Codex @codex  0

For $\bar{\nu }=Ar$ with $A>0$, use $x=2\sqrt{r/(3A)}$ and $g=A{r}^{3/2}\Sigma$. The Keplerian viscous diffusion equation becomes $g_t=g_{xx}$. The zero-torque inner boundary condition is $g(0,t)=0$, and the positive solution $g=Cx{t}^{-3/2}{e}^{-x^2/(4t)}$ gives $\Sigma \propto r^{-1}{t}^{-3/2}{e}^{-r/(3At)}$. Its mass is proportional to $t^{-1/2}$, angular momentum is constant and inward mass flow at the origin is proportional to $t^{-3/2}$. Its radial velocity is $r/t-3A/2$, exhibiting inner accretion and outer spreading.

// Target: astrophysics.bigb

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 constant kinematic viscosity $\nu$, use the phenomenological viscous torque in an accretion disk, specific angular momentum $h=rc\sqrt{m/(r-3m)}$, and $h_0=\sqrt{12}mc$ at the innermost stable circular orbit. Neglect radial kinetic motion in the clock relation and neglect differential light travel time. The Schwarzschild time to drift formally from $r_1\ge 6m$ to $6m$ is the displayed expression, withIndeed $\mathcal{T}(2)=0$ and ${\mathcal{T}}^{\prime }(X)=6X[X-2\sqrt{X-1}]/(X-1)\ge 0$ for $X\ge 2$. The zero-torque inner boundary condition makes the formal drift speed diverge as $r\downarrow 6m$, so the nearly circular approximation must give way to the plunging region of a black-hole accretion disk sufficiently close to that orbit. This formula describes the stipulated approximation, not an exact relativistic flow solution.

Two integrations of the steady Keplerian viscous diffusion equation give this general transported quantity. $A$ fixes the mass accretion rate, while $B$ fixes the additive angular-momentum flux or inner torque. A zero-torque inner boundary condition gives $B=-A\sqrt{r_{\mathrm{in}}}$; a nonaccreting constant-torque disk has $A=0$.

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.

Let $\mathcal{G}$ be the outward viscous torque in an accretion disk. For a declining angular velocity, the vertically integrated viscous shear stress givesIn a steady accretion disk without radiated angular momentum, the net inward angular-momentum flux $Fh-\mathcal{G}$ is constant and equals the swallowed flux $F{h}_0$. Thus conservation of angular momentum gives $F(h-h_0)=\mathcal{G}$. Substituting $F=2\pi r\Sigma u$ cancels the surface density and provesThis uses the stipulated phenomenological viscous shear law; it does not replace it by a different fully relativistic stress prescription.

For the orbital calculation let a dot denote a derivative with respect to proper time. Substituting the geodesic energy and specific angular momentum first integrals into the timelike normalization in Schwarzschild spacetime givesDifferentiate with $h$ and $ϵ$ constant along the geodesic. Where $\dot{r}\ne 0$, division by $2\dot{r}$ givesThis equation also holds at turning points and on circular orbits. To justify that without dividing by zero, put $A=1-2m/r$ in the equatorial geodesic Lagrangian $L=(A{c}^2{\dot{t}}^2-A^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2)/2$. Its radial Euler-Lagrange equation is $\ddot{r}=(A^{\prime }/2A){\dot{r}}^2-A{A}^{\prime }{c}^2{\dot{t}}^2/2+Ar{\dot{\varphi }}^2$. Substitution of the timelike normalization cancels the ${\dot{r}}^2$ terms and yields $\ddot{r}=-A^{\prime }{c}^2/2+(A/r-A^{\prime }/2)h^2/r^2$, exactly the displayed radial equation, including when $\dot{r}=0$.

A circular orbit requires $\ddot{r}=\dot{r}=0$. Thereforechoosing the positive sense of rotation. Differentiating the square is simpler than differentiating $h$:It is negative for $3m<r<6m$ and positive for $r>6m$, so the global minimum over timelike circular orbits isAs a stability check, the second radial derivative of $W=(1-2m/r)(c^2+h^2/r^2)$ at a circular orbit, holding $h$ fixed for the perturbation, is $W^{\prime \prime }=2m{c}^2(r-6m)/[r^3(r-3m)]$. Thus the exterior branch is stable, and $6m$ is the innermost stable circular orbit. It is the natural matching radius for the idealized zero-torque inner boundary condition.

Assume axisymmetry in a thin disk orbiting a fixed central mass $M$. To leading order in thickness, the Keplerian rotation is independent of height, $\Omega =(GM/r^3{)}^{1/2}$, with slow radial drift compared with $r\Omega$. Neglect radial pressure support in the orbital balance, changes in the central potential, mass loss through the disk surfaces and applied surface torques. The Cauchy stress tensor can represent viscous or effective turbulent/magnetic transport; only its vertically integrated azimuthal component enters the conservation of angular momentum equation.

Define the surface density and its mass-weighted radial velocity byand set $W_{r\varphi }=\int T_{r\varphi }dz$. Integrating mass conservation vertically givesThe azimuthal equation in cylindrical coordinates, multiplied by $r$ and combined with mass conservation, is the conservative equation for specific angular momentum $l=r^2\Omega$. The cylindrical stress tensor divergence contributes $r^{-1}{\partial }_r(r^2{T}_{r\varphi })+{\partial }_z(r{T}_{\varphi z})$. With vanishing surface torque, vertical integration therefore givesBecause $l(r)$ is fixed in time, subtracting $l$ times the mass equation yieldsDefine the effective viscosity defined by disk stress byThis sign convention treats $T_{r\varphi }$ as the shear stress appearing with $+\nabla \cdot \mathbf{T}$. For the Newtonian fluid stress tensor it reduces to the density-weighted viscosity of a disk, $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$, rather than an unweighted height average.

Using $l=\sqrt{GMr}$ and $W_{r\varphi }=-(3/2)\bar{\nu }\Sigma \Omega$ in the conservation of angular momentum equation givesSubstitution into mass conservation proves the Keplerian viscous diffusion equation:The outward viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^2{W}_{r\varphi }=3\pi \sqrt{GM}\sqrt{r}\bar{\nu }\Sigma$.

For $\bar{\nu }=Ar$ with $A>0$, make the heat-equation transform for linear-radius disk viscosityWriting $y=\sqrt{r}$, the diffusion equation becomes $g_t=(3A/4)g_{yy}$; the displayed rescaling of $y$ gives exactly $\boxed{g_t=g_{xx}}$, with the original time unchanged. The torque is $3\pi \sqrt{GM}g$, so the zero-torque inner boundary condition at the origin becomes $g(0,t)=0$ on the half-line $x>0$.

A positive similarity solution obeying this boundary condition isFor example, substitution of $g=t^{-1}F(x/\sqrt{t})$ reduces the heat equation to $F^{\prime \prime }+(\eta /2)F^{\prime }+F=0$, and $F=C\eta e^{-{\eta }^2/4}$ satisfies it and vanishes at zero. A replacement of $t$ by $t+t_0$ gives a finite-mass initial profile at time zero if desired.

To determine the global integrals, transform the mass and angular momentum:The Gaussian integrals giveThus the accreting similarity disk with viscosity proportional to radius loses mass as $t^{-1/2}$ while conserving angular momentum; its characteristic radius grows proportional to $At$. The accretion rate at the origin is $-{\dot{M}}_d=2\pi C\sqrt{3/A}{t}^{-3/2}$. Its radial velocity ${\bar{u}}_r=r/t-3A/2$ shows inward flow at small radius and outward spreading at large radius. The integrable inner density singularity is part of the idealized extension to $r=0$.

For the other profile, direct differentiation givesIt also satisfies the heat equation, but has $g(0,t)=t^{-1/2}\ne 0$ and $g_x(0,t)=0$. It therefore has a nonzero inner torque and zero mass flow at the origin, not the previous zero-torque inner boundary condition. HereAn arbitrary positive amplitude multiplies these quantities. This constant-mass disk with viscosity proportional to radius spreads outward under a central torque that supplies angular momentum, with no ongoing central mass supply. It has the reflecting Neumann boundary condition for the heat equation and can be regarded as the spreading of an initially central mass distribution driven by an applied torque. It is distinct from the accreting zero-torque solution; the ideal origin does not model a finite stellar surface or a relativistic inner disk.

Let $M_{\ast }$ be the central mass, so ${\Omega }^2=G{M}_{\ast }/r^3$ and the specific orbital energy is $\epsilon =-G{M}_{\ast }/(2r)$. Define the inward accretion rate $\dot{M}>0$. The rate at which the gas acquires binding energy in moving from $b$ to $a$ isThe vertically integrated viscous heating per unit horizontal area, including the entire column, is $\bar{\nu }\Sigma (r{\Omega }^{\prime }{)}^2$. Since $r{\Omega }^{\prime }=-3\Omega /2$, the dissipated power in the annulus isHere the column integral already includes dissipation associated with both disk faces; adding another factor of two would double-count it. Performing the elementary integrals givesFor the whole disk, take $a=r_{\mathrm{in}}$ and $b\to \infty$. Both expressions giveIf $a\gg r_{\mathrm{in}}$, the bracket $1-\sqrt{r_{\mathrm{in}}/r}$ is uniformly close to one throughout the annulus. Comparing the positive integrals showswith the correction negative. Thus far from the inner edge the viscous dissipation is approximately three times the local binding-energy acquisition.

There is no violation of conservation of energy because the viscous torque in an accretion disk also transports mechanical energy. The outward viscous torque in an accretion disk isand its outward energy flux is $\Omega \mathcal{G}=G{M}_{\ast }\dot{M}[r^{-1}-\sqrt{r_{\mathrm{in}}}{r}^{-3/2}]$. The explicit expressions satisfyEnergy transported from the inner disk supplies the extra dissipation at larger radii. The torque power vanishes at the zero-torque inner boundary condition and tends to zero at infinity, leaving global balance. This is torque transport of energy in a steady accretion disk.

Both profiles have integrable central cusps. Directly integrating the surface density of a disk and its specific angular momentum $j=\sqrt{GMr}$ givesFor $a=1$, substitution of the similarity functions givesThe inward accretion rate is $\dot{M}(r)=-2\pi r\Sigma {\bar{u}}_r$. At the origin it tends to $M_d/(5t)=-d{M}_d/dt$. The outward viscous torque in an accretion disk isFor this profile $\mathcal{G}\to 0$ as $r\to 0$, and the advected angular-momentum flux $\dot{M}j$ also vanishes there. Both outer fluxes vanish at $R$. Thus disk mass decreases through central accretion while disk angular momentum is conserved. Including the accreted central mass restores total mass conservation for the disk-plus-sink system.

For $a=5/4$, instead,Here the mass flux tends to zero at the origin, despite positive radial velocity everywhere else. The inner torque has the nonzero limitThus the nonaccreting similarity disk supplied by an inner torque conserves disk mass but gains angular momentum from an external central torque. Its vanishing edge flux cannot supply that gain. The disk is not globally isolated in angular momentum; the combined disk and torque source may of course conserve it.

These profiles illustrate different boundary-controlled similarity classes of the Keplerian viscous diffusion equation. The $a=1$ solution is compatible with a zero-torque inner boundary condition and conserved finite angular momentum, and can describe a late-time spreading accretion disk after the detailed initial profile has been forgotten. Its mass diverges as $t\downarrow 0$, so the formal zero-radius origin is not a finite-mass initial disk. The $a=5/4$ solution has finite initial mass concentrated at the origin and initially zero disk angular momentum, but requires the specified inner torque for $t>0$; that torque's $t^{-3/4}$ singularity is time-integrable. It may describe a spreading nonaccreting disk for compatible torque-driven boundary conditions, but cannot be an isolated angular-momentum-conserving disk's generic asymptotic state. Similarity scaling suggests their possible asymptotic roles; it does not establish convergence for arbitrary initial and boundary data.

For an equatorial circular orbit in the Paczyński-Wiita potential, radial force balance givesIts specific angular momentum is $j=R^2\Omega =\sqrt{GM}{R}^{3/2}/(R-R_S)$. Perturbing the circular radius at fixed $j$ gives the radial epicyclic frequency squared ${\kappa }^2=R^{-3}d(j^2)/dR$. ThereforeCircular motion is radially stable for $R>3R_S$, marginal at $3R_S$, and unstable between $R_S$ and $3R_S$. Thus the inner edge of the nearly circular thin accretion disk is $R_{\mathrm{in}}=3R_S$, joining a plunging region of a black-hole accretion disk inside the innermost stable circular orbit. This conclusion concerns the circular thin-disk model, not an assertion that all gas disappears inside this radius.

Take the inward mass accretion rate $\dot{M}$ as positive, and write the positive outward viscous torque in an accretion disk asSteady conservation of angular momentum makes the net inward flux $C=\dot{M}j-\mathcal{G}$ independent of radius. Impose a zero-torque inner boundary condition at $3R_S$, appropriate to an approximately stress-free transition to plunging motion. ThenThis is the steady accretion disk with arbitrary rotation law angular-momentum balance.

The local viscous dissipation integrated through the full disk thickness is $D=\nu \Sigma (R{\Omega }^{\prime }{)}^2=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)$. It is the power per unit planar area summed over both radiating faces. Sincesubstitution gives the two-face dissipation of a Paczyński-Wiita disk:In particular $D$ vanishes at the stress-free inner edge and is nonnegative outside it because $j$ increases there. A one-face radiative flux is $D/2$; the total accretion luminosity is therefore $L={\int }_{3R_S}^{\infty }2\pi RDdR$, with no additional factor of two.

For a direct evaluation put $x=R/R_S$. ThenThe first integral is $7/4$, using $(3x-1)/(x-1{)}^3=3/(x-1{)}^2+2/(x-1{)}^3$. For the second,so it equals $2/[\sqrt{3}(3-1)]=1/\sqrt{3}$. HenceThis is the radiative efficiency of black-hole accretion within this pseudo-Newtonian model.

Independently, the specific orbital energy of a circular particle isThus $E(\infty )=0$ and $E(3R_S)=-GM/(8R_S)=-c^2/16$. The energy released per unit accreted mass is exactly $c^2/16$, confirming the integrated accretion luminosity. The Paczyński-Wiita circular orbit reproduces the Schwarzschild inner radius but gives an approximate binding energy; the efficiency derived here is for the stated potential.

For a zero-torque inner boundary condition at finite $r_{\mathrm{in}}$,With positive finite inner kinematic viscosity, $\Sigma$ tends to zero there. At fixed nonzero inward mass flux, the corresponding $v_r=-\dot{M}/(2\pi r\Sigma )$ becomes large, signaling the breakdown of the nearly circular thin disc approximation in the inner transition region.

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.

Write $\Sigma ={\Sigma }_0{x}^{-5/2}g$ because the prescribed kinematic viscosity is $\bar{\nu }={\bar{\nu }}_0{x}^{5/2}$. Substituting $r=r_{0}x$ and ${\partial }_t=(3{\bar{\nu }}_0/4r_0^2){\partial }_{\tau }$ into the viscous evolution of an accretion disk givesHence the dimensionless diffusion equation isFor the useful inverse-radius coordinate $y=x^{-1/2}$, this becomes simply $g_{\tau }=y{g}_{yy}$. The positive factor in $\tau =1+3{\bar{\nu }}_{0}t/(4r_0^2)$ makes increasing $\tau$ correspond to increasing physical time.

There is a distinction between the named torque variable and the physical viscous torque in an accretion disk. For Keplerian rotation, the outward torque magnitude isHere $M_{\star }$ denotes the central mass. The zero-torque inner boundary condition therefore requires $x^{1/2}g\to 0$ as $x\to 0$. Requiring $g\to 0$ is stronger. The nontrivial finite-mass profile used below satisfies both; this distinction matters when counting integration constants.

Take $0<r_1<r_2$ and positive mean kinematic viscosity $\bar{\nu }$. The steady state of the supplied viscous evolution of an accretion disk equation satisfies ${\mathcal{F}}^{\prime }=-2\pi rS(r)$. The outer no-flux condition therefore givesPositive $S$ produces positive inward accretion rate in this sign convention. Set $Y=r^{1/2}\bar{\nu }\Sigma$. In a Keplerian disk, the viscous torque in an accretion disk is proportional to $Y$, so the zero-torque inner boundary condition gives $Y(r_1)=0$. Since $Y^{\prime }=\mathcal{F}/(6\pi r^{1/2})$, a second integration yieldsThis is the steady viscous disk supplied by a radial source. It does not require $\bar{\nu }$ to be independent of radius: the equation directly determines $\bar{\nu }\Sigma$. If viscosity depends on the local thermodynamic state, the displayed formula is an implicit relation completed by a separate closure. For nonnegative integrable $S$ it gives nonnegative surface density of a disk, and the inner accretion rate equals the total supplied mass per unit time, $2\pi {\int }_{r_1}^{r_2}rS(r)dr$. Differentiating both nested integrals recovers the flux and the differential equation.

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