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Unformatted text preview: 1417 Chapter 37 1. From the time dilation equation t = t (where t is the proper time interval, 1 1 2 / , and = v / c ), we obtain F H G I K J 1 2 t t . The proper time interval is measured by a clock at rest relative to the muon. Specifically, t = 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. (a) The roundtrip (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 ) and 1000 years according to the clocks on Earth, which measure t . We solve Eq. 377 for : 2 2 1y 1 1 0.99999950. 1000y t t 1418 CHAPTER 37 (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 should be admitted that this is a fairly subtle question that has occasionally precipitated debates among professional physicists. 4. Due to the timedilation 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 the Lorentz factor (Eq. 378). Letting T denote the age of the father, then the conditions of the problem require T i = t i daughter + 25.00 y , T f = t f daughter – 25.00 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 ): 54 = 4 = 13.5 = 0.9973. 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. 377 to relate t to the proper lifetime of the particle t : 2 2 2 1 1 0.992 0.992 1 / t v d t t t c c v c which yields t = 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. 379 is 8.0 s....
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This document was uploaded on 04/22/2011.
 Spring '11

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