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Let a doppler radar be at rest and a car that is moving with speed $v$ away from the radar. The radar is continuously sending a monochromatic wave with frequency $\omega_0$. This wave is reflected back off the car back to the radar. What is the frequency $\omega_r$ of the reflected wave perceived by the radar (without taking into account relativistic effects)? I have two very long calculations which seem both correct. One tells me $\omega_r = \dfrac{c-v}{c+v}\omega_0$ and the other $\omega_r = \dfrac{c-3v}{c-v}\omega_0$. Which one is correct? Are they both wrong?

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  • $\begingroup$ Did you mean to instead say 'perceived by the car'? The correct relation is $\omega_r = \left(\frac{1-\beta}{1+\beta}\right)^{\frac{1}{2}} \omega_0$ where $\beta =\frac{v}{c}$ - see here. $\endgroup$ – diracula Mar 12 '17 at 12:08
  • $\begingroup$ @diracula No. I meant perceived by the radar. The emitted wave is sent how reflects on the car back to the radar. This reflected wave has a different frequency $\omega_r$ than the one that is emitted ($\omega_0$). My question is what is $\omega_r$? $\endgroup$ – Corvinus Mar 12 '17 at 12:12
  • $\begingroup$ If the wave is being reflected off the car you should mention it in the question. Also you've added 'without taking into account relativistic effects', but electromagnetic radiation is relativistic so this would give a (significantly) incorrect answer. $\endgroup$ – diracula Mar 12 '17 at 12:19
  • $\begingroup$ @diracula Yes, you're right. I've edited my question and added that the wave is reflected off the car. As for relativistic effects, I know not taking them into account is going to introduce some error but because $v << c$, I suppose that this effect can be neglected. $\endgroup$ – Corvinus Mar 12 '17 at 12:28
  • $\begingroup$ True, time dilation is a second order effect so you could get the correct first order result by neglecting it. I hadn't considered that, sorry. $\endgroup$ – diracula Mar 12 '17 at 12:40
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The simplest way to handle this problem is to replace the transmitter part of the radar gun with its virtual image created by its reflection off the back of the car.

If the car is moving away from the radar gun with a velocity $v$, then the reflected radar waves are perceived as coming from a virtual image of the transmitter that is travelling away from the receiver at a velocity of $2v$.

Just put this velocity into the standard Doppler shift equation for EM waves. Be very careful to follow any sign conventions for the velocity of the source...

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  • $\begingroup$ Very interesting! Could you elaborate more on the transmitter having a speed of $2v$? I do not get this part. $\endgroup$ – Corvinus Mar 12 '17 at 13:45
  • $\begingroup$ In one second the distance from the radar gun to the car increases by $v$ metres; the distance from the car (as mirror) to the image of the radar gun also increases by $v$ metres. The total increase in $1$ second is $2v$ metres... $\endgroup$ – DJohnM Mar 12 '17 at 15:37

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