Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

When a time-harmonic source and a receiver move relative to each other, the frequency of the received wave differs from the frequency of the emitted one. This is the Doppler effect, named after C. Doppler (1803–1853), who described it for sound and light in 1842.

A moving source

Let the source T move with the constant velocity $v$ , at the angle $θ$ to the line from T to the receiver R, which is at rest (Figure 6.1.20, left). The peak value of the wave emitted by T at $t = 0$ , when the distance is $r_{1}$ , arrives at R after

t_{1} = \frac{r _{1}}{c} seconds. (6.1.65)

One period $Δ t$ later T emits the next peak. By then it has moved $v Δ t$ metres from its original location, to the distance $r_{2}$ from R (Figure 6.1.20, right), and this peak arrives at R at

t_{2} = Δ t + \frac{r _{2}}{c} seconds. (6.1.66)

The interval between the two receptions, $Δ t^{'} = t_{2} - t_{1}$ , is the period seen by the receiver.

The geometry of the Doppler effect for a moving source. Left: at t=0 the source T, moving with \vec v at the angle \theta to the line towards the receiver R, emits a crest from the distance r_1. Right: one period later it emits the next crest from the distance r_2, having moved v\,\Delta t. — **Figure 6.1.20.** The geometry of the Doppler effect for a moving source. Left: at $t = 0$ the source T, moving with $v$ at the angle $θ$ to the line towards the receiver R, emits a crest from the distance $r_{1}$ . Right: one period later it emits the next crest from the distance $r_{2}$ , having moved $v Δ t$ .

By the law of cosines in the triangle of the right-hand figure,

t_{2} = Δ t + \frac{1}{c} r_{1}^{2} + (v Δ t)^{2} - 2 r_{1} v Δ t cos θ \approx Δ t + \frac{r _{1}}{c} 1 - \frac{2 v Δ t}{r _{1}} cos θ, (6.1.67)

where $(v Δ t)^{2}$ was dropped because the source moves only a tiny distance in one period, $v Δ t ≪ r_{1}$ . For the same reason the second term under the root is small, and with $1 - x \approx 1 - x /2$ for $x ≪ 1$ ,

t_{2} Δ t^{'} \approx Δ t + \frac{r _{1}}{c} (1 - \frac{v Δ t}{r _{1}} cos θ), = t_{2} - t_{1} = Δ t (1 - \frac{v}{c} cos θ) . (6.1.68)

The frequency of the original wave is $f = 1/Δ t$ , and the frequency of the received wave is

f^{'} = \frac{1}{Δ t ^{'}} = \frac{1}{Δ t ( 1 - \frac{v}{c} cos θ )} \approx f (1 + \frac{v}{c} cos θ), (6.1.69)

the last step using $1/ (1 - x) \approx 1 + x$ for $x = \frac{v}{c} cos θ ≪ 1$ .

That means that if the source is getting closer, $θ \in (0, π /2)$ , the frequency increases. This is the reason for the blue shift of approaching objects in astronomy. If the source moves away, $θ \in (π /2, π)$ , the frequency decreases, the red shift of receding galaxies. At $θ = π /2$ the source moves across the line of sight and, to first order, there is no shift at all. Figure 6.1.21 gives the picture: the wavefronts emitted at equal time steps crowd together ahead of the moving source and spread out behind it. This mechanism constitutes the basis of the continuous-wave (CW) radar.

**Figure 6.1.21.** Wavefronts emitted at equal time steps by a source moving to the right. Each circle is centred on the point where the source was when it emitted that front. Ahead of the source the fronts are closer together, a shorter wavelength and a higher frequency, behind it they are farther apart.

Press Play below to follow the $437 MHz$ beacon of a satellite at $7.5 km/s$ on an overhead pass at $500 km$ (Problem 6.1.5), and watch the shift sweep through zero at closest approach at $164 Hz/s$ on a straight track, or at $152 Hz/s$ with the toggle for the orbit round the Earth.

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A moving observer, and the radar echo

Let now T be at rest and the receiver R move with the speed $v$ at the angle $θ$ to the line from R to T. The crests, spaced by $λ = c / f$ , sweep past R with the relative speed $c + v cos θ$ , so R counts

f^{'} = \frac{c + v cos θ}{λ} = f (1 + \frac{v}{c} cos θ), (6.1.70)

the same result as for the moving source, to first order in $v / c$ . For electromagnetic waves only the relative velocity of source and observer can matter, by special relativity, and the exact head-on formula $f^{'} = f (1 + v / c) / (1 - v / c)$ differs from both classical expressions only in terms of order $(v / c)^{2}$ . Expanding, $(1 + x) / (1 - x) = 1 + x + \frac{1}{2} x^{2} + \dots$ , against $1 + x + x^{2} + \dots$ for the moving source, Eq. (6.1.69), and $1 + x$ for the moving observer, Eq. (6.1.70). At the speeds of cars, aircraft and satellites, $v / c \leq 3 \times 1 0^{- 5}$ , the difference is far below anything that matters.

A radar echo combines both cases. A target moving with $v$ at the angle $θ$ to the line of sight first receives the radar's wave as a moving observer, at $f_{1} = f (1 + \frac{v}{c} cos θ)$ , then re-radiates $f_{1}$ as a moving source (Figure 6.1.22):

f_{2} \approx f_{1} (1 + \frac{v}{c} cos θ) \approx f (1 + \frac{2 v}{c} cos θ) . (6.1.71)

The Doppler effect

To first order in $v / c$ , a moving source or a moving observer shifts the frequency to

f^{'} \approx f (1 + \frac{v}{c} cos θ), (6.1.72)

where $θ$ is the angle between the velocity and the line towards the other party. The echo of a target moving at $v$ is shifted by the two-way Doppler shift

f_{D} = f_{2} - f \approx \frac{2 v cos θ}{c} f = \frac{2 v cos θ}{λ} . (6.1.73)

In plain words

Moving towards the wave, or moving the source towards you, squeezes the crests together in time. One trip changes the frequency by the fraction $v cos θ / c$ , a radar echo makes the trip twice.

**Figure 6.1.22.** The radar echo. The target receives the frequency shifted once, as a moving observer, and sends it back shifted again, as a moving source: the echo carries twice the one-way shift.

The two-way shift is used wherever speed is to be measured without contact: police radar (Application 6.1.5), weather radar, which sees rain moving towards or away from it, aircraft and missile tracking, the orbit determination of satellites (Problem 6.1.5), and the Doppler log of ships. Ultrasound Doppler in medicine uses the same law with the speed of sound in place of $c$ .

Quick check

A car approaches an X-band speed gun ( $10.525 GHz$ ) head-on at $25 m/s$ ( $90 km/h$ ). What is the Doppler shift of the echo?

$f_{D} = 2 v f / c = 2 \times 25 \times 10.525 \times 1 0^{9} /3 \times 1 0^{8} = 1754 Hz$ , an audio tone. The relative shift is $2 v / c = 1.7 \times 1 0^{- 7}$ .

Common mistake · One-way or two-way

A source or an observer in motion gives the one-way shift $f v cos θ / c$ . A radar looking at a moving reflector gives twice that. Mixing the two halves or doubles every speed derived from a measured shift. The angle is always measured between the velocity and the line of sight, so motion across the line of sight gives no first-order shift.

Application 6.1.5 · Police radar

The physics. A continuous-wave (CW) radar transmits a single frequency $f$ . The car reflects it with the two-way Doppler shift, and the receiver multiplies (mixes) the echo with a sample of the transmitted wave: the difference frequency is the Doppler tone,

f_{D} = \frac{2 v cos θ}{c} f . (6.1.74)

The shift is tiny, $f_{D} / f \approx 2 \times 1 0^{- 7}$ at $30 m/s$ , but after mixing it is an audio tone of a few kilohertz, counted or Fourier-transformed in a fraction of a second (Figure 6.1.23). Stationary objects return exactly $f$ and are filtered out. Table 6.1.3 gives the shift per unit speed in the three radar bands used by the police.

The cosine error. The gun measures only the component $v cos θ$ of the velocity along its beam, so the reading is always low: by $1.5 %$ at $θ = 1 0^{\circ}$ and by $6 %$ at $2 0^{\circ}$ (Worked example 6.1.5). The error always favours the driver, and the officer aims far ahead along the road to keep $θ$ small.

Table 6.1.3. Doppler shift of a police radar per unit speed of the target, $2 f / c$ , head-on.

band	$f$	$f_{D}$ per km/h
X	$10.525 GHz$	$19.5 Hz$
K	$24.150 GHz$	$44.7 Hz$
Ka	$34.7 GHz$	$64.3 Hz$

**Figure 6.1.23.** Police radar. Left: the beam of a K-band gun meets the approaching car at the angle $θ$ to its motion. Right: the spectrum at the receiver, the strong leakage of the transmitter at $f$ and the weak echo at $f + f_{D}$ . Mixing the two produces the audio tone $f_{D}$ .

6.1.7 The Doppler effect

A moving source

A moving observer, and the radar echo