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Unformatted text preview: 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. We solve the time dilation equation for the time elapsed (as measured by Earth observers): t t 2 1 0 9990 ( . ) where t = 120 y. This yields t = 2684 y 3 2.68 10 y. ﾻ ﾻ 1491 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 Lorentz factor (Eq. 378). 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. 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. Thus, that equation (which describes the curve in Fig. 3723) becomes t = = If we set = 0.98 in this expression, we obtain approximately 40 s for t . 7. (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 (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 should be admitted that this is a fairly subtle question which has occasionally precipitated debates among professional physicists....
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This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.
 Fall '08
 Many
 Physics

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