Sound and the Doppler Effect

Beginner

Watch a duck paddle across a still pond. It makes new ripples all the time, but each one spreads from the spot where the duck was when it made it. Ahead of the duck the ripples crowd together; behind it they lie far apart. Sound from a moving siren does exactly the same in air, and that is what the blue rings in the animation show.

Your ear hears pitch as how often crests arrive. The listener in front gets crests close together, arriving often, so the siren sounds higher. The listener behind gets them spread out and arriving rarely, so it sounds lower. The siren itself never changes its note: the driver hears it normally all the time. This change of pitch caused by motion is the Doppler effect.

It works the other way round too. Run towards a still bell and you meet its crests sooner than if you stood still, so the note sounds slightly higher; run away and it drops.

Two mistakes are common. The pitch does not keep rising as the ambulance comes nearer: it stays high the whole time it approaches, stays low the whole time it leaves, and drops as it passes. Coming nearer only makes it louder, which is a different effect.

What if the source is faster than its own waves? Then it leaves all its crests behind, and they pile up along a V, like the wake of a speedboat. For sound in air this pile-up is a sonic boom, heard all the time a plane flies faster than sound, not only at the moment it reaches that speed.

Check yourself: a car sounds its horn as it drives away from you. Is the note you hear higher or lower than the one the driver hears? Answer: lower.

Intermediate

What we measure, and from where

A siren rushing past drops in pitch, yet the siren itself never changes its note. Sound in air is a travelling pattern of compressions and rarefactions. A source that vibrates $f$ times per second, its frequency in hertz (Hz), sends out one crest every period $T=1/f$. The crests travel at the speed of sound $v$, set by the air, not by the source: about 343 m/s at 20 °C. Neighbouring crests are one wavelength $\lambda =v/f$ apart, 0.85 m for a 400 Hz note in air at 340 m/s.

A listener counts something else: the crests that reach the ear each second. This heard frequency $f^{\prime }$ is what the ear perceives as pitch, and when nothing moves, $f^{\prime }=f$. The Doppler effect is the difference caused by motion. Every speed is measured relative to the air: $v_s$ for the source and $v_o$ for the observer, the listener, both moving along the line that joins them.

A moving source changes the wavelength

Once a crest has left the source, it spreads at $v$ from the point of emission, whatever the source does next. Let the source move towards a listener at $v_s$. During one period the last crest runs $vT$ ahead while the source follows it by $v_{s}T$, so in front of the source the crests are packed closer:
${\lambda }^{\prime }=\frac{v-v_s}{f}\text{(in front of the source)}$

Behind the source the spacing is $(v+v_s)/f$. The crests still travel at $v$, so a still listener in front receives $v/{\lambda }^{\prime }=fv/(v-v_s)$ crests per second, more than $f$, and a listener behind receives $fv/(v+v_s)$, fewer.

A moving listener changes the pitch in another way. A source at rest keeps the ordinary spacing $\lambda =v/f$, and a listener walking towards it at $v_o$ meets its crests at the relative speed $v+v_o$, so $f^{\prime }=f(v+v_o)/v$. Source motion changes the wavelength in the air; listener motion changes only how often the crests are met.

The Doppler formulas

For motion along the line that joins source and listener, with every speed measured relative to the air, the three basic cases are these:

• $f^{\prime }=\frac{fv}{v-v_s}\text{(source approaching)}$
• $f^{\prime }=\frac{fv}{v+v_s}\text{(source receding)}$
• $f^{\prime }=\frac{f(v+v_o)}{v}\text{(observer approaching)}$

When both move, the effects multiply: $f^{\prime }=f(v+v_o)/(v-v_s)$ for approach, and a speed directed away changes its sign. Check every answer: approach raises the frequency, retreat lowers it, and if nothing moves relative to the air, $f^{\prime }=f$. A car driving just ahead of an ambulance at its speed $u$ hears $f(v-u)/(v-u)=f$, like the ambulance driver.

Worked example: the siren in the animation

The animation exaggerates the motion so that the bunching is easy to see. The siren emits $f=400\mathrm{Hz}$ and moves at $v_s=170m/s$, half of $v=340m/s$. At rest its crests would be $\lambda =340/400=0.85\mathrm{m}$ apart. In front of it and behind it,
${\lambda }^{\prime }=\frac{340-170}{400}=0.425\mathrm{m},f^{\prime }=\frac{340}{0.425}=800\mathrm{Hz}$
$\frac{v+v_s}{f}=\frac{340+170}{400}=1.275\mathrm{m},f^{\prime }=\frac{340}{1.275}\approx 267\mathrm{Hz}$

These are the numbers on the screen. The purple readout gives 800 Hz for the purple listener ahead, the green readout 267 Hz for the green listener behind, and the labels in the bottom corners show the wavelengths, rounded to 0.43 m and 1.28 m. The blue rings are half the still-air gap apart in front and one and a half times it behind, and the orange-red $v_s$ arrow shows the siren's velocity.

Each side checks itself: $f^{\prime }{\lambda }^{\prime }=800\cdot 0.425=340m/s$; the crests keep their speed and only their spacing changes. A listener whom this siren passes hears its note fall by the factor $(v+v_s)/(v-v_s)=3$, an octave and a fifth.

Worked example: an ambulance passing by

An ambulance with a 700 Hz siren drives at 25 m/s (90 km/h) on a day when $v=343m/s$. Standing close to the road, you hear
$f^{\prime }=\frac{700\cdot 343}{343-25}\approx 755\mathrm{Hz},f^{\prime }=\frac{700\cdot 343}{343+25}\approx 652\mathrm{Hz}$
before and after it passes: a drop of about 103 Hz, a ratio of $368/318\approx 1.16$, more than a whole tone (1.12).

Now drive towards the ambulance yourself at 20 m/s. Before you meet, both of you approach; afterwards both recede:
$f^{\prime }=\frac{700\cdot (343+20)}{343-25}\approx 799\mathrm{Hz},f^{\prime }=\frac{700\cdot (343-20)}{343+25}\approx 614\mathrm{Hz}$
The drop grows to about 185 Hz, because both motions now contribute.

Finally, swap the roles. Driving at 25 m/s towards a still 700 Hz alarm, you hear $700\cdot 368/343\approx 751\mathrm{Hz}$, about 4 Hz less than the moving ambulance gave: small at road speeds, but real, because only the source's motion changes the wavelength in the air.

Reading the wavefront picture

Each blue ring in the animation is one crest, centred on the point where the siren was when it emitted it, not where the siren is now. The centres lie in a row along the road, the youngest and smallest ring closest to the siren. The spacing is smallest straight ahead, largest straight behind, and changes smoothly in between.

A listener beside the road hears this pattern sweep past. While the siren is far up the road, the crests reaching her are almost as squeezed as those straight ahead; once it is far down the road, almost as stretched as those behind. In between the pitch slides down, the more suddenly the closer she stands to the road. The rings show pitch, not loudness: the sound also grows louder as the siren approaches, because its power is spread over smaller spheres.

Common mistakes

• The pitch does not rise as the source comes nearer. A siren driving straight at you at a steady 25 m/s stays at 755 Hz; only the loudness grows, by about 6 dB each time the distance halves.
• Source and observer motion are not interchangeable. In air at 340 m/s, a 500 Hz siren approaching at 30 m/s is heard at 548 Hz; a listener approaching a still siren at 30 m/s hears 544 Hz.
• Speeds are measured relative to the air. A 10 m/s wind changes the speed of sound along it to 353 or 333 m/s, but if source and listener are at rest, every crest takes the same time to arrive and the pitch stays at $f$.
• Motion does not change the source: the ambulance driver, moving with the siren, hears its true note, 400 Hz in the animation and 700 Hz in the ambulance.
• Do not add the speed to the frequency: a 500 Hz source approaching at 30 m/s gives 548 Hz, not 530 Hz. Approach must land above $f$ and retreat below it, here at 459 Hz.

Where you meet it

Speed guns measure the Doppler shift of a microwave echo, Doppler ultrasound shows blood flowing in an artery, weather radar maps the winds inside storms, and bats track insects by the pitch of their echoes. Astronomers read the motion of stars from the shifts of their spectral lines, red for receding and blue for approaching, and find planets around other stars from the wobble they cause.

The next lesson, Loudness, Intensity and Decibels, deals with the other change you hear as a siren approaches, its growing loudness. Interference and Superposition explains the beats that instruments use to measure small shifts, and Special Relativity shows how the Doppler effect works for light, which has no air to measure speeds against.

Advanced

Counting crests: a moving source

A siren of frequency $f$ drives at speed $v_s$ through still air, in which sound travels at speed $v$, and a listener $C$ stands still. Why does $C$ hear a pitch $f^{\prime }$ different from $f$? The cleanest answer comes from counting crests. A pitch is the number of pressure crests that arrive per second, so all we need is the time between the arrivals of two neighbouring crests. Figure 1 sets up that count.

Legend

$C$	Listener at rest in still air.
$S_n$	Where the siren is when it sends crest $n$.
$S_{n+1}$	Where it is when it sends crest $n+1$, one period later.
$T$	Period of the siren, $T=1/f$.	$\mathrm{s}$
$v_{s}T$	Distance the siren moves in one period, along its velocity.	$\mathrm{m}$
$r_n$	Length of the path of crest $n$, from $S_n$ to $C$.	$\mathrm{m}$
$r_{n+1}$	Length of the path of crest $n+1$, from $S_{n+1}$ to $C$.	$\mathrm{m}$
$\theta$	Angle between the siren's velocity and the line to $C$, at the moment of emission.	$\mathrm{rad}$
$v_{s}T\cos \theta$	Component of the step $v_{s}T$ along the line to $C$: how much shorter the second path is.	$\mathrm{m}$

Think of the siren as a clock that sends out one crest every period $T=1/f$. Crest number $n$ leaves from the point $S_n$ at time $nT$. Crest $n+1$ leaves one period later from $S_{n+1}$, because in that period the siren has moved a distance $v_{s}T$ along the road: the arrow from $S_n$ to $S_{n+1}$ in Figure 1.

Once a crest has left, the siren no longer matters to it: it runs through the air at speed $v$, in a straight line to $C$. Crest $n$ covers the path $r_n$ and crest $n+1$ the path $r_{n+1}$, so they arrive at
$t_n=nT+r_n/v,t_{n+1}=(n+1)T+r_{n+1}/v$
The time between these arrivals is the period the listener hears:
$T^{\prime }=t_{n+1}-t_n=T+\frac{r_{n+1}-r_n}{v}$

So everything hangs on how much the second path differs from the first. Look at the dashed line in Figure 1, dropped from $S_{n+1}$ at right angles onto the first path. The step $v_{s}T$ is tiny compared with the distance to $C$ (one period lasts a few milliseconds), so the two paths are almost parallel. The second path is then shorter than the first by the stretch of the first path between $S_n$ and the foot of that dashed line. That stretch is the component of the step along the line to $C$, namely $v_{s}T\cos \theta$, where $\theta$ is the angle at $S_n$ between the siren's velocity and the line to $C$. Hence
$r_{n+1}-r_n=-v_{s}T\cos \theta$

Putting this into $T^{\prime }$ gives $T^{\prime }=T(1-(v_s/v)\cos \theta )$. Frequency is one over the period, so
$f^{\prime }=\frac{f}{1-(v_s/v)\cos \theta }$

Three directions show what the formula says. Straight ahead, $\theta =0$, it gives $fv/(v-v_s)$: the crests are squeezed together and the pitch rises. This case is exact at any distance, because both paths then lie on the same line. Straight behind, $\theta =180^{\circ }$, it gives $fv/(v+v_s)$: the crests are stretched apart and the pitch falls. Side-on, $\theta =90^{\circ }$, there is no shift, because a step at right angles to the line of sight does not change the distance to $C$ to first order.

Two points are easy to miss. First, only the radial component $v_s\cos \theta$, the part of the velocity along the line to the listener, counts; motion across that line does nothing. Second, $\theta$ is the angle at the moment of emission, not at the moment of hearing. The crest you hear now set off a travel time $r/v$ ago, at the so-called retarded time, when the siren was somewhere else. The roadside example below shows that this delay is easy to hear.

Counting phase: a moving observer

Now swap the roles: the siren stands still and the listener moves. The air then carries the same evenly spaced crests as for a listener at rest; nothing about the wave changes. What changes is how often the moving listener runs into the crests. A space-time picture, Figure 2, makes this count visible.

Legend

$x$	Position along the line from the source to the listener; the source is at $x=0$.	$\mathrm{m}$
$t$	Time.	$\mathrm{s}$
$T$	Time between two crests sent by the source: the period.	$\mathrm{s}$
$T^{\prime }$	Time between two crests met by the walking listener, here $2T/3$.	$\mathrm{s}$
$\lambda$	Wavelength: distance between neighbouring crests, $\lambda =vT$.	$\mathrm{m}$
$v_o$	Speed of the listener walking towards the source, here $v/2$.	$m/s$

In Figure 2 position $x$ runs to the right and time $t$ upwards, and the source sits at $x=0$, on the vertical axis. It sends out a crest every period $T$. Each crest moves to the right at speed $v$, so in the picture it is a straight sloping line, advancing one wavelength $\lambda =vT$ per period. That is why the crest lines are parallel and rise by $T$ for every $\lambda$ they cover.

A listener at rest is a vertical line in this picture, and she crosses one crest line every $T$. The listener of Figure 2 walks towards the source at $v_o=v/2$, so her track leans towards the axis, into the crest lines, and she cuts through them more often. She and the crests close in on each other at $v+v_o$, and the crests are $\lambda$ apart, so the time between crests becomes $T^{\prime }=\lambda /(v+v_o)=T/(1+v_o/v)=2T/3$, marked $T^{\prime }$ in the figure. In the time $4T$ in which a resting listener meets four crests, she meets six.

The same count in symbols uses the phase of the wave. Write the wave as a function of $\omega t-kx$, with angular frequency $\omega =2\pi f$ and wavenumber $k=\omega /v=2\pi /\lambda$. The phase grows by $2\pi$ from one crest to the next, so counting radians of phase is counting crests. A listener who starts at $x_0$ and walks towards the source is at $x=x_0-v_{o}t$, so the phase she experiences is
$\omega t-k(x_0-v_{o}t)=(\omega +k{v}_o)t-k{x}_0$
It grows at the rate $\omega +k{v}_o$, hence
${\omega }^{\prime }=\omega +k{v}_o,f^{\prime }=f\left(1+\frac{v_o}{v}\right)$
The extra $k{v}_{o}t$ radians in a time $t$ are the extra crests she walks into: the lean of her track in Figure 2. Walking away, $v_o$ changes sign and the pitch drops.

Source or observer: not the same shift

The two results look alike, but they are not the same. Give both cases one speed $u$ and write $\beta =u/v$. A source moving towards you at $u$ multiplies the frequency by
$\frac{1}{1-\beta }=1+\beta +{\beta }^2+\dots$
while you moving towards the source at $u$ multiply it by $1+\beta$. The two agree to first order in $\beta$ and differ by ${\beta }^2/(1-\beta )$, so the difference only matters when $u$ is a sizeable fraction of $v$.

At road speeds it is small but measurable. At 30 m/s, with sound at 340 m/s, $\beta \approx 0.088$ and the difference is 0.85 % of $f$: a 500 Hz siren is heard at 548.4 Hz when it moves and at 544.1 Hz when the listener moves, 4.3 Hz apart. At the animation's Mach 0.5, $u=v/2$, the difference is dramatic: a listener racing at 170 m/s towards a 400 Hz siren at rest hears 600 Hz, while a siren racing at 170 m/s towards a listener at rest is heard at 800 Hz.

Why the air picks out a frame

This asymmetry is no paradox. The air is a real, physical medium, and the frame in which it is at rest is special: "moving" in these formulas always means moving relative to the air.

To see how a frequency changes between frames, describe the same wave from a frame moving at velocity $u$ along $x$. Its position coordinate is $x^{\prime }$, with $x=x^{\prime }+ut$, and it keeps the same clock. The phase becomes
$\omega t-kx=(\omega -ku)t-k{x}^{\prime }$
The wavenumber is still $k$, so the wavelength is the same for every observer; only the frequency changes, to $\omega -ku$. A listener walking towards the source has $u=-v_o$, which gives back $\omega +k{v}_o$.

A moving source fits the same rule. In the source's own frame the air streams backwards past it at $v_s$, like a wind, so its sound runs forward relative to it at only $v-v_s$. Since it sends $f$ crests per second, they are spaced by $\lambda =(v-v_s)/f$, and that wavelength is shared by all frames. In the air the crests travel at $v$, so a listener at rest in the air hears $v/\lambda =fv/(v-v_s)$, exactly as the crest count found.

When both move, at any angles, the two factors multiply:
$f^{\prime }=f\frac{v+v_o\cos \phi }{v-v_s\cos \theta }$
Here $\theta$ is the angle at emission between the source's velocity and the line to the observer, as in Figure 1, and $\phi$ is the angle between the observer's velocity and the direction to the source.

The pitch curve heard beside the road

A siren that drives past does not jump from high to low: its pitch slides down a smooth curve. Working out that curve uses both points of the first section, the emission angle and the travel delay. Figure 3 shows the result.

Legend

$t$	Time.	$\mathrm{s}$
$f^{\prime }$	Frequency heard by $C$.	$\mathrm{Hz}$
$f$	Frequency of the siren (dashed line).	$\mathrm{Hz}$
$t_1$	Moment when the siren is level with $C$.	$\mathrm{s}$
$t_2$	Moment when $f^{\prime }$ equals $f$: the crest sent side-on at $t_1$ arrives.	$\mathrm{s}$
$d/v$	Time sound needs to cross the gap $d$ from the road to $C$.	$\mathrm{s}$
$d{f}^{\prime }/dt$	Slope of the curve at $t_1$: how fast the heard pitch falls.	$Hz/s$

Figure 3 plots the pitch $f^{\prime }$ that the listener $C$ hears against time $t$. The numbers are the animation's, scaled down to a small picture: $C$ stands $d=50$ length units from the road, sound covers 70 units per second and the siren 35, so $v_s/v=0.5$, as for the animation's 170 and 340 m/s. The siren's own frequency is $f=400\mathrm{Hz}$, the dashed line, and the siren is level with $C$ at $t_1=4\mathrm{s}$.

Each point of the curve takes two steps. For a time of hearing $t$, first find the emission time $t_e$ of the crest arriving then: the one for which the siren's distance to $C$ at $t_e$ equals the distance the crest has run since, $v(t-t_e)$. Then evaluate $f/(1-(v_s/v)\cos \theta )$ with the angle $\theta$ at $t_e$. The curve starts at 788 Hz at $t=0$, a little below the head-on 800 Hz, because that crest left when the line to $C$ was already 10° off the road. It falls to 279 Hz by $t=8\mathrm{s}$.

Now look at the middle of the curve. The siren is level with $C$ at $t_1$, yet the curve crosses the dashed line $f^{\prime }=f$ only later, at $t_2$. The heard pitch equals $f$ for the crest sent side-on, at $\theta =90^{\circ }$; that crest left at $t_1$ and still had to cross the gap $d$ to the listener. Hence
$t_2=t_1+d/v=4+50/70\approx 4.71\mathrm{s}$
which is the bracket $d/v$ in Figure 3. You hear the siren pass a moment after it has passed.

How fast does the pitch fall? The heard pitch depends on $t$ only through the emission time, so by the chain rule $d{f}^{\prime }/dt=(d{f}^{\prime }/d{t}_e)/(dt/d{t}_e)$, and both factors are simple.

For the arrival time, $t=t_e+r/v$, where $r$ is the siren's distance to $C$ at emission. That distance shrinks at the radial speed, $dr/d{t}_e=-v_s\cos \theta$, so $dt/d{t}_e=1-(v_s/v)\cos \theta$.

For the angle, let $X=v_s(t_1-t_e)$ be the distance the siren still has to go to be level with $C$. Then $\cos \theta =X/\sqrt{X^2+d^2}$, whose derivative with respect to $X$ is $d^2/(X^2+d^2{)}^{3/2}={\sin }^3\theta /d$. Since $dX/d{t}_e=-v_s$, it follows that $d(\cos \theta )/d{t}_e=-(v_s/d){\sin }^3\theta$.

Differentiating $f^{\prime }=f/(1-(v_s/v)\cos \theta )$ gives $d{f}^{\prime }/d{t}_e=f(v_s/v)(d\cos \theta /d{t}_e)/(1-(v_s/v)\cos \theta {)}^2$. Dividing by $dt/d{t}_e$ adds one more power of the denominator:
$\frac{d{f}^{\prime }}{dt}=-f\frac{v_s}{v}\frac{v_s}{d}\frac{{\sin }^3\theta }{{\left(1-(v_s/v)\cos \theta \right)}^3}$

At the crossing $t_2$, where $\theta =90^{\circ }$, this is $-f(v_s/v)(v_s/d)=-400\cdot 0.5\cdot 35/50=-140Hz/s$. The tangent drawn at $t_1$, labelled $d{f}^{\prime }/dt$, is steeper still: it is the steepest slope of the whole curve, as the worked problem below shows. The formula also says how the slide scales with distance. The slope goes as $1/d$, so twice as far from the road the same drop takes twice as long; on the road itself, $d\to 0$, the slide becomes a sudden step from 800 to 267 Hz.

Worked problem: the moment the siren passes

The most striking moment is when the siren is level with you. What pitch do you hear at that instant, and is the pitch falling fastest then? Figure 4 draws the geometry of that instant.

Legend

$C$	Listener beside the road.
$S_e$	Where the siren was when it sent the crest that now reaches $C$.
$S$	Where the siren is now, level with $C$.
$d$	Distance from the road to $C$.	$\mathrm{m}$
$\tau$	Time the crest has been travelling since it left $S_e$.	$\mathrm{s}$
$v\tau$	Distance the crest has travelled, from $S_e$ to $C$.	$\mathrm{m}$
$v_s\tau$	Distance the siren has moved in that time, from $S_e$ to $S$.	$\mathrm{m}$
$\theta$	Emission angle: between the siren's velocity and the line from $S_e$ to $C$.	$\mathrm{rad}$

At $t_1=4\mathrm{s}$ the siren is at $S$, level with $C$. The crest reaching $C$ at that instant, the dashed arc, left from $S_e$ a time $\tau$ earlier. In that time the crest travelled $v\tau$ from $S_e$ to $C$, the hypotenuse of the triangle, while the siren moved $v_s\tau$ along the road from $S_e$ to $S$. The third side is the gap $d$ from the road to $C$, at right angles to the road, and the angle at $S_e$ is the emission angle $\theta$.

Step 1, the pitch. In this right triangle the cosine of $\theta$ is the side along the road over the hypotenuse:
$\cos \theta =\frac{v_s\tau }{v\tau }=\frac{v_s}{v}$
The time $\tau$ cancels, so this holds whatever $C$'s distance from the road. The Doppler formula then gives
$f^{\prime }=\frac{f}{1-v_s^2/v^2}$
which for $v_s/v=0.5$ is $400/0.75\approx 533\mathrm{Hz}$, the value marked at $t_1$ in Figure 3, from a crest emitted at $\theta =60^{\circ }$. Pythagoras says how long ago it left: $(v\tau {)}^2=(v_s\tau {)}^2+d^2$, so $\tau =d/\sqrt{v^2-v_s^2}=50/\sqrt{70^2-35^2}\approx 0.82\mathrm{s}$.

Step 2, the steepest slope. In the slope formula only the angle varies, and it enters as the cube of the ratio $\sin \theta /(1-(v_s/v)\cos \theta )$, so the slope is steepest where that ratio is largest. The derivative of the ratio with respect to $\theta$ is
$\frac{\cos \theta -v_s/v}{{\left(1-(v_s/v)\cos \theta \right)}^2}$
which vanishes at $\cos \theta =v_s/v$, the very angle of Step 1. So the pitch falls fastest at the instant the siren is level with you. There $\sin \theta =\sqrt{1-v_s^2/v^2}$ and the ratio equals $1/\sqrt{0.75}$, against 1 at $\theta =90^{\circ }$, so the slope is $140/0.75^{3/2}\approx 216Hz/s$: the tangent in Figure 3.

Step 3, a real ambulance. For the ambulance of the intermediate level (700 Hz, 25 m/s, $v=343m/s$), $v_s/v\approx 0.073$. As it passes you hear $700/(1-0.073^2)\approx 703.7\mathrm{Hz}$, barely above its own pitch. Its pitch falls from about 755 Hz on approach to about 652 Hz as it leaves, a drop of 103 Hz. Standing 10 m from its path, the steepest slope is about $700\cdot 0.073\cdot 25/10\approx 128Hz/s$ (the factor $(1-0.073^2{)}^{3/2}$ differs from 1 by less than 1 %), so the drop takes roughly 0.8 s. That is of the order of $2d/v_s$, the time the ambulance takes to cover twice your distance from its path, during which the line to you swings through most of its angles.

Faster than its own waves: the Mach cone

What happens when the source catches up with its own sound? As $v_s\to v$, the crests sent ahead pile up into a single front. Once $v_s>v$, the source outruns every crest it sends.

Picture the crests at one instant. A crest emitted a time $\tau$ ago is a sphere of radius $v\tau$, centred where the source was then, a distance $v_s\tau$ behind its present position. Older crests are larger spheres centred further back, and all of them touch one cone whose tip is on the source. A line from the tip that just touches one sphere is at right angles to the radius at the touching point, which makes a right triangle with hypotenuse $v_s\tau$ and side $v\tau$ opposite the tip. The half-angle $\alpha$ of the cone therefore obeys $\sin \alpha =v\tau /(v_s\tau )$:
$\sin \alpha =\frac{v}{v_s}\text{(Mach cone)}$

The ratio $v_s/v$ is the Mach number: at Mach 2, $\alpha =30^{\circ }$; at Mach 1.5, about 42°. The faster the source, the narrower the cone.

The cone is hidden in the Doppler formula too. Its denominator $1-(v_s/v)\cos \theta$ vanishes when $\cos \theta =v/v_s=\sin \alpha$, the direction at right angles to the cone's surface: all the crests sent that way arrive at the same moment. That pile-up is the shock wave, and it sweeps across the ground as a sonic boom for as long as the aircraft stays supersonic, not only at the moment it reaches the speed of sound. For directions closer to straight ahead the denominator turns negative, which means crests sent later arrive earlier. So after the cone has passed, a listener hears two streams of sound at once, one of them backwards.

Any object faster than the waves of its medium drags such a cone, whether it makes a sound of its own or not: a bullet; a boat on shallow water, where long waves all travel at $\sqrt{gh}$ for a depth $h$; or a charged particle in water moving faster than light does there, $c/n\approx 0.75c$, whose Cherenkov radiation makes the water of reactor pools glow blue.

Light, echoes and radar

Light needs no medium, so there is no air to pick out a frame: only the relative velocity of source and observer can matter, and the two ways of counting must give the same answer. They do, once time dilation is included: a clock moving at speed $\beta c$ ticks slowly by the factor $\sqrt{1-{\beta }^2}$, as seen from the frame it moves through.

First count in the observer's frame, with the source approaching at $\beta =u/c$. Its slowed clock emits at $f\sqrt{1-{\beta }^2}$, and the moving-source factor $1/(1-\beta )$ comes on top: $f\sqrt{1-{\beta }^2}/(1-\beta )$. Now count in the source's frame, where the observer approaches. She meets crests at $f(1+\beta )$ per second of source time, but her own clock runs slow, so per second of her time she counts $f(1+\beta )/\sqrt{1-{\beta }^2}$. Since $1-{\beta }^2=(1-\beta )(1+\beta )$, both reduce to
$f^{\prime }=f\sqrt{\frac{1+\beta }{1-\beta }}$

Behind this agreement lies the rule found for sound. The phase $\omega t-kx$ counts crests, and all observers agree on a count, so $\omega$ and $k$ change between frames together, just as $t$ and $x$ do: for sound by the Galilean change of frame, for light by the Lorentz transformation. Time dilation alone also causes a shift with no counterpart for sound, the transverse Doppler effect: light emitted when the source is level with you arrives at $f\sqrt{1-{\beta }^2}$, below $f$. For a source receding at $\beta =0.1$, $f^{\prime }/f=\sqrt{0.9/1.1}\approx 0.905$, so its red hydrogen line at 656.3 nm is seen at 725.6 nm.

An echo from a moving target is shifted twice. First the target acts as a moving observer and receives $f(v+u)/v$; then it sends that sound back as a moving source, which multiplies by $v/(v-u)$. For a target approaching at $u$,
$f^{\prime \prime }=f\frac{v+u}{v-u}\approx f\left(1+\frac{2u}{v}\right)$
where the approximation holds for $u$ much smaller than $v$. The same double shift applies when the source itself drives at a fixed wall: had the animation's siren car driven at 170 m/s straight at a wall, its driver would hear the echo at $400\cdot 510/170=1200\mathrm{Hz}$.

Radar and medical ultrasound use the approximation. They mix the echo with the outgoing signal and measure the beat between them, $\Delta f=f^{\prime \prime }-f\approx 2fu/v$, which gives the target's speed. A 24.125 GHz traffic radar sees a car at 30 m/s with $\Delta f=2fu/c\approx 4.8\mathrm{kHz}$, and a 5 MHz ultrasound probe sees blood moving at 0.5 m/s (sound travels at 1540 m/s in tissue) with about 3.2 kHz.

Limits of validity

The formulas assume a uniform, still medium with a single wave speed. In air $v$ rises with temperature and shifts with the wind, and gradients of either bend sound away from the straight line of sight. They also assume that the source's velocity and direction change little during one period, 2.5 ms for a 400 Hz siren.

They are linear, too: they treat sound as small pressure swings, about 0.02 Pa on top of 100 000 Pa of atmospheric pressure for speech, which add up without changing the speed of sound. Near $v_s=v$ this fails. The piled-up crests form a strong compression that travels faster than ordinary sound, and the infinite $f^{\prime }$ of the formula becomes a shock front, curved near the nose and straightening into the Mach cone farther out. A sonic boom carries tens of pascals, usually heard as a double bang from the nose and the tail.

Dispersion, a wave speed that depends on wavelength, changes the cone as well: on deep water, where long waves outrun short ones, a ship's wake keeps a half-angle of about 19.5° over a wide range of speeds.

For sound, relativistic corrections are of order $(v_s/c{)}^2$, about $3\times 10^{-13}$ even at 170 m/s, so the Galilean formulas are exact in practice, while the source–observer asymmetry, of order $(v_s/v{)}^2$, is real. For light, the first-order $\Delta f/f\approx 2u/c$ is ample for radar, where $\beta \approx 10^{-7}$, but at $\beta =0.5$ the exact one-way factor is $\sqrt{3}\approx 1.73$, neither the source formula's 2 nor the observer formula's 1.5.

Where this leads

Special Relativity turns the source–observer puzzle into the Lorentz transformation of frequency and wavelength, and Electromagnetic Waves covers the radio waves and light that radar and astronomy use. Beats between echo and signal belong to Interference and Superposition, the loudness of an approaching siren to Loudness, Intensity and Decibels. The thermal motion of atoms broadens spectral lines by a few parts per million, linking Kinetic Theory of Gases with Atomic Models: the Bohr Atom, while Gravity as Curved Spacetime describes a different shift, of clocks in a gravitational field, not to be confused with the Doppler effect.