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Here $\beta =p_{\mathrm{rad}}/p_{\mathrm{gas}}$ is the radiation-to-gas pressure ratio, not plasma beta. Write the opacity as ${\kappa }_{\mathrm{op}}$ to distinguish it from the radial epicyclic frequency. The radiation pressure is $p_{\mathrm{rad}}=4\sigma T^4/(3c)$, so radiative diffusion can be written asSince $\beta$ is independent of height, $p_{\mathrm{rad}}=\beta p/(1+\beta )$. Substitution of hydrostatic equilibrium, $p^{\prime }=-\rho {\Omega }^{2}z$, givesDifferentiating and using the viscous heating equation, together with $r{\Omega }^{\prime }=-q\Omega$, yieldsThus the effective dynamic viscosity $\rho \nu$ is independent of height. If the surfaces are at $z=\pm H$, the vertically integrated kinematic viscosity satisfiesUse the previous result and the specified Eddington accretion rate convention,to obtain the full thicknessThe constant-pressure-ratio vertical disk model therefore has thickness proportional to the inward accretion rate. This is a full surface-to-surface thickness, rather than a density disk scale height. Its use as a thin disk requires $2H\ll r$.

Let $V=U-c$ and take $k>0$ without loss of generality. The Boussinesq approximation replaces density by ${\rho }_0$ in inertia and pressure acceleration, while retaining the density perturbation in buoyancy. Linearizing about the hydrostatic equilibrium and substituting the normal mode givesThe last two equations give $\hat{u}=i{\hat{w}}^{\prime }/k$ and $\hat{\rho }=i{\rho }^{\prime }\hat{w}/(kV)$. The horizontal momentum equation givesSubstituting in the vertical momentum equation yields the Taylor–Goldstein equation:The buoyancy frequency is real for stable density stratification. The impermeable walls impose $\hat{w}(\pm L)=0$.

The usual Pressure-opacity zones of a Shakura--Sunyaev thin disk, in order of increasing cylindrical radius, are an inner region dominated by radiation pressure and electron-scattering opacity, a middle region dominated by gas pressure and electron-scattering opacity, and an outer region dominated by gas pressure and the Kramers opacity law, often approximated by free-free opacity. The boundaries depend on mass, mass accretion rate, and the alpha disk viscosity parameter. At sufficiently large radius, accretion-disk self-gravity or changes in ionization can invalidate this three-zone model.

For the inner region writeHere $F$ is the standard thin-disk dissipation flux through one face, and $R_{\ast }$ is the innermost stable circular orbit treated as a zero-torque inner boundary. The given factor $1/2$ in the dissipation rate is appropriate to one face. For Keplerian rotation, it is $(9/8)\nu \Sigma {\Omega }_K^2$.

Let $H$ denote the disk scale height or vertical half-thickness, not the full thickness. In vertical hydrostatic equilibrium, the leading vertical acceleration is ${\Omega }_K^{2}z$, so a one-zone estimate givesIn the inner zone, $P_c\simeq P_{\mathrm{rad},\mathrm{c}}=a_{\mathrm{rad}}{T}_c^4/3$. For an optically thick medium with constant electron-scattering opacity, radiative diffusion gives the emergent one-face fluxEquivalently, combine $F=-c({\kappa }_{\mathrm{es}}\rho {)}^{-1}d{P}_{\mathrm{rad}}/dz$ with vertical pressure balance. Equating the radiative flux to the dissipated flux yields the radiation-supported inner-disk height:Order-unity vertical-profile factors depend on the precise height convention. The central mass cancels at fixed physical mass accretion rate, because both vertical gravity and local dissipation are proportional to $G{M}_{\mathrm{BH}}/R^3$.

If ${\dot{m}}_{\mathrm{Edd}}=L_{\mathrm{Edd},\mathrm{es}}/({\eta }_0{c}^2)$ and $r_g=G{M}_{\mathrm{BH}}/c^2$, this can also be writtenThe radiation-supported height rises from the formal zero at $R_{\ast }$ toward the approximately constant value $H_0=3{\kappa }_{\mathrm{es}}\dot{m}/(8\pi c)$. That zero is not a reliable description of the physical plunging region: the radiation-dominated approximation and the Newtonian zero-torque formula fail very close to the inner edge.

Beyond the inner zone, the height grows slowly with radius rather than staying on that plateau forever. Far from the inner boundary, the standard gas-pressure scalings give $H\propto R^{21/20}$ in the electron-scattering zone and $H\propto R^{9/8}$ in the Kramers zone. These follow from $H\propto T_c^{1/2}{R}^{3/2}$ with $T_c\propto R^{-9/10}$ and $R^{-3/4}$, respectively. A schematic profile is

vertical half-thickness H
  ^
  |                                               / outer gas/Kramers
  |                                          ____/
  |                                     ____/ middle gas/scattering
  |              ______________________/
  |          ___/ inner radiation/scattering plateau
  |       __/
  |     _/
  +----|----------------------------------------------> radius R
      R_*              zone boundaries are schematic

The height formula also tests the thin disk assumption. For a nonrotating hole with $R_{\ast }=6r_g$, the largest $H/R$ in this inner approximation occurs at $R=(9/4)R_{\ast }$. At $\dot{m}={\dot{m}}_{\mathrm{Edd}}$ and ${\eta }_0=0.1$, it is about $0.37$, so a disc this close to the conventional Eddington accretion rate is only marginally thin. Slim accretion disk effects and advected heat can then matter.

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.

In a static spherical star, hydrostatic equilibrium and the Poisson equation giveor equivalently $r^{-2}d(r^{2}g)/dr=4\pi G\rho$.

Let $\mathbf{\xi }$ be the fluid displacement field, so the velocity perturbation is $\delta \mathbf{u}={\partial }_t\mathbf{\xi }$. Linearizing the ideal-fluid momentum equation about the static state and cancelling the background hydrostatic terms givesConservation of mass says that the Lagrangian density perturbation is ${\Delta }_L\rho =-\rho \nabla \cdot \mathbf{\xi }$. The relation between Eulerian and Lagrangian fluid perturbations and adiabatic compression gives, with $\Delta =\nabla \cdot \mathbf{\xi }$,while linearized self-gravity gives $\boxed{{\nabla }^2\delta \Phi =4\pi G\delta \rho }$.

For the stated spherical harmonic displacement, the radial divergence is $r^{-2}d(r^2{\tilde{\xi }}_r)/dr$, while $\nabla {\tilde{\xi }}_h$ is radial and orthogonal to the angular gradient of $Y_l^m$. HenceEquating the radial and horizontal coefficients of $Y_l^m$ and $\nabla Y_l^m$, and applying the separated Laplacian in spherical coordinates to $\widetilde{\delta \Phi }(r)Y_l^m$, gives

Define the stellar buoyancy frequency byEliminating $\tilde{\Delta }$ between the density and pressure perturbations givesSubstitution in the radial equation, followed by use of $dp/dr=-\rho g$, yieldsRegular spherical profiles near the center have $\rho ={\rho }_c+O(r^2)$ and $p=p_c+O(r^2)$, while $m(r)=4\pi {\rho }_c{r}^3/3+O(r^5)$ and hence $g=O(r)$. Both logarithmic gradients in the definition of $N^2$ are $O(r)$, so $N^2=A{r}^2+O(r^4)$. A Sun-like radiative central stratification is stable, making $A>0$.

In a thin atmosphere in hydrostatic equilibrium, take gravity $g$ and mean molecular mass $\mu m_H$ as constant. The ideal gas equation of state and hydrostatic balance giveFor $P>1\mathrm{mbar}$,so the atmospheric lapse rate follows from the chain rule:It vanishes in the assumed isothermal region above $1\mathrm{mbar}$. If natural logarithms are used instead, the factor $\ln 10$ is absent and the fitted numerical value of $A$ changes accordingly.

Assume Newtonian spherical hydrostatic equilibrium, neglect rotation and thermal density changes, and require continuous pressure at the layer boundary. Writeso the mantle mass profile is $M(r)=C+D{r}^3$. Integrating $dP/dr=-G{\rho }_{2}M(r)/r^2$ inward from $P(R_p)=P_0$ gives the boundary pressure

For a homologous stellar sequence, mass conservation, hydrostatic equilibrium, and the ideal gas equation giveIntegrating the given stellar energy-generation rate over a fixed dimensionless profile givesTherefore

Radiative diffusion in a star can be writtenDividing by hydrostatic equilibrium giveswhich is constant by assumption. Since both $P_{\mathrm{rad}}$ and $P$ vanish at the surface, integration gives $P_{\mathrm{rad}}=(1-\beta )P$ with constant $\beta$.

Eliminating $T$ between the gas and radiation equations of state now givesThus the star is an $n=3$ stellar polytrope. PutThe structure equations become the Lane-Emden equationThe surface is the first zero ${\xi }_1\simeq 6.89685$. Writingthe Lane-Emden mass formula givesThereforewhereThe additional radiative-equilibrium relation is $1-\beta =\nu L/(4\pi cGM)$.

Write the mean interior density asSincethe assumed outward decrease of $\bar{\rho }$ implies $\rho \le \bar{\rho }$. In mass coordinates, hydrostatic equilibrium isFor every interior mass $m\le m_r$, monotonicity givesIntegrating from the centre and using ${\int }_0^{m_r}{m}^{-1/3}dm=3m_r^{2/3}/2$ proves

Let $\mathcal{R}$ be the gas constant per mole. At the centre,Eliminating $T_c$ gives the Eddington quartic relation in its central form,

At the surface, set $P=0$ and $m_r=M$ in the upper pressure bound. Cubing it givesThe function $(1-\beta )/{\beta }^4$ decreases strictly as $\beta$ increases on $0<\beta <1$. Define ${\beta }^{\ast }$ by equality in the resulting bound:Then ${\beta }_c\ge {\beta }^{\ast }$, orand rearrangement gives exactly

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.

Let $y=\beta /(1-\beta )=P_g/P_{\mathrm{rad}}$. The two pressure laws giveand henceThe radiative equation can be writtenDividing it by hydrostatic equilibrium and using $L_r/M_r=L/M$ givesSince $P=P_{\mathrm{rad}}(1+y)$ and $P_{\mathrm{rad}}\propto T^4$,Withthe reciprocal of the preceding moment equation becomesMultiplication by $y$ proves

The first equation is vertical hydrostatic equilibrium: the pressure gradient balances the vertical gravity of the central mass, ${\Omega }^{2}z$, and the disk's own potential ${\Phi }_d$. The second is the plane-parallel Poisson equation for disk self-gravity. The third balances the vertical increase of radiative flux against local viscous heating in a Keplerian alpha disk, and the fourth is the optically thick radiative diffusion law.

With $P/\rho \sim c_s^2$, $\Sigma \sim \rho H$, and all terms in hydrostatic balance comparable,The first comparison gives $H\sim c_s/\Omega$; inserting it into the second gives $\Omega c_s\sim G\Sigma$. Henceup to the order-one constants deliberately omitted by the scaling argument.

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

Assume an ideal-gas atmosphere of constant mean molecular mass and approximately constant gravity. If temperature decreases linearly with altitude, write $dT/dz=-\Gamma$. Combining the ideal-gas law with hydrostatic equilibrium givesThe endpoint conditions determine the power law directly:

LetThe enclosed mass isThis is continuous at the core boundary. Hydrostatic equilibrium gives $dP/dr=-Gm(r)\rho (r)/r^2$, with $P(R_p)=0$. DefineSince $F^{\prime }(r)=Gm(r)(A-br)/r^2$, the mantle pressure isInside the uniform core, the shell theorem gives $m(r)=4\pi {\rho }_c{r}^3/3$, so

Assume a thin isothermal atmosphere, so $g\simeq G{M}_p/R_p^2$ and the surface area are constant through it. Integrating hydrostatic equilibrium from base pressure $P_0$ to negligible top pressure gives atmospheric column mass $P_0/g$. ThereforeEquivalently, $M_{\mathrm{atm}}\sim 4\pi R_p^2{\rho }_{0}H$, with ${\rho }_0=P_0\mu m_H/(k_B{T}_0)$ and $H=k_B{T}_0/(\mu m_{H}g)$; the explicit temperature dependence cancels.

The four stellar structure equations aretogether with the perfect-gas equation of stateHere $a_{\mathrm{rad}}$ is the radiation constant, $ϵ$ is the specific stellar energy-generation rate, and $\kappa$ is the opacity.

Put $x=r/R$. Integrating the prescribed density gives the enclosed massSince $m_R=M$,Integrating hydrostatic equilibrium inward from $P(R)=0$ givesThe ideal-gas law then givesIn particular,

Suppose first that the gas-pressure fraction $\beta =P_g/P$ is spatially constant, with $0<\beta <1$. This is a sufficient condition for a stellar polytrope. Sincethe ratio $P_{\mathrm{rad}}/P_g=(1-\beta )/\beta$ givesThereforewithFor an ordinary positive polytropic index one also assumes $0\le s<3$.

The radiative diffusion in a star equation can be written asBecause $P_{\mathrm{rad}}=(1-\beta )P$ and $\beta$ is constant, hydrostatic equilibrium givesEquating them and using $\eta =(L_r/L)/(m_r/M)$ yields

Vertical hydrostatic equilibrium givesDefine vertically integrated pressure and the density-weighted disk scale height byMultiply the hydrostatic equation by $z$ and integrate. Since $zp\to 0$ at both boundaries, integration by parts givesTherefore