ch39

CHAPTER 39

Relativity 1* · You are standing on a corner and a friend is driving past in an automobile. Both of you note the times when the car

passes two different intersections and determine from your watch readings the time that elapses between the two events. Which of you has determined the proper time interval?

By definition, the proper time is measured by the clock in the rest frame of the car, i.e., by the clock in the car. 2 · The proper mean lifetime of pions is 2.6 × 10-8 s. If a beam of pions has a speed of 0.85c, (a) what would their

mean lifetime be as measured in the laboratory? (b) How far would they travel, on average, before they decay? (c) What would your answer be to part (b) if you neglect time dilation?

(a) Use Equs. 39-13 and 39-7 (b) ∆x = v∆t (c) Neglecting time dilation, ∆t = 2.6×10-8 s

γ = s. ; ? ?t =. = ./ 1094490185011 82 −×−

∆x = 0.85×3×108×4.94×10-8 m = 12.6 m

∆x = 0.85×3×108×2.6×10-8 m = 6.63 m 3 · (a) In the reference frame of the pion in Problem 2, how far does the laboratory travel in a typical lifetime of

2.6×10-8 s? (b) What is this distance in the laboratory’s frame? (a) ∆x = v∆tπ (b) ∆x′ = γ∆x

∆x = 6.63 m

∆x′ = 12.6 m 4 · The proper mean lifetime of a muon is 2 µs. Muons in a beam are traveling at 0.999c. (a) What is their mean

lifetime as measured in the laboratory? (b) How far do they travel, on average, before they decay? (a) Use Equs. 39-13 and 39-7 (b) ∆x = v∆t

γ = sµ . ; ? ?t =. = / 7443722)999.0(11 2−

∆x = 0.999×3×108×44.7×10-6 m = 13.4 km 5* · (a) In the reference frame of the muon in Problem 4, how far does the laboratory travel in a typical lifetime of

2 µs? (b) What is this distance in the laboratory’s frame? (a) ∆x = v∆tµ (b) ∆x′ = γ∆xµ

∆xµ = 0.999×3×108×2×10-6 m = 599.4 m

∆x′ = 13.4 km

Chapter 39 Relativity

6 · Jay has been posted to a remote region of space to monitor traffic. Toward the end of a quiet shift, a spacecraft goes by, and he measures its length using a laser device, which reports a length of 85 m. He flips open his handy reference catalogue and identifies the craft as a CCCNX-22, which has a proper length of 100 m. When he phones in his report, what speed should Jay give for this spacecraft?

From Equ. 39-14, γ = Lp /L; solve for V γ = 100/85; m/s . c = . = /V = c 10581527011 82 ×− γ

7 · A spaceship travels to a star 95 light-years away at a speed of 2.2×108 m/s. How long does it take to get there (a)

as measured on earth and (b) as measured by a passenger on the spaceship? (a) As measured on earth, ∆t = ∆x/V (b) Use Equ. 39-13; ∆tp = ∆t/γ

∆t = (95 c.y)/[(2.2/3) c] = 129.5 y

γ = 1.47; ∆tp = (129.5/1.47) y = 88 y 8 · The mean lifetime of a pion traveling at high speed is measured to be 7.5×10-8 s. Its lifetime when measured at

rest is 2.6×10-8 s. How fast is the pion traveling? Use Equ. 39-13; solve for V

γ = 7.5/2.6; m/s . c = . = ? /V = c 10812938011 82 ×−

9* · A meterstick moves with speed V = 0.8c relative to you in the direction parallel to the stick. (a) Find the length of

the stick as measured by you. (b) How long does it take for the stick to pass you? (a) Use Equ. 39-14 (b) ∆t = L/V

γ = 1/0.6; L = (1 m)/γ = 0.6 m

∆t = (0.6 m)/0.8c = 2.5 ns 10 · The half-life of charged pions, π+ and π −, is 1.8×10-8 s; i.e., in the rest frame of the pions if there are N pions at

time t = 0, there will only be N/2 pions at time t = 1.8×10-8 s. Pions are produced in an accelerator and emerge with a speed of 0.998c. How far do these particles travel in the laboratory before half of them have decayed?

Use Equs. 39-13 and 39-7; ∆x = V∆t γ = 15.8; ∆t = γ × 1.8 × 10-8 s = 28.44 × 10-8 s; ∆x = 85.1 m 11 ·· A friend of yours who is the same age as you travels to the star Alpha Centauri, which is 4 light-years away and

returns immediately. He claims that the entire trip took just 6 y. How fast did he travel? D′ = 2L/γ and ∆t′ = D′/V = 2L/γV 6 y = (8 c.y)/γV; 0.752 = (c/V)2(1 - V 2/c2) = c2/V 2 - 1;

V = 0.8c 12 ·· Two spaceships pass each other traveling in opposite directions. A passenger in ship A, which she knows to be 100

m long, notes that ship B is moving with a speed of 0.92c relative to A and that the length of B is 36 m. What are the lengths of the two spaceships as measured by a passenger in ship B?

1. Find LA′ = LAp/γ 2. Find LBp = γLB′

γ = 2.55; LA′ = 39.2 m LBp = 2.55×36 m = 91.8 m

13* ·· In the Stanford linear collider, small bundles of electrons and positrons are fired at each other. In the laboratory’s

frame of reference, each bundle is about 1 cm long and 10 µm in diameter. In the collision region, each particle has an


energy of 50 GeV, and the electrons and positrons are moving in opposite directions. (a) How long and how wide is each bundle in its own reference frame? (b) What must be the minimum proper length of the accelerator for a bundle to have both its ends simultaneously in the accelerator in its own reference frame? (The actual length of the accelerator is less than 1000 m.) (c) What is the length of a positron bundle in the reference frame of the electron bundle?

(a) 1. Use Equ. 39-25 to find γ 2. Find proper length of electron bundle; Lep = γLe (b) 1. Find length of accelerator in electron frame 2. Set Lacc,e = Lp and solve for Lacc,p (c) Find length of positron bundle in electron frame

γ = 50×103/0.511 = 9.785×104 Lep = 978.5 m; width unchanged = 10 µm Lacc,e = Lacc,p/γ Lacc,p = (978.5 m)γ = 9.57×107 m Lpos = (1 cm)/γ = 1.02×10-7 m = 0.102 µm

14 · Use the binomial expansion

(1 + x)n = 1 + nx + n(n -1)2

2x 0 + . . . ≈ 1 + nx

to derive the following results for the case when V is much less than c, and use the results when applicable in the following problems:

(a) cV +

2

2

21

1≈γ

(b) cV -

2

2

21

11

≈γ

(c) cV -

2

2

211

11 ≈≈−γ

γ

(a) Using the binomial expansion, and setting β = V/c, γ = (1 - β 2)-1/2 = 1 + 1/2β 2 + … ≈ 1 + 1/2V 2/c2. (b) 1/γ = (1 - β 2)1/2 = 1 - 1/2β 2 + … ≈ 1 - V 2/c2. (c) From part (a) it follows that γ - 1 ≈ 1/2V 2/c2. Note that the results given above apply only if β = V/c << 1. 15 ·· Show that when V << c the transformation equations for x, t, and u reduce to the Galilean equations. Use Equ. 39-11 for x′, replace γ by (1 + 1/2V 2/c2), and retain only the lowest non-vanishing term in V/c. We then have x′ = (1 + 1/2V 2/c2)(x - Vt) = x - Vt to lowest order in V/c.

Repeat the procedure for t′ = γ(t - Vx/c2) [Equ. 39-12]. We obtain t′ = (1 + 1/2V 2/c2)(t - Vx/c2) = t to lowest order in V/c. ux′ = (ux - V)/(1 - Vux/c2) [Equ. 39-19a]. If V/c << 1, the denominator is approximately equal to 1 and ux′ = ux - V, which is the Galilean transformation equation. Using γ = (1 + 1/2V 2/c2), the denominators of Equs. 39-19b and c reduce to 1 to lowest order in V/c, so uy′ = uy and uz′ = uz, as expected. 16 ·· Supersonic jets achieve maximum speeds of about (3×10-6)c. (a) By what percentage would you see a jet traveling

at this speed contracted in length? (b) During a time of 1 y = 3.15×107 s on your clock, how much time would elapse


on the pilot’s clock? How many minutes are lost by the pilot’s clock in 1 y of your time? (a) 100(Lp - L)/Lp = 100(1 - 1/γ) ≈ 50V 2/c2 (b) ∆t′ = ∆t/γ ≈ ∆t(1 - 1/2V 2/c2) = ∆t - ∆tV 2/2c2

V/c = 3×10-6; contraction = 4.5×10-10 % Time lost = 3.15×107(9×10-12/2) = 0.142 ms = 2.36 µmin Time on pilot’s clock = ( 3.15×107 - 1.42×10-4) min

17*·· How great must the relative speed of two observers be for the time-interval measurements to differ by 1%? (See

Problem 14.) (∆t - ∆t′)/∆t = 1 - 1/γ ≈ 1/2V 2/c2 1/2V 2/c2 = 0.01; c 0.02 = V = 0.141c = 4.24×107 m/s

18 ·· A spaceship of proper length L′ = 400 m moves past a transmitting station at a speed of 0.76c. At the instant that

the nose of the ship passes the transmitter, clocks at the transmitter and in the nose of the ship are synchronized to t = t′ = 0. The instant that the tail of the ship passes the transmitter a signal is sent and subsequently detected by the receiver in the nose of the ship. (a) When, according to the clock in the ship, is the signal sent? (b) When, according to the clock at the transmitter, is the signal received by the spaceship? (c) When, according to the clock in the ship, is the signal received? (d) Where, according to an observer at the transmitter, is the nose of the spaceship when the signal is received?

Let S be the reference frame of the ship and S′ be that of the earth (transmitter station). Let event A be the emission of the light pulse and event B the reception of the light pulse at the nose of the ship. (a) In both S and S′ the pulse travels at the speed c. Thus, tA = 400/0.76c s = 1.754 µs. (c) The time of travel of the pulse to the nose is 400/c s = 1.333 µs. Thus the pulse arrives at tB = 3.09 µs according to the clock in the ship. (b), (d) To find the time and location of event B in frame S′, use Equs. 39-12 and 39-11, respectively. Note that here V, the velocity of reference frame S′ relative to S, is -0.76c. (b) Evaluate γ and use Equ. 39-12; V = -0.76c (d) Use Equ. 39-11 with V = -0.76c

γ = 1.54; tB′ = 1.56(3.09×10-6 + 0.76×400/c) = 6.40 µs xB′ = 1.56(400 + 0.76c×3.09×10-6) m = 1723 m

19 ·· A beam of unstable particles emerges from the exit slit of an accelerator with a speed of 0.89c. Particle detectors

3.0 and 6.0 m from the exit slit measure beam intensities of 2×108 particles/cm2⋅s and 5×107 particles/cm2⋅s, respectively. (a) Find the proper half-life of the particles. (b) Determine the beam intensity at the exit slit of the accelerator. (c) The accelerator is adjusted so that the particles emerge from the exit slit with a speed of 0.96c. The beam intensity at the farther detector is again 5×107 particles/cm2⋅s. Find the beam intensity at the exit slit of the accelerator.

(a) 1. Find ∆t to travel 3 m in lab 2. I6/I3 = 1/4 = 1/22; ∆t = 2τ1/2 3. Use Equ. 39-13 to find τ1/2,p (b) I0 = 4I3 (c) 1. Find ∆t to travel 6 m 2. Find ∆tp and express as aτ1/2,p 3. I0 = I6×2a

∆t = (3 m)/(0.89c) = 1.124×10-8 s

τ1/2 = 0.562×10-8 s

γ = 2.193; τ1/2,p = 0.562×10-8/2.193 s = 2.56 ns

I0 = 8×108 particles/cm2.s

∆t = (6 m)/(0.96c) = 2.083×10-8 s

γ = 3.57; ∆tp = 0.583×10-8 s = 2.277τ1/2,p

I0 = 5×107×22.277 part./cm2.s = 2.42×108 particles/cm2.s


20 ·· Show that if ux′ and V in Equation 39-18a are both less than c, then ux is less than c. (Hint: Let ux′ = (1 - ε1)c and V = (1 - ε2)c, where ε1 and ε2 are small positive numbers that are less than 1.) We let ux′ = (1 - ε1)c and V = (1 - ε2)c. Then Equ. 39-18a becomes

2121

21

21

21

)(2)(2

)1)(1(1)(2

εεεεεε

εεεε

++−+−=

−−++−

=c

u x <10; Q.E.D.

21* ··· Two events in S are separated by a distance D = x2 - x1 and a time T = t2 - t1. (a) Use the Lorentz transformation

to show that in frame S′, which is moving with speed V relative to S, the time separation is t2′ - t1′ = γ(T -VD/c2). (b) Show that the events can be simultaneous in frame S′ only if D is greater than cT. (c) If one of the events is the cause of the other, the separation D must be less than cT, since D/c is the smallest time that a signal can take to travel from x1 to x2 in frame S. Show that if D is less than cT, t2′ is greater than t1′ in all reference frames. This shows that if the cause precedes the effect in one frame, it must precede it in all reference frames. (d) Suppose that a signal could be sent with speed c′ > c so that in frame S the cause precedes the effect by the time T = D/c′. Show that there is then a reference frame moving with speed V less than c in which the effect precedes the cause.

(a) Use Equ. 39-12. t2′ - t1′ = γ[(t2 - t1) - (V/c2)(x2 - x1)] = γ(T - VD/c2), where T = t2 - t1 and D = x2 - x1. (b) Events 1 and 2 are simultaneous in S′ if t2′ = t1′ or (T - VD/c2) = 0. Since V ≤ c, D ≥ cT. (c) If D < cT then t2′ > t1′, and the events are not simultaneous in S′. (d) If D = c′T > cT then T - VD/c2 = T[1 - (V/c)(c′/c)] = t2′ - t1′. In this case, t2′ - t1′ could be negative, i.e., t2′ could be less than t1′, or the effect could precede the cause. 22 · If event A occurs before event B in some frame, might it be possible for there to be a reference frame in which

event B occurs before event A? Yes; see Problem 21. 23 · Two events are simultaneous in a frame in which they also occur at the same point in space. Are they

simultaneous in other reference frames? Yes; see Problem 21. 24 ·· Two observers are in relative motion. In what circumstances can they agree on the simultaneity of two different

events? They will agree only if the two events occur at the same point in space; see Problem 21. Problems 25 through 29 refer to the following situation: An observer in S′ lays out a distance L′ = 100 light-minutes between points A′ and B′ and places a flashbulb at the midpoint C′. She arranges for the bulb to flash and for clocks at A′ and B′ to be started at zero when the light from the flash reaches them (see Figure 39-13). Frame S′ is moving to the right with speed 0.6c relative to an observer C in S who is at the midpoint between A′ and B′ when the bulb flashes. At the instant he sees the flash, observer C sets his clock to zero. 25* ·· What is the separation distance between clocks A′ and B′ according to the observer in S? L′ is the proper distance Lp; use Equ. 39-14 γ = 1.25; L = 80 c.min 26 ·· As the light pulse from the flashbulb travels toward A′ with speed c, A′ travels toward C with speed 0.6c. Show

that the clock in S reads 25 min when the flash reaches A′. (Hint: In time t, the light travels a distance ct and A′ travels


0.6ct. The sum of these distances must equal the distance between A′ and the flashbulb as seen in S.) 1. Find x, the location of the light flash in S at time t 2. Find location of A′ in S at time t 3. Total distance covered by light = 1.6ct - L/2; find t

x = -ct (for light flash moving toward A′) xA = -L/2 + 0.6ct

t = L/3.2c = (80 c.min)/3.2c = 25 min 27 ·· Show that the clock in S reads 100 min when the light flash reaches B′, which is traveling away from C with speed

0.6c. (See the hint for Problem 26.) 1. Find x, location of the light flash in S at time t 2. Find location of B′ in S at time t 3. Find time when flash reaches B′ as observed in S

x = ct (for light flash moving toward B′) xB = L/2 + 0.6ct

0.4ct = L/2; t = (80 c.min)/0.8c = 100 min 28 ·· The time interval between the reception of the flashes at A′ and B′ in Problems 26 and 27 is 75 min according to

the observer in S. How much time does he expect to have elapsed on the clock at A′ during this 75-min interval? Use Equ. 39-13 to find ∆tp, the proper time in S′ ∆tp = (75 min)/γ = (75/1.25) min = 60 min 29*·· The time interval calculated in Problem 28 is the amount that the clock at A′ leads that at B′ according to the

observer in S. Compare this result with LpV/c2. The time interval calculated in Problem 28 is 60 min LpV/c2 = 100×0.6 min = 60 min, as expected 30 ·· In frame S, event B occurs 2 µs after event A, which occurs at ∆x = 1.5 km from event A. How fast must an

observer be moving along the +x axis so that events A and B occur simultaneously? Is it possible for event B to precede event A for some observer?

From Problem 21b, tA′ = tB′ if ∆t = V∆x/c2; find V Yes, tB′ will be less than tA′ if V > 0.4c

V = (2×10-6×9×1016/1.5×103) m/s = 1.2×108 m/s = 0.4c

31 ·· Observers in reference frame S see an explosion located at x1 = 480 m. A second explosion occurs 5 µs later at x2

= 1200 m. In reference frame S′, which is moving along the +x axis at speed V, the explosions occur at the same point in space. What is the separation in time between the two explosions as measured in S′?

1. From Equ. 39-11, ∆x′ = γ(∆x - V∆t); set ∆x′ = 0 2. From Problem 21a, ∆t′ = γ(∆t - V∆x/c2)

V = ∆x/∆t = (1200 - 480)/5×10-6 m/s = 1.44×108 m/s

γ = 1.14; ∆t′ = 1.14(5×10-6 - 1.15×10-6) s = 4.39 µs 32 · How fast must you be moving toward a red light (λ = 650 nm) for it to appear green (λ = 525 nm)?

λ0/λ = f/f0 = α = V/c V/c +

−11

; 1 + 1

= cV

2

2

αα −

α = 1.238; V = 0.21c

33* · A distant galaxy is moving away from us at a speed of 1.85×107 m/s. Calculate the fractional redshift (λ′ -λ0)/λ0 in

the light from this galaxy.


(λ - λ0)/λ0 = λ/λ0 - 1 = V/c 1V/c + 1

− - 1 V/c = 0.185/3 = 0.0617; λ/λ0 - 1 = 0.0637

34 · Sodium light of wavelength 589 nm is emitted by a source that is moving toward the earth with speed V. The

wavelength measured in the frame of the earth is 620 nm. Find V.

From Problem 32, 1 + 1

= cV

2

2

αα −

, where α = λ0/λ α = 0.95; V = -0.0512c; source is receding, not approaching the earth.

35 · A student on earth hears a tune on her radio that seems to be coming from a record that is being played too fast.

She has a 33-rev/min record of that tune and determines that the tune sounds the same as when her record is played at 78 rev/min, that is, the frequencies are all too high by a factor of 78/33. If the tune is being played correctly, but is being broadcast by a spaceship that is approaching the earth at speed V, determine V.

From Problem 32, 1 + 1

= cV

2

2

αα −

, where α = f/f0 α = 78/33 = 2.364; V = 0.696c

36 ·· Derive Equation 39-16a for the frequency received by an observer moving with speed V toward a stationary

source of electromagnetic waves. In the rest frame of the source, the number of waves encountered by the observer in a time interval ∆ts is n = (c + V)∆ts/λ = (c + V)f0∆ts/c = f0(1 + V/c)∆ts. This time interval in the rest frame of the observer is given by

∆to = ∆ts/γ. The frequency noted by the observer is fo = n/∆to = γ(1 + V/c)f0 = f c/V 1

V/c + 1022− =

V/c 1V/c + 1

−

f0 = f V/c 1

c/V 10

22

−−

; Q.E.D.

37* · Herb and Randy are twin jazz musicians who perform as a trombone–saxophone duo. At the age of twenty,

however, Randy got an irresistible offer to join a road trip to perform on a star 15 light-years away. To cele brate his bounteous luck, he bought a new vehicle for the trip—a deluxe space-coupe which could do 0.999c. Each of the twins promises to practice diligently, so they can reunite afterward. Randy’s gig goes so fabulously well, however, that he stays for a full 10 years before returning to Herb. After their reunion, (a) how many years of practice will Randy have? (b) how many years of practice will Herb have?

(a) 1. Find ∆ttravel in Randy’s frame, ∆ttravel (R) 2. Total time for Randy is ∆ttravel (R) + 10 y (b) 1. Find ∆ttravel in Herb’s frame, ∆ttravel (H) 2. Total time for Herb is 10 y + ∆ttravel (H)

γ = 22.37; ∆ttravel (R) = (30.03 y)/22.37 = 1.34 y Randy’s years of practice = 11.34 y

∆ttravel (H) = (30 c.y)/(0.999c) = 30.03 y Herb’s years of practice = 40.04 y


38 ·· A clock is placed in a satellite that orbits the earth with a period of 90 min. By what time interval will this clock

differ from an identical clock on earth after 1 y? (Assume that special relativity applies and neglect general relativity.) 1. Use Equ. 11-19 and Problem 11-55 to find V 2. Use Problem 15c and Equ. 39-13

V = ve / 2 = 7.92×103 m/s = 2.64×10-4c Time difference = (3.15×107×6.97×10-8/2) s = 1.10 s

39 ·· A and B are twins. A travels at 0.6c to Alpha Centauri (which is 4 c⋅y from earth as measured in the reference

frame of the earth) and returns immediately. Each twin sends the other a light signal every 0.01 y as measured in her own reference frame. (a) At what rate does B receive signals as A is moving away from her? (b) How many signals does B receive at this rate? (c) How many total signals are received by B before A has returned? (d) At what rate does A receive signals as B is receding from her? (e) How many signals does A receive at this rate? (f) How many total signals are received by A? (g) Which twin is younger at the end of the trip, and by how many years?

(a) Use Doppler effect, Equ. 39-16b (b) 1. Find time for one way trip in frame SA 2. Number of signals received by B = number of signals sent by A, namely f0∆tA (c) 1. Find time in frame SA for round trip 2. Number of signals received by B = number of signals sent by A, namely f0∆TA (d) Proceed as in part (a) (e) Number of signals received by A = fA∆tA

(f) 1. Find fA,r, rate of signals received by A on return (g) trip; use Equ. 39-16a

2. Find number of signals received on return trip 3. Find total number of signals received by A (g) 1. A sent 1067 signals at 100 per year 2. B sent 1333 signals at 100 per year 3. Find age difference

f0 = 100 y-1; fB = y 1−− 50 =

0.6 + 10.6 1

f 0

∆tA = ∆tB/γ; γ = 1.25; ∆tB = 4/0.6 = 6.67 y; ∆tA = 5.33 y Number of signals received by B = 533

∆TA = 2∆tA = 10.67 y Number of signals received by B = 1067 fA = 50 y-1

Number of signals received by A = 50×5.33 = 267 fA,r = 200 y-1 Number of signals = 200×5.33 = 1066 Total signals received by A = 1066 + 267 = 1333 Age of A = 1067/100 = 10.67 y Age of B = 13.33 y A is 2.66 y younger than B.

40 · A light beam moves along the y′ axis with speed c in frame S′, which is moving to the right with speed V relative

to frame S. (a) Find the x and y components of the velocity of the light beam in frame S. (b) Show that the magnitude of the velocity of the light beam in S is c.

(a) Find ux and uy using Equs. 39-18a and b

(b) Find u + u 2y

2x = u

ux = V; uy = c/γ, where γ = 1/ c/V 1 22−

u = c = )c/V (1c + V 2222 − ; speed of light is the

same in all reference frames. 41* · A spaceship is moving east at speed 0.90c relative to the earth. A second spaceship is moving west at speed 0.90c

relative to the earth. What is the speed of one spaceship relative to the other?


Let S be the earth reference frame and S′ be that of the ship traveling east (positive x direction). Then in the reference frame S′, the velocity of S is directed west, i.e., V = -ux. Now apply the velocity transformation equation, Equ. 39-19a, to determine the speed of the other ship in the reference frame S′.

ux′ = c/u + 1

u2 =

c/ Vu 1V u

22x

x2

x

x

−−

ux = -0.9c; ux′ = 0.994c

42 ·· Two spaceships are approaching each other. (a) If the speed of each is 0.6c relative to the earth, what is the

speed of one relative to the other? (b) If the speed of each relative to the earth is 30,000 m/s (about 100 times the speed of sound), what is the speed of one relative to the other?

(a) See Problem 41; c/u + 1

u2 = u 22

x

xx ′

(b) (V/c)2 = 10-8; use Galilean transformation

ux′ = 1.2c/1.36 = 0.882c ux′ = 60,000 m/s

43 ·· A particle moves with speed 0.8c along the x″ axis of frame S″, which moves with speed 0.8c along the x′ axis

relative to frame S′. Frame S′ moves with speed 0.8c along the x axis relative to frame S. (a) Find the speed of the particle relative to frame S′. (b) Find the speed of the particle relative to frame S.

1. Use Equ. 39-18a to find ux′ in terms of ux″; here V of S′ relative to S″ is 0.8c 2. Now find ux; V of S relative to S′ is 0.8c

ux′ = 1.6c/1.64 = 0.9756c ux = (0.9756 + 0.8)c/(1 + 0.9756×0.8) = 0.997c

44 · The approximate total energy of a particle of mass m moving at speed u << c is (a) mc2, (b) 2

1 0mu2,

(c) cmu, (d) 21 mc2, (e) 2

1 cmu.

(a) 45*· Find the ratio of the total energy to the rest energy of a particle of rest mass m0 moving with speed (a) 0.1c, (b) 0.5c, (c) 0.8c, and (d) 0.99c. (a), (b), (c), (d) From Equ. 39-25, E/E0 = γ (a) γ = 1.005; (b) γ = 1.15; (c) γ = 1.67; (d) γ = 7.09 46 · A proton (rest energy 938 MeV) has a total energy of 1400 MeV. (a) What is its speed? (b) What is its

momentum? (a) From Problem 41, γ = E/E0; solve for u (b) From Equ. 39-21, p = γm0u

γ = 1.49; u = c γ 21/ 1 − = 0.742c

p = 1.49(938 MeV/c2)(0.742c) = 1037 MeV/c 47 · How much energy would be required to accelerate a particle of mass m0 from rest to (a) 0.5c, (b) 0.9c, and (c) 0.99c? Express your answers as multiples of the rest energy. (a), (b), (c) Energy needed = K = (γ - 1)m0c2 (a) γ = 1.155, K = 0.155E0; (b) K = 1.29E0; (c) K = 6.09E0


48 · If the kinetic energy of a particle equals its rest energy, what error is made by using p = m0u for its momentum? If K = E0 then γ = 2 (see Equ. 39-23). Therefore, prel = 2m0u, whereas pclass = m0u. The error is m0u or 50% of prel. 49* · What is the energy of a proton whose momentum is 3m0c?

Use E = )cm( + p c 20

2 (see Problem 52) E = 10 m0c2 = 2.97 GeV

50 ·· A particle with momentum of 6 MeV/c has total energy of 8 MeV. (a) Determine the rest mass of the particle.

(b) What is the energy of the particle in a reference frame in which its momentum is 4 MeV/c? (c) What are the relative velocities of the two reference frames?

(a) E0 = cp E 222 − (see Problem 52)

(b) E = )E( + cp 20

22 ; E0 is an invariant

(c) 1. Find ua and ub; use Equs. 39-21 and 39-25 2. Use Equ. 39-19a and solve for V

E0 = 5.29 MeV E = 6.63 MeV ua = (6/8)c = 0.75c; ub = (4/6.63)c = 0.603c V = (0.75 - 0.603)c/(1 - 0.75×0.603) = 0.268c

51 ·· Show that

du cu - 1m =

c / u - 1

um d2

2 2 / -3

022

0

Use q

dudq

p dudp

q =

qp

dud

2

−

. Thus

c/u 1c/u 1)/c/um( + m c/u 1

= c/u 1

um dud

22

222200

22

22

0

−−−

−.

This simplifies to m0/(1 - u2/c2)3/2; Q.E.D. 52 ·· Use Equations 39-21 and 39-25 to derive the equation .)cm( + cp = E

220

222

Square Equ. 39-25: .ucm + cm = c/u 1

u + ccm = c/u 1

cm = E

22220

42022

2222

022

4202 γ

−−

But from Equ. 39-21 we

have .cp = ucm 2222220 γ We therefore obtain the desired result, namely E 2 = (m0c2)2 + p2c2.

53* ·· Use the binomial expansion (Equation 39-27) and Equation 39-28 to show that when pc << m0c2, the total energy is

given approximately by

m2p

+ cm E0

22

0≈

From Equ. 39-28 we have E = )E( + cp 20

22 = m0c2 cm/p + 1 220

2 . When p/m0c << 1 we can expand the square

root, retaining only the first two terms. Thus, E = m0c2[1 + 1/2(p2/m02c2)] = m0c2 + p2/2m0; Q.E.D.

54 ·· (a) Show that the speed u of a particle of mass m0 and total energy E is given by


E)cm(

- 1 = cu

2

220

2 / 1

and that when E is much greater than m0c2, this can be approximated by

E2

)cm( - 1

cu

2

220≈

Find the speed of an electron with kinetic energy of (b) 0.51 MeV and (c) 10 MeV.

(a) See Problem 52. We have 1 - u2/c2 = (m0c2/E)2 and so u/c = .)/Ecm( 1 220− If E >> m0c2 we can expand the

square root, and keeping only the first two terms, we have u/c = 1 - (m0c2 )2/2E 2; Q.E.D. (b) Use the result of (a) and E = E0 + K (c) Here E ≅ 20E0 >> E0; use approximation for u

u = c ¼ 1 − = 0.866c

u = c(1 - 1/800) = 0.999c 55 ·· The rest energy of a proton is about 938 MeV. If its kinetic energy is also 938 MeV, find (a) its momentum and

(b) its speed.

(a) From Equ. 39-28, pc = E E 20

2 −

(b) See Problem 54(b)

p = 3 0E0/c = 1625 MeV/c

u = 0.866c

56 ·· What percentage error is made in using 2

1 m0u2 for the kinetic energy of a particle if its speed is (a) 0.1c and

(b) 0.9c? (a) 1. From Equ. 39-23, K = E0(γ - 1) 2. % error = 100(K - 1/2m0u2)/K (b) 1. Repeat as in part (a) 2. % error = 100(K - 1/2m0u2)/K

γ = 1.00504; K = 0.00504m0c2; 1/2m0u2 = 0.005m0c2 Error = (4/5)% = 8%

γ = 2.294; K = 1.294m0c2; 1/2m0u2 = 0.405m0c2 Error = 68.7%

57* ·· The K0 particle has a rest mass of 497.7 MeV/c2. It decays into a π − and π+, each with rest mass 139.6 MeV/c2.

Following the decay of a K0, one of the pions is at rest in the laboratory. Determine the kinetic energy of the other pion and of the K0 prior to the decay.

We shall first consider the decay process in the center of mass reference frame and then transform to the laboratory reference frame in which one of the pions is at rest. 1. Write the conditions for energy conservation in CM 2. Since one of the pions is at rest in the lab frame, γ = 1.78 for the transformation to the lab frame; find K of K0 3. Find the total initial energy in the lab frame 4. Kπ = E -2m0πc2

1.78 = m = ;cm 00

2 m2/m2 = c K2

K 00 ππ γγ

KK = 0.78×497.7 MeV = 389.5 MeV E = (497.7 + 389.5) MeV = 887.2 MeV Kπ = (887 - 2×139.6) MeV = 608 MeV

58 ·· The sun radiates energy at the rate of about 4×1026 W. Assume that this energy is produced by a reaction whose

net result is the fusion of 4 H nuclei to form 1 He nucleus, with the release of 25 MeV for each He nucleus formed.


Calculate the sun’s loss of rest mass per day. 1. Find number of reactions per second, N 2. Find ∆m per reaction 3. ∆M = 8.64×104N∆m

E/reaction = 4×10-12 J; N = 1038 s-1

∆m = 4×10-12/(3×108 )2 = 4.44×10-29 kg

∆M = 3.84×1014 kg 59 ·· Two protons approach each other head on at 0.5c relative to reference frame S′. (a) Calculate the total kinetic

energy of the two protons as seen in frame S′. (b) Calculate the total kinetic energy of the protons as seen in reference frame S, which is moving with speed 0.5c relative to S′ such that one of the protons is at rest.

(a) In S′, K = 2(γ - 1)E0; evaluate K (b) 1. Find u and γ of moving proton in S; V = 0.5c 2. Find K of moving proton; K = (γ - 1)E0

γ = 1.155; K = (2×0.155×938) MeV = 291 MeV u = c/(1 + 0.25) = 0.8c; γ = 1.67 K = 0.67×938 MeV = 628 MeV

60 ·· An antiproton p 0 has the same rest energy as a proton. It is created in the reaction

p + p + p + p p + p → . In an experiment, protons at rest in the laboratory are bombarded with protons of

kinetic energy KL, which must be great enough so that kinetic energy equal to 2m0c2 can be converted into the rest energy of the two particles. In the frame of the laboratory, the total kinetic energy cannot be converted into rest energy because of conservation of momentum. However, in the zero-momentum reference frame in which the two initial protons are moving toward each other with equal speed u, the total kinetic energy can be converted into rest energy. (a) Find the speed of each proton u such that the total kinetic energy in the zero-momentum frame is 2m0c2. (b) Transform to the laboratory’s frame in which one proton is at rest, and find the speed u′ of the other proton.

(c) Show that the kinetic energy of the moving proton in the laboratory’s frame is KL = 6m0c2. (a) We need K = E0 for each proton; find γ and u (b) Use Equ. 39-19a with V = -u, and ux = -u

(c) 1. Note that γ′ = [1 - (4 3 /7)2]-1/2 = 7

E0 = (γ - 1)E0; γ = 2; u = 0.866c (see Problem 55) ux′ = -1.732c/1.75 = -0.990c K′ = (γ′ - 1)E0 = 6m0c2

61* ··· A particle of rest mass 1 MeV/c2 and kinetic energy 2 MeV collides with a stationary particle of rest mass 2 MeV/c2. After the collision, the particles stick together. Find (a) the speed of the first particle before the collision, (b) the total energy of the first particle before the collision, (c) the initial total momentum of the system, (d) the total kinetic energy after the collision, and (e) the rest mass of the system after the collision.

(a) E = K + E0 = γE0; u/c = 1/ 1 2γ−

(b) E = γE0

(c) p2c2 = E 2 - E02; /c E E = p 2

02 −

(d), (e) 1. From energy conservation, Ef = Ei 2. pf = pi; Ef

2 = pf 2c2 + Ef0

2 3. Kf = Ef - Ef0

γ = 3; u = c 8/9 = 0.943c

E = 3E0 = 3 MeV

p = 8 E0/c = 2.828 MeV/c

Ef = 5 MeV Ef0 = 4.123 MeV = m0f c2; m0f = 4.123 MeV/c2 Kf = 0.877 MeV

62 · A set of twins work in an office building. One works on the top floor and the other works in the basement.

Considering general relativity, which one will age more quickly?


(a) They will age at the same rate. (b) The twin who works on the top floor will age more quickly. (c) The twin who works in the basement will age more quickly. (d) It depends on the speed of the office building. (e) None of these is correct. (b) 63 ··· A horizontal turntable rotates with angular speed ω. There is a clock at the center of the turntable and one at a

distance r from the center. In an inertial reference frame, the clock at distance r is moving with speed u = rω. (a) Show that from time dilation according to special relativity, time intervals ∆t0 for the clock at rest and ∆tr for the moving clock are related by

c2 2

2ωr- t

t - t 2

0

0r ≈∆

∆∆ if rω << c

(b) In a reference frame rotating with the table, both clocks are at rest. Show that the clock at distance r experiences a pseudoforce Fr = mrω2 in this accelerated frame and that this is equivalent to a difference in gravitational potential

between r and the origin of φ r -φ0 = 21 r2ω2. Use this potential difference in Equation 2 to show that in this frame the

difference in time intervals is the same as in the inertial frame.

(a) For rω /c << 1, 1/γ ≈ 1 - r2ω 2/2c2 (see Problem 14). From Equ. 39-13, .c2

r = t

t t2

22

0

0r ω−∆

∆−∆

(b) The pseudoforce is given by FP = -ma, where a is the acceleration of the non-inertial reference frame (see p. 129). In this case, a = rω 2, the centripetal acceleration directed inward. Hence, FP = Fr = mrω 2. If we now associate a potential with this pseudoforce, using the standard definition φ = -(1/m)∫Fr dr, the potential difference

between the center and a point at r is -1/2r2ω 2. Then, using Equ. 2, p. 1273, .c2

r = t

t t2

22

0

0r ω−∆

∆−∆

64 · True or false: (a) The speed of light is the same in all reference frames. (b) Proper time is the shortest time interval between two events. (c) Absolute motion can be determined by means of length contraction. (d) The light-year is a unit of distance. (e) Simultaneous events must occur at the same place. (f) If two events are not simultaneous in one frame, they cannot be simultaneous in any other frame. (g) If two particles are tightly bound together by strong attractive forces, the rest mass of the system is less than the

sum of the masses of the individual particles when separated. (a) True (b) True (c) False (d) True (e) False (f) False (g) True 65* · An observer sees a system consisting of a mass oscillating on the end of a spring moving past at a speed u and

notes that the period of the system is T. Another observer, who is moving with the mass–spring system, also measures its period. The second observer will find a period that is (a) equal to T, (b) less than T, (c) greater than T, (d) either


(a) or (b) depending on whether the system was approaching or receding from the first observer, (e) There is not sufficient information to answer the question.

(b) 66 · The Lorentz transformation for y and z is the same as the classical result: y = y′ and z = z′. Yet the relativistic

velocity transformation does not give the classical result uy = uy′ and uz = uz′. Explain. Although ∆y = ∆y′, the ∆t ≠ ∆t′. Consequently, uy = ∆y/∆t ≠ ∆y′/∆t′ = uy′. 67 · A spaceship departs from earth for the star Alpha Centauri, which is 4 light-years away. The spaceship travels at

0.75c. How long does it take to get there (a) as measured on earth and (b) as measured by a passenger on the spaceship?

(a) As measured on earth ∆t = L/u (b) Use Equ. 39-13 to find ∆tp

∆t = (4 y.c)/(0.75c) = 5.33 y

γ = 1.51; ∆tp = 3.53 y 68 · The total energy of a particle is twice its rest energy. (a) Find u/c for the particle. (b) Show that its momentum is

given by p = 3 0m0c.

(a) E = γE0 (see Equ. 39-25); solve for u/c (b) Use Equ. 39-28; p2c2 = (γ2 - 1)E0

2 γ = 2; u/c = γ 21/ 1 − = 0.866

p = cm3 0

69* · How fast must a muon travel so that its mean lifetime is 46 µs if its mean lifetime at rest is 2 µs?

Find γ from Equ. 39-13; then u/c = γ 21/ 1 − γ = 23; u = 0.999c

70 · A distant galaxy is moving away from the earth with a speed that results in each wavelength received on earth

being shifted such that λ′ = 2λ0. Find the speed of the galaxy relative to the earth.

λ0/λ = f/f0 = α = V/c 1V/c + 1

− ; 1 + 1

= cV

2

2

αα −

α = 2; V = 0.60c

71 · How fast must a meterstick travel relative to you in the direction parallel to the stick so that its length as measured

by you is 50 cm?

Use Equ. 39-14 and u/c = γ 21/ 1 − γ = 2; u = 0.866c

72 · Show that if V is much less than c, the doppler shift is given approximately by ∆f/f ≈ ±V/c.

∆f/f0 = (f - f0)/f0 = f/f0 - 1 = V/c 1V/c + 1

− - 1. Expanding numerator and denominator using the binomial expansion

gives, to lowest order in V/c, ∆f/f0 = (1 + 1/2V/c)(1 + 1/2V/c) - 1 = 1 + V/c - 1 = V/c. The sign depends on whether the source and receiver are approaching or receding. Here we have assumed that they are approaching.


73* ·· If a plane flies at a speed of 2000 km/h, for how long must it fly before its clock loses 1 s because of time dilation? ∆t - ∆tp = ∆t(1 - 1/γ) ≈ ∆tV 2/2c2 (see Problem 14) ∆t = [(2×9×1016)/(555.5)2] s = 5.83×1011 s ≈ 1.85×104 y 74 ·· The radius of the orbit of a charged particle in a magnetic field is related to the momentum of the particle by p = BqR 39-41

This equation holds classically for p = mu and relativistically for p = m0u/ c / u - 1 22 . An electron with kinetic

energy of 1.50 MeV moves in a circular orbit perpendicular to a uniform magnetic field B = 5×10-3 T. (a) Find the radius of the orbit. (b) What result would you obtain if you used the classical relations p = mu and K = p2/2m?

(a) 1. Use Equ. 39-28; pc = E E 20

2 − = KE2 + K 02

2. R = p/Bq

(b) R = /Bq; 2mK m = 9.11×10-31 kg; K = 2.4×10-13 J

p = 1.94 MeV/c = 1.04×10-21 kg.m/s R = 1.04×10-21/(5×10-3×1.6×10-19) m = 1.30 m R = 0.827 m

75 ·· Oblivious to economics and politics, Professor Spenditt proposes building a circular accelerator around the earth’s

circumference using bending magnets that provide a magnetic field of magnitude 1.5 T. (a) What would be the kinetic energy of protons orbiting in this field in a circle of radius RE? (See Problem 74.) (b) What would be the period of rotation of these protons?

(a) 1. Find p = REBq

2. ;E E + cp = K 020

22 − for pc >> E0, K ≈ pc

(b) E ≈ pc and u ≈ c; T = 2πRE /c

p = 1.53×10-12 kg.m/s = 2.87×109 MeV/c K = 2.87×109 MeV T = 0.133 s

76 ·· Frames S and S′ are moving relative to each other along the x and x′ axis. Observers in the two frames set their

clocks to t = 0 when the origins coincide. In frame S, event 1 occurs at x1 = 1.0 c⋅y and t1 = 1 y and event 2 occurs at x2 = 2.0 c⋅y and t2 = 0.5 y. These events occur simultaneously in frame S′. (a) Find the magnitude and direction of the velocity of S′ relative to S. (b) At what time do both these events occur as measured in S′? (a) See Problem 21; ∆t - V∆x/c2 = 0; solve for V; note that ∆t = t2 - t1 = -0.5 y

(b) Use Equ. 39-12 to find t1′ = t2′

V/c = c/[(1 c.y)/(-0.5 y)] = -0.5; V = -0.5c; S′ moves in the negative x direction

γ = 1.155; t1′ = t2′ = 1.73 y 77* ·· An interstellar spaceship travels from the earth to a distant star system 12 light-years away (as measured in the

earth’s frame). The trip takes 15 years as measured on the ship. (a) What is the speed of the ship relative to the earth? (b) When the ship arrives, it sends a signal to the earth. How long after the ship leaves the earth will it be before the earth receives the signal?

(a) 1. ∆t′ = L′/u = L/γu

2. Solve for u/c; u/c = )u/c( + 1

)u/c(2

2

γγ

(b) T = L/u + L/c

γu = (12 c.y)/(15 y) = 0.8c

u/c = 0.625 = 0.64/1.64 ; u = 0.625c

T = (12/0.625 + 12) y = 31.2 y


78 ·· The neutral pion π 0 has a rest mass of 135 MeV/c2. This particle can be created in a proton–proton collision: p + p → p + p + π 0 Determine the threshold kinetic energy for the creation of a π0 in a collision of a moving and stationary proton. (See

problem 60.) Here we can use the result given in Problem 85. We have Σmin c2 = 2×938 MeV = 1876 MeV; Σmfin c2 = (1876 + 135) MeV = 2011 MeV; mtarget c2 = 938 MeV. We obtain, using the expression given in Problem 85, Kth = [(1876 + 2011)(2011 - 1876)/1876] MeV = 280 MeV. Below, we follow the procedure employed in the solution of Problem 60. (a) 1. Use energy conservation to find γ in CM frame

2. Find u/c; u/c = 1/ 1 2γ−

(b) Transform to the lab frame and find u′ (c) Find Klab of proton; Klab = (γlab - 1)E0

2γ(938 MeV) = 2011 MeV; γ = 1.072 u = 0.36c u′ = 0.72c/(1 + 0.362) = 0.637c

γlab = 1.30; Klab = 281 MeV

79 ·· A rocket with a proper length of 1000 m moves in the +x direction at 0.6c with respect to an observer on the

ground. An astronaut stands at the rear of the rocket and fires a bullet toward the front of the rocket at 0.8c relative to the rocket. How long does it take the bullet to reach the front of the rocket (a) as measured in the frame of the rocket, (b) as measured in the frame of the ground, and (c) as measured in the frame of the bullet?

(a) In rocket frame, ∆t = ∆tp = Lp/u (b) 1. Use Equ. 39-18a to find uground 2. Find u′, speed of bullet relative to rocket as seen from the ground 3. Find L = Lp/γ and ∆tground = L/u′ (c) 1. Find L′, length of rocket in bullet frame 2. ∆tbullet = L′/u

∆t = (1000/0.8×3×108 ) s = 4.17 µs uground = 1.4c/(1 + 0.48) = 0.946c Relative to rocket, as seen from the ground, u′ = (0.946 - 0.6)c = 0.346c

γ = 1.25; ∆tground = (800/0.346×3×108 ) s = 7.71 µs

γ = 1.67; L′ = 600 m

∆tbullet = (600/0.8×3×108 ) s = 2.5 µs 80 ··· In a simple thought experiment, Einstein showed that there is mass associated with electromagnetic radiation.

Consider a box of length L and mass M resting on a frictionless surface. At the left wall of the box is a light source that emits radiation of energy E, which is absorbed at the right wall of the box. According to classical electromagnetic theory, this radiation carries momentum of magnitude p = E/c (Equation 32-13). (a) Find the recoil velocity of the box such that momentum is conserved when the light is emitted. (Since p is small and M is large, you may use classical mechanics.) (b) When the light is absorbed at the right wall of the box, the box stops, so the total momentum remains zero. If we neglect the very small velocity of the box, the time it takes for the radiation to travel across the box is ∆t = L/c. Find the distance moved by the box in this time. (c) Show that if the center of mass of the system is to remain at the same place, the radiation must carry mass m = E/c2.

(a) Since momentum is conserved, E/c + Mv = pi = 0. Therefore, v = -E/Mc. (b) Distance moved by the box is d = vL/c = -EL/Mc2 (c) Let x = 0 be at the center of the box and let the mass of the photon be m. Then initially the center of mass is at xCM = -1/2mL/(M + m). When the photon is absorbed at the other end of the box then the center of mass is at xCM = [-MEL/Mc2 + m(1/2L - EL/Mc2)]/(M + m). Since no external forces act on the system, the two expressions


for xCM must be equal. Solving for m gives m = E/[c2(1 - E/Mc2)]. But E/Mc2 << 1 since the box macroscopic, i.e., Mc2 ~ 1016 J whereas E = hf < 1 J for any reasonable values of f. Consequently, m = E/c2. 81* ··· A rocket with a proper length of 700 m is moving to the right at a speed of 0.9c. It has two clocks, one in the nose

and one in the tail, that have been synchronized in the frame of the rocket. A clock on the ground and the nose clock on the rocket both read t = 0 as they pass. (a) At t = 0, what does the tail clock on the rocket read as seen by an observer on the ground? When the tail clock on the rocket passes the ground clock, (b) what does the tail clock read as seen by an observer on the ground, (c) what does the nose clock read as seen by an observer on the ground, and (d) what does the nose clock read as seen by an observer on the rocket? (e) At t = 1 h, as measured on the rocket, a light signal is sent from the nose of the rocket to an observer standing by the ground clock. What does the ground clock read when the observer receives this signal? (f) When the observer on the ground receives the signal, he sends a return signal to the nose of the rocket. When is this signal received at the nose of the rocket as seen on the rocket?

We shall use the following notation: S is the ground reference frame, S′ is the reference frame of the rocket, and V = 0.9c is the speed of the rocket relative to S. We denote by T and N the tail and nose of the rocket, respectively. (a) 1. Write the initial conditions in S′ 2. Find xT using length contraction 3. Find tT′ at t = 0 using Equ. 39-12 (b) Find the time for rocket to move a distance L′ (c) tN = ∆t′ = (2.59 - 2.1) µs (d) tN′ = tT′ (clocks are synchronized in S′) (e) 1. Find ∆t, time the signal is sent; use Equ. 39-13 2. Find ∆ttravel, time of travel of signal to ground 3. Find trec, time the signal is received on ground (f) 1. Find ∆x when the signal is sent 2. In S, signal travels at 0.1c relative to rocket; find time t when signal reaches rocket. 3. Use Equ. 39-13 to find tN′

tN′ = 0, xN′ = 0, and tT′ = 0, xT′ = -L′ = -700 m xT = -L′/γ tT′ = γ(-VxT/c2) = VL′/c2 = 0.9×700/c = 2.1 µs tT′ = L′/V = 700/0.9×3×108 s = 2.59 µs tN = 0.49 µs tN′ = 2.59 µs

∆t = 2.294×1 h = 2.294 h

∆ttravel = ∆x/c = 2.294×0.9 h = 2.065 h trec = (2.294 + 2.065) h = 4.36 h

∆x = (4.36 h)(0.9c) = 3.924 c.h

∆t = (4.36 h)(0.9c/0.1c) = 39.24 h t = (39.24 + 3.924) h = 43.16 h tN′ = (43.16/2.294) h = 18.8 h

82 ··· An observer in frame S standing at the origin observes two flashes of colored light separated spatially by ∆x =

2400 m. A blue flash occurs first, followed by a red flash 5 µs later. An observer in S′ moving along the x axis at speed V relative to S also observes the flashes 5 µs apart and with a separation of 2400 m, but the red flash is observed first. Find the magnitude and direction of V.

1. ∆t′ = γ(∆t - V∆x/c2) (see Problem 21a); set ∆t′ =-∆t and solve for V 2. Substitute numerical values

-∆t/γ = ∆t - V∆x/c2;

∆t2(1 - V 2/c2) = ∆t2 - 2V∆x∆t/c2 + V 2∆x2/c4 V = 2(∆x/∆t)/[1 + (∆x/c∆t)2] V = 2.70×108 m/s = 0.9c; positive x direction.

83 ··· Reference frame S′ is moving along the x′ axis at 0.6c relative to frame S. A particle that is originally at x′ = 10 m

at 0 =’ t x is suddenly accelerated and then moves at a constant speed of c/3 in the -x′ direction until time

cm/ 60 =’ t2 , when it is suddenly brought to rest. As observed in frame S, find (a) the speed of the particle, (b) the

distance and direction the particle traveled from t′1 to t′2, and (c) the time the particle traveled.


(a) Use Equ. 39-16a; ux′ = -c/3 (b) 1. Find x2′; ∆t′ = t2′ - t1′ = 60 m/c = 200 ns 2. Find x2 using Equ. 39-9 3. Find x1 and ∆x = x2 - x1 (c) 1. Find t1 and t2 2. ∆t = t2 - t1

ux = (3/5 - 1/3)c/(1 - 0.2) = c/3 x2′ = 10 - (60 m/c)(c/3) = -10 m

γ = 1.25; x2 = 1.25(-10 + 36) m = 32.5 m x1 = 1.25(10 m) = 12.5 m; ∆x = 20 m t1 = 1.25(6 m/c) = 7.5 m/c = 25 ns t2 = 1.25(60 m/c - 6 m/c) = 225 ns

∆t = 200 ns 84 ··· In reference frame S the acceleration of a particle is a = ax i + ay j + az k. Derive expressions for the acceleration

components ax', ay', and az' of the particle in reference frame S' that is moving relative to S in the x direction with velocity V.

From Equ. 39-19a we have )c/Vu (1

du )cV)(V/ u( + du )c/Vu (1 =

c/Vu 1V u d = du 22

x

x2

xx2

x2

x

xx

−−−

−

−′

du )c/Vu (1)c/V (1

= x22x

22

−−

.

From Equ. 39-10, dt′ = γd(t - Vx/c2) = γdt - (γV/c2)dx = γdt - (γV/c2)(dx/dt)dt = γ(1 - Vux/c2)dt. We now obtain

ax′ =

a = dtdu

)c/Vu (1 )c/V (1

= dt

du33

xx32

x

22x

δγγ −−

′′

, where δ = (1 - Vux/c2). Proceeding in exactly the same manner,

one obtains

a)c/ Vu( +

a = a 33

x2

y

22y

yδγδγ

′ and an identical expression for az′ with z replacing y.

85* ··· When a projectile particle with kinetic energy greater than the threshold kinetic energy Kth strikes a stationary

target particle, one or more particles may be created in the inelastic collision. Show that the threshold kinetic energy of the projectile is given by

Here Σmin is the sum of the rest masses of the projectile and target particles, Σmfin is the sum of the rest masses of the final particles, and mtarget is the rest mass of the target particle. Use this expression to determine the threshold kinetic energy of protons incident on a stationary proton target for the production of a proton–antiproton pair; compare your result with that of Problem 60. In solving this problem we shall adopt the convention of the problem statement and use m to denote rest masses rather than relativistic masses. Let mi denote the mass of the incident (projectile) particle. Then Σmin = mi + mtarget. Consider now the situation in the center of mass reference frame. At threshold we have E 2 -p2c2 = Σmfin c2. Note that this is a relativistically invariant expression. In the laboratory frame, the target is at rest so Etarget = Et = Et,0. We can therefore

m

mmmmK

target

2infinfinin

th2

c) -( ) + ( =

∑∑∑∑


write (Ei + Et,0)2 - pi2c2 = (Σmfin c2)2. For the incident particle, Ei

2 - pi2c2 = Ei,0

2 and Ei = Ei,0 + Kth, where Kth is the threshold kinetic energy of the incident particle in the laboratory frame. We can now express Kth in terms of the rest energies: (Et,0 + Ei,0)2 + 2KthEt,0 = (Σmfin c2)2. But Et,0 + Ei,0 = Σmin c2 and Et,0 = mtarget c2. Solving for Kth we obtain

m

mmmmK

target

2infinfinin

th2

c ))( + ( =

Σ−ΣΣΣ

For the creation of a proton - antiproton pair in a proton - proton collision, Σmin = 2mp, Σmfin = 4mp, and mtarget = mp. The above expression then gives Kth = (6×2/2)mp c2 = 6mp c2, where mp denotes the rest mass of a proton. 86 ··· A particle of rest mass M0 decays into two identical particles of rest mass m0, where m0 = 0.3M0. Prior to the

decay, the particle of rest mass M0 has an energy of 4M0c2 in the laboratory. The velocities of the decay products are along the direction of motion of M0. Find the velocities of the decay products in the laboratory.

We shall solve the problem for the general case of a particle of rest mass M0 decaying into two identical particles each of rest mass m0. In the center of mass reference frame, M0c2 = 2mc2 = 2γm0c2. Solving for u/c we obtain

u/c = )M/m(2 1 200− , where u is the speed of each of the decay particles in the CM frame. Next we determine

the speed V of the laboratory frame relative to the CM frame. The energy of the particle of rest mass M0 is γCMM0c2,

where γCM = 1/ c/V 1 22− and V/c = βCM = 1/ 1 γ 2CM− We can now use the velocity transformation,

Equ. 39-18a to determine ulab, the speeds of the decay products in the laboratory reference frame. The result is

cu (u/c) 1u/c

= CM

CMlab

ββ±

±. The ± refers to the fact that one of the decay particles will travel in the direction

of M0, the other in the direction opposite to that of M0. In the present instance, γCM = 4, βCM = 0.968, 2m0/M0 = 0.6, and u/c = 0.8. We find ulab = 0.996c and ulab = 0.775c. 87 ··· A stick of proper length Lp makes an angle θ with the x axis in frame S. Show that the angle θ ′ made with the x′

axis in frame S′, which is moving along the +x axis with speed V, is given by tan θ ′ = γ tan θ and that the length of the stick in S′ is

′ θθ

γ + L = L 22

2

1/2

sincos1

p

In its reference frame, the stick has x and y components Lpx = Lp cos θ and Lpy = Lp sin θ. Only Lpx is Lorentz contracted to Lx′ = Lpx/γ, so the length in the reference frame S′ is L′ = (Lx′2 + Ly′2)1/2 = Lp (cos2 θ/γ2 + sin2 θ)1/2. The angle that L′ makes with the x′ axis is given by tan θ ′ = Ly′/Lx′ = sin θ/(cos θ/γ) = γ sin θ/cos θ = γ tan θ. 88 ··· Show that if a particle moves at an angle θ with the x axis with speed u in frame S, it moves at an angle θ ′ with

the x′ axis in S′ given by

V/u) - (

= θγ

θθ cos

sintan ′

The angle of u′ with the x′ axis is tan θ ′ = V/u) (

=

V) (u u

= V u

c/Vu 1

)c/Vu (1u

= ’u

’u

x

2x

2x

y

x

y

−−−−

− θγθ

θγθ

γ cossin

cossin

.


89* ··· For the special case of a particle moving with speed u along the y axis in frame S, show that its momentum and

energy in frame S′ are related to its momentum and energy in S by the transformation equations. Compare these equations with the Lorentz transformation for x′, y′, z′, and t′. These

e

c

Vp -

cE

= cE’

;’p =’ p ,p =’ p ,c

VE - p =’ P 2

xzzyy2xx γγ quations show that the quantities px, py, pz, and

E/c transform in the same way as do x, y, z, and ct.

In S, ux = uz = 0, uy = u; px = pz = 0, py = γu m0u, and E = γu m0c2. Here, γu = c/u 11/ 22− . Then,

applying the velocity transformation equations we find, in S′, ux′ = -V, uy′ = u/γ, uz′ = 0. This gives u′2 = V 2 + u2(1 - V 2/c2) = V 2 + u2 - V 2u2/c2 and (1 - u′2/c2) = 1 - V 2/c2 - u2/c2 + V 2u2/c4 = (1 - V 2/c2)(1 - u2/c2). In

S′ the momentum components are px′ = γ′m0ux′, py′ = γ′m0uy′, and pz′ = 0, where

γ′ = c/V 1/ 22u −− γ = c/’u 11/ 22 . In terms of the parameters in S, px′ = -γu m0V/ c/V 1 22− = -γEV/c2, where

c/V 11/ = 22−γ . Since px = 0, px′ = γ(px - EV/c2). In terms of the parameters in S, py′ = py and pz′ = pz.

E′ = γ′m0c2 = γγu m0c2 = γE = γ(E - Vpx/c) and E′/c = γ(E/c - Vpx/c2). Note that E′2 - p′2c2 = E02 (see Problem

91), which demonstrates that E0 is a relativistic invariant. Also, comparison with Equs. 39-11 and 39-12 shows that the components of p and E/c transform as do the components of r and ct. 90 ··· The equation for the spherical wavefront of a light pulse that begins at the origin at time t = 0 is x2 + y2 + z2 -(ct)2 = 0. Using the Lorentz transformation, show that such a light pulse also has a spherical wavefront in frame S′ by showing that x′2 + y′2 + z′2 -(ct′)2 = 0 in S′. The Lorentz transformation was derived on the basis of the postulate that the speed of light is c in any inertial reference frame. Thus, if the clocks in S and S′ are synchronized at t = t′ = 0, then it follows from the Einstein postulate that r2 = c2t2 and r′2 = c2t′2 or r2 - c2t2 = 0 = r′2 - c2t′2. In other words, the quantity s2 = r2 - c2t2 = 0 is a relativistic invariant, which can also be written as x2 + y2 + z2 - c2t2 = 0. Using the Lorentz transformation equations for x, y, z, and t we have x′2 + y′2 + z′2 - (ct′)2 = γ2(x2 - 2Vxt + V 2t2) + y2 + z2 - γ2(c2t2 - 2Vxt + V 2x2/c2). The terms linear in x cancel; the terms in x2 combine to give γ2x2(1 - V 2/c2) = x2; the coefficients of the terms in (ct)2 give γ2(V 2/c2 - 1) = -1. Thus, r2 - c2t2 = r′2 - c2t′2 as required by the Einstein postulate. 91 ··· In Problem 90, you showed that the quantity x2 + y2 + z2 -(ct)2 has the same value (0) in both S and S′. Such a

quantity is called an invariant. From the results of Problem 89, the quantity )(E/c - p + p + p 22z

2y

2x must also be an

invariant. Show that this quantity has the value -m0c2 in both the S and S′ reference frames. From Equ. 39-28, p2c2 - E2 = -E0

2 or p2 - (E/c)2 = -m02c2. Since m0 is the rest mass, i.e., the mass of the particle in

its rest frame, it is a constant. It follows that p2 - (E/c)2 must be a relativistic invariant. Also, in Problem 89 we saw that the components of p and the quantity E/c transform like the components of r and the quantity ct. In Problem 90 we demonstrated that r2 - (ct)2 is a relativistic invariant. Consequently, p2 - (E/c)2 must also be relativistically invariant. 92 ··· Two identical particles of rest mass m0 are each moving toward the other with speed u in frame S. The particles

collide inelastically with a spring that locks shut (Figure 39-14) and come to rest in S, and their initial kinetic energy is transformed into potential energy. In this problem you are going to show that the conservation of momentum in


reference frame S′, in which one of the particles is initially at rest, requires that the total rest mass of the system after

the collision be c/u - 1/m2 220 . (a) Show that the speed of the particle not at rest in frame S′ is

)c/u + 2u/(1 =u’ 22 0and use this result to show that

c/u + 1c/u - 1

= c’u

- 122

22

2

2

(b) Show that the initial momentum in frame S′ is )c/u - u/(1m2 =p’ 220 0. (c) After the collision, the composite

particle moves with speed u in S′ (since it is at rest in S). Write the total momentum after the collision in terms of the

final rest mass M0, and show that the conservation of momentum implies that c/u - 1/m2 = M 2200 .

(d) Show that the total energy is conserved in each reference frame.

(a) 1. Use Equ. 39-19a, where V = -u

2. Express c’u

12

2

− in terms of u and c

(b) Write p′ = γ′m0u′; γ′ =

1/ c/’u 1 22−

(c) pf′ = γM0u = p′; γ = 1/ c/u 1 22−

(d) 1. Write Ei and Ef in S 2. Write Ei′ and Ef′ in S′

u′ = 2u/(1 + u2/c2)

c/u + 1c/u 1

= )c/u + (1)c/u (1

= c/u + c/u2 + 1

c/u4 1

22

22

222

222

4422

22 −−−

p′ = c/u 1

um2 =

c/u + 12u

c/u 1c/u + 1

m 220

2222

22

0 −−

M0 = 2m0/γ(1 - u2/c2) = 2m0γ Ei = 2m0γc2; Ef = M0c2 = 2m0γc2; Ei = Ef Ei′ = m0c2 + γ′m0c2 = 2m0c2/(1 - u2/c2) Ef′ = γM0c2 = 2m0c2/(1 - u2/c2); Ei′ = Ef′

ch39

Documents

length of b

vt b x

time t

reference frame

electron frame

length of accelerator

c relative

ship b