Supersonic sources and the Mach cone

Take a source emitting f₀ = 500 Hz while moving at vₛ = 0.90 v through air with v = 340 m s⁻¹. Each front expands at 340 m s⁻¹ about the point it left, and the next front is emitted 2.00 ms later from a point 0.612 m further on. Ahead of the source the spacing between successive fronts is therefore (v − vₛ)/f₀ = 34/500 = 0.068 m against a resting wavelength of 0.68 m: the fronts are compressed tenfold, and a fixed listener ahead counts them at f₀ v/(v − vₛ) = 5000 Hz. That is the Doppler shift, and it is the subsonic face of this lesson's construction. Push vₛ towards v and the spacing ahead shrinks as λ(1 − M) → 0 while the heard frequency climbs without bound. At M = 1 exactly, every front the source has ever emitted is tangent to one plane through its nose, and the formula's divergence is the linear model announcing that it has run out of room. Beyond M = 1 the fronts cross, so a new object has to be found.

Write the family of fronts as F(x, y, τ) = (x − vₛ τ)² + y² − v²(t − τ)² = 0. An envelope is a curve touched by every member, and the standard recipe for one is to solve F = 0 together with ∂F/∂τ = 0, holding x, y and t fixed while τ alone varies. Differentiating gives −2vₛ(x − vₛ τ) + 2v²(t − τ) = 0, so the touching member satisfies vₛ(x − vₛ τ) = v²(t − τ). Measure X = x − vₛ t from the source's present position and T = t − τ since emission; then x − vₛ τ = X + vₛ T and the condition reads T = vₛ X/(v² − vₛ²). For a supersonic source T > 0 forces X < 0: the touching happens behind. Substitute back into F = 0 and the algebra collapses to y² = v² X²/(vₛ² − v²) = X²/(M² − 1). That is a pair of straight lines through the source at tan μ = 1/√(M² − 1), equivalently sin μ = 1/M; rotate about the axis and it is the cone. At M = 2.00, tan μ = 1/√3 and μ = 30.0°. For M < 1 the right-hand side is negative and there is no real envelope at all, which is the calculus agreeing with the picture of nested fronts.

When does a ground observer hear anything? Put the aircraft at height h, now, above the origin, and the observer s behind it under the track. Sound emitted a time T ago left the aircraft MvT further back, so it reaches the observer now if (MvT − s)² + h² = (vT)². That is a quadratic in vT: (M² − 1)(vT)² − 2Ms(vT) + s² + h² = 0, with roots vT = [Ms ± √(s² − (M² − 1)h²)]/(M² − 1). Real roots need s ≥ h√(M² − 1): the observer must be inside the cone, and outside it nothing has arrived. Inside there are two. With h = 10.0 km, M = 2.00 and s = 25.0 km the discriminant is √(625 − 300) km = 18.0 km, so vT = (50.0 ± 18.0)/3 km and T = 75.6 s or 35.5 s at v = 300 m s⁻¹. The older signal left when the aircraft was 45.4 km back, 20.4 km before reaching the observer; the younger left 21.3 km back, 3.7 km after passing. One arrives from ahead, one from behind, together. As s falls to h√(M² − 1) = 17.3 km the discriminant vanishes and the two roots coalesce: the cone is the locus of double roots, the envelope condition of the previous section met from the observer's side.

A cone of half-angle μ, apex at height h, axis horizontal, cuts the ground in a conic. With x measured back along the track from beneath the aircraft and y across it, a point on the cone obeys √(y² + h²) = x tan μ, so y² = x²/(M² − 1) − h². That is a hyperbola: its vertex sits on the track h√(M² − 1) behind the aircraft, and far from the track its wings straighten onto asymptotes at ±μ, so the boom front on the ground is a V whose opening is the cone angle itself. The front sweeps along with the aircraft at vₛ, and a point at lateral offset y is reached when the aircraft is √(M² − 1)√(h² + y²) past abeam, a lag of Δt = √(M² − 1)√(h² + y²)/(Mv). At M = 1.50 and h = 12.0 km with v = 295 m s⁻¹, the vertex trails by 12.0 × 1.118 = 13.4 km and the on-track lag is 13 420/(1.50 × 295) = 30.3 s; a village 9.0 km to the side has slant distance 15.0 km to the track, so its front is 16.8 km behind and its boom comes 37.9 s after abeam. In a uniform atmosphere the hyperbola never ends, so the carpet would be infinitely wide; the real edge is set by refraction (last section).

Rays are normal to wavefronts, and the boom front is the cone, so the ray reaching an on-track observer at the instant of the boom is perpendicular to the cone's edge. That ray makes angle 90° − μ with the flight path, and it left the track from a point h tan μ = h/√(M² − 1) before the overhead point, travelling a slant distance h/cos μ. Take M = 2.00, h = 10.0 km and v = 300 m s⁻¹: the emission point is 5.77 km short of overhead, the path is 11.5 km, and the sound takes 38.5 s to arrive. In those 38.5 s the aircraft has flown 23.1 km, which puts it 17.3 km past overhead — precisely h√(M² − 1), the vertex of the ground trace from the previous section. The two pictures agree because they must: the retarded-time quadratic of section three has its double root at vT = M h/√(M² − 1) = 11.5 km, the same slant path. So there are three lengths a student must keep separate: the emission point is 5.77 km before overhead, the aircraft is 17.3 km past it, and the sound has travelled 11.5 km. Dividing h by v, 33.3 s, answers none of those questions.

In linear theory the field of a moving point source is a sum over its retarded times, each term carrying 1/|1 − M cos θ|, θ the angle between flight direction and the ray to the observer. On the cone the ray is at 90° − μ, so cos θ = sin μ = 1/M and the denominator vanishes: linear acoustics predicts infinite pressure on the surface it has just located. An envelope of wavefronts is a caustic: where the small-amplitude premise of the wave equation fails, not where pressure really diverges. A real boom's finite jump comes from nonlinear steepening — a pressure rise travels faster than a fall, so the waveform shocks up into an N-wave — and its size, of order 50 to 100 Pa under an airliner, depends on mass, length and height in ways the cone angle does not contain. Nor is the atmosphere uniform: sound speed rises towards the warm ground, rays curve upward, and the shallow lateral ones turn over before landing, which bounds the carpet to a few tens of kilometres. Correspondence holds: as M → 1⁺ the half-angle opens to 90°, the stacked plane of the first section; as M → ∞ it closes to zero.