# Why isn't the Doppler effect (for sound waves) symmetric with respect to source and receiver?

The asymmetry comes from the medium. In the first case the source is at rest with respect to the medium and in the second case the receiver is at rest with respect to the medium.

These two cases are not physically equivalent. In the case of the source at rest with respect to the medium the wavelength of the wave is isotropic, but not when it is moving with respect to the medium.

As Dale has said, the asymmetry is due to the medium in which waves propagate. The propagation is tied to the medium (the propagation speed is only $$v_s$$ with respect to this medium), so it is not simply the relative velocities of the source and observer that matters.

In the relativistic version of the Doppler effect, however, the situation is entirely symmetric, because the speed of light is the same in all inertial frames. For motion in 1D, the observer measures a frequency of $$f = \sqrt{\frac{1+v/c}{1-v/c}}f_0$$ where $$v$$ is the velocity with which the source is approaching the observer and vice versa, and $$c$$ is the speed of light.

When thinking about the Doppler shift, I think it's important to decouple the wave from its source, that is: the source is not a property of the wave itself.

So, in the relativistic case, there isn't any rest frame. All motion is relative, so that velocity symmetry is mandatory. A photon on its own does not have a rest frame, nor does it have an intrinsic frequency/wavelength. The Doppler shift formula is thus a relation between the photon as seen in two frames, and that relation can only depend on the relative velocity.

For sound, the medium defines a preferred rest frame. A phonon (or sound wave) has a well defined frequency/wavelength in the absence of a source or an observer. Since the wave with well defined peaks and troughs exists moving through the medium (at $$v$$) with frequency $$f$$, and you move through that medium at $$v_r$$, you're going to intercept more or fewer peaks per unit time according to:

$$f_r = f( v\pm v_r)$$

The reception is completely decoupled from the emission. If it's a (collimated) A440 in air ($$f_s=440\,$$Hz), that wave-train is no different from an octave lower A ($$f_s=220\,$$Hz)being emitted from a platform receding at $$v/2$$. Per the other answers explanations, the wave in the medium has:

$$f = f_s/(v\pm v_s)$$

If you chain these two relations together, you get:

$$f_r = f\frac{ v\pm v_r}{v\mp v_s}$$

There is no velocity symmetry, and none should be expected because the medium defines an absolute rest frame.