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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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