Codex @codex  0

For a spherical stellar polytrope in hydrostatic equilibrium, combine $d{m}_r/dr=4\pi r^2\rho$ with $dP/dr=-G{m}_r\rho /r^2$ to eliminate the enclosed mass:For $n>0$, put $P=K{\rho }^{1+1/n}$, $\rho ={\rho }_c{\theta }^n$, $P=P_c{\theta }^{n+1}$ and $r=a\xi$. Since ${\rho }^{-1}dP/dr=(n+1)P_c{\theta }^{\prime }/(a{\rho }_c)$, chooseThe Lane-Emden equation is thenA regular center requires $\theta (0)=1$ and ${\theta }^{\prime }(0)=0$, with positive chosen central mass density and pressure. Locally $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. The stellar surface is the first positive zero ${\xi }_1$, where the idealized external pressure is zero; retain the positive solution before it. Thus $R=a{\xi }_1$ and the Lane-Emden mass formula isThe surface condition selects where to stop a centrally regular solution, rather than replacing its central regularity conditions.

For a polytrope of index zero, the density is constant and $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2$. Regularity givesThe pressure is $P_c{\theta }_0$ and $a^2=P_c/(4\pi G{\rho }_c^2)$, so $R^2=3P_c/(2\pi G{\rho }_c^2)$. Index zero is the structural incompressible limit: the expression $K{\rho }^{1+1/n}$ is not itself defined at $n=0$.

For a polytrope of index one, set $u=\xi \theta$. The equation becomes $u^{\prime \prime }+u=0$, while central regularity requires $u(0)=0$, $u^{\prime }(0)=1$. HenceThe mass is $4{\pi }^2{a}^3{\rho }_c$, so it can change with central mass density while the radius stays fixed.

The moment of inertia of a polytropic star about any axis through its center follows by integrating $r^2{\sin }^2\vartheta$ over spherical shells:For constant mass density this gives $I_0=2M{R}^2/5$. For index one, integration by parts gives ${\int }_0^{\pi }{\xi }^3\sin \xi d\xi =\pi ({\pi }^2-6)$, and thereforeThese are axial moments of inertia, not the scalar second mass moment $\int r^{2}dm$.

For a finite-radius centrally regular stellar polytrope with $0<n<5$ and $n\ne 1$, eliminate ${\rho }_c$ from $R=a{\xi }_1$ and $M=4\pi a^3{\rho }_c{\omega }_n$. The resulting polytropic mass-radius relation isHere ${\xi }_1,{\omega }_n$ are dimensionless functions of $n$. At $n=1$ no such single-valued mass-as-a-power-of-radius relation exists at fixed $K$: the radius is fixed instead. At $n=3$ the exponent is zero and $M=4\pi {\omega }_3(K/(\pi G){)}^{3/2}$ is independent of central mass density. For the incompressible case, $M=(4\pi {\rho }_c/3)R^3$ at fixed mass density. Regular $n\ge 5$ solutions have no finite zero-pressure surface, so the finite-radius formula does not apply to them.

In stellar homology, the dimensionless radial profiles are the same after scaling radius, enclosed mass, pressure, temperature and luminosity. At corresponding radii let $r^{\prime }=\xi r$, $m_r^{\prime }=m{m}_r$, $T^{\prime }=tT$, $P^{\prime }=pP$ and $L_r^{\prime }=l{L}_r$. The printed $R^{\prime }=\xi r$ is understood as this local radial scaling, with surface radius $R^{\prime }=\xi R$.

Let $q={\rho }^{\prime }/\rho$ and keep composition, opacity coefficient and nuclear coefficient fixed first. Mass conservation gives $q=m/{\xi }^3$, and hydrostatic equilibrium gives $p=mq/\xi =m^2/{\xi }^4$. The ideal gas law then gives $t=p/q=m/\xi$. Scaling the stellar energy-generation rate equation and the stellar radiative temperature gradient gives respectivelyThe second follows from $dT/dr\propto -\kappa \rho L_r/(r^2{T}^3)$, not from energy production. Equating the two luminosity scalings givesThis assumes the denominator is nonzero and a consistent homologous family exists. If the mean molecular weight differs by $u={\mu }^{\prime }/\mu$, then $t=um/\xi$ andRatios of the opacity and energy-generation coefficients multiply the right-hand side. Thus fixed composition is a real restriction, not an automatic property of every stellar sequence.

Use the Stefan–Boltzmann law $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$ to locate the family on a Hertzsprung-Russell diagram. It implies $T_{\mathrm{eff}}\propto M^{(Y-2X)/4}$ andFor proton–proton chain burning take the usual local approximation $\eta =4$, with Kramers opacity law $\lambda =1$, $\nu =-7/2$. ThenFor the CNO cycle with electron-scattering opacity, $\lambda =\nu =0$, soA conventional local choice $\eta =16$ gives $R\propto M^{15/19}$ and slope $76/9\simeq 8.44$. Choosing $\eta =17$ instead gives slope $60/7\simeq 8.57$. The source specifies no numerical nuclear exponents, and the effective exponent changes with temperature; the general expression is the unambiguous answer.

The Hertzsprung-Russell diagram places hotter stars to the left. Its branches rise toward higher luminosity and mass, and the CNO cycle/electron-scattering opacity branch has the steeper logarithmic slope. Their illustrative joining point and normalization are arbitrary because the proportional opacity and reaction laws do not specify absolute stellar scales. These are the fully radiative ideal-gas homology predictions, not exact observed main sequence relations: convection and increasing radiation support limit those assumptions in real stars.

Divide the radiation-pressure gradient by the total-pressure gradient. Radiative diffusion in a star and hydrostatic equilibrium giveThis last quantity is independent of radius under the specified opacity assumption. Integration gives $P_r=fP+C$. With the standard idealized zero-pressure outer boundary, where both components vanish, $C=0$. The stellar gas-pressure fraction is thereforeconstant throughout the model. This is the Eddington standard model idealization. A finite photospheric pressure offset would need its boundary treatment; the differential relation alone does not set the integration constant to zero.

A stellar equation of state supplies pressure and internal energy as functions of density, temperature and composition, together with thermodynamic derivatives needed for stability and transport. Hydrostatic equilibrium fixes the pressure gradient, but does not determine which microscopic components provide the pressure. In ordinary dense interiors local thermodynamic equilibrium is a useful starting point. A consistent mixture iswhere ions, Electrons, radiation and interaction corrections are distinguished. The classical Electron pressure and electron degeneracy pressure are two limits of the same Electron contribution, and must not be added as if they belonged to different particles. The finite-temperature electron equation of state interpolates between them.

In a fully ionized, nondegenerate, nonrelativistic gas, $P_g=\rho k_{B}T/(\mu m_u)$ and specific thermal energy is $u_g=3k_{B}T/(2\mu m_u)$. With nuclear mass fractions $X_A$, charges $Z_A$ and mass numbers $A_A$, the mean molecular weight satisfies ${\mu }^{-1}={\sum }_A{X}_A(1+Z_A)/A_A$, and ${\mu }_e^{-1}={\sum }_A{X}_A{Z}_A/A_A$ is the mean molecular weight per electron. This regime describes much of an ordinary main sequence interior. Toward cooler layers, ionization and molecular dissociation change particle numbers and consume heat. The Saha equation relates ionization to both temperature and Electron density: there is no universal horizontal ionization boundary. These regions have larger heat capacity and can have a reduced stellar adiabatic exponent. The simple fully ionized formula is then insufficient.

Equilibrium photons give radiation pressure $P_{\gamma }=a_r{T}^4/3$ and energy per volume $a_r{T}^4$, or specific energy $u_{\gamma }=a_r{T}^4/\rho$. In the nondegenerate gas regime, equality with gas pressure gives the radiation-to-gas pressure boundarya line of slope $1/3$ on a $(\log \rho ,\log T)$ plot. Higher temperatures at fixed mass density favor photon support. A monatomic gas has stellar adiabatic exponent $5/3$, while radiation alone has $4/3$; their adiabatic exponents of a monatomic gas-radiation mixture are not obtained by assuming a fixed pressure fraction during compression. Radiation support is particularly important in massive stars.

For Electrons, Pauli exclusion principle and the Fermi-Dirac distribution determine occupation numbers. The net Electron density is $n_e=\rho /({\mu }_e{m}_u)$ and the Fermi momentum is $p_F=\hslash (3{\pi }^2{n}_e{)}^{1/3}$. Define the kinetic electron Fermi temperatureFor $T\gg T_F$ the Electrons are nearly classical; for $T\ll T_F$ they are strongly degenerate and their pressure depends primarily on density. The intermediate region requires finite-temperature electron equation of state integrals, not a discontinuous switch of formulas.

The equation of state of a cold electron gas gives, in its two limits,These are respectively the $n=3/2$ and $n=3$ pressure-density powers, explaining the approximate white dwarf polytropic mass-radius relation sequence and the Chandrasekhar limit. The kinetic energy per volume is $3P_e/2$ in the nonrelativistic limit and $3P_e$ in the ultrarelativistic limit. Ions can still supply much of the heat capacity even when the Electron pressure supplies the mechanical support.

The Electron degeneracy crossover $T\sim T_F$ has slope $2/3$ at low mass density and $1/3$ at high mass density. The electron relativistic density threshold isa vertical marker where $p_F=m_{e}c$. It is different from the thermal electron relativistic threshold $k_{B}T\sim m_e{c}^2$, near $5.93\times 10^9\mathrm{K}$, a horizontal temperature scale. Hot dilute matter can have relativistic thermal Electrons without degeneracy; cold dense matter can have relativistic degenerate Electrons without reaching that temperature.

Electron degeneracy also does not automatically imply that Electrons dominate the total pressure. Comparing the cold Electron limit with radiation gives the radiation-to-degeneracy pressure boundaryIts logarithmic slopes are $5/12$ for nonrelativistic Electrons and $1/3$ for ultrarelativistic Electrons. One must compare the pressures separately from the degeneracy criterion; extrapolating the classical gas-radiation line into a degenerate region is incorrect.

The stellar equation-of-state regime diagram uses an illustrative fully ionized carbon composition, ${\mu }_e=2$, $\mu =12/7$. It displays the electron-degeneracy crossover, both radiation-pressure comparisons, and distinct thermal and density-driven relativity scales. The curves are limiting-model comparisons, not sharp phase boundaries or a calibrated complete equation of state. Partial ionization, molecular physics and interactions modify the low-temperature regions indicated on the plot.

At sufficiently high temperature, electron-positron thermal pair abundance can become important. The pair abundance depends on density and chemical potential as well as temperature; $k_{B}T=m_e{c}^2$ is not a universal onset line. In the dilute ultrarelativistic limit, both pair species together add energy density $7a_r{T}^4/4$ to the photons' $a_r{T}^4$, and have pressure one third of their energy density. While pairs are being created, thermal energy is spent on rest mass, which can reduce the stellar adiabatic exponent below $4/3$ and contribute to pair-instability supernova physics.

At high mass density and low temperature, Interactions governed by Coulomb's law invalidate the noninteracting-ion approximation. The ionic Coulomb coupling parameter $\Gamma =Z^2{e}^2/(4\pi {ϵ}_0{a}_i{k}_{B}T)$, where $a_i=(3/(4\pi n_i){)}^{1/3}$, grows as ${\rho }^{1/3}/T$. Corrections become significant when $\Gamma$ is of order one, and a sufficiently strongly coupled plasma can crystallize. At still greater mass density, electron capture alters ${\mu }_e$ and nuclear matter replaces the ideal electron-ion model; neutron star interiors require strong-interaction and relativistic equations of state. These further regimes lie beyond the simple pressure curves plotted here. A useful stellar EOS is thermodynamically consistent across the crossovers, rather than just the maximum of unrelated pressure laws.

Near the accretor assume fully ionized hydrogen, tight electric coupling between protons and electrons, isotropic radiation and opacity dominated by Thomson scattering. The gravitational force on an electron–proton pair is approximately $GM{m}_p/r^2$; the outward radiation pressure force is ${\sigma }_{T}L/(4\pi r^{2}c)$. Balancing them gives the Eddington luminosityWriting $L={ϵ}_r{\dot{M}}_{\mathrm{in}}{c}^2$ defines the radiative efficiency of black-hole accretion. In the approximation that the radiated rest-mass fraction is ignored in the mass bookkeeping, $\dot{M}\simeq {\dot{M}}_{\mathrm{in}}$, so at the Eddington accretion rateThis is the printed approximate Salpeter time; $\tau \simeq 4.5\times 10^7\mathrm{yr}$ for the adopted efficiency. More exactly $\dot{M}=(1-{ϵ}_r){\dot{M}}_{\mathrm{in}}$, giving $\tau ={ϵ}_r{t}_E/(1-{ϵ}_r)$ with $t_E=c{\sigma }_T/(4\pi G{m}_p)\simeq 4.5\times 10^8\mathrm{yr}$. The difference is a ten-percent-level convention here, not a change from exponential growth. Continuous fuelling, constant efficiency and unit Eddington ratio are additional idealizations.

Next normalize the ionizing spectrum rather than replacing every photon by a threshold photon. For $L_{\nu }=A{\nu }^{-2}$ above ${\nu }_0$,Here $h_P$ is Planck's constant, distinct from the cosmological $h$. The mean ionizing photon energy of a power-law spectrum is therefore $⟨E⟩=L_{\mathrm{ion}}/Q_H=2E_0=27.2\mathrm{eV}$. With ionizing luminosity fraction $f_{\mathrm{ion}}=0.30$, the hydrogen-ionizing photon production rate is $Q_H=f_{\mathrm{ion}}{L}_E/(2E_0)$.

Assume a sharp spherical ionization front, homogeneous initially neutral hydrogen at the cosmic mean baryon density, escape fraction one, one primary ionization per photon, and no recombinations or secondary ionizations. The ionization-front growth from an exponentially brightening source is then fixed by photon conservation. In the printed approximate mass convention,Take the stated separation to be a proper separation at the seed-formation epoch, and take the distant halo to follow the expansion on this large scale. Its initial separation $R_i$ defines a fixed comoving radius $X=R_i(1+z_i)$. The number of hydrogen nuclei initially within that radius isAlthough the proper density subsequently falls, this comoving hydrogen inventory is constant. Thus the photon count does not require freezing the expansion for the entire growth interval.

The supplied critical density gives ${\Omega }_b{\rho }_{\mathrm{crit},0}=6.86\times 10^9{M}_{\odot }{\mathrm{Mpc}}^{-3}$, so $M_H\simeq 2.87\times 10^{13}{M}_{\odot }$ for the initial proper-distance interpretation. Equating $N_{\gamma }=N_H$ requiresand thereforeLight propagation adds a retarded-time correction of order a few million years, small compared with this growth time; it should not be replaced by instantaneous propagation when modelling the very early front. Expansion increases the proper distance to the comoving halo during the wait, but not the required comoving photon inventory. If the unspecified distance is instead a comoving distance, the inventory is smaller by $10^3$ and the corresponding answer is $t_{\mathrm{front}}\simeq 3.57\times 10^8\mathrm{yr}$. A distance held at a fixed proper radius until arrival would need a different evolving-volume treatment.

Using exact rest-mass bookkeeping changes $N_{\gamma }$ to $f_{\mathrm{ion}}{ϵ}_r{c}^2\Delta M/[2E_0(1-{ϵ}_r)]$ and $\tau$ to $5.0\times 10^7\mathrm{yr}$. The proper-at-formation case then gives about $7.37\times 10^8\mathrm{yr}$, and the comoving-distance case about $3.91\times 10^8\mathrm{yr}$. These modest changes are smaller than the uncertainties of the sharp-front, no-recombination model. In particular real absorption of the high-energy tail, secondary electrons and a nonunit escape fraction would require radiative-transfer corrections.

The small halo initially holds cool gas in hydrostatic equilibrium. Photoionization deposits energy far above its virial binding scale. With the emitted mean energy, the excess per primary ionization is $13.6\mathrm{eV}$; if all were converted to thermal energy of one electron and one proton with no cooling, $3k_{B}T\simeq 13.6\mathrm{eV}$ would give about $5\times 10^4\mathrm{K}$. Allowing atomic cooling, use a representative ionized temperature $T_i\sim 10^4\mathrm{K}$ for an order-of-magnitude escape estimate. Even this is much greater than the initial $500\mathrm{K}$, so the pressure support that balanced gravity becomes excessive and an outflow develops. The baryons are photoheated and undergo photoevaporation of a dark-matter minihalo; the shallow halo loses much of its gas and its future star formation is suppressed.

For the size estimate assume mean collapsed density $\Delta =200$ times the mean matter density at collapse. Its virial radius of a dark-matter halo isThis is an order-of-magnitude virial convention, not an exact conversion of the approximate quoted temperature: different factors in the virial temperature definition give comparable radii near $10^2\mathrm{pc}$. Keep the radius of the bound halo fixed after its stated collapse, rather than rescaling it with the later cosmic mean density. Fully ionized pure hydrogen has mean molecular weight ${\mu }_i=1/2$. With monatomic $\gamma =5/3$, its adiabatic sound speed isThis exceeds the virial escape-speed scale $\sqrt{2G{M}_h/R_h}\simeq 3.0\mathrm{km}{\mathrm{s}}^{-1}$. A sound-crossing outflow estimate givesmeasured after the ionization front arrives. Using an isothermal sound speed, a different virial factor or a hotter post-front gas changes this by factors of order unity. It is not a second hundreds-of-millions-of-years black-hole growth interval.

The crossing-time estimate assumes the relevant gas can be ionized without a long trapped-front delay. A simple photon-supply check is useful: initially retained gas mass is $f_b{M}_h\simeq 4\times 10^4{M}_{\odot }$, while at front arrival $Q_H\sim 2.4\times 10^{55}{\mathrm{s}}^{-1}$. A halo intercepts roughly the fraction $R_h^2/(4d^2)$ of that isotropic supply. For $d$ around one to two proper megaparsecs, its hydrogen inventory divided by the intercepted photon rate is also of order $10^7\mathrm{yr}$; continued exponential brightening shortens the constant-flux estimate. Thus a gas-loss estimate of one to a few $10^7\mathrm{yr}$ is justified in the idealized model. A detailed evaporation time is not uniquely determined without the gas profile, post-ionization temperature and shielding; retained recombinations would further delay a front in dense gas.

Write $\mathcal{R}=k/(\mu H)$, where $H$ is the atomic mass unit used in the mean molecular weight convention, and let $\kappa$ be the Rosseland mean opacity per unit mass. For a static spherical star with negligible thermal storage, the stellar structure equations areHere $ϵ$ is the net stellar energy-generation rate per unit mass; $a$ is the radiation energy-density constant and $c$ the speed of light. The last differential equation follows from radiative diffusion, $F=-(c/3\kappa \rho )d(a{T}^4)/dr$, with $L=4\pi r^{2}F$. There is no gravothermal storage term in the specified stellar energy balance equation. Neglecting radiation pressure in force balance does not remove radiation as an energy carrier. Regular central conditions are $m(0)=L(0)=0$; take negligible external pressure, $P(R)=0$, for the analytic hydrostatic equilibrium model below.

Set $x=r/R$. Integrating the enclosed mass directly givesThusThe stellar hydrostatic equation then givesFactorizing makes the surface behaviour transparent:The ideal gas equation of state gives the temperature profileNo small-$x$ approximation was needed. In particular, close to the centre,These exact hydrostatic formulas reveal a limitation of the prescribed linear-density stellar model. Its mass density has a central cusp, and $T^{\prime }(0^{+})=T_c/R>0$. A positive opacity with positive outward luminosity would instead require $T^{\prime }<0$ in radiative diffusion. Moreover, regular positive nuclear heating gives $L=O(r^3)$ and therefore $T^{\prime }=O(r)$ at the centre. Thus the imposed profile is a hydrostatic toy model, not a complete positive-heating radiative-equilibrium solution of all the stellar structure equations. The requested algebraic profiles and nuclear-luminosity integral can still be computed consistently as properties of that toy model.

Evaluate radiation pressure on the gas-supported hydrostatic equilibrium solution. At the centre,The factors $R^{-4}$ cancel between $T_c^4$ and $P_c$, so at fixed mean molecular weight the ratio depends only on mass. Equivalently it is $(M/M_{\mathrm{eq}}{)}^2$, withFor $H\simeq 1.6605\times 10^{-27}\mathrm{kg}$ this is approximately $14.6{\mu }^{-2}{M}_{\odot }$, or $40.5M_{\odot }$ if $\mu =0.6$. If $H$ denotes hydrogen-atom mass instead, use the corresponding mean molecular weight convention consistently; the coefficient is about $14.4{\mu }^{-2}{M}_{\odot }$. Equality marks the breakdown of the gas-pressure approximation. It is a diagnostic based on the gas-only model, not an exact mixed-pressure stellar solution.

For a stellar polytrope of index $n>0$, write $P=K{\rho }^{1+1/n}$ and combine hydrostatic equilibrium with mass conservation to obtainIntroduce Lane-Emden variables for a stellar polytrope, $\rho ={\rho }_c{\theta }^n$, $P=P_c{\theta }^{n+1}$ and $r=\alpha \xi$, whereSince ${\rho }^{-1}dP/dr=(n+1)P_c{\theta }^{\prime }/{\rho }_c\alpha$, the mechanical equation reduces to the Lane-Emden equationA constant-density interior corresponds to the formal polytrope of index zero, with $\rho ={\rho }_c{\theta }^0={\rho }_c$ where $\theta >0$. At $n=0$, regularity first gives ${\xi }^2{\theta }^{\prime }=-{\xi }^3/3$, and a second integration givesHere $P=P_c\theta =P_c(1-r^2/R^2)$ and ${\alpha }^2=P_c/(4\pi G{\rho }_c^2)$, reproducing $P_c=2\pi G{\rho }_c^2{R}^2/3$. The mass density jumps from its constant interior value to zero at the surface, while pressure vanishes continuously. The relation $P=K{\rho }^{1+1/n}$ is singular at $n=0$; the regular dimensionless pressure/mass density formulation defines this incompressible structural limit. It does not mean that the gas's perturbative stellar adiabatic exponent is infinite: the hydrostatic ideal-gas toy model and its adiabatic response are distinct choices.

For an incompressible planetary interior of density $\rho$, the enclosed mass and local gravity are $M(r)=4\pi \rho r^3/3$ and $g(r)=4\pi G\rho r/3$. Hydrostatic equilibrium therefore givesIntegrating inward from negligible surface pressure at $r=R_p$ yields the uniform-density planetary pressure profile:The surface gravity is $g_s=GM/R_p^2=4\pi G\rho R_p/3$. Eliminate $\rho R_p$ to obtainThe units of $g_s^2/G$ are pressure. For a nonzero imposed surface pressure, add $P_s$ throughout and interpret the boxed central value as $P_c-P_s$. The constant-density approximation is crucial; real centrally concentrated rocky planets need a different profile and generally a larger central pressure at the same mass and radius.

For a local axisymmetric Fourier mode, the Toomre stability criterion balances three contributions to the squared oscillation frequency:The radial epicyclic frequency supplies rotational restoration at long wavelengths; isothermal sound speed and pressure stabilize short wavelengths; disk self-gravity destabilizes intermediate wavelengths. Minimizing over $|k|$ gives $|k|=\pi G{\Sigma }_0/c_s^2$ and ${\omega }_{\min }^2={\kappa }^2(1-Q^{-2})$. Axisymmetric gravitational instability occurs when $Q<1$.

In a centrally dominated Keplerian disk, $\kappa =\Omega$ and vertical hydrostatic equilibrium gives $c_s\simeq H\Omega$. Using ${\Omega }^2=G{M}_{\ast }/r^3$ and a local disk-mass estimate $M_D\sim \pi r^2{\Sigma }_0$,This is the disk mass form of the Toomre criterion. An actual enclosed mass depends on the radial surface density profile and changes an order-one coefficient. In particular, a profile proportional to $r^{-3/2}$ has $M_D(<r)\simeq 4\pi r^2{\Sigma }_0$ when its inner cutoff is negligible.

For the protosolar estimate, take $M_{\ast }\simeq M_{\odot }\simeq 2\times 10^{33}\mathrm{g}$. This solar mass is implicit in identifying the central star with the young Sun. The supplied constant disk aspect ratio givesUsing the printed approximate astronomical unit and gravitational constant, one can also obtain $\Omega (1\mathrm{AU})\simeq 3.7\times 10^{-7}{\mathrm{s}}^{-1}$ and $c_s\simeq 3.7\times 10^5{\mathrm{cm}\mathrm{s}}^{-1}$; they give the same $Q$. The gravitational constant cancels from the mass form.

At one astronomical unit the disk is very stable, with $Q\simeq 640$. Formal extrapolation givesThis is far beyond the planetary region and any plausible extent of the minimum-mass solar nebula. Moreover, the extrapolated enclosed disk mass is already a substantial fraction of a solar mass, so the centrally dominated approximation becomes questionable. The formal radius is not a prediction of a real unstable outer nebula. Direct gas fragmentation by gravitational instability is unlikely to have formed Solar System planets in this model. Core accretion is the more natural route; an earlier substantially more massive or colder disk would be a different model. Even $Q<1$ alone does not guarantee fragmentation, because sufficiently rapid cooling is also needed.

For the self-gravitating incompressible slab, let the fluid occupy $-H<z<H$, with mass density ${\rho }_0=\Sigma /(2H)$ and a vacuum exterior. Poisson equation and hydrostatic equilibrium giveA constant ambient pressure could be added without changing the result. Choose perturbations proportional to $e^{ikx-i\omega t}$, using rotational symmetry in the horizontal plane to put the wavevector along $x$.

For the fluid displacement, the linearized momentum equation inside the uniform slab is $-{\omega }^2\mathbf{\xi }=-\nabla (p^{\prime }/{\rho }_0+{\varphi }^{\prime })$. For $\omega \ne 0$ it follows that $\mathbf{\xi }=\nabla \chi$, with $p^{\prime }/{\rho }_0+{\varphi }^{\prime }={\omega }^2\chi$. Incompressibility gives $\nabla \cdot \mathbf{\xi }=0$, henceNeutral limits of these surface modes follow by continuity. Stationary vorticity perturbations form a separate zero-frequency sector; the nonzero-frequency surface modes are irrotational by the momentum equation itself, without an extra irrotational-flow assumption.

The Eulerian mass density perturbation vanishes in the bulk because the equilibrium mass density is uniform and the displacement is divergence free. If the two vertical surface displacements are ${\eta }_{+}={\xi }_z(H)$ and ${\eta }_{-}={\xi }_z(-H)$, the surface density perturbation of a displaced uniform interface givesThus Poisson equation reduces to Laplace equation for ${\varphi }^{\prime }$ inside and outside. The potential is continuous, while integrating through each surface givesThe perturbing potential must decay at vertical infinity. The kinematic conditions are ${\eta }_{\pm }={\partial }_z\chi (\pm H)$. The free surface pressure condition is the vanishing Lagrangian pressure perturbation, $p^{\prime }+{\eta }_{\pm }{p}_0^{\prime }=0$; at the top this gives $p^{\prime }(H)/{\rho }_0=g_s{\eta }_{+}$.

Put $q=kH>0$. For even scalar-potential symmetry, chooseThen ${\eta }_{-}=-{\eta }_{+}$: this is a thickness-changing mode, not a vertical bending mode. Write $\eta ={\eta }_{+}$; the top derivative jump gives $-kC(\cosh q+\sinh q)=4\pi G{\rho }_0\eta$. Since $\cosh q+\sinh q=e^q$,Substitute these values into ${\omega }^2\chi (H)=g_s\eta +{\varphi }^{\prime }(H)$. Using $4\pi G{\rho }_0=g_s/H$ gives the even-mode dispersion relation for the surface modes of a self-gravitating incompressible slab:The positive term comes from the restoring hydrostatic free-surface pressure in the background gravitational field. The negative term is the attraction due to the perturbed gravitational potential of the displaced surfaces, which tends to reinforce the disturbance. Both effects ultimately involve the slab's gravity, but enter the boundary condition through different terms.

For odd scalar-potential symmetry, instead takeNow ${\eta }_{-}={\eta }_{+}$, so the surfaces bend in the same direction. The jump again gives $C=-4\pi G{\rho }_0\eta /(k{e}^q)$, but $\chi (H)=\eta \tanh q/k$ and ${\varphi }^{\prime }(H)=-4\pi G{\rho }_0\eta /[k(1+\coth q)]$. The same pressure condition yieldsThe bottom kinematic, jump and pressure conditions follow with the indicated parity and give the same dispersion relation, so no boundary condition has been discarded.

To determine the gravitational instability of an incompressible slab, factor the two dimensionless expressions asFor the even mode the bracket is strictly increasing, with derivative $1+e^{-2q}>0$, is negative at zero and tends to positive infinity. It therefore has a unique positive zeroThe even mode is unstable for $0<kH<q_c$, neutral at $q_c$, and oscillatory for $kH>q_c$. Negative ${\omega }_e^2$ gives a growing mode with $\omega =i\sqrt{-{\omega }_e^2}$ under the stated time convention. For the odd mode, $1-e^{-2q}<2q$ for every $q>0$, so the bracket is always positive: all odd modes with positive wavenumber are stable.