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The Mach angle of a supersonic flow with Mach number $M>1$ is $\mu =\arcsin (M^{-1})$. 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 obeysThe 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 $s=-\tau >0$. Its circle has center $(-Us,0)$ 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 $Us$ 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 $c=340\mathrm{m}{\mathrm{s}}^{-1}$ and a source moving through it at $U=680\mathrm{m}{\mathrm{s}}^{-1}$. Then $M=2$ and $\mu =\arcsin (1/2)=30^{\circ }$. An emission one second earlier is now a sphere centered $680\mathrm{m}$ behind the source with radius $340\mathrm{m}$. 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 $-1/(M\sqrt{M^2-1})$ in radians, and its limits are $90^{\circ }$ as $M\downarrow 1$ and $0$ as $M\to \infty$.

Take a steady moving point source in uniform resting gas, with constant sound speed $c$ and source velocity $U{\mathbf{e}}_x$. Place the source at the origin at the current time $t=0$. In the linear acoustics model, an emission at an earlier time $\tau <0$ has a spherical wavefront centered at $(U\tau ,0,0)$ with radius $-c\tau$. An axial cross-section is the circle in the displayed equation. It is centered at the old release position, not the source's present position. The circles describe the geometry of weak acoustic waves, not a measured shock wave or its pressure jump.

For $M=0.5$, every old circle contains the present source because its center is $U|\tau |$ from the origin while its radius is the larger distance $c|\tau |$. It reaches ahead of the source. For $M=2$, the rightmost point of every old circle is $(U-c)\tau <0$: the source outruns all of its old wavefronts. The continuously emitted wavefronts have a common geometric envelope, the Mach cone. The plotted discrete release times sample this continuous family; rotating the axial section about the motion axis recovers the three-dimensional construction.

A sonic boom is the sound associated with shock waves from a body moving at a supersonic Mach number. The weak-wave Mach cone describes a propagation geometry rather than the finite pressure signature determining the sound strength.