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Longitudinal Doppler Effect

  1. Oct 23, 2013 #1
    A and B leave from a common point and travel in opposite directions with
    relative speed v. When B’s clock shows that a time T has elapsed, he (B)
    sends out a light signal. When A receives the signal, what time does his (A’s)
    clock show? Answer this question by doing the calculation entirely in (a) A’s
    frame, and then (b) B’s frame.

    (Y = gamma)

    a)

    In A's frame, when A's clock reads YT, B's clock reads T. This means B is at a distance YTv from A. When B emits the photon, the photon takes time YTv/c to reach A in A's frame. Thus, the total time for A is YT(1 + v/c).

    b)

    In B's frame, when his clock reads T, A is at a distance Tv away. Then, B emits a photon which travels at a speed of (c-v) relative to A in B's reference frame. Thus, the time taken for the photon to reach A is Tv/(c-v). Thus the total time this takes according to B is T + Tv/(c-v) = T(1 + v/(c-v)). By time dilation, in A the total time is YT(1 + v/(c-v)).

    I get these two different answers. Does anyone know what I'm doing wrong?
     
  2. jcsd
  3. Oct 23, 2013 #2

    stevendaryl

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    Staff Emeritus
    Science Advisor

    Your calculation in B's frame is wrong. You got it correct, that the time, according to B's frame, for the photon to reach A is T_arrive = T(1+v/(c-v)) = T/(1 - v/c). But in B's frame, A's clock is running slower, so the elapsed time on A's clock is T'_arrive = T_arrive/Y = T/(Y (1-v/c)).

    That's the same as your calculation in A's frame, since

    T/(Y (1-v/c)) = YT (1+v/c)
     
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