Consider at stationnary radar at the origin ##z=0## and a target (speed ##v##) moving along the ##z## axis and(adsbygoogle = window.adsbygoogle || []).push({}); awayfrom the radar. The radar is sending plane waves (frequency ##f_i##) to the target and they come back to the radar (the radar is then both an emitter and a receiver). I am interested in theexactexpression of the frequency ##f_r## of the received wave (the one that bounces back off the target) viewed by the "receiver part" of the radar.

(I did not learn special relativity) : Because the radar is the observer (and not the moving target), I cannot make directly use of the relativistic doppler shift formula. My goal then is to convert my situation into a source/moving-observer one.

Here is my attempt to the problem

Let's call the ##\tau(t)## the time taken for the emitted wave to go hit the target and come back to the radar. ##\dfrac{\tau(t)}{2}## is then the time taken for the wave to hit the target and is given by this equation :

$$c \left(\dfrac{\tau(t)}{2}\right) = d(t) + v\left(\dfrac{\tau(t)}{2}\right)$$

where ##c## is the speed of light, ##d(t) = d_0+vt## the distance radar-target. We then find :

$$\tau(t) = \dfrac{2d(t)}{c-v} = \dfrac{2d(0)}{c-v} + \dfrac{2vt}{c-v} $$

I now imagine a equivalent situation in which there is no more target but another radar moving with speed ##v_r## which is going to act like the receiver (recall that in the first situation, these two radars were considered as a single entity). The time taken by the emitted wave to hit the receiver is : $$\tau(t) = \dfrac{d_r(t)}{c-v_r} = \dfrac{d_r(0)}{c-v_r} + \dfrac{vt}{c-v_r}$$ where ##d_r(t)## is the distance from the stationnary radar (##z=0##) to the receiving radar. To make those two situations equivalent, I have to equate the two expressions for ##\tau(t)## : $$\dfrac{2d(0)}{c-v} + \dfrac{2vt}{c-v} = \dfrac{d_r(0)}{c-v_r} + \dfrac{vt}{c-v_r}$$ I find $$v_r = \dfrac{2cv}{c+v}$$ The frequency observed by the radar is then (by the relativistic doppler formula): $$f_r = f_i \sqrt{\dfrac{1-v_r/c}{1+v_r/c}} = f_i\sqrt{\dfrac{(1 - \frac{2 v}{c + v})}{(1+\frac{2 v}{c + v})}} = f_i\sqrt{\dfrac{c - v}{c + 3 v}}$$ Is my reasoning correct? Did I forget to make use of special relativity when I calculated ##\tau(t)##? Does the radar hear the same frequency in both situations? (in other words, is it sufficient to equate the two expressions for ##\tau(t)## to make the two situations equivalent from the receiver standpoint?).

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# I Relativistic doppler shift and radar

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