Radar sensors make use of the Doppler effect to measure the radial velocity of an object. The radar's Tx antennas emit an electromagnetic wave which travels to the moving objects, is reflected and the radar's Rx antennas detect the incident wave. Due to the movement of the measured object, the received wave has a different frequency.

According to wikipedia (https://en.wikipedia.org/wiki/Doppler_effect#Radar), the shift in frequency $\Delta f$ is given by: $$\Delta f = \frac{2 \Delta v_r}{c} f_0,$$ where $\Delta v_r$ is the relative radial velocity between radar sensor and object, $c$ is the speed of light and $f_0$ is the radar's frequency. A derivation of this equation is for example given in this script about radar sensors (page 21 and following): https://www.ei.ruhr-uni-bochum.de/media/ei/lehrmaterialien/39/a715b063167d904ec4a9a5cea2a1a54d4defc115/RuhrUni_Scriptum.pdf

What bugs me know is the following: The derivation is entirely classifcal. No time dilation or relativistic effects enter the picture. My understanding is that for electromagnetic waves (which do not need a medium for propagation), the equation for the relativistic Doppler effect has to be used: $$f_r = \sqrt{ \frac{1-\beta}{1+\beta} } f_s$$ with $\beta = v/c$.

But this equation (with its square root) does not match at all with the mentioned formula for the Doppler radar. For me it appears like that a classical method is used for the derivation of the radar's frequency shift, just like it would have been done for a sonar sensor (which uses sound waves and not EM waves). Why is this correct?

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    $\begingroup$ Please Taylor expand the relativistic formula. $\endgroup$
    – Semoi
    Commented Feb 13, 2021 at 11:33
  • $\begingroup$ So Taylor expansion of $\sqrt{(1-\beta)/(1+\beta)}$ around $\beta=0$ yields $1-\beta$ plus terms of the order of $\beta^2$. Except for a factor of two (which just comes from the two-way effect), this could explain it. So it's just the approximation for small (compared to $c$) velocities? $\endgroup$
    – Merlin1896
    Commented Feb 13, 2021 at 11:37
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    $\begingroup$ This is what we should expect, isn't it? If we are travelling with a small speed the classical formulas should be correct. Why don't you post your findings as an answer? $\endgroup$
    – Semoi
    Commented Feb 13, 2021 at 12:13

1 Answer 1


In the case of radar velocity measurement by doppler shifts, note that the velocity of the object being measured is of order ~10 to 1000 m/s. This is so slow compared to the speed of the radar pulse itself that relativistic corrections do not have to be made in order to get a velocity measurement accurate enough to justify writing someone a speeding ticket or to shoot down a fighter jet.


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