ChR ISM
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ChR ISM


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T, the time between the arrival of 
signals on Earth is related to the 
proper time interval T0 (the time 
between signals in Al\u2019s frame of 
reference) through the time-dilation 
equation: 
 
( )2201 cv
TT \u2212= 
Substituting for T and simplifying 
yields: 
( )
0
22
Bert 600.1
1
T
cv
f
\u2212= 
 
Substitute numerical values and 
evaluate Bertf : 
 
( )
( )( ) 1
2
Bert y 50y 0100.0600.1
600.01 \u2212=\u2212=f 
Chapter R 
 
 
1068 
 
(b) Express the number of signals 
N received by Bert in terms of 
the number of signals sent by Al: 
 
Al0 tfN \u394= 
Use the time dilation equation to 
express the elapsed time in Al\u2019s 
frame of reference in terms of the 
elapsed time in Bert\u2019s frame (the 
proper elapsed time): 
2
BertAl 1\u394\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtt 
 
Substitute for Al\u394t to obtain: 
 
2
Bert0 1\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtfN (1) 
 
Find Bert\u394t : y667.6
0.600
y00.4
\u394 Bert =\u22c5= c
ct 
 
Substitute numerical values in 
equation (1) and evaluate N: ( )( )
533
600.01y 667.6y100
2
1
=
\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212= \u2212
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 \u394= 
Because 
2
BertAl 1\u394\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtt , 
the time in Al\u2019s frame for the 
round trip Al\u394T is given by: 
 
2
BertAlAl 1\u3942\u3942\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212==
c
vttT 
 
 
 
Substituting for Al\u394T in the 
expression for N yields: 
 
2
Bert0 1\u3942 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtfN 
 
Substitute numerical values and 
evaluate N: 
 
( )( )
3
2
1
1007.1
600.01y667.6y1002
×=
\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212= \u2212
c
cN
 
 
Special Relativity 
 
 
1069
(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
\u2212=f 
 
(e) Express the number of signals N 
received by Al: 
 
AlAl tfN \u394= 
Substitute for Alt\u394 to obtain: 2
BertAl 1\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtfN 
 
Substitute numerical values and 
evaluate N: 
 
( )( )
267
600.01y667.6y0.50
2
1
=
\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212= \u2212
c
cN
 
 
(f) Express the total number of 
signals received by Al: 
 
return
returnoutboundtot
267 N
NNN
+=
+=
 
The number, returnN , of signals 
received by Al on his return trip is 
given by: 
2
Bertreturn Al,
Alreturn Al,return
1\u394
\u394
\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
=
c
vtf
tfN
 
 
Substitute for returnN to obtain: 2
Bertreturn Al,tot 1\u394267 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212+=
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
\u2212=f 
 
Substitute numerical values and evaluate totN : 
 
( )( ) 321tot 1033.11334600.01y667.6y200267 ×==\u239f\u23a0\u239e\u239c\u239d\u239b\u2212+= \u2212 c cN 
 
(g) Their age difference is: 
 
AlBert ttt \u394\u2212\u394=\u394 
Chapter R 
 
 
1070 
 
Substitute numerical values to 
obtain: 
y2.67
y 100
1067
y 100
1334
\u394 11 =\u2212= \u2212\u2212t 
and 
Bert.an younger thy 2.67 is Al 
 
Relativistic Energy and Momentum 
 
39 \u2022 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
\u2212
=
\u2212
= 
 
Solve for the ratio of E to E0 to 
obtain: 
2
2
0 1
1
c
vE
E
\u2212
= 
 
(a) Evaluate 0EE for v = 0.100c: 
 ( ) 01.1100.01
1
2
2
100.00
=
\u2212
=\u23a5\u23a6
\u23a4
=
c
cE
E
cv
 
 
(b) Evaluate 0EE for v = 0.500c: 
 ( ) 15.1500.01
1
2
2
500.00
=
\u2212
=\u23a5\u23a6
\u23a4
=
c
cE
E
cv
 
 
(c) Evaluate 0EE for v = 0.800c: 
 ( ) 67.1800.01
1
2
2
800.00
=
\u2212
=\u23a5\u23a6
\u23a4
=
c
cE
E
cv
 
 
(d) Evaluate 0EE for v = 0.990c: 
 ( ) 09.7990.01
1
2
2
999.00
=
\u2212
=\u23a5\u23a6
\u23a4
=
c
cE
E
cv
 
 
 
Special Relativity 
 
 
1071
40 \u2022 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 2 2 2 2 4E p c m c= + (Equation R-17), and once we have p we can solve for the 
speed v using v/c = pc/E (Equation R-16). 
 
(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 += \u21d2
c
EE
p
2
0
2 \u2212= 
 
Substitute numerical values and 
evaluate p: ( ) ( )
c
c
c
p
MeV/ 1040
MeV/ 1039
Mev 938Mev 1400 22
=
=
\u2212=
 
 
(a) Use Equation R-16 to express the 
speed of the proton: E
pc
c
v = \u21d2 c
E
pcv \u239f\u23a0
\u239e\u239c\u239d
\u239b= 
 
Substitute numerical values and 
evaluate v: 
( ) cccv 742.0
MeV 1400
MeV/ 1039 =\u239f\u23a0
\u239e\u239c\u239d
\u239b= 
 
41 \u2022 [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
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=
\u2212
\u2212
=
 
 
Chapter R 
 
 
1072 
(a)Substitute numerical values and 
evaluate K(0.500c): 
 
( ) ( )
2
2
2
155.0
1
500.01
1500.0
mc
mc
cc
cK
=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=
 
 
(b) Substitute numerical values and 
evaluate K(0.900c): 
 
( ) ( )
2
2
2
29.1
1
900.01
1900.0
mc
mc
cc
cK
=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=
 
 
(c) Substitute numerical values and 
evaluate K(0.990c): ( ) ( )
2
2
2
09.6
1
990.01
1990.0
mc
mc
cc
cK
=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=
 
 
42 \u2022 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 \u2212=\u2212 (1) 
From Equation R-14 we have: 
 ( )
( )
2
2
2
2
2
1
1
1
1
mc
cv
mc
cv
mcK
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=
\u2212
\u2212
=
 
 
For K = E0: 
( ) KcvK \u239f
\u239f
\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
= 1
1
1
2 
or 
( ) 21
1
2
=
\u2212 cv
 
 
Special Relativity 
 
 
1073
From Equation R-10, the relativistic 
momentum of the particle is: 
 
( ) mvcv
mvp 2
1 2
rel =\u2212
= 
Substitute in equation (1) and 
simplify to obtain: 
%50
2
11
rel
=\u2212=\u2212
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 \u2212=
\u2212
= 
Because 1 1 22 <\u2212 cv , the non-relativistic expression is always low compared to 
the relativistically correct expression for momentum. 
 
43 \u2022 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 \u2022\u2022 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 
 
 
1074