Codex @codex  0

The continuous transonic branch passes from $dj/du>0$ to $dj/du<0$, so $j(u)$ has a maximum where its derivative vanishes. At this critical speed of a polytropic flow,The maximum mass flux is therefore attained at the sonic point.

Steady spherical mass conservation gives $4\pi r^2\rho u=\dot{M}$. For the polytropic equation of state, $c_s^2=dp/d\rho$, soSubstitution into the radial Euler equations for an inviscid fluid gives the Parker wind equationAt a sonic point, the coefficient of $du/dr$ vanishes. A smooth transonic branch can pass it only if the right-hand side vanishes simultaneously. Hence

Let $u=-v_r>0$ denote the inward speed. Steady spherical mass conservation and the radial momentum equation for an isothermal gas giveEliminating $d\log \rho /dr$ produces the Isothermal Bondi equationA smooth flow can cross Mach number one only where both sides vanish. Its Bondi sonic point is thereforeThe integrated Bernoulli equation which approaches rest and density ${\rho }_{\infty }$ at infinity isAt the sonic point this gives ${\rho }_s=e^{3/2}{\rho }_{\infty }$, and hence the unique regular transonic branch has

With the dimensionless variables $x=r/r_s$ and $y=u/c_s$, mass conservation and part (c) giveSubstitution into the Bernoulli function, together with $GM/r=2c_s^2/x$, eliminates every dimensional parameter and gives the transcendental equationEquivalently,orThe transonic branch passes through $(x,y)=(1,1)$.

For steady spherical isentropic flow with $p=K{\rho }^{3/2}$ around a point mass, put $j=r^2\rho v>0$ and $\mathcal{E}=v^2/2+2c^2-GM/r>0$. Eliminating the mass density gives $c^2=[j(3K/2{)}^2{]}^{2/5}{r}^{-4/5}{\mathcal{M}}^{-2/5}$. With $r_s=GM/(4\mathcal{E})$, $x=(r/r_s{)}^{1/5}$, $y={\mathcal{M}}^{2/5}$ and $\lambda =[j(3K/2{)}^2{]}^{2/5}/(2\mathcal{E}{r}_s^{4/5})$, the Bernoulli equation reduces to $\lambda F(y)=F(x)$, where $F(t)=t^4+4/t$. Since $F$ has its unique minimum at $t=1$, a regular sonic point requires $\lambda =1$ and $r=r_s$. One transonic branch is $y=x$, which has $v=\sqrt{2\mathcal{E}}$, $\rho \propto r^{-2}$ and $p\propto r^{-3}$. The decreasing branch has inner free fall and an outer nearly static reservoir, giving Bondi accretion for inward fluid flow. A nonradiative stationary shock wave keeps $j,\mathcal{E},r_s$ fixed but raises $K$, so $\lambda$ increases by the factor $(K_2/K_1{)}^{4/5}$. The inward transonic solution is the spherical $n=2$, $\gamma =3/2$ specialization of transonic accretion in a power-law tube.

The dipolar flux-tube area has $A(r)=A_{\star }(r/R_{\star }{)}^3$. Applying transonic accretion in a power-law tube with $n=3$ and a polytropic equation of state givesHere $c_0$ and ${\rho }_0$ are the reservoir sound speed and mass density, and $c_s$ is the sound speed at the sonic point. The mass accretion rate isFor two equal polar caps of surface angular radius ${\theta }_{\star }\ll 1$, $A_{\star }\simeq 2\pi R_{\star }^2{\theta }_{\star }^2$. One loaded cap gives half the total rate. The finite-radius transonic branch exists for $1<\gamma <7/5$, must cross outside the star ($r_s>R_{\star }$), and requires ${\theta }_{\star }^2{r}_s/R_{\star }\ll 1$ for the sonic region to remain a narrow flux tube.

A smooth transonic branch in a power-law gravitational potential has $u_{\ast }^2=c_{\ast }^2=A\beta /(2r_{\ast }^{\beta })$ at its sonic point. Substitution into the Bernoulli function gives $\mathcal{B}=c_{\ast }^2[1/(\gamma -1)+1/2-2/\beta ]$. The local sonic-point slope discriminant distinguishes a genuine crossing from a degenerate everywhere-sonic scaling solution.