Mach angle ( $μ = arcsin (M^{- 1})$ )

The Mach angle of a supersonic flow with Mach number

M > 1

μ = arcsin (M^{- 1})

. Infinitesimal pressure disturbances lie on the corresponding Mach cone, and a weak oblique shock approaches this angle as its strength tends to zero.

The Mach angle is the half-angle between a Mach cone generator and its backward motion axis for a steady point source in a uniform three-dimensional medium with isotropy and constant sound speed. It obeys

sin μ = \frac{c}{U} = \frac{1}{M}, 0 < μ < \frac{π}{2}, M > 1.

(1)

The relation concerns weak acoustic waves in linear acoustics. It is not the angle of a finite oblique shock produced by finite turning of the fluid flow; that angle also depends on the deflection and the gas properties.^[ref]

For a short proof, choose an old emission in the moving-point acoustic wavefronts construction and set

s = - τ > 0

. Its circle has center

(- U s, 0)

and radius

cs

. A tangent line from the present source at the origin meets the circle at a right angle to its radius. The resulting right triangle has hypotenuse

U s

and opposite side

cs

, giving the displayed sine ratio. The emission age cancels: every wavefront has the same tangent directions, so a continuum of expanding spheres produces one simple Mach cone.

In an original example, stipulate uniform gas with sound speed

c = 340 m s^{- 1}

and a source moving through it at

U = 680 m s^{- 1}

. Then

M = 2

and

μ = arcsin (1/2) = 3 0^{\circ}

. An emission one second earlier is now a sphere centered

680 m

behind the source with radius

340 m

. The selected sound speed defines this example; it is not a constant for air at every temperature. The Mach angle decreases as

M

increases: its derivative is

- 1/ (M M^{2} - 1)

in radians, and its limits are

9 0^{\circ}

M ↓ 1

and

0

M \to \infty

Figure 1.
Mach half-angle decreases as the gas-relative speed ratio increases
. Original plot of the Mach angle $μ = arcsin (1/ M)$ for $M > 1$ , with the constructed $M = 2$ , $μ = 3 0^{\circ}$ example. The open point at $M = 1$ marks a limit, not a finite supersonic cone. This weak-wave prediction gives neither a finite-deflection oblique shock angle nor measured sonic boom strength. **Details**.

Table of contents
- Mach cone Mach angle

Mach cone ( $sin θ = 1/ M$ )

 0  0

The Mach cone forms the geometric envelope of acoustic wave wavefronts from a source moving at a supersonic Mach number

M > 1

. In the source reference frame the ray-velocity sphere is centered at

- M c_{0} e_{x}

and has radius

c_{0}

. Its tangent rays make angle

sin^{- 1} (1/ M)

to the backward flight direction; all other emitted acoustic rays lie inside that Mach cone.

The Mach cone is the geometric envelope of the spheres in moving-point acoustic wavefronts. To obtain its axial boundary without a diagram, put

s = - τ > 0

and impose the geometric envelope conditions

F = 0

and

\partial_{s} F = 0

F (x, y, s) = (x + U s)^{2} + y^{2} - c^{2} s^{2} .

(1)

The partial derivative condition gives

x = - (U^{2} - c^{2}) s / U

. Substituting into

F = 0

gives

y^{2} = c^{2} (U^{2} - c^{2}) s^{2} / U^{2}

, hence

y = \pm \frac{- x}{M ^{2} - 1}, x < 0, y^{2} + z^{2} = \frac{- x}{M ^{2} - 1} q u a d in three dimensions .

(2)

For

M = 2

, the Mach cone section has slope magnitude

1/ 3

, consistent with a Mach angle of

3 0^{\circ}

. If emission begins at a finite time, the reached region is limited by the oldest wavefront; the infinite cone describes the steady geometric boundary obtained with an indefinitely long source history. In a uniformly moving gas, advection shifts each emission's sphere by the gas velocity times its age; working with the source's relative velocity through the gas restores the same construction.