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The Doppler effect formula is $$f = \frac{(v\pm v_r)}{(v\mp v_s)}f_0$$ where $f$ and $f_0$ are the observed and emitted frequency, respectively, and $v, v_r$ and $v_s$ the speed of the waves, receiver and source, respectively (all relative to the medium.) The numerator has $+$ if the receiver moves towards the source, and the denominator has $-$ if the source moves towards the receiver.

Now, assume the receiver and the source are approaching each other. My intuition tells me that there should be no physical difference whether it is the source which is moving towards the receiver, or vice versa. Actually, the only difference between these cases shall be the coordinate system chosen. But the equation above suggests otherwise because one velocity is in the denominator and the other one in the numerator.

For example, take $v=2$ and receiver and source approaching each other at a speed $u = 1$ (disregarding units.) This, I may say, happens because either $v_s = 0$ and $v_r = 1$ (towards the source), or $v_s = 1$ (towards the receiver) and $v_r = 0$. In the first case, we get $$ f = \frac{2+1}{2} f_0 = 1.5 f_0 $$

In the second case: $$ f' = \frac{2}{2-1} f_0 = 2 f_0 $$

Where does this asymmetry come from? Or why is my intuition wrong?

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    $\begingroup$ why do you expect there to be symmetry in the first place? $\endgroup$
    – JEB
    Jun 13, 2020 at 2:31
  • $\begingroup$ @JEB Doppler effect in electromagnetic waves is pretty symmetric. $\endgroup$
    – fraxinus
    Jun 13, 2020 at 14:47
  • $\begingroup$ @fraxinus Because the Doppler effect for light is a fundamentally different phenomenon from the Doppler effect for mechanical waves. Light is not a wave in a medium. $\endgroup$ Jun 14, 2020 at 14:33

4 Answers 4

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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.

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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.

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  • $\begingroup$ Implying the wrong formula in the question? $\endgroup$ Jun 14, 2020 at 19:30
  • $\begingroup$ No, the equation in the question is perfectly valid for sound waves when the source and receiver are traveling at non-relativistic speeds relative to air. The relativistic version in my answer applies to electromagnetic waves, for which there is no special propagation medium. $\endgroup$
    – Puk
    Jun 14, 2020 at 19:53
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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.

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This asymmetry is best understood analyzing the case when the source moves towards the receiver approximately at wave propagation speed $c$. Then the Doppler ratio will be $$ \frac ff_0 = \frac {c}{c-c} = \infty $$

It means that receiver will not register any waves until the source will arrive fully at the receiver, and then the receiver will experience an immediate huge shock wave with infinite frequency. It looks like:

Enter image description here

I thought very long about the best analogy to this situation. I think this would be best understood by a tsunami, which is caused, say, by an under-water earthquake. The earthquake pressure wave (source) travels towards the coast at the sea waves' speed, and thus each wavefront adds to another, raising the forward wave amplitude, until it crashes the coast with enormous shock wave power.

Now imagine you want to get the same tsunami collision energy only by moving yourself (receiver) in the ordinal sea waves. How fast should you be going? Probably your speed should be very high, so that you could be hitting many ordinal waves per second. Returning back to mathematics, for getting the same infinite Doppler ratio, but just in case of the receiver moving towards the source, the equation must be:

$$ \frac ff_0 = \frac {c+\infty}{c} = \infty $$

In other words, you need to move at infinite speed.

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    $\begingroup$ The Tsunami analogy probably adds more confusion that it removes, since the velocity of surface waves on water depends on the water depth, and for large amplitude waves the velocity also depends on the amplitude. Those two facts make it completely different from the usual Doppler effect for sound transmission through the air. $\endgroup$
    – alephzero
    Jun 14, 2020 at 0:08
  • $\begingroup$ For average people it's far more easy to imagine collecting sea waves energy than collecting sound wave energy, thus Tsunami analogy is more approachable for understanding mentioned asymmetry. Your mentioned differences can be ignored, cause in the OP question about Doppler effect asymmetry, it doesn't adds anything of value. $\endgroup$ Jun 14, 2020 at 16:49

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