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! 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 . ! ! ! ! ! !