= Solution
\b[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>,
$$
\boxed{\frac{DQ}{Dt}=\frac{d}{dt}Q(\mathbf X(t),t)
=\frac{\partial Q}{\partial t}+\mathbf u\cdot\nabla Q.}
$$
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.
\b[The forces on a parcel.] A material volume $dV$ has mass $\rho\,dV$. The pressure forces on its boundary sum to $-\nabla p\,dV$, while the <gravitational acceleration> is $-\nabla\psi$. <Newton's second law> therefore gives the <Euler equations for an inviscid fluid>,
$$
\boxed{\partial_t\mathbf u+(\mathbf u\cdot\nabla)\mathbf u
=-\frac1\rho\nabla p-\nabla\psi.}
$$
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.
\b[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 function>
$$
h_b(p)=\int_{p_{\rm ref}}^p\frac{dp'}{\rho(p')},
\qquad \nabla h_b=\frac{\nabla p}{\rho}.
$$
Dot 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 gives
$$
\mathbf u\cdot\nabla\left(\frac12u^2+h_b+\psi\right)=0.
$$
Thus the <Bernoulli function> is
$$
\boxed{H=\frac12u^2+\int\frac{dp}{\rho}+\psi
=\text{constant along each streamline}.}
$$
Every term has units of energy per mass, $\mathrm{m^2\,s^{-2}}$. Different streamlines can have different constants.
\b[The spherical accretion equation.] Write the inward speed as $u(r)>0$, so the velocity vector is $-u(r)\widehat{\mathbf r}$ and the positive <mass accretion rate> is
$$
\dot M=4\pi r^2\rho u,\qquad
\frac{d\ln\rho}{dr}=-\frac2r-\frac{d\ln u}{dr}.
$$
Assume a prescribed central potential $\psi=-GM/r$ and negligible gas self-gravity. The steady radial Euler equation is
$$
u\frac{du}{dr}=-\frac1\rho\frac{dp}{dr}-\frac{GM}{r^2}.
$$
For 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 obtain
$$
\boxed{(u^2-c_s^2)\frac{d\ln u}{dr}
=\frac{2c_s^2}{r}-\frac{GM}{r^2}
=\frac{2c_s^2}{r}\left(1-\frac{GM}{2c_s^2r}\right).}
$$
A smooth <sonic point> requires the right side to vanish where $u=c_s$, hence
$$
\boxed{r_s=\frac{GM}{2c_s(r_s)^2},\qquad u(r_s)=c_s(r_s).}
$$
For 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>.
\b[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. Thus
$$
\frac12u^2+c_s^2\ln\frac{\rho}{\rho_\infty}-\frac{GM}{r}=0,
$$
or
$$
\boxed{u^2=2c_s^2\ln\frac{\rho_\infty}{\rho}+\frac{2GM}{r}.}
$$
The logarithm has a dimensionless argument. Constant temperature during compression generally requires heat removal, so this is <Isothermal Bondi accretion>, rather than adiabatic accretion with $\gamma>1$.
\b[The critical accretion rate.] Select the smooth reservoir-connected <transonic branch>. Evaluate the last relation at $u_s=c_s$ and $GM/r_s=2c_s^2$:
$$
c_s^2=2c_s^2\ln\frac{\rho_\infty}{\rho_s}+4c_s^2,
\qquad
\boxed{\rho_s=e^{3/2}\rho_\infty.}
$$
Mass conservation then gives the <isothermal Bondi accretion rate>,
$$
\boxed{\dot M=4\pi r_s^2\rho_sc_s
=\pi e^{3/2}\frac{G^2M^2\rho_\infty}{c_s^3}.}
$$
The units are $\mathrm{kg\,s^{-1}}$. Equivalently $\dot M=4\pi\lambda G^2M^2\rho_\infty/c_s^3$, with $\lambda=e^{3/2}/4\simeq1.12$.
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.
\Image[/past-exam-of-the-astrophysical-fluid-dynamics-course-of-the-university-of-cambridge/2018-b3-isothermal-accretion.png]
{title=Original numerical transonic isothermal accretion profiles, normalized by the sonic radius and outer density; the speed becomes supersonic inside the sonic point}
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