!
1-5
This is a case of dilation. T = " T# in this problem with the proper time T " = T0
) # v& 2 ,
T = +1 " % ( .
+* $ c ' -.
)+ # L 2 &2 -+
0
in this case T = 2T0 , v = *1" %
( .
+, $ L0 ' +/
1-6
"1 2
# * L -2&
L0 ) v = c %1" , / (
%$ + L0 . ('
)+ # L 2 &2 -+
v = *1" % 0 ( .
,+ $ L0 ' /+
L=
12
# 0 13 &
= %1" 2 5 (
$ 1 44 '
2
v ) #T & ,
T0 / = +1 " % 0 ( .
c *+ $ T ' .-
12
# 0 13 &
= %1" 2 5 (
$ 1 44 '
This is a case of length contraction. L =
#
v2 &
L = %1" 2 (
c '
$
1-8
"1 2
12
;
12
therefore v = 0.866c .
L"
in this problem the proper length L " = L0 ,
#
12
; in this case L =
L0
,
2
12
therefore v = 0.866c .
L"
#
12
1 L %
v2 (
= = '1 $ 2 *
" L# & c )
% + L .2(
v = c '1 $ - 0 *
'& , L# / *)
!
1-9
L earth =
L earth
!
1-10
(a)
1-12
% + 75 . 2 (
= c '1 $ 0 *
'& , 100 / *)
12
= 0.661c
L"
#
$
v2 '
= L " &1# 2 )
c (
%
12
[
2 12
, L " , the proper length so L earth = L = L 1" ( 0.9 )
" = # " $ where " =
!
!
12
(
" =# 1 $ %
2 $1 2
)
'
v2 *
= & # ) 1$ 2 ,
c +
(
(
)(
$1 2
(
2 $1 2
]
= 8.33 - 10$8 s
)
(a)
70 beats/min or "t# =
(b)
"t = # "t$ = 1% ( 0.9 )
1
min
70
1)
( + min = 0.032 8 min beat or the number of beats per
' 70 *
2 %1 2 &
]
)[
= 2.6 - 10$8 s 1 $ ( 0.95)
d = v" = ( 0.95) 3 # 108 8.33 # 108 s = 24 m
minute " 30.5 " 31.
= 0.436L .
v
and
c
(b)
[
]
1-16
$ (1 # v c )1 2 '
v
)"
For an observer approaching a light source, " ob = &
. Setting " = and
& (1 + v c )1 2 ) source
c
%
(
after some algebra we find,
2
"=
2
# 2source $ #2obs (650 nm) $ ( 550 nm)
=
= 0.166
#2source + #2obs (650 nm )2 + ( 550 nm) 2
(
)
v = 0.166c = 4.98 % 107 m s (2.237 mi h) (m s)
v + u"
1-20
u=
1-21
u"X =
1-23
(a)
2
1 + vu" c
=
uX # v
0.90c + 0.70c
2
1+ (0.90c ) ( 0.70c ) c
= 1.11 % 108 mi h .
= 0.98c
0.50c # 0.80c
2 = #0.50c
1# u Xv c
1# ( 0.50c)( 0.80c ) c
Let event 1 have coordinates x1 = y1 = z1 = t 1 = 0 and event 2 have coordinates
x2 = 100 mm , y2 = z2 = t2 = 0 . In S " , x1" = # (x1 $ vt1 ) = 0 , y"1 = y1 = 0 , z"1 = z1 = 0 ,
2
=
$1
#1 2
$ v2 '
+
% v( .
2
and t1" = # -t1 $ ' 2 * x1 0 = 0 , with " = &1 # 2 )
and so " = 1# ( 0.70)
&
)
c (
,
c
/
%
In system S " , x2" = # (x2 $ vt2 ) = 140 m , y"2 = z"2 = 0 , and
[
]
#1 2
= 1.40 .
+
% v ( . (1.4 )( $0.70) (100 m)
t2" = # -t2 $ ' 2 * x2 0 =
= $0.33 µs .
&c ) /
,
3.00 1 108 m s
(b)
"x# = x2# $ x1# = 140 m
(c)
!
1-26
!
Events are not simultaneous in S " , event 2 occurs 0.33 µs earlier than event 1.
12
* $ v ' 2The observed length of an object moving with speed v is L = L" ,1 # & ) / with L " being
,+ % c ( /.
the proper length. For the two ships, we know that L2 = L1 , L2" = 3L"1 and v1 = 0.35c . Thus
2
* $ v '2# v2 &
2
2
2
2
2
2
,
/
L2 = L1 and 9L1" 1 # & ) = L1" 1 # ( 0.35) , giving 9 " 9% ( = 0.877 5 , or v2 = 0.95c .
$ c'
,+ % c ( /.
20.0 ly
"x
=
= 21.05 Speedo’s
In the Earth frame, Speedo’s trip lasts for a time "t =
v
0.950 ly yr
"t
= 21.05 yr 1 $ 0.95 2 = 6.574 yr
age advances only by the proper time interval: "t p =
#
( )
1-35
[
]
20.0 ly
"x
v2
1# 2 =
1# 0.752 = 17.64 yr .
v
0.750
ly
yr
c
! Speedo has landed on Planet X and is waiting for his brother, he ages by
While
during his trip. Similarly for Goslo, "t p =
!
!
20.0 ly
0.20 ly
2
"
1 " 0.75 = 17.64 yr .
0.750 ly yr 0.950 ly yr
!
!
Then Goslo ends up older by 17.64 yr " ( 6.574 yr + 5.614 yr) = 5.45 yr .
!
!
!
!
!
!