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Unformatted text preview: 1451 Chapter 37 1. From the time dilation equation ∆ t = γ ∆ t (where ∆ t is the proper time interval, γ β = 1 1 2 / , and β = v / c ), we obtain β = 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 μ s. We are also told that Earth observers (measuring the decays of moving muons) find ∆ t = 16.000 μ s. Therefore, 2 2.2000 s 1 0.99050. 16.000 s μ β μ = = 2. (a) We find β from γ β = 1 1 2 / : ( 29 2 2 1 1 1 1 0.14037076. 1.0100000 β γ = = = (b) Similarly, ( 29 2 1 10.000000 0.99498744. β = = (c) In this case, ( 29 2 1 100.00000 0.99995000. β = = (d) The result is ( 29 2 1 1000.0000 0.99999950. β = = 3. 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 : ( 29 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. 1452 CHAPTER 37 4. From the value of ∆ t in the graph when β = 0, we infer than ∆ t o in Eq. 379 is 8.0 s. Thus, that equation (which describes the curve in Fig. 3723) becomes 2 2 8.0 s 1 ( / ) 1 t t v c β ∆ ∆ = = . If we set β = 0.98 in this expression, we obtain approximately 40 s for ∆ t . 5. 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. ≈ × 6. 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 = γ (4.000 y) where γ 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 , 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 γ can be determined (and consequently v ): 44 = 4 γ ⇒ γ = 11 ⇒ β = 2 30 11 = 0.9959. 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 β : 2 2 1y 1 1 0.99999950. 1000y t t β ∆ = = = ∆ (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 1453 should be admitted that this is a fairly subtle question which has occasionally precipitated...
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics

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