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The Doppler Effect is typically formulated as follows: $$f' = \dfrac{v \pm v_o}{v \mp v_s} \cdot f$$

The reason for the frequency increasing for observer moving towards source seems clear enough. It is because the observer is traveling against the wave motion and so experiences more cycles per second.

On the other hand, if the source is moving, the scenario is very different.

It seems like it might be due to the speed of sound being a constant in the medium (let's say air). The source creates sound by pressurizing and depressurizing the air over and over again. Each pressurization follows the next with a frequency equal to $f$. Let's assume a spherical sound wave. In the forward direction (the source moving toward the observer), the direction of the propagation of these pressurizations is the same as that of the source's motion. In the backward direction (the source moving away from the observer), the direction of the propagation of these pressurizations is the opposite of that of the source's motion. So the distance (wavelength) between pressurizations is actually greater or less from the standpoint of the observer by a factor of $\frac{v}{v \pm v_s}$.

Is that explanation correct?

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    $\begingroup$ that is exactly how it works. $\endgroup$
    – paulina
    Commented Aug 19 at 14:07
  • $\begingroup$ You lost me on the spherical waves when an angle never appeared. If you're being pedantic, why not make them plane waves from a big flat diaphragm? $\endgroup$
    – JEB
    Commented Aug 19 at 15:35

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Yes, indeed, the physics of Doppler effect for electromagnetic waves and material waves is somewhat different. The reason is indeed that the speed of sound in respect to the reference frame of the medium (i.e., the air) is nearly constant, and a moving observer would perceive a wave as having higher or lower speed, because the actual oscillating particles are moving with a higher/lower speed in respect to the observer.

The other way to put it is that the reference frame of the medium and that of the observer are not equivalent. The same is true for the source moving in respect to the medium.

Finally, the last sentence in the Q seems to imply that the source is emitting plane waves. While we do often consider plane waves in theoretical discussions, it is very rare for a realistic finite source to emit such waves. A spherical or a cylindrical wave is a better approximation, and then the sound emitted forward in respect to the direction of motion propagates at higher frequency than the sound emitted backward. A train passing by is the classical example, although somewhat outdated. I suppose the same effect could be observed from an ambulance or a police car speeding by, but it might be somewhat obscured by the low-frequency cycle of the siren.

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  • $\begingroup$ Thanks! That's helpful. I was considering the waves as planar, just for the sake of simplification, but the question has been updated to reflect your comment in the last paragraph. $\endgroup$
    – DanBM
    Commented Aug 19 at 14:33
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    $\begingroup$ Any asymmetry in the formula between Tx and Rx (i.e., it is not invariant under source/receiver interchange) means there's an absolute rest frame in the problem. $\endgroup$
    – JEB
    Commented Aug 19 at 15:37
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    $\begingroup$ @JEB Good point. In this sense, the non-relativistic discussions of the Doppler effect on the basis of Galilean invariance are actually misleading. But the true and "Galilean" expressions for frequency shift agree to the first order, whereas the typical shifts are small. $\endgroup$
    – Roger V.
    Commented Aug 19 at 15:49
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    $\begingroup$ @RogerV. I'm not familiar with a Galilean invariance derivation. I'll check it out. My favorite SR derivation is derived from requiring the photon emission/detection's momentum transfer leaves the source/receiver on mass shell. Sort of the inverse of Einstein's photon-in-box argument. maybe. $\endgroup$
    – JEB
    Commented Aug 19 at 16:06

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