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The Doppler effect is a change between emitted and received frequency caused by motion of the source or receiver. An approaching siren can send more wave crests per second to a microphone than the source emits per second; a receding siren sends fewer. The source need not change its own oscillation. The Doppler shift is $\Delta f=f_o-f_s$, where $f_s>0$ is emitted frequency and $f_o>0$ is received frequency, both measured in hertz.

For acoustic waves, motion must be specified relative to the material carrying the wave. The speed of sound is measured in that material's rest reference frame. Moving the source changes the spacing of successive wavefronts; moving the receiver changes how quickly it crosses those wavefronts. These two mechanisms give different exact factors in the classical Doppler effect for sound, even for the same closing relative velocity.

This makes a microphone a motion probe: a stable tone supplies a repeating clock, and changes in arrival spacing reveal the source's motion toward or away from the receiver. An acoustic Doppler measurement can test the prediction directly. A passing source also illustrates why geometry matters: its speed can stay constant while the received frequency changes continuously, as in Doppler frequency along a trajectory.

The following sound model assumes linear acoustics in uniform stationary material with isotropy, constant speed of sound and negligible wave dispersion. Source and receiver speeds are below the speed of sound, and there is a single direct propagation path. Air motion requires transforming source and receiver velocities into the air's rest reference frame; a nonuniform flow requires a more general propagation model. At the sonic threshold the simple periodic-arrival picture fails; moving-point acoustic wavefronts and the Mach cone describe the relevant geometry, while finite shock waves require further physics. Light in vacuum instead requires the relativistic Doppler effect: the acoustic formula cannot be converted into its exact optical counterpart just by replacing the speed of sound with the speed of light.

A geometric envelope is a curve tangent to the members of a smoothly varying family of curves. For a regular implicit family $F(x,y,s)=0$, candidate contact points also satisfy the partial derivative condition ${\partial }_{s}F=0$; regularity is needed to turn these candidates into an actual envelope. The Mach cone is a three-dimensional analogue for expanding spheres.

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 $M\to \infty$.

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 $-M{c}_0{\mathbf{e}}_x$ 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 $s=-\tau >0$ and impose the geometric envelope conditions $F=0$ and ${\partial }_{s}F=0$ onThe partial derivative condition gives $x=-(U^2-c^2)s/U$. Substituting into $F=0$ gives $y^2=c^2(U^2-c^2)s^2/U^2$, henceFor $M=2$, the Mach cone section has slope magnitude $1/\sqrt{3}$, consistent with a Mach angle of $30^{\circ }$. 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.

The Mach number is the ratio of fluid flow speed to local sound speed.

More precisely, the Mach number is the dimensionless quantitywhere $U$ is the magnitude of the body's relative velocity through the surrounding gas, and $c>0$ is the local sound speed measured in the gas rest reference frame. For a fluid flow past a stationary body, the same Mach number uses the local fluid speed in the body's rest reference frame. A Mach number below, equal to, or above one means subsonic flow, sonic flow, or supersonic flow, respectively. The denominator is a property of the local gas: its sound speed depends on temperature and composition.^[ref]

The relative velocity matters even when the body is at rest in a wind tunnel. A uniform change of reference frame subtracts the same velocity from body and gas, so their relative velocity and the Mach number stay unchanged under a Galilean transformation. Ground speed alone cannot supply $U$ when there is wind. For an original collinear example, a body moving at $680\mathrm{m}{\mathrm{s}}^{-1}$ over the ground through gas moving in the same direction at $170\mathrm{m}{\mathrm{s}}^{-1}$ has $U=510\mathrm{m}{\mathrm{s}}^{-1}$. With a stipulated sound speed $c=340\mathrm{m}{\mathrm{s}}^{-1}$, its Mach number is $1.5$, rather than $2$. A moving observer changes apparent acoustic wave arrival times, as in the radial Mach number, without changing the body's Mach number relative to the gas.

Why is the threshold one so useful? Acoustic waves carry small pressure disturbances at the sound speed. In subsonic flow, such information can travel upstream relative to the body; in a uniform supersonic flow, it is swept downstream. The Mach cone turns this difference in communication into visible geometry. Matching Mach number between a wind tunnel and flight is therefore essential to matching compressible flow effects.^[ref]

An actual supersonic wind tunnel experiment makes the distinction between wavefronts and shock waves concrete. In NASA Glenn's 1-by-1-foot wind tunnel, interchangeable converging-diverging nozzle blocks set the Mach number. A strut-mounted exhaust-nozzle model had a nosecone, an air plenum and transparent sidewalls for schlieren photography. Before installing the model, the researchers photographed the empty test section: two shock wave patterns originated at a wall joint after the nozzle block. This control guided model placement to keep reflected shock waves away from the plume-study region. The rig included internal total pressure and static pressure sensing and a traversing external static pressure probe. A baseline nozzle and previous pressure data supplied a planned validation comparison; the report distinguishes this plan from the observed empty-tunnel control. This establishes an apparatus origin for those visible shock waves, not a measurement of the Mach cone from an infinitesimal source. The reported finite shock wave angle cannot simply be inserted into the Mach angle formula to calibrate $M$: shock wave strength and geometry also matter.^[ref]

In conventional schlieren photography, a bright slit source and two concave mirrors send parallel light rays through the wind tunnel and focus them at a knife edge before an optical camera. Spatial changes in mass density cause refraction; the knife edge converts the deflection into image contrast, making shock waves visible. This is the general optical principle, rather than a specification of every component in the preceding rig. A photograph projects a three-dimensional fluid flow into two dimensions; contrast alone does not provide a calibrated pressure jump or acoustic wave amplitude.^[ref]

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.

The Mach number is $\mathcal{M}=u/c_s$, comparing the flow speed relative to a disturbance with the local sound speed. For a body moving supersonically, spherical acoustic disturbances emitted earlier have a tangent envelope with Mach cone half-angle $\sin \theta =1/\mathcal{M}$. Signals cannot propagate arbitrarily far upstream. Compressive characteristics can converge and steepen into a shock wave, within which microscopic dissipation prevents a multivalued fluid state; supersonic motion alone does not require every flow to contain a shock.

Work in the shock frame with steady normal velocities $u_1,u_2>0$. Integrating conservation of mass, normal momentum and total energy over a thin pillbox, neglecting gravity across the thin layer, gives the Rankine-Hugoniot conditions for a perfect gas:The energy equation includes pressure work through specific enthalpy. The shock is adiabatic if negligible heat escapes the layer, but its irreversible dissipation means it is not isentropic.

Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and ${\mathcal{M}}_1=u_1/\sqrt{\gamma p_1/{\rho }_1}$. Momentum conservation gives $p_2=p_1+{\rho }_1{u}_1^2(1-1/\chi )$. Substitution in the energy equation produces the unshocked root $\chi =1$ and the compressive rootSince ${\rho }_1{u}_1^2/p_1=\gamma {\mathcal{M}}_1^2$, the strong-shock limit has negligible upstream thermal pressure compared with ram pressure. The Strong-shock Rankine-Hugoniot conditions giveFor unchanged composition, the ideal gas law implies $T_2/T_1=(p_2/p_1)/\chi$, soFor $\gamma =5/3$, the compression tends to four and $T_2\simeq 3\bar{m}{u}_1^2/(16k_B)$, where $\bar{m}$ is mean particle mass.

With constant specific heat $c_v$, entropy production in a perfect-gas shock is measured by the change in specific entropyThe strong-shock expression is thereforeThe exact jump is positive for every compressive shock with ${\mathcal{M}}_1>1$, in accordance with the Second law of thermodynamics; the asymptotic expression is only for ${\mathcal{M}}_1\gg 1$. There is a normalization issue in the printed description: this logarithmic formula with constant $c_v$ describes entropy per unit mass. The true entropy density is $\rho s$, so its difference is ${\rho }_2{s}_2-{\rho }_1{s}_1$, rather than the same logarithmic jump with a constant heat capacity per unit volume. A material parcel's irreversible entropy increase should be assessed using $\Delta s$.

Downstream gas is denser and hotter. For optically thin collisional emission the loss per unit volume is typically $n_e{n}_i\Lambda (T)$, whereas stored thermal energy scales approximately as $nT$. Compression can therefore reduce the radiative gas cooling time; excitation and ionization at the higher temperature can open additional cooling channels. The actual efficiency still depends on the astrophysical cooling function, composition and optical depth. In a radiative shock with separated scales, the temperature first rises across a narrow dissipative layer, then decreases through a much broader cooling region of characteristic length ${\ell }_{\mathrm{cool}}\sim u_2{t}_{\mathrm{cool}}$. Subsonic cooling usually brings further compression with approximately constant pressure. The immediate microscopic shock thickness and the macroscopic cooling length are distinct.

Let state $f$ be the fully cooled gas, with $T_f=T_1$ and the same composition. Set $a^2=p_1/{\rho }_1$, the isothermal sound speed squared. Across the entire shock plus cooling layer, mass and momentum are still conserved, but gas energy alone is not. With ${\chi }_f={\rho }_f/{\rho }_1$,The nontrivial isothermal shock therefore hasIf the upstream Mach number were defined with $a$ instead, this would be ${\rho }_f/{\rho }_1={\mathcal{M}}_{\mathrm{iso}}^2$. Retaining the distinction avoids losing the factor $\gamma$.

The gas's specific entropy rises irreversibly at the hot shock and then decreases as escaping radiation removes heat. Since the final temperature equals the initial temperature,This decrease for the gas is consistent with the Second law of thermodynamics: the emitted radiation and its eventual surroundings carry entropy away. It does not make the microscopic compressive shock reversible. The figure illustrates both the hot entropy jump and the subsequent cooling; its spatial cooling profile is schematic because no cooling law has been specified.

Supersonic communication. A point source moving at speed $V>c_s$ outruns its small acoustic waves. After a time $t$, the source has moved $Vt$, while a disturbance emitted at its old position has radius $c_{s}t$. The envelope of these spherical wavefronts is the Mach cone, with half-angle $\mu$ satisfyingInformation is restricted to the downstream cone, so upstream gas cannot smoothly anticipate a supersonic obstacle. Finite compressions can also steepen because their characteristics converge. Such disturbances form a shock wave, across which the fluid variables change rapidly. A perfectly uniform supersonic flow need not contain a shock: the cone expresses the communication constraint, while a disturbance or compression supplies the reason for a discontinuity.

Conserved fluxes. Work in the shock frame, take velocities normal to the front, and integrate the conservation laws through a thin steady control volume of unit area. Mass conservation gives constant flux $j=\rho u$, soThe normal momentum flux is the advected part $\rho u^2$ plus the pressure stress $p$. With negligible body-force impulse across the thin front,For an ideal gas with constant adiabatic index $\gamma$, the specific internal energy is $e=p/[(\gamma -1)\rho ]$. The energy flux iswhere $h_{\mathrm{sp}}$ is the specific enthalpy. No radiation or other heat flux escapes in this adiabatic shock model. Conservation of the energy flux, divided by the nonzero mass flux, givesThese are the Rankine-Hugoniot conditions for a perfect gas. The shock is adiabatic but irreversible: its entropy increases, so an isentropic relation cannot be imposed across the discontinuity.

Strong-shock compression. Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and neglect $p_1$ and the upstream enthalpy compared with ${\rho }_1{u}_1^2$ and $u_1^2$. Momentum conservation gives $p_2={\rho }_1{u}_1^2(1-1/\chi )$. Substitute this into energy conservation:ThusThe root $\chi =1$ describes no shock. The compressive branch of the Strong-shock Rankine-Hugoniot conditions isIt also gives $p_2=2{\rho }_1{u}_1^2/(\gamma +1)$. A monatomic gas has $\gamma =5/3$, hence $\chi =4$.

The finite slab in its initial rest frame. Denote the shock's lab speed by $V_s$. The upstream gas is initially stationary, so $u_1=V_s$ and the downstream shock-frame speed is $u_2=V_s/4$. Transforming to the initial slab frame givesThis is planar shock compression of a finite slab. In the uniform post-shock approximation, breakout occurs at $t_b=L/V_s$. The original front face has then moved $(3V_s/4)t_b=3L/4$, while the shock reaches the originally stationary far face at $L$, leaving thickness $L/4$. Equivalently, the fixed mass per area $\rho L$ becomes $4\rho L_2$. The result assumes a maintained approximately constant shock speed and no important boundary rarefaction during the crossing; an unconfined finite cloud need not remain uniformly compressed.

Escape estimate and the sound-speed convention. Let $M_c$ be the slab mass, $W>0$ its binding-energy magnitude in the star's field, and $E_{\mathrm{th}}$ its initial internal energy. Treat the potential as nearly uniform across the slab. For the bulk-ballistic escape estimate, requireThe Mach-number convention in shock unbinding estimates matters here. With the usual upstream adiabatic sound speed, $c_{\mathrm{ad}}^2=\gamma p/\rho =5p/(3\rho )$, define ${\mathcal{M}}_{\mathrm{ad}}=V_s/c_{\mathrm{ad}}$. ThenIf instead the shock speed is normalized by the isothermal sound speed, $c_T^2=p/\rho$, then ${\mathcal{M}}_T^2=(5/3){\mathcal{M}}_{\mathrm{ad}}^2$ andThe original PDF's “around five” is consistent with the latter convention or an order-of-magnitude sound-speed estimate. It is not the exact coefficient obtained with the conventional adiabatic Mach number and the stated bulk kinetic-energy criterion. Here $W=-E_{\mathrm{grav}}$ if the potential is zero at infinity; using negative gravitational energy directly on the right would be incorrect. Subsequent conversion of shock-generated internal energy to expansion, external pressure, or radiative losses would change the physical escape threshold, so this numerical factor is not universal.

A plane wave in the rest frame of the medium has phase $\mathbf{k}\cdot \mathbf{x}-{\Omega }_0(\mathbf{k})t$. Under the Galilean transformation ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{U}t$, this becomes $\mathbf{k}\cdot {\mathbf{x}}^{\prime }-[{\Omega }_0(\mathbf{k})-\mathbf{U}\cdot \mathbf{k}]t$. ThereforeHere ${\mathbf{c}}_g$ is the group velocity in the moving frame.

For acoustic waves, ${\Omega }_0=c_0|\mathbf{k}|$ and take $\mathbf{U}=M{c}_0{\mathbf{e}}_x$. A steady wave has $\omega =0$, requiring $\cos \beta =1/M$, where $\beta$ is the angle of $\mathbf{k}$ to the flight direction. Its ray velocity is $c_0\hat{\mathbf{k}}-M{c}_0{\mathbf{e}}_x$, with backward component $-c_0(M^2-1)/M$ and transverse magnitude $c_0\sqrt{M^2-1}/M$. Thus the angle $\theta$ of the ray to the backward flight direction obeysThe axial symmetry gives the Mach cone behind the aircraft.

For unsteady emitted disturbances, all ray velocities lie on the sphere ${\mathbf{c}}_g=-\mathbf{U}+c_0\hat{\mathbf{k}}$. Since $M>1$, the origin lies outside this velocity sphere and its tangent cone has exactly the same semi-angle. All other rays are strictly inside the cone. Equivalently a signal emitted $\tau$ time units ago has a wavefront sphere centered at $-\mathbf{U}\tau$ with radius $c_0\tau$; these spheres are enclosed by their Mach-cone envelope. Consequently the unsteady wavefield is confined to the cone's interior, with the steady envelope on its boundary.

For capillary waves put $a=\sqrt{T/\rho }$, so ${\Omega }_0=a{k}^{3/2}$. The positive-frequency steady condition is $a\sqrt{k}=U\cos \beta$, requiring $\cos \beta >0$. ThenandIts endpoint traces the circlesince $c_{gx}/U=-1/4+(3/4)\cos 2\beta$ and $c_{gy}/U=(3/4)\sin 2\beta$. As $\beta$ runs through $(-\pi /2,\pi /2)$ this traces the circle except its leftmost limiting point, corresponding to the zero-wavenumber limit. The circle surrounds the origin, so every direction except the precisely backward axis is attained by a nonzero-wavenumber steady ray; the backward direction is approached as $k\downarrow 0$. Thus the closure of the steady ray field extends all around the insect, with no excluded angular sector such as the acoustic Mach-cone exterior.

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.

A subsonic leading edge has an incoming edge-normal flow component below the speed of sound. It can occur even when the free stream is supersonic: the relevant component is the normal Mach number for a swept leading edge,Using a quarter-chord sweep angle instead would project onto a different direction and can change the classification incorrectly.^[ref]

For an original example, $M=1.6$ and ${\Lambda }_{\mathrm{LE}}=60{}^{\circ }$ give $M_n=0.8$. If the same aircraft wing is exposed to $M=2.2$ with unchanged sweep, $M_n=1.1$ and the edge is instead a supersonic leading edge. This is a flow-condition classification, not an immutable geometrical name.

For $M>1$ and aft sweep between zero and $90{}^{\circ }$, the threshold isAt $M=1.6$, the boundary is about $51.32{}^{\circ }$. Equivalently, the edge's angle from the downstream streamwise direction is less than the Mach angle, placing it inside the apex Mach cone. The normal Mach number for a swept leading edge article derives the underlying vector projection; no assumption about a measured surface pressure is needed for that geometry.

Subsonic edge-normal inflow does not imply that every part of the three-dimensional flow is subsonic or free of shock waves. Thickness, angle of attack and interference with a fuselage affect the surface flow. The regime is useful precisely when kept separate from those further aerodynamic predictions.

A supersonic flow has Mach number greater than one, so the flow speed exceeds the local sound speed.

For uniform gas moving at $U{\mathbf{e}}_x$ past a resting body, even an acoustic wave directed against the fluid flow has body-frame velocity $(U-c){\mathbf{e}}_x$ downstream. More generally, an acoustic wave travelling along a unit vector $\mathbf{n}$ has body-frame velocity $U{\mathbf{e}}_x+c\mathbf{n}$; its $x$ component is at least $U-c>0$. All such weak signals move downstream and their possible directions fit inside a Mach cone.

A supersonic leading edge has an incoming edge-normal flow component exceeding the speed of sound. For a straight planar edge at leading-edge sweep angle ${\Lambda }_{\mathrm{LE}}$ in the convention of normal Mach number for a swept leading edge, the criterion isIt classifies a direction of the incoming velocity, rather than asserting that the same local Mach number holds over the whole aircraft wing.^[ref]

For an original example with $M=2$ and ${\Lambda }_{\mathrm{LE}}=30{}^{\circ }$, $M_n=\sqrt{3}\simeq 1.732$. The condition can persist at larger sweep: at $M=2.2$, even a $60{}^{\circ }$ leading-edge sweep gives $M_n=1.1$: large sweep by itself does not establish a subsonic leading edge.

For $M>1$, the Mach angle is $\mu =\arcsin (1/M)$. On an aft-swept half-wing the edge's angle from the downstream streamwise direction is $\theta =90{}^{\circ }-{\Lambda }_{\mathrm{LE}}$. Thusso the leading edge lies outside the downstream Mach cone from its apex in this ideal geometry. At $M=2$, $\mu =30{}^{\circ }$ whereas the example edge has $\theta =60{}^{\circ }$. Equality is the sonic boundary and belongs to neither strict inequality.

The classification helps select a high-speed flow model. It does not determine shock wave strength, drag or flow separation without the section shape and operating conditions. A measured pressure coefficient distribution or flow visualization can test those additional predictions; naming the regime alone cannot.