Chapter 26
Relativity
Problem Solutions
26.1
(a) Observers on Earth m easure the tim e for the astronauts to reach Alpha Centauri as
tE 4.42 yr . But these observers are m oving relative to the astronaut’ s internal
biological clock and hence, experience a d ilated version of the proper tim e interval
t p m easured on that clock. From tE
t p , w e find
tp
tE
tE 1
vc
2
4.42 yr
1
0.950
2
1.38 yr
(b) The astronauts are m oving relative to the span of space separating Earth and Alpha
Centauri. H ence, they m easure a length contracted version of the proper d istance,
Lp 4.20 ly . The d istance m easured by the astronauts is
L
26.2
Lp
Lp 1
vc
2
4.20 ly
1
0.950
2
1.31 ly
(a) The length of the m eter stick m easured by the observer m oving at speed v 0.900 c
relative to the m eter stick is
L
Lp
Lp 1
vc
2
1.00 m
1
0.900
2
0.436 m
(b) If the observer m oves relative to Earth in the d irection opposite the m otion of the
m eter stick relative to Earth, the velocity of the observer relative to the m eter stick is
greater than that in part (a). The m easured length of the m eter stick w ill be
less than 0.436 m und er these cond itions.
496
Relativity
26.3
497
If the m oving clock is observed to “ run at a rate one-half the rate of an id entical clock at
rest” , this m eans that an observer m oving relative to the clock m easures the period T of
the clock’ s oscillating m echanism to be d ouble the proper period T p m easured by an
observer at rest relative to the clock (i.e., the m oving observer sees the clock “ tick” half
as often). Thus, T
1
26.4
1
or
2
2
vc
1 1 (v c)2
2Tp , or
Tp
vc
2
1
4
1
2 , giving
3
4
and
vc
3 2
(a) The tim e for 70 beats, as m easured by the astronaut and an y observer at rest w ith
respect to the astronaut, is t p 1.0 min . The observer in the ship then m easures a
rate of 70 beats min .
(b) The observer on Earth m oves at v 0.90 c relative to the astronaut and m easures the
tim e for 70 beats as
t
tp
tp
1
vc
1.0 min
2
1- 0.90
2.3 min
2
This observer then m easures a beat rate of
26.5
(a)
t
tp
tp
1
26.6
vc
2
1
v
t
(c)
v
tp
8
0.98
0.98 3.0 108 m s
(b) d
d
2.6 10
s
2
1.3 10
7
s
2.6 10
8
1.3 10
0.98 3.0 108 m s
70 beats
2.3 min
7
30 beats min
s
38 m
s
7.6 m
(a) As m easured by observers in the ship (that is, at rest relative to the astronaut), the
tim e required for 75.0 pulses is t p 1.00 min .
The tim e interval required for 75.0 pulses as m easured by the Earth observer is
t
tp
rate
(1.00 min)
75.0
t
75.0 1
1 (0.500)2 , so the Earth observer m easures a pulse rate of
0.500
1.00 min
2
65.0 min
498
CH APTER 26
(b) If v 0.990c , then
t
1.00 min
tp
1
0.990
2
and the pulse rate observed on Earth is
75.0 1
75.0
t
rate
0.990
2
10.6 min
1.00 min
That is, the life span of the astronaut (reckoned by the total num ber of his
heartbeats) is m uch longer as m easured by an Earth clock than by a clock aboard
the space vehicle.
26.7
(a) To the observer on Earth the m uon appears to have a lifetim e of
4.60 103 m
d
v
t
8
0.990 3.00 10 m s
1
(b)
1
1
vc
2
1
0.990
1.55 10 5 s
7.09
2
(c) To an observer at rest w ith respect to the m uon, its proper lifetim e is
t
tp
1.55 10
7.09
5
s
2.19 10
6
s
(d ) The m uon is at rest relative to the observer traveling w ith the m uon. Thus, the
m uon travels zero d istance as m easured by this obser ver. H ow ever, d uring the
observed lifetim e of the m uon, this observer sees Earth m ove tow ard the m uon a
d istance of
d
v
0.990 3.00 108 m s
tp
2.19 10
6
s
6.50 102 m
650 m
(e) As the third observer travels tow ard the incom ing m uon, his speed relative to the
m uon is greater than that of the observer at rest on Earth. Thus, his observed
gam m a factor ( t
t p )is higher, and he measures the muon’s lifetime as longer
than that m easured by the observer at rest w ith respect to Earth.
26.8
t
tp
tp
1
vc
3.0 s
2
1- 0.80
2
5.0 s
Relativity
26.9
The proper length of the faster ship is three tim es that of the slow er ship Lpf
499
3Lps , yet
they both appear to have the sam e contracted length, L. Thus,
L
Lps 1
c 8
This gives v f
26.10
vs c
2
vs c
3Lps
2
1
8
vf c
0.350
3
2
, or 1
vs c
2
9 9 vf c
2
2
3
c
0.950 c
After tim e t 1 h 3600 s has elapsed on the Earth-based clock, the tim e that is
m easured to have elapsed on the second clock by an observer w ho is at rest relative to
that clock is t p
t
t . Thus, the observer on Earth jud ges the m oving clock to be
slow by an am ount
t
tp
t 1
1
t 1
1 (v c ) 2
t 1
1
1
(v c ) 2
2
t
(v c ) 2
2
In this case, w ith v 1000 km h , the apparent d elay is
3 600 s 1000 km h
103 m
2
3.000 108 m s 1 km
26.11
2
1h
3 600 s
1.54 10 9 s
The tracksid e observer sees the supertrain length -contracted as
L
Lp 1
vc
2
100 m
1
0.95
2
31 m
The supertrain appears to fit in the tunnel
w ith 50 m 31 m
26.12
19 m to spare
Length contraction occurs only in the d im ension parallel to the
m otion.
(a) The sid es labeled L2 and L3 in the figure at the right are
unaffected , but the sid e labeled L1 w ill appear contracted
giving the box a rectangular shape.
1.54 ns
500
CH APTER 26
(b) The d im ensions of the box, as m easured by the observer
m oving at v 0.80 c relative to it, are
26.13
2.0 m , L3
L2
L2 p
L1
L1 p 1
(a) Classically,
2
vc
2.0 m
p mv mv0
p 0.040 9.11 10
31
2.0 m , and
L3 p
1
0.80
2
1.2 m
0.040mc , or
m 0.940c 0.900c
kg 3.00 108 m s
1.09 10
23
kg m s
(b) By relativistic calculations, the change in m om entum is
mv f
p
mvi
1 (v f c )
2
1 (vi c)
m
2
0.940c
0.900c
1 (0.940)
2
1 (0.900) 2
or
p
9.11 10
31
kg 3.00 108 m s
0.940
0.900
1 (0.940)
1.89 10
26.14
22
2
1 (0.900)2
kg m s
(a) Classically,
p mv m 0.990c
1.67 10
27
kg 0.990 3.00 108 m s
4.96 10
19
kg m s
(b) By relativistic calculations,
p
m 0.990c
mv
1 (v c ) 2
3.52 10
18
1 (0.990)2
1.67 10
27
kg 0.990 3.00 108 m s
1 (0.990)2
kg m s
(c) N eglecting relativistic effects at such speed s w ould introd uce an appr oxim ate 86%
error in the result. No , neglecting these effects is unreasonable.
Relativity
26.15
Mom entum m ust be conserved , so the m om enta of the tw o fragm ents m ust ad d to zero.
Thus, their m agnitud es m ust be equal, or
p2
p1
1
2.50 10
m1v1
28
1
w hich red uces to 3.37 v c
(a)
ER
(b) E
mc 2
1.67 10
mc 2
27
939 MeV
26.17
0.950
(c)
KE
E ER
(a)
ER
mc2
(b) KE
(c)
E
ER
1
vc
1
vc
kg v
2
2
4.96 10
and yield s v
kg 3.00 108 m s
2
vc
2
28
28
kg c
kg c
0.285 c
1 MeV
1.60 10-13 J
939 MeV
2
3.01 103 MeV
3.01 GeV
3.01 103 MeV 939 MeV 2.07 103 MeV
0.15 kg 3.00 108
E ER
27
4.96 10
2
ER
ER
1
1
kg 0.893 c
0.893
1.67 10
For the heavier fragm ent,
26.16
501
1 ER
2
2.07 GeV
1.35 1016 J
1
1 (0.900)
2
KE 1.35 1016 J 1.75 1016 J
1 1.35 1016 J
3.10 1016 J
1.75 1016 J
502
26.18
CH APTER 26
The w ork d one accelerating a free particle from rest to speed v equals the kinetic energy
given that particle. Thus,
KE E ER
mc 2
For a proton, ER
Thus,
or
26.19
ER
v c 1
1 4.99
3 750 MeV
1.67 10
3 750 MeV
939 MeV
1
ER
27
2
1 MeV
1.60 10-13 J
1
E ER
vc
4.99
2
0.980 c
The nonrelativistic expression for kinetic energy is KE
expression is KE
939 MeV
1
giving
3.99
2
kg 3.00 108
(
1)ER
1
2
1)mc2 w here
(
mv2 , w hile the relativistic
1
1
vc
2
. Thus, w hen the
relativistic kinetic energy is tw ice the pred icted nonrelativistic value, w e have
1
1
vc
1 mc 2
2
2
1 2
mv
2
or
1
1
Squaring both sid es of the last result and sim plifying gives
Ignoring the trivial solution v c 0 , w e m ust have
This is a quad ratic equation of the form x2
x
quad ratic form ula gives
1
v
c
4
x 1 0 with x
2
v
c
1
2
v
c
v
c
v
c
vc
4
v
c
2
1
5
2
0.618
1
2
v c . Applying the
5
2
w hich yield s
2
1 0
v c , w e ignore the negative solution and find
x
v
c
2
2
Since x
2
v c 0.618
0.786 c
0
Relativity
26.20
The energy input to the electron w ill be W
1
W
1
(a) If v f
vf c
1
0.900 c and vi
vi c
1
(b) When v f
1
0.900
2
1
0.990 c and vi
1
w here
f
i
ER
) ER , or
0.511 MeV
2
0.582 MeV
0.900 c , w e have
1
0.990
(
0.511 Mev
2
0.500
1
W
ER
2
Ei
0.500 c , then
1
W
26.21
1
2
Ef
1
0.900
0.511 Mev
2
2.45 MeV
When a cubical box m oves at speed v, the d im ension parallel to the m otion is length
contracted and other d im ensions are unaffected . If all ed ges of the box had length L p
w hen at rest, the volum e of the m oving box is
V
Lp
Lp
Lp 1
vc
2
Vp 1
vc
2
The relativistic d ensity is
m
V
N ote that
m
Vp 1
m
V
mc 2
c2 V
8.00 g
vc
2
1.00 cm
3
1
0.900
2
18.4 g cm3
ER
as suggested in the problem statem ent.
c2 V
503
504
26.22
CH APTER 26
Let m1 be the m ass of the fragm ent m oving at v1
v2 0.987 c .
0.868 c , and m2 be the m ass m oving at
From conservation of m ass-energy,
m1c 2
E
1
0.868
m2c 2
2
1
giving 2.01m1 6.22 m2
0.987
3.34 10
27
2
mc 2
[1]
kg
The m om enta of the tw o fragm ents m ust ad d to zero, so the m agnitud es m ust be equal,
giving p1 p2 or 1m1v1
2 m2v2 . This yield s
2.01m1 0.868 c
6.22 m2 0.987 c , or m1 3.52 m2
[2]
Substituting Equation [2] into [1] gives
7.07 6.22 m2
Equation [2] then yield s m1
26.23
27
3.34 10
3.52 2.51 10
The relativistic total energy is E
1 1 v2 c2 . Thus, E 2 c2
E2
c2
p2
2
m2 c 2
kg , or m2
v2
ER
2
28
2.51 10
kg
kg
8.84 10
28
kg
mc2 and the m om entum is p
m2c2 and p2
2
28
m2 c 2 1 v 2 c 2
2
mv w here
m2v2 , so subtracting yield s
m2 c 2
Rearranging, this becom es
E2
c2
26.24
p2
m2c 2
E2
or
p2c2
m2 c 4
(a) Since the total relativistic energy of a particle is E KE ER KE mc2 , requiring
that total energy be conserved (i.e., total energy before reaction = total energy after
reaction) gives KEi
mi c 2
KE f
m f c 2 , w here mi is the total m ass of the particles
present before reaction, KEi is the total kinetic energy before reaction, m f is the
total m ass of the particles present after the reaction, and KE f is the final total kinetic
energy.
Relativity
505
(b) Using atom ic m asses from the tables in Append ix B of the textbook, the total m ass
of the initial particles is
mi
m235 U
m1n
92
235.043 923 u 1.008 665 u
236.052 588 u
0
(c) The total m ass of the particles present after the reaction is
mf
m148 La
m87 Br
57
m1n
35
0
147.932 236 u 86.920 71119 u 1.008 665 u
235.861612 u
(d ) The m ass that m ust have been converted to energy in the reaction is
m mi
mf
236.052 588 235.861612 u
(e) Assum ing that KEi
KE f
26.25
(a)
(b) L
26.26
mc2
0 , the total kinetic energy of the prod uct particles is then
0.190 976 u 931.494 MeV u c2
KEi
E
ER
20.0 GeV 103 MeV
0.511 MeV 1 GeV
Lp
3.00 103 m
3.91 104
1 1 (ve c)2
0
177.893 MeV
3.91 104
2
7.67 10
(a) For an electron m oving at ve
e
0.190 976 u
m
7.67 cm
0.750c , the gam m a factor is
1 1 (0.750)2
1.51
The kinetic energy of a particle is KE E ER ( 1)ER , so if the kinetic energy of
a proton ( ER, p 938.3 MeV ) equals that of the electron ( ER,e 0.511 MeV ), w e m ust
have
(
But
vp
p
p
1) ER, p
(
e
1) ER,e
or
1
p
1 1 (v p c)2 , so (vp c)2 1 1
c 1
1
2
p
c 1
1
1 2.78 10
4
2
1.51 1
2
p
0.511 MeV
938.3 MeV
1 2.78 10
4
and the speed of the proton m ust be
0.0236 c
506
CH APTER 26
(b) As above,
1.51 for an electron having speed ve
e
0.750 c .
The m om entum of a particle is
p
mc 2 (v c)
c
mv
E R (v c )
c
pc
or
ER
v
c
If the m om entum of a proton equals that of the electron, then p p c
vp
ER , p
p
ER ,e
e
c
pe c , or
ve
c
and
vp c
1
Thus, v p c
vp c
2
e
2
vp c
6.17 10
2
1
4
6.17 10
2
1.51
6.17 10
4
2
0.511 MeV
938.3 MeV
0.750
6.17 10
4
2
v p c , w hich yield s
2
4
6.17 10
ER ,e ve
ER , p c
4
2
3.81 10
7
and the speed of the proton m ust be
vp
26.27
KE
c 3.81 10
E ER
so
7
3.00 108 m s
3.81 10
7
1.85 105 m s
1 ER
1
KE
ER
1
1
vc
giving
2
1
v c 1
1 KE ER
(a) The speed of an electron having KE 2.00 MeV w ill be
ve
c 1
1
1 KE ER,e
2
c 1
1
1 2.00 0.511
2
0.979 c
(b) For a proton w ith KE 2.00 MeV , the speed is
vp
185 km s
c 1
1
1 KE ER,e
2
c 1
1
1 2.00 938
2
0.065 2 c
2
Relativity
26.28
(c)
ve v p
(a)
Yes . As the spring is com pressed , positive w ork is d one on it or energy is ad d ed
to it. Since m ass and energy are equivalent, m ass has been ad d ed to the spring and
its total m ass has increased .
(b)
m
m
0.979 c 0.065 2 c
1
2
PEs
c2
ER
c2
2 3.00 10 m s
KE
E ER
so
1
0.914 c
kx2
c2
kx2
2c 2
2
200 N m 0.15 m
8
26.29
2.5 10
2
17
kg
1 ER
KE
ER
1
1
(a) When KE q
vc
V
giving
2
1
v c 1
1 KE ER
500 eV , and ER
e 500 V
1
v c 1
2
6
2
939 MeV , this yield s
1.03 10 3 c
3.10 105 m s
1 500 eV 939 10 eV
(b) When KE
q
e 5.00 108 V
V
1 500 MeV 939 MeV
From E 2
500 MeV
1
v c 1
26.30
507
( pc)2
0.758 c
2
ER2 w ith E 5 ER , w e find that p ER 24 c
(a) For an electron,
p
(b) For a proton,
p
0.511 MeV
24
c
939 MeV
c
24
2.50 MeV c
4.60 103
MeV
c
4.60 GeV c
508
26.31
CH APTER 26
(a) Observers on Earth m easure the d istance to And rom ed a to be
d 2.00 106 ly (2.00 106 yr) c . The tim e for the trip, in Earth’ s fram e of
reference, is
t
30.0 yr
tp
1
vc
2
The required speed is then v
w hich gives
1.50 10
5
2.00 106 yr c
d
t
v
c
30.0 yr
1
vc
1- v c
2
2
Squaring both sid es of this equation and solving for v c yield s
v c 1 1 2.25 10
v
2.25 10
1
c
2
10
. Then, using the approxim ation 1 1 x 1 x 2 gives
10
1 1.12 10
10
1 mc2 , and
(b) KE
1
1
vc
1
2
1
1
1 1.12 10
10 2
2.25 10
10
6.67 104
Thus,
KE
(c)
cost
6.67 104 1 1.00 106 kg 3.00 108 m s
2
6.00 1027 J
KE rate
6.00 1027 J
1 kWh
3.60 106 J
$0.13 kWh
$2.17 10 20
509
Relativity
26.32
The clock, at rest in the ship’ s fram e of reference, w ill m easure a proper tim e of
t p 10 h before sound ing. Observers on Earth m ove at v 0.75 c relative to the clock
and m easure an elapsed tim e of
t
tp
tp
1
10 h
vc
2
1- 0.75
2
15 h
The observers on Earth see the clock m oving aw ay at 0.75c and com pute the d istance
traveled before the alarm sound s as
d
26.33
v
0.75 3.0 108 m s
t
15 h
3600 s
1h
1.2 1013 m
The length of the space ship, as m easured by observers on Earth, is L
Lp 1
vc
2
. In
Earth’ s fram e of reference, the tim e required for the ship to pass overhead is
t
L
v
1
v2
1
c2
Lp 1
vc
2
v
Lp
1
v2
1
c2
Thus,
2
or
t
Lp
3.00 108 m s
1
v
2
1.74 10
26.34
1
17
2.4 108
s
m2
2
m
s
0.75 10 6 s
300 m
2
c
3.00 108 m s
1.74 10
17
s2
m2
0.80 c
The solution to this problem is easier if w e start w ith part (b). First, com pute the speed
of the protons.
KE
1013 MeV
1
1
1.06 1010 , or ~ 1010
KE
1 ER , so
ER
939 MeV
Thus, v c 1 1
2
~ c 1 10
20
c
510
CH APTER 26
(b) The d iam eter of the galaxy, as seen in the proton’ s fram e of reference, is
L
Lp 1
Since 1 ly
L 10
v c
Lp
2
ly
105 ly
10
1010
5
ly
3.156 107 s 3.00 108 m s
1 yr c
5
~
1016 m
1 ly
1016 m
1 km
~ 108 km
3
10 m
(a) The proton sees the galaxy rushing by at v c . The tim e, in the proton’ s fram e of
reference for the galaxy to pass is
t
or
26.35
10
L 10 5 ly
~
v
c
5
yr c
10
c
5
yr
3.156 107 s
1 yr
316 s
t ~ 102 s
mv
Using the relativistic form , p
1
w e find the d ifference
p
vc
p from the classical m om entum , mv :
mv mv (
1)mv
(a) The d ifference is 1.00% w hen (
1
0.990
1
1
vc
2
1
(b) The d ifference is 10.0% w hen (
1
0.900
1
1
mv
2
vc
2
1
1)mv 0.010 0 mv :
vc
2
0.990
2
or
v
0.141c
or
v
0.436 c
1)mv 0.100 mv :
vc
2
0.900
2
Relativity
26.36
511
The kinetic energy gained by the electron w ill equal the loss of potential energy, so
KE q
V
e 1.02 MV
1.02 MeV
1
2
(a) If N ew tonian m echanics rem ained valid , then KE
w ould be
2 1.02 MeV 1.60 10
2 KE
m
v
(b) KE
-31
9.11 10
1 ER , so
KE
ER
1
1
13
J MeV
kg
1.02 MeV
0.511 MeV
mv2 , and the speed attained
5.99 108
m
s
2c
3.00
The actual speed attained is
v c 1 1
26.37
2
c 1 1 3.00
2
0.943 c
(a) When at rest, m uons have a m ean lifetim e of
tp
2.2 s . In a fram e of reference
w here they m ove at v 0.95 c , the d ilated m ean lifetim e of the m uons w ill be
tp
tp
1
2.2 s
vc
2
1
0.95
7.0 s
2
(b) In a fram e of reference w here the m uons travel at v 0.95 c , the tim e required to
travel 3.0 km is
t
If N0
N
d
v
3.0 103 m
8
0.95 3.00 10 m s
1.05 10
5
s 10.5 s
5.0 104 m uons started the 3.0 km trip, the num ber rem aining at the end is
N0 e
t
5.0 104 e
10.5 s 7.0 s
1.1 104
512
26.38
CH APTER 26
The w ork required equals the increase in the gr avitational potential energy, or
W GMSun m Rg . If this is to equal the rest energy of the m ass rem oved , then
mc 2
Rg
GM Sun m
or
Rg
6.67 10
11
GM Sun
c2
Rg
N m2 kg 2 1.99 1030 kg
8
3.00 10 m s
26.39
1.47 103 m
2
1.47 km
Accord ing to Earth-based observers, the tim es required for the tw o trips are
For Speed o:
TS
L0
vS
20.0 yr
0.950 c
21.05 yr
For Goslo:
TG
L0
vG
20.0 yr
0.750 c
26.67 yr
Thus, after Speed o land s, he m ust w ait and age at the sam e rate as planet -based
observers, for an ad d itional TS TG TS (26.67 21.05) yr 5.614 yr before Goslo
arrives.
The tim e required for the trip accord ing to Speed o’ s internal biological clock (w hich
m easures the proper tim e for his aging process d uring the trip) is
T0 S
TS
TS 1
vS c
2
21.05 yr
1
0.950
2
6.574 yr
When Goslo arrives, Speed o has aged a total of
tS
T0S
TS
6.574 yr 5.614 12.19 yr
The tim e required for the trip accord ing to Goslo’ s internal biological clock (and hence
the am ount he ages) is
tG
T0G
TG
TG 1
vG c
2
26.67 yr
1
0.750
2
17.64 yr
Thus, w e see that w hen he arrives, Goslo is older than Speed o, having aged an
ad d itional
tG
tS
17.64 yr 12.19 yr
5.45 yr
Relativity
26.40
(a)
t
tp
tp
1
(b) d
v
t
vc
15.0 yr
2
1
2
0.700
0.700 c 21.0 yr
513
21.0 yr
0.700 1.00 ly yr
21.0 yr
14.7 ly
(c) The astronauts see Earth flying out the back w ind ow at 0.700c :
d
v
tp
0.700 c 15.0 yr
0.700 1.00 ly yr
15.0 yr
10.5 ly
(d ) Mission control gets signals for 21.0 yr w hile the battery is operating and then for
14.7 yr after the battery stops pow ering the transm itter, 14.7 ly aw ay:
21.0 yr 14.7 yr 35.7 yr
26.41
N ote: Excess d igits are retained in som e steps given
below to m ore clearly illustrate the m ethod of
solution.
We are given that L 2.00 m , and
30.0 (both
m easured in the observer’ s rest fram e). The
com ponents of the rod ’ s length as m easured in the
observer’ s rest fram e are
and
Lx
L cos
2.00 m cos30.0
1.732 m
Ly
L sin
2.00 m sin30.0
1.00 m
The com ponent of length parallel to the m otion has been contracted , but the com ponent
perpend icular to the m otion is unaltered . Thus, Lpy Ly 1.00 m and
Lx
Lpx
1
1.732 m
vc
2
1
0.995
17.34 m
2
(a) The proper length of the rod is then
Lp
L2px
L2py
17.34 m
2
1.00 m
2
17.4 m
(b) The orientation angle in the rod ’ s rest fram e is
p
tan
1
Lpy
Lpx
tan
1
1.00 m
17.34 m
3.30