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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

Introduce the dimensionless variablesUsing $P=\Sigma H^2{\Omega }_z^2$, vertical hydrostatic equilibrium becomes the parameter-free equationwith normalizations $\int \tilde{\rho }d\tilde{z}=1$ and $\int \tilde{p}d\tilde{z}=1$. For an isothermal atmosphere, $p=c_s^2\rho$, and these conventions give the Gaussian distribution

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.

For a family in stellar homology, dimensionless radial profiles are fixed. Mass conservation and hydrostatic equilibrium then give the central scalingsThe ideal gas equation of state, with fixed mean molecular weight, consequently giveswhere fixed dimensional constants such as $G$ are suppressed in homology relations.

The proton–proton chain law $ϵ\propto \rho T^{7/2}$ gives the nuclear luminosityOn the other hand, radiative diffusion in a star givesUsing the Kramers' opacity law ${\kappa }_c\propto {\rho }_c{T}_c^{-7/2}$,Thermal equilibrium requires $L_{\mathrm{nuc}}=L_{\mathrm{rad}}$. Their common mass factor cancels, leaving $R^{-13/2}\propto R^{-1/2}$, and hence

This model captures the gas-pressure support, pp-chain burning, and strongly mass-dependent luminosity of the lower main sequence, including the Sun approximately. Its fully radiative assumption is an idealization: the Sun has a convective envelope, and sufficiently low-mass red dwarfs become largely or fully convective, so real radii are not exactly constant.

Plane-parallel hydrostatic equilibrium gives $dP/dz=-\rho g$, while $d\tau =-\kappa \rho dz$. HenceFor $\kappa ={\kappa }_0{P}^{\alpha -1}{T}^{4-4\beta }$,The grey-atmosphere relation from part (ii) can be writtenIntegrating from the zero-pressure top, where $T=T_0$, gives

The natural matching point to the stellar interior is the photosphere $\tau =2/3$, where $T=T_e$ and $T_e^4=2T_0^4$. Multiplying the preceding expression by ${\kappa }_0{T}^{4-4\beta }/g$ then yields the stellar surface boundary conditionThe local relation $L_r=4\pi r^2\sigma T^4$ identifies this matching temperature with the local effective temperature.

Because $H\ll R$ and $M_{\mathrm{env}}\ll M$, the envelope has nearly constant area $4\pi R^2$ and nearly constant surface gravity of a star $g=GM/R^2$. Plane-parallel hydrostatic equilibrium isSince $d\Sigma /dz=-\rho$, one has $dP/d\Sigma =g$. Taking the pressure to vanish with $\Sigma$ at the surface gives

The spherical stellar energy equation is $d{L}_r/dr=4\pi r^2\rho ϵ$. With $r\simeq R$ and $F=L_r/(4\pi R^2)$,Division by $d\Sigma /dr=-\rho$ givesLikewise, radiative diffusion in a star givesand therefore

At the midplane, radial hydrostatic equilibrium givesIf $d{P}_0/dr<0$, then $\Omega <{\Omega }_K$: the pressure gradient provides part of the inward force and the gas is Sub-Keplerian. For radial pressure scale comparable to $r$,Using $H\sim c_s/{\Omega }_K$ and ${\Omega }_K-\Omega \simeq ({\Omega }_K^2-{\Omega }^2)/(2{\Omega }_K)$,