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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.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

In the vertical equations, $\mu$ denotes dynamic viscosity, while ${\mu }_m$ is the mean molecular weight in units of the proton mass $m_p$. The symbol $k$ denotes the Boltzmann constant. For the intended hydrostatic equilibrium calculation, put $P=K{\rho }^{\gamma }$ with $K>0$, $\gamma >1$, and set $\mathcal{R}=k/({\mu }_m{m}_p)$. Dividing the hydrostatic equation by $\rho$ and integrating from the symmetric midplane givesThe specific enthalpy vanishes at the free surface. Hence the polytropic vertical structure in stellar gravity isOn $|z|<H$, the ideal gas relation gives $T=T_m(1-z^2/H^2)$ and $P=P_m(1-z^2/H^2{)}^{\gamma /(\gamma -1)}$; the vacuum extension sets $\rho =P=0$ outside. The formula needs $\gamma >1$ for a finite positive semi-thickness and assumes the harmonic vertical gravity remains valid through this height.

There is a genuine inconsistency if all four printed structure equations are imposed literally with $F=0$ and finite positive opacity: radiative diffusion then requires $dT/dz=0$, whereas the polytropic temperature above has a nonzero gradient away from the midplane. The intended result is valid for the hydrostatic polytropic equation of state with the radiative closure replaced or neglected, for example an adiabatic opaque limit. It is not an exact simultaneous solution of the stated finite-opacity radiative system. With that radiative system and a nonvacuum ideal gas, $\gamma =1$ instead gives an isothermal solutionwhich has no finite free surface. For $\gamma \ne 1$, constant $T$ and constant $K$ would force constant density, contradicting hydrostatic balance on a nontrivial vertical interval. This is the zero-radiative-flux obstruction for a polytropic ideal gas.

Use the Euler equations for an inviscid fluid with no magnetic or gravitational force. Let $w$ denote specific enthalpy. For a homentropic flow, $dw=dp/\rho$, so the pressure acceleration is $-\nabla w$. The velocity identitytherefore givesTaking the curl, using that the curl of a gradient is zero and that differentiation commutes for smooth fields, proves barotropic vorticity transport:More generally the same proof works for any barotropic fluid, replacing $w$ by a pressure potential ${\int }^{\rho }{p}^{\prime }(s)ds/s$. Here “isentropic” must supply this barotropic closure, for example through one common specific entropy throughout the fluid. Merely imposing entropy advection equation $Ds/Dt=0$ allows spatial specific entropy gradients; in that case the vorticity equation contains the additional baroclinic vorticity generation term ${\rho }^{-2}\nabla \rho \times \nabla p$.

Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio $\gamma >1$, and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is $e=p/[(\gamma -1)\rho ]$, and the specific enthalpy is $w=\gamma p/[(\gamma -1)\rho ]$. Integrating the local conservation laws across a thin stationary control volume makes the mass flux $\rho u$, momentum flux $p+\rho u^2$, and energy flux $u(\rho u^2/2+\gamma p/(\gamma -1))$ continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state $p=K{\rho }^{\gamma }$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.

In the Strong-shock Rankine-Hugoniot conditions, $p_1/({\rho }_1{u}_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r={\rho }_2/{\rho }_1$. The first two flux balances give $u_2=u_1/r$ and $p_2={\rho }_1{u}_1^2(1-1/r)$ to leading order. The energy balance becomesDiscarding the unchanged-state root $r=1$ gives the compressive shock compression ratio and the required limits:

Consider a spherical stellar polytrope with $P=K{\rho }^{1+1/n}$, constant $K$, $0<n<5$, zero surface pressure and finite radius $R$. Its specific enthalpy, measured from zero density, is $h={\int }_0^{P}d{P}^{\prime }/\rho =(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =\Phi (R)=-GM/R$. Integrating over mass therefore yieldsThe static stellar virial theorem gives $3\int PdV=-\Omega$, and eliminating the pressure integral proves the gravitational energy of a stellar polytrope:For an ideal gas with constant specific-heat ratio, the accompanying internal energy isTo express both energies using $\gamma$ alone one additionally identifies the model as an adiabatic stellar polytrope, so $n=1/(\gamma -1)$. ThenFor $\gamma =5/3$ these become $\Omega =-6G{M}^2/(7R)$ and $U=3G{M}^2/(7R)$. A general polytropic index need not be determined by the thermodynamic specific-heat ratio; without the adiabatic identification both $n$ and $\gamma$ remain.

Let $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ be the material derivative, and write $h$ for specific enthalpy. The ideal magnetohydrodynamic induction equation and Gauss's law for magnetism giveDotting the momentum equation with $\mathbf{B}$ eliminates the Lorentz force density. Therefore the cross-helicity density satisfiesThe first law of thermodynamics gives $dh=Tds+dp/\rho$ for specific entropy $s$, without requiring uniform entropy. ConsequentlySince $\nabla \cdot \mathbf{B}=0$, the first term on the right is a divergence. This proves the cross-helicity conservation law, with fluxFor the perfect gas used here, $h=\gamma p/[(\gamma -1)\rho ]$.

In the shock frame, integrate mass conservation, momentum conservation, and total-energy conservation across a thin interval containing the stationary normal shock wave. No mass, momentum, or energy is stored in the vanishingly thin interval, so each flux has the same value on both sides. For a perfect gas, the specific enthalpy is $h=\gamma p/[(\gamma -1)\rho ]$. The resulting Rankine-Hugoniot conditions for a perfect gas areThe momentum flux includes both transported momentum and pressure force. The energy flux includes kinetic energy and enthalpy, the latter accounting for internal energy and the work needed to push gas through the interval. Entropy production can occur inside the normal shock wave even though total energy is conserved.

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.

In the synchronously rotating reference frame, stationary fluid obeyswhere $\Psi$ is the sum of the two gravitational potentials and the centrifugal potential. The circular binary's Kepler third law gives ${\Omega }^2=G(M_1+M_2)/a^3=GM/a^3$, hence

Uniform composition and the assumed central condensation make star 1 approximately barotropic, so one may define the specific enthalpyHydrostatic balance becomes $\nabla (h+\Psi )=0$, or $h+\Psi =\text{constant}$ throughout the star. Since $h$ is a monotone function of $P$ and $\rho$, constant-$P$ and constant-$\rho$ surfaces are equipotential surfaces of $\Psi$.

The thermal binding scale defined by equating heated gas sound speed to circular Keplerian orbit speed. It is $r_g=GM/c_s^2$. The circular-orbit binding energy is $GM/(2r)$ per unit mass, while the gas specific enthalpy is of order $c_s^2$. A heated disk surface at radii comparable to or larger than $r_g$ can therefore drive photoevaporation. The actual launching radius can differ by a factor of order unity; $c_s$ is not the ballistic escape speed.

Under the hypotheses of steady planar ideal-MHD field-line invariants, the specific enthalpy relation and the velocity projection of momentum give $\mathbf{B}\cdot \nabla Q=0$. The horizontal total energy flux, including the Poynting vector, is $k{\mathbf{B}}_{h}Q$. The magnetic correction accounts for work exchanged between the fluid and the field.

A polytropic flow uses a polytropic equation of state. When $K$ and the adiabatic exponent are constant, a smooth streamline has specific enthalpy $h=\gamma K{\rho }^{\gamma -1}/(\gamma -1)$ and adiabatic sound speed $c_s^2=\gamma K{\rho }^{\gamma -1}$. Spatially varying entropy can make $K$ depend on streamline; the constant-$K$ spherical wind is a special case.

The specific heat capacity at fixed pressure and composition is $c_P=(\partial h/\partial T{)}_P$, where $h$ is specific enthalpy. Equivalently, it is heat capacity at constant pressure divided by mass. Under the process assumptions in heat capacity, it includes expansion work as well as the internal energy change. The sensible heat example uses this response separately for ice and water.

A stagnant mushy-layer model has no bulk advection, equal phase mass densities, pure solid, and local equilibrium on the liquidus. With liquid fraction $\varphi$, specific enthalpy $E=c_{p}T+L\varphi$ and bulk solute concentration $B=\varphi C$, neglecting solute diffusion inside the mushy layer gives $T_t+(L/c_p){\varphi }_t=\kappa T_{zz}$ and $(\varphi C{)}_t=0$. The latter fixes composition at each stationary material level; it does not automatically set every level to the initial liquid concentration when solute can diffuse ahead of the growing mush.

For steady flow through a fixed control volume with one inlet and one outlet, the first law of thermodynamics and conservation of mass giveHere $\dot{m}$ is the mass flow rate, $\dot{Q}$ incoming heat per unit time, ${\dot{W}}_s$ outgoing non-flow work per unit time, $w$ the flow speed, $z$ elevation and $g$ gravitational acceleration. The specific enthalpy $h=u+p/\rho$ includes both the transported internal energy per unit mass $u$ and the pressure work needed to push volume through the boundary; $\rho$ is mass density. The inlet supplies that pressure work, while the outlet consumes it to displace the downstream fluid.^[ref]

For an insulated stationary throttle, require negligible heat transfer, no shaft or other non-flow work, and negligible changes in bulk kinetic energy and gravitational potential energy. Then $h_2=h_1$. With unchanged composition this is also equality of molar enthalpy. The upstream and downstream states must admit equilibrium pressure, temperature and property values, although the constriction itself is irreversible and can be far from thermodynamic equilibrium. Start-up storage of energy in the walls and fluid, heat leakage, or a large outlet speed invalidates the reduced balance. A nozzle can convert specific enthalpy into bulk kinetic energy even when it is insulated.

The same accounting for a fixed transported amount is $U_2-U_1=p_1{V}_1-p_2{V}_2$, hence $U_1+p_1{V}_1=U_2+p_2{V}_2$. This explains why dropping the two pressure work terms would give the wrong conserved quantity.^[ref]