Angular momentum, not gravity, is what stops gas from falling in — and the molecular viscosity that could carry it away is twelve decades too slow. Four checks, of which two refute. The prescription that replaces molecular viscosity has one free parameter, and the same parameter measured in two classes of disc differs by a factor of ten to forty, while the simulations of the very instability Module 12 derives produce at most a fifth of what the observations need.
Module 9 built spherical accretion and then, in its own closing table, refused the object it had been aimed at. Sagittarius A*'s gas does not fall radially inward: it carries angular momentum, and it circularises near $100\,r_S$ — five decades inside the Bondi radius — where it must then transport that angular momentum outward before any mass can reach the centre. Module 9's words were that this "is Module 11's subject and nothing in this module can address it".
Module 12 then stated the difficulty in one line and left it. A Keplerian disc has $\Omega \propto R^{-3/2}$, so its specific angular momentum $\ell = \Omega R^2 \propto R^{1/2}$ increases outward, which by the Rayleigh criterion makes it stable. A stable disc transports no angular momentum and therefore does not accrete. Module 12 assumed the shear $q = 3/2$ that this statement rests on, and said in print that deriving it needs a disc model, which is this module.
The plan of this module is one problem and one number. The problem is that a disc is too stable. The number is how badly molecular viscosity fails to fix it: twelve decades, which is not a discrepancy to be argued about but a statement that some other physics carries the angular momentum. What replaces molecular viscosity is a parametrisation with one free number in it, and the last two sections are about how big that number is and whether it deserves to be called a constant.
Two things this module does not do, and both are promises a shipped page already made. It does not re-derive the magnetorotational instability: Module 8 lists it among what Module 12 builds, and Module 12 §9 builds it. What this module owes that section is the $q = 3/2$ it assumed, and §2 supplies it. And it does not build two-temperature or radiatively inefficient accretion flows, which Module 2 and Module 9 both promised here; §12 says so in a row of its own rather than in silence.
1. Gas with angular momentum cannot fall in
Take a parcel of gas at rest far from a mass $M$, and give it a small specific angular momentum $\ell$ about that mass. As it falls, $\ell$ is conserved, and the centrifugal term $\ell^2/2R^2$ in its effective potential grows as $R^{-2}$ while the gravitational term $-GM/R$ grows only as $R^{-1}$. There is therefore a radius at which the two balance and the infall stops:
Inside $R_{\rm circ}$ the gas cannot go without losing angular momentum. It does not stop moving — it orbits. What was a radial infall becomes a disc, and the question changes from what holds the gas up? to what lets it down?
Definition 1 (a geometrically thin accretion disc). An axisymmetric, vertically hydrostatic, differentially rotating flow around a central mass $M$, described by a surface density $\Sigma(R,t)$ and a scale height $H(R)$ with $H/R \ll 1$, in which the orbital motion is Keplerian to first order in $(H/R)^2$ and mass moves inward on a timescale long compared with the orbital one. The material does not free-fall; it spirals.
Module 9 gave the number that makes this concrete. Sagittarius A*'s gas circularises near $100\,r_S$, where $r_S = 2GM/c^2$ is the Schwarzschild radius. At the GRAVITY Collaboration mass of $4.30\times10^{6}\,M_\odot$ that $r_S$ is $1.2699\times10^{12}\ \mathrm{cm}$, so the circularisation radius is $1.2699\times10^{14}\ \mathrm{cm}$ and the specific angular momentum needed to produce it is $2.6920\times10^{23}\ \mathrm{cm^2\,s^{-1}}$ by (1.1).
Two worked discs carry the rest of this module. They are configurations and not objects, in the sense Module 7 §7.1 and Module 12 §9 use the phrase: each entry is a round choice inside the range its class occupies, and no named system was read.
disc
$M$
$R$
$T$
$\mu$
$\Sigma$
$\Omega$ (s⁻¹)
$c_T$ (km s⁻¹)
$H/R$
$n$ (cm⁻³)
dwarf nova, outburst
$0.6\,M_\odot$
$10^{10}$ cm
$3\times10^{4}$ K
0.6
$100\ \mathrm{g\,cm^{-2}}$
$8.9234\times10^{-3}$
20.39
0.02285
$1.752\times10^{17}$
protoplanetary, 10 au
$1\,M_\odot$
10 au
50 K
2.34
$10\ \mathrm{g\,cm^{-2}}$
$6.2960\times10^{-9}$
0.4215
0.044751
$1.534\times10^{11}$
Both satisfy $H/R \ll 1$, so Definition 1 is self-consistent on both rows — which §7 proves rather than assumes. A third configuration, the inner disc of a $10^{8}\,M_\odot$ active galactic nucleus, enters at §8 and is not in this table, because it has no independent temperature: §8 computes one for it from the accretion rate.
2. Disc kinematics, and the stability that is the problem
Everything in this section is kinematics. No pressure, no viscosity, no magnetic field: only what a point mass does to a ring of gas.
Proposition 1 (Keplerian rotation and the shear parameter). For a disc whose self-gravity is negligible and whose radial pressure gradient is small compared with gravity, centrifugal balance gives
$$ \Omega(R) \;=\; \sqrt{\frac{GM}{R^{3}}}, \qquad v_K = \Omega R = \sqrt{\frac{GM}{R}}, \qquad \ell = \Omega R^{2} = \sqrt{GMR}, \tag{2.1}$$
and therefore the shear parameter $q \equiv -\,\mathrm{d}\ln\Omega/\mathrm{d}\ln R$ takes the value $q = 3/2$ exactly.
Proof. Radial force balance on a ring in steady circular motion is $\Omega^2 R = GM/R^2$, which is (2.1). Taking logarithms, $\ln\Omega = \tfrac12\ln(GM) - \tfrac32\ln R$, so $\mathrm{d}\ln\Omega/\mathrm{d}\ln R = -3/2$ identically, independent of $M$ and of $R$. ∎
The generator differentiates $\Omega$ numerically rather than returning $3/2$, and an assertion requires the answer to be $3/2$ to better than $10^{-6}$; the run prints $1.5000000000$. That is deliberate: Module 12 §9 says in print that its $q = 3/2$ is assumed and not derived, so this module must produce it from something rather than name it.
Proposition 2 (the epicyclic frequency, and the Rayleigh criterion). For $\Omega \propto R^{-q}$ the epicyclic frequency, at which a displaced ring oscillates radially, is
$$ \kappa_{\rm ep}^{2} \;=\; \frac{1}{R^{3}}\frac{\mathrm{d}}{\mathrm{d}R}\!\left(R^{4}\Omega^{2}\right) \;=\; 2\,\Omega^{2}\,(2-q). \tag{2.2}$$
The flow is stable to axisymmetric displacements when $\kappa_{\rm ep}^2 \gt 0$, equivalently when $\mathrm{d}(\ell^2)/\mathrm{d}R \gt 0$, equivalently when $q \lt 2$. A Keplerian disc has $q = 3/2$, hence $\kappa_{\rm ep} = \Omega$ exactly and $\mathrm{d}(\ell^2)/\mathrm{d}R = GM$ at every radius.
Proof. Whatever holds the rotation law up, it acts as a central acceleration $g(R)$ directed inward, and circular motion at $R$ requires $g(R) = \Omega^2(R)\,R$. For a point mass $g = GM/R^2$, but the argument below needs only that $g$ depends on $R$ alone. Displace a ring from $R$ to $R+\xi$ at fixed specific angular momentum $\ell(R) = \Omega(R)R^{2}$. Its radial equation of motion is
$$ \ddot{\xi} \;=\; \frac{\ell(R)^{2}}{(R+\xi)^{3}} \;-\; g(R+\xi), $$
whose right-hand side vanishes at $\xi = 0$, because $\ell(R)^2/R^3 = \Omega^2R = g(R)$. Expanding to first order in $\xi$,
$$ \ddot{\xi} \;=\; -\left[\frac{3\ell(R)^{2}}{R^{4}} + g^{\prime}(R)\right]\xi \;\equiv\; -\,\kappa_{\rm ep}^{2}\,\xi, $$
so the displaced ring oscillates when the bracket is positive and runs away when it is negative. Now $3\ell^2/R^4 = 3\Omega^2$, and for $\Omega \propto R^{-q}$, $g = \Omega^2R \propto R^{1-2q}$ gives $g^{\prime} = (1-2q)\Omega^2$. Hence
$$ \kappa_{\rm ep}^{2} \;=\; 3\Omega^{2} + (1-2q)\Omega^{2} \;=\; (4-2q)\,\Omega^{2} \;=\; 2\,\Omega^{2}(2-q), $$
which is the right-hand side of (2.2). The middle expression of (2.2) is the same number written as a derivative: $R^{4}\Omega^{2} = \ell^{2} \propto R^{4-2q}$, so $R^{-3}\mathrm{d}(R^{4}\Omega^{2})/\mathrm{d}R = (4-2q)\Omega^{2}$. Since $\mathrm{d}(\ell^{2})/\mathrm{d}R = R^{3}\kappa_{\rm ep}^{2}$, the two stability statements are one statement, and both hold for $q \lt 2$. Setting $q = 3/2$ gives $\kappa_{\rm ep}^{2} = 2\Omega^{2}(2-3/2) = \Omega^{2}$; substituting Proposition 1’s $\Omega^{2} = GM/R^{3}$ then gives $\mathrm{d}(\ell^{2})/\mathrm{d}R = R^{3}\Omega^{2} = GM$ at every radius. ∎
The trivial limits, both of which the generator asserts. A rigidly rotating fluid has $q = 0$ and (2.2) gives $\kappa_{\rm ep} = 2\Omega$, which is the Coriolis result for a parcel in a rotating frame. At the Rayleigh boundary $q = 2$, (2.2) gives $\kappa_{\rm ep} = 0$ exactly: the restoring force vanishes and the displaced ring neither returns nor runs away. Both are checked to $10^{-14}$ on every run.
Three quantities of a Keplerian disc, each divided by its own value at the inner radius. Two fall outward and one rises, and the one that rises is the one the Rayleigh criterion tests. Because $\ell^2 = GMR$, the discriminant $\mathrm{d}(\ell^2)/\mathrm{d}R$ equals $GM$ at every radius — positive, and independent of radius. The disc is stable everywhere, and that is the problem this module exists to solve.
The result that sets up the rest of the module, and Fig. 1 is a picture of it. A Keplerian disc is stable to axisymmetric displacements at every radius, because $\kappa_{\rm ep}^{2} = \Omega^{2} \gt 0$ there. Axisymmetric instability would require $q \gt 2$, and a point mass produces $q = 3/2$ and can produce nothing else. Within the physics of this section — no pressure, no viscosity, no field — there is therefore no linear axisymmetric route by which rings exchange angular momentum, and the disc does not accrete. That is the narrowest form of the statement and the only form Proposition 2 proves; whether non-axisymmetric or nonlinear hydrodynamic routes exist is a separate question that this book does not settle, and §11 does not need it, because the mechanism it reaches for is magnetic and not hydrodynamic at all.
Two consequences are worth stating before moving on. First, $\kappa_{\rm ep} = \Omega$ for a Keplerian disc is a degeneracy, not a coincidence: it is why a Keplerian orbit closes on itself and does not precess, and it is why the radial and orbital periods of a planet are the same. Second, the same equality makes the Toomre criterion of §10 simpler than it looks in the galactic-dynamics literature, where $\kappa_{\rm ep}$ and $\Omega$ differ.
3. The anchor: the timescale gap
Module 1 §6 derived the transport coefficients of a collisional gas, and Module 1 §9 — its captures-and-misses table — says of the ideal-flow row that "§6 gives the coefficients; Module 11 needs them for accretion discs". This is where they are needed, and the answer they give is that they are not enough.
Module 1's Proposition 4 gives the dynamic viscosity of a gas of mass density $\rho$, thermal speed $v_{\rm th}$ and mean free path $\lambda$ as $\mu \simeq \tfrac13 \rho v_{\rm th}\lambda$, so the kinematic viscosity is
in which the density has cancelled. The coefficient $\tfrac13$ is Module 1's and is not a choice made here.
The thermal speed in (3.1) is the ion's, not the mixture's. Viscosity in a hydrogen plasma is carried by the ions, because at equal temperature and comparable mean free path the ratio of the ion to the electron contribution is $(\rho_i/\rho_e)(v_i/v_e) = \sqrt{m_p/m_e} = 42.9$. Using the mean molecular weight of the mixture, $\mu = 0.6$, would overstate $v_{\rm th}$ and with it $\nu_{\rm mol}$ by a factor of $1.2957$.
For the dwarf-nova row the midplane number density is $1.7524\times10^{17}\ \mathrm{cm^{-3}}$, the Coulomb logarithm is $5.0355$, and Module 1's Coulomb mean free path is $2.6770\times10^{-4}\ \mathrm{cm}$ — which is $1.1716\times10^{-12}$ of a scale height, so the gas is collisional beyond any doubt. The ion thermal speed is $25.1115\ \mathrm{km\,s^{-1}}$, and (3.1) gives
CHECK 1 — REFUTED. Molecular viscosity, by 12.01 decades. A dwarf nova in outburst decays on a timescale of days; the configuration here takes five days, $4.32\times10^{5}\ \mathrm{s}$. Against the molecular viscous time of (3.2) the ratio is $\mathbf{1.0330\times10^{12}}$, that is 12.01 decades. Equivalently, the $\alpha$ of §6 that $\nu_{\rm mol}$ corresponds to is $4.8098\times10^{-13}$, short of the measured $0.1$–$0.4$ by $11.32$ to $11.92$ decades. The viscous time exceeds the age of the universe; the disc it describes has never accreted anything.
Three quantities in this calculation could be argued about, and all three are priced rather than waved at. Using the mixture instead of the ion carrier is worth a factor of $1.2957$. Doubling the Coulomb logarithm from $5.0355$ to $10.0710$ is worth a factor of $0.5$. And (3.1) feeds Module 1’s formula the total particle density, where the scatterers are the ions and a fully ionised hydrogen plasma has half as many of those, which is worth another factor of $0.5$. Together they move the answer by $0.71$ of a decade against a gap of $12.01$. The third is the one that moves the headline, and it moves it downward: taking $n_i = n/2$ gives $11.71$ decades in place of $12.01$. This module prints $12.01$ because that is what Module 1 §6’s formula gives on the census row as tabled, and prints the other number here because a reader is entitled to it. Neither reading changes the verdict. Nothing inside the transport coefficient closes a gap of that size. Fig. 2 puts all five timescales on one axis.
Five timescales for one dwarf-nova configuration on a single logarithmic axis. The dynamical time is $1.1206\times10^{2}\ \mathrm{s}$ and the thermal time at $\alpha = 0.2$ is five times that; the observed outburst sits near $10^{5.6}\ \mathrm{s}$; and the viscous time computed from collisions alone sits $12.01$ decades beyond it, marked by the red brace. The $\alpha$-disc viscous time at $\alpha = 0.2$, $1.0732\times10^{6}\ \mathrm{s}$, lands near the observation — which is the whole content of §6.
The dwarf-nova parameters are a configuration and not an object. $M = 0.6\,M_\odot$, $R = 10^{10}\ \mathrm{cm}$, $T = 3\times10^{4}\ \mathrm{K}$, $\Sigma = 100\ \mathrm{g\,cm^{-2}}$ and a five-day outburst are round choices in the ranges their class occupies, and no named system was read for this module. King, Pringle & Livio's Sect. 2.1 rests on detailed light curves of SS Cyg, VW Hyi, U Gem, WZ Sge and Z Cam; none of those papers was fetched. A named system would change the third significant figure of $12.01$ and not the twelfth decade.
4. The viscous torque
If angular momentum is to move outward, something must exert a torque. In a fluid with kinematic viscosity $\nu$ the stress between neighbouring annuli is a shear stress, and it does exactly that.
Proposition 3 (the viscous torque). Let $G(R)$ be the torque that the disc outside radius $R$ exerts on the disc inside it. In a flow of kinematic viscosity $\nu$ and surface density $\Sigma$, that torque is
$$ G(R) \;=\; 2\pi R \cdot \nu\Sigma R \frac{\mathrm{d}\Omega}{\mathrm{d}R} \cdot R \;=\; 2\pi R^{3}\,\nu\Sigma\,\frac{\mathrm{d}\Omega}{\mathrm{d}R}. \tag{4.1}$$
For any $q \gt 0$ this is negative: the outer disc takes angular momentum from the inner disc, so angular momentum flows outward.
Proof. The two sides of a cylindrical surface at radius $R$ slide past each other with a velocity difference, per unit radial separation, of $R\,\mathrm{d}\Omega/\mathrm{d}R$: this is the rate of strain, with the rigid-rotation part removed, because a fluid in rigid rotation has no internal stress. The viscous shear stress is $\eta$ times the rate of strain, and integrating the dynamic viscosity $\eta = \rho\nu$ over the disc thickness replaces $\rho$ by $\Sigma$, giving a force per unit length of circumference $\nu\Sigma R\,\mathrm{d}\Omega/\mathrm{d}R$. Multiplying by the circumference $2\pi R$ gives a force and by the moment arm $R$ gives (4.1). For $\Omega \propto R^{-q}$ with $q \gt 0$, $\mathrm{d}\Omega/\mathrm{d}R \lt 0$, so $G \lt 0$. A negative torque on the inner disc slows it; by Newton’s third law the outer disc receives $-G \gt 0$ and is sped up. Angular momentum therefore crosses the surface outward at the rate $|G|$. ∎
The sign is worth dwelling on, because it is the opposite of the intuition that friction between orbits should drag the outer material inward. The inner disc rotates faster, so it is the inner disc that does work on the outer one. Angular momentum moves outward and mass moves inward, and each is what the other requires.
5. How a disc spreads
Proposition 4 (the surface-density diffusion equation). Conservation of mass and of angular momentum for an axisymmetric thin disc with a Keplerian rotation law give
$$ \frac{\partial\Sigma}{\partial t} \;=\; \frac{3}{R}\frac{\partial}{\partial R}\!\left[R^{1/2}\frac{\partial}{\partial R}\!\left(\nu\Sigma R^{1/2}\right)\right]. \tag{5.1}$$
This is a diffusion equation for $\Sigma$ with diffusion coefficient of order $\nu$. Whether it is linear depends entirely on the closure: with the $\alpha$ prescription of §6 at fixed temperature, $\nu = \alpha c_T^{2}/\Omega$ carries no $\Sigma$ and (5.1) is linear. It becomes nonlinear the moment the disc’s thermal balance ties the temperature back to $\Sigma$ — which is what the dwarf-nova outburst cycle of §3 is, and which this module does not build.
Proof. Mass conservation for an annulus of width $\mathrm{d}R$ carrying radial velocity $v_R$ is
$$ R\frac{\partial\Sigma}{\partial t} + \frac{\partial}{\partial R}\!\left(R\Sigma v_R\right) = 0. $$
Angular momentum conservation for the same annulus, with the torque of (4.1) acting on each face, is
$$ R\frac{\partial}{\partial t}\!\left(\Sigma R^{2}\Omega\right) + \frac{\partial}{\partial R}\!\left(R\Sigma v_R\cdot R^{2}\Omega\right) = \frac{1}{2\pi}\frac{\partial G}{\partial R}. $$
Expanding the second equation and using the first to eliminate $\partial\Sigma/\partial t$ leaves
$$ R\Sigma v_R \frac{\partial}{\partial R}\!\left(R^{2}\Omega\right) = \frac{1}{2\pi}\frac{\partial G}{\partial R}, $$
For $\Omega = \sqrt{GM/R^{3}}$ we have $R^{2}\Omega = \sqrt{GM}\,R^{1/2}$, so $\partial(R^{2}\Omega)/\partial R = \tfrac12\sqrt{GM}\,R^{-1/2}$, and $\mathrm{d}\Omega/\mathrm{d}R = -\tfrac32\sqrt{GM}\,R^{-5/2}$ turns (4.1) into $G = -3\pi\sqrt{GM}\,\nu\Sigma R^{1/2}$. Substituting both,
$$ \Sigma v_R \;=\; \frac{2}{\sqrt{GM}\,R^{1/2}}\cdot\frac{1}{2\pi}\frac{\partial G}{\partial R} \;=\; -\frac{3}{R^{1/2}}\frac{\partial}{\partial R}\!\left(\nu\Sigma R^{1/2}\right). $$
Substituting this into mass conservation gives $\partial\Sigma/\partial t = -R^{-1}\partial(R\Sigma v_R)/\partial R$, which is (5.1). ∎
The algebra in the last two lines of that proof is the standard one and is not this module's; the derivation is laid out in full in Pringle's 1981 review, which is cited here for the result only and was not fetched. Nothing numerical in this module depends on (5.1): it is here because §6's viscous time is the diffusion time of this equation, and a reader is entitled to see where that time comes from.
6. The α prescription, and what it is not
Section 3 established that molecular viscosity is twelve decades too small. Shakura and Sunyaev's response in 1973 was not to find a bigger viscosity but to parametrise one, and to bound it.
Definition 2 (the α prescription). The disc's effective kinematic viscosity is written
$$ \nu \;=\; \alpha\, c_T H, \tag{6.1}$$
with $c_T$ the isothermal sound speed and $H$ the scale height, and $\alpha$ a dimensionless number of order unity or less.
The argument behind the bound is short. Whatever carries the stress is a motion within the disc; if it is turbulence, its eddies cannot be larger than $H$ without leaving the disc, and cannot be faster than $c_T$ without shocking and dissipating. So $\nu \lesssim c_T H$, and $\alpha \lesssim 1$. That is the whole of it: the prescription is not a derivation and this module will not pretend otherwise.
Three things about Shakura and Sunyaev's own α that a reader who opens their paper will meet. Their p. 338 writes it as a sum, $\alpha = v_t/v_s + H^2/(4\pi\rho v_s^2)$, of a turbulent Mach number and a magnetic-to-thermal energy ratio — not as a single stress coefficient. Their $H$ is the magnetic field, in the older convention where $H$ denotes $B$; this book's $H$ is the scale height, and the two collide inside the founding paper of the subject. And their $v_s$ obeys $\rho v_s^2/2 = \tfrac32\rho kT/m_p + \varepsilon_r$, so with radiation neglected $v_s^2 = 3kT/m_p$ — the root-mean-square thermal speed, larger than this book's $c_T$ by $1.34$ at $\mu = 0.6$. Writing (6.1) with $c_T$ is a convention of this book, and $\alpha$ absorbs the factor.
Their own footnote on the same page is worth carrying too: $\alpha$ "is assumed to be constant along the disk in our calculations", and the disc's observational appearance — its spectrum and its surface temperature — "do not strongly depend on the chosen value of $\alpha$". §8 proves the second half of that sentence, and §11 is about what it costs.
7. Vertical structure, and what "thin" means
Proposition 5 (vertical hydrostatic balance). In a disc whose self-gravity is negligible, the vertical component of the central mass's gravity at height $z \ll R$ is $g_z = \Omega^2 z$, and hydrostatic balance with an isothermal equation of state gives
$$ \rho(z) \;=\; \rho_0\,\exp\!\left(-\frac{z^{2}}{2H^{2}}\right), \qquad H \;=\; \frac{c_T}{\Omega}, \qquad \Sigma \;=\; \sqrt{2\pi}\,\rho_0 H. \tag{7.1}$$
Proof. The vertical gravity of a point mass at cylindrical radius $R$ and height $z$ is $g_z = GMz/(R^2+z^2)^{3/2}$, which for $z \ll R$ is $GMz/R^3 = \Omega^2 z$. Vertical hydrostatic balance for an isothermal layer is $c_T^2\,\mathrm{d}\rho/\mathrm{d}z = -\rho\,\Omega^2 z$, whose solution is the Gaussian in (7.1) with $H = c_T/\Omega$. Integrating over all $z$ gives $\int\rho\,\mathrm{d}z = \sqrt{2\pi}\rho_0 H$. ∎
The factor is $\sqrt{2\pi} = 2.5066$ and not $2$. Writing $\Sigma = 2\rho_0 H$ — the constant-density slab — would put the midplane density $25.3$ per cent high. The generator integrates the Gaussian numerically over $\pm 20H$ on every run and asserts that the result is $\Sigma$ to one part in $10^{6}$.
The approximation $g_z \simeq \Omega^2 z$ has a price and it is small. At two scale heights on the dwarf-nova row, $z = 4.5698\times10^{8}\ \mathrm{cm}$ and $z/R = 0.04570$, the thin-disc form overstates the exact vertical gravity by 0.3134 per cent — and $95.45$ per cent of a Gaussian column lies inside $2H$. The error grows as $(z/R)^2$, so it is the aspect ratio and not the height that prices it.
Definition 3 (the thermal time). $t_{\rm th} \equiv 1/(\alpha\Omega)$, the time in which viscous dissipation supplies the disc’s own thermal energy, to within a factor of order unity. The factor is computable and is named here rather than hidden: the thermal energy per unit disc area is $U = \Sigma c_T^{2}/(\gamma-1)$, and the dissipation on both faces is $\tfrac94\nu\Sigma\Omega^{2} = \tfrac94\alpha\Sigma c_T^{2}\Omega$ once $\nu = \alpha c_T H$ and $H = c_T/\Omega$ are put in — a rate that §8 derives from Proposition 3 and that is quoted forward to here. The ratio is $U/(\tfrac94\alpha\Sigma c_T^{2}\Omega) = \tfrac{4}{9(\gamma-1)}(\alpha\Omega)^{-1}$, which is $\tfrac23(\alpha\Omega)^{-1}$ at $\gamma = 5/3$. This module writes $1/(\alpha\Omega)$ and carries the $2/3$ nowhere, because (7.3) uses only the ordering of the three times and no factor of order unity can change it. Every printed value of $t_{\rm th}$ in this module is $1/(\alpha\Omega)$ exactly.
Proposition 6 (the three timescales). With $\nu = \alpha c_T H$ and $H = c_T/\Omega$, the dynamical, thermal and viscous times of a thin disc stand in the ratio
$$ t_{\rm dyn} : t_{\rm th} : t_\nu \;=\; 1 \;:\; \frac{1}{\alpha} \;:\; \frac{1}{\alpha}\left(\frac{R}{H}\right)^{2}, \tag{7.3}$$
where $t_{\rm dyn} = 1/\Omega$, $t_{\rm th} = 1/(\alpha\Omega)$ and $t_\nu = R^2/\nu$. For a thin disc with $\alpha \lt 1$ the ordering is strict.
Proof. $t_\nu = R^2/\nu = R^2/(\alpha c_T H)$, and $c_T = \Omega H$ by Proposition 5, so $t_\nu = R^2/(\alpha\Omega H^2) = (R/H)^2/(\alpha\Omega)$. Dividing by $t_{\rm dyn} = 1/\Omega$ gives the third entry, and $t_{\rm th}/t_{\rm dyn} = 1/\alpha$ is immediate. Since $H/R \ll 1$ and $\alpha \lt 1$, each ratio exceeds one. ∎
For the dwarf-nova row at $\alpha = 0.2$ the three are $1.1206\times10^{2}$, $5.6032\times10^{2}$ and $1.0732\times10^{6}$ seconds, with $(R/H)^2 = 1.9154\times10^{3}$. That ordering is what makes a thin disc a well-posed object. It can be treated as vertically hydrostatic and in local thermal balance while it evolves radially, because the radial evolution is the slowest thing happening.
Proposition 7 (what "thin" means, and the answer to Module 3's "no rotation"). Dividing $H = c_T/\Omega$ by $R$ gives
$$ \frac{H}{R} \;=\; \frac{c_T}{v_K}. \tag{7.2}$$
A disc is thin exactly when it is cold in units of its own orbital speed.
Proof. $H = c_T/\Omega$ by Proposition 5 and $v_K = \Omega R$ by Proposition 1, so $H/R = c_T/(\Omega R) = c_T/v_K$. ∎
Module 3 §11 listed "no rotation" among its idealisations and said it "excludes centrifugal flattening and the rotational support of discs", pointing here. Equation (7.2) is that repayment, and it is worth saying why it and not something more obvious. A Keplerian disc satisfies $\Omega^2R^3/GM = 1$ exactly at every radius, but that identity is Kepler's law restated and says nothing about a disc; it is true of a single planet. What distinguishes a disc from Module 3's hydrostatic star is the ratio of the two support mechanisms: radially the disc is supported by rotation and by nothing else, vertically by pressure and by nothing else, and the aspect ratio (7.2) is the flattening. A sphere has $H/R = 1$; the dwarf-nova row has $0.02285$ and the protoplanetary row $0.044751$.
8. The steady disc and its temperature
Proposition 8 (the steady thin disc). In a steady state with a constant accretion rate $\dot M$ and a zero-torque inner boundary at $R_{\rm in}$,
$$ \nu\Sigma \;=\; \frac{\dot M}{3\pi}\left[1 - \sqrt{\frac{R_{\rm in}}{R}}\right] \;\equiv\; \frac{\dot M}{3\pi}\,f(R). \tag{8.1}$$
Proof. In steady state the results in the proof of Proposition 4 give $\dot M = -2\pi R\Sigma v_R$ constant, and the angular momentum equation integrates once to
$$ \dot M\,R^{2}\Omega \;+\; 2\pi R^{3}\nu\Sigma\,\frac{\mathrm{d}\Omega}{\mathrm{d}R} \;=\; \text{const}. $$
The constant is fixed at the inner edge, and the condition there is $G(R_{\rm in}) = 0$ and not $\mathrm{d}\Omega/\mathrm{d}R = 0$. The Keplerian law (2.1) has $\mathrm{d}\Omega/\mathrm{d}R = -\tfrac32\Omega/R$, which is non-zero at every radius, so no Keplerian disc can satisfy the second condition anywhere. A Keplerian disc meets the first through $\nu\Sigma \to 0$ at $R_{\rm in}$ — the material leaves the disc at a black hole's last stable orbit, or a boundary layer this module does not resolve carries $\Omega$ down to a slowly spinning star's surface value just inside $R_{\rm in}$ and takes $\mathrm{d}\Omega/\mathrm{d}R$ through zero there. With $G(R_{\rm in}) = 0$ the constant is $\dot M R_{\rm in}^{2}\Omega(R_{\rm in}) = \dot M\sqrt{GMR_{\rm in}}$. Substituting the Keplerian $\Omega$ and $G = -3\pi\sqrt{GM}\,\nu\Sigma R^{1/2}$ gives $\dot M\sqrt{GMR} - 3\pi\sqrt{GM}\,\nu\Sigma R^{1/2} = \dot M\sqrt{GMR_{\rm in}}$, and dividing through by $\sqrt{GM}\,R^{1/2}$ gives (8.1). ∎
The zero-torque inner boundary is an assumption and not a theorem. It is exactly right for a disc ending at a surface of vanishing stress and wrong for one threaded by a magnetic field anchored in a spinning central object, where the torque at $R_{\rm in}$ need not vanish and can even be negative. Every number in this section and the next inherits it.
Proposition 9 (the dissipation rate and the temperature profile). The viscous dissipation per unit disc area per side is
$$ D(R) \;=\; \frac{9}{8}\nu\Sigma\,\Omega^{2} \;=\; \frac{3GM\dot M}{8\pi R^{3}}\,f(R), \tag{8.2}$$
so if the disc radiates locally as a blackbody, $\sigma T_{\rm eff}^4 = D(R)$ and
$$ T_{\rm eff}(R) \;=\; \left[\frac{3GM\dot M}{8\pi\sigma R^{3}}f(R)\right]^{1/4} \;\xrightarrow[R \gg R_{\rm in}]{}\; \propto R^{-3/4}. \tag{8.3}$$
The profile has a maximum at $R = (49/36)\,R_{\rm in}$.
Proof. Take the annulus between $R$ and $R + \mathrm{d}R$. The torques on its two faces do work on it at the rate $\partial(\Omega G)/\partial R\,\mathrm{d}R$, and of that, the part $\Omega\,\partial G/\partial R\,\mathrm{d}R$ is the rotational energy that arrives with the angular momentum the torque delivers. The difference is dissipated:
$$ \frac{\partial(\Omega G)}{\partial R} - \Omega\frac{\partial G}{\partial R} \;=\; G\,\frac{\mathrm{d}\Omega}{\mathrm{d}R} \qquad\text{per unit radius.} $$
Substituting (4.1) gives $2\pi R^{3}\nu\Sigma(\mathrm{d}\Omega/\mathrm{d}R)^{2}$, positive because it is a square, as a dissipation must be. Dividing by the annulus area per unit radius, $2\pi R$, gives the dissipation per unit area of disc on both faces together as $\nu\Sigma\,(R\,\mathrm{d}\Omega/\mathrm{d}R)^{2}$. For the Keplerian law $R\,\mathrm{d}\Omega/\mathrm{d}R = -\tfrac32\Omega$, so the bracket squared is $\tfrac94\Omega^{2}$ and the both-faces rate is $\tfrac94\nu\Sigma\Omega^{2}$. Half emerges from each face, giving $D = \tfrac98\nu\Sigma\Omega^{2}$, and substituting (8.1) and $\Omega^{2} = GM/R^{3}$ gives the second form of (8.2). For the maximum, write $x = R/R_{\rm in}$, so $T_{\rm eff}^4 \propto x^{-3}(1-x^{-1/2})$. Differentiating, $-3x^{-4} + \tfrac72 x^{-9/2} = 0$, giving $x^{1/2} = 7/6$ and $x = 49/36 = 1.361111$. ∎
The generator finds the maximum numerically on a 200 001-point grid and gets $1.361108\,R_{\rm in}$ against the exact $49/36 = 1.361111$; an assertion requires agreement to $10^{-4}$. The factor of two in "per side" matters: dropping it moves $T_{\rm eff}$ by $2^{1/4}$, which is 19 per cent.
For the active-galactic-nucleus configuration — $M = 10^{8}\,M_\odot$, $R_{\rm in} = 6R_g = 8.8598\times10^{13}\ \mathrm{cm}$, and $\dot M = 1.3987\times10^{25}\ \mathrm{g\,s^{-1}} = 0.2220\,M_\odot\,\mathrm{yr^{-1}}$, which is a tenth of the Eddington rate at $\eta_{\rm rad} = 0.1$ — the profile runs as follows.
$R/R_g$
$T_{\rm eff}$ (K)
$\mathrm{d}\ln T_{\rm eff}/\mathrm{d}\ln R$
10
72 320.57
$-0.32044$
30
39 702.71
$-0.64887$
100
17 398.66
$-0.70945$
$10^{3}$
3 252.88
$-0.73950$
$10^{4}$
586.58
$-0.74686$
$10^{5}$
104.76
$-0.74902$
The maximum is at $8.166667\,R_g$, where $T_{\rm eff} = 75\,112.74\ \mathrm{K}$. Fig. 3 plots the profile and, in a panel of its own, the local slope that CHECK 2 tests.
CHECK 2 — CONFIRMED as a shape, and not as a digit. The exponent $-3/4$ contains no viscosity: $\nu$, $\alpha$ and $\Sigma$ all cancel between (8.1) and (8.2), leaving a statement of energy conservation alone. King, Pringle & Livio's Sect. 1 says exactly that, and says what it is measured against: the profile "is independent of the viscosity, being just a statement of energy conservation, and is in reasonable accord with both continuum spectra and eclipse mapping of cataclysmic variables". This module's index at $10^{5}R_g$ is $-0.74902$, which differs from $-0.75$ by $0.00098$. But that review prints no exponent and no error bar, so what is confirmed is the shape of the profile and not a digit of its slope. This is a weaker check than Module 10's solar-wind indices, which had 99 per cent confidence half-widths, and the module says so here rather than in a footnote.
The steady disc's effective temperature and the local slope of that curve, against radius in gravitational radii. The temperature vanishes at the inner edge because $f$ does, rises to a maximum at $(49/36)R_{\rm in} = 8.1667\,R_g$, and then falls; the slope approaches $-3/4$ from above and reaches $-0.74902$ at $10^{5}R_g$. Inside the marked radius the slope is positive and off the top of its own panel, so the curve starts where it enters.
9. What a disc radiates, and the price of a Newtonian book
Proposition 10 (the radiative efficiency of a thin disc). Integrating $2D(R)$ over the whole disc gives
$$ L \;=\; \frac{GM\dot M}{2R_{\rm in}}, \qquad\text{so}\qquad \eta_{\rm rad} \;\equiv\; \frac{L}{\dot M c^{2}} \;=\; \frac{R_g}{2R_{\rm in}}, \qquad R_g \equiv \frac{GM}{c^{2}}. \tag{9.1}$$
At the Schwarzschild innermost stable circular orbit, $R_{\rm in} = 6R_g$, this is exactly $1/12 = 0.083333$.
Proof. $L = \int_{R_{\rm in}}^{\infty} 2D(R)\,2\pi R\,\mathrm{d}R$, where the leading $2$ counts both faces and $2\pi R\,\mathrm{d}R$ is the area of the annulus. Substituting (8.2),
$$ L \;=\; \int_{R_{\rm in}}^{\infty} 2\cdot\frac{3GM\dot M}{8\pi R^{3}}f(R)\cdot 2\pi R\,\mathrm{d}R \;=\; \frac{3}{2}\,GM\dot M\int_{R_{\rm in}}^{\infty}\frac{f(R)}{R^{2}}\,\mathrm{d}R. $$
With $x = R/R_{\rm in}$ and $\mathrm{d}R = R_{\rm in}\,\mathrm{d}x$ the integral is $R_{\rm in}^{-1}\int_1^\infty x^{-2}\left(1-x^{-1/2}\right)\mathrm{d}x = R_{\rm in}^{-1}\left[1 - \tfrac23\right] = 1/(3R_{\rm in})$, so
$$ L \;=\; \frac{3}{2}\,GM\dot M\cdot\frac{1}{3R_{\rm in}} \;=\; \frac{GM\dot M}{2R_{\rm in}}. $$
Dividing by $\dot M c^{2}$ and using $R_g = GM/c^{2}$ gives (9.1); setting $R_{\rm in} = 6R_g$ gives $\eta_{\rm rad} = 1/12$. ∎
Half the binding energy, and not all of it. The specific binding energy of a circular orbit at $R_{\rm in}$ is $GM/2R_{\rm in}$, not $GM/R_{\rm in}$: the other half is the kinetic energy of the orbit, which the disc still carries when the material crosses the inner boundary and has not radiated.
This book is Newtonian, and six gravitational radii from a black hole is not a place where Newtonian gravity is right. Bardeen, Press and Teukolsky's eq. (2.21) gives the marginally stable orbit as $r_{\rm ms} = 6M$ for a non-rotating hole, and their eq. (2.12) gives the energy per unit rest mass of an equatorial circular orbit. Evaluated at $a = 0$ and $r = 6M$ that formula returns $4/(3\sqrt2) = \sqrt{8/9}$ exactly, so the relativistic efficiency is $1 - \sqrt{8/9} = 0.057191$. Their eq. (2.14), the extreme-Kerr simplification, at $r_{\rm ms} = M$ gives $1/\sqrt3$ and hence $1 - 1/\sqrt3 = 0.422650$.
Neither number is retyped from memory and neither is printed in that paper. The strings "8/9", "0.057" and "0.42" do not occur in Bardeen, Press and Teukolsky; the generator evaluates their formula. Their general expression has $r^{3/2} - 3Mr^{1/2} + 2aM^{1/2}$ under a square root, which at $a = M$ and $r = M$ is $1 - 3 + 2 = 0$ exactly, so evaluating it at the extreme-Kerr innermost orbit divides by zero — the first draft of the generator did precisely that and returned a NaN. Their own text warns of it: "Appearances are deceptive! ... The confusion is due to the subtle nature of the Boyer–Lindquist coordinates at $r = M$ for $a = M$." Their eq. (2.14) therefore has its own function, and an assertion requires the general one to stay non-finite there.
CHECK 3 — CONFIRMED against two independently published values. Shakura and Sunyaev's p. 339 states, in passing, that "in the case of Schwarzschild's metric $\eta \simeq 0.06$, in a Kerr black hole $\eta$ can attain 40 %". Against the two numbers computed here from a 1972 orbit formula, the ratios are $0.9532$ at zero spin and $1.0566$ at maximal spin. A formula published in 1972 and a sentence published in 1973 about the same two metrics agree to $4.7$ per cent at zero spin and $5.7$ per cent at maximal spin, and this module's generator reproduces both.
The price of a Newtonian book, and it is 45.7 per cent. At the same radius, $6R_g$, the Newtonian $1/12$ exceeds the relativistic $1 - \sqrt{8/9}$ by a factor of $1.4571$. The Newtonian calculation has no gravitational redshift and no relativistic correction to the binding energy of the orbit, and six gravitational radii from a black hole those are not small. Naming that number is the only honest way to use (9.1) here.
The Newtonian efficiency $\eta_{\rm rad} = R_g/2R_{\rm in}$ against the inner radius, with six values marked: four at $6R_g$ and two at $1R_g$. At $6R_g$, from the top down — the $0.1$ that module09.html:757 asserts, the Newtonian $1/12 = 0.083333$ that Proposition 10 derives and that lies on the curve, the $0.06$ Shakura and Sunyaev print, and the relativistic $0.057191$ computed from Bardeen, Press and Teukolsky’s eq. (2.12). At $1R_g$, the extreme-Kerr pair: $0.422650$ from their eq. (2.14) and the $0.40$ of Shakura and Sunyaev’s "can attain 40 %". The vertical gap at $6R_g$ between the curve and the relativistic point is the 45.7 per cent named above.
What this does to Module 9's factor of $2\times10^{7}$
Module 9 §8 refutes Bondi accretion onto Sagittarius A* in part by comparing an implied radiative efficiency with a thin disc's, and its Check 11 says: "A geometrically thin accretion disc radiates with $\eta_{\rm rad} \approx 0.1$. The implied $\eta_{\rm rad} \approx 4\times10^{-9}$ is smaller by a factor of $2\times10^{7}$." That $0.1$ was asserted there and is derived here, so the comparison can now be checked.
Module 9's own generator prints $L_X = 2.0\times10^{33}\ \mathrm{erg\,s^{-1}}$ and $\dot M c^2 = 4.543\times10^{41}\ \mathrm{erg\,s^{-1}}$, so its efficiency is $4.4024\times10^{-9}$. Dividing each candidate thin-disc efficiency by that:
efficiency
where it comes from
factor
to one figure
$0.1$
asserted at module09.html:757
$2.2715\times10^{7}$
$2\times10^{7}$
$1/12 = 0.083333$
Proposition 10, Newtonian
$1.8929\times10^{7}$
$2\times10^{7}$
$0.057191$
Schwarzschild, from BPT eq. (2.12)
$1.2991\times10^{7}$
$1\times10^{7}$
$0.06$
Shakura & Sunyaev p. 339
$1.3629\times10^{7}$
$1\times10^{7}$
Module 9's printed factor survives the Newtonian derivation and would read $1\times10^{7}$ at the relativistic value. No number in Module 9 depends on the factor — it is the size of a refutation that stays a refutation on any row of that table — so Module 9 is left as it stands, and the repayment is this paragraph rather than an edit to a shipped page.
One thing the derivation does settle: $0.1$ is not a Schwarzschild thin-disc number. It lies above the Newtonian $1/12$ by $1.2000$ and above the relativistic value by $1.7485$. Yu and Tremaine's study of the local black-hole mass density against the integrated quasar light finds that the accounting closes if quasars convert mass to energy with $\varepsilon \simeq 0.1$, and they say themselves that a higher value, "$\varepsilon \sim 0.2$, ... is possible for thin-disk accretion onto a Kerr BH". A population that radiates at $0.1$ on average contains spinning holes. Fig. 4 puts all six values on one axis against the Newtonian curve.
Yu and Tremaine's $\varepsilon \simeq 0.1$ is a consistency value and not a measurement with an error bar: it is the efficiency that makes two independently estimated quantities agree, and their local density carries an $h_{0.65}^2$ scaling. Their $\varepsilon$ is this module's $\eta_{\rm rad}$; the symbol is not reused here because Module 9 already writes bare $\varepsilon$ for a different quantity, the ratio of a field to its equipartition value.
10. When a disc weighs too much
Everything so far has assumed that the disc's own gravity does not matter. Module 5 §8 promised the correction here, listing among its idealisations "no rotation", which "omits the stabilisation of long wavelengths by rotation in a disc".
Proposition 11 (the Toomre criterion for a gas disc). A rotating, isothermal, self-gravitating sheet has the dispersion relation
$$ \omega^{2} \;=\; c_T^{2}k^{2} \;-\; 2\pi G\Sigma k \;+\; \kappa_{\rm ep}^{2} \tag{10.1}$$
for axisymmetric disturbances of wavenumber $k$ in the plane, and is stable at every $k$ if and only if
$$ Q \;\equiv\; \frac{c_T\,\kappa_{\rm ep}}{\pi G\Sigma} \;\gt \; 1. \tag{10.2}$$
Proof. Equation (10.1) has three terms, and this module supplies one of them and imports the other two. The first two are the dispersion relation of a non-rotating self-gravitating sheet, $\omega^2 = c_T^2k^2 - 2\pi G\Sigma|k|$, which Module 5 carries as its Problem D2 (module05.html:798) and derives there from the sheet potential $\Phi_1 = -(2\pi G\Sigma_1/|k|)\,e^{ikx-|k||z|}$; it is quoted here for $k \gt 0$, and in this module’s $c_T$ in place of Module 5’s $c_s$, which is the same isothermal sound speed. The self-gravity term is linear rather than quadratic in $k$ because the perturbation’s gravity is sourced by a surface density and not by a volume density. The third term is the rotational restoring force: Proposition 2 showed that a ring displaced radially at fixed angular momentum oscillates at $\kappa_{\rm ep}$, and an axisymmetric disturbance in the plane displaces rings radially, so it adds $+\kappa_{\rm ep}^{2}$ to $\omega^{2}$ exactly as a spring added in parallel adds to an oscillator’s squared frequency. That addition is stated and not derived here, because deriving it needs the linearised fluid equations in a rotating frame, which no module of this book writes; it is imported on the same terms as the magnetorotational instability of §11. Given (10.1), the right-hand side is a parabola in $k$ with a minimum at $k_* = \pi G\Sigma/c_T^{2}$, where
$$ \omega^{2}(k_*) \;=\; \kappa_{\rm ep}^{2} \;-\; \frac{(\pi G\Sigma)^{2}}{c_T^{2}}. $$
Requiring this to be positive is $c_T\kappa_{\rm ep}/(\pi G\Sigma) \gt 1$, which is (10.2). At $Q = 1$ the minimum is exactly zero and the sheet is marginal at $k_*$ alone. ∎
The trivial limit, which the generator asserts. Set the rotation to zero. Then $\kappa_{\rm ep} = 0$, (10.1) becomes $\omega^2 = c_T^2k^2 - 2\pi G\Sigma k$, and that is the relation Module 5's Problem D2 derives (module05.html:798), unchanged — unstable for every $k \lt 2\pi G\Sigma/c_T^2$ and stable above it, with no wavenumber at which a non-rotating sheet is stable to all disturbances. In the same limit (10.2) gives $Q = 0$, which is $Q < 1$: always unstable. So rotation is the only thing in (10.2) that can stabilise a sheet at long wavelengths, which is exactly what module05.html:764 said it omitted. Setting a new term to zero and recovering the earlier module's result is the check the Module 12 editor pass asked every later dispersion-relation proof to carry; this is Module 11's.
The coefficient $\pi$ is derived above and is NOT Toomre's. Toomre's 1964 paper is about a disc of stars: his p. 1217 says he found it "a legitimate first approximation to ignore the interstellar gas and dust, and to concentrate here on the stability of a thin disk composed only of stars", and sends the gaseous case to a later paper. His own criterion, his eq. (65), is $\sigma_{u,\min} = 3.36\,G\mu/\kappa$ with $\mu$ the projected stellar density — the coefficient is $3.36$ and not $\pi$, a difference of 7 per cent. The criterion carries his name here because that is what it is called; the gas coefficient is this module's algebra, with Safronov's 1960 paper — which Toomre's own reference list carries — named as the precedent and not read.
Because a Keplerian disc has $\kappa_{\rm ep} = \Omega$ by Proposition 2, (10.2) simplifies, and substituting the steady-state $\Sigma$ of (8.1) with $\nu = \alpha c_T H$ and $H = c_T/\Omega$ gives a form in which the radius has disappeared:
Proposition 12 (Q for a steady α-disc).
$$ Q \;=\; \frac{3\alpha\,c_T^{3}}{G\dot M\,f(R)}. \tag{10.3}$$
Both $\Omega$ and $R$ cancel. The stability of a steady disc against its own gravity is a statement about its local temperature and two global numbers, and about nothing else.
Proof. From (8.1), $\Sigma = \dot M f/(3\pi\nu) = \dot M f\Omega/(3\pi\alpha c_T^2)$. Substituting into (10.2) with $\kappa_{\rm ep} = \Omega$:
$$ Q = \frac{c_T\Omega}{\pi G}\cdot\frac{3\pi\alpha c_T^{2}}{\dot M f\Omega} = \frac{3\alpha c_T^{3}}{G\dot M f}. \qquad\blacksquare$$
For the active-galactic-nucleus configuration at $\alpha = 0.1$, with the midplane temperature taken equal to $T_{\rm eff}$:
$R/R_g$
$T_{\rm eff}$ (K)
$\Sigma$ (g cm⁻²)
$Q$
100
17 398.66
$9.4357\times10^{6}$
$1.5934$
$10^{3}$
3 252.88
$1.9500\times10^{6}$
$1.0542\times10^{-1}$
$10^{4}$
586.58
$3.6159\times10^{5}$
$7.6346\times10^{-3}$
$10^{5}$
104.76
$6.5127\times10^{4}$
$5.6646\times10^{-4}$
$Q$ falls outward and passes through $1$ at $146.7\,R_g = 2.1669\times10^{15}\ \mathrm{cm} = 7.022\times10^{-4}\ \mathrm{pc}$.
That radius is a LOWER BOUND and must be read as one. A disc of optical depth $\tau = \kappa_{\rm op}\Sigma/2$, with $\kappa_{\rm op}$ the Rosseland mean opacity per unit mass and the half accounting for one side of the midplane, has $T_{\rm mid}^{4} = \tfrac34\tau\,T_{\rm eff}^{4}$. Whenever $\tau \gt 4/3$ — which is what "optically thick" means here, and which the AGN row satisfies by orders of magnitude at the surface densities tabled above — the true midplane temperature exceeds $T_{\rm eff}$, the true $c_T$ is larger, and by (10.3) the true $Q$ is larger at every radius, which moves the radius at which $Q = 1$ outward. Using $T_{\rm eff}$ therefore says only that a steady thin $\alpha$-disc around a $10^{8}\,M_\odot$ black hole is self-gravitating no closer in than $7\times10^{-4}$ pc. This module does not carry the optically thick correction, and a self-gravitating disc is not described by Proposition 8 at all, so the steady solution refutes itself somewhere outside that radius and the module does not say exactly where.
On the protoplanetary row, where self-gravity is the physics of a real class of object rather than a caveat on an extrapolation, $Q = 126.5629$ with the gas coefficient and $118.3360$ with Toomre's stellar $3.36$ — the two differ by 7 per cent and the module does not average them. The surface density at which that row would reach $Q = 1$ is $1265.6286\ \mathrm{g\,cm^{-2}}$, against the row's $10\ \mathrm{g\,cm^{-2}}$.
11. What supplies the stress, and how big α is
Definition 2 named a number and bounded it. It did not say what physical process carries the stress, and for twenty years nobody could. Module 12 §9 is the answer: a Keplerian disc threaded by the weakest imaginable magnetic field is violently unstable, and the magnetorotational instability grows at $\gamma_{\max} = (q/2)\Omega$.
That section assumed $q = 3/2$ and said so. Proposition 1 derives it, so $\gamma_{\max} = 0.7500000000\,\Omega$ is now a result of this book rather than an input to it. The instability itself is not re-derived here: Module 8 §6.3 lists it among what Module 12 builds, and deriving it twice would make one of the two modules redundant.
Proposition 13 (the containment bound). If the magnetorotational instability's saturated field is identified with equipartition at a fraction $\alpha$ of the gas pressure, $B^2 \sim 4\pi\alpha\rho c_T^2$, then the fastest-growing wavelength of Module 12 §9 satisfies
$$ \frac{\lambda_{\max}}{2H} \;=\; \frac{4\pi\sqrt{\alpha}}{\sqrt{15}}, \tag{11.1}$$
in which every disc property has cancelled. The mode fits inside the disc only if
$$ \alpha \;\lt \; \frac{15}{16\pi^{2}} \;=\; 0.094989. \tag{11.2}$$
Proof. $v_A = B/\sqrt{4\pi\rho} = c_T\sqrt{\alpha}$ from the stated identification. Module 12 §9 gives the fastest-growing mode at $kv_A = (\sqrt{15}/4)\Omega$, so $\lambda_{\max} = 2\pi v_A/[(\sqrt{15}/4)\Omega] = (8\pi/\sqrt{15})\sqrt{\alpha}\,c_T/\Omega$, and $c_T/\Omega = H$ by Proposition 5. Dividing by $2H$ gives (11.1); setting it to one and squaring gives (11.2). ∎
The first draft of the generator printed (11.1) as a two-row table over the census, and both rows read $1.026040$. They had to: the two discs are $6.06$ decades apart in midplane density and every disc property cancels out of the ratio. A table of identical rows pretends to be a census result and is not one, and the table was replaced by the bound. The identification $B^2 \sim 4\pi\alpha P$ is an order-of-magnitude one, not a result; simulations put the ratio of the Maxwell stress to $B^2$ nearer $0.3$ than $1$.
The measured $\alpha$ of dwarf novae, $0.1$–$0.3$, lies above the bound (11.2) by factors of $1.0528$ to $3.1583$; the protostellar $0.01$ lies inside it, at $0.1053$ of it. So the crude identification is self-consistent in cold discs and marginal in hot ones. That is a coincidence of two crude numbers and the module claims nothing more for it.
How big is α?
King, Pringle and Livio titled a 2007 review with that question. Every number in the table below is from that one review, so what follows cannot be a disagreement between two papers.
system
$\alpha$
kind
where, in King, Pringle & Livio (2007)
thin, fully-ionised discs
$0.1$–$0.4$
a range
abstract, and their Sect. 5
dwarf novae
$0.1$–$0.3$
a range
their Sect. 2.1
soft X-ray transients
$0.2$–$0.4$
a range
their Sect. 2.2, Dubus et al. (2001)
AGN optical variability
$0.01$–$0.03$
lower limits
their Sect. 2.3.1, Starling et al. (2004)
T Tauri discs, 10–100 au
$0.01$
a value
their Sect. 2.3.2, Hartmann et al. (1998)
FU Orionis outbursts
$0.001$–$0.003$
a range
their Sect. 2.3.3
MHD simulations, no imposed field
$\le 0.02$
a bound
their Sect. 5
CHECK 4 — REFUTED. α is not a constant of nature. Two gaps. (a) Across disc classes: the fully-ionised $0.1$–$0.4$ exceeds Hartmann et al.’s T Tauri $0.01$ by a factor of $10$ to $40$, which is $1.00$ to $1.60$ decades. The protostellar side is itself a range: against the same review’s FU Orionis value of $0.001$–$0.003$ the comparison is $33$ to $400$, or $1.52$ to $2.60$ decades. The narrower figure is quoted as the headline and the wider one beside it, rather than picking whichever is larger. The two classes differ in ionisation, so this gap has a candidate physical cause and is not a contradiction. (b) Observation against simulation, on the same class of disc: the observed $0.1$–$0.4$ against a simulated $\alpha \le 0.02$ is at least a factor of 5, at least 0.70 decades, and one-sided, because their Sect. 5 states the simulation side as a bound and not a value. Their own abstract calls it "an order of magnitude smaller".
Six estimates of $\alpha$, all from King, Pringle and Livio's 2007 review, on a logarithmic axis. A bar is a range and an arrow is a bound: the AGN row is a set of lower limits and points right, the simulation row is an upper bound and points left. The dashed line is the containment bound (11.2). Drawing either bound as a bar would make the figure state a value neither paper gives.
Gap (b) is the one the review is about, and it is uncomfortable for this book specifically. The simulations are of the magnetorotational instability that Module 12 §9 derives. The mechanism this book builds does not, in the calculations that solve it directly, reach the $\alpha$ that observations require. Two things keep that from being a refutation of the mechanism. The authors' Sect. 4 notes that the limitations of the simulations — restricted box sizes, boundary conditions, and above all the suppression of any globally imposed field — "would indeed tend to lead to underestimating the value of $\alpha$", so the side more likely to be wrong is the theoretical one. And the bound points down, so the true gap may be smaller than $5$ and cannot be larger than stated from that side alone.
What is refuted is not the prescription. Equation (6.1) remains a definition of $\alpha$, and a definition cannot be wrong. What is refuted is the reading of $\alpha$ as a constant of nature — and Shakura and Sunyaev never claimed it was one. Their bound is $\alpha \lesssim 1$ and nothing more, and their own footnote says the disc's observable appearance barely depends on it.
12. What this theory captures, and what it misses
The thin $\alpha$-disc captures a great deal for what it is: a one-parameter closure applied to a vertically averaged, axisymmetric, Keplerian flow. It produces the right topology of the problem — angular momentum outward, mass inward, energy released at three times the local rate far out — and a temperature profile that is independent of the closure and is what is observed. What it misses is a long list, and the interesting entries are the ones a shipped module of this book already asked for.
idealisation
what it costs
where it is repaid
$\alpha$ is constant, in space and in time
Shakura and Sunyaev assumed it and said so; §11 shows the same symbol takes values $10$ to $40$ apart in different disc classes. Nothing in this module tells you which value to use
Nowhere in this book
The zero-torque inner boundary
Propositions 8 to 10 all inherit it. A magnetically connected central object breaks it, and the bracket $f$ — hence $1/12$, hence the factor of three, hence the $(49/36)R_{\rm in}$ maximum — changes
Nowhere in this book
Newtonian gravity at $6R_g$
Measured in §9: the efficiency comes out $45.7$ per cent high
Not repaid; the correction is quoted from Bardeen, Press and Teukolsky and not derived, because the book has no general relativity
One temperature, and an optically thick, radiatively efficient disc
Module 2 §11 promises "two-temperature accretion flows in Module 11" (module02.html:686) and Module 9 §9 promises "two-temperature flows in Module 11" (module09.html:848) — without the word "accretion" — and Module 9 §8.3 names the radiatively inefficient flow as the defining reading of Sagittarius A*. This module does not build either. Such a flow is geometrically thick, violates Definition 1, and needs a collisionless two-fluid energy budget that Module 1 only sketched
Refused in print, here. No planned module builds it, and saying so is better than promising a repayment that never comes — the form Module 8 §9 used for oblique MHD shocks
Self-gravity neglected
§10 prices it: a steady AGN disc is self-gravitating no closer in than $7\times10^{-4}$ pc, and that is a bound because the midplane is hotter than $T_{\rm eff}$
Partly repaid in §10; the optically thick correction is not carried
No magnetic field in the disc's own structure
The field appears only through $\alpha$ and through §11's order-of-magnitude bound. Magnetic pressure, tension and winds are absent from Propositions 6 to 9
Module 12 for the physics; the coupling to disc structure is in no module of this book
Axisymmetric, vertically averaged, no radial pressure gradient
Excludes spiral structure, warps, and the transonic region near $R_{\rm in}$ where the Keplerian assumption of Proposition 1 fails
Nowhere in this book
Radiation carries no momentum
The AGN configuration runs at a tenth of Eddington, where radiation pressure is not negligible in the inner disc and $H/R$ is not small
Read as a whole, that table says the same thing twice. The two rows that cost most — a constant $\alpha$ and a zero-torque boundary — are both closures, not physics, and neither is repaid anywhere. A theory whose central parameter is measured to be non-constant and whose boundary condition is assumed is a useful theory and not a complete one, and the gap between $\alpha \le 0.02$ in simulation and $0.1$–$0.4$ in observation is where that incompleteness is visible.
13. Problems
C1. Starting from $\kappa_{\rm ep}^2 = R^{-3}\,\mathrm{d}(R^4\Omega^2)/\mathrm{d}R$, derive (2.2) for $\Omega \propto R^{-q}$. Evaluate $\kappa_{\rm ep}/\Omega$ at $q = 0$, $1$, $3/2$, $1.9$ and $2$, and state at which $q$ a disc first becomes Rayleigh-unstable.
Solution. $\Omega^2 \propto R^{-2q}$, so $R^4\Omega^2 \propto R^{4-2q}$ and $\mathrm{d}(R^4\Omega^2)/\mathrm{d}R \propto (4-2q)R^{3-2q}$; dividing by $R^3$ gives $\kappa_{\rm ep}^2 = (4-2q)\Omega^2 = 2\Omega^2(2-q)$. The ratios $\kappa_{\rm ep}/\Omega = \sqrt{2(2-q)}$ are $2.000000$, $1.414214$, $1.000000$, $0.447214$ and $0.000000$. Instability requires $\kappa_{\rm ep}^2 \lt 0$, hence $q \gt 2$: a point mass gives $3/2$ and cannot produce it.
C2. For the protoplanetary configuration — $M = 1\,M_\odot$, $R = 10$ au, $T = 50$ K, $\mu = 2.34$, $\Sigma = 10\ \mathrm{g\,cm^{-2}}$ — compute $c_T$, $\Omega$, $H$, $H/R$, $\rho_0$, and the viscous time at $\alpha = 0.01$.
Solution. $c_T = \sqrt{kT/\mu m_u} = 0.421497\ \mathrm{km\,s^{-1}}$ and $\Omega = \sqrt{GM/R^3} = 6.296043\times10^{-9}\ \mathrm{s^{-1}}$, so $H = c_T/\Omega = 6.694628\times10^{12}\ \mathrm{cm} = 0.447508$ au and $H/R = 0.044751$. The midplane density is $\Sigma/\sqrt{2\pi}H = 5.959140\times10^{-13}\ \mathrm{g\,cm^{-3}}$. With $\nu = \alpha c_T H = 2.821764\times10^{15}\ \mathrm{cm^2\,s^{-1}}$, $t_\nu = R^2/\nu = 7.931041\times10^{12}\ \mathrm{s} = 2.513195\times10^{5}\ \mathrm{yr}$. The $\alpha = 0.01$ is Hartmann et al.'s value for T Tauri discs at 10–100 au, quoted through King, Pringle and Livio Sect. 2.3.2, and this configuration sits at 10 au.
C3. Show that $T_{\rm eff}$ of (8.3) is maximised at $R = (49/36)R_{\rm in}$, and evaluate the maximum for the AGN configuration ($M = 10^{8}\,M_\odot$, $R_{\rm in} = 6R_g$, $\dot M = 1.3987\times10^{25}\ \mathrm{g\,s^{-1}}$).
Solution. With $x = R/R_{\rm in}$, $T_{\rm eff}^4 \propto x^{-3}(1 - x^{-1/2})$. Setting the derivative to zero: $-3x^{-4} + \tfrac72 x^{-9/2} = 0$, so $x^{1/2} = 7/6$ and $x = 49/36 = 1.361111$. In gravitational radii that is $8.166667\,R_g$, where $T_{\rm eff} = 75\,112.74$ K. At $R_{\rm in}$ itself $T_{\rm eff} = 0$ exactly, because $f$ vanishes there.
D1. Derive the steady-state relation (8.1) from Proposition 3, stating where the zero-torque inner boundary enters.
Solution. See the proof of Proposition 8. The boundary condition enters once, in fixing the integration constant, and the condition is $G(R_{\rm in}) = 0$ and not $\mathrm{d}\Omega/\mathrm{d}R = 0$: the Keplerian law has $\mathrm{d}\Omega/\mathrm{d}R = -\tfrac32\Omega/R$, which vanishes at no radius, so a Keplerian disc meets the zero-torque condition through $\nu\Sigma \to 0$ instead. With $G(R_{\rm in}) = 0$ the constant equals $\dot M\sqrt{GMR_{\rm in}}$. Any other inner boundary — a magnetically connected star, for instance — leaves $G(R_{\rm in}) \ne 0$, which changes that constant and with it the bracket $f$.
D2. Show that far from the inner edge the disc radiates three times the rate at which gravitational energy is released locally, and say where the excess comes from. Find the radius at which the two are equal.
Solution. The specific orbital energy is $-GM/2R$, so material moving inward through an annulus of width $\mathrm{d}R$ releases $GM\dot M\,\mathrm{d}R/(2R^2)$; the annulus has area $2\times 2\pi R\,\mathrm{d}R$ counting both faces, so the local release per unit area per side is $GM\dot M/(8\pi R^3)$. Against (8.2) the ratio is $3f(R)$, which tends to $3$. The ratios at $R/R_{\rm in} = 1.5$, $2$, $10$, $10^2$ and $10^4$ are $0.550510$, $0.878680$, $2.051317$, $2.700000$ and $2.970000$. Setting $3(1 - \sqrt{R_{\rm in}/R}) = 1$ gives $R = \tfrac94 R_{\rm in}$ exactly. Of the three units far out, one is released locally and two are carried outward to that radius by the viscous torque of Proposition 3 from material further in; inside $\tfrac94 R_{\rm in}$ the annuli radiate less than they release, because they are exporting energy rather than importing it.
D3. Derive (10.3) and show that $\Omega$ and $R$ cancel.
Solution. See the proof of Proposition 12. The cancellation happens because $\Sigma \propto \Omega/c_T^2$ in a steady $\alpha$-disc while $Q \propto c_T\Omega/\Sigma$, so the two factors of $\Omega$ divide out and no explicit $R$ survives except inside $f$.
K1. Compute the molecular viscous time of the protoplanetary configuration, and compare it with C2. Warning: the gas is molecular at 50 K, so the Coulomb mean free path of §3 does not apply. Use Module 1's order-of-magnitude neutral cross-section $\sigma \approx 10^{-15}\ \mathrm{cm^2}$.
Solution. With $n = 1.533623\times10^{11}\ \mathrm{cm^{-3}}$ the neutral mean free path is $\lambda = 1/(n\sigma) = 6.520507\times10^{3}\ \mathrm{cm}$, and $v_{\rm th}$ at 50 K with $\mu = 2.34$ is $0.672611\ \mathrm{km\,s^{-1}}$, so $\nu_{\rm mol} = \tfrac13 v_{\rm th}\lambda = 1.461922\times10^{8}\ \mathrm{cm^2\,s^{-1}}$ and $t_\nu = 1.530828\times10^{20}\ \mathrm{s} = 4.850902\times10^{12}\ \mathrm{yr}$. That is $1.930173\times10^{7}$ times C2's answer, or $7.29$ decades. Using the Coulomb formula here would be meaningless: at 50 K in molecular hydrogen there are no free charges to deflect, and the formula's $(kT)^2/n e^4$ describes a process that is not occurring.
K2. Using Module 9's accretion rate and black-hole mass for Sagittarius A*, compute the luminosity a thin disc would produce and compare it with the observed X-ray luminosity. Check that your answer reproduces the efficiency of Proposition 10.
Solution. Module 9's generator gives $\dot M c^2 = 4.543\times10^{41}\ \mathrm{erg\,s^{-1}}$, so $\dot M = 5.054769\times10^{20}\ \mathrm{g\,s^{-1}}$; its mass is $8.5502\times10^{39}\ \mathrm{g}$, so $R_{\rm in} = 6R_g = 3.809709\times10^{12}\ \mathrm{cm}$. Then $L = GM\dot M/2R_{\rm in} = 3.785833\times10^{40}\ \mathrm{erg\,s^{-1}}$, and $L/\dot Mc^2 = 0.083333 = 1/12$ as Proposition 10 requires. Against the observed $L_X = 2.0\times10^{33}\ \mathrm{erg\,s^{-1}}$ the ratio is $1.892917\times10^{7}$, or $7.28$ decades — the same refutation Module 9 prints as a factor of $2\times10^{7}$, reached by the luminosity route instead of the efficiency route.
K3. Find the value of $\alpha$ at which the magnetorotational instability's fastest-growing wavelength equals the full disc thickness $2H$, under the equipartition identification $B^2 \sim 4\pi\alpha\rho c_T^2$. Compare it with the measured values of §11, and say what the comparison is and is not worth.
Solution. By (11.2) the value is $15/(16\pi^2) = 0.094989$. The dwarf-nova range $0.1$–$0.3$ exceeds it by factors of $1.0528$ to $3.1583$; the protostellar $0.01$ is inside it, at $0.1053$ of it. So the crude identification is self-consistent in cold discs and marginal in hot ones — the same split CHECK 4 measures, which is suggestive and nothing more. Both sides are order-of-magnitude arguments: the saturated stress is not exactly $B^2/4\pi$, and $\alpha$ itself is measured to a factor of a few.
Notation
symbol
meaning
units
first use
$\alpha$
the disc viscosity parameter of (6.1). Module 12’s notation table already fixes this symbol for this meaning in print. Five other meanings exist in this book and none occurs here: Module 5's is a length scale, Module 6's the mixing-length parameter $\ell/H$, Module 10's the ratio $P_{\rm turb}/P_{\rm th}$, Module 2's, Module 9's and Module 12's a fitted density exponent, and Module 7's $\alpha_B$, $\alpha_S$ mixing coefficients. Shakura and Sunyaev's own $\alpha$ is a sum of two terms; see §6
—
Definition 2
$\Sigma$
the disc's surface density, $\int\rho\,\mathrm{d}z$. Module 5's $\Sigma$ is a sheet's surface density and Module 10's is Solomon et al.'s cloud value $170\,M_\odot\,\mathrm{pc^{-2}}$; both are module-local and neither occurs here
$\mathrm{g\,cm^{-2}}$
Definition 1
$\mu$
three meanings, and the table names all three because two of them are quotations. In the census table, in $c_T = \sqrt{kT/\mu m_u}$ and in Problem C2 it is the mean molecular weight, dimensionless, the Module 9 convention against $m_u$ and not against $m_{\rm H}$. In (3.1) it is the dynamic viscosity, kept in Module 1’s own glyph because (3.1) quotes that module’s Proposition 4 eq. (6.2), $\mu \simeq \tfrac13\rho v_{\rm th}\lambda$, and converting a symbol inside a quoted formula would misrepresent it; the quantity survives only as far as $\nu_{\rm mol} = \mu/\rho$ on the same line. In the warning at §10 it is Toomre’s projected stellar surface density, inside his eq. (65) and nowhere else. No equation in this module carries two of the three
the dynamic viscosity, written $\eta$ in the proof of Proposition 3 alone, where a shear stress is $\eta$ times a rate of strain and the disc integral replaces $\rho$ by $\Sigma$. It is the same physical quantity as $\mu$ in (3.1), and the two glyphs are deliberate: $\mu$ where Module 1 is quoted, $\eta$ where a stress is written. Module 10’s $\eta$ is the Kolmogorov dissipation scale and does not occur here
$\mathrm{g\,cm^{-1}\,s^{-1}}$
Proposition 3
$\ell$
specific angular momentum $\Omega R^2$, and nothing else in this module.Module 6's $\ell$ is a mixing length and Module 10's a separation inside the inertial range; neither occurs here. Module 12 §9 writes $\ell = \Omega R^2$ in the same sense and carries no notation row for it, so this row serves the whole book
$\mathrm{cm^2\,s^{-1}}$
(1.1)
$\Omega$, $v_K$
angular velocity and orbital speed, both Keplerian unless said otherwise. Bare $\Omega$ throughout: every equation in this module carries it, so a subscript would be noise on every line. Module 12’s notation table rows the subscripted $\Omega_\odot$ and $\Omega_{\rm K}$, and that module writes bare $\Omega$ in two senses — a star’s rotation rate in its §5 (module12.html:364) and a disc’s angular velocity in its §9 — so the bare glyph is not new here. Module 3’s $\Omega$ is a gravitational potential energy and does not occur here
$\mathrm{s^{-1}}$, $\mathrm{cm\,s^{-1}}$
(2.1)
$q$
the shear parameter $-\mathrm{d}\ln\Omega/\mathrm{d}\ln R$. Module 12 §9 takes it as $3/2$ and assumed; Proposition 1 derives it
—
Proposition 1
$\kappa_{\rm ep}$
the epicyclic frequency, always subscripted. The disc literature writes it $\kappa$; Module 12 reserved the subscripted form against the thermal conductivity and this module inherits the reservation. Module 4's $\kappa$ is an evanescent decay rate
$\mathrm{s^{-1}}$
Proposition 2
$c_T$
the isothermal sound speed $\sqrt{kT/\mu m_u}$, Module 9's convention. Shakura and Sunyaev's $v_s$ obeys $\rho v_s^2/2 = \tfrac32\rho kT/m_p + \varepsilon_r$ and is larger by $1.34$ at $\mu = 0.6$; $\alpha$ absorbs the factor
$\mathrm{cm\,s^{-1}}$
Definition 2
$H$
the disc scale height $c_T/\Omega$, Module 6's book-wide meaning. Shakura and Sunyaev's $H$ is the magnetic field — see §6
cm
Proposition 6
$\nu$
kinematic viscosity, the Module 7 and Module 10 convention, unchanged. $\nu_{\rm mol}$ carries its subscript wherever the molecular value is meant
$\mathrm{cm^2\,s^{-1}}$
(3.1)
$\dot M$, $R_{\rm in}$, $f$
the accretion rate, the inner radius, and the bracket $1 - \sqrt{R_{\rm in}/R}$
$\mathrm{g\,s^{-1}}$, cm, —
Proposition 8
$\eta_{\rm rad}$, $R_g$
the radiative efficiency $L/\dot Mc^2$ and the gravitational radius $GM/c^2$. Never bare $\eta$ for an efficiency: bare $\eta$ is the dynamic viscosity, rowed above, and the bare $\eta$ that appears in §9 is inside the quotation of Shakura and Sunyaev’s own sentence and is theirs, not this module’s. Yu and Tremaine write $\varepsilon$ for this quantity; that symbol is not reused here because Module 9 already uses bare $\varepsilon$ for a field ratio
—, cm
Proposition 10
$Q$
the Toomre parameter (10.2), with the gas coefficient $\pi$, derived in Proposition 11 and not Toomre's; his own coefficient is $3.36$ and is for a disc of stars
—
Proposition 11
$t_{\rm dyn}$, $t_{\rm th}$, $t_\nu$
the dynamical, thermal and viscous times of (7.3)
s
Proposition 5
Sources
Shakura, N. I. & Sunyaev, R. A. (1973), "Black holes in binary systems. Observational appearance", A&A24, 337–355. Read from the ADS scan, which has no text layer: pages 338 and 339 were rendered at 190 dpi and read as images. Their p. 338 gives $\alpha = v_t/v_s + H^2/(4\pi\rho v_s^2)$, the bound $\alpha \lesssim 1$, and the footnote on the weak dependence of the disc's appearance on $\alpha$; their p. 339 gives $\eta \simeq 0.06$ for Schwarzschild and "can attain 40 %" for Kerr.
King, A. R., Pringle, J. E. & Livio, M. (2007), "Accretion disc viscosity: how big is alpha?", MNRAS376, 1740–1746, arXiv:astro-ph/0701803. Read in full. Supplies every entry of §11's table, the statement that $T \propto R^{-3/4}$ is viscosity-independent and in accord with CV spectra and eclipse mapping, and the simulation bound $\alpha \le 0.02$.
Bardeen, J. M., Press, W. H. & Teukolsky, S. A. (1972), "Rotating black holes: locally nonrotating frames, energy extraction, and scalar synchrotron radiation", ApJ178, 347–369. Read from the ADS scan. Their eq. (2.21) gives $r_{\rm ms} = 6M$ at $a = 0$ and $r_{\rm ms} = M$ at $a = M$; their eqs. (2.12) and (2.14) give the orbit energy. They print no efficiency; the two values in §9 are computed here from their formulae.
Yu, Q. & Tremaine, S. (2002), "Observational constraints on growth of massive black holes", MNRAS335, 965–976, arXiv:astro-ph/0203082. Read in full. Their $\varepsilon \simeq 0.1$ is the efficiency that closes the Soltan accounting, not a measurement with an error bar.
Toomre, A. (1964), "On the gravitational stability of a disk of stars", ApJ139, 1217–1238. Read from the ADS scan. Cited for the criterion's name and for his stellar coefficient $3.36$ (his eq. 65) only. His p. 1217 states that he ignores the gas and treats a disc of stars, and sends the gaseous case to a later paper; the coefficient $\pi$ used here is derived in Proposition 11.
Safronov, V. S. (1960), Ann. d'Astrophysique23, 979. The gas-disc precedent for (10.2). Not read; the volume and page are from Toomre's own reference list, and the result is derived here rather than quoted.
Pringle, J. E. (1981), "Accretion discs in astrophysics", ARA&A19, 137–162. Not read. Cited for the standard derivation of (5.1) only; no number in this module comes from it.
Hartmann, L. et al. (1998); Dubus, G. et al. (2001); Starling, R. L. C. et al. (2004); Smak, J. (1998, 1999); Cannizzo, J. K. (2001a, 2001b); Schreiber, M. R. et al. (2003, 2004); Buat-Ménard, V. et al. (2001); Hawley, J. F. & Krolik, J. H. (2001). None read. All are quoted through King, Pringle & Livio (2007), which collects them, and the route is named wherever a number from one of them appears.
Balbus, S. A. & Hawley, J. F. (1991), ApJ376, 214–233. The magnetorotational instability. Not read, and not needed here: Module 12 §9 derives the result this module cites.