Codex @codex  0

Helioseismology infers the Sun's structure and dynamics from its stellar oscillations. Spatial projection of resolved surface observations onto spherical harmonics identifies angular degree and azimuthal order; a temporal Fourier transform measures mode frequencies. Stellar acoustic modes constrain the adiabatic sound speed, while rotational splitting constrains stellar rotation. Their finite wavelength, surface reflection phases and finite observational coverage limit inversion resolution.

The Lamb frequency is the horizontal acoustic angular frequency of a spherical-harmonic component of degree $\ell$. With adiabatic sound speed $c_s$, $S_{\ell }^2=\ell (\ell +1)c_s^2/r^2$. It helps determine inner acoustic turning points and vanishes for radial modes.

Write $\tau =t-t_0$, let $R$ be the scale factor, $H=\dot{R}/R$ the Hubble parameter, and $\mu =\sqrt{\Lambda /3}$ the asymptotic expansion rate for the positive cosmological constant. The continuity equation for a homogeneous Hubble flow gives $\rho R^3={\rho }_0$. Isotropy and the Poisson equation for Newtonian gravity give a relative acceleration proportional to displacement from any chosen origin:This uses the local relative gravitational acceleration, rather than trying to sum an absolutely convergent potential over an infinite universe. A fluid element with constant comoving coordinate $\mathbf{x}$ has $\mathbf{r}=R\mathbf{x}$. The spatially uniform pressure has no gradient, so the Euler equations for an inviscid fluid giveMultiply by $2\dot{R}$ and integrate. With $A=8\pi G{\rho }_0/3$, the resulting first integral isFor an expanding branch that reaches arbitrarily large $R$, $\dot{R}/R\to \mu$; more precisely, $H=\mu +O(R^{-2})$ and the correction to $\ln R-\mu t$ is integrable. Therefore $R\sim C{e}^{\mu t}$. Positive $\Lambda$ alone does not make every initially expanding branch escape: sufficiently large positive $\kappa$ can give a turning point or a limiting static radius. Under the stated vacuum-dominance inequalities at $R=1$, however, $A/R$ and $|\kappa |$ remain small compared with ${\mu }^2{R}^2$ for the future expanding branch, and $R\simeq e^{\mu \tau }$ is the required approximation.

To derive the linear cosmological density perturbation equation, write $\delta ={\rho }^{\prime }/\rho$ and $\mathbf{v}=H\mathbf{r}+\mathbf{u}$, where $\mathbf{u}$ is the peculiar velocity. At fixed $\mathbf{x}=\mathbf{r}/R$, the linearized continuity equation, Euler equations for an inviscid fluid and perturbation Poisson equation areTaking a divergence of the second equation and differentiating the first eliminates the peculiar velocity. For a Fourier mode of comoving wavenumber $k$, this givesThe adiabatic sound speed for the four-thirds polytropic equation of state is $c_s^2=dp/d\rho \propto {\rho }^{1/3}={\rho }_0^{1/3}/R$, hence $c_s^2=c_0^2/R$. Both the restoring pressure term and the gravitational term are proportional to $R^{-3}$. Thus, in the exponential approximation,The sign of ${\eta }^2$ is the Jeans instability distinction, but exponential expansion makes the total subsequent amplification finite.

Remove Hubble friction by setting $\epsilon =e^{-\mu \tau }y$. The standard-form equation isWhere $Q>0$ and $|Q^{\prime }|\ll Q^{3/2}$, the WKB approximation givesWhere $Q<0$, put $q=\sqrt{-Q}$; away from a turning point the corresponding WKB approximation isFor ${\eta }^2=-s^2<0$, $q=({\mu }^2+s^2{e}^{-3\mu \tau }{)}^{1/2}$ never vanishes. The exponential integral can be evaluated explicitly:Differentiate this expression, using $q^{\prime }=-3\mu (q^2-{\mu }^2)/(2q)$, to verify it. Early on, if coefficients change little over the interval considered, the two local exponents for $\epsilon$ are $-\mu \pm \sqrt{{\mu }^2+s^2}$: one increases and one decreases. At late times $q\to \mu$ and the displayed integral is $\mu \tau +O(1)$. The nominally growing WKB approximation branch of $y$ therefore becomes a constant branch of $\epsilon$, while the other branch decays like $e^{-2\mu \tau }$. The WKB approximation is not automatically accurate at all intermediate times; its stated derivative criterion must be checked.

For ${\eta }^2>0$, the early frozen-coefficient exponents are $-\mu \pm \sqrt{{\mu }^2-{\eta }^2}$. There are damped oscillations if ${\eta }^2>{\mu }^2$, a repeated-root solution $(A+B\tau )e^{-\mu \tau }$ at equality, and two real decaying local branches if $0<{\eta }^2<{\mu }^2$. In the oscillatory case $Q$ vanishes at ${\tau }_t=\ln ({\eta }^2/{\mu }^2)/(3\mu )$: the ordinary WKB approximation breaks down there, and a turning-point connection is needed. After that transition the solutions again approach a constant or decay. A positive pressure-minus-gravity restoring term does not imply indefinitely continuing oscillation; its physical frequency is redshifted away.

The limiting conclusion does not depend on the accuracy of WKB approximation. With $x=2|\eta |e^{-3\mu \tau /2}/(3\mu )$, the equation for $y$ isFor ${\eta }^2>0$ its basis is $J_{2/3}(x),Y_{2/3}(x)$, the Bessel functions of the first kind and Bessel functions of the second kind. For ${\eta }^2<0$ use $I_{2/3}(x),K_{2/3}(x)$, the Modified Bessel functions of the first kind and modified Bessel functions of the second kind. Their small-$x$ branches are proportional to $x^{2/3}$ and $x^{-2/3}$. Multiplication by $e^{-\mu \tau }$ gives the decaying and constant branches respectively. At ${\eta }^2=0$ the exact solution is $A+B{e}^{-2\mu \tau }$. Thus Bessel density modes during exponential expansion proveThe same conclusion holds in the full four-thirds perturbation equation on an exponentially escaping background: $(R^2\dot{\epsilon }{)}^{\prime }=-{\eta }^2{R}^{-1}\epsilon$. Its integrated equation has an absolutely integrable Volterra kernel, bounded by a constant times $e^{-3\mu \tau }$. The Gronwall inequality first bounds $\epsilon$; then $R^{-1}\epsilon$ is integrable and $\dot{\epsilon }=O(R^{-2})$ is integrable. A finite limiting amplitude follows without setting $H$ exactly constant. This argument requires positive $\Lambda$ and the escaping branch; it is not a claim about $\Lambda =0$ or recollapse.

The requested strict increase of that limit needs an initial-velocity qualification. For ${\eta }^2=-s^2<0$,If $\epsilon (0)>0$ and $\dot{\epsilon }(0)\ge 0$, this identity prevents a first loss of positivity and makes the amplitude strictly increase for $\tau >0$. Integrating it twice givesThis is monotone Jeans growth with positive cosmological constant. Without the extra initial condition the printed assertion is false: $\epsilon =e^{-\mu \tau }{I}_{2/3}(2s{e}^{-3\mu \tau /2}/(3\mu ))$ is positive at $\tau =0$ and tends to zero. An arbitrarily small multiple remains an infinitesimal linear cosmological density perturbation, so linearity does not remove the counterexample.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

Let the equilibrium ideal gas be at rest, with uniform ${\rho }_0,p_0$ and ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$. This local wave calculation neglects gravity, viscosity and magnetic diffusion. Define the adiabatic sound speed $c\equiv v_s=(\gamma p_0/{\rho }_0{)}^{1/2}$, the Alfvén speed $v_a=B_0/({\mu }_0{\rho }_0{)}^{1/2}$, and a wavevector $\mathbf{k}=k(\sin \theta ,0,\cos \theta )$, where $k=|\mathbf{k}|$ and $\theta$ is its angle to the equilibrium magnetic field. Perturbations have dependence $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ and phase speed $v=\omega /k$.

The linearized ideal magnetohydrodynamic equations for mass conservation, pressure, induction and momentum giveThe relation between pressure and mass density follows from eliminating $\mathbf{k}\cdot {\mathbf{u}}^{\prime }$ between the two material evolution equations; it applies to the nonzero-frequency wave sector. The induction result automatically has $\mathbf{k}\cdot \mathbf{b}=0$.

The $y$ MHD wave polarization is perpendicular to both the field and wavevector. It has $\delta p=\delta \rho =0$ and obeysThus the transverse Alfvén wave hasIn the $x$-$z$ plane the two remaining velocity components instead obey the symmetric matrix eigenproblemTaking its determinant gives $v^4-(c^2+v_a^2)v^2+c^2{v}_a^2{\cos }^2\theta =0$. Solving this quadratic equation in $v^2$ derives both magnetoacoustic wave phase speeds,The plus sign is the fast magnetosonic wave, the minus sign the slow magnetosonic wave. The radicand is nonnegative, since it equals $[(c^2-v_a^2{)}^2+4c^2{v}_a^2{\sin }^2\theta ]/4$. Both squared speeds are nonnegative. Each nonzero mode has forward and backward propagation; an advected entropy disturbance is a separate zero-frequency mode in this static equilibrium.

Work in a gas region with $\rho ,p>0$ and physical adiabatic exponent $\gamma >1$. With $\xi =x/t$, each time derivative is $-\xi /t$ times differentiation in $\xi$, and each spatial derivative is $1/t$ times it. The continuity equation, Euler momentum equation and the entropy advection equation for pressure reduce toTheir coefficient determinant is $(u-\xi )[(u-\xi {)}^2-c^2]$, where the adiabatic sound speed is $c=\sqrt{\gamma p/\rho }$. If the determinant is nonzero, all derivatives vanish and the solution is uniform.

The apparent third characteristic $u=\xi$ does not give a smooth fan: on such an interval $u^{\prime }=1$, whereas continuity requires $\rho u^{\prime }=0$, impossible for $\rho >0$. It cannot support an isolated smooth entropy variation either. Subtracting $\gamma$ times the logarithmic density equation from the logarithmic pressure equation gives $(u-\xi )(\ln (p/{\rho }^{\gamma }){)}^{\prime }=0$. A nonzero continuous entropy derivative would force $u=\xi$ on an interval and give the same contradiction. Thus the specific entropy is constant throughout each smooth branch.

Every nonconstant branch therefore lies in one acoustic characteristic family. Let $\sigma =\pm 1$ denote it, with $\xi =u+\sigma c$, so $u-\xi =-\sigma c$. The constant entropy means $p=K{\rho }^{\gamma }$, $K>0$, and $c^{\prime }/c=(\gamma -1){\rho }^{\prime }/(2\rho )$. Continuity gives $u^{\prime }=\sigma c{\rho }^{\prime }/\rho =2\sigma c^{\prime }/(\gamma -1)$. Differentiating $\xi =u+\sigma c$ then provesIn particular, $u-2\sigma c/(\gamma -1)=J$ is a constant Riemann invariant. All nonconstant smooth local solutions have the explicit formon an interval where $c>0$. Along the left-going family ($\sigma =-1$) sound speed decreases with $\xi$; along the right-going family ($\sigma =+1$) it increases. These are the two self-similar ideal-gas rarefaction branches. Together with arbitrary uniform positive states they exhaust the smooth local solutions. A global solution can join them piecewise at fan edges, where derivatives generally jump, and can also have discontinuities; those do not constitute additional smooth branches.

Let $c_L=(\gamma p_L/{\rho }_L{)}^{1/2}$ and $c_R=(\gamma p_R/{\rho }_R{)}^{1/2}$, with positive pressures and densities. A nontrivial single rarefaction wave has constant specific entropy and constant Riemann invariant, so for one family label $\sigma =\pm 1$ the necessary matching conditions areThe final inequality says that the fan opens rather than compresses. Its edge speeds are ${\xi }_L=u_L+\sigma c_L$ and ${\xi }_R=u_R+\sigma c_R$, and their difference is $\sigma (c_R-c_L)(\gamma +1)/(\gamma -1)>0$. The profiles in part (a) then interpolate the two end states and supply a sufficient construction. For the left family, $c_R<c_L$, hence ${\rho }_R<{\rho }_L$ and $p_R<p_L$; for the right family all three inequalities reverse. In both cases $u_R>u_L$. These are the single-rarefaction matching conditions for an ideal-gas Riemann problem.

For a concrete nontrivial example, take $\gamma =5/3$ andThe adiabatic sound speeds are $c_L=1$ and $c_R=1/2$, with common $K=3/5$ and left-family invariant $u+3c=3$. The fan edges are ${\xi }_L=-1$ and ${\xi }_R=1$. The complete self-similar solution is uniform at these stated left and right values outside the fan, while inside $-1<\xi <1$,It matches both states continuously, has $u^{\prime }=3/4=2/(\gamma +1)$, and has constant specific entropy. For fixed $x\ne 0$, its $t\downarrow 0$ limit is the required initial step.

Assume a nonrotating, nonmagnetic equilibrium, inviscid adiabatic perturbations, and no dissipation. Let $g={\varphi }^{\prime }(R)>0$, so hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use $\mathbf{\xi }{e}^{i\omega t}$ for the fluid displacement. To avoid a source ambiguity, write Eulerian fluid perturbations as $p_1,{\rho }_1,{\varphi }_1$, and use $\Delta$ for quantities following the displaced material element, as in Eulerian and Lagrangian fluid perturbations. They are related by $\Delta p=p_1+{\xi }_R{p}^{\prime }$ and similarly for mass density and potential.

The printed identity needs two corrections: its compression denominator must be $\gamma p$, not $\gamma \rho$, and the pressure and potential appearing in this form of the identity must be Eulerian fluid perturbations. The printed denominator has the wrong dimensions for an energy density. The claimed Lagrangian interpretation does not produce the displayed buoyancy-compression split; the following derivation specifies the consistent version.

Linearized mass conservation, the adiabatic relation, and the Poisson equation giveConsequently $p_1=-\gamma p\nabla \cdot \mathbf{\xi }-{\xi }_R{p}^{\prime }$. Linearizing the Euler equations for an inviscid fluid about the static star gives the linear adiabatic stellar oscillation equationsRequire regularity at the center, a vanishing Lagrangian surface pressure, and the usual matching to a decaying exterior gravitational potential. Because $p$ and $\rho$ vanish at the surface, the pressure boundary term vanishes for regular displacements; compactly supported trials also suffice below. The perturbed Newtonian gravitational potential solves Laplace's equation outside the star and must be included there when its field energy is computed.

Multiply the momentum equation by $-{\mathbf{\xi }}^{\ast }$ and integrate. The pressure term is $-{\int }_V{p}_1\nabla \cdot {\mathbf{\xi }}^{\ast }dV$. Let $d=\nabla \cdot \mathbf{\xi }$. The adiabatic relation and hydrostatic equilibrium implyThe pressure and equilibrium-gravity terms then combine pointwise aswhere $N^2=g[(1/\gamma )p^{\prime }/p-{\rho }^{\prime }/\rho ]$ is the stellar buoyancy frequency. The potentially troublesome cross terms cancel exactly.

For the perturbed self-gravity term, integration by parts givesThus the corrected stellar displacement energy identity isThe right side is a real quadratic form. With these conservative boundary conditions, the force operator is a self-adjoint operator. Its Rayleigh quotient characterizes squared mode frequencies, and a negative admissible trial quotient proves an unstable spectral component; it does not require the trial displacement itself to be an exact eigenfunction.

For a generic compressive disturbance of scale $\ell$ in a region with slowly varying equilibrium coefficients, estimate ${\rho }_1\sim \rho \xi /\ell$, $p_1\sim \gamma p\xi /\ell$ and ${\varphi }_1\sim 4\pi G{\rho }_1{\ell }^2$. The self-gravity and compression energy densities have orderswith adiabatic sound speed $c_s^2=\gamma p/\rho$. Under the plausible stellar estimate $c_s^2\sim G\bar{\rho }{a}^2$, global modes can have comparable gravity and pressure contributions, while their ratio for small-scale compressive modes is of order $(\rho /\bar{\rho })(\ell /a{)}^2$. Thus perturbations with $\ell \ll c_s/\sqrt{4\pi G\rho }$ admit the Cowling approximation. This estimate compares two nonzero compressive terms; it cannot alone justify neglecting gravity for a trial deliberately chosen to have $p_1=0$.

For an explicit convective trial, choose $c<d<e<b$ and a nonzero smooth bumpLet $Y$ be a normalized spherical harmonic of degree $\ell \ge 1$, let $L=\ell (\ell +1)$, and let ${\nabla }_{\Omega }$ be the unit-sphere gradient. The pressure-free localized stellar convective trial isIt is supported away from the center and surface, so it satisfies every needed mechanical boundary condition. Since ${\Delta }_{\Omega }Y=-LY$,Therefore $p_1=-\gamma p\nabla \cdot \mathbf{\xi }+\rho g{\xi }_R=0$ exactly. Its buoyancy energy is ${\int }_d^e\rho N^2{R}^2|f{|}^{2}dR<0$, whereas its kinetic norm ${\int }_d^e\rho R^2(|f{|}^2+L|h{|}^2)dR$ is positive. In the Cowling approximation this gives a negative Rayleigh quotient, and hence $N^2<0$ in an interior shell implies convective instability. Retaining perturbed self-gravity only adds a nonpositive term, so the same trial proves instability in the full conservative problem too.

There is also a quantitative small-scale check for this very trial. Its mass density amplitude is ${\rho }_1=(\rho N^2/g)fY$, independent of $\ell$. Write ${\rho }_1=s(R)Y$ and ${\varphi }_1=F(R)Y$. The radial Poisson equation givesBy the Cauchy-Schwarz inequality, $A\le 4\pi G(\int R^4|s{|}^{2}dR{)}^{1/2}(A/L{)}^{1/2}$, and so the angular-degree bound on perturbed stellar self-gravity isIt tends to zero with horizontal scale $R/\ell$, while the strictly negative buoyancy integral is fixed. This justifies the small-scale neglect of gravity even though the trial's compression term vanishes.

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.

The circular Keplerian orbit has speed $v_K=(GM/r{)}^{1/2}$. Equating it with the heated gas sound speed givesThis is the photoevaporative gravitational radius, where thermal and orbital binding energies have the same order of magnitude. A circular orbit has specific mechanical energy $-GM/(2r)$. Heating adds thermal energy and, in a fluid outflow, available specific enthalpy of order $c^2$. For example, if $c$ is the adiabatic sound speed, an ordinary ideal gas has enthalpy $c^2/(\gamma -1)$; for $\gamma =5/3$ this is $3c^2/2$. At $r\ge r_g$, this more than compensates the circular-orbit binding energy. Equivalently the hot hydrostatic scale height satisfies $H_{\mathrm{hot}}/r\sim c/v_K\gtrsim 1$, so a thin bound surface layer cannot be maintained. With continued irradiation, the gas can expand into a thermal wind: photoevaporation removes disk material.

The condition is a thermal binding scale, rather than an assertion that the sound speed equals the ballistic escape speed, which is $\sqrt{2}{v}_K$. Detailed wind launching can change the numerical critical radius by factors of order unity. Using exactly the supplied numerical estimates, $c=10^6\mathrm{cm}{\mathrm{s}}^{-1}$, and therefore