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A Keplerian accretion disk with a zero-torque inner boundary condition and local blackbody cooling has $T_{\mathrm{eff}}^4=3GM\dot{M}(1-\sqrt{r_{\mathrm{in}}/r})/(8\pi \sigma r^3)$. The Stefan–Boltzmann law uses the flux through one face; the standard thin-disk dissipation flux includes both faces. The scale $T_{\mathrm{in}}=[3GM\dot{M}/(8\pi \sigma r_{\mathrm{in}}^3){]}^{1/4}$ is a normalization, not the actual temperature at the stress-free inner edge.

For a Keplerian accretion disk with $\bar{\nu }\propto r^{5/2}$, dimensionless $g=\bar{\nu }\Sigma /({\bar{\nu }}_0{\Sigma }_0)$ satisfies $g_{\tau }=4x^{5/2}{g}_{xx}+6x^{3/2}{g}_x$. In $y=x^{-1/2}$ this is $g_{\tau }=y{g}_{yy}$. The ansatz $g=G(y/\tau )$ gives $G=A+B{e}^{-y/\tau }$; finite disk mass selects $A=0$. Its nonzero profile has logarithmically infinite disk angular momentum on an infinite radial domain.

Integrating the Keplerian accretion disk diffusion equation gives $dJ/dt=6\pi \sqrt{G{M}_{\star }}[r^{3/2}{\partial }_r(\bar{\nu }\Sigma ){]}_{\mathrm{inner}}^{\mathrm{outer}}$ when the angular-momentum integral and boundary limits are legitimate. An inner zero-torque inner boundary condition alone does not eliminate the outer angular-momentum flux. For divergent total $J$, use finite-radius cutoffs before making a conservation claim.

Far from the inner edge of a steady Keplerian accretion disk, the inner angular-momentum flux is negligible compared with $\dot{M}j$. Since $\Omega \propto R^{-3/2}$, the viscous torque in an accretion disk becomes $\mathcal{G}=3\pi \nu \Sigma j$. The steady balance $\mathcal{G}\simeq \dot{M}j$ therefore gives $\nu \Sigma =\dot{M}/(3\pi )$. With the stated surface density of a disk,Steady conservation of mass gives $\dot{M}=-2\pi R\Sigma v_R$, and consequentlyBoth relations use the far-inner-edge approximation, rather than extrapolating the stress-free boundary profile to the inner edge.

For the trace passive scalar, put $\sigma =\Sigma C$. Substitute this into the contaminant conservation equation and expand its advective flux:The braces multiplying $C$ vanish by background conservation of mass. Hence the concentration satisfiesThe small contaminant fraction ensures it does not alter the ambient disk dynamics. The surface-density model already averages vertically; an axisymmetric background and ring-like injection give $C=C(R,t)$. More generally, rapid differential orbital motion winds azimuthal structure and facilitates mixing, but radial-only dependence remains a symmetry or mixing assumption, not a property of every initial injection.

For the axisymmetric advection-diffusion equation, cylindrical geometry givesHere $k=\zeta \nu$ and $k\Sigma =\zeta {\nu }_0{\Sigma }_0$ are independent of radius. ThusLet $Y=\ln (R/R_{\ast })$, with any fixed reference length $R_{\ast }$. Since $C_Y=R{C}_R$ and $C_{YY}=R^2{C}_{RR}+R{C}_R$, this is the tracer diffusion in a power-law accretion disk equationAssume $\zeta >0$, ${\nu }_0>0$, and write $v=3{\nu }_0/2$, $D=\zeta {\nu }_0$. The change $X=Y+vt$ turns this into the heat equation $C_t=D{C}_{XX}$. The radial Dirac delta function must be transformed with its Jacobian:Convolution with the Gaussian heat kernel givesThis full-line logarithmic solution extends the far-edge power-law model to $0<R<\infty$. An actual finite inner boundary would require its own boundary condition and modified kernel.

The fraction must be computed with the contaminant mass measure, rather than with $CdR$. IndeedThe total contaminant mass is the conserved quantity $M_c=2\pi {\Sigma }_0{C}_0/R_0$. Its normalized logarithmic radius is therefore a normal distribution with mean $Y_0-vt$ and variance $2Dt$. Integrating this tail gives, for $r>0$,The complementary error function accounts for the one-time mass fraction outside the specified radius.

For an outward target $r_1>1$, put $\ell =\ln r_1>0$. Since $\mathrm{erfc}$ is decreasing, maximize $F$ by minimizing $z(t)=(\ell +vt)/\sqrt{4Dt}$. Its derivative has the sign of $vt-\ell$, so the unique maximum occurs at $t_{\ast }=\ell /v=2\ln r_1/(3{\nu }_0)$. At this time $z(t_{\ast })=\sqrt{v\ell /D}$. Thus the printed expression is correctly derived as the maximum instantaneous exterior fraction:

It is not, however, the probability of ever reaching the radius. A tracer can pass through the target and later return; its history is not counted by a one-time exterior fraction. The underlying logarithmic diffusion process iswhere $B_t$ is standard Brownian motion. To compute its actual hitting probability, first put absorbing levels at $-A$ and $\ell$ relative to $Y_0$. Conditioning on the first small time step gives the backward equation $D{h}^{\prime \prime }-v{h}^{\prime }=0$ for the probability of reaching the upper level first, with $h(-A)=0$, $h(\ell )=1$. Solving givesLetting the lower level tend to $-\infty$ exhausts all finite-time upward-hitting paths. Therefore the correct eventual probability in this idealized model isFor example, if $\ln r_1=2\zeta /3$, the maximum exterior fraction is $\frac{1}{2}\mathrm{erfc}(1)\simeq 0.07865$, whereas the actual hitting probability is $e^{-1}\simeq 0.36788$. This is an explicit counterexample to the last inference in the question and an instance of maximum occupation is not a hitting probability.

For $r_1=1$, the target is the initial radius and the hitting probability is one, while ${\sup }_{t>0}F(1,t)=1/2$ with no attained maximum. For $0<r_1<1$, the inward drift ensures eventual hitting with probability one, and ${\sup }_{t>0}F(r_1,t)=1$ as $t\downarrow 0$. The square-root expression involving $\ln r_1$ is therefore only meaningful for the outward-target case, and even there it describes occupancy rather than first passage.

A passive disk is heated predominantly by intercepted stellar radiation, rather than by its own viscous dissipation. A rough luminosity criterion, ignoring geometric factors, isThe intrinsic accretion disk luminosity is of order $GM\dot{M}/(2R_{\ast })$, with additional accretion energy potentially released at the stellar surface. The local test is more precise: absorbed irradiation must dominate the viscous flux at the radius being considered. For an ordinary steady Keplerian accretion disk, sufficiently far outside its inner edge,where $\alpha$ is the irradiation grazing angle of a disk. Reflection, very small incidence angles and shadowing can invalidate a simple total-luminosity comparison. The following calculation assumes an illuminated, absorbing surface.

At a point $(R,z_s)$ on the upper surface, the outward unit normal in the meridional plane is proportional to $(-z_s^{\prime },1)$, and a light ray from the point source is directed along $(R,z_s)$. Its flux through the upper face is the incident inverse-square flux times minus their scalar product:For a thin disk with small surface slope this becomesThe tangent slope and ray slope differ by $\alpha \simeq z_s^{\prime }-z_s/R$. Positive incidence requires disk flaring, $d(z_s/R)/dR>0$, and an unobstructed ray from the star. A negative value of the formula would mean that direct illumination of this face is absent, not that it receives a negative heating flux. The hierarchy $z_s\gg R_{\ast }$ makes the finite angular size of the star negligible here.

Let $a=\mathcal{R}/\mu$, so the ideal gas equation of state gives the isothermal sound speed $c_s^2=a{T}_c$. In the stellar vertical gravitational potential, hydrostatic equilibrium has vertical acceleration $-{\Omega }_K^{2}z$, where ${\Omega }_K^2=GM/R^3$. Integration of $c_s^2{\partial }_z\rho =-\rho {\Omega }_K^{2}z$ gives a Gaussian atmosphere with disk scale heightConsequently $q=z_s/R=fH/R$ satisfies $T_c=GM{q}^2/(a{f}^{2}R)$. Substitute this into the stipulated Stefan–Boltzmann law $F_d=\sigma T_c^4$ and the irradiation flux:Here $f$ specifies surface height in units of disk scale height and is unrelated to the wind-energy fraction in the preceding solution. This separable differential equation integrates exactly toIf the thin-disc solution is formally matched to $q(R_M)=1$, then $B=1-R_0^2/R_M^2$, givingIt follows that the flaring passive disk asymptotic isThis applies in the thin intermediate region $R\ll \min (R_0,R_M)$, in addition to the point-source and stellar-gravity assumptions. The corrections are of relative order $R^2/R_0^2+R^2/R_M^2$. In particular, the printed hierarchy $R\ll R_M$ must still be accompanied by the initial thin-disc condition $q\ll 1$: if $R_M\gg R_0$, radii larger than $R_0$ cannot satisfy that approximation. The matching at $q=1$ is an extrapolation used to set an integration constant; the derivation is not a valid treatment of a thick surface at $R_M$ itself.

Identify the stellar surface temperature with its effective radiating temperature, so that luminosity of a spherical blackbody gives $L_{\ast }=4\pi R_{\ast }^2\sigma T_{\ast }^4$. The given stellar surface scale height has $H_{\ast }^2=a{T}_{\ast }{R}_{\ast }^3/(GM)$, hence $GM/(a{T}_{\ast }{R}_{\ast })=(R_{\ast }/H_{\ast }{)}^2$. Substituting into $R_0$ yieldsThus a small stellar surface scale height leaves a wide interval above $R_{\ast }$ and below $R_0$ in which the thin disk flaring law can apply. The point-source approximation imposes the stronger lower cutoff obtained from $z_s\gg R_{\ast }$:This cutoff is still much smaller than $R_0$ when $R_0\gg R_{\ast }$. The full interval is therefore $R_{\ast }^{7/9}{R}_0^{2/9}\ll R\ll \min (R_0,R_M)$, whenever these bounds overlap. An assumed thickening radius must lie above the lower cutoff for all of the original approximations to hold simultaneously. The stellar luminosity-temperature relation is essential to the final comparison; independent arbitrary values of $L_{\ast }$ and $T_{\ast }$ would not imply it.

Take the accretion rate $\dot{M}>0$ to mean inward flow. In a steady state, conservation of mass makes the inward mass flux independent of radius, so $\mathcal{F}=\dot{M}$. Integrating its radial expression givesFor Keplerian rotation, $\Omega =(G{M}_{\ast }/r^3{)}^{1/2}$ and the viscous torque in an accretion disk is $\mathcal{G}=3\pi \bar{\nu }\Sigma \sqrt{G{M}_{\ast }r}$. The zero-torque inner boundary condition therefore sets $\bar{\nu }\Sigma =0$ at $r_{\ast }$. It fixes $C=-\dot{M}\sqrt{r_{\ast }}/(3\pi )$, giving the steady density profileThis is the Keplerian accretion disk solution on $r_{\ast }<r<r_{\mathrm{out}}$. It determines the kinematic viscosity–surface density product; a separate viscosity closure is needed to turn it into an explicit power law. If the kinematic viscosity is finite and nonzero at the inner edge, the surface density tends to zero there.

The standard thin-disk dissipation flux, summed over both faces, follows by substituting the Keplerian accretion disk profile into the viscous heating rate:The Stefan–Boltzmann law gives $\mathcal{C}=2\sigma T^4$ because there are two emitting faces. Local radiative equilibrium consequently gives the effective-temperature profile of a zero-torque diskHere $\sigma$ is the Stefan-Boltzmann constant. The scale $T_{\mathrm{in}}$ is not the temperature exactly at the inner boundary: the zero-torque inner boundary condition makes that formal temperature zero. Maximizing $T^4$ shows that the maximum effective temperature of a zero-torque disk occurs at $r=49r_{\ast }/36$, with $T_{\max }\simeq 0.488T_{\mathrm{in}}$.

At equal central mass and accretion rate, characteristic effective temperatures scale as $r_{\ast }^{-3/4}$. Thus, comparing corresponding values of $r/r_{\ast }$,The neutron-star disk is about 180 times hotter. The Planck law and Wien displacement law move its characteristic emission to about 180 times higher frequency, or 180 times shorter wavelength. A white dwarf disk commonly emits in optical and ultraviolet bands, while the hotter neutron star disk can emit in X-rays. Absolute bands require an actual accretion rate; the relative shift follows directly from the stated scaling.

A cataclysmic variable is a close semidetached binary in which a white dwarf accretes from a cool, usually low-mass donor star filling its Roche lobe. In an ordinary hydrogen-rich system the donor star is often near the lower main sequence. In a nonmagnetic cataclysmic variable (a disk-fed CV), the white dwarf's magnetic field is too weak to control the flow over the disk. This leaves a characteristic disk-fed geometry, summarized in the original schematic below. NASA's introduction to cataclysmic variables describes the basic components.

Roche-lobe overflow passes through the inner Lagrange point $L_1$. A nearly ballistic stream bends in the rotating frame and strikes the outer accretion disk, producing an accretion-disk stream-impact hot spot. Its retained angular momentum prevents direct radial infall. Viscous evolution of an accretion disk transports angular momentum outward while gas moves inward through a nearly Keplerian accretion disk. Close to the white dwarf, gas slows from orbital rotation toward the stellar rotation in an accretion-disk boundary layer. The disk, accretion-disk stream-impact hot spot, accretion-disk boundary layer, white dwarf and donor star all contribute to the spectrum and, where the orbital inclination permits, to the eclipses of an eclipsing binary. In particular, “nonmagnetic” describes the accretor's control of the flow; it does not imply that the donor star cannot sustain a magnetic field.

The gravitational power available from accretion is approximately $L_{\mathrm{acc}}=G{M}_{\mathrm{WD}}{\dot{M}}_{\mathrm{acc}}/R_{\mathrm{WD}}$. For a slowly rotating white dwarf and a thin steady Keplerian accretion disk, the specific energy changes from approximately zero far out to $-G{M}_{\mathrm{WD}}/(2R_{\mathrm{WD}})$ at the inner disk. Roughly half the available power is radiated by the accretion disk; the rest is released in the accretion-disk boundary layer as the orbital kinetic energy is dissipated. Stellar rotation and departures from a steady thin disk change this partition.

A classical nova has a different energy source. Transferred hydrogen accumulates on the white dwarf; compression heats the base of its envelope until temperature-sensitive hydrogen burning accelerates. Electron degeneracy pressure initially weakens the expansion response to heating, helping a thermonuclear runaway develop. The envelope subsequently expands and ejects material, producing a large optical outburst followed by a decline as the ejecta expand and residual burning ends. The white dwarf usually survives, so continued accretion can build another fuel layer. A rough recurrence scale is the ignition-envelope mass divided by the mean accretion rate; both this mass and the rate vary strongly among systems. The event is an envelope eruption, and the retained fraction is not automatically unity. Starrfield, Iliadis and Hix's nova calculations explains this nuclear mechanism. A classical nova is powered by unstable nuclear burning on the white dwarf.

A dwarf nova undergoes recurrent, shorter brightenings powered principally by enhanced gravitational accretion. The hydrogen-ionization disk instability creates cold, mostly neutral and hot, ionized branches of the disk's accretion-disk thermal S-curve, separated by unstable equilibria. In quiescence the cool disk stores matter because inward transport is slow. Once a critical surface density is reached, a heating transition puts the disk into a hotter, more state with higher effective viscosity: the inward accretion rate and luminosity rise and the disk drains. A cooling transition returns it to quiescence, completing the cycle. A persistent increase in the donor star's transfer rate is not required. Sufficiently high transfer rates can keep the disk on its hot stable branch, giving a nova-like variable rather than ordinary disk cycles. Lasota's disk-instability analysis and Hameury's disk-instability review develop this picture. A dwarf-nova outburst is a disk instability, not a white-dwarf thermonuclear explosion. The names classify mechanisms and need not identify permanently distinct binaries: a nova-producing binary can also possess an unstable disk between nuclear eruptions.

The cataclysmic-variable orbital-period distribution is not smooth. For ordinary hydrogen-rich cataclysmic variables, prominent features are a cataclysmic-variable period gap around two to three hours, a cataclysmic-variable period minimum near eighty minutes, and an accumulation near that minimum. These are population features rather than absolute exclusions. Selection effects matter: luminous high-accretion rate systems are easier to find than faint evolved systems. Helium-transferring binaries have a different period range and are not described by the hydrogen-rich minimum. Gänsicke and collaborators' period-minimum study documents the observed accumulation.

The Roche-lobe-filling period-density relation makes the orbital period a measure of donor structure. Combining $R_d\simeq 0.46a(M_d/M{)}^{1/3}$ with Kepler's third law givesAs the donor star loses mass, its stellar radius response exponent $\zeta =d\log R_d/d\log M_d$ impliesA donor with $\zeta >1/3$ evolves toward shorter orbital periods. When its effective response falls below $1/3$, continued mass loss instead lengthens the orbital period: this is the cataclysmic-variable period bounce. A very low-mass donor may be substellar and increasingly affected by electron degeneracy pressure; the ideal degenerate scaling $R_d\propto M_d^{-1/3}$ illustrates the reversal. The precise cataclysmic-variable period minimum depends on thermal disequilibrium and the strength of orbital angular momentum loss. Near a turning point $|\dot{P}|$ is small; for an approximately steady evolutionary flow of systems, the number per period interval scales as $N(P)\propto 1/|\dot{P}|$, explaining the accumulation. Knigge, Baraffe and Patterson's donor-based evolutionary study relates the donor star's response to these period features.

Long-term transfer is driven mainly by losses of orbital angular momentum, rather than by disk outbursts. Magnetic braking of a binary star removes the cool donor star's spin through a magnetized stellar wind. Tidal synchronization makes the orbit replenish that spin, so the wind extracts orbital angular momentum. This is usually the dominant standard driving mechanism above the cataclysmic-variable period gap. Gravitational-wave emission from a binary system supplies a baseline loss, especially important below the gap. For a weak-field, slowly moving circular orbit, the circular gravitational-wave inspiral givesBecause ordinary transfer from the lighter donor star tends to expand its Roche lobe if orbital angular momentum is conserved, an external loss is needed to sustain contact. In the conservative contact approximation, the binary mass-transfer contact equation readswhere the stellar radius response exponent must match the evolutionary timescale and additional donor expansion has been neglected. The positive denominator on the stable branch makes ${\dot{J}}_{\mathrm{ext}}<0$ drive ${\dot{M}}_d<0$. Nova ejecta can introduce additional nonconservative binary mass transfer. Knigge's evolutionary discussion describes the standard loss mechanisms and their limitations.

In the disrupted magnetic braking model, relatively rapid mass loss above the cataclysmic-variable period gap keeps the donor star inflated relative to stellar thermal equilibrium. When the donor approaches the fully convective star transition, the model postulates a substantial reduction in magnetic braking of a binary star. The donor star can contract within its Roche lobe, suppressing Roche-lobe overflow near the upper edge of the gap. Gravitational-wave emission from a binary system continues to shrink the detached binary; near the lower edge the Roche lobe again reaches the donor radius and transfer resumes. No mass transfer is needed during the detached crossing. The density relation predicts a radius ratio $(3/2{)}^{2/3}\simeq 1.31$ between contact at three and two hours if the masses remain nearly fixed, illustrating the required inflation before detachment. The torque reduction is a model ingredient, not a claim that all fully convective stars lose their magnetic fields. Zorotovic and collaborators' detached-binary study tests the predicted detached population in the gap.

The standard formation channel starts with an initially wider binary star containing two main sequence stars. The initially more massive component evolves first and becomes a giant. Unstable Roche-lobe overflow can engulf the companion in a common envelope. Drag causes inward orbital motion, releasing two-body orbital energy and transferring angular momentum to the envelope. If the envelope is expelled before merger, a close detached binary survives, containing the exposed core, which becomes a white dwarf, and the lower-mass companion. This explains how a binary becomes much tighter than the giant progenitor's radius would have allowed. The outcome depends on envelope binding and the efficiency of energy deposition, summarized approximately by the common-envelope energy formalism; ejection is not guaranteed. Ivanova and collaborators' common-envelope analysis discusses the relevant physics and uncertainties.

Subsequent magnetic braking of a binary star and gravitational-wave emission from a binary system shrink the detached binary until its companion fills its Roche lobe. Stable Roche-lobe overflow, for a suitable binary mass ratio and stellar radius response exponent, then creates a cataclysmic variable. Its later nuclear eruptions, disk cycles and secular orbital period evolution occur on different timescales. The formation sequence is a wide binary, envelope ejection, a close detached white-dwarf binary, and angular-momentum-driven contact.

Interpret the outer Dirac delta function as a boundary source carrying its full mass, or as the limit of a narrow source lying just inside $r_2$. Then ${\int }_u^{r_2}vS(v)dv=r_2{S}_0$ for $r_1<u<r_2$, andThus the interior is the usual steady Keplerian accretion disk with a constant inward accretion rate. The product $\bar{\nu }\Sigma$ vanishes at the stress-free inner edge, and the surface density of a disk does too if the viscosity has a finite nonzero boundary value; far from that edge $\bar{\nu }\Sigma$ approaches $\dot{M}/(3\pi )$. Its density need not itself be constant because the viscosity may vary.

The outer-edge source of a viscous disk requires care about the boundary trace. The flux just inside the injecting edge is $\mathcal{F}(r_2^{-})=\dot{M}$, while the no-flux value outside the source is $\mathcal{F}(r_2^{+})=0$; the source produces the intervening jump. A simultaneously continuous interior condition $\mathcal{F}(r_2^{-})=0$ would contradict nonzero steady injection. A symmetrically regularized delta integrated only up to its center can instead give half its full weight; that would mean a different supplied mass, not a different interior solution for the same $\dot{M}$. The physically normalized one-sided limiting source removes this convention ambiguity.

Use $\dot{M}>0$ for inward mass accretion rate, and $T_m>0$ for magnetic torque adding angular momentum to the disk. Combining the continuity equation with conservation of angular momentum identifies the inward mass flux as $(dh/dr{)}^{-1}d\mathcal{G}/dr$. In a steady Keplerian accretion disk, this is the supplied constant $\dot{M}$, soThe boundary condition is $\mathcal{G}(r_{\mathrm{in}})=T_m$. Since $d\Omega /dr=-3\Omega /(2r)$, the viscous torque in an accretion disk is $\mathcal{G}=3\pi \bar{\nu }\Sigma h$. Hence the steady Keplerian accretion disk with an inner torque hasIn particular, $\Sigma (r_{\mathrm{in}})=\lambda \dot{M}/(3\pi \bar{\nu })$. For a physical steady solution with this boundary condition, $\lambda \ge 0$.

For $0\le \lambda \ll 1$, the surface density of a disk rises from a small inner value toward $\dot{M}/(3\pi \bar{\nu })$: this is the usual Keplerian accretion disk with an approximately zero-torque inner boundary condition. For $\lambda \gg 1$, the inner surface density of a disk is large and decreases approximately as $r^{-1/2}$, giving a torque-dominated accretion disk. This approximation applies where $r/r_{\mathrm{in}}\ll (\lambda -1{)}^2$; sufficiently far out, any fixed finite $\lambda$ again approaches the same constant asymptote.

The constant net inward angular-momentum flux isFor $\lambda >1$, net angular momentum moves outward. The strong-torque profile is therefore decretion disk-like in its stress-dominated structure, but the specified mass flux remains inward: a genuine net decretion disk requires an outward mass flux, not merely large $\lambda$.

At $r_{\mathrm{in}}=r_m$, the Maxwell stress tensor estimate givesUsing the magnetospheric truncation radius scaling in part (b), all dimensional parameters cancel:If the displayed $4\pi$ is retained and the scaling radius is assigned unit coefficient, the estimate is $1/(4\pi )$; retaining $\Sigma \simeq 2H\rho$ and exact free-fall speed throughout gives $\sqrt{2}/(4\pi )$. Those prefactors are not controlled by the scaling argument. In particular, it does not generate a parametrically large $\lambda \gg 1$.

In a steady state, mass conservation makes $\mathcal{F}$ independent of radius. Taking the accretion rate $\dot{M}>0$ for inward flow gives $\mathcal{F}=-\dot{M}$. The conservation of angular momentum equation then makes $-\dot{M}h+\mathcal{G}$ constant. The zero-torque inner boundary condition fixes this constant to $-\dot{M}{h}_{\mathrm{in}}$, soThe viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^3\bar{\nu }\Sigma {\Omega }^{\prime }=2\pi qh\bar{\nu }\Sigma$. Combining these expressions gives the steady accretion disk with arbitrary rotation lawFor $x=r/R_g$, the Paczyński-Wiita circular orbit formulas giveConsequentlyIn particular, $f(6)=0$ and $f\to 1$ far from the black hole, recovering the outer Keplerian accretion disk.

Inside the innermost stable circular orbit, the gas enters the plunging region of a black-hole accretion disk. Its rapid inward motion leaves little time for stresses to exchange angular momentum, motivating the zero-torque inner boundary condition. This is a thin-disc approximation; a strong magnetic stress could change it.

Matter supplied from very large radius has negligible specific orbital energy, while matter crossing the inner edge carries ${\epsilon }_{\mathrm{in}}=-\eta c^2$. With zero inner torque, no energy is supplied by a stress at that edge. If the heat released outside it escapes by radiative transfer, rather than being lost through inward advection, the integrated conservation of energy balance gives

Inside the innermost stable circular orbit, nearly circular motion is unstable and gas enters the plunging region of a black-hole accretion disk. Its inflow time becomes shorter than the time on which internal stress can communicate angular momentum back to the disk, motivating the zero-torque inner boundary condition at $r_{\mathrm{ISCO}}$.

In a steady state, the given diffusion equation impliesThe constant mass supply fixes $C=\dot{M}/(6\pi )$, and a second integration givesZero torque means $\bar{\nu }\Sigma =0$ at $r=r_{\mathrm{ISCO}}$, so $C_0=r_{\mathrm{ISCO}}^{1/2}$. Thus the Keplerian accretion disk relation is

For a Keplerian accretion disk, $\Omega =(G{M}_{\mathrm{BH}}/R^3{)}^{1/2}$, soHere $F$ is the dissipation summed over the two faces. The total luminosity isThis is the binding energy per unit mass of a circular orbit at the inner edge, multiplied by the accretion rate. The zero-torque inner boundary condition ensures that no additional mechanical work enters from smaller radii.

The luminosity emitted outside radius $R$ isWriting $y=(R_{\mathrm{ISCO}}/R{)}^{1/2}$, the condition $L(>R)=L/2$ becomes $3y^2-2y^3=1/2$. Its physical root is $y=1/2$, and henceHalf of a thin disk's luminosity therefore comes from only the innermost factor four in radius, so strong-gravity effects and the inner-boundary condition strongly influence its observed spectrum.

For a perfect gas, $c_s^2\sim kT/({\mu }_m{m}_p)$. At the margin of the Toomre stability criterion,Integrating viscous heating through the disk and applying radiative diffusion givesEquating these fluxes yieldsThe kinematic viscosity is $\bar{\nu }\sim \alpha c_s^2/\Omega$, soBecause a Keplerian accretion disk has $\Omega \propto r^{-3/2}$,

In a Keplerian accretion disk, $u_{\varphi }=R\Omega =(GM/R{)}^{1/2}$. Substitution into the steady azimuthal Navier--Stokes equation and cancellation of the common Keplerian factors givesSince $R^2\Omega \propto R^{1/2}$ and $d\Omega /dR=-3\Omega /(2R)$, this reduces to the standard viscous drift formulaThe minus sign describes inward drift when $\nu \Sigma R^{1/2}$ increases outward.

An optically thick annulus radiates as a blackbody from both faces, soFor a steady Keplerian accretion disk with a zero-torque inner boundary condition,ConsequentlyandAway from the inner edge, $T\propto R^{-3/4}$.

Ignoring inclination and distance factors, the multitemperature blackbody disk spectrum isSet $x=h\bar{\nu }/(k_{B}T)$. Since $T\propto R^{-3/4}$, $RdR\propto {\bar{\nu }}^{-8/3}{x}^{5/3}dx$, whereas the Planck function contributes ${\bar{\nu }}^3/(e^x-1)$. In the stated intermediate-frequency range the radial endpoints become $0$ and $\infty$, leaving a frequency-independent convergent integral. Therefore

Define the surface density of a disk, outward mass flux, internal torque, and surface magnetic torque byThe sign convention makes $\mathcal{G}$ positive for outward angular momentum transport in an ordinary Keplerian accretion disk. Vertical integration of mass conservation gives

The specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal equation by $r$, use the continuity equation to put its left-hand side in conservative form, and integrate over $z$. The assumed decay removes the vertical mass and viscous fluxes, whereas the magnetic surface stress remains:Subtracting $h$ times the integrated mass equation yieldsSince $F_M=-(dh/dr{)}^{-1}({\partial }_r\mathcal{G}-r\mathcal{T})$, substitution in mass conservation gives the required one-dimensional advection-diffusion equation

Write $y=\nu \Sigma$. A steady inward accretion rate has $F_M=-\dot{M}$, so the integrated angular-momentum equation isFor $\beta =0$, $\mathcal{T}=0$ and thereforeafter imposing the zero-torque inner boundary condition at $r_{\mathrm{in}}$. In a Keplerian accretion disk, $h\propto r^{1/2}$ and $\mathcal{G}=3\pi hy$, so