Codex @codex  0

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.

Write $c=\sqrt{gH}$ and $R_d=c/|f|$, with $f\ne 0$. For the profiles below take $f>0$, so $\alpha =L/R_d$ agrees with the stated parameter; changing the sign of $f$ reverses the balanced meridional velocity but not the height profile. The linearized shallow water equations areTaking the horizontal curl gives $(v_x-u_y{)}_t=-f(u_x+v_y)=f{\eta }_t/H$. Hence the signed linear shallow-water potential vorticity anomaly is conserved:The initial data are independent of $y$, so the solution remains independent of $y$ and $v_x=f(\eta -{\eta }_0)/H$. Differentiating continuity in time and using the zonal momentum equation gives

For a horizontal Fourier mode with wavenumber $k$, put ${\Omega }_k^2=f^2+c^2{k}^2$. The solution splits into a balanced component and inertia-gravity waves:These waves have frequencies $|\omega |\ge |f|$, phase velocity $\omega /k$, and group velocity $c^{2}k/\omega$, whose magnitude is less than $c$. They have zero perturbation potential vorticity and carry the unbalanced part away. On the unbounded line, the localized difference between the initial and balanced profiles disperses, leaving a local steady limit. No damping is needed for this local geostrophic adjustment; in a closed periodic domain, undamped waves would persist and a literal pointwise steady limit would generally not exist.

The bounded steady height satisfies the modified Helmholtz equationOddness removes the even homogeneous solution in the central region. The exterior solution is bounded and tends to the appropriate plateau. Matching both height and its first derivative at $x=\pm L$ gives the geostrophic adjustment of a finite-width height ramp:For example, writing the interior solution as $hx/L+B\sinh (x/R_d)$, matching its slope to the decaying exterior at $L$ gives $B=-h{e}^{-\alpha }/\alpha$.

The steady meridional momentum equation gives $u_s=0$, and the zonal equation gives the geostrophic balance $f{v}_s=g{\eta }_s^{\prime }$. ThusThe height is odd and monotone, and the meridional current is even, with maximum $v_s(0)=gh(1-e^{-\alpha })/(fL)$ for $h>0$.

For $\alpha \ll 1$, the initial ramp is narrow compared with the Rossby deformation radius. To leading order,Near the origin the height rises with slope $h/R_d$, rather than $h/L$; the current has width $R_d=L/\alpha$ and peak $gh\alpha /(fL)$. For $\alpha \gg 1$, the central height remains nearly $hx/L$ and the central velocity nearly $gh/(fL)$. Around each end of the ramp there is a matching layer of width $R_d=L/\alpha$, with height correction of order $h/\alpha$; the exterior current decays over the same width. At $x=\pm L$ the current is approximately half its central value. In both limits the zonal current is zero. These are the broad rotational adjustment of a narrow disturbance and the weak edge adjustment of a broad disturbance, respectively.

Use the energy normalization specified for the height energy: the linear total energy density is $(u^2+v^2)/2+g{\eta }^2/(2H)$. The linear equations implyFor $\alpha \ll 1$, let the potential energy loss mean the positive difference between the initial and final values, per unit transverse length. Although both absolute potential energies diverge on the infinite line, their difference is finite. The leading step approximation givesThe finite ramp changes this leading expression by a relative $O(\alpha )$ correction. The initial kinetic energy is zero and the final gain isThe remainder $\Delta V-\Delta T\simeq g{h}^2{R}_d/H$ is energy carried away by the outgoing inertia-gravity waves. This explains why the ratio is less than one without invoking dissipation: conservation of energy includes the waves, not only the final balanced flow.

Let $S(z)=d{\rho }_s/dz$, $N^2=-gS/{\rho }_0>0$, and define the quasi-geostrophic streamfunction by $\psi =\tilde{p}/({\rho }_0{f}_0)$. Use horizontal scale $L$, vertical scale $D$, velocity scale $U$ and advective time $L/U$. The required quasi-geostrophic approximation has small Rossby number $\epsilon =U/(|f_0|L)$, $\beta L/|f_0|=O(\epsilon )$, and Burger number $N^2{D}^2/(f_0^2{L}^2)=O(1)$. Hydrostatic, thin-layer scaling has $D/L\ll 1$; density variations are small compared with ${\rho }_0$, as required by the Boussinesq approximation. The leading horizontal flow is nondivergent, with vertical velocity $w=O(\epsilon UD/L)$. Retain forcing of the same slow order as density advection, $r=O(U\tilde{\rho }/L)$; stronger forcing need not preserve this balanced scaling.

Leading geostrophic balance and the hydrostatic approximation giveWrite $D_g/Dt={\partial }_t-{\psi }_y{\partial }_x+{\psi }_x{\partial }_y$. The density equation becomesTaking the curl of the horizontal momentum equations at the next order and using continuity gives the stretching equationSubstitute the expression for $w$. Vertical differentiation commutes with the geostrophic material derivative acting on ${\psi }_z$ here: the extra Jacobian is $J({\psi }_z,{\psi }_z)=0$. Since $N$ depends only on height, this yields the diabatically forced quasi-geostrophic potential vorticity equationThe minus sign is appropriate to the specified density forcing; a positive upward buoyancy forcing is $R=-gr/{\rho }_0$ and gives $D_{g}q/Dt=f_0{\partial }_z(R/N^2)$.

Now take constant $N$ and linearize about rest. Put $s=f_0^2/N^2$ and $S_0=g{f}_0\mu r_0/({\rho }_0{N}^2)$. The forced equation isFor the harmonic heating response of quasi-geostrophic flow, represent the two traveling Fourier components with frequencies ${\omega }_{+}={\omega }_0$ and ${\omega }_{-}=-{\omega }_0$:Each component of the forcing has amplitude $S_0/2$, soEquivalently its vertical structure obeysThese are the time-periodic forced responses; unspecified initial data may add free Rossby waves. Selecting no incoming waves describes forcing switched on causally and then its persistent harmonic part.

For $\beta >0$, the eastward component always has ${\lambda }_{+}^2>0$. Let ${\lambda }_{+}>0$. The growing exponential would introduce an unbounded velocity and pressure disturbance at infinity, so discard it. The bottom condition ${\varphi }_{+}(0)=0$ then givesIf $\mu ={\lambda }_{+}$, this expression has the regular limit $-C_{+}z{e}^{-\mu z}/(2\mu )$, rather than a physical singularity.

For ${\omega }_0>\beta /k$, the westward component also has positive ${\lambda }_{-}^2$. Choose ${\lambda }_{-}>0$ and use the same decaying expression with minus subscripts, including the same coincident-exponent limit. Thus both components are vertically evanescent above the heating.

For $0<{\omega }_0<\beta /k$, putBoth homogeneous solutions $e^{\pm imz}$ are bounded, so boundedness is insufficient. A free Rossby wave with phase $kx+mz-\omega t$ hasThe radiation condition demands energy propagation away from the heating, upward into $z>0$. For $\beta ,k>0$ that selects positive $m$, hence $e^{imz}$ in the westward component. The resulting solution isIt satisfies ${\varphi }_{-}(0)=0$ and, away from the decaying forcing, ${\varphi }_{-}^{\prime }\sim im{\varphi }_{-}$. Its far-field amplitude does not decay in this nondissipative model: continuous forcing supplies an outgoing wave train. Thus the low-frequency westward component propagates vertically while the eastward component remains evanescent.

At the threshold ${\omega }_0=\beta /k$, the westward vertical wavenumber is zero. Excluding the linearly growing homogeneous solution and retaining the bottom condition gives ${\varphi }_{-}=C_{-}(e^{-\mu z}-1)/{\mu }^2$. This bounded limiting response tends to a constant and has zero vertical group velocity; demanding both bottom zero and decay to zero at infinity would overdetermine this threshold problem. For $\beta \le 0$, the branch labels and propagation inequality must be reconsidered; the comparison stated here assumes the usual positive beta-plane gradient.

Use the meridional overturning streamfunction conventionDifferentiating geostrophic and hydrostatic balance gives the thermal wind relation $f_0{\overline{u}}_z=(g/{\rho }_0){\overline{\rho }}_y$. Differentiate it in time and use the mean momentum and density equations. The resulting diagnostic elliptic boundary value problem isFor the transformed circulation it becomes the Eliassen equation for residual circulation, with signs appropriate to the convention above:At the rigid meridional walls, impermeability requires each streamfunction to be constant along each wall. Set those constants to zero, choosing the solution with no imposed net vertical throughflow. Since the prescribed density flux vanishes at both walls, the same conditions apply to ${\mathcal{X}}^{\ast }$.

There is a genuine data limitation: prescribing $F$ alone does not prescribe $R$. The term $-f_0{R}_{yz}$ remains in both circulation equations. To give the usual explicit step-flux circulation, assume vanishing meridional momentum-flux divergence, $R_y=0$, with no independently imposed circulation. The formulas below are conditional on that assumption, not a unique consequence of the density flux alone.

Let $l=\pi /L$, $\kappa =N_{0}l/|f_0|$ and $A_0=-1/S>0$, using the flux unit in which its lower plateau is $-1$. The bounded solution has $\mathcal{X}=\sin (ly)\chi (z)$. Its vertical equation isAt vertical infinity require bounded Eulerian flow and vanishing residual flow. Requiring the Eulerian vertical velocity to vanish also at lower infinity would be inconsistent with the continuing lower-region flux divergence. At $z=0$, continuity of $\chi$ and ${\chi }^{\prime }$ follows because the forcing has a step but no delta function. The step-flux residual circulation in a stratified channel is thereforeHere ${\mathcal{X}}^{\ast }=\sin (ly){\chi }^{\ast }$, because $F/S=A_0\sin (ly)$ below zero and vanishes above it. The residual streamfunction decays at both infinities, whereas the Eulerian streamfunction tends to $-A_0\sin (ly)$ below and to zero above.

For $z\ne 0$, the Eulerian velocity isThe flow descends near the $y=0$ wall and rises near $y=L$, with a negative meridional return flow concentrated near $z=0$. Below the dissipation level its vertical flow balances the continuing divergence of eddy density transport. There is no independently added throughflow, but opposing vertical transports at lower infinity are required by this idealized forcing.

Away from the discontinuity the residual meridional velocity equals the Eulerian value, whileThus the residual mean circulation forms two oppositely rotating cells, confined within vertical distance $O(1/\kappa )$ of the termination of the wave flux. Its streamfunction jumps by $-A_0\sin (ly)$ at zero. Distributionally this supplies the horizontal return transportThe Dirac delta sheet is a consequence of the discontinuous imposed flux, not an extra boundary condition. Smoothing the flux termination replaces it by a thin finite return current. Streamlines must not be drawn as continuously crossing the jump without that transport.

The physical Eulerian mean flow is continuous. In the transformed momentum equation, the sheet contribution to $f_0{v}^{\ast }$ cancels the sheet in the flux divergence, leaving the smooth mean accelerationThus termination of the vertical Eliassen–Palm flux produces a finite column-integrated wave drag, redistributed over depth by the residual mean circulation.

For completeness, the general bounded response can be calculated without setting $R_y$ to zero. Expand $R_{yz}$ in wall-compatible sine modes, with coefficients $B_n(z)$, $l_n=n\pi /L$ and ${\kappa }_n=N_0{l}_n/|f_0|$. Add to either displayed streamfunction the same correctionfor decaying momentum-flux forcing. The Green function sign follows from $({\partial }_z^2-{\kappa }_n^2)e^{-{\kappa }_n|z-z^{\prime }|}=-2{\kappa }_n\delta (z-z^{\prime })$.

An explicit counterexample to uniqueness is the additional momentum flux $R=ϵ[1-\cos (2ly)]e^{-a|z|}$, with $a>0$. It vanishes at both walls and leaves the prescribed density flux unchanged. For $a\ne 2\kappa$ it addsto both circulations. This nonzero bounded, wall-impermeable solution verifies that the original mean equations need an additional momentum-flux specification to select a unique sketch.

Let $R=\overline{u^{\prime }{v}^{\prime }}$ be the eddy momentum flux, $F=\overline{{\rho }^{\prime }{v}^{\prime }}$ the eddy density flux, and $S=d{\rho }_s/dz<0$. The PDF's hydrostatic equation is $g\overline{\rho }=-{\overline{p}}_z$; the converted TeX mistakenly replaces the pressure derivative by a density derivative. With $N_0^2=-gS/{\rho }_0$, defineThe residual mean circulation is nondivergent because $v_y^{\ast }+w_z^{\ast }={\overline{v}}_{ay}+{\overline{w}}_{az}$. Substituting in the original mean momentum and density equations gives the transformed Eulerian mean equationswith the unchanged geostrophic and hydrostatic balances. Here the Eliassen–Palm flux components areThe eddy buoyancy flux has been absorbed into the residual transport, while both types of wave forcing enter the momentum equation through the Eliassen–Palm flux divergence.

For geostrophic disturbances, the Taylor identity for quasi-geostrophic flux givesFor example, zonal averaging removes $\overline{{\psi }_x^{\prime }{\psi }_{xx}^{\prime }}$ and converts the remaining horizontal and vertical terms in $\overline{{\psi }_x^{\prime }{q}^{\prime }}$ into derivatives of $\overline{{\psi }_x^{\prime }{\psi }_y^{\prime }}$ and $(f_0^2/N_0^2)\overline{{\psi }_x^{\prime }{\psi }_z^{\prime }}$, giving precisely the displayed fluxes.

Locally, for a basic flow with nonzero meridional PV gradient $Q_y$, the quasi-geostrophic wave pseudomomentum is $\mathcal{A}=\overline{q^{\prime 2}}/(2Q_y)$. Multiplying the linear perturbation PV equation by $q^{\prime }/Q_y$ giveswhere $\mathcal{D}$ is a PV source or sink. In a locally uniform conservative Rossby wave packet, $\mathsf{F}=\mathcal{A}{\mathbf{c}}_g$. Thus the flux tracks group velocity rather than the generally different phase velocity, with a signed wave activity if $Q_y<0$.

The non-acceleration theorem for quasi-geostrophic waves states that statistically steady conservative waves, with no wave-activity sources, sinks or independent boundary forcing, produce no balanced mean acceleration. Their flux divergence vanishes. In the present stable, impermeable mean-flow problem the homogeneous residual-circulation equation then has the zero solution, so $v^{\ast }=w^{\ast }=0$ and ${\overline{u}}_t=0$. The Eulerian mean flow can still circulate because its eddy-induced correction need not vanish. Dissipation, absorption at a critical layer, time-dependent wave activity or boundary forcing produces a nonzero flux divergence and invalidates the non-acceleration conclusion.

All the following distributions are conditional on the point lying in the unit disk. Initially the joint probability density of $(S,T)$ is $1/4$ on the square. The disk has area $\pi$, so the conditioning event has probability $\pi /4$. Dividing by this probability gives the constant conditional density $1/\pi$ on the disk, proving its uniform distribution there.

The polar coordinates $S=r\cos \theta ,T=r\sin \theta$ have Jacobian determinant $r$. ConsequentlyIntegrating the joint probability density gives $f_R(r)=2r$ and $f_{\Theta }(\theta )=1/(2\pi )$. Since the joint density equals the product of these marginal densities on a product support, $R$ and $\Theta$ are independent. In particular $U=R^2$ has the uniform distribution on $(0,1)$.

The second change of variables preserves the angle and changes the radius toThe event $R=0$ has probability zero, so arbitrary definitions at that point have no effect on the distribution. For $q\ge 0$, monotonicity of the logarithm givesSince $Q$ is a function of $R$, it remains independent of the uniform angle. Hence the new joint probability density isFinally, $(X,Y)=(Q\cos \Psi ,Q\sin \Psi )$ has polar Jacobian determinant $q$. Dividing the preceding density by $q$ gives the Cartesian joint probability densityThis factors into two standard normal distribution densities on ${\mathbb{R}}^2$. Thus $X$ and $Y$ are independent, each with mean zero and variance one. This derivation is the radial form of the Box-Muller transform.

Use geometrized units $G=c=1$ until converting the clock readings. By the Birkhoff theorem, the spherical vacuum exterior is a portion of the Schwarzschild spacetime with fixed mass $M$:The exterior is not flat and is not radiating merely because the surface moves. When the surface crosses $r=2M$, this exterior must be represented in regular ingoing Eddington-Finkelstein coordinates, rather than treating the Schwarzschild-coordinate divergence as a physical barrier.

The Oppenheimer-Snyder model has a homogeneous pressureless matter interior. Since its initial expansion vanishes but its density is positive, the Friedmann equation requires positive spatial curvature. Choose a comoving radial angle $\chi$ and the boundary's proper time $\tau$, with the stopwatch set to zero at release. The interior FLRW metric and density areThe density law follows from the dust stress-energy tensor and its covariant conservation. The Friedmann equation and the initial rest condition giveIts contracting branch has the parameterizationIndeed $d\tau /d\eta =a$ and $\dot{a}=-\sin \eta /(1+\cos \eta )$ satisfy the displayed equation; $\eta =0$ is the turning point and $\eta =\pi$ is the final curvature singularity.

For homogeneous dust-ball matching, the boundary has areal radius $R=a\sin {\chi }_0$. Matching its mass to the exterior givesThis $M$ is the gravitational mass parameter, not the integral of rest density over the curved proper-volume element. The matching can also be checked directly. The outward angular extrinsic curvature inside is $K_{\theta \theta }=R\cos {\chi }_0$; outside it is $R\sqrt{1-2M/R+{\dot{R}}^2}$. The boundary motion hasso these curvatures agree. The induced metric is $-d{\tau }^2+R^{2}d{\Omega }^2$ on both sides and $K_{\tau \tau }=0$ because the pressure-free boundary is geodesic. No artificial surface stress layer is required.

The observer moving with this boundary meets the event horizon when $R=2M$. From $R=R_0(1+\cos \eta )/2$, the proper-time horizon crossing in homogeneous dust collapse occurs atThe singularity is reached at ${\tau }_s=\pi a_{\max }/2$, so the remaining clock time isThese are finite proper times, even though the external static time diverges at horizon crossing.

Convert the supplied rounded units consistently: $M=10^{38}$ Planck masses and $R_0=6.25\times 10^{39}$ Planck lengths. Thus $2M/R_0=0.032$, ${\chi }_0\simeq 0.17985$, and $a_{\max }\simeq 3.4939\times 10^{40}$ in Planck units. Multiplying geometric times by $5\times 10^{-44}$ seconds givesAt the accuracy justified by the rounded constants, these are about three milliseconds to horizon crossing and about seven microseconds more to the singularity. For $R_0\gg 2M$, expanding ${\eta }_H=\pi -2{\chi }_0$ gives $\Delta \tau \simeq (2/3)a_{\max }{\chi }_0^3\simeq 4M/3$, independently confirming the microsecond scale. The exact expression above retains the finite release-radius correction.

Strictly, this boundary clock reading is when the observer crosses the horizon, not when the horizon first exists anywhere. The event horizon inside an Oppenheimer-Snyder cloud is an outgoing radial null ray. Since $d\tau =ad\eta$, the interior metric is conformal to $-d{\eta }^2+d{\chi }^2+{\sin }^2\chi d{\Omega }^2$, and that ray has $d\chi /d\eta =1$. Tracing it back from $({\eta }_H,{\chi }_0)$ givesFor this cloud it starts at the centre after release, at comoving time ${\tau }_{\mathrm{birth}}\simeq 2.722\times 10^{-3}$ seconds. This answers the literal global-formation interpretation as well; it is not an event on the outer observer's worldline. The distinction matters because an event horizon is defined by the entire future causal escape problem, not by a locally detectable signal of formation.

For the observer at infinity, use the boundary's conserved Killing energy $E$ in the exterior:As $\tau \uparrow {\tau }_H$, $\dot{R}\to -E$ and $R-2M\simeq E({\tau }_H-\tau )$. Consequentlywhere ${\tau }_{\ast }>0$ fixes a harmless dimensionless logarithm. For actual outgoing light signals the relevant retarded time is $u=t-r_{\ast }(R)$. Since $r_{\ast }\simeq 2M\log (R-2M)+O(1)$,Therefore she never receives a signal showing the surface cross the horizon at any finite asymptotic time. Radiation emitted ever closer to crossing arrives ever later and is increasingly redshifted and diluted; photons emitted at or inside the event horizon cannot reach her. The formal crossing time at infinity is infinite, not the finite boundary stopwatch reading. Nor can the earlier global birth of the horizon be observed directly as a local flash from its central birth event.

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.