Resolução do tipler-vol.1-Chap R SM

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Chapter R 
Relativity 
 
Conceptual Problems 
 
1 • You are standing on a corner and a friend is driving past in an 
automobile. Each of you is wearing a wristwatch. 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. Have either of you 
determined the proper time interval? Explain your answer. 
 
Determine the Concept In the reference frame of the car both events occur at the 
same location (the location of the car). Thus, your friend’s watch measures the 
proper time between the two events. 
 
2 • In Problem 1, suppose your friend in the car measures the width of the 
car door to be 90 cm. You also measure the width as he goes by you. 
(a) Does either of you measures the proper width of the door? Explain your 
answer. (b) How will your value for the door width compare to his? (1) Yours will 
be smaller, (2) yours will be larger, (3) yours will be the same, (4) you can’t 
compare the widths, as the answer depends on the car’s speed. 
 
Determine the Concept The proper length of an object is the length of the object 
in the rest frame of the object. The proper length of a meter stick is one meter. 
 
(a) Because the door is at rest in the reference frame of the car, its width in that 
frame is its proper width. If your friend measures this width, say by placing a 
meter stick against the door, then he will measure the proper width of the door. 
 
(b) In the reference frame in which you are at rest, the door is moving, so its 
width is less than its proper width. To measure this width would be challenging. 
(You could measure the width by measuring the time for the door to go by. The 
width of the door is the product of the speed of the car and the time.) 
 
3 • [SSM] If event A occurs at a different location than event B in some 
reference frame, might it be possible for there to be a second reference frame in 
which they occur at the same location? If so, give an example. If not, explain why 
not. 
 
Determine the Concept Yes. Let the initial frame of reference be frame 1. In 
frame 1 let L be the distance between the events, let T be the time between the 
events, and let the +x direction be the direction of event B relative to event A. 
Next, calculate the value of L/T. If L/T is less than c, then consider the two events 
in a reference frame 2, a frame moving at speed v = L/T in the +x direction. In 
frame 2 both events occur at the same location. 
 
1051 
 Chapter R 
 
 
1052 
4 • [SSM] If event A occurs prior to event B in some frame, might it be 
possible for there to be a second reference frame in which event B occurs prior to 
event A? If so, give an example. If not, explain why not. 
 
Determine the Concept Yes. Let L and T be the distance and time between the 
two events in reference frame 1. If L ≤ cT, then something moving at a speed 
less than or equal to c could travel from the location of event A to the location of 
event B in a time less than T. Thus, it is possible that event A could cause event 
B. For events like these, causality demands that event A must precede event B in 
all reference frames. However, if L > cT then event A cannot be the cause of 
event B. For events like these, event B does precede event A in certain reference 
frames. 
 
5 • [SSM] Two events are simultaneous in a frame in which they also 
occur at the same location. Are they simultaneous in all other reference frames? 
 
Determine the Concept Yes. If two events occur at the same time and place in 
one reference frame they occur at the same time and place in all reference frames. 
(Any pair of events that occur at the same time and at the same place in one 
reference frame are called a spacetime coincidence.) 
 
6 • Two inertial observers are in relative motion. Under what 
circumstances can they agree on the simultaneity of two different events? 
 
Determine the Concept We will refer to the two events as event A and event B. 
Assume that in the reference frame of the first observer there is a stationary clock 
at the location of each event, with clock A at the location of event A and clock B 
at the location of event B, and that the two clocks are synchronized. Because the 
two events are simultaneous in this frame, the readings of the two clocks at the 
time the events occur are the same. Also, event A and the reading of clock A at 
the time of event A are a spacetime coincidence, so all observers must agree with 
that clock reading. In like manner, event B and the reading of clock B at the time 
of event B are a spacetime coincidence. If observer B is moving parallel with the 
line joining the two clocks then the clocks readings will differ by Lv/c2 in the 
reference frame of B, where L is the distance between the clocks in the reference 
frame of observer A. This means that observer B will agree that the two clock 
readings at the times of the events are the same, but will not agree that the events 
occurred at the same time unless L = 0. 
 
7 • The approximate total energy of a particle of mass m moving at speed 
v << c is (a) mc2 + 
1
2 mv
2, (b) mv2, (c) cmv, (d) 
1
2 mc
2, (e) cmv. 
 
 
 
Relativity 
 
 
1053
Picture the Problem We can use the expression for the total relativistic energy 
of the particle to express its energy in terms of its speed and expand the radical 
factor binomially to find the correct approximate expression for the total energy 
when v << c. 
 
Express the total relativistic energy 
of the particle as the sum of its 
kinetic energy and the rest energy: 
 21
2
2
2
2
2
2
2
1
1
−
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
−
=+=
c
vmc
c
v
mcmcKE
 
 
Expand the radical factor binomially 
to obtain: 
. for 
......1
2
2
12
2
2
12
2
2
2
12
cvmvmc
mvmc
c
vmcE
<<+≈
++=⎟⎟⎠
⎞
⎜⎜⎝
⎛ ++=
and )(a is correct. 
 
8 • True or false: 
 
(a) The speed of light is the same in all reference frames. 
(b) The proper time interval 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) For two events to form a space-time coincidence they must occur at the same 
 place. 
(f) If two events are not simultaneous in one frame, they cannot be simultaneous 
in any other frame. 
 
(a) True. This is Einstein’s second postulate. 
 
(b) True. The time between events that happen at the same place in a reference 
frame is called the proper time and the time interval Δt measured in any other 
reference frame is always longer than the proper time. 
 
(c) False. Absolute motion cannot be detected. 
 
(d) True. A light-year is the distance light travels (in a vacuum) in one year. 
 
(e) True. For two events to form a space-time coincidence, they must not only 
occur at the same place but must also occur at the same location. 
 
 Chapter R 
 
 
1054 
 
(f) False. The fact that two events are not simultaneous in one frame tells us 
nothing about their simultaneity in any other frame. 
 
9 •• (a) Show that pc has dimensions of energy. (b) There is a geometrical 
interpretation of Equation R-17 based on the Pythagorean Theorem. Draw a 
picture of a triangle illustrating this interpretation. 
 
(a) Express the dimensions of pc 
and simplify them to obtain: [ ][ ]
[ ]E
T
ML
T
LM
T
L
T
LMcp
==
⎟⎠
⎞⎜⎝
⎛⋅=⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛ ⋅=
2
2
2
 
 
(b) The geometrical interpretation, 
based on the Pythagorean Theorem, 
of Equation R-17 is shown to the 
right: 
 
E pc
2mc 
 
10 •• A wad of putty of mass m1 strikes and sticks to a second wad of putty 
of mass m2 which is initially at rest. Do you expect that after the collision the 
combined putty
mass will be (a) greater than, (b) less than, (c) the same as 
m1 + m2. Explain your answer. 
 
Determine the Concept Because the collision is inelastic, kinetic energy is 
converted into mass energy. ( )a is correct. 
 
11 •• [SSM] Many nuclei of atoms are unstable; for example, 14C, an 
isotope of carbon, has a half-life of 5700 years. (By definition, the half-life is the 
time it takes for any given number of unstable particles to decay to half that 
number of particles.) This fact is used extensively for archeological and 
biological dating of old artifacts. Such unstable nuclei decay into several decay 
products, each with significant kinetic energy. Which of the following is true? 
(a) The mass of the unstable nucleus is larger than the sum of the masses of the 
decay products. (b) The mass of the unstable nucleus is smaller than the sum of 
the masses of the decay products. (c) The mass of the unstable nucleus is the 
same as the sum of the masses of the decay products. Explain your choice. 
 
Determine the Concept Because mass is converted into the kinetic energy of the 
fragments, the mass of the unstable nucleus is larger than the sum of the masses of 
the decay products. ( )a is correct. 
Relativity 
 
 
1055
12 •• Positron emission tomography (PET) scans are common in modern 
medicine. In this procedure, positrons (a positron has the same mass, but the 
opposite charge as an electron) are emitted by radioactive nuclei that have been 
introduced into the body. Assume that an emitted positron, traveling slowly (with 
negligible kinetic energy) collides with an electron traveling at the same slow 
speed in the opposite direction. They undergo annihilation and two quanta of light 
(photons) are formed. You are in charge of designing detectors to receive these 
photons and measure their energy. (a) Explain why you would expect these two 
photons to come off in exactly opposite directions. (b) In terms of the electron 
mass me , how much energy would each photon have? (1) less than mc2, 
(2) greater than mc2, (3) exactly mc2. Explain your choice. 
 
Determine the Concept 
(a) The linear momentum of the electron-positron pair was zero just before their 
annihilation and the emission of the two photons and therefore the linear 
momentum of the two photons must be zero. 
 
(b) ( )3 is correct. Each photon has the rest mass energy, mec2, equivalent of an 
electron. Because there are two photons and they have the same momentum, they 
share equally, each getting half the available initial rest energy of 2mec2. 
 
Estimation and Approximation 
 
13 •• [SSM] In 1975, an airplane carrying an atomic clock flew back and 
forth at low altitude for 15 hours at an average speed of 140 m/s as part of a time-
dilation experiment. The time on the clock was compared to the time on an atomic 
clock kept on the ground. What is the time difference between the atomic clock 
on the airplane and the atomic clock on the ground? (Ignore any effects that 
accelerations of the airplane have on the atomic clock which is on the airplane. 
Also assume that the airplane travels at constant speed.) 
 
Picture the Problem We can use the time dilation equation to relate the elapsed 
time in the frame of reference of the airborne clock to the elapsed time in the 
frame of reference of the clock kept on the ground. 
 
Use the time dilation equation to 
relate the elapsed time Δt according 
to the clock on the ground to the 
elapsed time Δt0 according to the 
airborne atomic clock: 
 
2
0
1
Δ
Δ
⎟⎠
⎞⎜⎝
⎛−
=
c
v
tt (1) 
Because v << c, we can use the 
approximation x
x 2
11
1
1 +≈− 
to obtain: 
2
2 2
11
1
1 ⎟⎠
⎞⎜⎝
⎛+≈
⎟⎠
⎞⎜⎝
⎛− c
v
c
v
 
 Chapter R 
 
 
1056 
 
Substitute in equation (1) to obtain: 
 
0
2
0
0
2
2
1
2
11
t
c
vt
t
c
vt
Δ⎟⎠
⎞⎜⎝
⎛+Δ=
Δ⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛+=Δ
 (2) 
where the second term represents the 
additional time measured by the clock 
on the ground. 
 
Evaluate the proper elapsed time 
according to the clock on the 
airplane: 
 
( ) s1040.5
h
s3600h15 40 ×=⎟⎠
⎞⎜⎝
⎛=Δt 
 
Substitute numerical values and 
evaluate the second term in equation 
(2): 
( )
ns9.5s1089.5
s1040.5
m/s10998.2
m/s140
2
1
Δ
9
4
2
8
≈×=
×⎟⎟⎠
⎞
⎜⎜⎝
⎛
×=
−
t'
 
 
14 •• (a) By making any necessary assumptions and finding certain stellar 
distances, estimate the speed at which a spaceship would have to travel for its 
passengers to make a trip to the nearest star (not the Sun!) and back to Earth in 1.0 
Earth years, as measured by an observer on the ship. Assume the passengers make 
the outgoing and return trips at constant speed and ignore any effects due to the 
spaceship stopping and starting. (b) How much time would elapse on Earth during 
their roundtrip? Include 2.0 Earth years for a low-speed exploration of the planets 
in the vicinity of this star. 
 
Picture the Problem We can use the length-contraction equation 
( )20 1 cvLL −= and vLt =Δ to estimate the speed of the spaceship. Note that 
there is no time dilation effect during the two-year exploration period. Take the 
distance to the nearest sun to be 4.0 c⋅y. 
 
(a) According to the travelers, the 
elapsed time for the trip is: 
 
( )
v
cvL
v
Lt
2
0 1Δ
−== 
 
Multiplying both sides of the 
equation by c yields: 
 
( ) ( )
cv
cvL
v
cvcL
tc
2
0
2
0 11Δ
−=−= 
 
Because cΔt = 1.0 c⋅y and 
L0 = 4.0 c⋅y: ( )cv
cv 210.4
0.1
−= 
 
Relativity 
 
 
1057
Solve for v to obtain: 
ccv 97.0
17
16 == 
 
(b) Express the elapsed time as the 
sum of the travel and exploration 
times: 
 
nexploratio wayl1
nexploratiotravel
ΔΔ2
ΔΔΔ
tt
ttt
+=
+=
 (1) 
In the Earth frame-of reference: 
 y 12.40.97
y 0.4
Δ 0 way1 =⋅== c
c
v
Lt 
 
Substitute numerical values in 
equation (1) and evaluate Δt: ( ) y 2.10y 0.2y 12.42Δ =+=t 
 
15 •• (a) Compare the kinetic energy of a moving car to its rest energy. 
(b) Compare the total energy of a moving car to its rest energy. (c) Estimate the 
error made in computing the kinetic energy of a moving car using non-relativistic 
expressions compared to the relativistically correct expressions. Hint: Use of the 
binomial expansion may help. 
 
Picture the Problem We can compare these energies by expressing their ratios. 
Assume that the speed of the car is 30 m/s (≈ 67 mi/h). 
 
(a) Express the ratio of the kinetic 
energy of a moving car to its rest 
energy: 
 
2
2
2
2
1
2 2
1 ⎟⎠
⎞⎜⎝
⎛==
c
v
mc
mv
mc
K 
 
For a car moving at 30 m/s: 
 
%105
m/s 10998.2
m/s 30
2
1
13
2
82
−×≈
⎟⎠
⎞⎜⎝
⎛
×=mc
K
 
 
(b) Express the ratio of the total 
energy of a moving car to its rest 
energy: 
500 000 000 000 000.1
122
2
2
≈
+=+=
mc
K
mc
mcK
mc
E
 
 
(c) The error made in computing the 
kinetic energy of a moving car using 
the non-relativistic expression 
compared to the relativistically 
correct expression is given by: 1
Δ
rel
rel-non
rel
relrel-non
rel
−=
−=
K
K
K
KK
K
K
 (1) 
 
From Equation R-14, the relativistic 
kinetic energy is given by: 2
2
2
rel 11
2
1
mc
c
vK ⎥⎥⎦
⎤
⎢⎢⎣
⎡
−⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
−
 (2) 
 Chapter R 
 
 
1058 
Expand 
2
1
2
2
1
−
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
c
v binomially to obtain: 
 
4
4
2
2
4
4
2
2
2
2
8
3
2
11...
8
3
2
111
2
1
c
v
c
v
c
v
c
v
c
v ++≈+++=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
−
 
 
Substituting in equation (2) and
simplifying yields: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
⎟⎟⎠
⎞
⎜⎜⎝
⎛+=
⎥⎦
⎤⎢⎣
⎡ +=
2
2
rel-non
2
2
22
2
4
4
2
2
rel
4
31
8
3
2
1
8
3
2
1
c
vK
c
vmvmv
mc
c
v
c
vK
 
 
Substitute for Krel in equation (1) 
and simplify to obtain: 
1
4
31
1
4
31
Δ
1
2
2
2
2
rel-non
rel-non
rel
−⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
−
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
−
c
v
c
vK
K
K
K
 
 
Expanding 
1
2
2
4
31
−
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
c
v binomially yields: 
 
( ) ( )( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛−≈+⎟⎟⎠
⎞
⎜⎜⎝
⎛−−+⎟⎟⎠
⎞
⎜⎜⎝
⎛−+=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−
2
22
2
2
2
21
2
2
4
31...
4
3
2
21
4
311
4
31
c
v
c
v
c
v
c
v 
 
Substitute for 
1
2
2
4
31
−
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
c
v in the 
expression for 
rel
Δ
K
K to obtain: 
 
2
2
2
2
rel 4
31
4
31Δ
c
v
c
v
K
K =−⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= 
Relativity 
 
 
1059
 
Substitute numerical values and 
evaluate 
rel
Δ
K
K : 
( )
( )
15
28
2
rel
105.7
m/s 10998.2
m/s 30
4
3Δ
−×=
×=K
K
 
 
Length Contraction and Time Dilation 
 
16 • The proper average (or mean) lifetime of a pion (a subatomic particle) 
is 2.6 × 10–8 s. (A neutral pion has a much shorter lifetime. See Chapter 41.) A 
beam of pions has a speed of 0.85c relative to a laboratory. (a) What would be 
their mean lifetime as measured in that laboratory? (b) On average, how far 
would they travel in that laboratory before they decay? (c) What would be your 
answer to Part (b) if you had neglected time dilation? 
 
Picture the Problem The pion beam enters the laboratory at a point called the 
entrance port (EP). A typical pion in the beam disintegrates at a second point 
called the mean disintegration location (MDL). Event A occurs when a typical 
pion enters the room at EP, and event B occurs when the pion disintegrates at 
MDL. (a) In the reference frame of the beam these two events occur at the same 
location, so the time Δt0 between these two events in this frame is the proper time 
between the events. We can use the time dilation formula to calculate the time Δt 
between these events in the reference frame of the laboratory, a reference frame 
that moves with speed v = 0.85c relative to the reference frame of the beam. 
(b) In the reference frame of the room a typical pion travels at speed v for time 
Δt. 
 
(a) Sketch a typical pion in the 
reference frame of the pion beam. 
In this frame the typical pion is at 
rest and both EP and MDL are 
moving at speed v. Event A occurs 
when EP is next to a typical pion, 
and event B occurs when MDL 
reaches the pion at the instant the 
pion disintegrates: 
 
 
EP MDL
Reference frame of beam
Typical pion at rest
v v
 
 
 
Sketch a typical pion in the reference 
frame of the laboratory. In this 
frame the typical pion moves at 
speed v. Event A occurs when a 
typical pion enters the room at EP, 
and event B occurs when the pion 
reaches MDL at the instant the pion 
disintegrates: 
 
EP MDL
Laboratory reference frame 
Typical pion in motion
v
 
 Chapter R 
 
 
1060 
Use the time-dilation equation to 
relate the mean lifetime Δt of a 
typical pion in the laboratory frame 
to the proper mean lifetime Δt0 in the 
reference frame of the beam: 
 
0
2
1
tt
v
c
ΔΔ =
⎛ ⎞− ⎜ ⎟⎝ ⎠
 
Substitute numerical values and 
evaluate Δt: 
 
s 109.4
s 1094.4
85.01
s 106.2
Δ
8
8
2
8
−
−
−
×=
×=
⎟⎠
⎞⎜⎝
⎛−
×=
c
c
t
 
 
(b) Express the distance a typical 
pion will travel in the lab frame 
before it decays in terms of its speed 
and its mean lifetime in the lab 
frame: 
 
x v tΔ = Δ 
Substitute numerical values and 
evaluate Δx: 
 
( )( ) m 31s 1094.485.0Δ 8 =×= −cx 
(c) Neglecting time dilation, 
Δt = 2.6 × 10−8 s, and: ( )( ) m .66s 106.285.0Δ 8 =×= −cx 
 
17 • [SSM] In the reference frame of a pion in Problem 16, how far does 
the laboratory travel in 2.6 × 10–8 s? 
 
Picture the Problem We can use πtvx Δ=Δ , where Δtπ is the proper mean 
lifetime of the pions, to find the distance traveled by the laboratory frame in a 
typical pion lifetime. 
 
The average distance the laboratory 
will travel before the pions decay is 
the product of the speed of the pions 
and their proper mean lifetime: 
 
πtvx Δ=Δ 
Substitute numerical values and 
evaluate Δx: 
 
( )( )
m6.6
m 63.6s106.285.0Δ 8
=
=×= −cx
 
 
18 • The proper average (or mean) lifetime of a muon (a subnuclear 
particle) particle is 2.20 μs. Muons in a beam are traveling at 0.999c relative to a 
laboratory. (a) What is their lifetime as measured in that laboratory? (b) On 
average, how far do they travel in that laboratory before they decay? 
Relativity 
 
 
1061
Picture the Problem We can express the mean lifetimes of the muons in the 
laboratory in terms of their proper lifetimes using ( )20 1ΔΔ cvtt −= . The 
average distance the muons will travel before they decay is related to their speed 
and mean lifetime in the laboratory frame according to .tvx Δ=Δ 
 
(a) Use the time-dilation equation to 
relate the mean lifetime of a muon in 
the laboratory Δt to their proper 
mean lifetime Δt0: 
 
2
0
1
Δ
Δ
⎟⎠
⎞⎜⎝
⎛−
=
c
v
tt 
Substitute numerical values and 
evaluate Δt: 
 
s49
s2.49
0.9991
s1020.2
Δ
2
6
μ
μ
=
=
⎟⎠
⎞⎜⎝
⎛−
×=
−
c
c
t
 
 
(b) The average distance muons 
travel before they decay in terms of 
their speed and mean lifetime in the 
laboratory frame of reference is: 
 
tvx Δ=Δ 
Substitute numerical values and 
evaluate Δx: 
( )( ) km15s2.49999.0Δ == μcx 
 
19 • In the reference frame of the muon in Problem 20, how far does the 
laboratory travel in 2.20 μs? 
 
Picture the Problem We can use ,πtvx Δ=Δ where Δtπ is the proper mean 
lifetime of the pions, to find the distance traveled by the laboratory frame in a 
typical pion lifetime and ( )20 1ΔΔ cvxx −= to find this distance in the 
laboratory’s frame. 
 
Express the average distance the 
laboratory will travel before the pions 
decay in terms of the speed of the 
pions and their proper mean lifetime: 
 
πtvx Δ=Δ 
Substitute numerical values and 
evaluate Δx: 
 
( )( ) m659s20.2999.0Δ == μcx 
 
 
 Chapter R 
 
 
1062 
20 • You have been posted in a remote region of space to monitor traffic. 
Toward the end of a quiet shift, a spacecraft goes by and you measure its length 
using a laser device. This device reports a length of 85.0 m. You flip open your 
handy reference catalogue and identify the craft as a CCCNX-22, which has a 
proper length of 100 m. When you phone in your report, what speed should you 
give for this spacecraft? 
 
Picture the Problem The measured length L of the spacecraft is related to its 
proper length L0 and its speed according to ( )220 1 cvLL −= . We can solve this 
equation for v as a function of c, L, and L0. 
 
Express the length L of the 
spacecraft in terms of its proper 
length L0: 
 
2
2
0 1 c
vLL −= ⇒
2
0
1 ⎟⎟⎠
⎞
⎜⎜⎝
⎛−=
L
Lcv 
Substitute numerical values and 
evaluate v: ( )
m/s1058.1
m100
m0.851m/s10998.2
8
2
8
×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−×=v
 
 
21 • [SSM] A spaceship travels from Earth to a star 95 light-years away 
at a speed of 2.2 × 108 m/s. How long does the spaceship take to get to the star 
(a) as measured on Earth and (b) as measured by a passenger on the spaceship? 
 
Picture the Problem We can use Δx = vΔt to find the time for the trip as 
measured on Earth and ( )20 1ΔΔ cvtt −= to find the time measured by a 
passenger
on the spaceship. 
 
(a) Express the elapsed time, as 
measured on Earth, in terms of the 
distance traveled and the speed of the 
spaceship: 
 
v
xt Δ=Δ 
Substitute numerical values and 
evaluate Δt: 
 
y103.1
y129
Ms31.56
y1s1009.4
y
m10461.9
m/s102.2
y95
Δ
2
9
15
8
×=
=××=
⋅
×××
⋅=
c
ct
 
 
Relativity 
 
 
1063
(b) A passenger on the spaceship 
will measure the proper time: 
 
2
0 1ΔΔ ⎟⎠
⎞⎜⎝
⎛−=
c
vtt 
 
Substitute numerical values and 
evaluate the proper time: ( ) ( )( )
y88
m/s10998.2
m/s102.21y129Δ 28
28
0
=
×
×−=t
 
 
22 • The average lifetime of a beam of subatomic particles called pions (see 
Problem 16 for details on these particles) traveling at high speed is measured to 
be 7.5 × 10–8 s. Their average lifetime at rest is known to be 2.6 × 10–8 s. How 
fast is this pion beam traveling? 
 
Picture the Problem We can express the mean lifetime of the pion in the 
laboratory in terms of its proper lifetime using ( )20 1ΔΔ cvtt −= and solve for v 
to find its speed. 
 
Use the time-dilation equation to 
relate the mean lifetime of a muon in 
the laboratory Δt to their proper 
mean lifetime Δt0: 
 
2
2
0
1
Δ
Δ
c
v
tt
−
= ⇒
2
01 ⎟⎠
⎞⎜⎝
⎛
Δ
Δ−=
t
tcv 
Substitute numerical values and evaluate v: 
 
( ) m/s108.2
s105.7
s106.21m/s10998.2 8
2
8
8
8 ×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
×
×−×= −
−
v 
 
23 • A meterstick moves with speed 0.80c 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? 
 
Picture the Problem We can find the measured length L of the meterstick using ( )220 1 cvLL −= and the time it takes to pass you using L = vΔt. 
 
(a) Express the length L of the 
meterstick in terms of its proper 
length L0: 
 
2
2
0 1 c
vLL −= 
Substitute numerical values and 
evaluate L: ( ) ( ) cm608.01m0.1 2
2
=−=
c
cL 
 Chapter R 
 
 
1064 
(b) Express the time it takes for the 
meterstick to pass you in terms of its 
apparent length and speed: 
 
v
Lt =Δ 
Substitute numerical values and 
evaluate Δt: ns5.28.0
m60.0
Δ ==
c
t 
 
24 • Recall that the half-life is the time it takes for any given amount of 
unstable particles to decay to half that amount of particles. The proper half-life of 
a species of charged subatomic particles called pions is 1.80 × 10–8 s (See 
Problem 16 for details on pions.) . Suppose a group of these pions are produced 
in an accelerator and emerge with a speed of 0.998c. How far do these particles 
travel in the accelerator’s laboratory before half of them have decayed? 
 
Picture the Problem We can express the distance the pions will travel in the 
laboratory using and find their half-life in the accelerator laboratory 
using 
tvx ΔΔ =
( )20 1ΔΔ cvtt −= . 
 
Express the average distance the 
pions will travel before decaying in 
terms of their speed proper mean 
lifetime: 
 
tvx ΔΔ = 
Use the time-dilation equation to 
relate the mean lifetimes of the pions 
in the accelerator laboratory Δt to 
their proper mean lifetime Δt0: 
 
2
0
1
Δ
Δ
⎟⎠
⎞⎜⎝
⎛−
=
c
v
tt 
Substitute for to obtain: tΔ
 2
0
1
Δ
Δ
⎟⎠
⎞⎜⎝
⎛−
=
c
v
tvx 
 
Substitute numerical values and 
evaluate Δx: 
( )( ) m2.85
998.01
s1080.1998.0
Δ
2
8
=
⎟⎠
⎞⎜⎝
⎛−
×=
−
c
c
cx 
 
25 •• [SSM] Your friend, who is the same age as you, travels to the star 
Alpha Centauri, which is 4.0 light-years away, and returns immediately. He 
claims that the entire trip took just 6.0 y. What was his speed? Ignore any 
accelerations of your friend’s spaceship and assume the spaceship traveled at the 
same speed during the entire trip. 
Relativity 
 
 
1065
Picture the Problem To calculate the speed in the reference frame of the friend, 
who is named Ed, we consider each leg of the trip separately. Consider an 
imaginary stick extending from Earth to Alpha Centauri that is at rest relative to 
Earth. In Ed’s frame the length of the stick, and thus the distance between Earth 
and Alpha Centauri, is shortened in accord with the length contraction formula. 
As three years pass on Ed’s watch Alpha Centauri travels at speed v from its 
initial location to him. 
 
Sketch the situation as it is in your 
reference frame. The distance 
between Earth and Alpha C. is the 
rest length L0 of the stick discussed 
in Picture the Problem: 
 
Alpha C.
Your reference frame
v
Earth
Ed moving
0L
 
 
Sketch the situation as it is in Ed’s 
reference frame. The distance 
between Earth and Alpha C. is the 
length L of the moving stick 
discussed in Picture the Problem: 
 
Alpha C.
Your friend Ed's reference frame
v
Earth
Ed at rest
L
v
 
 
The two events are Ed leaves Earth 
and Ed arrive at Alpha Centauri. In 
Ed’s frame these two events occur at 
the same place (next to Ed). Thus, 
the time between those two events 
Δt0 is the proper time between the 
two events. 
 
Δt0 = 3 y 
 
The distance L traveled by Alpha 
Centauri in Ed’s frame during the 
first three years equals the speed 
multiplied by the time in Ed’s frame: 
 
0L v t= Δ 
 
 
The distance between Earth and 
Alpha Centauri in Ed’s frame is the 
contracted length of the imaginary 
stick: 
 
2
2
0 1 c
vLL −= 
Equate these expressions for L to 
obtain: 
 
02
2
0 1 tvc
vL Δ=− 
 
 Chapter R 
 
 
1066 
Substituting numerical values yields: ( ) ( )y 0.31y 0.4 2
2
v
c
vc =−⋅ 
or 
2
2
2
2
16
91
c
v
c
v =− ⇒ 1
16
25
2
2
=⎟⎟⎠
⎞
⎜⎜⎝
⎛
c
v 
 
Solving for v gives: cv 80.0= 
 
26 •• Two spaceships pass each other traveling in opposite directions. A 
passenger in ship A knows that her ship is 100 m long. She 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? 
 
Picture the Problem We can use the relationship between the measured length L 
of the spaceships and their proper lengths L0 to find the lengths of the two 
spaceships as measured by a passenger in ship B. 
 
Relate the measured length LA of 
ship A to its proper length: 
 
2
A,0A 1 ⎟⎠
⎞⎜⎝
⎛−=
c
vLL 
 
Substitute numerical values and 
evaluate LA: 
 
( ) m3992.01m100 2A =⎟⎠
⎞⎜⎝
⎛−=
c
cL 
 
Relate the proper length L0,B of ship 
B to its measured length LB: 
 
2
B
B,0
1 ⎟⎠
⎞⎜⎝
⎛−
=
c
v
LL 
 
Substitute numerical values and 
evaluate : B,0L
m92
92.01
m36
2B,0
=
⎟⎠
⎞⎜⎝
⎛−
=
c
c
L 
 
27 •• Supersonic jets achieve maximum speeds of about 3.00 × 10–6c. (a) By 
what percentage would a jet traveling at this speed contract in length? (b) During 
a time of exactly one year or 3.15 × 107 s on your clock, how much time would 
elapse on the pilot’s clock? How much time is lost by the pilot’s clock in one year 
of your time? Assume you are on the ground and the pilot is flying at the specified 
speed for the entire year. 
 
 
 
Relativity 
 
 
1067
Picture the Problem We can use the relationship between the measured length L 
of the jet and its proper length to express the fractional change in the length of the 
jet traveling at its maximum speed. In Part (b) we can express the elapsed time on 
the pilot’s clock Δt0 in terms of the elapsed time Δt on your clock. 
 
(a) Express the fractional change in 
length of the jet: 
 
00
0 1
L
L
L
LL −=− 
Relate L to L0: 2
0 1 ⎟⎠
⎞⎜⎝
⎛−=
c
vLL ⇒
2
0
1 ⎟⎠
⎞⎜⎝
⎛−=
c
v
L
L 
 
Substitute for 0LL to obtain: 
2
2
0
0 11
c
v
L
LL −−=− (1) 
 
 
2
2
4
4
2
2
2
2
2
11
...
8
3
2
111
2
1
c
v
c
v
c
v
c
v
−≈
++−=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
 
 
Expand 2
2
1
c
v− binomially to 
obtain: 
 
Substituting in equation (1) yields: 
2
2
2
2
0
0
2
1
2
111
c
v
c
v
L
LL =⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=− 
 
Substitute numerical values and 
evaluate the jet’s fractional change in 
length: 
 
( )
%1050.4
1000.3
2
1
10
2
26
0
0
−
−
×=
×≈−
c
c
L
LL
 
 
(b) Express the elapsed time on the 
pilot’s clock Δt0 in terms of the 
elapsed time Δt on your clock: 
 t
c
vt
c
vtt
c
vt
Δ
2
1
Δ
2
11ΔΔ1Δ
2
2
2
2
2
2
0
2
1
−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −≈⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
 
where the second term represents the 
time lost on the pilot’s clock. 
 
 Chapter R 
 
 
1068 
 
Substitute numerical values and evaluate the elapsed time on the pilot’s clock 
in 1 y = 3.15 × 107 s: 
 ( ) ( ) y 00.1s1015.31000.3
2
11Δ
2
11Δ 72
26
2
2
0 =×⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×−=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
−
c
ct
c
vt 
 
The time lost by the pilot’s clock in 
time Δt of your time is given by: 
 
t
c
vtt Δ
2
1
ΔΔ 2
2
0 =− 
Substitute numerical values and evaluate 0ΔΔ tt − for 1 y = 3.15 × 107 s: 
 ( ) ( ) ( ) min .372
s 60
min 1s 421s1015.31000.3
2
1
ΔΔ 72
26
0 μμ =⎟⎠
⎞⎜⎝
⎛=××=−
−
c
ctt 
 
28 •• The proper mean lifetime of a muon (see Problems 18 and 19 for 
details) is 2.20 μs. Consider a muon, created in Earth’s atmosphere, speeding 
toward the surface 8.00 km below, at a speed of 0.980c. (a) What is the 
likelihood that it will survive its trip to ground before decaying? The probability 
of a muon decaying is given by P = 1− e−Δt / τ , where Δt is the time interval as 
measured in the reference frame in question. (b) Calculate this from the point of 
view of an observer moving with the muon. Show that, from the point of view of 
an observer on Earth, the answer is the same. 
 
Picture the Problem We can use the given probability function to find the 
probability of the muon’s survival from both the point of view of an Earth 
observer and the muon. To do so, we’ll need to use the time-dilation relationship 
in (a) and the length-contraction relationship in (b). 
 
(a) The probability of a muon 
decaying is τteP Δ1 −−= , so the 
probability of survival is: 
 
τteP Δsurvival
−= (1) 
From the point of view of an 
observer on Earth: 
 
( )( )
s 229.27
m/s 10998.2980.0
m 8000
Δ 8
μ=
×== v
Lt
 
 
The mean lifetime of the muon is 
given by: 
 ( )
s 055.11
980.01
s 20.2
1
22
0
μ
μττ
=
−
=
⎟⎠
⎞⎜⎝
⎛−
=
c
v 
 
Relativity 
 
 
1069
Substitute numerical values in 
equation (1) and evaluate Psurvival: %5.8
s 055.11
s 229.27
== − μ
μ
eP 
 
(b) From the point of view of the 
muon: 
 
v
Lt =Δ (2) 
The muon’s reference frame is the 
one in which it is at rest and Earth is 
rushing toward it at 0.980c. Express 
the thickness of Earth’s atmosphere 
as measured in the muon’s frame of 
reference: 
 
22
0 1 cvLL −= 
Substituting for L in equation (2) 
yields: 
 v
cvL
t
22
0 1Δ
−= 
 
Substitute numerical values and 
evaluate Δt: 
 
( )
( )( ) s 419.5m/s 10998.2980.0 980.01m 8000Δ 8
2
μ=×
−=t
 
Substituting numerical values in 
equation (1) yields: %5.8s .202
s .4195
survival ==
− μ
μ
eP 
a result that agrees, to two significant 
figures, with the result obtained in (a). 
 
29 •• A spaceship commander traveling to the Magellanic Clouds, travels at 
a uniform speed of 0.800c. When leaving the Kuiper belt, whose outer edge is 
50.0 AU from Earth (Note: 1 AU = 150,000,000 km, and represents the average 
distance between Earth and the Sun; AU = astronomical unit), he sends a 
message to ground control, Houston, Texas, saying that he is fine. Fifteen 
minutes later (according to him) he realizes he has made a typo, so he sends a 
correction. How much time passes at Houston between the receipt of his initial 
message and the second message? 
 
Picture the Problem Let ΔT be the proper time interval between event A, the 
sending of the first message, and event B, the sending of the correction. These 
two events take place on the ship, so the time interval between the two events it 
the time interval in the reference frame of the ship. Thus, ΔT = 15 min. Let ΔT′ 
be the time interval between these same two events in the reference frame of 
Houston. Then the time between the arrival of the two messages in Houston is 
. travel
messages
between '' TTT Δ+Δ=Δ
 
The time between the arrival of the 
two messages in Houston is: 
 
travel
messages
between '' TTT Δ+Δ=Δ 
 
 Chapter R 
 
 
1070 
Because Houston is moving at speed 
v = 0.800c relative to the ship: ( )22spaceship1' cv
T
T −
Δ=Δ 
 
Substituting for 'TΔ gives: 
( ) travel22spaceshipmessagesbetween '1 Tcv
T
T Δ+−
Δ=Δ (1) 
 
During the time interval 'TΔ the ship 
travels a distance: 
 
'ΔΔ Tvx = 
The reply has to travel the distance 
Δx farther than the first message and 
this takes time . The reply 
travels at speed c, so: 
travel'TΔ
travel'Tcx Δ=Δ 
 
 
Equating these expressions for ∆x 
gives: 
 
travel'' TcTv Δ=Δ ⇒ ''travel Tc
vT Δ=Δ 
 
Substituting for 'TΔ yields: 
( )22spaceshiptravel 1' cv
T
c
vT −
Δ=Δ 
 
( ) ( )
( ) ⎟⎠⎞⎜⎝⎛ +−
Δ=
−
Δ+−
Δ=Δ
c
v
cv
T
cv
T
c
v
cv
T
T
1
1
11
22
spaceship
22
spaceship
22
spaceship
messages
between
 
 
Substitute for and simplify to 
obtain: 
travel'TΔ
Substitute numerical values and 
evaluate : 
messages
betweenTΔ
 
( ) ( )
min 45
800.01
800.01
min 15
2messages
between
=
+
−
=ΔT
 
 
The Relativity of Simultaneity 
 
30 • According to Jamal: (a) what is the distance between the flashbulb and 
clock A? (b) How far does the flash travel to reach clock A? (c) How far does 
clock A travel while the flash is traveling from the flashbulb to it? 
 
Picture the Problem The proper distance L0 between the flashbulb and clock A is 
50.0 c⋅min and is related to the distance L measured by Jamal according to 
( )20 1 cvLL −= . Because clock A and the flashbulb are at rest relative to each 
other, we can find the distance between them using the same relationship with 
L0 = 50.0 c⋅min. 
Relativity 
 
 
1071
 
(a) Express L in terms of L0: 2
0 1 ⎟⎠
⎞⎜⎝
⎛−=
c
vLL 
 
Substitute numerical values and 
evaluate L: 
 
( )
min0.40
600.01min0.50
2
⋅=
⎟⎠
⎞⎜⎝
⎛−⋅=
c
c
ccL
 
 
(b) Jamal sees the flash traveling 
away from him at 1.0c and clock A 
approaching at 0.6c. So: 
 
min0.25
600.1
min0.40
Δ =⋅=
c
ct 
and so 
min0.25 ⋅= cL 
 
(c) Express the distance d clock A 
travels while the flash is traveling to 
it in terms of the speed of clock A 
and the time Δt it takes the flash to 
reach it: 
 
⎟⎠
⎞⎜⎝
⎛=Δ=
c
Lvtvd 
Substitute numerical values and 
evaluate L: 
( )
min0.15
min0.25600.0
⋅=
⎟⎠
⎞⎜⎝
⎛ ⋅=
c
c
ccd
 
 
31 •• According to Jamal, how long does it take the flash to travel to clock 
A, and what does clock C read as the flash reaches clock A? 
 
Picture the Problem As the light pulse from the flashbulb travels toward clock A 
with speed c, clock A travels toward clock C with speed 0.600c. The sum of the 
distances traveled by the flash and clock
A must equal the distance separating 
them as seen in Jamal’s frame of reference. 
 
Express the sum of the distances 
between clock A and the flashbulb 
as seen in Jamal’s frame of 
reference: 
 
 600.0 21 l=+ ctct ⇒ ct 200.3
l= 
where is the distance separating 
clocks A and B. 
l
Express l in terms of : 0l 2
0 1 ⎟⎠
⎞⎜⎝
⎛−=
c
vll 
 
 Chapter R 
 
 
1072 
Substitute for l to obtain: 
 
2
0 1
200.3
⎟⎠
⎞⎜⎝
⎛−=
c
v
c
t l 
 
Substitute numerical values and 
evaluate t: 
 
min25
600.01
3.200c
min100 2
=
⎟⎠
⎞⎜⎝
⎛−⋅=
c
cct
 
 
Because clock C starts from zero 
when the flashbulb passes next to it, 
its reading when the flash reaches 
clock A will equal the time required 
for the flash to travel to clock A. 
Hence clock C will read: 
min25=Ct 
 
32 •• Show that clock C reads 100 min as the light flash reaches clock B, 
which is traveling away from clock C with speed 0.600c. 
 
Picture the Problem As the light flash from the flashbulb travels toward clock B 
with speed c, clock B travels away from clock C with speed 0.600c. The 
difference between these distances must equal the distance between clock B and 
the flashbulb as seen in Jamal’s frame of reference. 
 
Express the difference between the 
distances traveled by the flash of 
light and clock B as seen in Jamal’s 
frame of reference: 
 
 600.0 21 Lctct =− ⇒ c
Lt
800.0
= 
where L is the separation of clocks A 
and B. 
Use the length contraction equation 
to express L in terms of L0: 
2
0 1 ⎟⎠
⎞⎜⎝
⎛−=
c
vLL 
 
Substitute for L to obtain: 2
0 1
800.0
⎟⎠
⎞⎜⎝
⎛−=
c
v
c
Lt 
 
Substitute numerical values and 
evaluate t: 
min100
600.01
800.0
min100 2
=
⎟⎠
⎞⎜⎝
⎛−⋅=
c
c
c
ct
 
 
Relativity 
 
 
1073
33 •• According to Jamal, the reading on clock C advances from 25 min to 
100 min between the reception of the flashes by clocks A and B in Problems 31 
and 32. According to Jamal, how much will the reading on clock A advance 
during this 75-min interval? 
 
Picture the Problem We can find the elapsed time on the clock at A (the proper 
time) from the elapsed time on the clock at B using ( )20 1ΔΔ cvtt −= . 
 
Express the elapsed proper time Δt0 
on clock A in terms of the elapsed 
time Δt measured by Jamal: 
2
0 1ΔΔ ⎟⎠
⎞⎜⎝
⎛−=
c
vtt 
 
 
Substitute numerical values and 
evaluate Δt0: ( )
min60
600.01min75Δ
2
0
=
⎟⎠
⎞⎜⎝
⎛−=
c
ct
 
 
34 •• The advance of clock A calculated in Problem 33 is the amount that 
clock A leads clock B according to Jamal. Compare this result with vL0/c2, where 
v = 0.600c. 
 
Picture the Problem We can compare the time calculated in Problem 33 with 
L0v/c2 by evaluating their ratio. 
 
Express the ratio r of L0v/c2 to Δt0: 
 
2
2
0
2
2
0
0
2
0
1Δ
1Δ
Δ
⎟⎠
⎞⎜⎝
⎛−
=
⎟⎠
⎞⎜⎝
⎛−
==
c
vtc
vL
c
vt
c
vL
t
c
vL
r
 
 
Substitute numerical values and 
evaluate r: 
 
( )( )
( )
1
600.01min75
600.0min100
2
2
=
⎟⎠
⎞⎜⎝
⎛−
⋅=
c
cc
ccr 
 
35 •• [SSM] In an inertial reference frame S, event B occurs 2.00 μs after 
event A and 1.50 km distant from event A. How fast must an observer be moving 
along the line joining the two events so that the two events occur simultaneously? 
 Chapter R 
 
 
1074 
For an observer traveling fast enough is it possible for event B to precede event 
A? 
 
Picture the Problem Because event A is ahead of event B by Lpv/c2 where v is 
the speed of the observer moving along the line joining the two events, we can use 
this expression and the given time between the events to find v. 
 
Express the time Δt between events 
A and B in terms of , v, and c: pL 2
p
c
vL
t =Δ ⇒ 
p
2
L
tcv Δ=
 
 
( ) ( )
( )
c
c
sv
400.0
m/s10998.2
m/s1020.1
km50.1
00.2m/s10998.2
8
8
28
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
××=
×= μ
 
 
Substitute numerical values and 
evaluate v: 
( )
p
AB
2
L
ttcv −= 
 
Rewrite the expression for v in terms 
of tA and tB yields: 
Express the condition on tB − tB A if 
event B is to precede event A: 
 
0AB <− tt and ( )
p
AB
2
L
ttcv −> 
 
Substitute numerical values and 
evaluate v: 
( ) ( )
c
v
400.0m/s1020.1
km50.1
s00.2m/s10998.2
8
28
=×=
×> μ
 
 
Event B can precede event A provided v > 0.400c. 
 
36 •• A large flat space platform has an x axis painted on it. A firecracker 
explodes on the x axis at x1 = 480 m, and a second firecracker explodes on the x 
axis 5.00 μs later at x2 = 1200 m. In the reference frame of a train traveling 
alongside the x axis at speed v relative to the platform, these two explosions occur 
at the same place on that axis. What is the separation in time between the two 
explosions in the reference frame of the train? 
 
Picture the Problem Let both explosions occur at the front end of the train—
where there is a clock fastened to the train. In the reference frame of the train, 
both explosions occur at the same place. The clock is stationary in this frame, so 
it does not run slow. Thus, in this frame the amount the clock reading advances is 
also the time T0 between the two events. In the reference frame of the platform 
the clock (and the train) moves with speed v = L/T, where L = x2 – x1 and 
Relativity 
 
 
1075
T = 5.00 μs. In this frame the clock is moving so it runs slow, advancing by 
only ( )221 cvT − during time T. All observers must agree with both the clock 
reading when the first explosion occurs (a spacetime coincidence), and with the 
clock reading when the second firecracker goes off (also a spacetime 
coincidence). Thus, all observers agree with the amount the clock reading 
changes by in the interval between the two events. 
 
 
 
Equate the expressions for the 
amount the clock advances to obtain: 
 
( )220 1 cvTT −= 
Substitute L/T for v to obtain: 
 
2
0 1 ⎟⎠
⎞⎜⎝
⎛−=
cT
LTT 
 
Substitute numerical values and evaluate T0: 
 
( ) ( )( ) s39.4s1000.5m/s102.998 m480m1200100.5
2
680 μμ =⎟⎟⎠
⎞
⎜⎜⎝
⎛
××
−−= −sT 
 
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 
perform on a star 15 light-years away. To celebrate his good fortune, he bought a 
new vehicle for the trip—a deluxe space-coupe that travels at 0.99c. Each of the 
twins promises to practice diligently, so they can reunite afterward. However, 
Randy’s gig goes so well that he stays for a full 10 years before returning to 
Herb. After their reunion, (a) how many years of practice will Randy have had; 
(b) how many years of practice will Herb have had? 
 
Picture the Problem Randy’s clocks, including his biological clocks, are always 
at rest in Randy’s reference frame, so they advance at the same rate that time 
advances in Randy’s frame. The same is true for Herb and the clocks in his 
reference frame. Think of a long imaginary stick, at rest relative to Earth, with 
one end at Earth and the other at the star. The rest length of this stick is 15 c·y. 
The time Randy has to practice is the time for the trip in Randy’s reference frame, 
and the time Herb has to practice is the time for Randy’s trip in Herb’s reference 
frame. 
 
 Chapter R 
 
 
1076 
Sketch the situation as it is in Herb’s 
reference frame. The distance 
between Earth and the star is the rest 
length L0 of the stick (discussed in 
Picture the Problem): 
Star
Herb's reference frame
v
Earth
Randy moving
0L
 
 
Sketch the situation as it is in 
Randy’s reference frame. The
distance between Earth and the star 
is the length L of the moving stick 
(discussed in Picture the Problem): 
 
Star
Randy's reference frame
v
Earth
Randy at rest
L
v
 
 
To find the time T0 in Randy’s 
reference frame for the star to travel 
the distance L we apply the formula 
distance equals speed multiplied by 
time. However, to do this we must 
find L, which is related to L0 by the 
length contraction formula: 
 
0vTL = ⇒ v
c
vL
v
LT
2
2
0
0
1−
== 
 
Substitute numerical values and 
evaluate T0: 
( )
y 137.2
99.0
99.01y15 2
0 =−⋅= c
c
T 
 
To find the time T in Herb’s 
reference frame for Randy to travel 
to the star we apply the formula 
distance equals speed multiplied by 
the time: 
 
vTL =0 ⇒ v
LT 0= 
 
 
Substitute numerical values and 
evaluate T: y 152.150.99
y15 =⋅=
c
cT 
 
(a) The time for the trip in Randy’s 
frame is ( ) y 14y 2.1372 y 10 =+ . 
 
(b) The time for the trip in Herb’s 
frame is ( ) y 04y 15.1522 y 10 =+ . 
While Randy is at the Star the two 
reference frames are at rest relative 
to each other. Thus, the time that 
Randy is at the star is the same in 
both frames. The calculations for the 
return trip give the same result as for 
the outgoing trip. 
 
 
 
 
Relativity 
 
 
1077
38 •• Al and Bert are twins. Al travels at 0.600c to Alpha Centauri (which is 
4.00 c⋅y from Earth as measured in the reference frame of Earth) and returns 
immediately. Each twin sends the other a light signal every 0.0100 y as measured 
in his own reference frame. (a) At what rate does Bert receive signals as Al is 
moving away from him? (b) How many signals does Bert receive at this rate? 
(c) How many total signals are received by Bert before Al returns to Earth? (d) At 
what rate does Al receive signals as Bert is moving away from him? (e) How 
many signals does Al receive at this rate? (f) How many total signals are received 
by Al before Al returns to Earth? (g) Which twin is younger at the end of the trip 
and by how many years? 
 
Picture the Problem (a) In Bert’s reference frame each successive signal from Al 
travels an additional distance equal to vT, where v is Al’s speed and T is the 
interval between successive signals in Al’s reference frame. We also need to take 
into account the dilation of the time intervals measured on Earth due to Al’s 
motion. In Parts (b) and (c) we can use Al0 tfN Δ= and 22BertAl 1ΔΔ cvtt −= to 
find the number of signals received by Bert. In Parts (d) and (e) we can proceed 
similarly to find the number of signals received by Al and the total number he 
receives before he returns to Bert. Finally, we can use the number of signals each 
received to find the difference in their ages resulting from Al’s trip. 
 
(a) The rate at which Bert receives 
signals is the reciprocal of the 
time between the arrival of 
successive signals on Earth: 
Bertf
 
signalsbetween 
Bert
1
T
f = 
delaysignalsbetween TTT += 
or, because T
c
vTT 600.0delay == , 
TTTT 600.1600.0signalsbetween =+= 
The time between the arrival of 
successive signals on Earth is the 
sum of the time T between signals in 
Al’s frame of reference and the delay 
introduced by the fact that Al travels 
a distance vT between signals: 
 
Substitute for to obtain: signalsbetween T
 T
f
600.1
1
Bert = 
 
T, the time between the arrival of 
signals on Earth is related to the 
proper time interval T0 (the time 
between signals in Al’s frame of 
reference) through the time-dilation 
equation: 
( )2201 cv
TT −= 
 Chapter R 
 
 
1078 
( )
0
22
Bert 600.1
1
T
cv
f
−= 
 
Substituting for T and simplifying 
yields: 
Substitute numerical values and 
evaluate : Bertf
( )
( )( ) 1
2
Bert y 50y 0100.0600.1
600.01 −=−=f 
 
(b) Express the number of signals 
N received by Bert in terms of 
the number of signals sent by Al: 
 
Al0 tfN Δ= 
Use the time dilation equation to 
express the elapsed time in Al’s 
frame of reference in terms of the 
elapsed time in Bert’s frame (the 
proper elapsed time): 
2
BertAl 1ΔΔ ⎟⎠
⎞⎜⎝
⎛−=
c
vtt 
 
Substitute for to obtain: AlΔt
 
2
Bert0 1Δ ⎟⎠
⎞⎜⎝
⎛−=
c
vtfN (1) 
 
Find : BertΔt y667.6
0.600
y00.4
Δ Bert =⋅= c
ct 
 
Substitute numerical values in 
equation (1) and evaluate N: ( )( )
533
600.01y 667.6y100
2
1
=
⎟⎠
⎞⎜⎝
⎛−= −
c
cN
 
 
(c) Express the number of signals N 
received by Bert in terms of the 
number of signals sent by Al before 
he returns: 
 
Al0 TfN Δ= 
Because 
2
BertAl 1ΔΔ ⎟⎠
⎞⎜⎝
⎛−=
c
vtt , 
the time in Al’s frame for the 
round trip is given by: AlΔT
2
BertAlAl 1Δ2Δ2Δ ⎟⎠
⎞⎜⎝
⎛−==
c
vttT 
 
 
 
Substituting for in the expression 
for N yields: 
AlΔT 2
Bert0 1Δ2 ⎟⎠
⎞⎜⎝
⎛−=
c
vtfN 
Relativity 
 
 
1079
Substitute numerical values and 
evaluate N: 
 
( )( )
3
2
1
1007.1
600.01y667.6y1002
×=
⎟⎠
⎞⎜⎝
⎛−= −
c
cN
 
 
(d) Proceed as in Part (a) to find the 
rate at which Al receives signals as 
Bert is moving away from him: 
 
1
Al y0.50
−=f 
 
(e) Express the number of signals N 
received by Al: 
 
AlAl tfN Δ= 
Substitute for to obtain: AltΔ 2
BertAl 1Δ ⎟⎠
⎞⎜⎝
⎛−=
c
vtfN 
 
Substitute numerical values and 
evaluate N: 
 
( )( )
267
600.01y667.6y0.50
2
1
=
⎟⎠
⎞⎜⎝
⎛−= −
c
cN
 
 
(f) Express the total number of 
signals received by Al: 
 
return
returnoutboundtot
267 N
NNN
+=
+=
 
The number, , of signals 
received by Al on his return trip is 
given by: 
returnN
2
Bertreturn Al,
Alreturn Al,return
1Δ
Δ
⎟⎠
⎞⎜⎝
⎛−=
=
c
vtf
tfN
 
 
Substitute for to obtain: returnN 2
Bertreturn Al,tot 1Δ267 ⎟⎠
⎞⎜⎝
⎛−+=
c
vtfN 
 
Proceeding as in Part (a), find the 
rate at which signals are received by 
Al on the return trip: 
1
return Al, y200
−=f 
 
Substitute numerical values and evaluate : totN
 
( )( ) 321tot 1033.11334600.01y667.6y200267 ×==⎟⎠⎞⎜⎝⎛−+= − c cN 
 
 Chapter R 
 
 
1080 
(g) Their age difference is: AlBert ttt Δ−Δ=Δ 
 
y2.67
y 100
1067
y 100
1334
Δ 11 =−= −−t 
and 
Bert.an younger thy 2.67 is Al 
Substitute numerical values to 
obtain: 
 
Relativistic Energy and Momentum 
 
39 • Find the ratio of the total energy to the rest energy of a particle of mass 
m moving with speed (a) 0.100c, (b) 0.500c, (c) 0.800c, and (d) 0.990c. 
 
Picture the Problem We can use Equation R-15 to find the ratio of the total 
energy to the rest energy for the given particle. 
 
From Equation R-15: 
 
2
2
0
2
2
2
11
c
v
E
c
v
mcE
−
=
−
= 
 
Solve for the ratio of E to E0 to 
obtain: 
2
2
0 1
1
c
vE
E
−
= 
 
(a) Evaluate 0EE for v = 0.100c: 
 ( ) 01.1100.01
1
2
2
100.00
=
−
=⎥⎦
⎤
=
c
cE
E
cv
 
 
(b) Evaluate 0EE for v = 0.500c: 
 ( ) 15.1500.01
1
2
2
500.00
=
−
=⎥⎦
⎤
=
c
cE
E
cv
 
 
(c) Evaluate 0EE for v = 0.800c: 
 ( ) 67.1800.01
1
2
2
800.00
=
−
=⎥⎦
⎤
=
c
cE
E
cv
 
 
(d) Evaluate 0EE for v = 0.990c: 
 ( ) 09.7990.01
1
2
2
999.00
=
−
=⎥⎦
⎤
=
c
cE
E
cv
 
 
Relativity 
 
 
1081
4
40 • A proton (rest energy 938 MeV) has a total energy of 1400 MeV. 
(a) What is its speed? (b) What is its momentum? 
 
Picture the Problem The rest energy E0 is equal to mc2. We are given E0 and E, 
where E0 = 938 MeV and the total energy E = 1400 MeV. (The total energy is the 
rest energy plus the kinetic energy).
We can find the momentum p of the proton 
using (Equation R-17), and once we have p we can solve for the 
speed v using v/c = pc/E (Equation R-16). 
2 2 2 2E p c m c= +
 
(b) Use Equation R-17 to relate the 
momentum to the total energy and 
the rest energy: 
42222 cmcpE += and 20 mcE =
so 
2
0
222 EcpE += ⇒
c
EE
p
2
0
2 −= 
 
 
Substitute numerical values and 
evaluate p: ( ) ( )
c
c
c
p
MeV/ 1040
MeV/ 1039
Mev 938Mev 1400 22
=
=
−=
 
 
(a) Use Equation R-16 to express the 
speed of the proton: E
pc
c
v = ⇒ c
E
pcv ⎟⎠
⎞⎜⎝
⎛= 
 
Substitute numerical values and 
evaluate v: 
( ) cccv 742.0
MeV 1400
MeV/ 1039 =⎟⎠
⎞⎜⎝
⎛= 
 
41 • [SSM] How much energy would be required to accelerate a particle 
of mass m from rest to (a) 0.500c, (b) 0.900c, and (c) 0.990c? Express your 
answers as multiples of the rest energy, mc2. 
 
Picture the Problem We can use Equation R-14 to find the energy required to 
accelerate this particle from rest to the given speeds. 
 
From Equation R-14 we have: 
 ( )
( )
2
2
2
2
2
1
1
1
1
mc
cv
mc
cv
mcK
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
=
−
−
=
 
 
 Chapter R 
 
 
1082 
(a) Substitute numerical values and 
evaluate K(0.500c): 
 
( ) ( )
2
2
2
155.0
1
500.01
1500.0
mc
mc
cc
cK
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
=
 
 
(b) Substitute numerical values and 
evaluate K(0.900c): 
 
( ) ( )
2
2
2
29.1
1
900.01
1900.0
mc
mc
cc
cK
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
=
 
 
(c) Substitute numerical values and 
evaluate K(0.990c): ( ) ( )
2
2
2
09.6
1
990.01
1990.0
mc
mc
cc
cK
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
=
 
 
42 • If the kinetic energy of a particle equals its rest energy, what 
percentage error is made by using p = mv for its momentum? Is the non-
relativistic expression always low or high compared to the relativistically correct 
expression for momentum? 
 
Picture the Problem We can use Equations R-10 and R-14 to express the error 
made in using p = mv for the momentum of the particle when K = E0. 
 
The error in using p = mv for the 
momentum of the particle is given 
by: 
 
relrel
rel 1
p
p
p
pp −=− (1) 
From Equation R-14 we have: 
 ( )
( )
2
2
2
2
2
1
1
1
1
mc
cv
mc
cv
mcK
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
=
−
−
=
 
 
For K = E0: 
( ) KcvK ⎟
⎟
⎠
⎞
⎜⎜⎝
⎛
−
−
= 1
1
1
2 
or 
( ) 21
1
2
=
− cv
 
 
Relativity 
 
 
1083
From Equation R-10, the relativistic 
momentum of the particle is: 
 
( ) mvcv
mvp 2
1 2
rel =−
= 
Substitute in equation (1) and 
simplify to obtain: 
%50
2
11
rel
=−=−
mv
mv
p
p 
 
The ratio of the non-relativistic 
momentum of a particle to its 
relativistic momentum is given by: 
 
22
22
rel
1
1
cv
cv
p
p
p
p −=
−
= 
Because 1 1 22 <− cv , the non-relativistic expression is always low compared to 
the relativistically correct expression for momentum. 
 
43 • What is the total energy of a proton whose momentum is 3mc? 
 
Picture the Problem We can use Equation R-17 to find the total energy of any 
proton whose momentum is given. See Problem 42 for the rest energy of a proton. 
 
The total energy, momentum, and 
rest energy of the proton are related 
by Equation R-17: 
 
( )22222 mccpE += 
Substitute for the momentum of the 
proton: 
 
( ) ( )
424242
22222
109
3
cmcmcm
mccmcE
=+=
+= 
 
Solving for E yields: 210mcE = 
 
Substitute for m and evaluate E: ( ) GeV97.2MeV/93810 22 == ccE 
 
44 •• Using a spreadsheet program or graphing calculator, make a graph of 
the kinetic energy of a particle with rest energy of 100 MeV for speeds between 0 
and c. On the same graph, plot 221 mv by way of comparison. Using the graph, 
estimate at about what speed the non-relativistic expression is no longer a good 
approximation to the kinetic energy. As a suggestion, plot the energy in units of 
MeV and the speed in the dimensionless form v/c. 
 
Picture the Problem We can create a spreadsheet program to plot both the 
classical and relativistic kinetic energy of the particle. 
 
 Chapter R 
 
 
1084 
The relativistic kinetic energy of 
the particle is given by: ( ) ⎟
⎟
⎠
⎞
⎜⎜⎝
⎛
−
−
= 1
1
1
2
2
icrelativist
cv
mcK 
 
A spreadsheet program to graph Krelativistic and Kclassical is shown below. The 
formulas used to calculate the quantities in the columns are as follows: 
 
Cell Formula/Content Algebraic Form 
v/c + 0.05 A8 A7+0.05 
B7 0.5*$B$3*A7^2 2
2
1 mv 
C7 $B$3*(1/((1−A7^2)^0.5)−1)
( ) ⎟
⎟
⎠
⎞
⎜⎜⎝
⎛
−
−
1
1
1
2
2
cv
mc 
 
 
 A B C 
1 
2 
3 mc2= 100 MeV 
4 
5 
v/c 
2
2
1 mv ( )
2
21
1 mc
cv ⎟
⎟
⎠
⎞
⎜⎜⎝
⎛
−
6 Kclassical Krelativistic
7 0.00 0.00 0.00 
8 0.05 0.13 0.13 
9 0.10 0.50 0.50 
10 0.15 1.13 1.14 
 
23 0.80 32.00 66.67 
24 0.85 36.13 89.83 
25 0.90 40.50 129.42 
26 0.95 45.13 220.26 
 
The solid curve is the graph of the relativistic kinetic energy. The relativistic 
formula, represented by the continuous curve, begins to deviate from the classical 
curve around v/c ≈ 0.4. 
 
Relativity 
 
 
1085
0
50
100
150
200
250
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v /c
E
 (M
eV
)
Kclassical
Krelativistic
 
 
45 •• [SSM] (a) Show that the speed v of a particle of mass m and total 
energy E is given by ( ) 212 221 ⎥⎥⎦
⎤
⎢⎢⎣
⎡ −=
E
mc
c
v and that when E is much greater than 
mc2, this can be approximated by ( )2 2221 Emccv −≈ . Find the speed of an electron 
with kinetic energy of (b) 0.510 MeV and (c) 10.0 MeV. 
 
Picture the Problem We can solve the equation for the relativistic energy of a 
particle to obtain the first result and then use the binomial expansion subject to 
E >> mc2 to obtain the second result. In Parts (b) and (c) we can use the first 
expression obtained in (a), with E = E0 + K, to find the speeds of electrons with 
the given kinetic energies. See Table 39-1 for the rest energy of an electron. 
 
(a) The relativistic energy of a 
particle is given by Equation R-15: 
 2
2
2
1
c
v
mcE
−
= 
 
Solving for v/c yields: 
 
( ) 21
2
22
1
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −=
E
mc
c
v (1) 
 
Expand the radical expression binomially to obtain: 
 
( ) ( ) sorder term-higher 
2
111 2
22
2
22
+−=−=
E
mc
E
mc
c
v 
 
 Chapter R 
 
 
1086 
Because the higher-order terms are 
much smaller than the 2nd-degree 
term when E >> mc2: 
 
( )
2
22
2
1
E
mc
c
v −≈ 
 
( )
2
22
1
E
mccv −= 
 
(b) Solve equation (1) for v: 
Because E = E0 + K: 
( ) 2
0
2
0
2
0
1
111
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−=+−=
E
K
c
KE
Ecv 
 
For an electron whose kinetic 
energy is 0.510 MeV: 
 
( )
c
cv
866.0
MeV0.511
MeV510.01
11MeV510.0 2
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−=
 
(c) For an electron whose kinetic 
energy is 10.0 MeV: 
 
( )
c
cv
999.0
MeV0.511
MeV0.101
11MeV0.10 2
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−=
 
 
46 •• Use the binomial expansion and Equation R-17 to show that when 
pc << mc2, the total energy is given approximately by E ≈ mc2 + p2/(2m). 
 
Picture the Problem We can solve Equation R-17 for E and factor from 
under the resulting radical to obtain an expression to which we can apply the 
binomial expansion to write the radical expression as a power series. Finally, we 
can invoke the condition that pc << mc
( )22mc
2
to complete the argument that the total 
energy is given approximately by ( )mpmcE 222 +≈ . 
 
( )2222 mccpE += 
 
From Equation R-17 we have: 
Factor under the radical and simplify to obtain: ( )22mc
 
( ) ( ) ( ) 2
2
2
22
22
2
22
22
22 111
mc
pmc
mc
cpmc
mc
cpmcE +=+=⎥⎥⎦
⎤
⎢⎢⎣
⎡ += 
Relativity 
 
 
1087
Expand the radical factor binomially to obtain: 
 
sorder term-higher 
2
111 22
221
22
2
++=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
cm
p
cm
p 
 
For pc << mc2: 
22
221
22
2
2
111
cm
p
cm
p +≈⎟⎟⎠
⎞
⎜⎜⎝
⎛ + 
 
Substituting for 
21
22
2
1 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
cm
p and 
simplifying yields: energy kinetic classical energy rest 
22
11
2
2
22
2
2
+=
+=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +=
m
pmc
cm
pmcE
 
47 •• Derive the equation E2 = p2c2 + m2c4 (Equation R-17) by eliminating v 
from Equations R-10 and R-16. 
 
Picture the Problem We can solve Equation R-16 for v and substitute in 
Equation R-10 to eliminate v. Simplification of the resulting expression leads 
to ( ) .22222 mccpE +=
 
221 cv
mvp −= 
 
Express the relativistic momentum of 
a particle using Equation R-10: 
From Equation R-16 we have: 
E
pc
c
v = ⇒
E
pcv
2
= 
 
( )
2
22
2
1
c
Epc
E
pcm
p
−
= 
or 
2
22
2
1
1
E
cpE
mc
−
= 
 
Substitute for v and simplify to 
obtain: 
Multiply both sides of the 
equation by 2
22
1
E
cpE − : 
 
2
2
22
1 mc
E
cpE =− 
 Chapter R 
 
 
1088 
Square both sides of the equation: ( )222 222 1 mcEcpE =⎟⎟⎠
⎞
⎜⎜⎝
⎛ − 
 
Solve for E2 to obtain Equation R-17: ( )22222 mccpE += 
 
48 •• 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. 
 
Picture the Problem (a) We can solve the relation for total energy, momentum, 
and rest energy of the proton for its momentum and evaluate this expression for 
K = E0. In Part (b) we can equate the expression for the momentum of the proton 
under the conditions that exist in this problem with the general expression for the 
relativistic momentum of the proton and solve the resulting equation for the speed 
of the proton. 
 
(a) Relate the total energy, 
momentum, and rest energy of a 
proton: 
 
( )22222 mccpE += 
( )
c
KEK
c
EKE
c
EE
p
0
2
2
0
2
0
2
0
2
2+=
−+=−=
 
 
Solving for p yields: 
If K = E0: 
c
E
c
EE
p EK 0
2
0
2
0 32
0
=+== 
 
( )
c
c
p EK
GeV/ 62.1
MeV 9383
0
=
== 
 
Substitute numerical values and 
evaluate p: 
(b) In Part (a) we established that 
the momentum of the proton is 
given by: 
 
c
Ep EK 0
3
0
== 
 
Substituting for E0 and simplifying 
yields: mcc
mcp EK 3
3 2
0
=== 
 
Relativity 
 
 
1089
2
2
1
c
v
mcp
−
= 
 
The relativistic momentum of the 
proton is also given by: 
Equating these expressions gives: 
 
2
2
1
3
c
v
mcmc
−
= 
 
Solve for v to obtain: 
ccv 866.0
2
3 == 
 
49 •• What percentage error is made in using 221 mv for the kinetic energy of 
a particle if its speed is (a) 0.10c and (b) 0.90c? 
 
Picture the Problem We can use the expressions for the classical and relativistic 
kinetic energies of a particle to obtain a general expression for the fractional error 
made in using 221 mv for the kinetic energy of a particle as a function of its speed. 
 
The percentage error made in using 
2
2
1 mv for the kinetic energy of a 
particle is given by: 
rel
classical
rel
classicalrel
rel
1
K
K
K
KK
K
K −=−=Δ 
 
 
The relativistic kinetic energy of the 
particle is given by Equation R-14: 
 
( )
( )
2
2
2
2
2
rel
1
1
1
1
mc
cv
mc
cv
mcK
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
=
−
−
=
 
 
Substitute for Krel in the expression for 
relK
KΔ and simplify to obtain: 
 
( ) ( )
2
2
2
2
2
2
2
1
rel
1
1
12
1
1
1
1
1
c
cv
v
mc
cv
mv
K
K
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
−=Δ 
 
 Chapter R 
 
 
1090 
(a) Evaluate f for v = 0.10c: 
( ) ( )
( )
%75.0
1
10.01
12
10.0110.0
2
2
2
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
−=
c
ccf
 
(b) Evaluate f for v = 0.90c: 
 
( ) ( )
( )
%69
1
90.01
12
90.0190.0
2
2
2
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−
−
−=
c
ccf
 
 
General Problems 
 
50 • A spaceship departs from Earth for the star Alpha Centauri, which is 
4.0 c⋅y away in the reference frame of Earth. 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? 
 
Picture the Problem We can find the duration of the trip, as measured on Earth, 
using the definition of average speed; that is, Δt = L/v. The elapsed time measured 
by the passenger is the proper time and is related to Δt through the time dilation 
equation (Equation R-3). 
 
(a) Relate the time Δt for the trip as 
measured on Earth to its length L and 
the speed u of the spaceship: 
 
v
Lt =Δ 
y3.5y33.5
75.0
y0.4
Δ ==⋅=
c
ct 
 
Substitute numerical values and 
evaluate Δt: 
(b) Use Equation R-3, the time 
dilation equation, to express the 
proper time measured by a 
passenger: 
 
2
0 1ΔΔ ⎟⎠
⎞⎜⎝
⎛−=
c
vtt
 
 
Substitute numerical values and 
evaluate Δt0: ( ) ( ) y5.375.01y33.5Δ 2
2
p =−= c
ct 
Relativity 
 
 
1091
51 • The total energy of a particle is three times its rest energy. (a) Find v/c 
for the particle. (b) Show that its momentum is given by mcp 8= . 
 
Picture the Problem (a) We can solve the expression for the relativistic energy 
of the particle for v/c and evaluate this expression for E = 3E0. In Part (b) we can 
solve the expression relating the total energy, momentum, and rest energy of the 
particle for p and evaluate it for E = 3E0 to show that its momentum is given by 
p = 8 mc. 
 
(a) The relativistic energy of the 
particle is given by Equation R-15: 22
0
22
2
11 cv
E
cv
mcE −=−= 
 
Solving for v/c yields: 
 2
2
01
E
E
c
v −= 
 
For E = 3E0: 
943.0889.0
9
1 2
0
2
0 ==−=
E
E
c
v 
 
(b) Equation R-17 relates the total 
energy, momentum, and rest energy 
of the particle: 
 
( ) 202222222 EcpmccpE +=+= 
Solving for p yields: 
 c
EE
p
2
0
2 −= 
 
Evaluating p for E = 3E0 yields: ( )
mc
c
mc
c
E
c
EE
p
8
883 20
2
0
2
0
=
==−=
 
 
52 •• A spaceship travels past Earth moving at 0.70c relative to Earth. 5.0 
minutes after the spaceship passes closest to Earth, a message is sent from 
Houston, Texas, to the craft. (Neglect all effects of the rotational motion of 
Earth.) (a) How long does it take for the signal to arrive? (b) The spaceship and 
the control center agree on the time when the ship passes closest to Earth. Five 
minutes after the message is received aboard the ship, a return message is sent by 
the ship back to Houston. What is the time interval in Houston between the time 
their message was sent, and the time the return message is received? 
 
 
 
 Chapter R 
 
 
1092 
Picture the Problem The pictorial representation summarizes the information 
given in the problem statement. Note that the spaceship, moving at 0.70c, has a 
5.0 min head start. This means that, in the Earth frame of reference, the message 
is received by the spaceship after a time Δt has passed since the spaceship passed 
closest to Earth. In Part (b) we’ll have to determine the time-dilated interval
of 
5.0 min aboard the spaceship. 
Houston
Spaceship position
when signal is sent
Spaceship position 
when signal is received
Δ
 cv 70.0= cv 70.0= cv 70.0=
x = v t = c( t + t )Δ Δ delay
Δ delayv t v( t − t )delayΔ Δ
c( t − t )delayΔ Δ
 
 
(a) The distance Δx from Earth to the 
ship is given by: 
 
( )delayΔΔ ttcx Δ−= 
 
tvx ΔΔ = 
 
Δx is also given by: 
Equate these expressions to 
obtain: 
 
( ) tvttc Δ=Δ− delayΔ ⇒
c
v
t
t
−
Δ=Δ
1
delay 
 
min 71
min 67.16
70.01
min 0.5
Δ
=
=
−
=
c
ct
 
 
Substitute numerical values and 
evaluate Δt: 
(b) Express the total time from the 
sending of the original message 
(5.0 min after the closest passage) to 
the receipt of the return message: 
 
min 0.5ΔΔΔ return ship totot −+= ttt (2) 
 
Relativity 
 
 
1093
 
According to observers on Earth, 
the time between receipt of the 
message and its return is: 
 
( )
min 001.7
70.01
min 0.5
1
min 0.5
Δ
22return
=
−
=
⎟⎠
⎞⎜⎝
⎛−
=
c
v
t
 
and 
min23.67
min 001.7min 67.16Δ ship to
=
+=t
 
 
In Earth’s frame of reference, the 
spaceship is at a distance from Earth 
given by: 
 
ship toship toship to Δ70.0ΔΔ tctvx == 
Substitute numerical values and evaluate : ship toΔx
 
( )( ) m 10980.2
min
s 60min 3.672m/s 10998.270.0Δ 118ship to ×=⎟⎠
⎞⎜⎝
⎛ ××=x 
 
The time for the signal to return to 
Earth from this distance is: 
 min17s994
m/s 102.998
m 10980.2Δ
Δ 8
11
messagereturn 
==
×
×==
c
xt 
 
Substitute numerical values in 
equation (2) and evaluate Δttot: min 13
min 0.5min 17min 19Δ tot
=
−+=t
 
 
53 •• [SSM] Particles called muons traveling at 0.99995c are detected at 
the surface of Earth. One of your fellow students claims that the muons might 
have originated from the Sun. Prove him wrong. (The proper mean lifetime of the 
muon is 2.20 μs.) 
 
Picture the Problem Your fellow student is thinking that the time dilation factor 
might allow muons to travel the 150,000,000,000 m from the Sun to Earth. You 
can discredit your classmate’s assertion by considering the mean lifetime of the 
muon from Earth’s reference frame. Doing so will demonstrate that the distance 
traveled during as many as 5 proper mean lifetimes is consistent with the 
origination of muons within Earth’s atmosphere. 
 
 
 
 
 
 
 Chapter R 
 
 
1094 
The distance, in the Earth frame of reference, a muon can travel in n mean 
lifetimes τ is given by: 
 
( ) ( )
( )
( )222
0
111 cv
ccvn
cv
nv
cv
dnd
−
=
−
=
−
= ττ 
 
Substitute numerical values for v, c, and τ and simplify to obtain: 
 
( )( )( )
( ) ( )nnd km 0.6699995.01
s 20.2m/s 10998.299995.0
2
8
=
−
×= μ 
 
In 5 lifetimes a muon would travel a 
distance: 
 
( )( ) km 3305km 0.66 ==d , a distance 
approximating a low-Earth orbit. 
In 100 lifetimes, d ≈ 6600 km, or approximately one Earth radius. This relatively 
short distance should convince your classmate that the origin of the muons that 
are observed on Earth is within our atmosphere and that they certainly are not 
from the Sun. 
 
54 •• (a) How tall is Mount Everest in a reference frame traveling with a 
cosmic ray muon that is traveling straight down, relative to Earth, at 0.99c? Take 
the height of Mount Everest according to an Earth-based observer to be 8846 m. 
(b) How long does it take the muon to travel the height of the mountain from the 
reference frame traveling with the muon? (c) How long does it take the muon to 
travel the height of the mountain from the Earth-based reference frame? 
 
Picture the Problem Apply the length contraction equation to find the height of 
Mount Everest in the frame of reference traveling with the muon and use the 
relationship between the distance an object travels, its speed, and the elapsed time 
to solve Parts (b) and (c). 
 
(a) The height h of Mount Everest in 
the frame of reference traveling with 
the muon is given by: 
 
( ) 021 hcvh −= 
where h0 is the height of Mount Everest 
in Earth’s frame of reference. 
 
Substitute numerical values and 
evaluate h: 
 
( ) ( )
km 2.1
km 248.1km 846.899.01 2
=
=−=h
 
 
 
 
 
Relativity 
 
 
1095
(b) In the frame of reference 
traveling with the muon, the 
mountain is traveling toward it at a 
speed of 0.99c. Hence the time to 
travel the height of the mountain is 
given by: 
 
mountain
framemuon Δ v
ht = 
Substitute numerical values and 
evaluate Δt: ( )
s 2.4
m/s 10998.299.0
km 248.1
Δ 8framemuon 
μ=
×=t 
 
(c) From the Earth-based reference 
frame, the time it takes the muon to 
travel the height of the mountain is 
the ratio of the height of the 
mountain to the speed of the muon: 
 
muon
0
frameearth Δ v
ht = 
Substitute numerical values and 
evaluate : frameearth Δt ( )( )
s 30
m/s 10998.299.0
km 846.8
Δ 8frameearth 
μ=
×=t 
 
55 ••• A gold nucleus has a radius of 3.00 × 10−14 m, and a mass of 197 amu. 
(1 amu has a rest energy of 932 MeV.) During experiments at Brookhaven 
National Laboratory, these nuclei are routinely accelerated to a kinetic energy of 
3.35 × 104 GeV. (a) How much less than the speed of light are they traveling? 
(b) At these energies, how long does it take them to travel 100 m in the laboratory 
frame? 
 
Picture the Problem We can apply Equation R-15, the expression for the total 
relativistic energy of a particle, to find how much less than the speed of light the 
gold nuclei are traveling. The time it takes the nuclei to travel 100 m in the 
laboratory frame can be found using the distance, rate, and time equation. 
 
(a) Express the difference between 
the speed of light and the speed of 
the gold nucleus: 
 
vcv −=Δ (1) 
The total relativistic energy of a gold 
nucleus is given by Equation R-15: ( )2
2
2
1 cv
mcmcK
−
=+ 
 Chapter R 
 
 
1096 
Solving for ( )21 cv− yields: 
 ( )
K
mc
K
mc
cv 2
2
2
1
1
+
=− (2) 
 
Evaluate 
K
mc2 to obtain: 
 
( )( )
11048.5
GeV 1035.3
MeV/amu 932amu 197
3
4
2
<<×=
×=
−
K
mc
 
 
Because mc2 << K, equation (2) can 
be approximated by: 
 
( )
K
mccv
2
21 ≈− (3) 
Solve for v/c to obtain: 
 
22
1 ⎟⎟⎠
⎞
⎜⎜⎝
⎛−=
K
mc
c
v 
 
Expanding the radical binomially 
yields: 
 
sorder termhigher 
2
111
2222 2
1
+
⎟⎟⎠
⎞
⎜⎜⎝
⎛−=⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
K
mc
K
mc
 
 
Because mc2 << K: 
 
2222
2
111
2
1
⎟⎟⎠
⎞
⎜⎜⎝
⎛−≈⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
K
mc
K
mc 
and 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−=
22
2
11
K
mccv 
 
Substituting for v in equation (1) 
yields: 
c
K
mc
K
mcccv
22
22
2
1
2
11Δ
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−−=
 
 
Substitute numerical values and evaluate Δv: 
 
( ) km/s .504m/s 10998.2
GeV 1035.3
MeV/amu 932amu 197
2
1
Δ 8
2
4 =×⎟⎠
⎞⎜⎝
⎛
×
×=v 
 
Relativity 
 
 
1097
(b) Because the nuclei are, to three 
significant figures, traveling at the 
speed of light: 
s 334.0
m/s102.998
m 100Δ
Δ 8 μ=×=≈ c
xt 
 
 
56 ••• Consider the flight of a beam of neutrons produced in a nuclear 
reactor. These neutrons have kinetic energies of up to 1.00 MeV. The rest energy 
of a neutron is 939 MeV. (a) What is the speed of 1.00 MeV neutrons? Express 
your answer in terms of v/c. (b) If the average lifetime of such a neutron
is 15.0 
min, what is the maximum length of a beam of such neutrons (in a vacuum, in 
the absence of any interactions between the neutrons and other material)? 
Estimate this maximum range by calculating the length corresponding to five 
mean lifetimes. After five mean lifetimes only e−5 or 0.007 (0.7%) of the 
neutrons remain. (c) Compare this to the range of so-called ″thermally 
moderated″ neutrons, whose kinetic energies are around 0.025 eV. Express your 
answer as a percentage. That is, what percent of the 1.00 MeV-neutron range is 
the thermally moderated-neutron range? (Note that our assumption of a vacuum 
continues, however in reality neutrons of this energy interact readily with matter, 
such as air or water, and ″real″ ranges are much shorter.) 
 
Picture the Problem We can apply ( )222 1 cvmcmcK −=+ , the expression 
for the total relativistic energy of a particle, to find the speed of 1.00 MeV 
neutrons. We can use the equation relating distance, rate, and time to estimate 
their maximum beam length. Finally, we can use the non-relativistic expression 
for their kinetic energy to find the range of ″thermally moderated″ neutrons. 
 
(a) The total relativistic energy of a 
neutron is given by Equation R-15: 
 
( )2
2
2
1 cv
mcmcK
−
=+ 
 
 
Solving for v/c yields: 
 
2
21
11
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
+
−=
mc
Kc
v 
 
Substitute numerical values and 
evaluate v/c: 
%61.4
MeV 939
MeV 00.11
11
2
=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
+
−=
c
v 
 
 Chapter R 
 
 
1098 
 
(b) Express the maximum 
expected beam length of the 
neutrons as a function of the 
number n of lifetimes: 
 
tc
c
vnx ΔΔ ⎟⎠
⎞⎜⎝
⎛= (1) 
Evaluate this expression for n = 5: 
 
( )( )( ) m 106.22
min
s 60min 0.15m/s 102.9980461.05Δ 108 ×=⎟⎠
⎞⎜⎝
⎛ ××=x 
 
(c) Express the ratio r of the 
range of ″thermally moderated″ 
neutrons to the range of 1.00 
MeV neutrons: 
 
MeV 1.00
moderated thermally 
Δ
Δ
x
x
r = (2) 
Because the energy of the 
″thermally moderated″ 
neutrons is 0.025 eV, we can 
use the non-relativistic equation 
for their kinetic energy: 
 
2
2
2
12
2
1 c
c
vmmvK ⎟⎠
⎞⎜⎝
⎛== ⇒ 22mc
K
c
v = 
Substitute for v/c in equation 
(1) to obtain: tcmc
Knx Δ2Δ 2moderated thermally = 
 
Substitute numerical values and evaluate : moderated thermally Δx
 
( ) ( )( ) m 1084.9
min
s 60min 0.15m/s 102.998
MeV 939
eV 025.025Δ 68moderated thermally ×=⎟⎠
⎞⎜⎝
⎛×=x 
 
Substitute numerical values in 
equation (2) and evaluate r: %016.0m1022.6
m 1084.9
10
6
=×
×=r 
 
57 ••• You and Ernie are trying to fit a 15-ft-long ladder into a 10-ft-long 
shed with doors at each end. You suggest to Ernie that you open the front door to 
the shed and have him run toward it with the ladder at a speed such that the 
length contraction of the ladder shortens it enough so that it fits in the shed. As 
soon as the back end of the ladder passes through the door, you will slam it shut. 
(a) What is the minimum speed at which Ernie must run to fit the ladder into the 
shed? Express it as a fraction of the speed of light. (b) As Ernie runs toward the 
shed at a speed of 0.866c, he realizes that in the reference frame of the ladder, it 
is the shed which is shorter, not the ladder. How long is the shed in the rest frame 
of the ladder? (c) In the reference frame of the ladder is there any instant that 
Relativity 
 
 
1099
both ends of the ladder are simultaneously inside the shed? Examine this from 
the point of view of relativistic simultaneity. 
 
Picture the Problem Let the letter ″L″ denote the ladder and the letter ″S″ the 
shed. We can apply the length contraction equation to the determination of the 
minimum speed at which Ernie must run to fit the ladder into the shed as well as 
the length of the shed in rest frame of the ladder. 
 
(a) Express the length LL of the shed 
in Ernie’s frame of reference in 
terms of its proper length LL,0: 
 
2
0L,L 1 ⎟⎠
⎞⎜⎝
⎛−=
c
vLL ⇒ 2
L,0
2
L1
L
Lcv −= 
 
Substitute numerical values and 
evaluate v: 
 
ccv 75.0
ft15
ft101
2
=⎟⎟⎠
⎞
⎜⎜⎝
⎛−= 
 
(b) Express the length LS of the shed 
in the rest frame of the ladder in 
terms of its proper length LL,0: 
 
2
2
0S,S 1 c
vLL −=
 
 
Substitute numerical values and 
evaluate LS: 
 
( ) ft0.5866.01ft10 2S =⎟⎠
⎞⎜⎝
⎛−=
c
cL 
 
(c) No. In your rest frame, the back end of the ladder will clear the door before the 
front end hits the wall of the shed, while in Ernie’s rest frame, the front end will 
hit the wall of the shed while the back end has yet to clear the other door. 
 
Let the ladder be traveling from left to right. To ″explain″ the simultaneity issue 
we first describe the situation in the reference frame of the shed. In this frame the 
ladder has length LL = 7.5 m and the shed has length LS,0 = 10 m. We (mentally) 
put a clock at each end of the shed. Let both clocks read zero at the instant the 
left end of the ladder enters the shed. At this instant the right end of the ladder is 
a distance = 2.5 m from the right end of the shed. At the instant the 
right end of the ladder exits the shed both clocks read Δt, where 
LS, 0 − LL
( ) ns63.9Δ LS,0 =−= vLLt . There are two space-time coincidences to consider: 
the left end of the ladder enters the shed and the clock at the left end of the shed 
reads zero, and the right end of the ladder exits the shed and the clock on the right 
end of the shed reads ( ) ns63.9LS,0 =− vLL . In the reference frame of the ladder 
the two clocks are moving to the left at speed v = 0.866c. In this frame the clock 
on the right (the trailing clock) is ahead of the clock on the left by 
ns, 9.282S,0 =cvL so when the clock on the right reads 9.63 ns the one on the left 
reads –19.3 ns. This means the left end of the ladder is yet to enter the shed when 
 Chapter R 
 
 
1100 
the right end of the ladder is exiting the shed. This is consistent with the assertion 
that in the rest frame of the ladder, the ladder is longer than the shed, so the entire 
ladder is never entirely inside the shed.