The shift a mirror gives twice

A moving reflector does two jobs and gets paid for both.

It receives the incoming wave while moving, which shifts what it hears. It then re-radiates what it received while still moving, which shifts what it sends. The echo therefore carries the shift twice, and the size of that double shift is the whole of what a speed radar measures.

The interesting part is that the two shifts are not the same operation, so the round trip is not a square. Getting that right is the difference between an instrument that reads correctly at every speed and one that reads correctly at slow ones.

Two shifts, two different functions #

The first rung on this ladder makes the point that a moving source and a moving observer produce different formulae in a medium, even though both produce a shift and both reduce to the same thing at low speed. The reflector uses one of each.

As a receiver, the target runs into the wavefronts. In a time $t$ the wave brings $ft$ crests past a fixed point, and the target intercepts an extra $vt/\lambda$ of them by moving into them. The frequency it experiences is

$$f_1 = f_0\,\frac{c+v}{c}.$$

As a source, the target emits at $f_1$ while moving forward, so each successive crest is emitted from a point closer to the observer and the crests are packed into a shorter distance. The frequency received back is

$$f_2 = f_1\,\frac{c}{c-v} = f_0\,\frac{c+v}{c-v}.$$

The first operation multiplies by something; the second divides by something else. Only their product is symmetric under swapping approach for recession, and it is the product that a radar sees.

At a closing speed of 90 km/h in air the echo returns at $1.1541$ times the transmitted frequency, where a single shift would give $1.0729$ and the square of a single shift would give $1.1511$. The discrepancy is small and it is systematic, and it is proportional to $(v/c)^2$ — which is exactly the term that has to be got right for anything moving at a serious fraction of the wave speed.

It is worth seeing where the difference between the two operations actually lives. Expand the round trip in powers of $v/c$:

$$\frac{c+v}{c-v} = 1 + \frac{2v}{c} + \frac{2v^2}{c^2} + \frac{2v^3}{c^3} + \cdots$$

while the square of the one-way observer shift is $1 + 2v/c + v^2/c^2$. The two agree in the constant and in the first-order term — which is why nobody notices — and differ at second order by a factor of two. A supersonic aircraft measured acoustically, or a satellite measured optically to the precision that navigation now requires, is exactly where that factor stops being invisible.

There is also a case where the two operations cannot be combined at all: a reflector receding faster than the wave travels. As $v \to c$ the denominator vanishes and the returned frequency diverges for an approaching target; for a receding one at $v > c$ in a medium there is no echo, because the wave never catches up. Sound has such targets routinely and light has none, and the difference is the whole of why the relativistic version below has to look different.

The measurement is a beat, not a frequency #

Nobody measures the returned frequency directly. A $24$ GHz radar looking at a car sees a shift of a few kilohertz, and measuring a few kilohertz on top of twenty-four gigahertz means measuring one part in ten million of a quantity that also drifts.

Two frequencies a little apart, added, give a rapid oscillation at their average and a slow swelling at their difference. Mixing the echo with the transmission produces exactly that, and only the difference carries the speed — which is why a radar does not measure a frequency at all. It measures a beat of a few hundred hertz against a carrier of ten gigahertz, and the ratio of those two numbers is the reason the method works with ordinary electronics.

So the echo is mixed with the transmission. The product of two sinusoids contains their sum and their difference, and a filter throws the sum away; what is left is a signal at the difference frequency, which is the shift itself. That is the beat, and it converts a measurement of one part in $10^7$ into a measurement of an audio tone.

The arithmetic is unusually tidy. To first order in $v/c$ the beat is

$$\Delta f = \frac{2v}{\lambda},$$

so the instrument’s constant is set by its wavelength alone. At $24.125$ GHz the wavelength is $12.4$ mm, and every metre per second of closing speed produces $161$ hertz. A car at 90 km/h returns just over four kilohertz — the medium’s own speed never enters, because the wavelength already contains it — — a tone in the middle of a telephone’s range, which is why the earliest speed radars had a loudspeaker rather than a display, and why an operator learned to recognise a lorry by its sound.

The mixing is worth a further sentence because of what it costs. Multiplying two signals produces the difference and the sum, and it also produces the difference with its sign removed: a beat of four kilohertz is the same tone whether the target is approaching or receding. Recovering the sign requires a second mixer fed with the transmission shifted by a quarter cycle, so that the two outputs are in quadrature and their relative phase says which way. Cheap radars leave that out and cannot tell approach from recession; every radar that draws a picture has it.

The component, and the part that is lost #

A beam measures the shift along itself and knows nothing about the rest.

If the target’s velocity makes an angle $\theta$ with the beam, only $v\cos\theta$ enters the formula. A radar at $15°$ off the line of travel reads $87$ per cent of the speed, at $30°$ it reads $87$ per cent of nothing much less, and at $60°$ it reads exactly half. At $90°$ it reads zero, and a car crossing directly in front of a radar is invisible to it however fast it is going.

The error is one-sided. The cosine of anything is at most one, so the reading is never high, and the bias cannot be removed by averaging or by care. It also cannot be recovered from the measurement: one beam returns one number, and one number cannot separate a fast target at a large angle from a slow one straight ahead.

The practical answers are all about supplying the missing information from somewhere else. A police radar is aimed along the road and its operator is trained to keep the angle small, so the error is quadratic in the angle and therefore forgiving. A traffic radar that must look across lanes uses two beams and takes a ratio. A weather radar sweeps and infers the wind field from how the measured component varies with bearing, which is a tomography rather than a measurement. An ultrasound machine measuring blood flow displays the angle the operator has told it to assume, and the reading scales with the secant of that angle — which is why the number on the screen is a statement about the operator as much as about the patient.

Range and speed cannot both be sharp #

A continuous tone gives the speed and says nothing about the distance, because there is nothing in it to time. A pulse gives the distance, by its round-trip delay, and gives the speed less well the shorter it is.

A short pulse contains a wide band of frequencies, and a frequency shift smaller than that width cannot be told from the pulse’s own spread. That is the whole of the trade: range resolution wants a short pulse and velocity resolution wants a long one, and the product of the two uncertainties is fixed. Every radar design is a choice of where to sit on that curve, and pulse compression is the trick for sitting in two places at once.

A pulse of duration $T$ has a spectral width of about $1/T$. A Doppler shift smaller than that is buried in the pulse’s own bandwidth, so the smallest resolvable speed is

$$v_{\min} \approx \frac{\lambda}{2T}.$$

Meanwhile the range is resolved to about $cT/2$. Multiply them: the product of the range resolution and the velocity resolution is $\lambda c/4$, independent of the pulse. Shortening the pulse improves one and spoils the other, exactly in proportion, and there is no waveform that improves both.

That is the same relation between a duration and a spectral width that constrains a wavepacket in quantum mechanics, arrived at with nothing quantum in the argument — the constant is different and the structure is identical, because both are statements about Fourier transforms rather than about physics.

What modern radars do about it is not to beat the trade but to spread it. A long pulse with its frequency swept across a band is long in time — good for the Doppler measurement — and yet compresses to a short effective pulse when correlated against the transmitted sweep, because its bandwidth is large. The product is unchanged; what changes is that the energy is spread over a long transmission rather than concentrated in a short one, which is what makes the return detectable at all.

A radar pays the spreading loss twice — once going out and once coming back — so the returned power falls as the fourth power of the range. Doubling the range therefore costs a factor of sixteen in transmitter power, which is why radar range grows so slowly with everything else and why the useful gains have come from integrating many pulses rather than from bigger transmitters.

One more consequence of the two-way path deserves stating, because it is the reason radar is hard rather than the reason it is useful. The transmitted wave spreads on the way out and the scattered wave spreads on the way back, so the returned power falls as the inverse fourth power of the range rather than the second. Doubling the useful range costs a factor of sixteen in transmitted power, or in antenna area, or in integration time — and the fourth power is why radar sets are large and why their range figures move so little between generations.

The relativistic version, where the asymmetry disappears #

Light has no medium, so the distinction between a moving source and a moving observer has nothing to attach itself to, and the two shifts must be the same function. They are:

$$\frac{f'}{f} = \sqrt{\frac{1+\beta}{1-\beta}},$$

and the round trip is the square of it, $(1+\beta)/(1-\beta)$ — which is the same expression as the acoustic round trip with $c$ the speed of light.

That coincidence of form is not a coincidence. The classical round trip is (observer factor) × (source factor); the relativistic one-way shift is the geometric mean of those two factors, which is exactly what removing the medium’s preferred frame requires. Whether the source or the receiver is “really” moving stops being a question, and the only thing left is a single factor that depends on the relative speed.

Flashes sent every second by one observer and received every $k$ seconds by another define a single ratio by an experiment rather than by a transformation — and that ratio is enough to generate the whole of special relativity. Time dilation, length contraction and the composition of velocities are all algebra on $k$, which is why the relativistic Doppler formula is not a correction to the classical one but the more fundamental of the two.

The square root has a name in this context: it is Bondi’s $k$, and an entire construction of special relativity is built out of it and out of the fact that a round trip multiplies by it twice. So the acoustic radar and the relativistic derivation are performing the same operation on the same object, and the reason the round trip is the natural quantity in both is that a round trip is what an experiment can actually do without agreeing about distant clocks.

What a shift is worth as a measurement #

There is a reason to prefer measuring a speed this way rather than by timing a displacement, and it is not convenience.

Timing a displacement means measuring two positions and subtracting, so the precision of the speed is the precision of a position divided by the interval between the measurements — and both errors are present, so a short interval is punished twice. A Doppler measurement asks for a frequency instead, and frequency is the quantity physics can measure best: a beat of four kilohertz counted for a second is known to a part in four thousand from the counting alone, and counted for ten seconds to a part in forty thousand.

The comparison is stark in orbit determination. Ranging to a spacecraft by pulse timing gives its distance to a few metres; measuring the Doppler shift of its carrier over a few minutes gives its line-of-sight speed to a fraction of a millimetre per second, which after integration constrains the orbit far better than the ranging does. Every deep-space navigation solution is dominated by the Doppler data, and the two-way link — an uplink shifted once and a downlink shifted again, with the spacecraft coherently multiplying the received carrier — is exactly the double shift of this essay, arranged so that no clock aboard the spacecraft need be trusted.

What the target has to be, and usually is not #

Every line above treats the reflector as a rigid object with one velocity. Real targets are not.

A cloud of scatterers with a spread of velocities returns a band rather than a line, and the width of the band measures the distribution. That is a nuisance for a radar trying to track a single object and it is the measurement itself for a weather radar, where the spread reports turbulence — and for a Doppler lidar, where the thermal width of a gas’s velocity distribution reports its temperature. The same broadening is noise in one instrument and signal in another.

A car is a collection of surfaces with slightly different velocities, and the wheels are going twice the car’s speed at the top and nothing at the bottom, so the return is a smear with a strong line in it rather than a line. That smear is a nuisance for measuring the car’s speed and it is a signature: the rotating parts of a helicopter produce sidebands whose spacing is the blade rate, and identifying an aircraft from those sidebands is a discipline of its own.

A cloud of scatterers is worse and more useful. Rain returns a band whose width is the spread of drop velocities, so a weather radar reads turbulence from the width and wind from the centre. Blood returns a band whose upper edge is the peak flow in the vessel. A gas returns a band whose width is the thermal speed, which is how a spectral line measures a temperature and, at a different scale, how a moving source’s whole spectrum is transformed — the same double-shift arithmetic applied to emission rather than reflection.

There is a second failure of the rigid picture that is easy to miss and matters for the fastest instruments. The derivation assumes the target’s velocity is constant over the round trip. A target accelerating at $a$ changes its Doppler shift during the measurement, and a measurement lasting $T$ therefore sees a chirp of $2aT/\lambda$ hertz rather than a tone. For a car braking hard and a millisecond measurement that is a fraction of a hertz and irrelevant; for a manoeuvring aircraft and a tenth-second integration it is hundreds of hertz and smears the line into uselessness. The response is to fit a chirp rather than a tone, which recovers the acceleration as a bonus — and which is why long integrations and agile targets are not compatible without a model of the motion.

And the reflector has to reflect. A surface that absorbs returns nothing; a surface angled away returns its energy somewhere else. The shape of what comes back is a subject in itself and it is not a Doppler question at all, which is why the two halves of a radar’s design — how much comes back, and at what frequency — are studied by people who rarely talk to each other.

Two histories, one instrument #

Doppler’s paper of 1842 was about starlight and was mostly wrong about what it explained — he thought the colours of binary stars were the effect, which they are not — and the acoustic case was settled experimentally three years later by Buys Ballot, who put trumpeters on a train.

The reflection version has a different lineage and a specific date. Continuous-wave radar could detect a moving object from the beat it produced against the transmission long before pulse radar could measure a range, and the discovery was accidental: in 1922 Taylor and Young at the US Naval Aircraft Radio Laboratory noticed their signal fading as a ship passed between transmitter and receiver. What they had was a bistatic Doppler detector, and the thing that made it valuable twenty years later was not that it detected aircraft but that it detected moving aircraft against a background of hills and buildings that returned vastly more energy and returned it unshifted.

That is still the reason the technique matters. A radar looking down at a low-flying aircraft receives a return from the ground thousands of times stronger than the one from the target; separating them by range is hopeless, because they are at the same range. Separating them by Doppler shift is easy, because the ground is not moving and the aircraft is. Every airborne radar is built around that filter, and the trade it makes — the blind speeds at which a target’s shift coincides with the ground’s, the clutter notch that hides slow targets — is a trade about the arithmetic in this essay.

The ladder from here #

Later rungs on this anchor: the range–Doppler ambiguity function, which is the proper description of what a waveform can and cannot resolve; continuous-wave frequency-modulated radar, where range and speed are recovered together from one sweep; Doppler tomography, which reconstructs a velocity field from projections and is what a weather radar network is doing; and the acousto-optic case, where the moving reflector is a sound wave in a crystal and the shifted light is used to steer a beam.

The neighbouring ladders are the note that changes on approach, which is one shift rather than two; the cone the source leaves behind, where the source outruns its own shifts; and everything from an exchange of pulses, where the round-trip factor stops being an instrument’s constant and becomes the foundation of a theory.