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For steady spherical polytropic flow, write $u=u_r$ and use the polytropic equation of state $p=K{\rho }^{\gamma }$, with $K>0$. Mass conservation and radial Euler momentum equation areThe mass equation gives ${\rho }^{\prime }/\rho =-2/r-u^{\prime }/u$. Substitute it into momentum balance to obtainAt a sonic point the derivative coefficient vanishes. A smooth finite-slope solution must make the numerator vanish there too:Finally $dp/\rho =d[\gamma K{\rho }^{\gamma -1}/(\gamma -1)]$. Integrating momentum gives the Bernoulli equationThis is kinetic energy plus specific enthalpy plus gravitational potential per unit mass. The polytropic stellar wind selects a particular transonic branch of these equations.

The outward mass loss rate is $\dot{M}=4\pi r^2\rho u>0$. Put $\beta =4/(\gamma +1)$, $\alpha =4(\gamma -1)/(\gamma +1)$ and $y=\rho r^{\beta }$. Both terms in the Bernoulli equation then have the same radial scaling: $\beta (\gamma -1)=4-2\beta =\alpha$. Eliminate $u$ and multiply by $r^{\alpha }$ to obtainFor positive terminal speed, $C>0$. Differentiation gives $F^{\prime }=r^{\alpha -2}[\alpha Cr-(1-\alpha )GM]$. It has an interior minimum precisely when $0<\alpha <1$, or $1<\gamma <5/3$, atAt a sonic point, the Bernoulli equation and $GM/r_c=2c_{s,c}^2$ give $C=c_{s,c}^2(5-3\gamma )/[2(\gamma -1)]$, recovering this radius. The limiting $C=0$, $\gamma =5/3$ case has no isolated interior minimum selected by these formulas.

The function $G$ diverges at both ends of $y>0$ and has a unique minimum whereThis condition is $u_c^2=c_{s,c}^2$. Thus a smooth crossing requires minimum matching at a polytropic sonic point: $F(r_c)=G(y_c)$. If the minimum of $F$ is lower, an interval has no real positive solution; if it is higher, the two $y$ branches stay separate and cannot cross the sonic value. When the minima agree, their positive second derivatives give $G^{\prime \prime }(y_c)(y-y_c{)}^2\sim F^{\prime \prime }(r_c)(r-r_c{)}^2$, allowing a finite-slope branch to pass between them. The wind with finite positive terminal speed takes the lower-$y$ branch outside $r_c$, because $\rho \sim \dot{M}/(4\pi \sqrt{2C}{r}^2)$ and $\beta <2$ make $y\to 0$.

At the matching point,Insert these into $\dot{M}=4\pi r_c^2{\rho }_c{c}_{s,c}$:The positive-$C$, $1<\gamma <5/3$ range is part of this finite-radius transonic wind result, rather than an unrestricted formula for every $\gamma >1$.

To recover the printed dispersion relation, interpret the surface maintained flat as an imposed flat upper boundary for the perturbation, so its normal velocity is zero. This is the rigid-lid approximation. Work with perturbation velocity potentials ${\varphi }_{-}=A{e}^{ky}{e}^{ikx+\sigma t}$ below the interface and ${\varphi }_{+}=B\cosh [k(y-h)]e^{ikx+\sigma t}$ in the upper layer. They solve the Laplace equation, decay as $y\to -\infty$, and obey ${\partial }_y{\varphi }_{+}(h)=0$.

The linear kinematic boundary conditions at the interface give $(\sigma +ikU){\eta }_0=kA$ and $\sigma {\eta }_0=-kB\sinh (kh)$. Equality of pressures, using the linear Bernoulli equation, gives $(\sigma +ikU)A=\sigma B\cosh (kh)$. The equal fluid densities cancel the interface gravitational restoring term. Eliminating $A,B,{\eta }_0$ yieldsWith $q=\tanh (kh)$, the roots areFor every $k>0$ one root has positive real part. Thus the vortex sheet has a Kelvin-Helmholtz instability at every wavelength in this model. If $kh\gg 1$, $q\to 1$ and the growing rate is $Uk/2$. If $kh\ll 1$, $q\sim kh$ and it is $Uk\sqrt{kh}$.

There is a physical qualification to the wording. A dynamically deformable free surface at finite gravity is not exactly a flat lid. Its linear conditions combine to ${\sigma }^2{\varphi }_{+}(h)+g{\partial }_y{\varphi }_{+}(h)=0$. Writing the upper potential as $B\cosh (ky)+C\sinh (ky)$ and using the same interface conditions instead givesThe stated quadratic follows in the strong-gravity flat-surface limit, or with the flat condition imposed exactly. It is not the exact dispersion relation of a freely moving upper surface at arbitrary finite $g$. This makes explicit the boundary interpretation needed for the requested formula.

Use the standard ideal-fluid assumptions of incompressible flow, inviscid flow, gravity $-g{\mathbf{e}}_y$, and no surface tension. Irrotational flow gives ${\mathbf{u}}_j=\nabla {\phi }_j$ in each layer, and incompressibility gives Laplace equationThe rigid-wall boundary conditions are ${\phi }_{1y}=0$ at $y=h_1$ and ${\phi }_{2y}=0$ at $y=-h_2$. At the moving interface, each fluid has the same normal velocity as the interface:The dynamic condition is continuity of pressure, $p_1=p_2$ at $y=\eta$. Bernoulli equation in each fluid reads ${\phi }_{jt}+\frac{1}{2}|\nabla {\phi }_j{|}^2+p_j/{\rho }_j+gy=C_j(t)$. Choosing the potential gauges to subtract a common reference pressure, the interface condition becomesTangential velocities need not agree across an inviscid interface.

For constant fluid density $\rho$, steady Euler equations for an inviscid fluid under a conservative force per unit mass $-\nabla \Phi$ read $(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla p/\rho -\nabla \Phi$. Here $\mathbf{u}$ is velocity, $p$ pressure and $\Phi$ potential per unit mass. Dotting with $\mathbf{u}$ givesThis is the Bernoulli equation. The constants can differ between streamlines. If density varies, the same derivation for a barotropic fluid replaces $p/\rho$ by ${\int }^{p}d{p}^{\prime }/\rho (p^{\prime })$; constant density is the appropriate water approximation here.

For the draining cone, the radius at height $h$ is $h$ and the horizontal area is $\pi h^2$. Both the free surface and the jet are at atmospheric pressure; their gravitational potential difference is approximately $gh$. The speed of the descending surface is smaller than the jet speed by the area ratio ${ϵ}^2/h^2$, so it can be neglected when $h\gg ϵ$. The Bernoulli equation then gives Torricelli's law, $U\simeq \sqrt{2gh}$. Conservation of mass givesHence the leading drainage time of a conical tank isRemoving the tip places the aperture at height of order $ϵ$, and the quasi-steady calculation fails in the last layer of that depth. Integrating down only to $h=O(ϵ)$ changes the displayed leading time by relative order $(ϵ/h_0{)}^{5/2}$; replacing the actual head $h-ϵ$ by $h$ introduces a larger, but still vanishing, relative correction of order $ϵ/h_0$. The formula is the ideal inviscid estimate under the question's stated approximation, not an exact law for the final transient or a real contracted jet.

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.

Let $F(x,z,t)=z-\eta (x,t)$. The interface is a material surface for both fluids: its material derivative vanishes on $F=0$. With $u_j={\partial }_x{\varphi }_j$ and $w_j={\partial }_z{\varphi }_j$, this givesThus the kinematic boundary condition says that particles stay on the moving interface. The tangential velocities can differ; each side has its own material derivative. Replacing both derivatives by one background speed would be incorrect.

The dynamic boundary condition for an inviscid interface is pressure continuity, because the fluids have no viscous normal stress and there is no surface tension. The unsteady Bernoulli equation in each layer of irrotational flow readsEquating these pressures at $z=\eta$ gives the required dynamic boundary condition for an inviscid interface. Choose the potential gauges so that ${\rho }_1{C}_1={\rho }_2{C}_2$, or set both constants to zero relative to a common reference pressure. For example, ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ removes the distinct uniform-stream Bernoulli constants without changing either velocity. Material-interface kinematics and pressure continuity supply the two matching conditions, with the Bernoulli gauge understood.

Let $u(t)$ be the signed air speed along the straw towards the bubble. Conservation of mass in a pipe for the incompressible plug flow givesThe unsteady Bernoulli equation along the straw, whose open end is at atmospheric pressure, givesby the Young–Laplace equation. Substitution yields

Set $w=a^2\dot{a}$. Since $a\dot{w}=-C$, the chain rule gives $wdw/da=-Ca$. The initial condition $\dot{a}(0)=0$ impliesWriting $a=a_0\sin \theta$ and integrating gives the implicit solutionAt disappearance $a=0$ and $\theta =0$, so

For an incompressible flow, $\nabla \cdot \mathbf{u}=0$. If the flow is also irrotational, then locally $\mathbf{u}=\nabla \varphi$ for a velocity potential, and therefore

The boundary conditions areThe fluid region is not simply connected, so a potential may change by the constant $\kappa$ after one circuit while its gradient remains single-valued. Superposing uniform flow, the cylinder doublet, and the circulation gives the potential flow around a circular cylinder with circulationIts velocity components are

On $r=a$, $u_r=0$ and, with $\lambda =\kappa /(4\pi Ua)$,The Bernoulli equation gives $p=p_{\infty }+\rho (U^2-u_{\theta }^2)/2$. The pressure force per unit length on the cylinder isThe $x$ component vanishes by symmetry, while ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ givesin agreement with the Kutta–Joukowski theorem.

A stagnation point satisfies $u_r=u_{\theta }=0$. On the cylinder this means $\sin \theta =\lambda$. Thus $|\lambda |<1$ gives two surface stagnation points, $|\lambda |=1$ gives one coincident surface point, and $|\lambda |>1$ gives none on the surface. Away from the cylinder, $u_r=0$ requires $\cos \theta =0$. Solving $u_{\theta }=0$ then gives one physical exterior root when $|\lambda |>1$:At $|\lambda |=1$ this root lies on $r=a$ and agrees with the single surface point.

In the narrow polar flux tube, the magnetic field lines are almost radial and the transverse width is much smaller than the radial scale. The Lorentz force density has no component along the magnetic field, so the longitudinal magnetically channelled accretion is hydrodynamic. Let $v(r)>0$ denote inward speed. Mass conservation, the Euler equations for an inviscid fluid, and the polytropic equation of state giveUsing the adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$ and $A\propto r^3$,The Bernoulli equation iswhere the right-hand side comes from matching to a nearly stationary reservoir with negligible gravitational potential. The fixed magnetic field provides transverse confinement; neglecting magnetic forces in this one-dimensional equation concerns their longitudinal component.

A regular sonic point must make both sides of the flow equation vanish:Substituting into the Bernoulli equation yieldsFor $\gamma >1$, positive finite sound speed therefore requires the critical adiabatic index for dipolar accretionTo check that this gives real regular crossings, differentiate the flow equation at the sonic point and put $x=r_s{v}_s^{\prime }/c_s$. The transonic accretion in a power-law tube calculation givesThe minus sign gives the transonic branch whose Mach number increases inward. At $\gamma =7/5$ the positive-energy reservoir cannot match a finite sonic point; for larger $\gamma$ the required $c_s^2$ is negative. A physical surface-crossing solution also requires $r_s>R_{\star }$ and validity of the narrow flux tube approximation up to the sonic region.

Take $u(r)>0$ to be the inward radial speed and select the regular critical solution for a gas reservoir of density ${\rho }_{\infty }$ and adiabatic sound speed $c_{\infty }$. Other mathematically subsonic steady solutions need a different inner boundary condition. For the polytropic equation of state, let $\gamma =1+1/n$, soMass conservation and the Bernoulli equation giveThe gas is at rest at infinity, so its kinetic and gravitational terms vanish there.

Differentiate these two conserved quantities, using $d\log \rho /dr=-2/r-u^{\prime }/u$. Eliminating the density derivative yields the polytropic Bondi accretion equationAt a regular sonic point, both sides vanish:The Bernoulli equation at that point reduces to $n{c}_{\infty }^2=(n-3/2)c_c^2$. For $n>3/2$, equivalently $1<\gamma <5/3$,Since $c_s^2\propto {\rho }^{1/n}$, the critical density is ${\rho }_c={\rho }_{\infty }[2n/(2n-3){]}^n$. Evaluate the conserved mass accretion rate at the sonic point:whereThe isothermal limit $\gamma \to 1$ gives $\lambda =e^{3/2}/4$, agreeing with Isothermal Bondi accretion.

For $\gamma =5/3$, $n=3/2$ and the Newtonian critical radius tends to zero while $c_c$ diverges. The limiting critical solution approaches Mach number one only as $r\to 0$; it has no finite-radius Newtonian sonic transition. This is a failure of the Newtonian inner approximation, not an obstruction to accretion onto a black hole. A physical Schwarzschild black hole has a finite event horizon, and the regular relativistic inflow becomes supersonic outside it.

The singular-looking factor in the rate has a finite limit. Writing $ϵ=n-3/2$, its logarithm is $ϵ\log [n/ϵ]\to 0$, so the critical Bondi accretion rate for gamma equals five thirds isThis is also the leading cold-reservoir limit of relativistic spherical accretion; an exact rate for relativistically hot gas requires the relativistic equations, not this Newtonian expression.

For completeness, the finite relativistic sonic point can be seen without assigning it an arbitrary inner radius. Let $b=c_s^2/c^2$ denote the relativistic squared sound speed and $g=\gamma -1$. For an isentropic relativistic gas with $p=K{\rho }_{\mathrm{rest}}^{\gamma }$, its dimensionless specific enthalpy is $h=g/(g-b)$. Conservation of rest-mass flux and relativistic Bernoulli energy gives the critical relationsFor $g=2/3$ and $c_{\infty }\ll c$, expansion gives $b_{\infty }\simeq (9/4)b_c^2$, henceHere the asymptotic relativistic and Newtonian sound speeds agree to leading order. This finite-radius transition explains why the gas can reach the horizon despite the Newtonian $r_c=0$ result.

Using $\gamma =\lambda$ and the Pythagorean trigonometric identity,so the speed is independent of $\theta$. Since $\gamma =\pi /\alpha >1$,and the speed tends to zero at the vertex. The steady irrotational Bernoulli equation givesThus the fluid pressure is also independent of $\theta$, withIn particular, the pressure difference between the vertex and $r_0$ is

Equation (1) is Laplace equation for the velocity potential. It expresses incompressibility, $\nabla \cdot \mathbf{u}=0$, together with $\mathbf{u}=\nabla \varphi$. Equation (2) is the linearized dynamic free-surface condition: the unsteady Bernoulli equation and constant atmospheric pressure balance the potential acceleration against the gravitational restoring term $g\zeta$. Equation (3), with the printed $c$ understood as $\zeta$, is the kinematic free-surface condition: the surface moves with the vertical fluid velocity. Equations (4) and (5) are no-penetration boundary conditions at the rigid bottom and cylindrical sidewall, respectively.

Irrotationality gives ${\mathbf{u}}_i=\nabla {\varphi }_i$ and ${\nabla }^2{\varphi }_i=0$. At $z=\zeta$, the exact kinematic conditions areand pressure continuity gives $p_1=p_2$. Bernoulli equation in each fluid is

Linearize at $z=0$ and write $\zeta =Z{e}^{i(kx-\omega t)}$. Decay at infinity givesThe kinematic conditions give $-i\omega Z=-k{A}_1=k{A}_2$, while pressure continuity givesElimination yieldsStable waves require the lower fluid to be denser, ${\rho }_2>{\rho }_1$; otherwise this is the Rayleigh-Taylor instability.

A velocity potential satisfying uniform flow at infinity and no radial flow through $r=a$ isThereforeOn the cylinder, $u_r=0$ and $u_{\theta }=-2U\sin \theta$. The steady Bernoulli equation, compared with the uniform flow at infinity, givesHencethe pressure in potential flow around a circular cylinder.