The Doppler Effect and Shock Waves
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the concept of wave speed relative to a medium and the basic geometry of circular wavefronts emitted by a moving source.
Why this matters
The Doppler effect is the primary diagnostic tool for measuring motion in astronomy, medicine, and radar.
While the wave equation describes how disturbances propagate, the Doppler effect describes how those disturbances are perceived when the source or observer moves. This phenomenon is not merely a curiosity; it is the fundamental mechanism behind the measurement of stellar and galactic motions, the operation of police speed guns, and the formation of sonic booms. Understanding the vector nature of this shift, and the geometric limit where the source exceeds the wave speed, bridges the gap between simple harmonic motion and complex fluid dynamics.
The Moving Source Intuition
Imagine a boat moving across a calm lake. It drops a pebble into the water every second. If the boat is stationary, the ripples form perfect concentric circles centered on the boat. However, if the boat moves forward, each new ripple is centered slightly ahead of the previous one. The ripples in front of the boat are compressed together, while those behind are stretched apart. An observer standing on the shore will hear the "splash" (or see the ripple crest) more frequently from the front and less frequently from the back. This compression of wavefronts corresponds to a higher frequency, and the stretching corresponds to a lower frequency. This geometric argument holds regardless of the specific type of wave, provided the medium is stationary relative to the ground.
General Vector Formulation
In a general three-dimensional scenario, the source and observer may move in arbitrary directions. The key insight from the cited derivation (D-073) is that only the component of velocity along the line connecting the source and observer (the line of sight) affects the frequency shift. Let \(\hat{n}\) be the unit vector pointing from the source to the observer. The observed frequency \(f_o\) is related to the source frequency \(f_s\) by the ratio of the effective wave speed relative to the observer and the effective emission rate relative to the medium.
Here, \(c\) is the wave speed in the medium. The term \(\vec{v}_o\cdot\hat{n}\) is the component of the observer's velocity along \(\hat{n}\), that is, directly away from the source. If the observer moves towards the source, \(\vec{v}_o\) is opposite to \(\hat{n}\), the dot product is negative, the numerator becomes \(c + |\vec{v}_o\cdot\hat{n}|\) and \(f_o\) increases, as intuition suggests. Similarly, if the source moves towards the observer, \(\vec{v}_s\) is in the direction of \(\hat{n}\), so \(\vec{v}_s\cdot\hat{n}\) is positive, making the denominator \(c - |\vec{v}_s\cdot\hat{n}|\), which also increases \(f_o\). This vector form is crucial for applications like radar, where the target's velocity is rarely aligned perfectly with the transmitter.
The Sonic Boom and Mach Cones
When the source velocity \(v_s\) exceeds the wave speed \(c\), the denominator in the Doppler formula can approach zero or become negative, indicating a breakdown of the standard "wavefront arriving one by one" picture. Geometrically, the source is outrunning its own wavefronts. The wavefronts emitted at earlier times form spheres that the source has left behind. These spheres share a common tangent envelope, forming a cone. This is the Mach cone.
The half-angle \(\mu\) of this cone is determined by the geometry of the right-angled triangle formed by the source's path, the radius of the oldest wavefront, and the line of tangency. If the source has traveled a distance \(v_s t\), the oldest wavefront has radius \(c t\). The sine of the angle \(\mu\) is the ratio of the opposite side (radius) to the hypotenuse (source path):
Introducing the Mach number \(M = v_s/c\), we can write this as \(\sin\mu = 1/M\). For \(M > 1\), \(\mu\) is real and less than 90 degrees. As \(M\) increases, \(\mu\) decreases, meaning the cone becomes narrower. The "boom" heard by an observer is the sudden arrival of this coherent shock front, which carries a large amount of energy concentrated in a thin shell.
Energy and Pressure Discontinuities
In the subsonic regime (\(M < 1\)), the pressure disturbances are smooth and propagate outward. In the supersonic regime, the pressure changes discontinuously across the Mach cone. This discontinuity is a shock wave. The strength of the shock, and thus the loudness of the sonic boom, depends on the Mach number and the altitude of the source. The energy of the wave is not lost but is compressed into the conical surface. This concentration of energy is why supersonic aircraft produce such distinct and powerful acoustic events, whereas subsonic aircraft produce a continuous, lower-intensity hum.
The derivation
The Doppler Effect and Shock Fronts
For a scalar wave of fixed speed \(c\) in a medium at rest, a source of proper frequency \(f_s\) and an observer in uniform motion are related by \(f_o = f_s\,\dfrac{c-\vec{v}_o\cdot\hat{n}}{c-\vec{v}_s\cdot\hat{n}}\), where \(\hat{n}\) is the unit vector pointing from source to observer along the propagation ray, so only line-of-sight velocity components shift the frequency. When the source outruns the wave, \(v_s > c\), the emitted spheres acquire a common tangent envelope — a Mach cone of half-angle \(\mu\) with \(\sin\mu = c/v_s = 1/M\).
A common misconception is that the Doppler shift depends on the absolute velocity of the source relative to the ground. In reality, for sound waves, it depends on the velocity relative to the medium. If the wind blows, the effective wave speed \(c\) changes relative to the ground, altering the shift even if the source and observer are stationary relative to each other. This medium-dependence distinguishes sound from light, where the relativistic Doppler effect depends only on the relative velocity between source and observer. C
Worked example
Problem. A jet flies level at an altitude \(h = 3.0\ \mathrm{km}\) with speed \(v_s = 680\ \mathrm{m/s}\) through air in which the speed of sound is \(c = 340\ \mathrm{m/s}\). Find the Mach number, the half-angle \(\mu\) of the Mach cone, and how far past an observer on the ground the jet is when the observer hears the boom.
- Mach number: \(M = v_s/c = 680/340 = 2.0\).
- Mach angle: \(\sin\mu = 1/M = 0.50\), so \(\mu = 30^\circ\).
- The cone reaches the observer when the line from the jet to the observer makes the angle \(\mu\) with the flight path, so the horizontal distance is \(x = h/\tan\mu = 3000/\tan 30^\circ \approx 5.2 \times 10^{3}\ \mathrm{m}\).
- Time after the jet passed overhead: \(x/v_s \approx 5196/680 \approx 7.6\ \mathrm{s}\).
Units check. \(M\) is a ratio of speeds and is dimensionless; \(x\) is metres divided by a pure number; \(x/v_s\) is \(\mathrm{m}/(\mathrm{m/s}) = \mathrm{s}\).
Three quick tests
- If a source moves towards a stationary observer at half the speed of sound, what is the ratio of the observed frequency to the source frequency?
Answer
Using \( f_o = f_s \frac{c}{c-v_s} \), with \( v_s = 0.5c \), \( f_o = f_s \frac{c}{0.5c} = 2 f_s \). The ratio is 2. - What happens to the Mach cone angle as the Mach number increases?
Answer
Since \(\sin\mu = 1/M\), as \(M\) increases, \(\sin\mu\) decreases, so \(\mu\) decreases. The cone becomes narrower. - Does the Doppler effect for sound depend on the motion of the medium (wind)?
Answer
Yes. The wave speed \(c\) is relative to the medium. If the wind blows, the effective speed relative to the ground changes, altering the frequency shift.
Problem set for PU-103 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.