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For steady Isothermal Bondi accretion from gas at rest at infinity, define $\lambda =\dot{M}{c}_s^3/(4\pi G^2{M}^2{\rho }_{\infty })$, $x=r/r_s$ and $y=u/c_s>0$, where $r_s=GM/(2c_s^2)$. The Bernoulli function and mass conservation give $y^2/2-\ln y=2\ln x+2/x-\ln (4\lambda )$. The left side has minimum $1/2$ at $y=1$, and the radius-dependent part on the right has minimum two at $x=1$. Existence over all positive radii therefore requires the displayed bound. Equality permits the smooth accelerating-inward transonic branch and gives the usual isothermal Bondi accretion rate. Smaller positive rates have an everywhere-subsonic reservoir-connected branch, and zero rate gives a hydrostatic atmosphere. An absorbing inner boundary inside the sonic radius selects the critical inflow; steady isothermal equations and the outer density alone do not specify that inner condition.

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.

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.

Take $u(r)>0$ to be the inward radial speed and select the regular critical solution for a gas reservoir of density ${\rho }_{\infty }$ and adiabatic sound speed $c_{\infty }$. Other mathematically subsonic steady solutions need a different inner boundary condition. For the polytropic equation of state, let $\gamma =1+1/n$, soMass conservation and the Bernoulli equation giveThe gas is at rest at infinity, so its kinetic and gravitational terms vanish there.

Differentiate these two conserved quantities, using $d\log \rho /dr=-2/r-u^{\prime }/u$. Eliminating the density derivative yields the polytropic Bondi accretion equationAt a regular sonic point, both sides vanish:The Bernoulli equation at that point reduces to $n{c}_{\infty }^2=(n-3/2)c_c^2$. For $n>3/2$, equivalently $1<\gamma <5/3$,Since $c_s^2\propto {\rho }^{1/n}$, the critical density is ${\rho }_c={\rho }_{\infty }[2n/(2n-3){]}^n$. Evaluate the conserved mass accretion rate at the sonic point:whereThe isothermal limit $\gamma \to 1$ gives $\lambda =e^{3/2}/4$, agreeing with Isothermal Bondi accretion.

For $\gamma =5/3$, $n=3/2$ and the Newtonian critical radius tends to zero while $c_c$ diverges. The limiting critical solution approaches Mach number one only as $r\to 0$; it has no finite-radius Newtonian sonic transition. This is a failure of the Newtonian inner approximation, not an obstruction to accretion onto a black hole. A physical Schwarzschild black hole has a finite event horizon, and the regular relativistic inflow becomes supersonic outside it.

The singular-looking factor in the rate has a finite limit. Writing $ϵ=n-3/2$, its logarithm is $ϵ\log [n/ϵ]\to 0$, so the critical Bondi accretion rate for gamma equals five thirds isThis is also the leading cold-reservoir limit of relativistic spherical accretion; an exact rate for relativistically hot gas requires the relativistic equations, not this Newtonian expression.

For completeness, the finite relativistic sonic point can be seen without assigning it an arbitrary inner radius. Let $b=c_s^2/c^2$ denote the relativistic squared sound speed and $g=\gamma -1$. For an isentropic relativistic gas with $p=K{\rho }_{\mathrm{rest}}^{\gamma }$, its dimensionless specific enthalpy is $h=g/(g-b)$. Conservation of rest-mass flux and relativistic Bernoulli energy gives the critical relationsFor $g=2/3$ and $c_{\infty }\ll c$, expansion gives $b_{\infty }\simeq (9/4)b_c^2$, henceHere the asymptotic relativistic and Newtonian sound speeds agree to leading order. This finite-radius transition explains why the gas can reach the horizon despite the Newtonian $r_c=0$ result.

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

Put $Y=y^2$. On the inner supersonic branch, the equation from part (d) isIts asymptotic expansion as $x\to 0^{+}$ beginsTaking the positive square root givesSince $x=r/r_s$ and $y=u/c_s$,The leading term is the free-fall speed $\sqrt{2GM/r}$; the logarithmic term is the first pressure correction to the inner Isothermal Bondi accretion flow.