A and B leave from a common point and travel in opposite directions with(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Longitudinal Doppler Effect

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