ps1soln

# E when the observer is moving towards it on the way

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Unformatted text preview: use 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= 1−v f0 1+v (4) and hence receives ∆τB T 1−v √ = f0 1 − v 2 (5) 2 2 1+v On the way back, since B is moving towards A, the doppler shift formula ﬂips and f f 1+v √ f0 1 − v 2 1−v ∆τB T = 2 2 (6) Adding those two together, the total number of counts B receives is f0 T , which makes sense; every signal that A emitted during his proper time T is captured by B . 1 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 vT T + 2 2 1−v f0 1+v (7) vT T − 2 2 1+v f0 1−v (8) and on the way back, he receives meaning that in total A receives √ f0 T 1 − v 2 , (9) which again makes sense: B was broadcasting at...
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## This document was uploaded on 02/28/2014.

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