= Solution
Let $u=u_r<0$ for accretion. The steady spherical <continuity equation> and radial <Euler equations for an inviscid fluid> give
$$
\frac{d}{dr}(r^2\rho u)=0,\qquad
u\frac{du}{dr}=-\frac1\rho\frac{dp}{dr}-\frac{d\Phi}{dr},\qquad
\dot M=-4\pi r^2\rho u>0.
$$
For a <barotropic fluid>, $dp/dr=c_s^2\,d\rho/dr$. Dividing <continuity equation> by $r^2\rho u$ gives $\rho'/\rho=-2/r-u'/u$. Eliminating this derivative yields
$$
\boxed{\frac{u^2-c_s^2}{u}\frac{du}{dr}=\frac{2c_s^2}{r}-\Phi'(r).}
$$
At a smooth transonic critical point, the coefficient of $du/dr$ vanishes. A finite derivative then requires the right side to vanish simultaneously:
$$
\boxed{u_c=-c_{s,c},\qquad \Phi'(r_c)=\frac{2c_{s,c}^2}{r_c}.}
$$
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 gives
$$
\left(2+\frac C{c_s^2}\right)(u_c')^2
+\frac{4C}{r_cu_c}u_c'+\frac{4C+2c_s^2}{r_c^2}+\Phi''(r_c)=0.
$$
This 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 equation>
$$
\boxed{\frac12u^2+h(\rho)+\Phi=B.}
$$
Changing 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 is
$$
(r_c-r_G)^2=r_Br_c,
$$
whose two roots are $r_G+\tfrac12r_B(1\pm\sqrt{1+\beta})$. Only the plus root lies outside $r_G$. Therefore
$$
\boxed{r_c=r_G+\frac{GM}{4c_s^2}(1+\sqrt{1+\beta})
=\frac{GM}{4c_s^2}\bigl(1+\beta/2+\sqrt{1+\beta}\bigr)>r_G.}
$$
For this isothermal case $C=0$, and the slope relation reduces to
$$
(u_c')^2=\frac{GM}{(r_c-r_G)^3}-\frac{c_s^2}{r_c^2}
=\frac{c_s^2(r_c+r_G)}{r_c^2(r_c-r_G)}>0.
$$
The 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 $u_c=-c_s$,
$$
\rho_c=\rho_0\exp\left[\frac{GM}{c_s^2(r_c-r_G)}-\frac12\right]
=\rho_0\exp\left[\frac4{1+\sqrt{1+\beta}}-\frac12\right].
$$
Substitution into $\dot M=4\pi r_c^2\rho_cc_s$ gives
$$
\boxed{\dot M=\frac{\pi\rho_0(GM)^2}{4c_s^3}
\bigl(1+\beta/2+\sqrt{1+\beta}\bigr)^2
\exp\left[\frac4{1+\sqrt{1+\beta}}-\frac12\right].}
$$
This 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 $r_G/r_B\to0$, the result tends to the <isothermal Bondi accretion rate> $\pi e^{3/2}\rho_0(GM)^2/c_s^3$, a useful normalization check.
Back to article page
