CH-37 - Chapter 37 1 From the time dilation equation t =...

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Chapter 37 1. From the time dilation equation t = t 0 (where t 0 is the proper time interval, 1 1 2 / , and = v / c ), we obtain F H G I K J 1 0 2 t t . The proper time interval is measured by a clock at rest relative to the muon. Specifically, t 0 = 2.2000 s. We are also told that Earth observers (measuring the decays of moving muons) find t = 16.000 s. Therefore, 2 2.2000 s 1 0.99050. 16.000 s 2. (a) We find from 1 1 2 / : ( ) 2 2 1 1 1 1 0.14037076. 1.0100000 (b) Similarly, ( ) 2 1 10.000000 0.99498744. (c) In this case, ( ) 2 1 100.00000 0.99995000. (d) The result is ( ) 2 1 1000.0000 0.99999950. 3. We solve the time dilation equation for the time elapsed (as measured by Earth observers): t t 0 2 1 0 9990 ( . ) where t 0 = 120 y. This yields t = 2684 y 3 2.68 10 y. 1491
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4. Due to the time-dilation effect, the time between initial and final ages for the daughter is longer than the four years experienced by her father: t f daughter t i daughter = ( 4.000 y ) where is Lorentz factor (Eq. 37-8). Letting T denote the age of the father, then the conditions of the problem require T i = t i daughter + 20.00 y and T f = t f daughter – 20.00 y . Since T f T i = 4.000 y, then these three equations combine to give a single condition from which can be determined (and consequently v): 44 = 4 = 11 = =0.9959. 5. In the laboratory, it travels a distance d = 0.00105 m = vt , where v = 0.992 c and t is the time measured on the laboratory clocks. We can use Eq. 37-7 to relate t to the proper lifetime of the particle t 0 : ( ) 2 2 0 0 2 1 1 0.992 0.992 1 / t v d t t t c c v c � � � � � � which yields t 0 = 4.46 10 –13 s = 0.446 ps. 6. From the value of t in the graph when = 0, we infer than t o in Eq. 37-9 is 8.0 s. Thus, that equation (which describes the curve in Fig. 37-23) becomes t = = If we set = 0.98 in this expression, we obtain approximately 40 s for t . 7. (a) The round-trip (discounting the time needed to “turn around”) should be one year according to the clock you are carrying (this is your proper time interval t 0 ) and 1000 years according to the clocks on Earth which measure t . We solve Eq. 37-7 for : 2 2 0 1y 1 1 0.99999950. 1000y t t (b) The equations do not show a dependence on acceleration (or on the direction of the velocity vector), which suggests that a circular journey (with its constant magnitude centripetal acceleration) would give the same result (if the speed is the same) as the one described in the problem. A more careful argument can be given to support this, but it 212 CHAPTER 37
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should be admitted that this is a fairly subtle question which has occasionally precipitated debates among professional physicists.
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