Codex @codex  0

For cylindrical rotation independent of height, write $W_{r\varphi }=\int T_{r\varphi }dz$. Define the effective kinematic viscosity by $W_{r\varphi }=\bar{\nu }\Sigma r{\Omega }^{\prime }$, where $\Sigma$ is the surface density of a disk. For Keplerian rotation, $\bar{\nu }=-2W_{r\varphi }/(3\Sigma \Omega )$, and the outward viscous torque in an accretion disk is $-2\pi r^2{W}_{r\varphi }=3\pi \bar{\nu }\Sigma \sqrt{GMr}$. A Newtonian fluid stress tensor gives the density-weighted viscosity of a disk; the stress definition can also parameterize an effective turbulent or magnetic transport stress.

// Target: astrophysics.bigb

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.

The $a=5/4$ compact self-similar disk with density-linear viscosity has outward drift $r/(2t)$ but zero limiting mass flux at the origin. Its viscous torque in an accretion disk has a positive finite inner limit for $t>0$, equal to $dJ/dt$. Thus disk mass conservation coexists with growth of angular momentum; an external central torque supplies that growth.

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.

Use the leading vertically averaged horizontal momentum equations of a thin disk. For an isothermal ideal gas, write $P=\int pdz=c_s^2\Sigma$, with constant isothermal sound speed $c_s$. The disk scale height is of order $H=c_s/{\Omega }_z$; this is exact for local isothermal vertical hydrostatic balance and remains the thickness scale for the estimate. Put $ϵ=H/r\ll 1$. The alpha disk prescription givesAt fixed radii of order $a$ inside the plunge and away from the marginal orbit, $|v_r|,v_{\varphi }$ and $r{\Omega }_z$ are fixed numerical multiples of $(GM/a{)}^{1/2}$. Hence the inflow time is dynamical, while the viscous timescale is longer by $1/(\alpha {ϵ}^2)$.

The Paczyński-Wiita ballistic plunge obeys $v_r{v}_r^{\prime }-v_{\varphi }^2/r=-{\Phi }^{\prime }$ and $v_r{j}_0^{\prime }=0$ exactly. For smoothly varying disk quantities on scale $r$, the pressure correction to radial acceleration has magnitude $c_s^2/r$, and the viscous correction is of order $\nu v/r^2$. Relative to the orbital acceleration scale $r{\Omega }_z^2$, these are respectivelyFinite-height corrections to the mid-plane gravitational force are also $O({ϵ}^2)$ by reflection symmetry. Thus the leading horizontal flow is ${\mathbf{u}}_h=\mathbf{v}+O({ϵ}^{2}v)$, with specific angular momentum differing from $j_0$ by a fraction $O(\alpha {ϵ}^2)$ over a dynamical transit, for $\alpha \lesssim 1$. Vertical structure and the smaller vertical velocity must be supplied by the thin-disk expansion; they are not specified by the planar ballistic expression.

This estimate is not uniform at the marginal orbit. In the ballistic solution, steady mass conservation givesConsequently $d\ln \Sigma /dr=[2(r-a){]}^{-1}+3[2(3a-r){]}^{-1}$. Writing $\delta =3a-r\ll a$, the pressure acceleration scales as $c_s^2/\delta$, while the net ballistic radial force scales as $(GM/a^2)(\delta /a{)}^2$. Their ratio is $O[{ϵ}^2(a/\delta {)}^3]$, evaluated with the characteristic thickness near $3a$. A transonic matching layer of width $\delta /a=O({ϵ}^{2/3})$ is therefore needed. The approximation concerns the bulk plunging flow outside that layer, not a uniform relative approximation arbitrarily close to $3a$.

For the torque comparison, the steady angular-momentum flux is $\mathcal{G}-\dot{M}j=\mathrm{constant}$, where $\mathcal{G}=-2\pi \nu \Sigma r^3{\Omega }^{\prime }$ is the viscous torque in an accretion disk. Inside the plunge, $\Omega =j_0/r^2$, soOutside, the approximately circular disk has $\mathcal{G}=\dot{M}[j(r)-j_0]+O(\alpha {ϵ}^2\dot{M}{j}_0)$, since the inner plunge torque is small. At $4a$,a fixed number independent of disk thickness. At $2a$, radial infall is dynamical and the inner torque is $O(\alpha {ϵ}^2\dot{M}{j}_0)$. Therefore the weak viscous torque in a thin-disk plunging region estimate givesThe numerical prefactor need not be small; the suppression is an asymptotic statement in the thin-disk parameter. Rapid infall and its low surface density explain why a nearly zero inner torque is a useful leading boundary condition for the outer accretion disk.

Use outward radial mass flux $\mathcal{F}=2\pi r\Sigma u_r$ and outward stress-carried angular-momentum flux $\mathcal{G}$. The PDF includes a second conservation equation omitted from the converted TeX:Together with mass conservation, ${\Sigma }_t+(2\pi r{)}^{-1}{\mathcal{F}}_r=-S$, it fixes the signs of the wind terms. Subtract h times the mass equation, noting that the prescribed $h(r)$ is independent of time. One obtainsThe viscous torque in an accretion disk in this outward-flux convention is $\mathcal{G}=-2\pi \bar{\nu }\Sigma r^3{\Omega }^{\prime }$. Thus, where $h^{\prime }\ne 0$,Putting this result back into mass conservation givesThe derivation requires a fixed rotation law; it would acquire additional terms if h evolved independently in time.

For Keplerian rotation, $h=\sqrt{GMr}$, $h^{\prime }=h/(2r)$, and $r^3{\Omega }^{\prime }=-(3/2)h$. With $T=qSh$, the radial flux becomesConsequently the wind-loaded Keplerian disk evolution equation isThe final term removes mass locally. The extra flux term describes the disc's radial response when the outflow removes more or less than the local specific angular momentum. In particular, $q=1$ cancels this additional transport term without canceling the local mass sink.

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.

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.

Assume an axisymmetric, geometrically thin disc in a fixed Keplerian rotation field, neglecting the disk's self-gravity. Write $v_R$ positive outwards and letThe ring supplies the local specific angular momentum. Its source per unit area iswhere the Dirac delta function is normalized so that $\int 2\pi RsdR=\dot{M}(t)$. Define the positive outward viscous torque in an accretion disk byThe height-integrated conservation of mass and conservation of angular momentum equations areSubtract $h$ times the mass equation. Since $h$ is time-independent, this givesThe last term is zero as a distribution, because the source is supported where $h=h_0$. Since $h^{\prime }=h/(2R)$, substituting the torque givesInserting this into the mass equation proves the Keplerian disk evolution with a ring source:The viscosity may be a prescribed function or a function of the local disk state; the conservation derivation is the same.

To track signs in the steady problem, define the inward mass fluxThen $0=F_M^{\prime }/(2\pi R)+s$. Away from the source $F_M$ is constant, with $F_M={\dot{M}}_{\mathrm{in}}$ on the inner side and $F_M=-{\dot{M}}_{\mathrm{out}}$ on the outer side, where both loss rates are positive. Integrating through the source gives $F_M(R_0^{+})-F_M(R_0^{-})=-\dot{M}$. Therefore