2.5 — The Doppler Effect and Shock Waves

An ambulance goes past and the siren drops in pitch as it passes. Everyone has heard it. The usual explanation — "the sound waves get squashed in front" — is right as far as it goes, and it hides something important: a source moving towards you and an observer moving towards the source give different answers, even at the same relative speed. They are two different physical situations, and only one of them changes the wavelength.

Christian Doppler predicted the effect in 1842 from theory alone. In 1845 the Dutch meteorologist Christophorus Buys Ballot tested it by hiring a locomotive, loading it with trumpet players holding a steady note, and stationing musicians with perfect pitch beside the track to write down what they heard as it passed. It remains one of the more charming experiments in the history of physics, and it confirmed Doppler exactly.

A moving source. Each wavefront is emitted from wherever the source was at the time, so the fronts bunch together ahead of it and spread out behind. The medium's speed is unchanged — only the spacing is. Image: Wikimedia Commons.

Case 1: the source moves, the observer stands still

A source emits at frequency f_s, so it produces one wavefront every T = 1/f_s seconds. Each front then travels outwards at v, the speed set by the medium.

Here is the point everything turns on: each front is emitted from a different place, because the source has moved between emissions.

Suppose the source moves towards you at v_s. It emits a front, then in time T that front has travelled vT towards you — but the source has also travelled v_sT towards you before emitting the next one. So the gap between the two fronts is:

\lambda' = vT - v_sT = (v-v_s)T = \frac{v-v_s}{f_s}

The wavelength is genuinely shorter. The waves in the air really are bunched up.

You are stationary, so you receive them at speed v, and the frequency you hear is:

f_o = \frac{v}{\lambda'} = \boxed{f_s\left(\frac{v}{v-v_s}\right)}

For a receding source, the source moves away between emissions and the fronts are stretched, so the sign flips:

f_o = f_s\left(\frac{v}{v+v_s}\right)

Case 2: the observer moves, the source stands still

Now nothing has changed about the waves themselves. The source is still, so the fronts are evenly spaced at the normal \lambda = v/f_s and stay that way. The wavelength is unchanged.

What changes is how fast you run into them. Move towards the source at v_o and the fronts come at you at a closing speed of v+v_o. Frequency is closing speed divided by spacing:

f_o = \frac{v+v_o}{\lambda} = \frac{v+v_o}{v/f_s} = \boxed{f_s\left(\frac{v+v_o}{v}\right)}

And moving away:

f_o = f_s\left(\frac{v-v_o}{v}\right)

Putting them together, and how to get the signs right

\boxed{f_o = f_s\left(\frac{v \pm v_o}{v \mp v_s}\right)}

Rather than memorise which sign goes where, use one rule that never fails:

Any motion that brings the two closer together raises the pitch. Any motion that separates them lowers it. Choose the signs to make the fraction do that.

So if the observer is approaching, the numerator must get bigger: use +v_o. If the source is approaching, the denominator must get smaller: use -v_s.

The two cases are not the same, and here is the proof

Take v = 343 m/s, f_s = 1000 Hz, and a relative speed of 30 m/s.

Source approaching:

f_o = 1000\times\frac{343}{343-30} = 1000\times\frac{343}{313} = 1095.8\ \text{Hz}

Observer approaching:

f_o = 1000\times\frac{343+30}{343} = 1000\times\frac{373}{343} = 1087.5\ \text{Hz}

Different by 8 Hz. The relative speed is identical in both cases and the answers are not, because sound travels in a medium and the medium provides an absolute reference for "who is moving". The moving source changes the actual wavelength in the air; the moving observer does not.

That fact matters more than it looks. It is exactly why the light Doppler formula must be different: light has no medium, so there is no way even in principle to say which of the two is moving, and any correct formula can depend only on the relative speed. Chapter 6.3 derives it, and the result is

f_o = f_s\sqrt{\frac{1-\beta}{1+\beta}}, \qquad \beta = \frac{v_{\text{rel}}}{c}

symmetric in exactly the way the sound formula is not. The asymmetry above is a fingerprint of the medium, and its absence for light was one of the clues that there is no ether.

Worked example: the ambulance

An ambulance siren emits at 900 Hz and drives past you at 25 m/s. What do you hear before and after?

Approaching:

f_o = 900\times\frac{343}{343-25} = 900\times\frac{343}{318} = 970.8\ \text{Hz}

Receding:

f_o = 900\times\frac{343}{343+25} = 900\times\frac{343}{368} = 838.9\ \text{Hz}

The drop is 970.8 - 838.9 = 131.9 Hz, which at this pitch is about two and a half semitones. That is a large, unmistakable musical interval, which is why the effect is so obvious.

Worth noticing: at no point do you hear the true 900 Hz — except at the exact instant the ambulance is level with you, when its velocity is entirely sideways and none of it is towards or away from you. The drop is not gradual across the whole approach either; it is slow while the ambulance is far off and then sweeps rapidly through as it passes, because what matters is the component of velocity along the line to you.

Worked example: both moving

A train travelling at 30 m/s sounds a 500 Hz horn. A car travels at 20 m/s towards the train, on a parallel road. What does the driver hear?

Both motions bring them closer, so the numerator grows and the denominator shrinks:

f_o = 500\times\frac{343+20}{343-30} = 500\times\frac{363}{313} = 580.0\ \text{Hz}

Shock waves and the sonic boom

Look again at the source formula:

f_o = f_s\left(\frac{v}{v-v_s}\right)

As v_s \to v, the denominator goes to zero and the predicted frequency goes to infinity. Something has broken, and what has broken is the assumption that the source stays behind its own wavefronts.

At exactly v_s = v, the source keeps pace with every front it has ever emitted. All of them pile up on a single plane directly in front of it. That is not a very high frequency; it is a wall of superposed pressure — a shock wave.

Beyond v_s > v, the source outruns its own sound. The fronts it emitted earlier are left behind, and their expanding spheres have a common tangent surface: a cone trailing back from the aircraft.

The cone's half-angle follows from a right triangle. In time t the source travels v_st while the front it emitted at the start has expanded to radius vt. So:

\sin\theta = \frac{vt}{v_st} = \frac{v}{v_s} = \frac{1}{M}

where M = v_s/v is the Mach number, named after Ernst Mach, who photographed supersonic bullets in 1887 and saw the cone.

\boxed{\sin\theta = \frac{1}{M}}

At M = 1 the angle is 90° — a flat wall. At M = 2 it is 30°. At M = 3 it is 19.5°. Faster means a narrower cone, trailing further back.

Three things about sonic booms that are commonly misunderstood:

The boom is not made at the moment of breaking the sound barrier. The cone trails the aircraft continuously for as long as it is supersonic. It sweeps across the ground like a wake behind a boat, and you hear the boom when the cone reaches you — which may be long after the aircraft passed overhead.

There are usually two booms, not one. One shock forms at the nose and another at the tail, and at ground level they are typically separated by a tenth of a second or so, which the ear hears as a distinct double crack.

Nothing is "broken". The sound barrier was named when early aircraft met a wall of drag near Mach 1 and several broke up trying. The drag rise is real — shock waves form on the wing before the aircraft itself reaches Mach 1 and disturb the airflow badly — but it is an aerodynamic problem with an aerodynamic solution, which turned out to be thin swept wings and enough thrust. Chuck Yeager flew through it on 14 October 1947.

The same physics with light instead of sound gives Cherenkov radiation. Nothing outdoes light in vacuum, but light in water travels at c/1.33, and a fast particle from a reactor or a cosmic ray can exceed that. It leaves an optical shock cone, seen as the eerie blue glow in a reactor pool. Chapter 8.5 shows how Super-Kamiokande uses exactly that cone to work out which way a neutrino came from.

Where this shows up in your life

Speed cameras and weather radar bounce a radio wave off a moving object and measure the shift in the returned frequency. The wave is Doppler-shifted twice — once because the car is a moving observer receiving it, and again because the car is then a moving source re-radiating it — so the shift is doubled:

\Delta f = \frac{2v}{c}f_0

At 34 GHz and 30 m/s, \Delta f = 2\times30\times34\times10^9/3\times10^8 = 6800 Hz, which is a comfortably measurable audio-range beat and is why Doppler radar guns have a characteristic tone.

Doppler ultrasound measures blood flow the same way, from the shift in echoes off moving red blood cells. It is how a foetal heartbeat is found and how a surgeon checks whether an artery is narrowed, since a narrowing raises the flow speed by continuity (Chapter 1.11) and the raised speed shows directly as a larger shift.

Redshift is the single most important measurement in cosmology. Light from distant galaxies arrives stretched towards the red, and the amount tells you the recession speed. Edwin Hubble found in 1929 that the shift grows with distance, which is the expanding universe. Chapter 12.5 does that properly, including the important correction that cosmological redshift is not really a Doppler shift — the galaxies are not moving through space so much as space is expanding between them, and the wavelength stretches with it.

Exoplanets are found by watching a star wobble. A planet and its star orbit their common centre of mass, so the star moves towards and away from us slightly, and its spectral lines shift back and forth in step with the planet's year. The shifts are tiny — Jupiter moves the Sun by 12 m/s, Earth by 0.09 m/s — and modern spectrographs measure stellar velocities to under 1 m/s, which is walking pace, across tens of light years.

And a bat has to correct for it. A bat flying at 5 m/s emitting at 80 kHz receives echoes shifted by about 2.3 kHz, which would push the return outside the narrow band its ear is most sensitive to. Several species lower their emitted frequency in flight by exactly the amount needed to bring the echo back into that band, a behaviour called Doppler shift compensation. The bat is solving the equation in this chapter, continuously, in the dark.

What Part 2 established

Part 1 dealt with things that go somewhere. Part 2 dealt with things that go nowhere and repeat, and the two chapters that mattered most were the ones where a single differential equation turned out to describe a spring, a pendulum, a circuit and an atom, and where confining a wave between two boundaries forced it to take only certain discrete frequencies. Both of those ideas come back with full force in Part 7, where the standing-wave argument applied to an electron is what makes an atom have energy levels at all.

What the next Part fixes

Everything so far has been about things you can point at: a block, a string, a planet. Part 3 is about what happens when there are 10^{23} of them and you cannot possibly track each one — a room full of air, a cup of tea cooling, an engine. It turns out that giving up on the individual particles and asking only about averages produces laws that are more certain than the mechanics they came from, not less, and one of those laws is the only one in physics that knows the difference between the past and the future.