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Use the leading vertically averaged horizontal momentum equations of a thin disk. For an isothermal ideal gas, write $P=\int pdz=c_s^2\Sigma$, with constant isothermal sound speed $c_s$. The disk scale height is of order $H=c_s/{\Omega }_z$; this is exact for local isothermal vertical hydrostatic balance and remains the thickness scale for the estimate. Put $ϵ=H/r\ll 1$. The alpha disk prescription givesAt fixed radii of order $a$ inside the plunge and away from the marginal orbit, $|v_r|,v_{\varphi }$ and $r{\Omega }_z$ are fixed numerical multiples of $(GM/a{)}^{1/2}$. Hence the inflow time is dynamical, while the viscous timescale is longer by $1/(\alpha {ϵ}^2)$.

The Paczyński-Wiita ballistic plunge obeys $v_r{v}_r^{\prime }-v_{\varphi }^2/r=-{\Phi }^{\prime }$ and $v_r{j}_0^{\prime }=0$ exactly. For smoothly varying disk quantities on scale $r$, the pressure correction to radial acceleration has magnitude $c_s^2/r$, and the viscous correction is of order $\nu v/r^2$. Relative to the orbital acceleration scale $r{\Omega }_z^2$, these are respectivelyFinite-height corrections to the mid-plane gravitational force are also $O({ϵ}^2)$ by reflection symmetry. Thus the leading horizontal flow is ${\mathbf{u}}_h=\mathbf{v}+O({ϵ}^{2}v)$, with specific angular momentum differing from $j_0$ by a fraction $O(\alpha {ϵ}^2)$ over a dynamical transit, for $\alpha \lesssim 1$. Vertical structure and the smaller vertical velocity must be supplied by the thin-disk expansion; they are not specified by the planar ballistic expression.

This estimate is not uniform at the marginal orbit. In the ballistic solution, steady mass conservation givesConsequently $d\ln \Sigma /dr=[2(r-a){]}^{-1}+3[2(3a-r){]}^{-1}$. Writing $\delta =3a-r\ll a$, the pressure acceleration scales as $c_s^2/\delta$, while the net ballistic radial force scales as $(GM/a^2)(\delta /a{)}^2$. Their ratio is $O[{ϵ}^2(a/\delta {)}^3]$, evaluated with the characteristic thickness near $3a$. A transonic matching layer of width $\delta /a=O({ϵ}^{2/3})$ is therefore needed. The approximation concerns the bulk plunging flow outside that layer, not a uniform relative approximation arbitrarily close to $3a$.

For the torque comparison, the steady angular-momentum flux is $\mathcal{G}-\dot{M}j=\mathrm{constant}$, where $\mathcal{G}=-2\pi \nu \Sigma r^3{\Omega }^{\prime }$ is the viscous torque in an accretion disk. Inside the plunge, $\Omega =j_0/r^2$, soOutside, the approximately circular disk has $\mathcal{G}=\dot{M}[j(r)-j_0]+O(\alpha {ϵ}^2\dot{M}{j}_0)$, since the inner plunge torque is small. At $4a$,a fixed number independent of disk thickness. At $2a$, radial infall is dynamical and the inner torque is $O(\alpha {ϵ}^2\dot{M}{j}_0)$. Therefore the weak viscous torque in a thin-disk plunging region estimate givesThe numerical prefactor need not be small; the suppression is an asymptotic statement in the thin-disk parameter. Rapid infall and its low surface density explain why a nearly zero inner torque is a useful leading boundary condition for the outer accretion disk.

Reflection symmetry gives ${\Phi }_z(R,0)=0$. In a vertically stable thin disk, expand the vertical force as ${\Phi }_z(R,z)={\Omega }_z^{2}z+O(z^3)$, where ${\Omega }_z^2={\Phi }_{zz}(R,0)$ is the vertical epicyclic frequency squared. For local hydrostatic equilibrium and height-independent isothermal sound speed, $p=c_s^2\rho$ andThus $\rho ={\rho }_0\exp [-z^2/(2H^2)]$ with $H=c_s/{\Omega }_z$. Integrating the pressure and mass density through the disk gives the vertically integrated pressure of an isothermal disk:The force expansion is approximate in $H/R$, while $P=c_s^2\Sigma$ is exact for a height-independent sound speed.

In a nearly Keplerian disk, the vertical oscillation and radial epicyclic motion frequencies are close to the orbital frequency. The exact leading orbital differences are $\Omega -{\Omega }_z$ and $\Omega -\kappa$, soThus ${\omega }_1$ is the nodal precession rate of a slightly tilted orbit, and ${\omega }_2$ is the apsidal precession rate of a slightly eccentric orbit. Their signs distinguish prograde from retrograde precession. This also agrees with the sign convention in the complex tilt: a freely precessing ring obeys $W_t=i{\omega }_{1}W$.

For exactly Keplerian motion, ${\Omega }_z=\kappa =\Omega$, so ${\omega }_1={\omega }_2=0$ and $P/\Sigma =c_s^2$. In the local approximation, freeze the slowly varying coefficients and take $W,G\propto e^{i(kR-\omega t)}$. The inviscid warp equations of a nearly Keplerian disk becomeEliminating $G$ gives ${\omega }^2=k^{2}P/(4\Sigma )=k^2{c}_s^2/4$. HenceThese are bending waves of a Keplerian disk, carrying changes in orbital tilt. The local approximation needs wavelength small compared with $R$; the secular thin-disk warp equations also describe the long-vertical-wavelength regime $H\ll \lambda \ll R$.

For a nonspinning hole, $a=0$ gives ${\omega }_1=0$. A steady disk warp then satisfies $G_R=0$. The zero inner torque forces $G=0$ throughout, and the second warp equation, with positive $P$, forces $W_R=0$. The outer orientation fixes $W(R)=W_{\infty }$ everywhere. An apsidal precession term by itself therefore does not require a tilt warp.

For nonzero spin, set the time derivatives to zero. The two equations giveEliminating the internal torque yieldsUse ${\omega }_2=3\Omega /r$, ${\omega }_1=2a\Omega r^{-3/2}$, ${\Omega }^2{R}^3=GM$, and $P\simeq {\Omega }^2\Sigma H^2$. The first coefficient becomes $GM\Sigma H^{2}r/12$ and the second becomes $2aGM\Sigma r^{-3/2}$. ThereforeThis is the stationary disk warp equation under the stated weak-precession approximations. The inner condition is $W_R(R_{\mathrm{in}})=0$ when its coefficient is finite and nonzero, because $G=iGM\Sigma H^{2}r{W}_R/12$.

Let $R_g=GM/c^2$, so $R=R_{g}r$, and impose $H=ϵR$, $\Sigma \propto R^{-3/4}$. Converting both radial derivatives carefully to $r$ and dividing out the common positive factor givesor $W_{rr}+9W_r/(4r)+(24a/{ϵ}^2)r^{-9/2}W=0$. For $a>0$, introduceSince $x_r$ is proportional to $r^{-9/4}$, the first-derivative term cancels exactly, leaving $W_{xx}+W=0$. Thus $W=A\cos x+B\sin x$.

The outer boundary $r\to \infty$ is $x\to 0$, giving $A=W_{\infty }$. The specified inner-radius relation is exactly $x_{\mathrm{in}}=\pi$. The inner zero-torque condition $W_x(\pi )=0$ then gives $B=0$. Hence the stationary bending warp of a Kerr disk isThe tilt magnitude is $\beta (r)=|W_{\infty }||\cos x|$. Far out it tends to the imposed tilt. Moving inward, it decreases to zero atthen grows again with opposite signed tilt, reaching $W(r_{\mathrm{in}})=-W_{\infty }$. All tilt vectors lie on the same line in the complex $W$ plane. Their azimuth is $\arg W_{\infty }$ outside the node and differs by $\pi$ inside; azimuth is undefined at the aligned node itself. The tilt vector remains smooth there. Thus this is an oscillatory standing bending warp, with equal outer and inner tilt magnitudes and reversed inner orientation, rather than monotone inner alignment.

The prescribed real inner radius assumes $a>0$: for $a<0$, its square root is not real, so that particular boundary prescription does not apply. With a separately chosen positive inner radius and negative spin, put $x=4\sqrt{24|a|}{r}^{-5/4}/(5ϵ)$. The equation becomes $W_{xx}-W=0$, and the same outer tilt and zero inner torque giveThis negative-spin solution decreases smoothly toward the inner disk without a sign reversal. The specified positive-spin solution is the cosine profile above; the $a=0$ case is the flat solution already derived.

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.

A passive disk is heated predominantly by intercepted stellar radiation, rather than by its own viscous dissipation. A rough luminosity criterion, ignoring geometric factors, isThe intrinsic accretion disk luminosity is of order $GM\dot{M}/(2R_{\ast })$, with additional accretion energy potentially released at the stellar surface. The local test is more precise: absorbed irradiation must dominate the viscous flux at the radius being considered. For an ordinary steady Keplerian accretion disk, sufficiently far outside its inner edge,where $\alpha$ is the irradiation grazing angle of a disk. Reflection, very small incidence angles and shadowing can invalidate a simple total-luminosity comparison. The following calculation assumes an illuminated, absorbing surface.

At a point $(R,z_s)$ on the upper surface, the outward unit normal in the meridional plane is proportional to $(-z_s^{\prime },1)$, and a light ray from the point source is directed along $(R,z_s)$. Its flux through the upper face is the incident inverse-square flux times minus their scalar product:For a thin disk with small surface slope this becomesThe tangent slope and ray slope differ by $\alpha \simeq z_s^{\prime }-z_s/R$. Positive incidence requires disk flaring, $d(z_s/R)/dR>0$, and an unobstructed ray from the star. A negative value of the formula would mean that direct illumination of this face is absent, not that it receives a negative heating flux. The hierarchy $z_s\gg R_{\ast }$ makes the finite angular size of the star negligible here.

Let $a=\mathcal{R}/\mu$, so the ideal gas equation of state gives the isothermal sound speed $c_s^2=a{T}_c$. In the stellar vertical gravitational potential, hydrostatic equilibrium has vertical acceleration $-{\Omega }_K^{2}z$, where ${\Omega }_K^2=GM/R^3$. Integration of $c_s^2{\partial }_z\rho =-\rho {\Omega }_K^{2}z$ gives a Gaussian atmosphere with disk scale heightConsequently $q=z_s/R=fH/R$ satisfies $T_c=GM{q}^2/(a{f}^{2}R)$. Substitute this into the stipulated Stefan–Boltzmann law $F_d=\sigma T_c^4$ and the irradiation flux:Here $f$ specifies surface height in units of disk scale height and is unrelated to the wind-energy fraction in the preceding solution. This separable differential equation integrates exactly toIf the thin-disc solution is formally matched to $q(R_M)=1$, then $B=1-R_0^2/R_M^2$, givingIt follows that the flaring passive disk asymptotic isThis applies in the thin intermediate region $R\ll \min (R_0,R_M)$, in addition to the point-source and stellar-gravity assumptions. The corrections are of relative order $R^2/R_0^2+R^2/R_M^2$. In particular, the printed hierarchy $R\ll R_M$ must still be accompanied by the initial thin-disc condition $q\ll 1$: if $R_M\gg R_0$, radii larger than $R_0$ cannot satisfy that approximation. The matching at $q=1$ is an extrapolation used to set an integration constant; the derivation is not a valid treatment of a thick surface at $R_M$ itself.

Identify the stellar surface temperature with its effective radiating temperature, so that luminosity of a spherical blackbody gives $L_{\ast }=4\pi R_{\ast }^2\sigma T_{\ast }^4$. The given stellar surface scale height has $H_{\ast }^2=a{T}_{\ast }{R}_{\ast }^3/(GM)$, hence $GM/(a{T}_{\ast }{R}_{\ast })=(R_{\ast }/H_{\ast }{)}^2$. Substituting into $R_0$ yieldsThus a small stellar surface scale height leaves a wide interval above $R_{\ast }$ and below $R_0$ in which the thin disk flaring law can apply. The point-source approximation imposes the stronger lower cutoff obtained from $z_s\gg R_{\ast }$:This cutoff is still much smaller than $R_0$ when $R_0\gg R_{\ast }$. The full interval is therefore $R_{\ast }^{7/9}{R}_0^{2/9}\ll R\ll \min (R_0,R_M)$, whenever these bounds overlap. An assumed thickening radius must lie above the lower cutoff for all of the original approximations to hold simultaneously. The stellar luminosity-temperature relation is essential to the final comparison; independent arbitrary values of $L_{\ast }$ and $T_{\ast }$ would not imply it.

In a steady flow the ideal magnetohydrodynamic induction equation gives $\nabla \times (\mathbf{u}\times \mathbf{B})=0$. Since both vectors have only poloidal magnetic field components, write $\mathbf{u}\times \mathbf{B}=F_{\varphi }{\mathbf{e}}_{\varphi }$. Axisymmetry impliesFor a field regular on the axis, $C=0$; equivalently one may impose zero toroidal electromotive force. Under this physical condition, $\mathbf{u}$ is parallel to $\mathbf{B}$. Write $\rho \mathbf{u}=k\mathbf{B}$. Continuity and $\nabla \cdot \mathbf{B}=0$ then give $\mathbf{B}\cdot \nabla k=0$. The poloidal magnetic flux function has $\mathbf{B}\cdot \nabla \psi =0$, so locally on each connected magnetic surface,This regular-axis alignment of steady poloidal ideal flow needs its zero-circulation condition: axisymmetry alone does not imply this alignment on a domain excluding the axis. An explicit counterexample is available in a cylindrical annulus. Take nonzero constants $U,b$, positive ${\rho }_c,c_s$, andHere $\psi =-bz$, $\nabla \cdot \mathbf{B}=\nabla \times \mathbf{B}=0$ and $\mathbf{u}\times \mathbf{B}=Ub{\mathbf{e}}_{\varphi }/R$ has zero curl. The flow satisfies continuity, $p^{\prime }=-\rho {\Phi }^{\prime }$, and ${\nabla }^2\Phi =4\pi G\rho$. Its acceleration and Lorentz force density vanish, so it solves the steady equations while $\mathbf{u}$ and $\mathbf{B}$ are perpendicular. The nonzero toroidal circulation and the axial singularity explain why this counterexample is excluded from a regular polar accretion column. The remaining derivation uses the regular, aligned branch.

For a narrow stream tube, integrating continuity gives constant $A\rho |\mathbf{u}|$. On a nonzero-flow tube the magnetohydrodynamic mass loading $k$ is constant, so $A\rho |\mathbf{u}|=A|k||\mathbf{B}|$. Therefore $A|\mathbf{B}|$ is constant and $A\propto |\mathbf{B}{|}^{-1}$. This also follows directly from conserved magnetic flux through a flux tube.

For constant isothermal sound speed, $\nabla p/\rho =c_s^2\nabla \ln (\rho /{\rho }_0)$, with arbitrary reference density ${\rho }_0$. Project momentum along $\mathbf{B}$. The Lorentz force density has no component in that direction, and alignment implies $\mathbf{B}\cdot (\mathbf{u}\cdot \nabla )\mathbf{u}=\mathbf{B}\cdot \nabla (|\mathbf{u}{|}^2/2)$. Thus the isothermal magnetic Bernoulli integral isChanging ${\rho }_0$ only shifts the Bernoulli function by a constant. Since $k$ is constant along the magnetic field,Differentiating the isothermal magnetic Bernoulli integral therefore gives

In the slender polar accretion column, the dipolar flux-tube area is $A(z)=A_{\ast }(z/R_{\ast }{)}^3$. The decreasing axial field is a leading approximation, not an exactly solenoidal field $(0,0,B_z(z))$ throughout a cylinder. Indeed $\nabla \cdot \mathbf{B}={\partial }_z{B}_z\ne 0$ for that literal field. Near the axis a small radial component $B_R\simeq -(R/2){\partial }_z{B}_z=3R{B}_z/(2z)$ supplies the required radial divergence. Its magnitude is smaller by $R/z$, even though its divergence is leading order. Keeping this expanding flux tube while neglecting transverse forces yields the intended one-dimensional model.

Use the positive inward speed $v=-u_z$; the signed sonic velocity is $u_z=-c_s$. Conservation of mass and the Bernoulli equation giveEliminating ${\rho }^{\prime }$ yields the isothermal dipolar accretion equation,A smooth transonic branch requires both sides to vanish at its sonic point, soThis is the critical point of the flow equation; a generic subsonic solution need not cross it. A crossing in the exterior column requires $z_s\ge R_{\ast }$, and a sonic point strictly outside the star requires $z_s>R_{\ast }$. Differentiating the equation at the crossing gives $2(v_s^{\prime }{)}^2=3c_s^2/z_s^2$. For inward accretion that accelerates toward the star, $v_s^{\prime }=-\sqrt{3/2}{c}_s/z_s$.

The reservoir boundary condition fixes $ϵ=c_s^2\ln ({\rho }_{\infty }/{\rho }_0)$. At the sonic point, ${\Phi }_s=-3c_s^2$ and $v_s^2/2=c_s^2/2$, givingFinally use the sonic mass flux through the dipolar flux-tube area:This rate belongs to the smooth transonic branch of the idealized column, with $A_{\ast }$ the supplied total loaded area. The reservoir condition alone does not force every steady solution onto this branch. A real narrow dipolar column must match an outer flow where the slender approximation ceases to apply.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative. For any specific quantity $f$, conservation of mass givesDotting the momentum equation with $\Sigma \mathbf{u}$ produces the kinetic energy balanceThe Coriolis acceleration does no work because $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$. Since the shearing-sheet tidal potential is time-independent, its advected potential-energy density obeysFor the isothermal equation of state, define the barotropic energy density $U=c_s^2\Sigma \ln (\Sigma /{\Sigma }_{\mathrm{ref}})$, with fixed positive reference surface density ${\Sigma }_{\mathrm{ref}}$. The continuity equation impliesAdding these balances combines the two pressure terms into $-\nabla \cdot (P\mathbf{u})$. This establishes isothermal shearing-sheet energy conservation. Thus the conserved energy density and its flux areChoosing the density unit so that ${\Sigma }_{\mathrm{ref}}=1$ reproduces the printed logarithm. Changing the reference adds a multiple of the conserved mass to $E$ and its advective energy flux. The barotropic energy density is the mathematical energy of this fixed-isothermal sound speed closure; it is not the microscopic thermal energy of a thermally isolated gas. Maintaining an isothermal process can require heat exchange.

For a local axisymmetric Fourier mode, the Toomre stability criterion balances three contributions to the squared oscillation frequency:The radial epicyclic frequency supplies rotational restoration at long wavelengths; isothermal sound speed and pressure stabilize short wavelengths; disk self-gravity destabilizes intermediate wavelengths. Minimizing over $|k|$ gives $|k|=\pi G{\Sigma }_0/c_s^2$ and ${\omega }_{\min }^2={\kappa }^2(1-Q^{-2})$. Axisymmetric gravitational instability occurs when $Q<1$.

In a centrally dominated Keplerian disk, $\kappa =\Omega$ and vertical hydrostatic equilibrium gives $c_s\simeq H\Omega$. Using ${\Omega }^2=G{M}_{\ast }/r^3$ and a local disk-mass estimate $M_D\sim \pi r^2{\Sigma }_0$,This is the disk mass form of the Toomre criterion. An actual enclosed mass depends on the radial surface density profile and changes an order-one coefficient. In particular, a profile proportional to $r^{-3/2}$ has $M_D(<r)\simeq 4\pi r^2{\Sigma }_0$ when its inner cutoff is negligible.

For the protosolar estimate, take $M_{\ast }\simeq M_{\odot }\simeq 2\times 10^{33}\mathrm{g}$. This solar mass is implicit in identifying the central star with the young Sun. The supplied constant disk aspect ratio givesUsing the printed approximate astronomical unit and gravitational constant, one can also obtain $\Omega (1\mathrm{AU})\simeq 3.7\times 10^{-7}{\mathrm{s}}^{-1}$ and $c_s\simeq 3.7\times 10^5{\mathrm{cm}\mathrm{s}}^{-1}$; they give the same $Q$. The gravitational constant cancels from the mass form.

At one astronomical unit the disk is very stable, with $Q\simeq 640$. Formal extrapolation givesThis is far beyond the planetary region and any plausible extent of the minimum-mass solar nebula. Moreover, the extrapolated enclosed disk mass is already a substantial fraction of a solar mass, so the centrally dominated approximation becomes questionable. The formal radius is not a prediction of a real unstable outer nebula. Direct gas fragmentation by gravitational instability is unlikely to have formed Solar System planets in this model. Core accretion is the more natural route; an earlier substantially more massive or colder disk would be a different model. Even $Q<1$ alone does not guarantee fragmentation, because sufficiently rapid cooling is also needed.

Let $e$ be specific internal energy, $Q$ the heat supplied per unit volume per unit time, and $E=\rho (e+|\mathbf{u}{|}^2/2)$. For an inviscid perfect gas in a prescribed Newtonian gravitational potential $\Phi$, the fluid total-energy equation isThere is no gravitational term if no external body is present. For a time-independent $\Phi$, adding potential energy gives the equivalent conservative form ${\partial }_t(E+\rho \Phi )+\nabla \cdot [(E+p+\rho \Phi )\mathbf{u}]=Q$. Subtracting the kinetic-energy equation and using mass conservation givesFor a perfect gas with constant specific-heat ratio $\gamma$, $e=p/[(\gamma -1)\rho ]$. The isothermal equation of state therefore makes $e=c_s^2/(\gamma -1)$ constant. Hence the required heat supply and heat loss areCompression requires cooling; expansion requires heating. The isothermal sound speed $c_s$ differs from the adiabatic sound speed $\sqrt{\gamma }{c}_s$.

Define the isothermal Mach numbers by ${\mathcal{M}}_i=u_i/c_s$. Orient the normal to a stationary isothermal shock along the flow, so the mass flux $j={\rho }_1{u}_1={\rho }_2{u}_2>0$. Tangential velocity is continuous for this planar hydrodynamic shock. Normal momentum conservation givesFor a genuine discontinuity $u_1\ne u_2$, factorization gives $u_1{u}_2=c_s^2$. Thus the isothermal shock jump relations areAcross a thin shock, gravitational potential and tangential kinetic energy are unchanged. The specific enthalpy $h=\gamma c_s^2/(\gamma -1)$ is the same on both sides, so the energy removed per unit area per unit time isPositive net cooling requires $u_1>u_2$, equivalently ${\rho }_2>{\rho }_1$ and ${\mathcal{M}}_1>1>{\mathcal{M}}_2$. Cooling across an isothermal shock permits only compression shocks; the formal expansion discontinuity requires energy supplied by the surroundings. The continuous state $u_1=u_2$ is not a shock.

For steady radial flow, mass conservation gives $r^2\rho u_r=\text{constant}$. Differentiating it and eliminating the mass density gradient from radial momentum balance givesA regular sonic point has $|u_r|=c_s$, so the right-hand side must vanish there as well. ThereforeThis is the Bondi sonic point for accretion; the same critical radius appears in an isothermal Parker wind. A general steady solution need not pass through a sonic point, but a smooth transonic one must obey both critical conditions.

Put $x=r/r_s$ and $\mathcal{M}=|u_r|/c_s$ on a branch whose flow direction is fixed. The radial equation becomesIntegrating yields the integrated isothermal Bondi flow relation:For a transonic branch, $x=\mathcal{M}=1$ gives $C=-3/2$. Expansion about the critical point gives $(\mathcal{M}-1{)}^2=(x-1{)}^2+O((x-1{)}^3)$, with slope $-1$ for inflow supplied by gas at rest at infinity and slope $+1$ for a transonic outflow.

For Isothermal Bondi accretion, integrate radial momentum balance once more, using $u_r\to 0$ and $\rho \to {\rho }_{\infty }$ at infinity:At the Bondi sonic point, $GM/r_s=2c_s^2$ and $u_r^2=c_s^2$, so ${\rho }_s=e^{3/2}{\rho }_{\infty }$. The positive inward isothermal Bondi accretion rate is consequentlyThe critical mass density and rate are fixed by the regular transonic solution and the specified reservoir at infinity; an arbitrary subsonic solution does not share this accretion rate.

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.