This is a case of dilation. T = " T# in this problem with the proper time T " = T0
1-5
) # v& 2 ,
T = +1 " % ( .
+* $ c ' -.
)+ # L 2 &2 -+
0
in this case T = 2T0 , v = *1" %
( .
+, $ L0 ' +/
#
v2 &
L = %1" 2 (
c '
$
"1 2
# * L -2&
L0 ) v = c %1" , / (
%$ + L0 . ('
)+ # L 2 &2 -+
v = *1" % 0 ( .
+, $ L0 ' +/
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 =
1-6
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 .
The problem is solved by using time dilation. This is also a case of v << c so the binomial
1-7
% 2c2 ( "t # "t $ ) (
&
v2 )
v2 "t$
*
expansion is used "t = # "t$ % (1 + 2 +"t$ , "t # "t$ =
2 ; v='
" t$
2c
' 2c *
&
)
12
;
"t = (24 h day )( 3 600 s h) = 86 400 s ; "t = "t# $ 1 = 86 399 s ;
# 2( 86 400 s " 86 399 s) &
v=%
(
86 399 s
%$
('
!
1-8
L=
12
= 0.004 8c = 1.44 ) 106 m s .
L"
#
12
1 L %
v2 (
= = '1 $ *
" L# & c 2 )
!
!
% + L .2(
v = c '1 $ - 0 *
, L# / *)
!'&
1-10
(a)
12
% + 75 . 2 (
= c '1 $ 0 *
'& , 100 / *)
" = # " $ where " =
!
!
"1 2
(
" =# 1 $ %
2 $1 2
)
!
(b)
!
1-12
(a)
(
12
= 0.661c
v
and
c
'
v2 *
= & # ) 1$ 2 ,
c +
(
)(
$1 2
(
)[
= 2.6 - 10$8 s 1 $ ( 0.95)
)
d = v" = ( 0.95) 3 # 108 8.33 # 108 s = 24 m
1
70 beats/min or "t# =
min
70
2 $1 2
]
= 8.33 - 10$8 s
1)
( + min = 0.032 8 min beat or the number of beats per
' 70 *
2 %1 2 &
[
"t = # "t$ = 1% ( 0.9 )
(b)
]
minute " 30.5 " 31.
1-14
Only the x-component of L0 contacts.
(a)
v
L0
Ly
!0
Lx
L x [L0 cos #0 ]
%
L y " = L0 sin #0 $ Ly = L0 sin # 0
L x " = L0 cos # 0 $
2)
&
2
L = ((L x ) + Ly +
'
*
( )
12
&, L cos # / 2
)
2
0
= (. 0
1 + (L0 sin #0 ) +
%
0
('+*
&
)
,
v2 /
= L0 (cos2 # 0 . 1 2 2 1 + sin2 # 0 +
c 0
('
+*
As seen by the stationary observer, tan " =
(b)
1-16
For an observer approaching a light source, " ob
Ly
Lx
12
=
12
&
)
v2
= L0 (12 2 cos 2 #0 +
c
'
*
12
L0 sin " 0
= # tan " 0 .
L0 cos "0 #
$ (1 # v c )1 2 '
v
)"
. Setting " = and
=&
& (1 + v c )1 2 ) source
c
%
(
after some algebra we find,
2
2
#2
$ #2obs (650 nm) $ ( 550 nm)
" = 2source
=
= 0.166
# source + #2obs (650 nm )2 + ( 550 nm) 2
(
1-19
1-21
1-23
u XA
$1
)
v = 0.166c = 4.98 % 107 m s (2.237 mi h) (m s) = 1.11 % 108 mi h .
u XA # u XB
2 u XA
; 0.70c =
= "u XB ; u"XA = 0.7c =
2 or
1 # u XA u XB c 2
1 + ( u XA c)
0.70u 2XA " 2cu XA + 0.7c2 = 0 . Solving this quadratic equation one finds u XA = 0.41c
therefore u XB = "u XA = "0.41c .
uX # v
0.50c # 0.80c
u"X =
= #0.50c
2 =
1# u Xv c
1# ( 0.50c)( 0.80c ) c 2
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 ,
(a)
#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 , y2" = z2" = 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
1-31
In this case, the proper time is T0 (the time measured by the students using a clock at rest
relative to them). The dilated time measured by the professor is: "t = # T0 where
"t = T + t . Here T is the time she waits before sending a signal and t is the time required
for the signal to reach the students. Thus we have: T + t = " T0 . To determine travel time t,
realize that the distance the students will have moved beyond the professor before the
signal reaches them is: d = v(T + t) . The time required for the signal to travel this distance
" v% "
v%
d v
is: t = = (T + t) . Solving for t gives: t = $ 'T $1 ( '
#c& #
c&
c c
" v% "
v%
equation for (T + t ) yields: T + $ ' T$ 1 ( '
# c& #
c&
(1
1-37
"1 2
. Substituting this into the above
#
v&
= ) T0 , or T% 1 " (
$
c'
2
#
v&) # v& ,
expression for " this becomes: T = %1 " ( +1 " % ( .
$
c ' +* $ c ' .2
#
v&) # v& ,
T = T0 %1 " ( +1 " % ( .
$
c ' +* $ c ' .-
(1
"1
)#
v& #
v& ,
= T0 +% 1 " ( % 1 + ( .
c'$
c ' .+*$
"1
= ) T0 . Using the
"1 2
T0 , or
12
.
Einstein’s reasoning about lightning striking the ends of a train shows that the moving
observer sees the event toward which she is moving, event B, as occurring first. We may
take the S-frame coordinates of the events as ( x = 0 , y = 0 , z = 0 , t = 0 ) and ( x = 100 m ,
y = 0 , z = 0 , t = 0 ). Then the coordinates in S " are given by Equations 1.23 to 1.27. Event
A is at ( x " = 0 , y" = 0 , z" = 0 , t " = 0 ). The time of event B is:
%
%
v (
1
%
0.8c
(
80 m (
t " = # ' t $ 2 x* =
' 0 $ 2 (100 m)* = 1.667 '
* = $4.44 + 10$7 s .
8
&
)
c )
c
1 $ 0.8 2 &
& 3 + 10 m s )
!
The time elapsing before A occurs is 444 ns.
!
!
!
!
!
!
!
!