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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 $M\to \infty$.

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]

A nozzle shapes a passage to control the speed and direction of a fluid flow. In compressible flow, converging-diverging passages can accelerate a gas through a sonic flow throat into supersonic flow.

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.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

For the isothermal equation of state, $p=c_s^2\rho$, with constant isothermal sound speed $c_s$. A nonzero steady plane-parallel flow has constant mass flux $\rho u=j$. Choose the positive flow direction for convenience. The continuity equation then gives ${\rho }^{\prime }/\rho =-u^{\prime }/u$, and the steady Euler momentum equation becomesIntegrating and absorbing the reference velocity into the constant yields, for the Mach number $\mathcal{M}=u/c_s$,For the opposite flow direction use $\mathcal{M}=|u|/c_s$; the same relation holds.

A smooth sonic point has $u=c_s$, so its momentum equation requires ${\Phi }^{\prime }=0$. Moreover $f(\mathcal{M})={\mathcal{M}}^2/2-\ln \mathcal{M}$ satisfies $f^{\prime }(1)=0$, $f^{\prime \prime }(1)=2$ and has its minimum $f(1)=1/2$. Thus, if the flow reaches $\mathcal{M}=1$ at $x_{\ast }$,A sonic crossing is therefore possible at a maximum of the potential, withAt a nondegenerate maximum, the expansions give $(\mathcal{M}-1{)}^2=-{\Phi }^{\prime \prime }(x_{\ast })(x-x_{\ast }{)}^2/(2c_s^2)+\cdots$. Choosing either smooth sign gives a crossing slope ${\mathcal{M}}^{\prime }(x_{\ast })=\pm \sqrt{-{\Phi }^{\prime \prime }(x_{\ast })/(2c_s^2)}$, joining a subsonic flow to a supersonic flow, or the reverse. This establishes the plane-parallel isothermal sonic transition at a potential maximum. For a globally defined crossing, the same maximum must also be high enough for the Bernoulli relation to remain real over the intended domain.

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

Assume a stationary spherical outflow with speed $v>0$, mass-loss rate $\dot{M}>0$, an optically thick thermal radiation field and negligible radiation inertia compared with matter inertia. The stellar wind continuity equation givesUse the printed gas fraction $\beta =P_{\mathrm{gas}}/P$, with $P=P_{\mathrm{gas}}+P_{\mathrm{rad}}$, and assume constant mean molecular weight ${\mu }_{\mathrm{mol}}$. Put $\mathcal{R}=k_B/({\mu }_{\mathrm{mol}}{m}_u)$ and $a_g^2=\mathcal{R}T$ for the isothermal sound speed squared. Then $P_{\mathrm{gas}}=\rho a_g^2$ and $P_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$. The momentum equation including the radiation-pressure gradient isDo not also add a separate radiative force here: the diffusion radiation-pressure gradient already represents that force.

Using radiative diffusion in a star,Write $g=G{M}_r/r^2$ and $\Gamma =L_r/L_{\mathrm{crit}}=\kappa L_r/(4\pi cG{M}_r)$, the local ratio to the Eddington luminosity. Substituting continuity into the gas-pressure gradient gives the gas and radiation pressure stellar wind equationThe denominator changes sign at the sonic point $v=a_g$. A smooth transonic branch must have its right-hand numerator vanish there too. In a subsonic flow the denominator is negative, so acceleration requires a negative numerator; for a supersonic flow it requires a positive numerator. A regular subsonic-to-supersonic branch must therefore pass through a compatible common zero, with a finite slope obtained by differentiating the equations. Boundary conditions and that regularity requirement normally select the mass flux. Other branches can remain subsonic, remain supersonic or encounter a singular gradient. In the isothermal nonradiating limit the numerator is $2a_g^2/r-GM/r^2$, recovering the Parker wind equation and critical radius $GM/(2a_g^2)$.

To expose the role of $\beta$, rewrite the diffusion temperature gradient asHenceAt the stipulated local value $\Gamma =1$, with $0<\beta <1$, its numerator isThe gas-pressure contribution is not removed by setting the luminosity equal to the Eddington value. Instead the negative temperature gradient makes the thermal term in the numerator positive as well. A subsonic outflow then has $dv/dr<0$, and no common numerator/denominator zero is available for smooth sonic passage. This is the sonic-point obstruction at the Eddington luminosity.

Thus a subsonically launched wind cannot accelerate smoothly to supersonic speed under these diffusion assumptions at $L_r/L_{\mathrm{crit}}=1$. It is not a claim that every initially supersonic solution decelerates; those have positive denominator and accelerate in this equation. Even in the radiation-dominated limit $\beta \to 0$, the spherical thermal term remains positive for $a_g^2>0$. The singular pressureless limit is not a finite-temperature sonic crossing. Finally $a_g$ is the thermal sound scale of this diffusion-controlled stationary equation, not the full adiabatic sound speed of a tightly trapped gas-plus-radiation perturbation. Optically thin winds or different heating/driving mechanisms require another transport model.

Project the steady Euler equations for an inviscid fluid along a streamline. With no gravity,For an isentropic fluid, $dp=c_s^{2}d\rho$, soThe scalar mass flux is $j=\rho u$, and henceIt increases with speed in subsonic flow and decreases with speed in supersonic flow.

A sonic flow has Mach number exactly one: the local fluid speed relative to the body equals the local sound speed. In uniform gas flowing at $U=c$ past a resting body, an upstream-directed acoustic wave has zero body-frame velocity. A sonic point is a location where this equality holds. In the moving-point acoustic wavefronts construction, old spheres touch the source at the origin; the Mach angle tends to $90^{\circ }$ as the supersonic flow approaches this threshold.

A supersonic wind tunnel produces controlled supersonic flow in its test section. Its nozzle geometry and imposed pressure conditions set the test Mach number.