# How abruptly does the Doppler shift change sign?

(For context, I originally thought of this question in the context of electromagnetic Doppler shift, but I'm also curious if the same logic applies for acoustic Doppler shift.)

Assume you are watching an object approaching you at relativistic speeds, for example fast enough that the measured frequency of its emissions is shifted by $$10\%$$. The object is not on a collision course, but the point of closest approach is a reasonably short, i.e. non-relativistic distance away. If the object emits a continous wave radio signal, over what timescale does the measured frequency of that signal change as it passes the point of closest approach?

I cannot intuitively accept that it changes $$20\%$$ of a potentially very large number (e.g. $$1\text{ GHz}$$) instantly, because classical mechanics really dislikes discontinuities. But the transition between moving towards and moving away is in some sense instantaneous, given the boundary between the two is infinitesimal. What actually happens then?

The instantaneous change occurs when you consider the Doppler shift in only one dimension. In three dimensions you can consider the correction when the velocity vector and the separation vector are not parallel. Usually such corrections go like $$\cos\theta$$, where $$\theta$$ is the angle between the two vectors, but more complicated things are possible.

Years ago I sat down and computed the speeds for which acoustic Doppler shifts correspond to musical intervals. That gave me the superpower of being able to stand on a sidewalk, listen to the WEEE-ooom as a car drove past, and say to myself “a major third? They're speeding!” But because of the $$\cos\theta$$ dependence, the trick gets harder as you get further from the road.

• Would you be able to elaborate on the actual form of that correction? Is it just the component of velocity parallel to the separation? Oct 10 '20 at 22:06
• A superficial reading of Wikipedia says yes, $\Delta f(\theta) = (\Delta f)_\text{forward} \cos\theta$. But sometimes there is second-order funny business if you drill into the details; I don’t know whether any such exceptions would apply to your question.
– rob
Oct 10 '20 at 22:13
• IIRC a major third is 5:4. So $(v_0+v)/(v_0-v)=5/4$ which comes out to $v=v_0/9$. One ninth of the speed of sound. Hmm. That's speeding in most places but not quite everywhere :-). Oct 12 '20 at 5:57
• I made exactly the same calculation once. Also, it might give the OP a feeling for how quickly the sign can change. For sound, the most dramatic change that you are likely to experience is standing on a platform in a railway station as an express comes through without stopping. Standing close to the edge which enhance the effect but also the danger; don't try this at home. Oct 12 '20 at 9:24
• @JyrkiLahtonen Thanks for the sample calculation, whose result I didn't actually remember. The minor third and major second correspond to common US highway speed postings, but not for highways with sidewalks, which is probably why I allowed this particular superpower to atrophy.
– rob
Oct 12 '20 at 14:16

The object cannot occupy your same place as he passes you, so let us assume that the trajectory is a straight line that passes next to you. As the object approaches, the component of the velocity in your direction diminishes, to the point of being zero when the object is next to you. Thus the doppler effect will change continuously, from blue to zero to red shifted.

You can pretty well observe acoustic Doppler shift just by listening to moving objects emitting some more or less constant-frequency (i.e. tone) sound.

Motor cars (or even better, motorbikes) are pretty good for Doppler observations.

People are pretty much used to approach-passby-goaway sound pattern these objects make.

If you listen from a sidewalk, you hear a quick change to lower tone as a car passes by. The more distance from the road, the slower the change.

There is really no gross difference in EM Doppler effect.

Suppose you adopt a coordinate system such that you are at the origin, and the position of the object can be given by $$(x,y) = (vt,d_0)$$. Note that with the preceding, the object makes its closest approach of $$d_0$$ at $$t=0$$.The distance to the object is given by $$d=(v^2t^2+d_0^2)^{\frac 1 2}$$. This gives $$d' = \frac {v^2t}{(v^2t^2+d_0^2)^{\frac 1 2}}$$. It is this quantity, $$d'$$, that governs the doppler shift, not $$v$$. That is, the formula for the doppler shift is $$f = \frac {c}{c+d'}f_0$$, where $$f$$ is the observed frequency, $$c$$ is the velocity of the wave, and $$f_0$$ is the original frequency. Substituting in, we get $$f = \frac {c(v^2t^2+d_0^2)^{\frac 1 2}}{c(v^2t^2+d_0^2)^{\frac 1 2}+v^2t}f_0$$. At $$t=0$$, we get $$f = f_0$$; there is no doppler effect.

We can also take the reciprocal of both sides to get $$\lambda = \frac {c(v^2t^2+d_0^2)^{\frac 1 2}+v^2t}{c(v^2t^2+d_0^2)^{\frac 1 2}}\lambda_0$$, which can be rewritten as $$\lambda = \lambda_0 + \frac {v^2t}{c(v^2t^2+d_0^2)^{\frac 1 2}}\lambda_0$$ or $$\lambda = \lambda_0 + \frac {v^2t}{cd}\lambda_0$$. This can also be rewritten as $$\lambda = \lambda_0+\frac x d \frac v c \lambda_0$$

So if $$O$$ is your position and $$A$$ is the closest the object gets to you, and $$B$$ is the object's current position, you can take the distance $$AB$$, divide by the distance $$OB$$, multiply by the ratio between the object's distance and the wave velocity, and that will be the relative amount the wavelength is altered. This factor is changing the fastest at $$t=0$$. When $$t=0$$, $$\lambda' = \frac {v^2}{dc}\lambda_0$$

$$\frac x d$$ can also be expressed as the cosine of the angle the line $$OB$$ makes with the line $$OA$$.

The instantaneous frequency shift is proportional to the radial component of the velocity at the time the sound was emitted, and with opposite sign (moving inwards there is a positive frequency shift, which becomes negative when moving outwards). For a uniform linear motion, the scalar product of the relative position vector (from the receiver to the emitter) with the velocity vector is proportional to position along the trajectory, taking as origin the point of closest approach (if the receiver is at the origin and the emitter at $$(h,vt)$$ at instant $$t$$, for fixed $$h$$ and $$v$$, this scalar product is $$v^2t$$, so linear in$$~t$$). But to get the radial component of the velocity, one needs the scalar product with the normalised position vector (in the opposite direction to get the right sign), so one must divide by the distance between emitter and receiver, so one gets a dependence on $$t$$ proportional to $$-t/\sqrt{C^2+t^2}$$ for some constant $$C$$ (of dimension time). In the example one gets $$\frac{-v^2t}{\sqrt{h^2+v^2t^2}}=\frac{-vt}{\sqrt{(h/v)^2+t^2}},$$ so the constant is $$C=h/v$$, the time required for the emitter to traverse the same distance as the closest distance to the receiver. Just for a check, for large absolute value of $$t$$ the denominator tends to $$|t|$$, so the values lie between $$+v$$ and $$-v$$ (and there is a proportionality factor the maps this to the $$\pm$$ the actual maximal frequency shift). A caveat is in place though, this is computed as function of emission time, but the frequency is is heard at reception time; there is a monotonic but non-uniform mapping between those quantities. I'm not going to work out a formula, but during the approach phase time differences are compressed (which is indeed exactly the Doppler effect) so the decrease from the up-shifted frequency to the true frequency proceeds faster than the ensuing decrease from the true frequency to the down-shifted frequency. The constant $$C$$ gives the time scale at which most of the frequency shift takes place.

Let us consider a one dimensional simplification of this. There is a wave source moving to the right, emitting wavefronts to the left and to the right. When the source is to your left, you receive the (blueshifted) wavefronts that are moving to the right. Once the source has passed you and is now to your right, you receive the (redshifted) wavefronts that are moving to the left. This is how the Doppler shift changes abruptly when the source passes you at close distance.