A source emitting electromagnetic signals with proper frequency `omega_(0) = 3.0.10^(10)s^(-1)` moves at a constant velocity `v = 0.80c` along a stright line separated from a stationary observer `P` by a distance `l` (Fig.) Find the frequency of the singnals perceived by the observer at the moment when
(a) the source is at the point `O`,
(b) the observer sees it at the point `O`.

Explanation

(a) If light is received by the observer at `P` at the moment when the source is at `O`, it must have been emitted by the source when it was at `O'` and travelled along `O'P`. Then if `O'P = ct, O'O = vt`
and `cos theta = (v)/(c ) = beta`
In the frame of the observer, the frequency of the light is `omega` while its wave vector is
`(omega)/(c )(cos theta, sin theta,0)`
we can calculate teh value of `omega` by relating it to proper frequency `omega`. The relation is
`omega_(0) = (omega)/(sqrt(1-beta^(2))) (1-beta cos theta)`
(To derive the formula is this from it is easiest to note that `(omega)/(sqrt(1-v//c)^(2)) -(overset rarr(k).oversetrarr(v))/(sqrt(1-v^(2)//c^(2)))` is an invarant which takes the value `omega_(0)` in the rest frame of the source).
Thus `omega = (omega_(0)sqrt(1-beta^(2)))/(1-beta^(2)) = (omega_(0))/(sqrt(1-beta^(2))) = 5 xx 10^(10) sec^(-1)`
(b) For the light to be received at the instant observer see the source at `O`, light must be emitted when the observer is at `O` at `90^(@) = theta`
Then as before `omega_(0) = (omega_(0))/(sqrt(1-beta^(2)))` or `omega = omega_(0) sqrt(1-beta^(2)) = 1.8 xx 10^(10) sec^(-1)`
In this case the observer will receive light along `OP` and he will ''see'' that the source is at `O` even through the source will have moved ahead at the instant the light is received.