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I Relativistic doppler shift and radar

  1. Mar 19, 2017 #1
    Consider at stationnary radar at the origin ##z=0## and a target (speed ##v##) moving along the ##z## axis and away from 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 the exact expression 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.

    Here is my attempt to the problem
    (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.

    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?).
     
  2. jcsd
  3. Mar 19, 2017 #2

    PAllen

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    No, this is not correct. Hints:

    1) If you do this all in the emitter receiver frame, you need not use SR Doppler at all, just wave crest propagation to and from a moving mirror.

    2) If you do it in the mirror frame, you need to worry about SR Doppler twice, for receding emitter to mirror; and for reflected signal per mirror receding from original source.

    You should be able to work it out two ways from these hints, getting the same answer for both.


    Note, I am deliberately not giving you the extremely simple answer, so you get the satisfaction of deriving it yourself.
     
    Last edited: Mar 19, 2017
  4. Mar 20, 2017 #3

    PAllen

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    Since the OP hasn't come back, I will post the answer for the record, in case someone visits this thread. The received frequency is simply (c - v)/(c + v) times the initial frequency.
     
  5. Mar 20, 2017 #4

    PeterDonis

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    Shouldn't there be a square root there?
     
  6. Mar 20, 2017 #5

    George Jones

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    The Doppler shift has to be applied twice. First A is the transmitter and B is the receiver; then B is the transmitter and A is the receiver.
     
  7. Mar 20, 2017 #6

    PeterDonis

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    Ah, got it.
     
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