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In a nonzero steady one-dimensional flow with an isothermal equation of state with $p=c_s^2\rho$ and constant cross section, the continuity equation gives $\rho u=\text{constant}$. Eliminating density from momentum gives $(u-c_s^2/u)u^{\prime }=-{\Phi }^{\prime }$. With Mach number $\mathcal{M}=|u|/c_s$, integration yields ${\mathcal{M}}^2/2-\ln \mathcal{M}+\Phi /c_s^2=C$. The function $f(\mathcal{M})={\mathcal{M}}^2/2-\ln \mathcal{M}$ has minimum $1/2$ at $\mathcal{M}=1$, so a smooth transonic crossing at $x_{\ast }$ requires a local maximum of $\Phi$ and $C=1/2+\Phi (x_{\ast })/c_s^2$. At a nondegenerate maximum, $f=1/2+(\mathcal{M}-1{)}^2+\cdots$ gives ${\mathcal{M}}^{\prime }(x_{\ast })=\pm \sqrt{-{\Phi }^{\prime \prime }(x_{\ast })/(2c_s^2)}$.

The radial Mach number is the source velocity projected toward the observer, divided by the sound speed. It enters the moving-surface retarded Jacobian through $d|x-y(\tau )|/d\tau =-\hat{r}\cdot v$. The sign is directional: a source moving toward the observer has positive $M_r$ and compressed arrival times.

In a stationary, horizontally homogeneous wall boundary layer, Hamiltonian ray-tracing equations conserve $\omega$ and $k_x$. If $U$ increases away from the wall, downstream-oriented $k_x>0$ waves have decreasing positive intrinsic frequency and decreasing $k_y^2$. At $U=\omega /k_x-c_0$, the transverse wavenumber vanishes and the acoustic ray turns. For $U(0)=0$, a downstream launch angle ${\beta }_0$ above the wall can reach a uniform stream of Mach number $M_{\infty }$ only if $\cos {\beta }_0\le (1+M_{\infty }{)}^{-1}$. This uses the positive acoustic branch, constant sound speed and no intervening critical layer; a turning point itself requires a uniform local approximation rather than the ordinary WKB method.

A point source moving at angular speed $\Omega$ on a circle of radius $a$ has observer-direction Mach number $m=\Omega a\sin \iota /c_0$. In the far field its retarded phase satisfies $\Theta +\theta -\pi /2=\Omega (t-r/c_0)+m\sin \Theta$. For $m<1$ there is one retarded root, and a constant source strength gives ${\rho }^{\prime }=q\dot{\Theta }/(4\pi c_0^2\Omega r)$. For $m>1$, multiple retarded roots can occur; their contributions require $|1-m\cos \Theta |$ and a sum over roots. Coalescing roots produce a caustic in the ideal point-source model.

For a nonseparating small body in a smooth nearly inviscid eddy of speed $U$ and length $\ell$, the fluctuating resultant is an added mass or pressure-gradient force $F\sim {\rho }_0{a}^3{U}^2/\ell$. A uniform pressure integrates to zero over a closed surface. With time $\ell /U$, its acoustic dipole power divided by the eddy's acoustic quadrupole power is $(a/\ell {)}^6/M^2$, where $M=U/c_0$ is the Mach number. Equal powers therefore give the displayed threshold. An area-scaled separated drag force with the same eddy time instead gives $a/\ell \sim \sqrt{M}$; a body-scale time gives a different criterion. The threshold depends on the specified loading and time models.

A sonic point is a point at which the flow speed equals the local sound speed, so the Mach number is one.

A subsonic flow has Mach number less than one, so its speed is below the local sound speed.

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

Consider steady isentropic flow toward a Newtonian gravitational potential $-GM/r$ through a tube of cross-sectional area $A(r)=C{r}^n$. Write $v>0$ for the inward speed and use a polytropic equation of state with specific-heat ratio $\gamma >1$. Mass conservation and the Euler equations for an inviscid fluid givewhere $c_s$ is the adiabatic sound speed. A regular sonic point therefore has $v_s=c_s$ and $r_s=GM/(n{c}_s^2)$. The Bernoulli equation, matched to a nearly stationary reservoir with sound speed $c_0$, givesFor $n>1/2$, a finite positive sonic point requires $1<\gamma <(2n+1)/(2n-1)$. Differentiating the flow equation at that point, with $x=r_s{v}_s^{\prime }/c_s$, givesIts discriminant is $4n[(2n+1)-(2n-1)\gamma ]$, so the same bound permits real regular slopes. The branch on which the Mach number rises inward selects the negative sign inFor a spherical tube $n=2$, this recovers the $5/3$ threshold of Bondi accretion; for a dipolar flux-tube area, $n=3$ gives $7/5$. The tube approximation must remain valid between the reservoir matching region and the accretor.

In two dimensions, the outgoing harmonic Helmholtz equation kernel has magnitude proportional to $(k_{0}r{)}^{-1/2}$. Combining this spreading with the compact acoustic quadrupole source and $k_0\ell =O(M)$ changes the density-amplitude Mach number power to $7/2$ when geometric range is separated.