Model let the moving clock be in frame s and an

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Model: Let the moving clock be in frame S´ and an identical at-rest clock be in frame S. Solve: The ticks being measured are those of the moving clock. The interval between 2 ticks is measured by the same clock in S´—namely, the clock that is ticking—so this is the proper time: . t τ Δ = Δ The rest clock measures a longer interval t between two ticks of the moving clock. These are related by 2 1 t τ β Δ = Δ The moving clock ticks at half the rate of the rest clock when τ = 1 2 t . Thus 2 2 2 2 1 1 / 1/2 1 (1/2) 0.866 v c v c c β = = = =
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37.18. Solve: (a) The starting event is the astronaut leaving earth. The finishing event is the astronaut arriving at the star system. The time between these events as measured on earth is 4.5 ly 4.5 ly 5.0 y 0.9 0.9 ly/y t c Δ = = = (b) For the astronaut, the two events occur at the same position and can be measured with just one clock. Thus, the time interval in the astronaut’s frame is the proper time interval. 2 2 2 0.9 1 1 5.0 y 0.19 5.0 y 2.2 y v c t c c τ Δ = Δ = = = (c) The total elapsed time is the time for the astronaut to reach the star system plus the time for light to travel from the star system to the earth. The time is Δ t + 4.5 y = 5.0 y + 4.5 y = 9.5 y
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37.19. Model: Let S be the earth’s reference frame and S the rocket’s reference frame. Solve: (a) The astronauts measure proper time i . t τ Δ = Δ Thus ( ) ( ) 2 2 10 y 120 y 0.9965 1 1 t v c v c v c τ Δ Δ = = = (b) In frame S, the distance of the distant star is ( )( ) ( )( ) 0.9965 60 y 0.9965 ly y 60 y 59.8 ly x v t c Δ = Δ = = =
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37.20. Model: The earth’s frame is S and the airliner’s frame is S . S moves relative to S with velocity v . Also, assume zero acceleration/ deceleration times. The first event is when the airliner takes off and almost instantly attains a speed of v = 250 m/s. The second event is when the airliner returns to its original position after 2 days. It is clear that the two events occur at the same position in frame S and can be measured with just one clock. This is however not the case for an observer in frame S. Solve: (a) You have aged less because your proper time is less than the time in the earth frame. (b) In the S frame (earth), 6 4 2 5 10 m 4.0 10 s 250 m/s t × × Δ = = × In the S frame (airliner), ( ) ( ) 1 2 2 2 2 2 2 2 4 8 2 8 1 1 2 1 250 m/s 4.0 10 s 1.4 10 s 14 ns 2 2 3.0 10 m v t v c t c v t t c τ τ Δ = Δ ≈ Δ ⇒ Δ − Δ ≈ Δ = × = × = × You age 14 ns less than your stay-at-home friends.
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37.21. Model: The ground’s frame is S and the moving clock’s frame is S . Visualize: 1.0 d 1.0 ns 86400.000000001 s t Δ = + = and 1.0 d 86,400 s. t Δ = = Solve: We want to solve for v in t t γ Δ = Δ Using the binomial approximation for γ : 2 2 1 1 2 t v t c Δ + Δ 2 2 2 2 1 1 2 1 2 t v t c v t c t Δ Δ = Δ Δ 1.0 d 1.0 ns 1.0 ns 2 1 2 1 2 1.0d 1.0d t v c c c t Δ + = = = Δ 9 8 1.0 10 s (3.0 10 m s) 2 46 m s 86,400 s v × = × = Assess: This speed is about 100 mph, which is certainly doable. The calculation would be difficult without the
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