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No measured normalization of the neutral-hydrogen column-density distribution, ionizing field or gas temperature is supplied, so the paper's formulas do not determine numerical densities on their own. With observational data and suitable ionization corrections, the high-redshift Lyman-alpha forest gives a total gas density of order a few $10^{-2}$ relative to today's critical density, comparable to the cosmological baryon density ${\Omega }_b\sim 0.04$–$0.05$ for $H_0\sim 70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$. The conclusion is that a large fraction of baryons at $z\simeq 3$ reside in diffuse ionized gas, rather than that every baryon has been individually accounted for. Schaye's analysis and Kim and collaborators' survey support that interpretation under their stated gas and ionizing-background assumptions.

The neutral-gas reservoir traced predominantly by damped Lyman-alpha systems is much smaller, of order $10^{-3}$ in the same convention. It is broadly comparable in order of magnitude to today's luminous stellar matter, whose cosmic stellar mass density is a few $10^{-3}$, but exact comparisons depend on cosmology, helium correction and which stellar components are counted. Storrie-Lombardi, McMahon and Irwin's neutral-gas census finds especially close comparison with present stellar disks. Most high-redshift gas is ionized; the smaller neutral reservoir has a mass of roughly the same order as today's visible stellar population. This comparison does not imply that the measured neutral reservoir alone simply becomes all present-day stars: gas accretion, recycling, ionization changes and outflows continue during galaxy evolution.

The neutral gas density from an absorption distribution is comparatively robust for high columns where the hydrogen is mostly neutral, but saturated lines, dust-obscured quasars, incomplete surveys and small numbers of rare high-column systems affect the integral. The ionized gas density from an absorption distribution has additional systematic uncertainty in the cosmic ionizing background, temperature, absorber depth, helium treatment and small-scale density inhomogeneity. At low columns, line blending and incompleteness matter; at high columns, attenuation makes an optically thin ionization correction invalid. A single power law extrapolated outside the measured column range can also produce a misleading mass integral. Hot gas, molecular gas and gas missed by the selected absorption population are not included automatically. These are observational and modelling qualifications to a baryon census, not failures of the column-mass accounting itself.

Use natural units $\hslash =c=k_B=1$ and the unreduced Planck mass $M_P=G^{-1/2}$. Neutrino decoupling occurs when the weak interaction rate falls below the Hubble parameter. The weak-decoupling temperature estimate equates the given rate with the radiation-era expansion rate:Here $g_{\ast }$ counts the effective number of relativistic energy degrees of freedom, and the Fermi constant is $G_F\simeq 10^{-5}{\mathrm{GeV}}^{-2}$. The omitted reaction-rate coefficients make this an order-of-magnitude estimate: decoupling occurs at a temperature of order one MeV.

The charge-conserving channel is $n+{\nu }_e\leftrightarrow p+e^{-}$, together with $p+{\bar{\nu }}_e\leftrightarrow n+e^{+}$. The positron sign in the printed neutron-capture channel violates electric charge conservation. Chemical equilibrium requires ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. Negligible lepton chemical potentials therefore give ${\mu }_n\simeq {\mu }_p$. Since neutrons and protons are nonrelativistic particles, their Maxwell-Boltzmann distribution givesBoth nucleons have two spin states, and their masses agree closely in the prefactor. Thus neutron-proton chemical equilibrium impliesThe heavier neutron is suppressed by a Boltzmann factor.

For the deuterium equilibrium abundance, retain the paper's convention $B=m_D-m_n-m_p<0$; the positive nuclear binding energy is ${\epsilon }_D=-B\simeq 2.22\mathrm{MeV}$. Chemical equilibrium of $n+p\leftrightarrow D+\gamma$ gives ${\mu }_D={\mu }_n+{\mu }_p$ because the equilibrium photon chemical potential vanishes. Using the nonrelativistic Maxwell--Boltzmann number density for each species and dividing givesHere $g_D=3$ and $g_n=g_p=2$. With the baryon-to-photon ratio $\eta =n_B/n_{\gamma }$, the photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, and $m_D\simeq 2m_N$, $m_n\simeq m_p\simeq m_N$, this becomesIn this convention $X_D=n_D/n_B$ counts nuclei, so its contribution to the fraction of baryons in deuterium is $2X_D$.

The deuterium bottleneck explains why helium-4 formation waits until $T$ is far below ${\epsilon }_D$. At higher temperature, there are vastly more photons than baryons; even the dilute energetic tail can destroy newly formed deuterium. A crude estimate balances ${\eta }^{-1}{e}^{-{\epsilon }_D/T}$ against one, giving $T\sim {\epsilon }_D/\log ({\eta }^{-1})\sim 0.1\mathrm{MeV}$ for $\eta$ of order $10^{-9}$. The power-law factors in the deuterium equilibrium abundance and the nuclear reaction rates refine this estimate. Once stable deuterium accumulates, further reactions can efficiently place nearly all surviving neutrons into helium-4.

Write $r_N=n_n/n_p$ at the onset of nuclear burning. Each helium-4 nucleus requires two neutrons and contains four baryons. Neutron-limited helium synthesis therefore gives the primordial helium mass fractionFor a realistic rough cosmological weak freeze-out estimate, $r_f$ is about $1/6$ and free-neutron decay after freeze-out reduces it to about $1/7$, giving roughly one quarter of the baryonic mass in helium-4. The survival relation that conserves baryon number is $X_n(T_N)=r_f(1+r_f{)}^{-1}{e}^{-\Delta t/{\tau }_n}$; the corresponding $r_N$ is $X_n/(1-X_n)$.

The rough neutrino decoupling rate by itself does not determine that one-quarter value. Substituting $T_d=1.8\mathrm{MeV}$ literally into $e^{-Q/T_d}$ would give $r\simeq 0.49$ and $Y_4\simeq 0.65$ before neutron decay. Neutron-proton chemical equilibrium persists to a lower temperature, approximately $0.7$--$0.8\mathrm{MeV}$ once the actual conversion rates are included. The rate scaling fixes the MeV scale; accurate abundances need the conversion coefficients and the delay through the deuterium bottleneck. This sequence is described in David Tong's discussion of nucleosynthesis.

Increasing $Q$ reduces the equilibrium neutron-to-proton ratio, so the primordial helium abundance decreases. Holding the freeze-out temperature $T_f$ and the decay survival factor fixed, the primordial helium response to the neutron-proton mass difference isFor $Q\simeq 1.29\mathrm{MeV}$ and $T_f=0.7$--$0.8\mathrm{MeV}$ this is about $0.85$--$0.87$, a reduction of roughly fifteen percent. A larger $Q$ also changes neutron decay and detailed weak rates; their quantitative effects require more microscopic information. The sign from the Boltzmann factor is already clear.

Use natural units, assume negligible neutrino chemical potentials, and approximate neutrino decoupling as occurring before electron-positron annihilation in cosmology. At high temperature, weak interactions maintain a common temperature for neutrinos, antineutrinos, electrons, Positrons and photons. Since ${\Gamma }_{\nu }/H\propto T^3$, the reaction rate eventually loses against expansion. Decoupled relativistic neutrinos then free stream; their momenta redshift as $a^{-1}$, so their distribution retains a temperature parameter satisfying $a{T}_{\nu }=\mathrm{constant}$.

The still-coupled electromagnetic bath receives the entropy of annihilating electrons and Positrons. Its effective number of relativistic entropy degrees of freedom is initiallywhere the two photon polarizations count as bosonic states and the four electron-positron spin states have the fermion entropy weight $7/8$. After the pairs disappear, $g_{\ast s}^{\mathrm{EM}}=2$. Cosmological entropy conservation in this coupled bath gives $g_{\ast s}^{\mathrm{EM}}(a{T}_{\gamma }{)}^3=\mathrm{constant}$. The three already decoupled neutrino species are excluded from this bath's entropy count. At decoupling $T_{\nu }=T_{\gamma }$, henceSubsequent expansion preserves this ratio in the idealized history. Residual interactions during annihilation give small corrections to the instantaneous-decoupling approximation. If a relic species later becomes massive, $T_{\nu }$ here remains the redshifting momentum-distribution parameter; it need not be its nonrelativistic kinetic temperature.

A single ordinary neutrino species includes one populated neutrino helicity and one antineutrino helicity. For negligible chemical potential, integration of the Fermi-Dirac distribution givesTherefore the number density of a relativistic neutrino relative to photons, with the antiparticle included, isThe factor $3/4$ is the fermionic number-density weight, distinct from the energy and entropy weight $7/8$. This relation continues to hold after this species becomes nonrelativistic if its comoving particle number is conserved.

Under the proposed matter-dominated, spatially flat model, the Friedmann equation requires the present relic-neutrino energy density to be approximately the critical density. Using the unreduced Planck mass $M_P=G^{-1/2}$ givesThus the critical-density mass of a thermal relic neutrino isTaking the paper's rounded values, $T_{\gamma 0}=2\times 10^{-13}\mathrm{GeV}$, $H_0=h/(4.5\times 10^{41}{\mathrm{GeV}}^{-1})$ and $M_P\sim 10^{19}\mathrm{GeV}$, givesFor $0.5<h<1$, this is of order $30$--$100\mathrm{eV}$ with these approximate constants. More precise constants change the coefficient, not the requested mass scale. This mass is much larger than the present momentum scale $T_{\nu 0}\simeq 1.4\times 10^{-4}\mathrm{eV}$, validating the nonrelativistic energy density approximation. The result assumes that one thermal relic species dominates the matter density; it is conditional on that model rather than a measurement of a neutrino mass.

Buoyancy and the convective criterion. Consider a small fluid parcel displaced outward by $\delta r$. Assume that sound waves keep its pressure equal to the ambient pressure, while heat exchange is slow enough that its displacement is an adiabatic process. Composition is uniform, so the parcel and its surroundings have the same mean molecular weight. For an ideal gas, $P\propto \rho T$, and an adiabatic parcel obeys $P\propto {\rho }^{\gamma }$. If $\delta \ln P<0$ is the pressure change on moving outward, thenThe density excess of the parcel at its new location is thereforeIf $\nabla >(\gamma -1)/\gamma$, this excess is negative: the outward-displaced parcel is lighter and accelerates farther outward. A downward displacement similarly continues downward. Conversely, a smaller gradient gives a restoring buoyancy force. The Schwarzschild criterion is thus stability for $\nabla <{\nabla }_{\mathrm{ad}}$ and instability for $\nabla >{\nabla }_{\mathrm{ad}}$, where the adiabatic temperature gradient is ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma$. For a monatomic gas,A composition gradient would require the additional density effect in the Ledoux criterion; it is absent in this calculation.

Opacity and pressure. The supplied opacity is characteristic of a cool hydrogen-rich stellar atmosphere dominated by negative hydrogen ion opacity. Mostly neutral hydrogen can attach electrons supplied by partially ionized metals; bound-free and free-free absorption involving the resulting $H^{-}$ ions produces a steep positive temperature dependence. The weak density dependence and strong temperature sensitivity can be represented by local power-law fits. At fixed composition the given pressure form is equivalent to $\kappa \propto {\rho }^{1/2}{T}^{17/2}$, because $P\propto \rho T$. This is an approximate fit over the relevant cool-atmosphere regime, not an exact opacity law at all temperatures; molecular opacity or hydrogen ionization eventually changes the regime.

For a geometrically thin atmosphere take the surface gravity of a star to be constant, $g_s=GM/R^2$. Combining $dP/dr=-\rho g_s$ with $d\tau /dr=-\kappa \rho$ gives hydrostatic equilibrium in optical depth,The grey atmosphere relation gives $T^8=T_e^8(3\tau +2{)}^2/16$. With negligible external pressure, $P(0)=0$, integration yieldsAn imposed nonzero external pressure would add $P(0{)}^{3/2}$; the boundary assumption is what fixes the integration constant here.

The factor in parentheses is $3\tau /[2(3\tau +2)]$. HenceThe grey-atmosphere convection onset with eighth-power temperature opacity occurs when this reaches $2/5$, soThe radiative continuation below this boundary would be unstable. Efficient convection instead keeps the deeper star approximately adiabatic.

Matching to the convective interior. Write $P=K{\rho }^{5/3}$. Eliminating density using the ideal gas law givesat fixed composition. Since ${\tau }_c$ is a fixed number, the matching temperature $T_b$ is proportional to $T_e$, while the matching pressure satisfies $P_b^{3/2}\propto g_s{T}_e^{-8}$. ConsequentlyThe polytropic mass-radius relation for index $3/2$ gives $K\propto R{M}^{1/3}$. Raising the matching relation to the ninth power and using $g_s\propto M/R^2$ therefore givesFinally the Stefan–Boltzmann law $L=4\pi R^2{\sigma }_{\mathrm{SB}}{T}_e^4$ yields the Hayashi relation with eighth-power temperature opacity,Composition and the opacity normalization are fixed in this comparison. The entropy coefficient $K$ is spatially constant in each convective model, but need not be the same for stars with different masses or radii.

Thermal energy versus electrostatic repulsion. The mean translational kinetic energy of one nonrelativistic thermal proton isUsing the center-to-center touching separation $2r_p$, the Coulomb barrier estimate isThus $E_C/⟨E⟩\simeq 274$. Using $r_p$ rather than $2r_p$ as the rough interaction distance doubles the estimate but leaves the conclusion unchanged. Typical solar thermal energies are far below the classical barrier. The two essential ideas are the high-energy tail of the thermal distribution and quantum tunnelling through the Coulomb barrier: neither a mean-energy estimate nor a sharp classical cutoff describes the reacting population.

For two equal-mass protons, let $v$ be their relative speed. In the center-of-mass frame each has speed $v/2$, so their total collision energy isHere ${\mu }_r$ is the reduced mass, not the stellar mean molecular weight.

In the low-energy, nonresonant approximation, the geometrical quantum-wave factor in a collision cross-section is proportional to the squared de Broglie wavelength, hence to $1/E$. The slowly varying short-range reaction probability is absorbed into $S_0$; for proton–proton fusion it also includes the small weak-reaction probability. The rapidly varying exponential is the quantum tunnelling probability. For a Coulomb potential $V(r)=e^2/r$, its leading semiclassical exponent isThe integral follows from $r=(e^2/E){\sin }^2\theta$; its dimensionless integral is $\pi /2$. A finite nuclear radius modifies the short-distance factor absorbed into $S_0$. In this convention $E_B={\pi }^2{\mu }_r{e}^4/(2{\hslash }^2)$: it is a penetration-energy scale, distinct from the classical contact energy $E_C$.

Thermal averaging and pair counting. Each proton samples collision partners with number density $N$ and collision flux $v$, so its collision rate is $N⟨\sigma v⟩$. Multiplying by the number of protons counts each identical pair twice. The thermonuclear reaction rate is thereforeThe relative speed of two independent thermal protons has a Maxwell-Boltzmann velocity distribution with reduced mass $m_p/2$. Changing variables with $E=m_p{v}^2/4$ gives the normalized energy densitySubstituting $v=2\sqrt{E/m_p}$ and the collision cross-section causes the energy powers to cancel:

The Gamow peak and its Gaussian width. Set $\Phi (E)=E/(kT)+2\sqrt{E_B/E}$. It diverges at either endpoint and has its unique minimum whereThus the energy called $E_G$ in this question is the Gamow peak energy,In the stellar tunnelling regime $E_B\gg kT$,This establishes the requested hierarchy in its physical regime; it is not an inequality valid at arbitrary temperature.

At the minimum, $\Phi (E_G)=3E_G/(kT)$ and ${\Phi }^{\prime \prime }(E_G)=3/(2kT{E}_G)$. HenceThe Gaussian standard deviation is $\sqrt{2kT{E}_G/3}$, small compared with $E_G$ when $kT\ll E_G$. Extending the lower endpoint to $-\infty$ relative to the peak then givesCombining this with the thermal prefactor yields the Gaussian approximation to a nonresonant thermonuclear rate,At fixed reactant density and fixed $S_0,E_B$, comparison with the requested temperature form givesThe constant $\beta$ here has dimensions of temperature; it is not inverse temperature. Resonances, plasma screening or variation of $S(E)$ would require corrections to this particular approximation.

Mechanical balance and equipotentials. Let $\Phi$ be the Newtonian gravitational potential and $\varpi$ the cylindrical distance from the rotation axis. For rigid angular velocity $\Omega$, a fluid element at rest in the rotating frame has no Coriolis acceleration. The centrifugal acceleration is ${\Omega }^2\varpi \hat{\varpi }$, so hydrostatic equilibrium becomesUsing the Poisson equation for gravity and ${\nabla }^2(x^2+y^2)=4$ gives

On any regular connected equipotential surface, every tangent displacement satisfies $d\varphi =0$ and therefore $dP=0$. Locally $P=P(\varphi )$, with $dP/d\varphi =-\rho$. This makes $\rho$ a function of $\varphi$ as well. Equivalently, taking the curl of the hydrostatic equation gives $\nabla \rho \times \nabla \varphi =0$. Thus $P,\rho$ and ${\nabla }^2\varphi$ are constant on each surface. The last conclusion follows from the modified Poisson equation; it does not constrain the magnitude of the first derivative $|\nabla \varphi |$ to be constant there.

For uniform composition, an equation of state $P=P(\rho ,T)$ determines $T$ uniquely at fixed $P,\rho$ in the usual stellar regime. For the gas-plus-radiation equation, the derivative with respect to $T$ is positive. Hence $T=T(\varphi )$, and the opacity and radiative conductivity are also functions of $\varphi$.

Surface flux and the requested sketch. Define $f(\varphi )=-\chi dT/d\varphi$, positive for the outward radiative flux. ThenOn an outer equipotential surface, $f$ is constant and the outward normal is parallel to $\nabla \varphi$. Its effective gravity is $g=|\nabla \varphi |$, so $F_n=fg$. The Stefan–Boltzmann law $F_n={\sigma }_{\mathrm{SB}}{T}_e^4$ gives the radiative gravity-darkening law,A rapidly rotating star is oblate. Centrifugal support reduces the effective gravity near the equator, which is therefore cooler and dimmer; the poles are hotter. The original sketch below uses a subcritical Roche model of a uniformly rotating star to show this geometry and the corresponding temperature variation. The material temperature on an interior equipotential is not the same quantity as the emergent effective temperature defined by surface flux.

Why purely local radiative balance fails. Differentiating the flux givesOn one equipotential, $f,f^{\prime },{\nabla }^2\varphi$ and the proposed source $\rho ϵ(\rho ,T)$ are all constant, but $|\nabla \varphi {|}^2$ generally varies with latitude. Unless $f^{\prime }=0$ or there is another special cancellation, no single value of $\rho ϵ$ can match this flux divergence at all points on the surface. This is the radiative-equilibrium obstruction in a rotating barotropic star. Slow stellar meridional circulation can transport heat to compensate; it is distinct from abandoning hydrostatic balance at leading order.

Integrated energy balance with circulation. The steady continuity equation is $\nabla \cdot (\rho \mathbf{v})=0$. Combining the entropy–enthalpy relation with hydrostatic balance givesIntegrate the energy equation over $V$ and apply the divergence theorem:Uniform composition makes $h=h(\rho ,T)$ constant on the enclosing equipotential, and $\varphi$ is constant there by definition. Pull $(h+\varphi )$ out of the last integral; the remaining net mass flux is zero by steady continuity. Thus the equipotential luminosity conservation with stellar circulation isThis does not require $\mathbf{v}\cdot \hat{n}=0$ pointwise on $S$: inward and outward circulation may cancel in the net mass flux.

Finally, $f$ is constant on $S$ and the divergence theorem applied to the effective potential givesTherefore $-\chi dT/d\varphi =L/(4\pi Gm-2{\Omega }^{2}V)$. Since $dP/d\varphi =-\rho$,The mass $m$ is the mass enclosed by this surface, not necessarily the total stellar mass. The denominator is positive for the regular hydrostatic surfaces used here; near loss of centrifugal confinement the assumed stellar model must be reconsidered. The derivation uses slow circulation, retaining hydrostatic/equipotential thermodynamics to leading order while allowing a nonzero advective heat flux.

A spatially uniform pressure is a useful leading approximation when the sound speed makes the acoustic crossing time much shorter than the radiative cooling and thermal conduction times. The motions must remain at low Mach number, and gravity and other stresses must be negligible. Sound then redistributes pressure before appreciable thermal evolution occurs. Small pressure corrections can still drive the slow motion; imposing exactly zero pressure gradient in the full momentum equation at every order would be a stronger assumption.

The equation of state gives the specific internal energy of an ideal gas as $e=p/[(\gamma -1)\rho ]$. With $D/Dt={\partial }_t+v{\partial }_x$, the continuity equation is $D\rho /Dt=-\rho v_x$. Differentiating $e$ therefore yieldsInsert this into the internal energy equation, whose compressional term is $-p{v}_x$, and use spatially uniform pressure, so $Dp/Dt=p_t$. The result isFor temporally constant pressure put $A=\mu p/\mathcal{R}$, so the mass density is $\rho =A/T$. The continuity equation now gives $v_x=T^{-1}DT/Dt$. Multiplication of the last equation by $(\gamma -1)T/(\gamma p)$, with $\lambda ={\lambda }_0{T}^{\alpha }$, givesThis is the isobaric conductive cooling in mass coordinates transformation before changing coordinates.

There is a boundary assumption implicit in the proposed Lagrangian coordinate. For a mass integral from a fixed origin, integration of the continuity equation givesThus $m$ is material when the base mass flux vanishes, as appropriate for the stationary cold reservoir below. If the origin permits mass flux, replace $m$ by $m-{\int }^t\rho (0,s)v(0,s)ds$; this corrected label has zero material derivative. In a material mass coordinate, $D/Dt={\partial }_t{|}_m$ and ${\partial }_x=(A/T){\partial }_m$. In particularThe conduction coefficient therefore becomes $C{\lambda }_0$, with the same $C$ as the cooling coefficient. With $\tau =Ct$, the equation is

In the hot gas there is no volumetric radiative cooling. The initial temperature is uniform and the cold boundary introduces no finite mass length. The remaining nonlinear diffusion equation is invariant under $m\mapsto bm$, $\tau \mapsto b^2\tau$ with the same temperature scale. Its natural similarity solution is thereforeThe far-field condition also recovers the initial temperature at every fixed $m>0$ as $\tau \downarrow 0$. Substitution gives the ordinary differential equationFor $\alpha =1$, this reduces to $f^{\prime \prime }+\xi f^{\prime }/2=0$. Integrating first for $f^{\prime }$ gives $f^{\prime }=K\exp (-{\xi }^2/4)$. The two boundary conditions fix $K=1/\sqrt{\pi }$, hence the error-function isobaric cooling front isHere the error function has normalization $2/\sqrt{\pi }$, as printed in the original PDF; the converted TeX's $2/\pi$ is incorrect. The following original sketch shows the temperature profiles. The cooling layer broadens as $\sqrt{\tau }$, and at fixed positive mass label the temperature falls as time increases, while the remote hot gas remains at $T_0$.

For $\alpha =1$, the magnitude of the heat flux into the cold reservoir isAlthough the thermal conductivity tends to zero there, the physical temperature gradient diverges: their product has this finite limit. The stationary reservoir has no incoming advective energy flux and radiates the conducted energy immediately. Since $T_m(0,\tau )=T_0/\sqrt{\pi {\lambda }_0\tau }$, the radiated power per unit area isThus the requested exponent is $\boxed{k=1/2}$. The initial divergence is the usual idealization of an instantaneous temperature discontinuity in a diffusion equation.

Use the relative potential $\Psi ={\Phi }_{\ast }-\Phi$ and integrate the stated, untruncated Maxwell-Boltzmann velocity distribution over ${\mathbb{R}}^3$. The normalization cancels its three Gaussian integrals:The gravitational potential sign reverses under the relative-potential convention, so the spherical Poisson equation for Newtonian gravity isAlso ${\Psi }^{\prime }=-Gm(r)/r^2$, with $m^{\prime }=4\pi r^2\rho$. Differentiating the density relation yieldsAn ideal gas with constant temperature has $P=\rho k_{B}T/m_0$. Its hydrostatic equilibrium equation is precisely the same whenTogether with the same enclosed-mass equation, central density and boundary conditions, this proves the isothermal stellar and gas density equivalence. In dimensionless terms, $y=-[\Psi -\Psi (0)]/{\sigma }^2$ and $x=r\sqrt{4\pi G\rho (0)}/\sigma$ obey $x^{-2}(x^2{y}^{\prime }{)}^{\prime }=e^{-y}$ with $y(0)=y^{\prime }(0)=0$ in both systems.

This calculation uses the distribution as printed for all velocities. Imposing an additional cutoff at positive relative energy changes the integral to ${\rho }_1[e^W\mathrm{erf}\sqrt{W}-2\sqrt{W}/\sqrt{\pi }]$, where $W=\Psi /{\sigma }^2\ge 0$, and no longer gives an exactly isothermal gas profile. An untruncated self-gravitating isothermal sphere also needs an outer modification for finite total mass. The function $\mathrm{erf}$ here is the error function. Finally, the half-line Gaussian integral printed in the note lacks a factor: the correct value is ${\int }_0^{\infty }{e}^{-\alpha x^2}dx=\frac{1}{2}\sqrt{\pi /\alpha }$.

The density-weighted mean kinematic viscosity isusing $\nu =\alpha p/(\rho \Omega )$. Under the preceding rescaling, letUniqueness of the universal dimensionless solution makes $I_P$ a positive numerical constant independent of $\alpha ,\Omega ,\kappa ,\Sigma ,\mathcal{R}$. HenceFor Keplerian rotation, ${\Omega }^{-2/3}=(GM{)}^{-1/3}r$. The constant-opacity gas-pressure disk viscosity law is thereforeThe particular scaling convention changes the representation of $I_P$ but not this physical law. Its nontrivial powers come from balancing viscous heating, diffusive cooling and vertical support together, rather than assuming a fixed temperature or thickness.

Work at one radius and write $\mathcal{R}=k/({\mu }_m{m}_p)$. Use the surface density to fix the mass density normalization and introduce the height scaleChoose the dimensionless variablesThe pressure and temperature scales make the hydrostatic and ideal gas equations parameter-free. The flux scale fixes the heating coefficient at $9/4$, and the choice of $h$ makes the diffusion coefficient unity. Direct substitution gives the universal ordinary differential equationsFor example, the dimensionless coefficient in the temperature equation is $3\alpha \kappa {\Sigma }^2{\mathcal{R}}^4/(16\sigma {\Omega }^5{h}^6)=1$.

Let the upper free surface be at $\zeta ={\zeta }_s$, so its physical semithickness is $H=h{\zeta }_s$. Midplane symmetry and the radiative-zero disk surface conditions giveThe last condition is precisely $\Sigma ={\int }_{-H}^H\rho dz$. The unknown surface position and the two positive midplane values $P(0),\Theta (0)$ are determined as part of the normalized boundary value problem. Symmetry makes $P,D,\Theta$ even and $Q$ odd; $P^{\prime }(0)={\Theta }^{\prime }(0)=0$ already follow from the equations and are not additional independent boundary conditions.

The zero conditions do not mean zero surface flux. Integrating the heating equation gives $Q({\zeta }_s)=(9/4){\int }_0^{{\zeta }_s}Pd\zeta >0$. Near a regular surface with finite outgoing flux, dividing the hydrostatic equation by the temperature equation yields $dP/d\Theta \simeq {\zeta }_s{\Theta }^3/Q_s$. Thus $P\propto {\Theta }^4$, $D\propto {\Theta }^3$ and $\Theta \propto {\zeta }_s-\zeta$, so mass density also vanishes. The apparent singular quotient in the diffusion equation has a finite regular limit. All boundary conditions and equations are independent of the physical parameters: this is homologous vertical structure of a constant-opacity alpha disk.

The first equation is vertical hydrostatic equilibrium: the upward pressure gradient balances the central mass's vertical gravitational force. For $|z|\ll r$, the point-mass acceleration is $-GMz/(r^2+z^2{)}^{3/2}\simeq -{\Omega }_K^{2}z$, where Keplerian rotation gives ${\Omega }_K^2=GM/r^3$. The disc's self-gravity and vertical inertia are neglected.

The second equation is local thermal balance. The divergence of the vertical radiative flux equals viscous dissipation. In the convention of this alpha disk, $\nu =\alpha p/(\rho \Omega )$, and the dominant Keplerian shear has $rd\Omega /dr=-3\Omega /2$. ThereforeThe radiative flux increases from zero at the symmetric midplane to a positive outgoing value at the upper surface. Radial heat transport, thermal advection and rapid storage of thermal energy are excluded from this local balance.

The third equation is radiative diffusion, equivalentlyIt applies to optically thick material, with $\kappa$ the opacity per unit mass and $\sigma$ the Stefan-Boltzmann constant. Here constant electron-scattering opacity represents Thomson scattering; the formula uses a diffusive, nearly thermal radiation field. Positive upward flux requires temperature to decrease upwards.

The final equation is the ideal gas pressure law, with mean molecular weight ${\mu }_m$ in proton-mass units. Writing $\mathcal{R}=k/({\mu }_m{m}_p)$ gives $p=\mathcal{R}\rho T$. The pressure is gas pressure because radiation pressure is assumed negligible. The four equations express vertical support, local heating balance, diffusive cooling and the gas equation of state.

For a fixed column density or rest-equivalent-width threshold, the number of ordinary Lyman-alpha forest features per unit redshift generally increases towards higher redshift. Over much of $z\sim 1$–$4$, a rough description is $d\mathcal{N}/dz\propto (1+z{)}^{\gamma }$ with an index of order $2$–$3$, dependent on selection; the evolution is much slower at low redshift. Near reionization, blending and widespread opacity make a decomposition into individually detected lines increasingly inappropriate.

First separate physical evolution from geometrical path length. For comoving absorber density $n_c$ and proper interception area ${\sigma }_p$, absorber incidence and comoving number density givesEven a non-evolving population is not constant per unit redshift. Using absorption distance removes this particular geometrical effect.

The remaining redshift evolution of the Lyman-alpha forest reflects several competing processes. Cosmic expansion dilutes the gas; at fixed overdensity, photoionization equilibrium gives $n_{\mathrm{HI}}\propto \alpha (T)(1+z{)}^6/\Gamma$, so dilution can move many features below a fixed threshold. The cosmic ionizing background evolves as quasars, stellar sources and ionizing-photon absorption evolve; a decline of its intensity at late times partly offsets dilution. Photoheating changes temperature and radiative recombination, while gravitational instability redistributes gas into denser structures and hot shocked phases and changes its velocity field. Thus evolving line counts are not a direct count of conserved clouds; the density, thermal state, radiation field, geometry and observational selection must all be accounted for. The high-redshift increase and its physical interpretation are documented in the forest observations and models.