase in lifetime, then through what average distance could they travel in the laboratory inertial frame before disintegrating, if travelling at 0.99c in that frame? (c) Through what average distance do muons travel in the laboratory frame, if moving through that frame at 0.99c?

11. Calculate the relativistic momentum of a helium nucleus of

rest mass 6.65 10–27 kg, moving in the laboratory frame at 0.400c. Applying Inquiry Skills 12. Generate the data for the first two columns in Table 1 for

v values of c from 0.05 to 0.95, in intervals of 0.05. Use these

values to calculate tm and Lm. The use of a computer spreadsheet, such as Excel, is recommended. Table 1

7. In 2000, a 20-year-old astronaut left Earth to explore the

galaxy; her spaceship travels at 2.5 108 m/s. She returns in 2040. About how old will she appear to be? 8. A spaceship passes you at a speed of 0.90c. You find the

length of the ship in your frame to be 50.0 m. (The measurement is not easy: you have to arrange for a lamp to flash on the bow and the stern simultaneously in your frame, and then determine the distance between the flashes.) A few months later, the ship comes to the end of its mission and docks, allowing you to measure its length. (Now the measurement is easy: you can walk along the ship with a tape measure.) What length does your tape measure show?

v c

ts

tm

Ls

Lm

0.05

?

100 s

?

100 m

?

0.10

?

100 s

?

100 m

?

v2 1 c2

0.15

?

100 s

?

100 m

?

0.20

?

100 s

?

100 m

?

0.25

?

100 s

?

100 m

?

9. A spaceship goes past a planet at a speed of 0.80c relative

to the planet. An observer on the planet measures the length of the moving spaceship to be 40.0 m. The observer also finds that the planet has a diameter of 2.0 106 m. (a) The astronaut in the spaceship determines the length of the ship. What is this length? (b) The astronaut, while not in a position to actually measure the diameter of the planet, does succeed in computing a diameter indirectly. What is the resulting value?

NEL

(a) In what range of speeds is the time double the proper time? (b) At what speed has the length of the object contracted by 10%? (c) Based on the table, at what speed do relativistic effects become “noticeable”? What objects travel at these speeds? v (d) Plot graphs of tm and Lm versus c.

Einstein’s Special Theory of Relativity 579