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Take $\dot{M}>0$ to mean inward accretion rate, and let $S$ count loss through both faces per unit projected area of the accretion disk. An annulus of width $dR$ loses $2\pi RSdR$ to the wind. In steady state, its incoming mass flux at its outer edge equals the inward flux at its inner edge plus this loss. Hence mass conservation givesIf a loss rate is instead defined per individual face, it must be doubled before substituting for $S$ here.

Use Keplerian rotation ${\Omega }^2=GM/R^3$. The vertically integrated viscous dissipation, summed over both disk faces, isHere $\nu$ is the kinematic viscosity, and $\Sigma$ is the surface density of a disk; for height-dependent viscosity the product means $\int \rho \nu dz$. Material already in a circular orbit has specific orbital energy $v_K^2/2-GM/R=-GM/(2R)$. Giving it zero total binding energy therefore costs an additional $GM/(2R)$ per unit mass. The total launch speed is the escape velocity $\sqrt{2GM/R}$, but one must not charge the wind for the orbital kinetic energy it already possesses. Equating wind power to $f{Q}^{+}$ givesCombining this with mass conservation yields the wind-powered mass loss from a steady accretion disk relation

Define the outward viscous torque in an accretion disk asThe net inward angular momentum flux is $\dot{M}h-\mathcal{G}$. Because each unit mass lost to the wind carries the local specific angular momentum $h$, annular conservation of angular momentum readsIn the wind region, $\mathcal{G}=Rh{\dot{M}}^{\prime }/(3f)$. For constant $f$, differentiation and cancellation of $h$ givesThus the requested Cauchy-Euler differential equation is

For $f=1/3$, a power-law trial $\dot{M}\propto R^p$ gives $p^2+p/2-1/2=0$, with roots $p=1/2,-1$. Therefore $\dot{M}=A\sqrt{R}+B/R$ in $R_{\ast }\le R\le R_0$. A zero inner viscous torque in an accretion disk implies ${\dot{M}}^{\prime }(R_{\ast })=0$, so $B=A{R}_{\ast }^{3/2}/2$. There is no point mass sink at $R_0$, so the accretion rate is continuous there and equals ${\dot{M}}_0$. The complete result isIts derivative is nonnegative throughout the wind region, as required for a nonnegative loss rate $S$. The derivative can jump to zero outside $R_0$: the energy relation applies only inside the wind zone and does not impose ${\dot{M}}^{\prime }(R_0)=0$ on its inner limit. The viscous torque in an accretion disk remains continuous and extends outside via ${\mathcal{G}}^{\prime }={\dot{M}}_0{h}^{\prime }$.

At the inner boundary the exact accreted fraction isThe last expression is a leading-order asymptotic equality, not the exact finite-radius fraction. Most of the supplied mass is expelled before reaching the star when $R_{\ast }/R_0$ is small.

Let $\chi =H(F)$ use the Heaviside step function, and ${\delta }_s=|\nabla F|\delta (F)$, so that ${\mathbf{n}}_s=\nabla F/|\nabla F|$ points from the body into the fluid. The kinematic boundary condition gives $F_t+\mathbf{u}\cdot \nabla F=0$ on the surface. In the exterior, mass conservation and the Euler equations for an inviscid fluid areProducts with $\chi$ are understood distributionally, using ${\partial }_t\chi =F_t\delta (F)$ and ${\partial }_i\chi =F_i\delta (F)$. With ${\rho }^{\prime }=\rho -{\rho }_0$, the extended continuity equation isThe corresponding extended momentum equation isbecause its advective surface term is $\rho u_i(F_t+\mathbf{u}\cdot \nabla F)\delta (F)=0$. Differentiate continuity in time and substitute the divergence of momentum. Subtracting $c_0^2{\nabla }^2({\rho }^{\prime }\chi )$ gives the Ffowcs Williams-Hawkings equationwhere the inviscid Lighthill stress tensor isIf $p$ denotes total pressure, this tensor includes the constant ambient pressure $p_0$. Its contribution $p_0{\nabla }^2\chi$ cancels the corresponding surface term exactly. For radiation integrals it is convenient to subtract $p_0$ from both terms, making $T_{ij}=\rho u_i{u}_j+(p^{\prime }-c_0^2{\rho }^{\prime }){\delta }_{ij}$ with $p^{\prime }=p-p_0$. The displayed source identity is unchanged by this consistent pressure-gauge subtraction.

Assume no incoming sound and no free initial acoustic field. Put $R_y=|\mathbf{x}-\mathbf{y}|$, ${\tau }_y=t-R_y/c_0$, $Q={\rho }_0{u}_n{\delta }_s$, and $P_i=p^{\prime }{n}_{s,i}{\delta }_s$. Convolution with the retarded acoustic Green function gives the three source integrals, for an observer outside the body:These are the surface thickness, surface loading and volume quadrupole contributions. The retarded surface Dirac delta distributions must remain inside the integrals: integrating them over a moving retarded surface produces the moving-surface retarded Jacobian, and is not generally the same as simply substituting one retarded time into an instantaneous surface integral.

For an acoustic compact-source approximation, suppose the dominant sources have size $\ell$, source frequency scale $\omega$, and $\omega \ell /c_0\ll 1$. Also require $r=|\mathbf{x}|\gg \ell$ and the acoustic far field $\omega r/c_0\gg 1$. Then $R_y\simeq r-\mathbf{n}\cdot \mathbf{y}$, $1/R_y\simeq 1/r$, and the leading source integrals use the common time $\tau =t-r/c_0$. Slow source motion relative to $c_0$ is implicit in this leading retardation approximation. DefineThus $m$ is net displaced mass per unit time, $\mathbf{f}$ is the fluid pressure force on the body, and $S_{ij}$ is the integrated excess Lighthill stress tensor. The minus sign in $\mathbf{f}$ fixes the requested loading convention; the force on the fluid is opposite. On the radiating $1/r$ terms, ${\partial }_{x_i}\simeq -(n_i/c_0){\partial }_t$, givingSpatial derivatives of $1/r$ produce near-field terms of orders $r^{-2}$ and $r^{-3}$, omitted here. Neglecting source delays is a separate approximation from discarding those observer near-field terms. If a leading integrated source vanishes by symmetry, its first nonzero delay correction can be as important as another retained multipole.

For the translating sphere, write its center as $\mathbf{X}(t)=a_1\cos (\omega t)\mathbf{e}$, with $\mathbf{e}$ a unit vector, and its velocity as $\mathbf{V}=\dot{\mathbf{X}}$. Let $\mathcal{V}=4\pi a^3/3$, $\eta =a\omega /c_0\ll 1$ and $d=a_1/a\ll 1$. Impermeability gives $u_n=\mathbf{V}\cdot {\mathbf{n}}_s$. Since its volume is constant,in the leading compact formula. The individual local thickness sources have scale ${\rho }_0(a_1/r){\eta }^2$, but their net monopole cancels.

At first order in amplitude and leading order in $\eta$, the exterior incompressible velocity potential is $\varphi =-a^3\mathbf{V}\cdot {\mathbf{r}}_s/(2r_s^3)$, where ${\mathbf{r}}_s$ is measured from the mean center. It has ${\partial }_{r_s}\varphi =\mathbf{V}\cdot {\mathbf{n}}_s$ at $r_s=a$. Linear Bernoulli equation gives $p^{\prime }=-{\rho }_0{\varphi }_t=({\rho }_{0}a/2)\dot{\mathbf{V}}\cdot {\mathbf{n}}_s$ there. Since $\int n_{s,i}{n}_{s,j}dS=(4\pi a^2/3){\delta }_{ij}$,This is the added mass of a sphere. The explicit leading loading contribution isIt scales as ${\rho }_0(a_1/r){\eta }^3$. In homentropic small-amplitude flow the volume stress is quadratic to leading order, $T_{ij}=O({\rho }_0{V}^2)$ over volume $O(a^3)$, so the volume acoustic quadrupole has scaleIt is smaller than the linear dipole by $O(d\eta )=O(V/c_0)$ and can contain a second harmonic.

For completeness, the exact thickness-source integral has a nonzero first delay correction even though $m=0$. Its first spatial moment isExpand its integrand at ${\tau }_y=\tau +\mathbf{n}\cdot \mathbf{y}/c_0$. The leading nonzero translating-sphere acoustic thickness dipole is thereforeIt too scales as ${\rho }_0(a_1/r){\eta }^3$. Thus zero net-volume monopole does not mean zero thickness radiation. The full leading linear sound, summing thickness and loading, isThe leading compact equation gives the requested zero $\dot{m}$ and added-mass $\dot{f}$ terms; this extra delay calculation is necessary to obtain the complete linear field when its nominal monopole has cancelled.

Write $q={\partial }_z\bar{\rho }>0$ for the unstable, horizontally averaged density gradient. A turbulent displacement over a mixing length of order $W$ encounters a density contrast $\Delta \rho \sim Wq$. Balancing its buoyancy acceleration with the inertial acceleration givesA turbulent diffusivity for density is therefore $D_T=K{u}_{\ast }W=K{W}^2\sqrt{gq/\bar{\rho }}$, absorbing dimensionless factors into $K>0$. The upward density flux is $J_{\rho }=⟨w^{\prime }{\rho }^{\prime }⟩=-D_{T}q$: turbulent exchange carries denser water downwards and lighter water upwards. Defining $F_{\rho }=-J_{\rho }$ as the positive downward flux yields the buoyancy-driven turbulent density diffusion lawThis is a flux per unit horizontal area, with units of mass per area per time. The total density transport across a horizontal section is $W^2{F}_{\rho }$.

The assumptions are established high-Reynolds number turbulence, width-limited eddies, an instantaneous local balance between buoyancy and inertia, approximately constant $K$, negligible molecular diffusion, and zero mean vertical volume flux. The last condition follows from incompressibility and impermeable sidewalls in a closed tube. Density variations and the eddy statistics must also vary slowly enough for a local mixing-length closure to be meaningful.

Horizontal averaging of mass conservation, with upward $z$, gives ${\bar{\rho }}_t=-{\partial }_z{J}_{\rho }={\partial }_z{F}_{\rho }$. Differentiating once more supplies the requested evolution of the density gradient:Under the Boussinesq approximation, replace $\bar{\rho }$ in the inertial prefactor by the constant reference density ${\rho }_0$; retain the changing density gradient inside the flux.

Write the inward radial velocity as $u_R=-v(R)$, with $v>0$. Steady spherical mass conservation and the radial Euler equations for an inviscid fluid giveFor a barotropic fluid, define the sound speed $c_s^2=dp/d\rho >0$ and the barotropic enthalpy function by $dh=dp/\rho$. Choosing $\Phi (\infty )=0$, integration gives the Bernoulli equationThe continuity equation gives ${\rho }^{\prime }/\rho =-2/R-v^{\prime }/v$. Eliminating mass density from the momentum equation produces the barotropic spherical accretion equationThe sonic point is where the inward Mach number $v/c_s$ equals one. A smooth solution cannot cross the vanishing coefficient on the left with a nonzero right-hand side, so a regular sonic radius must obeyConversely a continuous flow that is subsonic at infinity and supersonic inward must cross unit Mach number; smooth passage selects a simultaneous zero of the two sides, hence a regular sonic point. The associated critical accretion branch, rather than an arbitrary subsonic solution with smaller flux, is selected by these outer and inner requirements.

For the polytropic equation of state, set $\Gamma =1+1/m$ to distinguish its exponent from any separately specified perturbation exponent. Assume $K>0$ and first take the usual positive index $m>0$. ThenFor the attractive mixed potential, assume $\lambda ,\mu >0$. The sonic condition and Bernoulli equation giveHence polytropic accretion in a mixed inverse-power potential reduces toFor $0<m\le 1/2$ all nonzero coefficients in the left side are nonnegative for $R_s>0$, with at least one strictly positive: there is no finite positive root. For $m>1/2$ the constant term is negative and the leading coefficient positive, giving exactly one positive root. Define $B=(2m-3)\mu /4$ and $C=(2m-1)\lambda /2$. ThenThis is the finite-radius transonic range for the two positive attractive terms.

The critical point has real distinct crossing slopes. To check this rather than just solve the sonic equalities, put $U=\lambda /R_s^2$, $V=\mu /(2R_s)$ and $x=R_s{v}_s^{\prime }/v_s$. Differentiating the barotropic spherical accretion equation at its simultaneous zero givesIts discriminant is $8m[2(2m-1)U+(2m-3)V]/(U+V)$. The sonic Bernoulli equation says $(2m-1)U+(2m-3)V=2m{c}_{\infty }^2$, so this discriminant is strictly positive for $m>1/2$. The accretion crossing is the slope for which $d\log (v/c_s)/d\log R<0$, namely the smaller root. The two slopes have opposite signs of this Mach-number derivative. This is the regular transonic saddle, not a singular vertical tangent.

The corresponding mass accretion rate follows without an additional arbitrary mass density normalization:One may equivalently replace ${\rho }_s$ by $[c_{s,s}^2/(\Gamma K){]}^m$. With a nonzero $\lambda$ term, inner free fall has $v\sim \sqrt{2\lambda }/R$ and continuity gives $\rho \propto R^{-1}$. Thus $v^2/c_s^2\propto R^{-2+1/m}$ increases without bound precisely in the positive-index range $m>1/2$, consistent with the required inner supersonic branch.

The assumptions on the parameters matter. If $\lambda =0$ and $\mu >0$, the finite sonic radius instead requires $m>3/2$, the usual Bondi accretion range; $m=3/2$ is the zero-radius limiting critical case. If $\mu =0$ and $\lambda >0$, the condition remains $m>1/2$. With arbitrary signs of $\lambda ,\mu$, one must use the quadratic and retain only positive roots with $c_{s,s}^2>0$ and real admissible crossing slopes; attraction was not explicitly given as a sign condition in the source.

If positive polytropic index was not intended as an implicit physical convention, there is a further mathematical range. For $K>0$, positive compressibility also permits $m<-1$, giving $0<\Gamma <1$. Both potential terms attractive then give a positive sonic root as well:The same mass accretion rate formula applies; its crossing-slope discriminant is positive. The range $-1<m<0$ has $c_s^2<0$ and no physical acoustic sonic point, $m=-1$ has constant pressure and zero sound speed, and $m=0$ does not define the printed equation of state. Thus the usual answer $m>1/2$ assumes $m>0$ and positive attractive coefficients; these conventions should not be confused with additional printed restrictions.

Assume a nonrotating, nonmagnetic equilibrium, inviscid adiabatic perturbations, and no dissipation. Let $g={\varphi }^{\prime }(R)>0$, so hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use $\mathbf{\xi }{e}^{i\omega t}$ for the fluid displacement. To avoid a source ambiguity, write Eulerian fluid perturbations as $p_1,{\rho }_1,{\varphi }_1$, and use $\Delta$ for quantities following the displaced material element, as in Eulerian and Lagrangian fluid perturbations. They are related by $\Delta p=p_1+{\xi }_R{p}^{\prime }$ and similarly for mass density and potential.

The printed identity needs two corrections: its compression denominator must be $\gamma p$, not $\gamma \rho$, and the pressure and potential appearing in this form of the identity must be Eulerian fluid perturbations. The printed denominator has the wrong dimensions for an energy density. The claimed Lagrangian interpretation does not produce the displayed buoyancy-compression split; the following derivation specifies the consistent version.

Linearized mass conservation, the adiabatic relation, and the Poisson equation giveConsequently $p_1=-\gamma p\nabla \cdot \mathbf{\xi }-{\xi }_R{p}^{\prime }$. Linearizing the Euler equations for an inviscid fluid about the static star gives the linear adiabatic stellar oscillation equationsRequire regularity at the center, a vanishing Lagrangian surface pressure, and the usual matching to a decaying exterior gravitational potential. Because $p$ and $\rho$ vanish at the surface, the pressure boundary term vanishes for regular displacements; compactly supported trials also suffice below. The perturbed Newtonian gravitational potential solves Laplace's equation outside the star and must be included there when its field energy is computed.

Multiply the momentum equation by $-{\mathbf{\xi }}^{\ast }$ and integrate. The pressure term is $-{\int }_V{p}_1\nabla \cdot {\mathbf{\xi }}^{\ast }dV$. Let $d=\nabla \cdot \mathbf{\xi }$. The adiabatic relation and hydrostatic equilibrium implyThe pressure and equilibrium-gravity terms then combine pointwise aswhere $N^2=g[(1/\gamma )p^{\prime }/p-{\rho }^{\prime }/\rho ]$ is the stellar buoyancy frequency. The potentially troublesome cross terms cancel exactly.

For the perturbed self-gravity term, integration by parts givesThus the corrected stellar displacement energy identity isThe right side is a real quadratic form. With these conservative boundary conditions, the force operator is a self-adjoint operator. Its Rayleigh quotient characterizes squared mode frequencies, and a negative admissible trial quotient proves an unstable spectral component; it does not require the trial displacement itself to be an exact eigenfunction.

For a generic compressive disturbance of scale $\ell$ in a region with slowly varying equilibrium coefficients, estimate ${\rho }_1\sim \rho \xi /\ell$, $p_1\sim \gamma p\xi /\ell$ and ${\varphi }_1\sim 4\pi G{\rho }_1{\ell }^2$. The self-gravity and compression energy densities have orderswith adiabatic sound speed $c_s^2=\gamma p/\rho$. Under the plausible stellar estimate $c_s^2\sim G\bar{\rho }{a}^2$, global modes can have comparable gravity and pressure contributions, while their ratio for small-scale compressive modes is of order $(\rho /\bar{\rho })(\ell /a{)}^2$. Thus perturbations with $\ell \ll c_s/\sqrt{4\pi G\rho }$ admit the Cowling approximation. This estimate compares two nonzero compressive terms; it cannot alone justify neglecting gravity for a trial deliberately chosen to have $p_1=0$.

For an explicit convective trial, choose $c<d<e<b$ and a nonzero smooth bumpLet $Y$ be a normalized spherical harmonic of degree $\ell \ge 1$, let $L=\ell (\ell +1)$, and let ${\nabla }_{\Omega }$ be the unit-sphere gradient. The pressure-free localized stellar convective trial isIt is supported away from the center and surface, so it satisfies every needed mechanical boundary condition. Since ${\Delta }_{\Omega }Y=-LY$,Therefore $p_1=-\gamma p\nabla \cdot \mathbf{\xi }+\rho g{\xi }_R=0$ exactly. Its buoyancy energy is ${\int }_d^e\rho N^2{R}^2|f{|}^{2}dR<0$, whereas its kinetic norm ${\int }_d^e\rho R^2(|f{|}^2+L|h{|}^2)dR$ is positive. In the Cowling approximation this gives a negative Rayleigh quotient, and hence $N^2<0$ in an interior shell implies convective instability. Retaining perturbed self-gravity only adds a nonpositive term, so the same trial proves instability in the full conservative problem too.

There is also a quantitative small-scale check for this very trial. Its mass density amplitude is ${\rho }_1=(\rho N^2/g)fY$, independent of $\ell$. Write ${\rho }_1=s(R)Y$ and ${\varphi }_1=F(R)Y$. The radial Poisson equation givesBy the Cauchy-Schwarz inequality, $A\le 4\pi G(\int R^4|s{|}^{2}dR{)}^{1/2}(A/L{)}^{1/2}$, and so the angular-degree bound on perturbed stellar self-gravity isIt tends to zero with horizontal scale $R/\ell$, while the strictly negative buoyancy integral is fixed. This justifies the small-scale neglect of gravity even though the trial's compression term vanishes.

For $\nu =1/2$, the gas-phase metallicity relation reduces to $Z=2y(1-M/M_0)$. Elimination of $M$ from mass conservation givesWith a fixed initial mass function, let $\kappa$ be the number of surviving low-mass tracer stars per unit formed stellar mass. Successive stars inherit the monotonically increasing gas-phase metallicity, so the cumulative and differential counts are respectivelywhere $Z_f=2y(1-M_f/M_0)$ is the abundance at the epoch when star formation stops. Above $Z_f$ the cumulative count is constant, and the differential count vanishes. In the complete gas-exhaustion limit, $Z_f\to 2y$ and $N_{\mathrm{total}}=2\kappa M_0$, so the normalized metallicity distribution function is $p_Z=1/(2y)$ on $0<Z<2y$; at a finite stopping epoch it is $1/Z_f$. The distribution is flat in linear metallicity. A logarithmic metallicity distribution instead satisfies $dN/d{\log }_{10}Z=(\ln 10)ZdN/dZ$ and is not flat.

Let $\psi (t)$ be the star formation rate, $M$ the remaining gas mass, and $S$ the mass put into stars. With zero returned bulk gas, pristine galactic gas inflow, and no galactic outflow, mass conservation givesA well-mixed gas reservoir loses existing metals at rate $Z\psi$ and gains newly synthesized metals at rate $y\psi$; the incoming gas brings none. ThusFor $0<\nu <1$, eliminate time and integrate:The proportional-infall model of galactic chemical evolution therefore gives, for initially pristine gas,The displayed form requires $Z_0=0$; more generally $Z=y/\nu +(Z_0-y/\nu )(M/M_0{)}^{\nu /(1-\nu )}$. Also $S=(M_0-M)/(1-\nu )$ for an initially star-free reservoir, so $M/M_0$ is not the gas fraction of a galaxy $M/(M+S)$. As a check, the $\nu \to 0$ limit is the closed-box model of galactic chemical evolution relation $Z=y\log (M_0/M)$; at $\nu =1$ the gas mass is constant and one must instead solve $\dot{Z}=(y-Z)\psi /M_0$. Retaining a nonzero stellar yield while setting bulk mass return to zero is the prescribed trace-metal approximation, not a literal claim that stellar ejecta have no mass.

Assume the physical coefficients $Q_a,Q_b$ are positive. Differentiate the catastrophic disruption threshold:The bracket is strictly increasing from a negative value to positive infinity, so there is one critical point and it is the global minimum. In fact $Q_D^{\ast }$ tends to infinity at both endpoints $D\to 0$ and $D\to \infty$. The strength-gravity disruption transition is thereforeAt this diameter, $b{Q}_b{D}_w^b=a{Q}_a{D}_w^{-a}$, so $Q_w^{\ast }=[(a+b)/b]Q_a{D}_w^{-a}$. This minimizes the full catastrophic disruption threshold; it need not be where its two contributions are equal.

For equal-density spherical planetesimals, the specific impact energy per unit target mass is $\frac{1}{2}(D_{\mathrm{im}}/D_w{)}^3{v}_{\mathrm{rel}}^2$. Setting this to $Q_w^{\ast }$ gives the catastrophic projectile diameterHere $v_{\mathrm{rel}}$ is the impact speed, the threshold is defined per target mass, and energy-coupling changes are not included. If projectile and target densities differ, multiply the diameter by $({\rho }_{\mathrm{target}}/{\rho }_{\mathrm{projectile}}{)}^{1/3}$. For $a=1/2$, $b=3/2$,so collecting the powers givesThe result presumes that this projectile diameter is available in the population.

For the dust, $n(D)dD$ is the total number in a diameter interval, not a number per unit volume. The mass normalization of a power-law size distribution with dust mass $M/2$ givesThe approximation uses $(D_{\min }/D_{\max }{)}^{4-\alpha }\ll 1$. Merely having a small diameter ratio does not guarantee this accuracy if $\alpha$ approaches four, so retain the exact denominator in that case.

With $\gamma >0$, the stated Poynting–Robertson drag law integrates directly:Thus dust arrives in order of increasing diameter. This is burst transport under inverse-size radial drag, assuming an instantaneous release at $r_0$ and no intervening destruction or changes of diameter.

Dust that reaches the planet's orbital radius has several possible fates. It can pass without a close encounter and continue spiralling towards the star; undergo resonant trapping of dust in an exterior mean-motion resonance, where resonant torque temporarily balances the drag; be scattered onto a different stellar Kepler orbit or ejected; or collide with the planet and be accreted. Resonant trapping of dust can increase orbital eccentricity until escape or an encounter becomes possible. Temporary planet-bound capture can become permanent if drag or another dissipative process removes sufficient energy. Arrival at a radial circle alone does not guarantee a collision with the planet at the correct longitude.

A more massive planet has a larger Hill sphere, stronger resonant perturbations and stronger planetary scattering. Capture into a mean-motion resonance is favoured when drift across it is slow compared with its libration timescale. Since $|\dot{r}|\propto D^{-1}$ here, larger grains drift more slowly and are generally easier to trap; small fast-drifting grains more readily cross a resonance. Grain radiation pressure can also move resonance locations, though the prescribed nearly circular drift model omits that additional change. For direct impact, a larger physical radius increases the geometric cross-section, and gravitational focusing gives, in the two-body encounter approximation,A compact massive planet can nevertheless scatter or eject grains rather than hit them: its encounter region can be much larger than its physical radius. Slower dust migration permits more repeated encounter opportunities. These competing effects mean neither planet mass nor grain diameter alone specifies an accretion probability.