# ch37 - 1 From the time dilation equation t = t0(where t0 is...

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1. From the time dilation equation t = γ t 0 (where t 0 is the proper time interval, γ β = 1 1 2 / , and β = v / c ), we obtain β = 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 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 § · = = ¨ ¸ © ¹

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2. (a) We find β from γ β = 1 1 2 / : ( ) 2 2 1 1 1 1 0.14037076. 1.0100000 β γ = = = (b) Similarly, ( ) 2 1 10.000000 0.99498744. β = = (c) In this case, ( ) 2 1 100.00000 0.99995000. β = = (d) The result is ( ) 2 1 1000.0000 0.99999950. β = =
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. ×

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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 = γ ( 4.000 y ) where γ 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 γ can be determined (and consequently v): 44 = γ 4 ¡ γ = 11 ¡ β = 2 30 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 § · = ¡ = = ¨ ¸ © ¹ which yields t 0 = 4.46 × 10 –13 s = 0.446 ps.

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6. From the value of t in the graph when β = 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 = t o 1 - ( v / c ) 2 = 8.0 s 1 − β 2 If we set β = 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 β : 2 2 0 1y 1 1 0.99999950. 1000y t t β § · § · = = = ¨ ¸ ¨ ¸ © ¹ © ¹ (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 which has occasionally precipitated debates among professional physicists.

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8. The contracted length of the tube would be L L = = = 0 2 2 1 300 1 0 999987 0 0153 β . . . m m. b g
9. (a) The rest length L 0 = 130 m of the spaceship and its length L as measured by the timing station are related by Eq. 37-13. Therefore, L = = 130 1 0 740 87 4 2 m m.

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• Fall '07
• PERELSTEIN,M

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