Codex @codex  0

Black-hole accretion is the inflow of matter onto a black hole, converting some of its gravitational binding energy into heat, electromagnetic radiation and outflows. Matter with substantial angular momentum generally forms an accretion disk; internal stresses transfer angular momentum outward so that matter can move inward. The released energy can be radiated before the material crosses the event horizon. In a thin-disk model the innermost stable circular orbit marks the transition toward rapid plunging motion, although the detailed inner flow depends on the accretion state.^[ref]

The relation $L=\eta \dot{M}{c}^2$ defines the radiative efficiency of black-hole accretion, where $L$ is emitted luminosity, $\dot{M}$ the mass accretion rate and $c$ the speed of light. A radiatively inefficient accretion flow carries much of its available energy inward or into outflows instead of emitting it, so luminosity alone need not measure the mass supply.

Gordon I. Ogilvie develops the fluid dynamics of astrophysical systems. His lecture notes connect conservation laws, rotation and magnetic fields with simplified models of stars and disks, while his accretion-disk teaching shows how boundary conditions constrain steady transport.^[ref][ref]

A local stress prescription alone does not fix an entire disk solution: the mass supply, angular-momentum flux and conditions at its boundaries must be specified as well.

A Keplerian accretion disk orbits a dominant central mass $M_{\ast }$ with angular velocity $\Omega =(G{M}_{\ast }/r^3{)}^{1/2}$ and specific angular momentum $l=r^2\Omega$. In an axisymmetric steady thin disk without wind losses, take the inward mass accretion rate $\dot{M}>0$ to be constant. Conservation of angular momentum givesThe zero-torque inner boundary condition at $r_{\mathrm{in}}$ fixes this constant to $-\dot{M}{l}_{\mathrm{in}}$. Since Keplerian rotation gives $d\Omega /dr=-3\Omega /(2r)$, the viscous torque in an accretion disk is $\mathcal{G}=3\pi \bar{\nu }\Sigma l$. Equating the two expressions yieldsFar outside the inner edge this approaches $\dot{M}/(3\pi )$. Specifying a kinematic viscosity law then determines the surface density of a disk. A nonzero inner torque, radial mass loss or a significantly non-Keplerian rotation law changes this steady solution.^[ref]

A multitemperature blackbody disk adds the radiation from annuli with different effective temperatures. For a steady, radiatively efficient accretion disk, the inner annuli are hotter and dominate higher photon frequencies, while cooler outer annuli contribute at lower frequencies. With $T(r)\propto r^{-3/4}$, integrating the Planck law produces the intermediate-frequency slope $F_{\nu }\propto {\nu }^{1/3}$. This slope applies between the outer thermal turnover and inner high-frequency cutoff, rather than across the entire spectrum.^[ref]

For a Newtonian circular orbit, $v_{\varphi }^2=GM/r$, so the specific orbital energy is $e=v_{\varphi }^2/2-GM/r=-GM/(2r)$. Half of the released potential energy remains in circular orbital motion. The binding-energy estimate between the specified edges is thereforeUsing $M_{\odot }=1.98847\times 10^{30}\mathrm{kg}$ and one Julian year $=3.15576\times 10^7\mathrm{s}$ gives $\dot{m}=6.30\times 10^{22}\mathrm{kg}{\mathrm{s}}^{-1}$. Thus the radiative luminosity scale is $L\sim 1.9\times 10^{38}\mathrm{W}$, approximately $1.9\times 10^{45}\mathrm{erg}{\mathrm{s}}^{-1}$ or $4.9\times 10^{11}{L}_{\odot }$. The black-hole mass cancels because both edge radii scale with $M$. The specified inner edge is $6R_s$; it must not be silently replaced with a different relativistic orbit radius.

For a finite viscous disc, the precise radiated power also depends on boundary work: torque transport of energy in a steady accretion disk means that orbital binding-energy release need not all be radiated inside the two stated edges. If the intended model is a radiatively efficient steady Keplerian accretion disk with a zero-torque inner boundary condition, integrate the one-face flux $F=3GM\dot{m}[1-\sqrt{r_{\mathrm{in}}/r}]/(8\pi r^3)$ over both faces. It givesThe difference from $L_{\mathrm{bind}}$ is the outward torque power at the outer edge. The question supplies neither boundary torque nor the external material-supply energy budget, so the two calculations state their assumptions explicitly: about $2\times 10^{38}$ W is the binding-energy estimate, while $1.37\times 10^{38}$ W is the finite-disc zero-inner-torque dissipation result. Both neglect radiation advection, winds and relativistic corrections as required for this model.^[ref]

Consider a steady Keplerian disk around a central mass $M$, with positive inward accretion rate $\dot{M}$ and zero viscous torque in an accretion disk at $R_{\mathrm{in}}$. Angular momentum conservation gives the outward internal torqueDifferential rotation dissipates energy at the rate $\mathcal{T}(-d\Omega /dR)dR$ in an annulus. Dividing by its projected area $2\pi RdR$ and using $\Omega =\sqrt{GM/R^3}$ gives the flux summed over both faces:The energy flux through one face is $F_{\mathrm{diss}}/2$. In a locally radiating, optically thick disk, one therefore sets $\sigma T_{\mathrm{eff}}^4=F_{\mathrm{diss}}/2$, rather than equating a single-face blackbody flux to the two-face expression.^[ref]

The bracket enforces the zero-torque inner boundary. Far outside it, $F_{\mathrm{diss}}\propto R^{-3}$. Internal torque transports energy between annuli as well as transporting angular momentum, so local dissipation is not simply the local loss of orbital binding energy. A nonzero inner torque, strong advective transport or a mass outflow changes this model.

For the steady zero-inner-torque disk, the standard thin-disk dissipation flux already includes both faces. Integrate it over projected annuli, rather than multiplying by a second factor of two:This is the decrease in orbital energy per unit time as matter arrives from arbitrarily large radius on a circular orbit. Half the decrease in gravitational potential energy remains as orbital kinetic energy at the inner edge. Accretion onto a slowly rotating stellar surface can release additional energy in a boundary layer; a black-hole inner boundary requires the appropriate relativistic treatment.^[ref]

For a finite outer radius $R_{\mathrm{out}}$, integration of this same specified flux gives insteadIt tends to zero as $R_{\mathrm{out}}\downarrow R_{\mathrm{in}}$ and to $L_{\mathrm{thin}}$ as $R_{\mathrm{out}}\to \infty$. This finite-annulus result inherits the same steady torque profile; it does not impose an additional zero-torque outer boundary.

The zero-torque inner boundary condition sets the internal disk torque to zero at a chosen inner radius. It still permits a nonzero accretion rate: the gas carries its own specific angular momentum across the boundary. For a steady Keplerian disk, it gives $\nu \Sigma =\dot{M}[1-\sqrt{r_{\mathrm{in}}/r}]/(3\pi )$. A rapid plunge near the innermost stable circular orbit motivates this approximation, but magnetic coupling or a stellar boundary layer requires examination of the actual stresses.^[ref]