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Choose the positive-frequency branch and real radial amplitude $u$. The azimuthal amplitude of the inertial-acoustic wave is $U_y=-i\Omega u/(2\omega )$, so taking real parts givesFor $|k|c_s/\Omega \ll 1$, $\omega =\Omega [1+O((k{c}_s/\Omega {)}^2)]$. Thus, to leading long-wavelength order,The quarter-cycle phase shift and the factor $1/2$ are the epicyclic motion relation in a Keplerian shearing sheet. This approximation drops the small frequency detuning; over times $|\omega -\Omega |t\gtrsim 1$ that phase approximation is not uniform. The associated fractional surface density of a disk perturbation is $\sigma /{\Sigma }_0=(ku/\omega )\cos (kx-\omega t)$, so linearization also requires $|ku/\omega |\ll 1$. The source's arbitrary real $u$ sets the wave phase/amplitude within that small-perturbation regime.

The razor-thin disk approximation uses vertically integrated pressure and surface density of a disk. To close the printed equations, interpret the barotropic equation of state as $P=P(\Sigma )$, or supply an equivalent fixed vertical conversion from the printed $\rho$ to $\Sigma$. Without that conversion, the literal $P=P(\rho )$ does not close this two-dimensional system. Set $c_s^2=(dP/d\Sigma {)}_0>0$, and write $\Sigma ={\Sigma }_0+\sigma$, $\mathbf{u}={\mathbf{u}}_0+{\mathbf{u}}^{\prime }$, $h^{\prime }=P^{\prime }/{\Sigma }_0=c_s^2\sigma /{\Sigma }_0$. Axisymmetric flow here means ${\partial }_y=0$, so background advection vanishes. The background shear enters through ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}_0=-3\Omega u_x^{\prime }{\mathbf{e}}_y/2$. The complete linearized equations areThese are the compressible Keplerian shearing sheet equations for an inertial-acoustic wave. No perturbed self-gravity is present in the given force law, so no gravitational potential perturbation is to be added. This closure is the barotropic closure of a razor-thin disk.

The radial velocity of a Kepler orbit follows from the vis-viva equation after subtracting the tangential component:Each radial interval is crossed twice per orbital period, so its fraction of time isThis is the phase-mixed radial probability of a Kepler orbit, normalized to one between $q$ and $Q$. Dividing by the annular area $2\pi rdr$ gives the phase-mixed comet surface density per comet:For $Q\gg a_p>q$,The printed density has the powers of $Q$ and $a_p$ interchanged. Its expression is too large by $Q/a_p$ in this limit. The normalized residence-time derivation fixes the corrected expression, which is also required for the later $Q^{1/2}$ ejection-time scaling. For a population of $N_c$ independent identical comets, multiply ${\Sigma }_c$ by $N_c$ to obtain a number surface density of a disk.

In the proper-eccentricity annulus, averaging over random proper phases cancels the proper vector in the first-order angular variation. The mean line density is proportional to $1-e_f\cos \theta$, while the full ring width is independent of angle. Thus at corresponding radial locationsThere is a forced-apocentre enhancement in number surface density of a disk, together with the proper-excursion edge profile.

For a narrow aligned eccentric ring with locally constant $e_f$, its width has the same first-order factor $1-e_f\cos \theta$ as its line density. Their ratio is constant: the constant-eccentricity aligned ring has no first-order surface-density contrast. It still contains more particles per unit longitude at apoapsis, spread across a wider area.

The more general surface density contrast in an eccentric ring depends on the eccentricity gradient. With a smooth number distribution per unit semi-major axis and aligned apsides,Hence decreasing $e_f(a)$ can also make the aligned ring apocentre-dense; for $e_f\propto a^{-1}$ it has the same first-order contrast as the proper-phase case. A positive gradient gives a pericentre enhancement. The problem leaves this profile unspecified, so there is no universal observational distinction based solely on the two apse densities.

Resolved ring width and a measured forced-eccentricity profile can distinguish the constant-width and similar-ellipse constructions. Interpreting observed brightness additionally requires the grain temperature, illumination, emissivity, and viewing geometry; brightness is not itself particle number density. Inferences from one contrast alone can therefore be degenerate.

Take $v>0$ to mean the magnitude of the inward radial velocity. In the Gaussian units used for the magnetic pressure, approximate the surface density of a disk by $\Sigma \simeq 2H\rho$. The mass accretion rate and free-fall speed then giveAt the magnetospheric truncation radius, balance the magnetic pressure of the magnetic dipole field with the specified ram pressure:ThusDiscarding the convention-dependent numerical factor givesThe magnetic pressure grows inward as $r^{-6}$, faster than the ram pressure $\propto r^{-5/2}$, so the field disrupts the flow inside this radius. The numerical coefficient depends on the vertical averaging and on treating the inflow as radial free fall, as stipulated here; a detailed thin disk boundary model need not have the same coefficient.

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

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

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

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

At fixed cylindrical radius, let $w=u_z$ and use the continuity equation ${\partial }_t\rho +{\partial }_z(\rho w)=0$. For the similarity solution $z=\xi \eta (t)$ and $\rho =\tilde{\rho }(\xi )/\eta$, the continuity equation reduces toHere $\eta$ is dimensionless and $\xi$ labels the initial height. Midplane symmetry sets the integration constant to zero, so the flow is a homologous vertical motion of an astrophysical disk:The hydrostatic approximation and ideal gas law becomeThe material derivatives are $D_{t}P=\tilde{P}\dot{\eta }$ and $D_t\rho =-\tilde{\rho }\dot{\eta }/{\eta }^2$, so the energy equation givesSeparation of variables requires a positive constant $K$ such thatIntegrating with $\eta (0)=1$ gives the slow cooling of an astrophysical disk similarity factorA closed set of profile equations is thereforeThese have the same polytropic vertical structure in stellar gravity as part (b). Writingwith $P_0=\mathcal{R}{\rho }_0{T}_0$, fixesReturning to physical height, the explicit profiles and evolving semi-thickness for $|z|\le H(t)$ areandOutside this moving surface $\rho =P=0$. The surface density of a disk is conserved because $\rho dz=\tilde{\rho }d\xi$. The column cools and contracts without changing its scaled profile: its midplane temperature falls as ${\eta }^2$, its mass density rises as ${\eta }^{-1}$ and its pressure falls as $\eta$.

The value $T_0$ determines $H_0$, but the mass density normalization also enters $C$. If the initial column is specifically the heated equilibrium of part (b), switching off its heating givesOtherwise $P_0$, or equivalently the conserved column mass, is additional initial data. The contraction timescale grows with $1+Ct$, so sufficiently slow initial evolution remains slow relative to the fixed vertical dynamical timescale of a disk. This is an exact solution of the reduced hydrostatic approximation, rather than the full momentum equation: its omitted vertical acceleration is $\ddot{\eta }\xi$, with $\ddot{\eta }/(\eta {\Omega }^2)=2(1+\beta )C^2/[(1+2\beta {)}^2{\Omega }^2(1+Ct{)}^2]$, which remains small and decreases if $C\ll \Omega$.

Analyze a local temperature perturbation at fixed radius and fixed surface density of a disk $\Sigma$. The thermal timescale of an accretion disk is shorter than its viscous timescale, so mass redistribution is negligible during the perturbation. Let $Q^{+}$ be viscous heating and $Q^{-}$ radiative cooling, measured consistently through one face or both faces, and define net cooling by $\dot{Q}=Q^{-}-Q^{+}$. A temperature increase is unstable if it decreases net cooling.

With gas pressure dominant, vertical hydrostatic equilibrium and the ideal gas law giveFor the alpha disk prescription with temperature-independent $\alpha$, $\nu =\alpha c_{s}H\propto T_c$, and the Keplerian rotation heating law isOne must not hold the steady-state $\dot{m}$ fixed while perturbing the temperature: the stress and heating respond through $\nu$, even though $\Sigma$ is temporarily fixed.

At an equilibrium temperature $T_0$ where $Q^{+}=Q^{-}=Q_0$, if $Q^{-}\propto T_c^q$ and $Q^{+}\propto T_c$, thenThe cooling exponent therefore decides the sign.

For hot, ionized, optically thin gas, the usual intended cooling model is thermal bremsstrahlung, whose volume emissivity scales as ${\rho }^2{T}_c^{1/2}$. Integrated vertically,Thus $q=0$, andA hotter annulus expands vertically, lowering its density enough to offset the direct increase in bremsstrahlung emission; viscous heating still rises.

For an optically thick medium with the Kramers' opacity law, $\kappa \propto {\rho }_c{T}_c^{-7/2}\propto T_c^{-4}$ at fixed $\Sigma$. Thermalized radiative diffusion givesIts falling opacity lets a hotter annulus radiate much more efficiently.

For an optically thick medium with dominant electron-scattering opacity, $\kappa$ is nearly independent of temperature. Provided enough true absorption is present to thermalize the radiation,This conclusion differs from the thermal instability of a radiation-pressure-dominated alpha disk, where the heating has a different temperature dependence. The question explicitly excludes that pressure regime.

The first conclusion needs its cooling assumption stated. $T>10^4\mathrm{K}$ alone does not guarantee that thermal bremsstrahlung dominates; atomic-line cooling, ionization changes, external illumination, or energy advection can change the answer. For a general optically thin emissivity ${\rho }^2\Lambda (T_c)$, the corresponding exponent isAccordingly, under the same fixed-$\Sigma$ alpha prescription,The unstable/stable/stable classification is the standard free-free/diffusion answer, rather than a universal result specified by temperature and optical depth alone.

Vertical hydrostatic equilibrium is established by sound waves over the disk scale height, soThe local heating-cooling balance is established on the thermal timeThe surface density of a disk evolves only as angular momentum is redistributed, on the viscous timeFor a thin disk, $H/r\ll 1$, so $t_{\mathrm{vert}}\lesssim t_{\mathrm{th}}\ll t_{\nu }$. The vertical and thermal equations can therefore be treated as local equilibria while $\Sigma$ evolves.

The standard Shakura--Sunyaev thin disk assumes: a steady state; axial symmetry; a geometrically thin disk $H/R\ll 1$; Newtonian, nearly Keplerian circular motion outside the inner edge; subsonic radial drift $|u_R|\ll u_{\varphi }$; vertical hydrostatic equilibrium; negligible disk self-gravity; an optically thick, locally thermal spectrum; local radiative balance between viscous heating and cooling; and a local alpha-viscosity stress with a zero-torque inner boundary near the innermost stable circular orbit. These assumptions also exclude dynamically dominant winds and large-scale external torques from the standard solution.

Start from the continuity equationIntegrate vertically, use axial symmetry and steadiness, and define the surface density of a disk $\Sigma =\int \rho dz$. Thenso the inward-positive accretion rate is

The standard Shakura--Sunyaev thin disk has three qualitative radial zones. Its hot inner part is dominated by radiation pressure and electron-scattering opacity; farther out, gas pressure overtakes radiation pressure while electron scattering can remain the main opacity; in the cool outer zone, gas pressure remains dominant and free-free opacity becomes important. The transition radii vary with $M$, $\dot{M}$, and $\alpha$.

For the inner zone, hold the surface density $\Sigma$ fixed during a local thermal perturbation. Vertical hydrostatic equilibrium givesso $H\propto T_c^4/\Sigma$. The alpha disk prescription then yieldswhereas optically thick radiative diffusion with nearly constant electron-scattering opacity givesAt equilibrium $Q^{+}=Q^{-}=Q_0$. For net cooling $\dot{Q}=Q^{-}-Q^{+}$,Thus the total-pressure alpha prescription predicts thermal instability of a radiation-pressure-dominated alpha disk: a temperature increase makes heating outrun cooling. The absence of ubiquitous, large-amplitude thermal limit cycles in luminous AGN light curves indicates that this local model omits stabilizing effects, plausibly magnetic pressure and stress, vertical advection, winds, or a stress law that does not simply track total pressure.

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

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

Define the surface density of a disk and outward radial mass flux byVertical integration of mass conservation eliminates the surface term because $\rho u_z\to 0$, and axisymmetry eliminates the azimuthal derivative. Hence

For $u_{\varphi }\simeq r\Omega (r)$, the specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal momentum equation by $r$, integrate vertically and azimuthally, and define the viscous torque in an accretion diskSubtracting $h$ times the mass equation from the integrated angular-momentum equation givesFor an axisymmetric circular flow, ${\Pi }_{r\varphi }=\mu rd\Omega /dr$. Ifthen

Combining the two conservation laws givesThe first term is the local rate of change of angular momentum per radial interval, $\mathcal{F}h$ is outward advective angular-momentum flux, and $\mathcal{G}$ is outward stress-carried angular-momentum flux.

With uniformly distributed apsidal orientations, the phase-mixed radial probability of a Kepler orbit gives the surface probability per cometFor $Q\gg r>q$, this is ${\pi }^{-2}{Q}^{-3/2}{r}^{-1/2}(1-q/r{)}^{-1/2}$. Multiply by the population size for a number surface density of a disk.

The surface density of a disk associated with a planar isotropic distribution is $S(\Phi )=2\pi {\int }_{\Phi }^{\infty }F(E)dE$. Differentiation gives $F(E)=-(2\pi {)}^{-1}dS/d\Phi {|}_{\Phi =E}$. A vanishing upper-limit boundary value recovers the original integral; monotonic decrease of $S(\Phi )$ ensures a nonnegative distribution.

A dilute dust component can be approximated as a pressureless fluid with velocity $\mathbf{v}$ and surface density of a disk ${\Sigma }_d$. In a gas velocity $\mathbf{u}$, linear drag force gives acceleration $-(\mathbf{v}-\mathbf{u})/t_s$ for positive aerodynamic stopping time $t_s$. Its equations include Coriolis acceleration, the shearing-sheet tidal potential, and ${\partial }_t{\Sigma }_d+\nabla \cdot ({\Sigma }_d\mathbf{v})=0$. Neglecting gas backreaction is justified only when the prescribed-gas approximation is appropriate. Strong orbit crossing or appreciable dust velocity dispersion can invalidate the single-valued pressureless-fluid description.

At an abruptly truncated razor-thin disk with nonzero edge surface density of a disk, the midplane radial force diverges logarithmically. Locally, mass occupies a half-plane and the inward force contains a nonzero angular factor multiplying $G{\Sigma }_{\mathrm{edge}}{\int }_{ϵ}^{\ell }ds/s$. The opposite exterior half-plane is absent, so its cancellation does not occur. Finite thickness or a taper taking the edge density to zero regularizes the force. Therefore a finite-edge galaxy rotation curve requires more than the ideal scale-free Mestel disk normalization.

For a positive inner magnetic torque $T_m$, define $\lambda =T_m/(\dot{M}{h}_{\mathrm{in}})$. Integrating steady conservation of angular momentum gives $\mathcal{G}=T_m+\dot{M}(h-h_{\mathrm{in}})$. With Keplerian rotation, this determines the surface density of a disk in the displayed formula.

A higher orbital line density at apoapsis does not always give a higher surface density of a disk: the local ring width also changes. In a narrow aligned eccentric ring, neglecting disc twist and keeping first-order orbital eccentricity, ${\Sigma }_{\mathrm{peri}}\propto 1+ade/da$ and ${\Sigma }_{\mathrm{apo}}\propto 1-ade/da$. Constant eccentricity therefore gives constant surface density to this order.