# &#34; ) - 2

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