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ps1soln - Physics 8.962 Problem Set 1 Solution 1 Lets call...

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Physics 8.962, Problem Set 1 Solution 1. Let’s call the twin that stays A and the twin on the spaceship B. We work in the ( x, t ) coordinate system attached to the planet, i.e. A’s coordinate system. A says that the trip took Δ t T . Now we want to find the proper time along B’s trajectory: we have the infinitesimal formula 2 = dt 2 - dx 2 . Along B’s trajectory x B ( t ) we have x B ( t ) = ( vt t < T 2 vT 2 - vt t > T 2 (1) Thus x B ( t ) = ± v , depending on the value of t ; so we find Δ τ B = Z dt s 1 - dx B dt 2 = T 1 - v 2 (2) Of course we could have done this immediately using the time dilation formula. Now, let’s count how many signals B receives during the trip. To do this we need the Doppler shift formula; recall that if a source emits signals at a frequency f 0 in its rest frame and an observer is moving towards it at velocity v , it observes a frequency f 0 f 0 = f 0 r 1 + v 1 - v (3) It is of course impossible to remember which way the ± go, but I know that I have it the right way around because this way it is blueshifted for positive v , i.e. when the observer is moving towards it. On the way out, B is moving away and so sees the signal at a frequency f 0 = r 1 - v 1 + v f 0 (4) and hence receives f 0 Δ τ B 2 = T 2 r 1 - v 1 + v f 0 1 - v 2 (5) On the way back, since B is moving towards A , the doppler shift formula flips and f 0 Δ τ B 2 = T 2 r 1 + v 1 - v f 0 1 - v 2 (6) Adding those two together, the total number of counts B receives is f 0 T , which makes sense; every signal that A emitted during his proper time T is captured by B . 1
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Now, let’s count how many signals A receives. The doppler shifts in this case are the same. However, A sees the receding Doppler shift not only for time T/ 2, but also for the time it takes for light to reach back from the distant planet, vT/ 2. Thus, from B going on the way out, A receives T 2 + vT 2 r 1 - v 1 + v f 0 (7) and on the way back, he receives T 2 - vT 2 r 1 + v 1 - v f 0 (8) meaning that in total A receives f 0 T 1 - v 2 , (9) which again makes sense: B was broadcasting at frequency f 0 for his entire proper time, which we worked out above was T 1 - v 2 . Indeed the number of counts B receives divided by the number of counts A received is γ , which makes sense as “number of counts” is a time telling mechanism. 2. Schutz 1.17 For this problem γ = 1 / 1 - . 8 2 = 5 / 3. (a) The friend on the barn measures the pole to be l pole γ = (20 m )( 3 5 ) = 12 m (10) (b) After the barn door is closed, in the friend’s frame, the front of the pole has 3 m to travel before hitting the front. Thus the time it takes is t = 3 m . 8 = 3 . 75 m (11) The interval between shutting the door and hitting the front is Δ s 2 = - Δ t 2 + Δ x 2 = - (3 . 75 m ) 2 + (15 m ) 2 = 210 . 938 m 2 (12) which is spacelike.
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