13. Radiation Hydrodynamics

Photons carry momentum, and a fluid that is lit from inside feels it. This module derives the Eddington limit that five shipped pages have been quoting, prices the composition hidden in its proton mass, and evaluates the one computation the book deferred by name. Four checks: one reproduction confirmed, a column-mean opacity confirmed to a factor of 1.1482, one refutation of the adiabatic assumption at the photosphere, and a grey atmosphere that holds to three per cent.

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Contents

Every module of this book so far has treated light as something a fluid emits and never as something that pushes on it. Module 1 §9 put that in its closing table in one line — "Photons neither carry momentum nor dominate the energy budget" — and named this module as the repayment. Eleven more debts in six other modules followed it. The sharpest are in Module 9, which used the Eddington luminosity $L_{\rm Edd} = 4\pi GMm_{\rm p}c/\sigma_{\rm T}$ in two problems and wrote in its notation table that the formula is "stated, not derived". Module 9 and Module 11 together send the reader here through five live links.

The destination is the Eddington limit, in §9. It does not come first, because its derivation needs three things the reader does not yet have: an opacity (§2), a radiative flux (§4), and the reason the Sun's own surface is not the simple electron-scattering gas the limit assumes (§6). A reader who arrived from module09.html:944 can go straight to §9 and come back for the definitions.

One thing this module does not do, and it says so where each debt is due. It does no atomic physics. The opacity of the solar photosphere is attributed, by a standard result this book does not derive, to the negative hydrogen ion, H⁻, and this book does not compute it; §6 takes the opacity from a published model atmosphere instead and says what that costs. Three debts are refused in §12, each in a row of its own, and each row says what the repayment would need.

1. Photons carry momentum

A photon of energy $E$ carries momentum $E/c$. A beam that delivers energy at a rate $F$ per unit area therefore delivers momentum at a rate $F/c$ per unit area, and a body that absorbs the beam feels a pressure $F/c$. For sunlight on the Earth that pressure is negligible. Inside a star the same accounting is not, and near a luminous compact object it decides whether gas can fall in at all. To do the accounting one needs to describe radiation that travels in every direction at once.

Definition 1 (the specific intensity and its moments). The specific intensity $I_\nu(\mathbf{r},\mathbf{n},t)$ is the energy carried in direction $\mathbf{n}$ per unit time, per unit area normal to $\mathbf{n}$, per unit solid angle and per unit frequency, in $\mathrm{erg\,s^{-1}\,cm^{-2}\,sr^{-1}\,Hz^{-1}}$. With $\theta$ the angle between $\mathbf{n}$ and a chosen axis $z$ and $\mathrm{d}\Omega$ the element of solid angle, its first three angular moments are $$ u_\nu = \frac{1}{c}\oint I_\nu\,\mathrm{d}\Omega, \qquad F_\nu = \oint I_\nu\cos\theta\,\mathrm{d}\Omega, \qquad P_\nu = \frac{1}{c}\oint I_\nu\cos^{2}\theta\,\mathrm{d}\Omega, \tag{1.1}$$ the radiation energy density per unit frequency, the flux along $z$, and the flux of $z$-momentum across a surface normal to $z$ — the radiation pressure. The mean intensity is $J_\nu = (1/4\pi)\oint I_\nu\,\mathrm{d}\Omega = c\,u_\nu/4\pi$. Quantities integrated over frequency drop the subscript: $u_{\rm rad} = \int u_\nu\,\mathrm{d}\nu$, $F = \int F_\nu\,\mathrm{d}\nu$, $P_{\rm rad} = \int P_\nu\,\mathrm{d}\nu$.

The second moment is written $P_\nu$ and never $K_\nu$, which is what most of the transfer literature writes, because $K$ is Module 3's polytropic constant. That $P_\nu$ is a pressure is the whole point of the first proposition.

Proposition 1 (the radiation pressure of an isotropic field). If $I_\nu$ does not depend on direction, then $P_\nu = u_\nu/3$ and $F_\nu = 0$. In thermal equilibrium at temperature $T$ the integrated energy density is $u_{\rm rad} = a_{\rm rad}T^{4}$, so $$ P_{\rm rad} \;=\; \frac{u_{\rm rad}}{3} \;=\; \frac{a_{\rm rad}T^{4}}{3}, \qquad a_{\rm rad} \;=\; \frac{4\sigma_{\rm SB}}{c}. \tag{1.2}$$ The factor $1/3$ is derived; the law $u_{\rm rad} = a_{\rm rad}T^4$ is stated.
Write $\mathrm{d}\Omega = \sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi = -\mathrm{d}\mu\,\mathrm{d}\phi$ with $\mu = \cos\theta$. For isotropic $I_\nu$ the three integrals in (1.1) are $\oint\mathrm{d}\Omega = 2\pi\int_{-1}^{1}\mathrm{d}\mu = 4\pi$, $\oint\mu\,\mathrm{d}\Omega = 2\pi\int_{-1}^{1}\mu\,\mathrm{d}\mu = 0$ and $\oint\mu^{2}\,\mathrm{d}\Omega = 2\pi\int_{-1}^{1}\mu^{2}\,\mathrm{d}\mu = 4\pi/3$. So $u_\nu = 4\pi I_\nu/c$, $F_\nu = 0$ and $P_\nu = 4\pi I_\nu/3c = u_\nu/3$. Integrating over $\nu$ gives $P_{\rm rad} = u_{\rm rad}/3$. What is stated and not derived is the Stefan–Boltzmann law for black-body radiation, $\int B_\nu(T)\,\mathrm{d}\nu = \sigma_{\rm SB}T^{4}/\pi$, where $B_\nu(T)$ is the Planck function; this book does no statistical mechanics of photons. Given it, the energy crossing a plane from one side of an isotropic black-body field is $\int_0^{2\pi}\!\!\int_0^1 B_\nu\,\mu\,\mathrm{d}\mu\,\mathrm{d}\phi = \pi B_\nu$ per unit frequency, which integrates to $\sigma_{\rm SB}T^4$, while $u_{\rm rad} = (4\pi/c)\int B_\nu\,\mathrm{d}\nu = 4\sigma_{\rm SB}T^{4}/c$. That is $u_{\rm rad} = a_{\rm rad}T^4$ with $a_{\rm rad} = 4\sigma_{\rm SB}/c$. ∎

The run evaluates $a_{\rm rad} = 7.565733\times10^{-15}\ \mathrm{erg\,cm^{-3}\,K^{-4}}$ from the exact SI value of $\sigma_{\rm SB}$. No earlier module fixed a symbol for the Stefan–Boltzmann constant: Module 3's notation table fixes $a_{\rm rad}$ only, and Module 11 §8 writes the constant as bare $\sigma$ (module11.html:360). The factor $1/3$ is a statement about geometry and not about photons. Its trivial limit is a parallel beam, $I_\nu \propto \delta(\mathbf{n}-\mathbf{n}_0)$ along $z$: then $\cos\theta = 1$ wherever the intensity is non-zero, the three integrals in (1.1) are equal, and $P_\nu = u_\nu$. A beam pushes three times harder than an isotropic field of the same energy density. Problem C1 asks for both cases and for a third, a field that fills one hemisphere uniformly and is empty in the other, which also has $P_\nu = u_\nu/3$ — a fact that §5 uses at the surface of a star.

2. The transfer equation

$\nu$ in this module is a frequency, and never the kinematic viscosity. A reader who has just finished Module 11 will read $\nu$ as a viscosity, because Module 7, Module 10 and Module 11 all write it so. In this module the subscript in $I_\nu$, $\kappa_\nu$, $j_\nu$, $B_\nu$ and the bare $\nu$ in $\nu_{\rm max}$ are a frequency, in Hz, as in Module 4 §4's $\nu_{\rm max}$ and $\nu_{\rm ac}$. No viscosity occurs anywhere in this module.
Definition 2 (opacity, emissivity and the source function). A ray crossing a path length $\mathrm{d}s$ of gas of density $\rho$ loses the fraction $\kappa_\nu\rho\,\mathrm{d}s$ of its intensity to absorption and scattering together; $\kappa_\nu$ is the mass extinction coefficient, or opacity, in $\mathrm{cm^{2}\,g^{-1}}$. The same gas adds $j_\nu\rho\,\mathrm{d}s$, where $j_\nu$ is the emissivity, in $\mathrm{erg\,s^{-1}\,g^{-1}\,sr^{-1}\,Hz^{-1}}$. Their ratio $S_\nu \equiv j_\nu/\kappa_\nu$ is the source function. In thermal equilibrium at temperature $T$, $S_\nu = B_\nu(T)$.

Opacity is never written bare $\kappa$ here. The bare glyph already has three meanings in this book — Module 1's thermal conductivity, Module 4's evanescent decay rate, and Module 12's $\kappa_\parallel$, $\kappa_\perp$ — so this module writes $\kappa_\nu$ at one frequency, $\kappa_{\rm R}$ for the Rosseland mean of §4, $\kappa_{\rm es}$ for electron scattering and $\kappa_{\rm F}$ for the flux mean of §9. And the emissivity is $j_\nu$ and not the $\eta_\nu$ of many texts, because module12.html:927 refuses a new bare meaning of $\eta$.

Proposition 2 (the transfer equation and its formal solution). In a static medium, along a ray with path length $s$, $$ \frac{\mathrm{d}I_\nu}{\mathrm{d}s} \;=\; \rho\left(j_\nu - \kappa_\nu I_\nu\right), \qquad\text{equivalently}\qquad \frac{\mathrm{d}I_\nu}{\mathrm{d}\tau_\nu} \;=\; S_\nu - I_\nu, \tag{2.1}$$ where $\mathrm{d}\tau_\nu = \kappa_\nu\rho\,\mathrm{d}s$ is accumulated in the direction of travel from the point where the ray enters, $\tau_\nu = 0$. Its solution is $$ I_\nu(\tau_\nu) \;=\; I_\nu(0)\,e^{-\tau_\nu} \;+\; \int_0^{\tau_\nu} S_\nu(t)\,e^{-(\tau_\nu - t)}\,\mathrm{d}t. \tag{2.2}$$ The first form of (2.1) is stated; (2.2) is derived.
What is stated. The first form of (2.1) is energy conservation for the photons in a thin cylinder along the ray: what leaves one end equals what entered the other, less what the gas removed, plus what it emitted. It omits the term $(1/c)\partial I_\nu/\partial t$, which is negligible whenever the light-crossing time is short against the time on which the medium changes, and it assumes the gas is at rest. What is derived. Dividing the first form by $\kappa_\nu\rho$ and using $\mathrm{d}\tau_\nu = \kappa_\nu\rho\,\mathrm{d}s$ and $S_\nu = j_\nu/\kappa_\nu$ gives the second. Multiply it by $e^{\tau_\nu}$: the left side becomes $e^{\tau_\nu}(\mathrm{d}I_\nu/\mathrm{d}\tau_\nu + I_\nu) = \mathrm{d}(I_\nu e^{\tau_\nu})/\mathrm{d}\tau_\nu$, so $\mathrm{d}(I_\nu e^{\tau_\nu})/\mathrm{d}\tau_\nu = S_\nu e^{\tau_\nu}$. Integrating from $0$ to $\tau_\nu$ gives $I_\nu(\tau_\nu)e^{\tau_\nu} - I_\nu(0) = \int_0^{\tau_\nu}S_\nu(t)e^{t}\,\mathrm{d}t$, and multiplying by $e^{-\tau_\nu}$ gives (2.2). ∎

The trivial limit. For a uniform source function $S_\nu$, (2.2) integrates to $I_\nu = I_\nu(0)e^{-\tau_\nu} + S_\nu(1 - e^{-\tau_\nu})$. At $\tau_\nu \ll 1$ this is $I_\nu(0) + (S_\nu - I_\nu(0))\tau_\nu$: the gas barely changes the ray. At $\tau_\nu \gg 1$ it is $S_\nu$, whatever came in: a thick enough slab forgets the light that entered it and shows only its own source function, which is $B_\nu(T)$ in thermal equilibrium. That is why a star looks like a black body of roughly its surface temperature. When the extinction includes scattering, $S_\nu$ contains the scattered light and depends on $I_\nu$ itself, so (2.2) is then an integral equation and not a solution; §5 meets exactly that.

3. Optical depth

Definition 3 (optical depth and the photon mean free path). The optical depth is $\mathrm{d}\tau_\nu = \kappa_\nu\rho\,\mathrm{d}s$. In an atmosphere it is measured inward from the observer, with $\tau_\nu = 0$ at the top; along a ray in Proposition 2 the same increment is accumulated in the direction of travel. The photon mean free path is $\lambda_{\rm ph} = 1/(\kappa_\nu\rho)$ at one frequency, or $1/(\kappa_{\rm R}\rho)$ for the Rosseland mean of §4.

The glyph $\tau$ is already the optical depth in this book: Module 6's notation table claims it "in $\tau = 2/3$ only", and this module takes it over in full. By (2.2) with no source, a ray survives an optical depth $\tau$ with probability $e^{-\tau}$, so the mean optical depth a photon travels before an interaction is $\int_0^\infty\tau e^{-\tau}\,\mathrm{d}\tau = 1$ and the mean distance is $\lambda_{\rm ph}$. In the ATLAS9 solar model of §6, at $\tau_{\rm R} = 2/3$, the photon mean free path is $84.2$ km.

Proposition 3 (the random walk). A photon that is scattered, or absorbed and re-emitted, in a random direction at every step, and that must cross a slab of optical depth $\tau \gg 1$, takes of order $$ N \;\approx\; \tau^{2} \tag{3.1}$$ steps to do it. This is an order of magnitude and is printed as one.
Let the photon take $N$ steps $\boldsymbol{\ell}_i$, each of length $\lambda_{\rm ph}$ on average and in an independent random direction. The net displacement is $\mathbf{d} = \sum_i\boldsymbol{\ell}_i$, and $\langle|\mathbf{d}|^{2}\rangle = \sum_i\langle|\boldsymbol{\ell}_i|^{2}\rangle + \sum_{i\ne j}\langle\boldsymbol{\ell}_i\cdot\boldsymbol{\ell}_j\rangle = N\lambda_{\rm ph}^{2}$, because independent random directions make every cross term average to zero. The slab has thickness $L = \tau\lambda_{\rm ph}$. Setting the root-mean-square displacement $\lambda_{\rm ph}\sqrt{N}$ equal to $L$ gives $N = \tau^{2}$. The factors of order unity that this sets aside — the spread of step lengths, whether the displacement is measured in one dimension or three, and the absorbing boundary — are what §4 fixes. ∎

The consequence is that a star is opaque in time as well as in intensity. The photon travels a total path $N\lambda_{\rm ph} = \tau^{2}\lambda_{\rm ph} = \tau L$, which is $\tau$ times longer than the straight line across. §4 turns this into a diffusion coefficient with its factor fixed.

4. The diffusion limit

Deep inside a star, where $\tau \gg 1$ over any distance on which the temperature changes, the radiation field is almost isotropic and almost black-body. The flux is then the small anisotropy that a temperature gradient forces on it, and it obeys a diffusion law. The one freedom is which average of $\kappa_\nu$ over frequency appears in it, and that is settled by a definition.

Definition 4 (the Rosseland mean opacity). $$ \frac{1}{\kappa_{\rm R}} \;\equiv\; \frac{\displaystyle\int_0^\infty \frac{1}{\kappa_\nu}\,\frac{\partial B_\nu}{\partial T}\,\mathrm{d}\nu}{\displaystyle\int_0^\infty \frac{\partial B_\nu}{\partial T}\,\mathrm{d}\nu}. \tag{4.1}$$ It is defined so that Proposition 4 holds with a single opacity. It is a harmonic mean, so frequencies where the gas is transparent dominate it, and it is weighted by $\partial B_\nu/\partial T$, which is where the diffusing flux is carried.
Proposition 4 (radiative diffusion). In a plane-parallel, static medium with $\tau \gg 1$ over a temperature scale height, the frequency-integrated flux along $z$ is $$ F \;=\; -\,\frac{c}{3\kappa_{\rm R}\rho}\,\frac{\mathrm{d}}{\mathrm{d}z}\!\left(a_{\rm rad}T^{4}\right) \;=\; -\,\frac{16\sigma_{\rm SB}T^{3}}{3\kappa_{\rm R}\rho}\,\frac{\mathrm{d}T}{\mathrm{d}z}. \tag{4.2}$$ The assumption that carries the proof is the Eddington closure, $P_\nu = u_\nu/3$, which Proposition 1 made exact for an isotropic field and which is here applied to one that is only nearly isotropic.
With $\mu = \cos\theta$ and $z$ the coordinate normal to the layers, $\mathrm{d}s = \mathrm{d}z/\mu$, so the first form of (2.1) is $\mu\,\partial I_\nu/\partial z = \rho(j_\nu - \kappa_\nu I_\nu)$. Multiply by $\mu$ and integrate over solid angle. On the left, $\oint\mu^{2}\,\partial I_\nu/\partial z\,\mathrm{d}\Omega = c\,\mathrm{d}P_\nu/\mathrm{d}z$ by (1.1). On the right, the emission of a gas at rest is isotropic, so $\oint j_\nu\mu\,\mathrm{d}\Omega = 0$, and $\oint\kappa_\nu I_\nu\mu\,\mathrm{d}\Omega = \kappa_\nu F_\nu$. The first moment of the transfer equation is therefore exact: $$ c\,\frac{\mathrm{d}P_\nu}{\mathrm{d}z} \;=\; -\,\kappa_\nu\rho\,F_\nu. $$ Now apply the closure $P_\nu = u_\nu/3$, and use that at $\tau\gg1$ the field is black-body to leading order, $u_\nu = 4\pi B_\nu(T)/c$, so that $\mathrm{d}P_\nu/\mathrm{d}z = (4\pi/3c)(\partial B_\nu/\partial T)\,\mathrm{d}T/\mathrm{d}z$. Then $$ F_\nu \;=\; -\,\frac{4\pi}{3\kappa_\nu\rho}\,\frac{\partial B_\nu}{\partial T}\,\frac{\mathrm{d}T}{\mathrm{d}z}. $$ The hinge is the order of the two operations. Integrating over frequency, $\mathrm{d}T/\mathrm{d}z$ and $\rho$ come outside, and what is left inside is $\int(1/\kappa_\nu)(\partial B_\nu/\partial T)\,\mathrm{d}\nu$, which by (4.1) is $(1/\kappa_{\rm R})\int(\partial B_\nu/\partial T)\,\mathrm{d}\nu$. The remaining integral is $\mathrm{d}/\mathrm{d}T$ of $\int B_\nu\,\mathrm{d}\nu = \sigma_{\rm SB}T^4/\pi$, that is $4\sigma_{\rm SB}T^{3}/\pi$. Hence $F = -(4\pi/3\kappa_{\rm R}\rho)(4\sigma_{\rm SB}T^{3}/\pi)\,\mathrm{d}T/\mathrm{d}z = -(16\sigma_{\rm SB}T^3/3\kappa_{\rm R}\rho)\,\mathrm{d}T/\mathrm{d}z$, the second form of (4.2). Since $a_{\rm rad} = 4\sigma_{\rm SB}/c$, $\mathrm{d}(a_{\rm rad}T^4)/\mathrm{d}z = (16\sigma_{\rm SB}T^{3}/c)\,\mathrm{d}T/\mathrm{d}z$, which turns the second form into the first. ∎

The trivial limit is an isothermal slab. With $\mathrm{d}T/\mathrm{d}z = 0$, (4.2) gives $F = 0$ at any opacity: a region at one temperature is filled with isotropic black-body radiation, and by Proposition 1 an isotropic field carries no flux. The other check is on the weighting. Had one averaged $\kappa_\nu$ directly instead of $1/\kappa_\nu$, a single opaque line would dominate the mean and choke the flux, which is the opposite of what happens: radiation leaks out through the transparent windows, and the harmonic mean (4.1) says so.

Proposition 5 (the two diffusivities). Written as $F = -D\,\mathrm{d}u_{\rm rad}/\mathrm{d}z$, (4.2) has photon diffusivity $$ D \;=\; \frac{c\,\lambda_{\rm ph}}{3}, \qquad \lambda_{\rm ph} = \frac{1}{\kappa_{\rm R}\rho}, \qquad\text{so}\qquad t_{\rm diff} \;\sim\; \frac{R^{2}}{D} \;=\; \frac{3\kappa_{\rm R}\rho R^{2}}{c}. \tag{4.3}$$ Written as a heat flux in a gas of specific heat $c_P$ at constant pressure, the same law is a diffusion of temperature with diffusivity $$ \chi \;=\; \frac{16\sigma_{\rm SB}T^{3}}{3\kappa_{\rm R}\rho^{2}c_P}. \tag{4.4}$$ The time $R^2/D$ is an order of magnitude and is printed as one.
With $u_{\rm rad} = a_{\rm rad}T^4$, the first form of (4.2) reads $F = -(c/3\kappa_{\rm R}\rho)\,\mathrm{d}u_{\rm rad}/\mathrm{d}z$, so $D = c/(3\kappa_{\rm R}\rho) = c\lambda_{\rm ph}/3$. For a diffusion equation $\partial u/\partial t = D\,\partial^{2}u/\partial z^{2}$, a disturbance of size $R$ decays on a time of order $R^{2}/D$; the exact coefficient depends on the shape of the disturbance, which is why this is an order of magnitude. Substituting $D$ gives $3\kappa_{\rm R}\rho R^{2}/c$. For (4.4), write the second form of (4.2) as $F = -K_{\rm rad}\,\mathrm{d}T/\mathrm{d}z$ with conductivity $K_{\rm rad} = 16\sigma_{\rm SB}T^3/(3\kappa_{\rm R}\rho)$. Energy conservation in a gas heated only by the divergence of this flux at constant pressure is $\rho c_P\,\partial T/\partial t = -\partial F/\partial z = \partial(K_{\rm rad}\,\partial T/\partial z)/\partial z$, which for locally uniform $K_{\rm rad}$ is a diffusion equation for $T$ with $\chi = K_{\rm rad}/(\rho c_P)$, which is (4.4). ∎

The factor $3$ in $t_{\rm diff}$ is the $1/3$ of Proposition 4, and it is the same $1/3$ as in Module 1 §6's kinetic-theory transport coefficients, $\mu \simeq \tfrac13\rho v\lambda$ (module01.html:363): a flux carried by carriers of speed $v$ and free path $\lambda$ has diffusivity $\tfrac13 v\lambda$, and for photons $v = c$. Proposition 3's random walk had no such factor, because it did not fix the geometry; this is the factor it set aside. The step-1 version of this module's generator carried the $3$ as "the standard three-dimensional correction", with no source; Gate D ruled that it be derived rather than sourced, and it is derived here.

$\chi$ is the diffusivity that Module 4 §7 named "the radiative diffusivity of the stellar gas" and did not evaluate. §7 evaluates it.

5. The grey atmosphere

At the surface of a star the diffusion law must fail, because $\tau$ goes to zero and the radiation field becomes one-sided. The simplest model that carries the radiation from the diffusion regime out to the surface is a grey atmosphere, in which $\kappa_\nu$ does not depend on frequency, in radiative equilibrium, in which every layer emits exactly the energy it absorbs so that the flux is constant with depth. That model has a closed-form temperature profile under the Eddington closure.

Proposition 6 (Eddington's grey atmosphere). A plane-parallel grey atmosphere in radiative equilibrium and local thermal equilibrium, under the Eddington closure $P_{\rm rad} = u_{\rm rad}/3$ and with no radiation incident from outside, carrying the flux $F = \sigma_{\rm SB}T_{\rm eff}^{4}$, has $$ T^{4}(\tau) \;=\; \tfrac34\,T_{\rm eff}^{4}\left(\tau + \tfrac23\right), \tag{5.1}$$ so that $T = T_{\rm eff}$ at $\tau = 2/3$. The exact grey solution, without the closure, replaces $2/3$ by the Hopf function $q_{\rm H}(\tau)$; that replacement is stated and not derived.
Grey means that $\kappa_\nu$ is the same at every $\nu$, so it equals its own Rosseland mean, $\kappa_\nu = \kappa_{\rm R}$. Let $\tau$ increase inward, $\mathrm{d}\tau = -\kappa_{\rm R}\rho\,\mathrm{d}z$ with $z$ outward, and $\mu = \cos\theta$ from the outward normal. The transfer equation (2.1) reads $\mu\,\mathrm{d}I/\mathrm{d}\tau = I - S$, frequency-integrated because $\kappa_\nu$ does not depend on $\nu$. Integrate over solid angle: $\mathrm{d}F/\mathrm{d}\tau = 4\pi(J - S)$. Radiative equilibrium says $F$ is constant, so $J = S$ at every depth. Multiply by $\mu$ and integrate: $c\,\mathrm{d}P_{\rm rad}/\mathrm{d}\tau = F$, so $P_{\rm rad} = F\tau/c + C$. The closure $P_{\rm rad} = u_{\rm rad}/3 = 4\pi J/3c$ converts this into $J(\tau) = (3F/4\pi)\tau + 3cC/4\pi$. The hinge is the surface condition, which fixes $C$. At $\tau = 0$ nothing comes in, and the Eddington approximation takes the outgoing intensity as uniform over the outward hemisphere, $I = I_0$ for $\mu \gt 0$ and $I = 0$ for $\mu \lt 0$. Then $J(0) = I_0/2$ and $F = 2\pi\int_0^1 I_0\mu\,\mathrm{d}\mu = \pi I_0$, so $J(0) = F/2\pi$. (Such a hemispheric field also has $P_{\rm rad} = u_{\rm rad}/3$, by Problem C1, so the surface condition does not contradict the closure.) Hence $J(\tau) = (3F/4\pi)(\tau + 2/3)$. In local thermal equilibrium $S = \int B_\nu\,\mathrm{d}\nu = \sigma_{\rm SB}T^{4}/\pi$, and $J = S$ gives $\sigma_{\rm SB}T^4/\pi = (3\sigma_{\rm SB}T_{\rm eff}^4/4\pi)(\tau + 2/3)$, which is (5.1). At $\tau = 2/3$ the bracket is $4/3$ and $T = T_{\rm eff}$. What is stated: the exact solution of $J = \Lambda[J]$, the integral equation that (2.2) becomes when $S = J$, is $J = (3F/4\pi)(\tau + q_{\rm H}(\tau))$; this module computes $q_{\rm H}$ numerically and does not derive it. ∎

The trivial limit is $\tau \to \infty$, and it closes a loop. Deep in the atmosphere (5.1) gives $\mathrm{d}T^4/\mathrm{d}\tau = \tfrac34T_{\rm eff}^4$. Put that into Proposition 4 in its optical-depth form, $F = (c/3)\,\mathrm{d}(a_{\rm rad}T^4)/\mathrm{d}\tau$: the flux is $(ca_{\rm rad}/3)(\tfrac34T_{\rm eff}^4) = (ca_{\rm rad}/4)T_{\rm eff}^4 = \sigma_{\rm SB}T_{\rm eff}^4$ exactly, because $ca_{\rm rad} = 4\sigma_{\rm SB}$. The grey atmosphere at depth is the diffusion law carrying its own flux. The run evaluates the ratio as $1.000000$ (Problem D2). This is a consistency check and cannot fail: the same moment equation and closure derived (5.1).

The Hopf function differs from $2/3$ by little. The run solves the exact grey problem and gets $q_{\rm H}(0) = 0.57730$ against the known $1/\sqrt3 = 0.57735$, and $q_{\rm H}(2/3) = 0.68876$. The closure therefore moves $T$ by $0.412$ per cent at $\tau = 2/3$, and by $3.0$ per cent at $\tau = 0.01$; the paragraph after CHECK 4 below uses both.

Which $T_{\rm eff}$. The check below uses $5777$ K, because the model atmosphere it is compared against was built at that value and says so in its header. Everywhere else this book uses the IAU nominal $5772$ K, which is $0.087$ per cent lower. Comparing a model to (5.1) at a $T_{\rm eff}$ it was not built for would put a spurious $0.087$ per cent into every row.

CHECK 4 — CONFIRMED, to 3.0 per cent in T. Eddington's grey profile (5.1) against the temperature structure of Castelli's ATLAS9 solar model, read at seven Rosseland optical depths:
$\tau_{\rm R}$ATLAS9 $T$ (K)grey $T$ (K)grey/ATLAS9
0.014733.54876.01.0301
0.15081.25030.60.9900
0.35455.15330.70.9772
2/35910.15777.00.9775
16228.36108.40.9807
26973.46870.00.9852
37481.87439.40.9943
The largest miss is $0.0301$ in $T$, which is $0.1259$ — 12.6 per cent — in $T^4$, and the flux goes as $T^4$. The band of 3.1 per cent was set after this table was computed, so this check measures the size of the grey model's error; it does not test a prediction made blind.

Fig. 1 draws both profiles from $\tau_{\rm R} = 0.01$ to $3$, with the table's seven levels as dots on the residual panel.

0.01 0.1 1 5000 5500 6000 6500 7000 7500 -3 0 +3 ATLAS9 model, T(τR) grey Eddington atmosphere, Teff = 5777 K 5910.1 K T = Teff τR = 2/3 ±3.1 per cent band grey/ATLAS9 − 1, per cent T (K) Rosseland optical depth τR
CHECK 4. Upper panel: temperature against Rosseland optical depth in the ATLAS9 solar model (solid) and in Eddington's grey atmosphere (5.1) at the model's own $T_{\rm eff} = 5777$ K (dashed). Lower panel: grey/ATLAS9 $-\,1$ in per cent, with the seven levels of the table as dots and the $\pm3.1$ per cent band shaded. The residual panel is the check. The vertical dotted line is $\tau_{\rm R} = 2/3$, where the grey model gives $T_{\rm eff}$ by construction and ATLAS9 gives $5910.1$ K.

What the check tests, and what it cannot separate. At $\tau = 2/3$ the grey model gives $T = T_{\rm eff}$ by construction, so the $2.30$ per cent miss there is the whole finding: ATLAS9's $\tau_{\rm R} = 2/3$ layer is $133.1$ K hotter than $T_{\rm eff}$. The exact grey solution raises $T(2/3)$ by only $0.412$ per cent, so $1.89$ of the $2.30$ per cent is not the Eddington closure. What it is, this check does not say: ATLAS9 has a frequency-dependent opacity and a mixing-length convection zone with $l/H = 1.25$, and both are absent from (5.1). At $\tau = 0.01$ the balance is the other way round. There the exact grey solution multiplies the grey temperature by $0.9697$, which would bring grey/ATLAS9 from $1.0301$ to $0.9988$: at the top of the atmosphere the closure, and not the grey opacity, is the error.

This check replaces the CHECK 4 that step 1 of this module's preparation designed — Eddington's standard-model $1 - \beta$ against the tabulated Sun, a ratio of $2.30$. That ratio is Module 3's result: its Problem K3 solution (module03.html:885) prints it and argues that it does not refute $n = 3$. It is cited in §8 and not re-checked here.

6. The anchor: the opacity of the solar photosphere

Module 6 §6 computed the photospheric scale height $H = 142.9$ km and density $\rho = 3.0631\times10^{-7}\ \mathrm{g\,cm^{-3}}$ (module06.html:496–:497) from four inputs: $T = 5772$ K, $\mu = 1.2250$, the IAU surface gravity $g = 2.7420\times10^{4}\ \mathrm{cm\,s^{-2}}$, and a gas pressure $P = 1.2\times10^{5}\ \mathrm{dyn\,cm^{-2}}$ at $\tau = 2/3$, which module06.html:490 says "comes from a solar atmosphere calculation". It never wrote down an opacity. Yet the photosphere is defined by one: it is the layer at optical depth $2/3$. Those inputs therefore imply an opacity, and the implication can be checked against a published model atmosphere.

The inference, and what it measures. Let $m$ be the mass per unit area above a level, the column mass. The optical depth of that column is $\tau = \int_0^m\kappa_{\rm R}\,\mathrm{d}m'$ whether or not the opacity varies, so $\tau/m$ is exactly the mass-weighted mean opacity of the column, $\bar\kappa_{\rm R}$. Hydrostatic balance, $\mathrm{d}P = g\,\mathrm{d}m$ with $P = 0$ at the top, gives $m = P/g$, and for an ideal gas $P/g = \rho H$ at any level. Demanding $\tau = 2/3$ at Module 6's level therefore gives

$$ \bar\kappa_{\rm R} \;=\; \frac{2/3}{\rho H} \;=\; \frac{(2/3)\,g}{P} \;=\; 0.15231\ \mathrm{cm^{2}\,g^{-1}}. \tag{6.1}$$

The number is from Module 6's printed $\rho$ and $H$; its inputs $P$ and $g$ give $0.15233$, and the difference is the rounding of the printed $\rho H = 4.3772$ against $P/g = 4.3764\ \mathrm{g\,cm^{-2}}$. $\mu$ and $T$ cancel: (6.1) is a statement about the pressure at $\tau = 2/3$, and nothing in it comes from convection. It is a mean over the column, and not the opacity at the bottom of it. The two agree only if $\kappa_{\rm R}$ is the same all the way up. By construction the photon mean free path at that mean opacity is $H/\tau = 1.5000\,H$.

The target is Castelli's ATLAS9 model of the solar atmosphere, which tabulates the column mass and the Rosseland mean at every depth together with the temperature and pressure; the run integrates $\tau_{\rm R} = \int\kappa_{\rm R}\,\mathrm{d}m$ down the column and rebuilds the density from the pressure, the electron density and the file's own abundances. ATLAS9 reaches $\tau_{\rm R} = 2/3$ at a column mass of $3.8123\ \mathrm{g\,cm^{-2}}$, so its column mean there is $\bar\kappa_{\rm R} = 0.17487\ \mathrm{cm^{2}\,g^{-1}}$; at $T = 5772$ K, $\tau_{\rm R} = 0.5430$, it is $0.15470$. ATLAS9 gives $\kappa_{\rm R} = 0.4447\ \mathrm{cm^{2}\,g^{-1}}$ at $\tau_{\rm R} = 2/3$, where $T = 5910.1$ K and $\rho = 2.6721\times10^{-7}\ \mathrm{g\,cm^{-3}}$, and $\kappa_{\rm R} = 0.3596$ at $T = 5772$ K, which it reaches at $\tau_{\rm R} = 0.5430$. As a test of the rebuild, the file's pressure satisfies hydrostatic balance $P = g\,m$ at its deepest layer to $0.9999$.

CHECK 2, THE ANCHOR — CONFIRMED as a column mean, to a factor 1.1482; REFUTED as a local opacity. Against ATLAS9's column mean, the inference (6.1) is low by 1.0157 at $T = 5772$ K and 1.1482 at $\tau_{\rm R} = 2/3$. The tolerance is the spread of the inputs themselves: Module 6's density is $1.1463$ times ATLAS9's at $\tau_{\rm R} = 2/3$, and ATLAS9's column mean differs between the two levels by $1.1304$; together that is $1.2958$. Both ratios lie inside it, and the run asserts it. By $m = P/g$ the $1.1482$ is Module 6's column mass over ATLAS9's, and Module 6's input pressure over ATLAS9's, $1.2\times10^{5}$ against $1.0447\times10^{5}$, is $1.1487$; the two differ by the rounding of Module 6's printed $\rho$ and $H$ and by the two gravities, $2.7420\times10^{4}$ against ATLAS9's $10^{4.4377}$. The check tests that borrowed pressure and nothing else. Read as the local opacity at $\tau = 2/3$, (6.1) is low by 2.361 to 2.920, because ATLAS9's $\kappa_{\rm R}$ rises from $0.0424$ at $\tau_{\rm R} = 0.01$ to $0.4447$ at $2/3$, and a mean over the column lies below the value at its base.

This verdict reverses the one the draft printed, and the tolerance was set after the number was known. The draft compared (6.1) with ATLAS9's local $\kappa_{\rm R}$, under a factor-2 tolerance fixed in advance, and printed a refutation. The editor's pass found that the two numbers are different quantities. The comparison above is the like-for-like one, and its tolerance, $1.2958$, is the same recipe as the draft's — Module 6's density ratio times the spread between the two levels — applied to the column-mean pair. It was computed after the $1.1482$ was known, so, like CHECK 4's band, it measures the agreement and is not a prediction made blind.

What is refuted is the reading of (6.1) as a local opacity, and Fig. 2 shows why. The inference averages $\kappa_{\rm R}$ over the whole column from $\tau_{\rm R} = 0$ to $2/3$, across which ATLAS9's $\kappa_{\rm R}$ rises by a factor of $10.48$ from $\tau_{\rm R} = 0.01$. It keeps rising below: from $0.3596$ at $\tau_{\rm R} = 0.543$ to $0.7274$ at $1$, a factor of $2.0228$ in $32.6$ km. A column mean cannot be recovered as a local value when the opacity changes that fast. §7 needs the local value, and that is why it takes $\kappa_{\rm R}$ from ATLAS9 and not from (6.1).

0.01 0.1 1 0.03 0.1 0.3 1 3 inferred column mean (6.1): 0.15231 fully ionised electron scattering, X = 0.7583: 0.34966 ATLAS9 κR(τR) 0.3596 at T = 5772 K 0.7274 at τR = 1 the averaged column ATLAS9 column mean τR/m κR (cm2 g−1) Rosseland optical depth τR
CHECK 2. ATLAS9's Rosseland mean opacity against Rosseland optical depth (solid), and ATLAS9's mean over the column above each depth, $\tau_{\rm R}/m$ (dotted), both axes logarithmic. Dashed lines: the column mean inferred from Module 6's pressure, (6.1), and the fully ionised electron-scattering opacity at $X = 0.7583$. The shaded band is the column that (6.1) averages, $\tau_{\rm R} = 0.01$ to $2/3$. The filled dots are the local $\kappa_{\rm R}$ at $T = 5772$ K, at $\tau_{\rm R} = 2/3$ ($0.4447$) and at $\tau_{\rm R} = 1$; the hollow dot is the column mean at $\tau_{\rm R} = 2/3$, $0.17487$. There the inference sits $1.1482$ below the column-mean curve and $2.920$ below the local curve. The electron-scattering line is not the photosphere's electron scattering, which is $5659.8$ times lower, $6.1780\times10^{-5}\ \mathrm{cm^2\,g^{-1}}$, off the bottom of the plot; see the text.

The limit of the target, stated in print. ATLAS9 is a model, with its own opacity distribution functions, its own convection and — per the file's own title card — Asplund et al. (2005) abundances; it is not a measurement. It is the right kind of number for this check, because it tabulates $\kappa_{\rm R}$ with the $T$ and $\rho$ at which it applies, and a check against it is a check against a published model and not against the Sun.

Why the photosphere is not electron-scattering

Electron scattering is the one opacity with no atomic physics in it: every free electron scatters with the Thomson cross-section $\sigma_{\rm T}$, whatever its energy (for $kT \ll m_{\rm e}c^2$). A fully ionised gas of hydrogen mass fraction $X$, the rest helium, has $n_{\rm e} = (\rho/2m_{\rm p})(1+X)$ electrons per unit volume — one per hydrogen nucleus, two per helium nucleus of mass $4m_{\rm p}$ — so

$$ \kappa_{\rm es} \;=\; \frac{\sigma_{\rm T}n_{\rm e}}{\rho} \;=\; \frac{\sigma_{\rm T}}{2m_{\rm p}}(1+X). \tag{6.2}$$

At $X = 1$ this is $\sigma_{\rm T}/m_{\rm p} = 0.39773\ \mathrm{cm^2\,g^{-1}}$, and at the Sun's present surface composition, $X = 0.7583$, it is $0.34966$. (The round formula $0.2(1+X)$ gives $0.3517$, which differs by $0.572$ per cent.) Heavier elements, fully ionised, contribute about one electron per two nucleon masses, like helium, so (6.2) holds for them too.

The photosphere is not fully ionised, and that, not the size of its opacity, is why electron scattering does not set it. At $5910$ K hydrogen is almost entirely neutral, and ATLAS9's own electron density at $\tau_{\rm R} = 2/3$ is $2.4815\times10^{13}\ \mathrm{cm^{-3}}$, supplied mostly by easily ionised metals. With those electrons, $\sigma_{\rm T}n_{\rm e}/\rho = 6.1780\times10^{-5}\ \mathrm{cm^2\,g^{-1}}$ — $5659.8$ times below the fully ionised value, and $7198.5$ times below ATLAS9's $\kappa_{\rm R}$. The opacity that fills that gap is, by the standard attribution this module does not derive, bound–free and free–free absorption by the negative hydrogen ion, H⁻, a hydrogen atom that has captured one of those scarce electrons. This book does no atomic physics, and H⁻ is refused here: its cross-sections are what ATLAS9 contains and what this module borrows.

The preparation run reached the same conclusion by a second route. For fully ionised gas every $\kappa_\nu \ge \kappa_{\rm es}$, so by (4.1) $\kappa_{\rm R} \ge \kappa_{\rm es}$, and a column that is fully ionised throughout has a column mean of at least $0.34966$. The column mean above $\tau_{\rm R} = 2/3$ is $0.15231$ by (6.1) and $0.17487$ in ATLAS9, both below that floor, so the column is not fully ionised. ATLAS9's local $\kappa_{\rm R} = 0.4447$ lies above the floor, which is allowed either way and decides nothing. The electron-density argument above and this bound agree.

7. Heat that leaks out of a sound wave

Module 4 §7 derived the Sun's acoustic modes on the assumption that every compression is adiabatic, and then wrote: "By (3.4) it requires $\omega\chi/c^2\ll1$, with $\chi$ now the radiative diffusivity of the stellar gas. That criterion is not evaluated for the Sun in this module. It is expected to hold deep in the star, where photons diffuse slowly, and to fail near the photosphere, where they escape; neither statement is computed here, and the size of the failure is a subject of Module 13" (module04.html:429). It is the book's one explicitly deferred computation, and it comes with a prediction this section can test at one end. In Module 4's criterion $c$ is the sound speed; in this module $c$ is the speed of light, so the sound speed is written $c_{\rm s}$ and the criterion is $\omega\chi/c_{\rm s}^2 \ll 1$.

Proposition 7 (the criterion is a ratio of two times). For a sound wave of cyclic frequency $\nu$, angular frequency $\omega = 2\pi\nu$, speed $c_{\rm s}$ and wavelength $c_{\rm s}/\nu$, in a gas of thermal diffusivity $\chi$, $$ \frac{\omega\chi}{c_{\rm s}^{2}} \;=\; \frac{2\pi}{t_{\rm diff}/\mathcal{P}}, \tag{7.1}$$ where $t_{\rm diff} = (c_{\rm s}/\nu)^2/\chi$ is the time for heat to diffuse across one wavelength and $\mathcal{P} = 1/\nu$ is the period. Adiabatic means $t_{\rm diff}/\mathcal{P} \gg 2\pi$.
$t_{\rm diff}/\mathcal{P} = [(c_{\rm s}/\nu)^2/\chi]\,\nu = c_{\rm s}^{2}/(\nu\chi)$. Then $2\pi/(t_{\rm diff}/\mathcal{P}) = 2\pi\nu\chi/c_{\rm s}^{2} = \omega\chi/c_{\rm s}^2$. ∎

So the criterion is not about a scale height at all. It asks whether heat has time to diffuse across a wavelength during one oscillation. If it does not, the compressions keep their heat and the wave is adiabatic; if it does, temperature differences smooth out within a cycle and the wave approaches isothermal. At order unity it is neither, and the wave loses energy to the radiation every cycle. The run checks (7.1) numerically to $10^{-9}$.

The inputs are $\nu_{\rm max} = 3090\ \mu$Hz, so $\omega = 1.941504\times10^{-2}\ \mathrm{rad\,s^{-1}}$; Module 4's photospheric sound speed $c_{\rm s} = 8.0810\times10^{5}\ \mathrm{cm\,s^{-1}}$; Module 6's $c_P = 1.6968\times10^{8}\ \mathrm{erg\,g^{-1}\,K^{-1}}$; and $\kappa_{\rm R}$, $\rho$ and $T$ from ATLAS9 at two levels. By (4.4), $\chi = 1.5027\times10^{13}\ \mathrm{cm^2\,s^{-1}}$ at $T = 5772$ K and $1.1587\times10^{13}$ at $\tau_{\rm R} = 2/3$.

CHECK 3 — the adiabatic assumption is REFUTED at the photosphere, and module04.html:429's prediction that it fails there is CONFIRMED.
level$\chi$ ($\mathrm{cm^2\,s^{-1}}$)$\omega\chi/c_{\rm s}^2$$t_{\rm diff}/\mathcal{P}$
$T = 5772$ K$1.5027\times10^{13}$0.446814.06
$\tau_{\rm R} = 2/3$$1.1587\times10^{13}$0.344518.24
The criterion is $\ll 1$, and $0.34$ to $0.45$ is not. Adiabatic wants the diffusion time across one wavelength to be many periods; at the photosphere it is $14$ to $18$.

What CHECK 2 moved. With the column-mean opacity (6.1), $0.15231$, used as if it were the local opacity, and Module 6's density, the same calculation gives $\chi = 2.398397\times10^{13}\ \mathrm{cm^2\,s^{-1}}$ and $\omega\chi/c_{\rm s}^2 = 0.7131$, a diffusion time across the $2615.2$ km wavelength of $2.8516\times10^{3}$ s against a period of $3.2362\times10^{2}$ s — $8.81$ periods. That value is superseded and is printed once, here, to show what the anchor check changed: the ATLAS9 opacity and density move the criterion from $0.7131$ to between $0.3445$ and $0.4468$, and the verdict with it from "order unity" to "a few tenths". Fig. 3 draws all three values.

0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.7131 inferred opacity, 0.15231 column mean, not local 0.4468 ATLAS9, T = 5772 K κR = 0.3596 0.3445 ATLAS9, τR = 2/3 κR = 0.4447 ≪ 1 the boundary, 1 ωχ/cs2 at νmax = 3090 μHz The shaded strip ends at 0.1 by convention: the criterion says much less than one and names no edge.
CHECK 3. The adiabaticity parameter $\omega\chi/c_{\rm s}^2$ at $\nu_{\rm max}$ for three opacities. The hollow marker, $0.7131$, uses the column-mean opacity (6.1) as a local one, which CHECK 2 shows it is not, and is superseded. The two filled markers use ATLAS9's $\kappa_{\rm R}$ and $\rho$ at two levels. The dashed line at $1$ is where, by (7.1), the diffusion time across a wavelength is $2\pi$ periods. The shaded strip marks "much less than one"; its edge at $0.1$ is a convention, because the criterion names none. No marker lies in it.

What is not computed. Module 4 §8 (module04.html:563) said the failure "damps the modes and shifts their frequencies". This check establishes the regime — that radiative leakage is not negligible at the photosphere: the parameter is $0.3445$ to $0.4468$, against a criterion of $\ll 1$ — and it does not compute the damping rate or the frequency shift. Those need the non-adiabatic oscillation equations, which this book does not write; §12 lists that debt as half paid. Module 4 expected the criterion to hold deep in the star, where $\rho$ is larger and $\chi \propto T^3/(\kappa_{\rm R}\rho^2)$ is smaller; this module does not evaluate it there.

8. A gas that contains radiation

Module 2 §11 listed $\gamma = 5/3$ among its idealisations and said it is wrong "for radiation-dominated interiors where $\gamma\to4/3$" (module02.html:688). This section pays that debt with the general statement: the adiabatic index of a mixture of an ideal monatomic gas and black-body radiation, at any gas fraction.

Let the total pressure be $P = P_{\rm gas} + P_{\rm rad}$ with $P_{\rm gas} = \rho k_BT/(\mu m_u)$ and $P_{\rm rad} = a_{\rm rad}T^{4}/3$, and let $\beta \equiv P_{\rm gas}/P$, Module 3's meaning exactly. The specific internal energy is written $e$, as in Module 2, and it is the gas's $\tfrac32P_{\rm gas}/\rho$ plus the radiation's $u_{\rm rad}/\rho$.

Proposition 8 (the first adiabatic index of a gas-plus-radiation mixture). With $e = 3P_{\rm gas}/(2\rho) + a_{\rm rad}T^{4}/\rho$, the adiabatic index $\Gamma_1 \equiv (\partial\ln P/\partial\ln\rho)_{\rm ad}$ is $$ \Gamma_1 \;=\; \beta \;+\; \frac{(4-3\beta)^{2}(\gamma-1)}{\beta + 12(\gamma-1)(1-\beta)}, \qquad \gamma = \tfrac53, \tag{8.1}$$ which is $5/3$ at $\beta = 1$ and $4/3$ at $\beta = 0$.
Work per unit mass with $v = 1/\rho$ and write $\mathcal{R} = k_B/(\mu m_u)$, so $P_{\rm gas} = \mathcal{R}T/v$ and $e = \tfrac32\mathcal{R}T + a_{\rm rad}T^{4}v$. The first law for an adiabatic change is $\mathrm{d}e + P\,\mathrm{d}v = 0$. Differentiating $e$, $$ \left(\tfrac32\mathcal{R} + 4a_{\rm rad}T^{3}v\right)\mathrm{d}T \;+\; \left(a_{\rm rad}T^{4} + \frac{\mathcal{R}T}{v} + \frac{a_{\rm rad}T^{4}}{3}\right)\mathrm{d}v \;=\; 0. $$ Express every term through $P$ and $\beta$: $\mathcal{R}T/v = \beta P$ and $a_{\rm rad}T^4 = 3(1-\beta)P$. The first bracket times $\mathrm{d}T$ is $Pv\,(\mathrm{d}T/T)\left[\tfrac32\beta + 12(1-\beta)\right]$, and the second bracket is $P\left[3(1-\beta) + \beta + (1-\beta)\right] = P(4 - 3\beta)$. With $\mathrm{d}v/v = -\mathrm{d}\ln\rho$, $$ \left.\frac{\mathrm{d}\ln T}{\mathrm{d}\ln\rho}\right|_{\rm ad} \;=\; \frac{4-3\beta}{\tfrac32\beta + 12(1-\beta)} \;=\; \frac{(4-3\beta)(\gamma-1)}{\beta + 12(\gamma-1)(1-\beta)}, $$ the second form multiplying top and bottom by $\gamma - 1 = 2/3$. Now differentiate the pressure itself: $\mathrm{d}P = P_{\rm gas}(\mathrm{d}\ln\rho + \mathrm{d}\ln T) + 4P_{\rm rad}\,\mathrm{d}\ln T$, so $$ \mathrm{d}\ln P \;=\; \beta\,\mathrm{d}\ln\rho \;+\; (4 - 3\beta)\,\mathrm{d}\ln T. $$ Dividing by $\mathrm{d}\ln\rho$ along the adiabat and substituting the previous line gives (8.1). Both limits. At $\beta = 1$ the fraction is $1\cdot(\gamma-1)/1$, so $\Gamma_1 = 1 + (\gamma - 1) = \gamma = 5/3$. At $\beta = 0$ it is $16(\gamma-1)/[12(\gamma-1)] = 4/3$, and $\Gamma_1 = 4/3$. ∎

(8.1) is Chandrasekhar's eq. (131), p. 57, of his 1939 Introduction to the Study of Stellar Structure, and he states both limits under it himself. His Table 1, p. 59, tabulates $\Gamma_1$ for $\gamma = 5/3$ at $1 - \beta = 0, 0.1, \ldots, 1$ to three decimals, and the run reproduces all eleven rows to $3.44\times10^{-4}$, inside the $5\times10^{-4}$ that three printed decimals allow. Selected values: $\Gamma_1 = 1.650997$ at $\beta = 0.99$, $1.562745$ at $0.9$, $1.425926$ at $0.5$ and $1.350228$ at $0.1$. Radiation lowers $\Gamma_1$ quickly: a radiation share of one per cent, $\beta = 0.99$, already takes it from $1.666667$ to $1.650997$.

Two different claims, and Module 2 was promised the first. (8.1) is a local statement: at any point, the adiabatic index is set by the local $\beta$. Module 3 §5's Proposition 6 (module03.html:366) is a global one — that a star with $\beta$ and $\mu$ uniform throughout is a polytrope $P = K\rho^{4/3}$ — and Module 3 §10 (module03.html:653) refutes that premise for the Sun. The two share the number $4/3$ and nothing else. module02.html:688 asked for the local statement, and Proposition 8 gives it at every $\beta$, with $4/3$ as its $\beta\to0$ limit.

At the Sun's centre, Module 3 §9 computed $1 - \beta_c = 6.1940\times10^{-4}$ from the tabulated model (module03.html:628); this module's run reproduces it as $6.193970\times10^{-4}$. By (8.1) that gives $\Gamma_1 = 1.665639$, short of $5/3$ by $1.0276\times10^{-3}$. Module 3's Problem K3 (module03.html:885) already sets Eddington's standard-model value, $1.4247\times10^{-3}$, against the tabulated one and finds it $2.30$ times larger; that is Module 3's result and is cited here, not re-checked.

Proposition 9 (where radiation pressure equals gas pressure). At temperature $T$, $P_{\rm rad} = P_{\rm gas}$ at the density $$ \rho_{\rm eq}(T) \;=\; \frac{a_{\rm rad}\,\mu m_u\,T^{3}}{3k_B}, \tag{8.2}$$ and radiation dominates the pressure where $\rho \lt \rho_{\rm eq}(T)$. At any point, $\rho/\rho_{\rm eq}(T) = \beta/(1-\beta)$.
Setting $\rho k_BT/(\mu m_u) = a_{\rm rad}T^4/3$ and solving for $\rho$ gives (8.2). At fixed $T$, $P_{\rm gas}$ is proportional to $\rho$ and $P_{\rm rad}$ does not depend on it, so $P_{\rm gas} \lt P_{\rm rad}$ exactly when $\rho \lt \rho_{\rm eq}$. Finally $P_{\rm gas}/P_{\rm rad} = \rho/\rho_{\rm eq}$ by the same proportionality, and $P_{\rm gas}/P_{\rm rad} = \beta/(1-\beta)$ by the definition of $\beta$. ∎

At $\mu = 0.832$, the value Module 3's Problem K3 uses, (8.2) gives $\rho_{\rm eq} = 2.5236\times10^{-5}\ \mathrm{g\,cm^{-3}}$ at $10^{6}$ K and $25.236$ at $10^{8}$ K. §11 draws the line.

9. The Eddington limit

Now the destination. A source of luminosity $L$ and mass $M$ sits at the centre, and ionised gas surrounds it. Gravity pulls each gram inward. The radiation streaming outward pushes each gram outward, by the momentum it deposits. Both forces fall as $1/r^2$.

Proposition 10 (the Eddington luminosity). Let gas at radius $r$ have flux-mean opacity $\kappa_{\rm F} \equiv (1/F)\int\kappa_\nu F_\nu\,\mathrm{d}\nu$, and let the source radiate $L$ isotropically. Gravity holds the gas only if $$ L \;\lt\; L_{\rm Edd} \;\equiv\; \frac{4\pi GMc}{\kappa_{\rm F}}, \tag{9.1}$$ independent of $r$. Electron scattering is the one opacity with no atomic physics in it and, by the argument of §6, a floor under the opacity of fully ionised gas at every frequency, so $\kappa_{\rm F} \ge \kappa_{\rm es}$. Taking $\kappa_{\rm F} = \kappa_{\rm es} = \sigma_{\rm T}(1+X)/(2m_{\rm p})$ gives the largest $L_{\rm Edd}$ such gas can have, and at $X = 1$ this is $$ L_{\rm Edd} \;=\; \frac{4\pi GMm_{\rm p}c}{\sigma_{\rm T}}, \tag{9.2}$$ which is the formula module09.html:944 states.
The flux at radius $r$ is $F = L/(4\pi r^2)$, and it carries momentum $F/c$ per unit area per unit time. By Proposition 2, a layer of thickness $\mathrm{d}s$ removes the fraction $\kappa_\nu\rho\,\mathrm{d}s$ of the flux at each frequency. Absorbed photons deposit their momentum; photons scattered symmetrically forward and backward, as Thomson scattering does, deposit it on average too. The momentum removed per unit volume per unit time is therefore $\rho\int\kappa_\nu F_\nu\,\mathrm{d}\nu/c = \kappa_{\rm F}\rho F/c$, and per unit mass the outward force is $\kappa_{\rm F}F/c = \kappa_{\rm F}L/(4\pi r^{2}c)$. Gravity per unit mass is $GM/r^2$ inward. The gas is held when $$ \frac{\kappa_{\rm F}L}{4\pi r^{2}c} \;\lt\; \frac{GM}{r^{2}}, $$ and the $r^2$ cancels, leaving (9.1). For electron scattering, $\sigma_{\rm T}$ does not depend on frequency, so $\kappa_{\rm F} = \kappa_{\rm es}$ whatever the spectrum, and (6.2) gives $\kappa_{\rm es}$. At $X = 1$, $\kappa_{\rm es} = \sigma_{\rm T}/m_{\rm p}$, and (9.1) becomes (9.2). The radiation pushes on the electrons and gravity pulls mainly on the protons; the electric field that holds a plasma neutral couples the two, which is why the force balance is taken per unit mass of plasma. ∎

The trivial limit is $\kappa_{\rm F} \to 0$, and (9.1) gives $L_{\rm Edd}\to\infty$: a gas that does not interact with the radiation feels no force from it, and there is no limit. The cancellation of $r$ is the content of the proposition. Without it there would be a critical radius for a given luminosity; with it there is a critical luminosity for a given mass, the same at every radius.

CHECK 1 — CONFIRMED as a reproduction. The finding is the composition. (9.2) reproduces every Eddington number the book already prints:
quantitycomputedin printratio
$L_{\rm Edd}(10\,M_\odot)$$1.257065\times10^{39}\ \mathrm{erg\,s^{-1}}$$1.257\times10^{39}$, module09.html:9131.000052
$\dot M_{\rm Edd}(10\,M_\odot)$, $\eta_{\rm rad} = 0.1$$1.398674\times10^{19}\ \mathrm{g\,s^{-1}}$$1.399\times10^{19}$0.999767
Bondi/Eddington3663.536640.999856
$L_{\rm Edd}(4.30\times10^{6}\,M_\odot)$$5.405380\times10^{44}\ \mathrm{erg\,s^{-1}}$$5.405\times10^{44}$, module09.html:9211.000070
$\dot M_{\rm Edd}$, Sgr A*$9.545153\times10^{-2}\,M_\odot\,\mathrm{yr^{-1}}$$9.545\times10^{-2}$1.000016
AGN disc, fraction of Eddington0.100002"a tenth", module11.html:666exact to 0.002 per cent
Eddington's bound for the Sun$k \lt 13061$$k \lt 13{,}200$, his p. 1150.9895
The finding. By (6.2) and (9.1), $L_{\rm Edd}$ scales as $2/(1+X)$. The $m_{\rm p}$ of module09.html:944 therefore means $X = 1$, pure ionised hydrogen. For accreted gas of protosolar composition, $X = 0.7261$, the limit for a $10\,M_\odot$ hole is $1.4565\times10^{39}\ \mathrm{erg\,s^{-1}}$, 1.1587 times Module 9's value — 15.9 per cent higher. At the Sun's present surface composition, $X = 0.7583$, it is $1.4299\times10^{39}$, $1.1375$ times.

Module 9 §10 wrote that the $m_{\rm p}$ "is a genuine proton mass and not a mean molecular weight" (module09.html:915). That is true, and it is also a statement about composition: a proton mass per electron is what pure ionised hydrogen has. Module 9's formula is correct for the gas it assumes, and nothing in Module 9 is changed. The three compositions are sourced: $X = 1$ is what the formula means; $X = 0.7583$ is the present solar surface of the BS2005-AGS,OP model, which Module 4 §4 already uses; and $X = 0.7261$ is the same model's initial composition, $1 - Y_i - Z_i$ from Bahcall, Serenelli and Basu's Table 1, which is the apt value for gas a black hole accretes because it has not had the Sun's helium settling. The 1.1 per cent against Eddington's own bound is not traced: his 1926 solar luminosity and mass were not read.

Eddington's own form

Eddington did not state a limiting luminosity, and he did not use electron scattering. In his Sect. 82, p. 115, of The Internal Constitution of the Stars he derives "an upper limit to the opacity $k$ in any star (perfect gas or not) for which $L$ and $M$ are known by observation", and gives its physical reading on the same page: "the radiation observed to be emitted must work its way through the star, and if there were too much obstruction it would blow up the star".

Proposition 11 (Eddington's bound on the opacity). In a star in hydrostatic equilibrium, if the gas pressure $p_{\rm G}$ increases inward, $\mathrm{d}p_{\rm G} \gt 0$ for an inward step, and if the luminosity per unit mass $L_r/M_r$ inside radius $r$ is no smaller than its surface value $L/M$, then the flux-mean opacity satisfies $$ k \;\lt\; \frac{4\pi cGM}{L}. \tag{9.3}$$ This is (9.1) read as a bound on the opacity of a star of known $L$ and $M$.
Hydrostatic balance for an inward step $\mathrm{d}r \lt 0$ gives the increase of total pressure $\mathrm{d}P = -\rho(GM_r/r^{2})\,\mathrm{d}r$. The frequency-integrated first moment from the proof of Proposition 4, $c\,\mathrm{d}P_{\rm rad} = -k\rho F\,\mathrm{d}r$ with $F = L_r/4\pi r^2$, gives the increase of radiation pressure. Their ratio is $\mathrm{d}P_{\rm rad}/\mathrm{d}P = kL_r/(4\pi cGM_r)$. Since $\mathrm{d}P = \mathrm{d}p_{\rm G} + \mathrm{d}P_{\rm rad}$ and $\mathrm{d}p_{\rm G} \gt 0$, $\mathrm{d}P_{\rm rad} \lt \mathrm{d}P$, so $kL_r/(4\pi cGM_r) \lt 1$. With $L_r/M_r \ge L/M$, $kL/(4\pi cGM) \le kL_r/(4\pi cGM_r) \lt 1$, which is (9.3). At the surface, $L_r = L$ and $M_r = M$, and (9.3) is (9.1) with $k$ for $\kappa_{\rm F}$: the same inequality. ∎

Eddington writes the ratio $(L_r/M_r)/(L/M)$ as a single symbol and argues that it exceeds one because energy generation is concentrated toward the centre; the symbol is not used here, because module12.html:927 refuses a new bare meaning of the glyph he chose. His "If both temperature and density increase the material pressure $p_G$ must increase" is the condition $\mathrm{d}p_{\rm G} \gt 0$. The name "Eddington limit" is kept because it is what the literature calls (9.1), with the qualification that Eddington stated it as a bound on $k$ and not with $\kappa_{\rm es}$. For the Sun (9.3) gives $k \lt 13061\ \mathrm{cm^2\,g^{-1}}$ with this book's constants; he gives $13{,}200$.

The limit's assumptions, each of which can fail. It assumes spherical symmetry, so that the flux at $r$ is $L/4\pi r^2$; a disc or a beam puts the flux where the gas is not. It assumes the opacity is electron scattering; any additional absorption lowers $L_{\rm Edd}$. And it assumes a steady state; a transient can exceed it while it blows the gas away. This module names no observed violation, because none was read.

10. Accretion at the limit

An accreting black hole radiates $L = \eta_{\rm rad}\dot Mc^{2}$, where the radiative efficiency $\eta_{\rm rad}$ is Module 9's notation. Module 9 defines the Eddington accretion rate as $\dot M_{\rm Edd} = L_{\rm Edd}/(\eta_{\rm rad}c^2)$ at $\eta_{\rm rad} = 0.1$ (module09.html:944).

Proposition 12 (growth at the Eddington rate). A hole that accretes at $\dot M_{\rm Edd}$ and radiates $\eta_{\rm rad}\dot Mc^2$ keeps $(1-\eta_{\rm rad})\dot M$, and its mass grows as $M \propto e^{t/t_{\rm e}}$ with $$ t_{\rm e} \;=\; \frac{\eta_{\rm rad}\,\kappa_{\rm es}\,c}{4\pi G\,(1-\eta_{\rm rad})}, \tag{10.1}$$ which does not depend on $M$.
The energy radiated, $\eta_{\rm rad}\dot Mc^2$ per unit time, leaves as photons and with it the rest mass $\eta_{\rm rad}\dot M$, so the hole gains $\mathrm{d}M/\mathrm{d}t = (1-\eta_{\rm rad})\dot M$. At $\dot M = \dot M_{\rm Edd} = L_{\rm Edd}/(\eta_{\rm rad}c^{2})$ with (9.1), $$ \frac{\mathrm{d}M}{\mathrm{d}t} \;=\; (1-\eta_{\rm rad})\,\frac{4\pi GMc}{\kappa_{\rm es}\,\eta_{\rm rad}c^{2}} \;=\; \frac{M}{t_{\rm e}}, $$ with $t_{\rm e}$ as in (10.1). The right side is linear in $M$, so $M = M_0e^{t/t_{\rm e}}$; $M$ cancelled out of $t_{\rm e}$. ∎

At $\eta_{\rm rad} = 0.1$ and $\kappa_{\rm es} = \sigma_{\rm T}/m_{\rm p}$ the run gives $M/\dot M_{\rm Edd} = 4.5049\times10^{7}$ yr — the Salpeter time, $\eta_{\rm rad}\kappa_{\rm es}c/4\pi G$, the same at $10\,M_\odot$ and at $10^{8}\,M_\odot$ to $3.33\times10^{-16}$ — and $t_{\rm e} = 5.0054\times10^{7}$ yr. The difference is the $(1-\eta_{\rm rad})$: the first number is the e-folding time if the hole kept everything that falls in, and it does not. Problem K1 asks how long a $10\,M_\odot$ seed takes to reach $10^{9}\,M_\odot$.

A sentence Module 9 §10 leaves unsaid. module09.html:913 prints that the Bondi-fed hole "would double its mass in $1.23\times10^{4}$ yr"; the run recomputes $M/\dot M = 1.2297\times10^{4}$ yr. That is exact at constant $\dot M$, and the page states no growth law. Under Bondi's $\dot M \propto M^{2}$ the doubling time halves, to $6.1484\times10^{3}$ yr; at an Eddington-limited $\dot M \propto M$ it is $\ln 2$ times $M/\dot M$, $8.5235\times10^{3}$ yr. This is not a defect in Module 9, which is not edited.

Which efficiency, and what it costs

$\dot M_{\rm Edd}$ carries $\eta_{\rm rad}$, and the book prints three values of it. Module 11 §9 derives $1/12$ for a Newtonian thin disc at the Schwarzschild innermost stable orbit and computes $0.057191$ from Bardeen, Press and Teukolsky's orbit formula.

$\eta_{\rm rad}$where$\dot M_{\rm Edd}(10\,M_\odot)$ ($\mathrm{g\,s^{-1}}$)against module09.html:913's print
0.1module09.html:944, assumed$1.3987\times10^{19}$$-0.0$ per cent
$1/12 = 0.083333$Module 11, Newtonian$1.6784\times10^{19}$$+20.0$ per cent
0.057191Module 11, Bardeen et al.$2.4456\times10^{19}$$+74.8$ per cent

(Against the computed value rather than the rounded print, the last row is $+74.9$ per cent.) This book keeps $\dot M_{\rm Edd}$ defined at $\eta_{\rm rad} = 0.1$, so no shipped number moves. The $0.1$ is a convention of the book and not a derived efficiency; the table shows what the convention costs.

What this section answers for module09.html:914. Module 9 found Bondi accretion onto a $10\,M_\odot$ hole in a molecular cloud running at $3664$ times $\dot M_{\rm Edd}$ and called the calculation self-inconsistent: "a calculation that includes radiation is required". That debt is paid as a bound and not as a flow. Proposition 10 shows that gas radiating at $\eta_{\rm rad} = 0.1$ cannot be held above $L_{\rm Edd}$, so a spherical flow that radiated at that efficiency above $\dot M_{\rm Edd}$ would push on its own inflow harder than gravity pulls it; whether and how such a flow settles is the radiative Bondi problem, which this module does not solve. Problem K2 prices the excess and finds the efficiency at which such a flow would sit exactly at the limit. This module does not solve radiative Bondi accretion — the flow in which radiation pressure and gravity act on the same infalling gas — and no module of this book does.

Sagittarius A* sits at the other end. Module 9 §10's Problem K3 (module09.html:921) puts its X-ray luminosity at $3.70\times10^{-12}$ of Eddington while its Bondi rate is $8.405\times10^{-5}$ of $\dot M_{\rm Edd}$, a gap of $2.2716\times10^{7}$, or $7.356$ decades; the efficiency it implies, $4.4021\times10^{-9}$, is the $\eta_{\rm rad} \approx 4\times10^{-9}$ that Module 9 §9 prints (module09.html:847), to a ratio of $1.101$. That refutation is Module 9's and is cited, not re-checked. Module 11 §8's AGN disc, at $1.3987\times10^{25}\ \mathrm{g\,s^{-1}}$ onto $10^{8}\,M_\odot$, runs at $0.100002$ of $\dot M_{\rm Edd} = 1.398674\times10^{26}\ \mathrm{g\,s^{-1}}$: the "tenth of Eddington" of module11.html:666 is exact to $0.002$ per cent. Fig. 4 puts the four cases on one axis.

10-12 10-10 10-8 10-6 10-4 10-2 100 102 104 Sgr A*, X-ray 3.70×10−12 the Sun 2.6772×10−5 AGN disc, Module 11 0.100002 Bondi hole, Module 9 3663.5 L = LEdd above: no steady state L/LEdd (filled circle: a measured luminosity; open diamond: Ṁ/ṀEdd read at ηrad = 0.1)
Four systems on one logarithmic axis of $L/L_{\rm Edd}$. Two kinds of point. Filled circles are measured luminosities over the Eddington luminosity: Sgr A*'s X-ray luminosity at $3.70\times10^{-12}$ and the Sun at $2.6772\times10^{-5}$ (with $\kappa_{\rm es}$ at $X = 0.7583$). Open diamonds are accretion rates over $\dot M_{\rm Edd}$, which equal $L/L_{\rm Edd}$ only if the gas radiates at $\eta_{\rm rad} = 0.1$: Module 11's AGN disc at $0.100002$ and Module 9's Bondi hole at $3663.5$ (printed there as $3664$). The Bondi hole lies above the line at $1$, which is Module 9's self-inconsistency drawn as a point.

11. Where radiation matters, and where it does not

Proposition 9 draws the boundary between gas-pressure and radiation-pressure support as a line $\rho_{\rm eq}\propto T^{3}$, which is a straight line of slope $3$ in the logarithms. Fig. 5 draws it with two points from this book.

104 105 106 107 108 10-13 10-10 10-7 10-4 10-1 102 the Sun's centre (tabulated) 1614.5 times the crossover at its own μ the photosphere, τR = 2/3 Prad/P = 2.95×10−5 Prad > Pgas: radiation dominates Pgas > Prad Prad = Pgas, slope 3 μ = 0.832 ρ (g cm−3) T (K)
The $(\rho, T)$ plane on logarithmic axes. The line is $P_{\rm rad} = P_{\rm gas}$, (8.2), at $\mu = 0.832$; below it, at lower density for the temperature, radiation dominates the pressure. The Sun's tabulated centre ($\rho_c = 150.50\ \mathrm{g\,cm^{-3}}$, $T_c = 1.5480\times10^{7}$ K, Module 3's BS2005-AGS,OP row) lies $1614.5$ times above the line's density at the centre's own $\mu = 0.82851$. The photosphere point is ATLAS9 at $\tau_{\rm R} = 2/3$, where $P_{\rm rad}/P = 2.9454\times10^{-5}$.

At the Sun's centre the line lies at $9.3219\times10^{-2}\ \mathrm{g\,cm^{-3}}$ for $\mu = 0.82851$, against the actual $150.50$: the centre is $1614.5$ times denser than the crossover, and by Proposition 9 that ratio is $\beta/(1-\beta)$, the same statement as $1 - \beta = 6.190\times10^{-4}$ computed from $\rho_c$ and $T_c$ at that $\mu$, which is $0.999382$ of Module 3's $6.1940\times10^{-4}$. Why radiation matters more in more massive stars follows from the slope. Along a sequence of stars of the same structure and composition, $\rho_c \propto M/R^3$, and the hydrostatic central pressure $P_c \propto GM^2/R^4$ with $P_c \propto \rho_cT_c$ gives $T_c \propto M/R$. Then $\rho_c/\rho_{\rm eq}(T_c) \propto (M/R^3)/(M/R)^3 = M^{-2}$: the radius cancels, and each factor of ten in mass moves the centre two decades toward the line. This is the scaling inside Eddington's quartic, whose $(1-\beta)/\beta^4 \propto M^2$ Module 3 solves in its Problem K3.

The Sun as a whole sits at $L_\odot/L_{\rm Edd} = 2.6772\times10^{-5}$, with $\kappa_{\rm es}$ at $X = 0.7583$ (Problem D3). Read it for what it is. By the proof of Proposition 11, at any radius the ratio of the radiative force to gravity is $\mathrm{d}P_{\rm rad}/\mathrm{d}P = \kappa_{\rm F}L_r/(4\pi cGM_r)$, and $L_\odot/L_{\rm Edd}$ is that ratio at the surface for an opacity of $\kappa_{\rm es}$. The photosphere's opacity is not $\kappa_{\rm es}$: its $\kappa_{\rm R} = 0.4447$ is $1.2719$ times larger, and its own electron scattering is $6.1780\times10^{-5}$ (§6). So $2.6772\times10^{-5}$ is the ratio for a fully ionised, electron-scattering surface. With $\kappa_{\rm R}$ in place of $\kappa_{\rm es}$ the ratio would be $3.405\times10^{-5}$, the same order; that is an estimate only, because $\kappa_{\rm F}$ is not $\kappa_{\rm R}$. What Module 3 §11 dropped when it wrote "Ideal gas (3.2), with radiation only in Proposition 6" (module03.html:795) is radiation pressure in the equation of state, and its size is $P_{\rm rad}/P = 1-\beta$: $6.1940\times10^{-4}$ at the centre, Module 3 §9's value (module03.html:628), and $2.9454\times10^{-5}$ at the photosphere. This module evaluates $1-\beta$ at those two points only. Radiation transport, which sets the temperature gradient through (4.2), is not a correction at all.

The radiating shock. Module 8 §3 (module08.html:336) computed that a $10\ \mathrm{km\,s^{-1}}$ shock in molecular gas at $10$ K that radiates away its heat compresses the gas by a factor of $2802$, against $3.993$ for an adiabatic shock at the same speed. That is radiation's other role in fluid dynamics — as an energy sink rather than a force — and it is Module 8's result, cited here and not re-derived.

12. What this module captures, and what it misses

The module captures radiation as a momentum carrier and as a heat carrier, in the two limits where each has a closed form: the optically thick diffusion law and the grey atmosphere, and the force balance on optically thin gas. What it misses are frequency dependence beyond the Rosseland mean, the time-dependent and moving-medium terms of the transfer equation, and all the atomic physics that sets real opacities.

idealisationwhat it costswhere it is repaid
Static medium; no $(1/c)\partial I/\partial t$, no Doppler or aberration termsExact for the Sun's photosphere to the ratio of flow speed to $c$; wrong for relativistic outflows and for radiation-driven winds, where the gas moves at a sizeable fraction of the speeds that set the opacityNowhere in this book
No atomic physics: H⁻ and every bound–free and line opacity borrowed§6 takes $\kappa_{\rm R}$ from ATLAS9, a model. CHECK 2 inherits its opacity distribution functionsRefused in print, here. This book does no atomic physics
Grey opacity and the Eddington closure in the atmosphereCHECK 4 measures the combined error: $3.0$ per cent in $T$, $12.6$ per cent in $T^{4}$. The closure alone is $0.412$ per cent at $\tau = 2/3$Not repaid; non-grey atmospheres need the atomic physics above
The adiabatic criterion evaluated, the damping not computed (module04.html:563)§7 confirms Module 4's prediction at the photosphere: radiative leakage is not negligible there, with the parameter $0.3445$ to $0.4468$ against a criterion of $\ll 1$. The deep-star half of the prediction is not evaluated. The damping rate and the frequency shift that Module 4 §8 names are not computedHalf paid. The non-adiabatic oscillation equations are in no module of this book
Spherical, steady, electron-scattering limit(9.1) bounds steady spherical accretion; it says nothing about a disc, a beam or a transientNowhere in this book
Radiative Bondi accretion not solved (module09.html:914)§10 pays the debt as a bound, and Problem K2 prices the excess; the flow itself is not computedPaid as a bound, not as a flow
Radiative heating of the solar wind (module02.html:685)Module 2 §11 split the wind's missing heating three ways and named this module for radiation. The radiative heating of a wind is not computed hereRefused in print, here
Radiative heating of dust (module05.html:759)Module 5 §8 named this module for the radiative heating that holds Barnard 68's dust temperature. That needs dust opacities, which the book does not carryRefused in print, here
The inner AGN disc with radiation pressure (module11.html:666)The disc's inner $P_{\rm rad}/P_{\rm gas}$ at a tenth of Eddington needs its vertical structure with radiation pressure in it, which Module 11 §12 did not build. The "tenth" itself is confirmed by CHECK 1Refused in print, here

Of the twelve debts the shipped pages printed against this module, eight are paid in full or as stated above: the charter row of Module 1 §9 by the module as a whole; module02.html:688 by Proposition 8; module03.html:795 by §11; module04.html:429 by CHECK 3; module09.html:847, :915 and :944 by Proposition 10 and CHECK 1; and module09.html:914 as a bound. One is half paid (module04.html:563) and three are refused in the table above.

13. Problems

C1. Show that a radiation field whose intensity $I_\nu$ does not depend on direction has $P_\nu = u_\nu/3$. Then compute $P_\nu/u_\nu$ for a parallel beam, $I_\nu \propto \delta(\mathbf{n} - \mathbf{n}_0)$ with $\mathbf{n}_0$ along $z$, and for a field that is uniform over the hemisphere $\cos\theta \gt 0$ and zero over the other.
Solution
For isotropic $I_\nu$, $P_\nu/u_\nu = \oint\cos^2\theta\,\mathrm{d}\Omega/\oint\mathrm{d}\Omega = (4\pi/3)/4\pi = 0.333333$; see the proof of Proposition 1. For the beam, all the intensity has $\cos\theta = 1$, so $P_\nu/u_\nu = 1.000000$: three times the isotropic value. For the hemisphere, $\int_0^1\mu^2\,\mathrm{d}\mu/\int_0^1\mathrm{d}\mu = 1/3$, so $P_\nu/u_\nu = 0.333333$ again, although its flux is not zero. The $1/3$ needs only that $\langle\cos^{2}\theta\rangle = 1/3$, not isotropy; this is why the surface condition of Proposition 6 does not contradict its closure.
C2. Compute $L_{\rm Edd}$ and $\dot M_{\rm Edd}$ for a neutron star of $M = 1.4\,M_\odot$, with $M_\odot = 1.9884\times10^{33}$ g, for fully ionised gas of hydrogen mass fraction $X = 1$ and $X = 0.7261$, the rest helium. Use the book's convention $\eta_{\rm rad} = 0.1$ in $\dot M_{\rm Edd} = L_{\rm Edd}/(\eta_{\rm rad}c^2)$.
Solution
$M = 2.783774\times10^{33}$ g. By (9.1) with (6.2), $L_{\rm Edd} = 1.759891\times10^{38}\ \mathrm{erg\,s^{-1}}$ at $X = 1$ and $2.039153\times10^{38}$ at $X = 0.7261$, a ratio of $1.158681 = 2/(1+X)$. The rates are $\dot M_{\rm Edd} = 1.958143\times10^{18}\ \mathrm{g\,s^{-1}}$ and $2.268864\times10^{18}\ \mathrm{g\,s^{-1}}$. The composition moves the limit by the factor $2/(1+X)$ and nothing else.
C3. In Eddington's grey atmosphere (5.1) with $T_{\rm eff} = 5772$ K, compute $T$ at $\tau = 0$, $2/3$ and $2$. The model is grey: $\kappa_\nu$ does not depend on frequency.
Solution
$T(0) = T_{\rm eff}(1/2)^{1/4} = 0.840896\,T_{\rm eff} = 4853.7$ K; $T(2/3) = 5772.0$ K by construction; $T(2) = T_{\rm eff}2^{1/4} = 6864.1$ K. The surface is not at zero temperature: it sees radiation from below and is heated by it. The exact grey solution lowers $T(0)$, because $q_{\rm H}(0) = 1/\sqrt3$ is less than $2/3$; the surface temperature falls by the factor $(q_{\rm H}(0)/(2/3))^{1/4}$.
D1. Derive (8.1) from the first law, $\mathrm{d}e + P\,\mathrm{d}(1/\rho) = 0$, with $e = 3P_{\rm gas}/(2\rho) + a_{\rm rad}T^{4}/\rho$ for a monatomic gas, $\gamma = 5/3$, and check both limits. Evaluate $\Gamma_1$ at $\beta = 1/2$.
Solution
See the proof of Proposition 8. The adiabatic temperature exponent $\mathrm{d}\ln T/\mathrm{d}\ln\rho$ is $0.666667 = \gamma - 1$ at $\beta = 1$ and $0.333333$ at $\beta = 0$, and $\Gamma_1 = \beta + (4 - 3\beta)\,\mathrm{d}\ln T/\mathrm{d}\ln\rho$ gives $1.666667$ and $1.333333$. At $\beta = 1/2$, $\Gamma_1 = 1.425926$; Chandrasekhar's Table 1 prints $1.426$ at $1 - \beta = 0.5$.
D2. Derive (4.2) from the first angular moment of the transfer equation, stating the Eddington closure $P_\nu = u_\nu/3$ as the assumption it is. Then apply the result to the grey atmosphere (5.1) at large $\tau$ and show that it carries exactly the flux $\sigma_{\rm SB}T_{\rm eff}^4$.
Solution
See the proof of Proposition 4; the closure enters once, in replacing $P_\nu$ by $u_\nu/3$, and it is exact only for $\langle\cos^2\theta\rangle = 1/3$. With $z$ outward and $\tau$ inward, $\mathrm{d}\tau = -\kappa_{\rm R}\rho\,\mathrm{d}z$, and (4.2) reads $F = (c/3)\,\mathrm{d}(a_{\rm rad}T^{4})/\mathrm{d}\tau$. From (5.1), $\mathrm{d}T^{4}/\mathrm{d}\tau = \tfrac34T_{\rm eff}^{4}$ at every depth, so $F = (ca_{\rm rad}/4)T_{\rm eff}^4 = \sigma_{\rm SB}T_{\rm eff}^4$. The run evaluates the ratio as $1.000000$.
D3. Show that Propositions 10 and 11 are one inequality. Use Proposition 11's condition that the gas pressure increases inward, $\mathrm{d}p_{\rm G} \gt 0$ for an inward step. Then evaluate Eddington's bound for the Sun, with $L_\odot = 3.828\times10^{33}\ \mathrm{erg\,s^{-1}}$ and $GM_\odot = 1.3271244\times10^{26}\ \mathrm{cm^3\,s^{-2}}$, and divide the Sun's surface electron-scattering opacity at $X = 0.7583$ by it.
Solution
At the surface, $L_r = L$ and $M_r = M$, and the proof of Proposition 11 gives $kL/(4\pi cGM) \lt 1$, which rearranges to $L \lt 4\pi GMc/k$: (9.1) with $\kappa_{\rm F} = k$. Proposition 10 reads the inequality as a bound on $L$ at given opacity; Proposition 11 reads it as a bound on the opacity at given $L$. For the Sun, $4\pi cGM_\odot/L_\odot = 13060.8\ \mathrm{cm^2\,g^{-1}}$. With $\kappa_{\rm es} = 0.34966$ the ratio is $2.6772\times10^{-5}$, which is $L_\odot/L_{\rm Edd}$: the two readings give the same number.
K1. A $10\,M_\odot$ black hole accretes at exactly the Eddington rate, with $\eta_{\rm rad} = 0.1$ and $\kappa_{\rm es} = \sigma_{\rm T}/m_{\rm p}$. The hole keeps the fraction $(1-\eta_{\rm rad})$ of what falls in. How long does it take to reach $10^{9}\,M_\odot$?
Solution
By (10.1), $t_{\rm e} = 5.0054\times10^{7}$ yr. The growth factor is $10^{8}$, which is $\ln 10^{8} = 18.420681$ e-folds, so the time is $9.2204\times10^{8}$ yr. Dropping the $(1-\eta_{\rm rad})$ — using the Salpeter time $4.5049\times10^{7}$ yr as the e-folding time — gives $8.2983\times10^{8}$ yr, 10 per cent short. The time does not depend on the seed's mass except through the logarithm of the growth factor.
K2. Module 9 §10's Bondi hole has $M = 10\,M_\odot$, with $M_\odot = 1.9884\times10^{33}$ g, and accretes $\dot M = 5.124\times10^{22}\ \mathrm{g\,s^{-1}}$ (module09.html:913). $\dot M_{\rm Edd}$ is defined at $\eta_{\rm rad} = 0.1$ and $L_{\rm Edd}$ by (9.2). What luminosity would the hole radiate at $\eta_{\rm rad} = 0.1$, how many times $L_{\rm Edd}$ is it, and at what $\eta_{\rm rad}$ would it sit exactly at $L_{\rm Edd}$?
Solution
$L = 0.1\,\dot Mc^2 = 4.6052\times10^{42}\ \mathrm{erg\,s^{-1}}$, which is $3663.5$ times $L_{\rm Edd} = 1.257065\times10^{39}$ — the same number as $\dot M/\dot M_{\rm Edd}$, because $\dot M_{\rm Edd}$ is defined at the same $0.1$. At $L = L_{\rm Edd}$ the efficiency is $L_{\rm Edd}/\dot Mc^2 = 2.7297\times10^{-5}$. A flow that radiates that inefficiently escapes the limit: it can accrete at the Bondi rate without radiation pressure stopping it. That is the regime Module 9 §9 puts Sgr A* in, at $\eta_{\rm rad} \approx 4\times10^{-9}$.
K3. Evaluate the adiabaticity parameter $\omega\chi/c_{\rm s}^2$ at the solar acoustic cutoff, $\nu_{\rm ac} = 4497\ \mu$Hz, with ATLAS9's values at $\tau_{\rm R} = 2/3$: $\kappa_{\rm R} = 0.4447\ \mathrm{cm^2\,g^{-1}}$, $\rho = 2.6721\times10^{-7}\ \mathrm{g\,cm^{-3}}$, $T = 5910.1$ K, $c_P = 1.6968\times10^{8}\ \mathrm{erg\,g^{-1}\,K^{-1}}$ and $c_{\rm s} = 8.081\times10^{5}\ \mathrm{cm\,s^{-1}}$.
Solution
By (4.4), $\chi = 1.1588\times10^{13}\ \mathrm{cm^2\,s^{-1}}$ (rounded inputs), and with $\omega = 2\pi\nu_{\rm ac}$ the parameter is $0.5014$. By (7.1) the diffusion time across a wavelength is $12.532$ periods. The parameter is linear in $\omega$, so at fixed $\chi$ it is $4497/3090 = 1.4553$ times its value at $\nu_{\rm max}$; against §7's $0.3445$ the ratio is $1.4554$, because the five rounded inputs above give $\chi = 1.1588\times10^{13}$ where §7's full-precision ATLAS9 values give $1.1587\times10^{13}$. At the cutoff, the highest frequency the atmosphere traps, the leakage is half a radian of heat per radian of oscillation.

Notation

symbolmeaningunitsfirst use
$I_\nu$, $J_\nu$, $u_\nu$, $F_\nu$, $P_\nu$specific intensity, mean intensity, energy density, flux and pressure (Definition 1); without $\nu$, integrated over frequency. $P_\nu$, not the literature's $K_\nu$, because $K$ is Module 3's polytropic constant. $J_\nu$ always carries its subscript; Module 8's bare $J$ is the Sedov energy integral and does not occur herevariousDefinition 1
$u_{\rm rad}$, $P_{\rm rad}$radiation energy density and pressure, subscripted because bare $u$ is the book's velocity$\mathrm{erg\,cm^{-3}}$Proposition 1
$a_{\rm rad}$radiation constant $4\sigma_{\rm SB}/c$, Module 3's symbol; never bare $a$$\mathrm{erg\,cm^{-3}\,K^{-4}}$Proposition 1
$\sigma_{\rm SB}$, $\sigma_{\rm T}$Stefan–Boltzmann constant and Thomson cross-section, always subscripted. Bare $\sigma$ is Module 1's collision cross-section, and Module 11 writes it both for that and for the Stefan–Boltzmann constant (module11.html:360, :692); this module subscripts both constants so that bare $\sigma$ never occurscgs; $\mathrm{cm^2}$Proposition 1; (6.2)
$\nu$frequency, bare and as a subscript. Module 7's, Module 10's and Module 11's kinematic viscosity does not occur hereHz§2
$\kappa_\nu$, $\kappa_{\rm R}$, $\kappa_{\rm es}$, $\kappa_{\rm F}$opacity at one frequency, Rosseland mean, electron scattering, flux mean. Never bare $\kappa$: Module 1's thermal conductivity, Module 4's evanescent decay rate and Module 12's $\kappa_\parallel$, $\kappa_\perp$ already claim it. Eddington's $k$ appears only in his own inequality (9.3)$\mathrm{cm^2\,g^{-1}}$Definition 2
$j_\nu$, $S_\nu$, $B_\nu(T)$emissivity (not $\eta_\nu$), source function, Planck function. $B_\nu$ always carries its subscript and argument, because bare $B$ is Module 12's magnetic fieldcgsDefinition 2
$\tau$, $\tau_\nu$, $\tau_{\rm R}$optical depth, Module 6's ruling; $\tau_{\rm R}$ the Rosseland optical depth. Module 1's collision time, Module 2's radial exponent, Module 4's acoustic radius and Module 12's $\tau_e$, $\tau_i$ do not occur—Definition 3
$\lambda_{\rm ph}$photon mean free path, $1/(\kappa_\nu\rho)$ at one frequency and $1/(\kappa_{\rm R}\rho)$ for the mean, of Module 1's $\lambda$ family. The acoustic wavelength in Proposition 7 is written $c_{\rm s}/\nu$ and gets no symbolcmDefinition 3
$D$, $t_{\rm diff}$photon diffusivity $c\lambda_{\rm ph}/3$ and diffusion time$\mathrm{cm^2\,s^{-1}}$, sProposition 5
$\chi$radiative heat diffusivity (4.4), as Module 4 set it up. Not the extinction coefficient that the transfer literature writes $\chi$$\mathrm{cm^2\,s^{-1}}$Proposition 5
$c$, $c_{\rm s}$speed of light; sound speed. Module 4 writes the sound speed $c$, and its criterion $\omega\chi/c^2$ is written $\omega\chi/c_{\rm s}^2$ here$\mathrm{cm\,s^{-1}}$(1.1); Proposition 7
$\mathcal{P}$the period $1/\nu$ of a sound wave, in Proposition 7 only; calligraphic because $P$ is a pressuresProposition 7
$q_{\rm H}(\tau)$the Hopf function, subscripted because bare $q$ is Module 11's shear parameter—Proposition 6
$T_{\rm eff}$effective temperature, $\sigma_{\rm SB}T_{\rm eff}^4 = F$. $5772$ K (IAU) except in CHECK 4, which uses ATLAS9's $5777$ KKProposition 6
$\beta$$P_{\rm gas}/P$, Module 3's meaning. Module 12's plasma beta does not occur—§8
$\Gamma_1$, $\gamma$first adiabatic index of the mixture, Module 4's Definition 1, the same quantity; $\gamma = 5/3$ the gas's own. $L/L_{\rm Edd}$ is written out and never $\Gamma$—Proposition 8
$e$specific internal energy, Module 2's symbol$\mathrm{erg\,g^{-1}}$Proposition 8
$X$hydrogen mass fraction, as Module 4 uses it: $1$, $0.7583$ (present solar surface) or $0.7261$ (protosolar)—(6.2)
$L_{\rm Edd}$, $\dot M_{\rm Edd}$, $\eta_{\rm rad}$Eddington luminosity, Eddington rate at $\eta_{\rm rad} = 0.1$, radiative efficiency, Module 9's notation unchanged$\mathrm{erg\,s^{-1}}$, $\mathrm{g\,s^{-1}}$, —Proposition 10
$t_{\rm e}$e-folding time of a hole growing at $\dot M_{\rm Edd}$, with $(1-\eta_{\rm rad})$ kepts, yrProposition 12
$L$, $F$luminosity and flux, Module 3's and Module 6's$\mathrm{erg\,s^{-1}}$, $\mathrm{erg\,cm^{-2}\,s^{-1}}$§1

Sources

  1. Eddington, A. S. (1926), The Internal Constitution of the Stars, Cambridge University Press. Read in the 1930 second impression (archive.org in.ernet.dli.2015.165852), whose note says the text is reprinted unchanged apart from a few misprints; the 1926 printing itself was not seen. The scan has no text layer; pages were rendered and read as images. His Sect. 82, p. 115, gives $k \lt 4\pi cGM/L$ and the Sun's $k \lt 13{,}200$; his eq. (83.4), p. 116, gives $L = 4\pi cGM(1-\beta)/k_0$.
  2. Chandrasekhar, S. (1939), An Introduction to the Study of Stellar Structure, University of Chicago Press (archive.org in.ernet.dli.2015.212456), read from rendered pages. Eq. (131), p. 57, is (8.1); Table 1, p. 59, is reproduced to $3.44\times10^{-4}$.
  3. Castelli, F., ATLAS9 solar model ap00t5777g44377k1asp.dat, $T_{\rm eff} = 5777$ K, $\log g = 4.4377$, fetched 2026-09-21 from https://wwwuser.oats.inaf.it/castelli/sun/ and stored verbatim as afd/data/castelli_atlas9_sun.dat. Supplies every ATLAS9 value in §5 to §7. A model, not a measurement.
  4. Castelli, F. & Kurucz, R. L. (2003), IAU Symp. 210, poster A20, arXiv:astro-ph/0405087. Describes the ATLAS9 opacity-distribution-function grid; it does not contain the solar file, which carries its own header.
  5. Bahcall, J. N., Serenelli, A. M. & Basu, S. (2005), "New Solar Opacities, Abundances, Helioseismology, and Neutrino Fluxes", ApJ 621, L85–L88. Table 1, row BS05(AGS,OP): $Y_i = 0.2599$, $Z_i = 0.0140$, hence the protosolar $X = 0.7261$. It prints no central density, temperature or $1-\beta$.
  6. Ferguson, J. W. et al. (2005), "Low-Temperature Opacities", ApJ 623, 585–596. Read; it prints no Rosseland mean at photospheric conditions, and no number here comes from it.
  7. The Hopf function is computed by this module's generator from the exact grey integral equation, and no table of it was read. The value $q_{\rm H}(0) = 1/\sqrt3$ against which it is checked is exact.

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