In the Doppler effect, a moving source crowds its wavefronts together ahead of it. What happens when the source moves as fast as the waves, or faster? The wavefronts pile up on top of each other into a cone, and the sound arrives all at once as a shock wave. A jet's sonic boom, the crack of a whip and the blue glow of Cherenkov light are all the same geometry.
Key idea
A source moving at $v_s$ faster than the wave speed $v$ leaves a cone of wavefronts with half-angle $\theta$ given by $$\sin\theta = \frac{v}{v_s} = \frac{1}{M},$$ where $M = v_s/v$ is the Mach number. The faster the source, the narrower the cone.
Why it works
Look at the wavefront the source made a time $t$ ago. It has grown to a circle of radius $vt$ around the point where it was made. Meanwhile the source has moved $v_st$ farther. Every older wavefront is a bigger circle around a point farther back, and all of them touch two straight lines through the source's present position. In the right triangle formed by the source's path ($v_st$, the hypotenuse) and the radius to the touching point ($vt$, at a right angle to the line), $$\sin\theta = \frac{vt}{v_st} = \frac{v}{v_s}.$$ The waves add up along the cone's surface, which carries a sudden jump in pressure.
Worked examples
Worked example: A cone's angle
A jet flies at Mach 2.0. Find the half-angle of its shock cone.
- $\sin\theta = \frac{1}{M} = \frac{1}{2.0} = 0.50$.
- $\theta = \sin^{-1}(0.50) = 30^\circ$.
Answer: $\theta = 30^\circ$
Worked example: When do you hear it?
A jet flies level at Mach 1.5, 8.0 km above you, where the speed of sound is 300 m/s. How long after it passes directly overhead do you hear the boom?
- $\sin\theta = 1/1.5$, so $\theta = 41.8^\circ$, and $v_s = 1.5 \times 300 = 450$ m/s.
- The cone trails the jet. It reaches you when the line from the jet to you makes angle $\theta$ with the flight path, so the jet is then a horizontal distance $x = \frac{h}{\tan\theta}$ past you.
- $x = \frac{8000\ \text{m}}{\tan 41.8^\circ} = \frac{8000}{0.894} = 8940$ m.
- Time: $\frac{8940\ \text{m}}{450\ \text{m/s}} = 19.9$ s. (We assumed one speed of sound all the way down.)
Answer: About 20 s after it passes overhead
Worked example: Light can do it too
Nothing outruns light in a vacuum, but light in water moves at $c/n$ with $n = 1.33$. An electron moves through water at $0.90c$. Find the half-angle of the cone of light wavefronts it leaves.
- Wave speed: $c/1.33 = 0.752c$. The electron is faster, so there is a cone.
- $\sin\theta = \frac{0.752c}{0.90c} = 0.835$, so $\theta = 56.6^\circ$.
Answer: $\theta \approx 57^\circ$: this is Cherenkov light, the blue glow in the water around a research reactor's core
Common mistake
Thinking a sonic boom happens only at the instant a plane passes the speed of sound. The cone travels with the plane for as long as it flies faster than sound, sweeping a path along the ground beneath it. Everyone along that path hears a boom when the cone passes them, at different times. Check with the geometry: the cone exists for every $v_s > v$, not just at $v_s = v$.
Common mistake
Using $\sin\theta = v_s/v$ upside down. The sine must be less than 1, and $v_s > v$, so it has to be $v/v_s$.