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In a steady axisymmetric barotropic fluid with purely azimuthal velocity $r\Omega {\mathbf{e}}_{\varphi }$ and a conservative body force, the barotropic enthalpy function obeys $\nabla (w+\Phi )=r{\Omega }^2{\mathbf{e}}_r$. Taking its curl yields the displayed cylindrical rotation law. The first integral is $w+\Phi -\int r{\Omega }^{2}dr=C$ in a connected fluid region. Baroclinic pressure and density need not satisfy this conclusion.

Steady spherical continuity equation gives $r^2\rho u=\text{constant}$. Combining this with Euler equations for an inviscid fluid and $c_s^2=dp/d\rho$ gives the displayed equation. Smooth transonic passage requires $u^2=c_s^2$ and ${\Phi }^{\prime }=2c_s^2/r$ simultaneously. The barotropic enthalpy function gives the integral $u^2/2+h+\Phi =\text{constant}$.

For steady inviscid motion of a barotropic fluid in a conservative gravitational potential $\Phi$, the Euler equations for an inviscid fluid imply the Bernoulli equationalong each streamline. The barotropic enthalpy function $h$ has units of energy per unit mass. For an isentropic ideal gas, $p=K{\rho }^{\gamma }$ gives $h=\gamma p/[(\gamma -1)\rho ]$ up to an additive constant. For constant density, the corresponding term is $p/\rho$. Pressure work converts between kinetic, enthalpy and gravitational energy.

A converging subsonic nozzle can accelerate a fluid at the expense of its enthalpy; in the incompressible horizontal approximation, higher speed accompanies lower static pressure. Together with conservation of mass, this explains a Venturi effect measurement. Compressible nozzle flow additionally needs its equation of state and sonic conditions. In a stellar wind, the potential term is $-GM/r$, so enthalpy or mechanical energy can be spent escaping gravity. A pressure-negligible marginal-escape wind has $v^2/2-GM/r=0$.

The constant need not be the same on different streamlines. Irrotational barotropic flow in a connected region gives a common value; a general rotational flow does not. A pump, viscosity, a radiating shock, or heat addition changes the relevant energy balance. In particular, adiabaticity alone does not allow one to drop the enthalpy term.

Assume an inviscid barotropic fluid acted on by a conservative body force $-\nabla \psi$. Define its barotropic enthalpy function byThe Euler equations for an inviscid fluid are then ${\partial }_t\mathbf{v}+(\mathbf{v}\cdot \nabla )\mathbf{v}=-\nabla (H+\psi )$. Using the vorticity $\mathbf{\omega }=\nabla \times \mathbf{v}$ and the vector identitygives ${\partial }_t\mathbf{v}+\nabla B=\mathbf{v}\times \mathbf{\omega }$, where $B=v^2/2+H+\psi$.

For steady flow, dotting this equation with $\mathbf{v}$ yields $\mathbf{v}\cdot \nabla B=0$. Thus Bernoulli's equation isThe constant need not agree between different streamlines. If the flow is also irrotational, the vector equation gives $\nabla B=0$, so one common constant applies throughout a connected flow region.

For an isentropic process with $p=K{\rho }^{\gamma }$, $H=\gamma p/[(\gamma -1)\rho ]$ up to an additive constant, equal to the thermodynamic specific enthalpy. For an externally maintained isothermal gas, $H=a^2\ln (\rho /{\rho }_{\mathrm{ref}})$; this pressure potential is not the constant thermodynamic enthalpy of a fixed-temperature ideal gas.

If steady flow is not assumed, a useful additional case is unsteady potential flow $\mathbf{v}=\nabla \varphi$. The same vector equation integrates spatially toGeneral unsteady barotropic flow does not possess the steady streamline constant without further conditions.

Following a parcel. Let $\mathbf{X}(t)$ be a fluid pathline, with $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The multivariable chain rule gives the material derivative,The first term measures change at fixed position; the advection term accounts for the parcel moving through spatial gradients. The derivative is also called the Lagrangian derivative.

The forces on a parcel. A material volume $dV$ has mass $\rho dV$. The pressure forces on its boundary sum to $-\nabla pdV$, while the gravitational acceleration is $-\nabla \psi$. Newton's second law therefore gives the Euler equations for an inviscid fluid,This neglects viscosity and other body forces. The left side is the parcel's acceleration, not just the time derivative at a fixed spatial location.

Bernoulli conservation. The claimed streamline result needs steady flow and a barotropic fluid, $p=p(\rho )$, or an equivalent enthalpy relation along each isentropic streamline. These hypotheses are necessary; the unsteady Euler equation alone does not imply the stated conservation law. Define the barotropic enthalpy functionDot the steady Euler equation with $\mathbf{u}$, using $\mathbf{u}\cdot [(\mathbf{u}\cdot \nabla )\mathbf{u}]=\mathbf{u}\cdot \nabla (u^2/2)$. This givesThus the Bernoulli function isEvery term has units of energy per mass, ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. Different streamlines can have different constants.

The spherical accretion equation. Write the inward speed as $u(r)>0$, so the velocity vector is $-u(r)\hat{\mathbf{r}}$ and the positive mass accretion rate isAssume a prescribed central potential $\psi =-GM/r$ and negligible gas self-gravity. The steady radial Euler equation isFor a barotropic equation of state, $dp/dr=c_s^{2}d\rho /dr$, with $c_s^2=dp/d\rho$. Eliminate the density derivative to obtainA smooth sonic point requires the right side to vanish where $u=c_s$, henceFor a nonisothermal gas this is an implicit relation involving the local sound speed. In the isothermal case $c_s$ is constant and this is the Bondi sonic point.

The isothermal integral. For $p=c_s^2\rho$, choose the enthalpy reference so that $h_b=c_s^2\ln (\rho /{\rho }_{\infty })$. The reservoir is at rest at infinity and $\psi (\infty )=0$, making its Bernoulli constant zero. ThusorThe logarithm has a dimensionless argument. Constant temperature during compression generally requires heat removal, so this is Isothermal Bondi accretion, rather than adiabatic accretion with $\gamma >1$.

The critical accretion rate. Select the smooth reservoir-connected transonic branch. Evaluate the last relation at $u_s=c_s$ and $GM/r_s=2c_s^2$:Mass conservation then gives the isothermal Bondi accretion rate,The units are $\mathrm{kg}{\mathrm{s}}^{-1}$. Equivalently $\dot{M}=4\pi \lambda G^2{M}^2{\rho }_{\infty }/c_s^3$, with $\lambda =e^{3/2}/4\simeq 1.12$.

The inner-boundary selection of isothermal Bondi flow is important: a standard absorbing accretor inside the sonic radius selects this critical solution. The steady equations and ${\rho }_{\infty }$ alone also admit subcritical subsonic branches supported by a different inner boundary; they do not by themselves fix a unique rate. The usual answer presumes transonic capture.

The centripetal acceleration of the purely rotating barotropic fluid is $-r{\Omega }^2{\mathbf{e}}_r$. For $p=K{\rho }^2$, the pressure force satisfies ${\rho }^{-1}\nabla p=2K\nabla \rho$, so its barotropic enthalpy function is $w=2K\rho$. The equilibrium equation is thereforeTaking its curl gives ${\partial }_z(r{\Omega }^2)=0$. Thus ${\Omega }^2$ is independent of z; for a continuous, consistently signed rotation law, so is $\Omega$. This proves the barotropic cylindrical rotation theorem. Integrating the remaining radial force yieldsThe constant C is common to a connected fluid region.

Interpret the semi-thickness as a free surface with $\rho (r,\pm H)=0$, taking the exterior pressure to be negligible. At $H=ϵr$, the gravitational potential on this surface is ${\Phi }_s=-GM/[r\sqrt{1+{ϵ}^2}]$. The first integral evaluated at the surface says $C={\Phi }_s(r)-\int r{\Omega }^{2}dr$. Differentiating along it givesand therefore $\boxed{{\Omega }^2=(1+{ϵ}^2{)}^{-1/2}GM/r^3}$. Radial pressure support makes this slightly slower than Keplerian rotation for nonzero thickness.

Subtracting the free-surface first integral from its value inside the disc gives the complete density profileIt is nonnegative for $K>0$. Integrating over height, with $y=z/r$, givesThus the surface density is independent of r in this quadratic-polytrope disk with constant aspect ratio. The factor $1/r$ in density is canceled by the scale r in the vertical extent. The free-surface assumption is essential to this determination; a prescribed nonzero boundary pressure would require its own boundary data. No radial edges are specified, so the formula describes the local equilibrium rather than a finite-mass disc extended to all radii.

Write the inward radial velocity as $u_R=-v(R)$, with $v>0$. Steady spherical mass conservation and the radial Euler equations for an inviscid fluid giveFor a barotropic fluid, define the sound speed $c_s^2=dp/d\rho >0$ and the barotropic enthalpy function by $dh=dp/\rho$. Choosing $\Phi (\infty )=0$, integration gives the Bernoulli equationThe continuity equation gives ${\rho }^{\prime }/\rho =-2/R-v^{\prime }/v$. Eliminating mass density from the momentum equation produces the barotropic spherical accretion equationThe sonic point is where the inward Mach number $v/c_s$ equals one. A smooth solution cannot cross the vanishing coefficient on the left with a nonzero right-hand side, so a regular sonic radius must obeyConversely a continuous flow that is subsonic at infinity and supersonic inward must cross unit Mach number; smooth passage selects a simultaneous zero of the two sides, hence a regular sonic point. The associated critical accretion branch, rather than an arbitrary subsonic solution with smaller flux, is selected by these outer and inner requirements.

For the polytropic equation of state, set $\Gamma =1+1/m$ to distinguish its exponent from any separately specified perturbation exponent. Assume $K>0$ and first take the usual positive index $m>0$. ThenFor the attractive mixed potential, assume $\lambda ,\mu >0$. The sonic condition and Bernoulli equation giveHence polytropic accretion in a mixed inverse-power potential reduces toFor $0<m\le 1/2$ all nonzero coefficients in the left side are nonnegative for $R_s>0$, with at least one strictly positive: there is no finite positive root. For $m>1/2$ the constant term is negative and the leading coefficient positive, giving exactly one positive root. Define $B=(2m-3)\mu /4$ and $C=(2m-1)\lambda /2$. ThenThis is the finite-radius transonic range for the two positive attractive terms.

The critical point has real distinct crossing slopes. To check this rather than just solve the sonic equalities, put $U=\lambda /R_s^2$, $V=\mu /(2R_s)$ and $x=R_s{v}_s^{\prime }/v_s$. Differentiating the barotropic spherical accretion equation at its simultaneous zero givesIts discriminant is $8m[2(2m-1)U+(2m-3)V]/(U+V)$. The sonic Bernoulli equation says $(2m-1)U+(2m-3)V=2m{c}_{\infty }^2$, so this discriminant is strictly positive for $m>1/2$. The accretion crossing is the slope for which $d\log (v/c_s)/d\log R<0$, namely the smaller root. The two slopes have opposite signs of this Mach-number derivative. This is the regular transonic saddle, not a singular vertical tangent.

The corresponding mass accretion rate follows without an additional arbitrary mass density normalization:One may equivalently replace ${\rho }_s$ by $[c_{s,s}^2/(\Gamma K){]}^m$. With a nonzero $\lambda$ term, inner free fall has $v\sim \sqrt{2\lambda }/R$ and continuity gives $\rho \propto R^{-1}$. Thus $v^2/c_s^2\propto R^{-2+1/m}$ increases without bound precisely in the positive-index range $m>1/2$, consistent with the required inner supersonic branch.

The assumptions on the parameters matter. If $\lambda =0$ and $\mu >0$, the finite sonic radius instead requires $m>3/2$, the usual Bondi accretion range; $m=3/2$ is the zero-radius limiting critical case. If $\mu =0$ and $\lambda >0$, the condition remains $m>1/2$. With arbitrary signs of $\lambda ,\mu$, one must use the quadratic and retain only positive roots with $c_{s,s}^2>0$ and real admissible crossing slopes; attraction was not explicitly given as a sign condition in the source.

If positive polytropic index was not intended as an implicit physical convention, there is a further mathematical range. For $K>0$, positive compressibility also permits $m<-1$, giving $0<\Gamma <1$. Both potential terms attractive then give a positive sonic root as well:The same mass accretion rate formula applies; its crossing-slope discriminant is positive. The range $-1<m<0$ has $c_s^2<0$ and no physical acoustic sonic point, $m=-1$ has constant pressure and zero sound speed, and $m=0$ does not define the printed equation of state. Thus the usual answer $m>1/2$ assumes $m>0$ and positive attractive coefficients; these conventions should not be confused with additional printed restrictions.

Let $u=u_r<0$ for accretion. The steady spherical continuity equation and radial Euler equations for an inviscid fluid giveFor a barotropic fluid, $dp/dr=c_s^{2}d\rho /dr$. Dividing continuity equation by $r^2\rho u$ gives ${\rho }^{\prime }/\rho =-2/r-u^{\prime }/u$. Eliminating this derivative yieldsAt a smooth transonic critical point, the coefficient of $du/dr$ vanishes. A finite derivative then requires the right side to vanish simultaneously:These are the sonic and regularity conditions. Otherwise the derivative is singular and the proposed smooth transonic passage fails. The local slope must also be a real root compatible with the desired branch. For example, defining $C=\rho d(c_s^2)/d\rho$ at the point and differentiating both sides givesThis explains why simultaneous vanishing is a necessary regularity condition, rather than an automatic proof that every potential admits a critical crossing.

Define the barotropic enthalpy function by $dh=dp/\rho$. The momentum equation integrates to the Bernoulli equationChanging the reference mass density in $h$ changes only the constant $B$. For an isothermal closure this logarithmic function is a barotropic barotropic pressure potential; it should not be confused with the constant thermodynamic specific enthalpy of an ideal gas held at fixed temperature.

For isothermal transonic accretion in the Paczyński-Wiita potential, assume $M>0$ and $c_s>0$, and define $r_B=GM/(2c_s^2)$ and $\beta =16c_s^2/c^2$. The critical equation iswhose two roots are $r_G+\frac{1}{2}{r}_B(1\pm \sqrt{1+\beta })$. Only the plus root lies outside $r_G$. ThereforeFor this isothermal case $C=0$, and the slope relation reduces toThe positive slope is the inward-accretion solution connecting a small inward speed at large radius to the supersonic inward branch. Thus the exterior critical point is a genuine nondegenerate transonic point.

Choose the reference mass density as ${\rho }_0$, so $h=c_s^2\log (\rho /{\rho }_0)$ and $B=0$ from the conditions at infinity. At $u_c=-c_s$,Substitution into $\dot{M}=4\pi r_c^2{\rho }_c{c}_s$ givesThis is the rate selected by the smooth transonic accretion solution. Arbitrary static or subsonic formal solutions are not assigned this rate merely by specifying conditions at infinity. As $r_G/r_B\to 0$, the result tends to the isothermal Bondi accretion rate $\pi e^{3/2}{\rho }_0(GM{)}^2/c_s^3$, a useful normalization check.

In the inertial frame the steady convective acceleration is $-R{\Omega }^2{\mathbf{e}}_R$. For a barotropic fluid, $\nabla p/\rho =\nabla h$, where $h$ is the barotropic enthalpy function. The equilibrium momentum equation therefore giveson each connected fluid region. Uniform rotation is essential: it makes the centrifugal acceleration derivable from the centrifugal potential $-{\Omega }^2{R}^2/2$.

The Cowling approximation sets the gravitational-potential perturbation to zero while retaining background gravity. The barotropic pressure force linearizes as $\delta (\nabla p/\rho )=\nabla \delta h$, with $\delta h=p^{\prime }/\rho =c_s^2{\rho }^{\prime }/\rho$. Denote its mode amplitude by $W$. The advective time derivative on a scalar mode is ${\partial }_t+\Omega {\partial }_{\varphi }$, so the stated positive-frequency exponential produces $i\sigma$ with $\sigma =\omega +m\Omega$. Linearizing the radial centrifugal term produces $-2\Omega v_{\varphi }$, while linearizing the azimuthal convective term produces $2\Omega v_R$. ThusThese are the pressure-gradient and Coriolis acceleration terms in the corotating-frame form. Independently, linearizing continuity equation givesThe equilibrium mass density is axisymmetric, so no additional azimuthal background-density derivative appears.

For $\sigma \ne 0$ and $D=4{\Omega }^2-{\sigma }^2\ne 0$, invert the horizontal momentum system:Inserting these into the continuity equation, dividing by $i\sigma$ and multiplying by $D$ givesThe two terms proportional to $\rho W_R$ cancel; this cancellation leaves precisely the background-density derivative in the last term. Since $-D/{\sigma }^2=1-4{\Omega }^2/{\sigma }^2$, this is exactlyThe singular frequency cases require the original velocity equations rather than this inversion.

In the low-frequency anelastic approximation for a rotating barotropic star, neglect the left side. Spherical background mass density satisfies ${\rho }_R/R={\rho }_z/z$, understood by its smooth limiting form on the equatorial plane. For $W=z{R}^m$,The remaining equation reduces toWriting $q=2\Omega /\sigma$, the bracket factors as $(m+1-q)(q+1)$. The regular desired branch is thereforeThe other algebraic factor corresponds to $\sigma =-2\Omega$, where the eliminated horizontal system is singular; it is not classified by this pressure-equation inversion. A constant-density background makes the bulk factor identically zero but does not invalidate the displayed solution.

For the regular branch, an explicit velocity check is particularly informative. Up to a common mode normalization,It satisfies $R{v}_R+z{v}_z=0$ and $\nabla \cdot \mathbf{v}=0$. Thus $\nabla \cdot (\rho \mathbf{v})=0$ for every spherical mass density profile, directly verifying the anelastic continuity condition. The motion is tangential to spherical shells and is the sectoral inertial mode of a slowly rotating barotropic star. For single-valued azimuthal modes, $m$ is a positive integer. With the source's exponential convention,For $m\ge 2$ the pattern is retrograde relative to the star but prograde in the inertial frame. This is a leading slow-rotation anelastic mode, not an exact solution of the compressible equations whose left-hand term was discarded.

For $p=K{\rho }^{1+1/m}$, the barotropic enthalpy function is $m{c}_s^2$. In $\Phi =-\lambda /R^2-\mu /R$, a regular sonic point obeys $c_{s,s}^2=\lambda /R_s^2+\mu /(2R_s)$, and its Bernoulli integral yields the displayed quadratic. For $K,\lambda ,\mu >0$ and the usual positive polytropic index, its unique positive root exists precisely when $m>1/2$. The mass accretion rate is $4\pi R_s^2{\rho }_{\infty }{c}_{s,s}(c_{s,s}^2/c_{\infty }^2{)}^m$. If indices with $0<1+1/m<1$ are admitted, $m<-1$ also permits a positive sonic root; negative compressibility indices do not.

In a connected inviscid uniformly rotating barotropic fluid, momentum balance makes the displayed sum constant. The pressure term is the barotropic enthalpy function. Perturbations experience Coriolis acceleration, and their rotating-frame frequency depends on the chosen exponential sign convention.