Stars, From Birth to Remnant Part 3: Energy Transport

Chapter 3: Energy Transport

Introduction

We closed Chapter 2 with the mechanical skeleton of a star: mass conservation, hydrostatic equilibrium, and the equation of state that tells us where the pressure comes from. This skeleton explains why a star does not collapse. But it does not touch one question at all: Why is the interior of a star hot, and how is that heat distributed inside?

The temperature at the centre of the Sun is about 15.7 million kelvin, while at its photosphere it is about 5800 K. What creates this difference is that the energy produced at the centre must flow outward. Heat flows from hot to cold, and through every layer of the star, every second, passes the entire energy produced inside that layer. The form of this flow determines the star's temperature profile. What is surprising is how slow the flow is. Light reaches the Earth from the Sun in eight minutes; yet the energy produced in the Sun's core today takes thousands of years to reach the surface. We will estimate this number ourselves shortly.

We use the word "transport" in the title in its general sense, that is, the transfer of energy from one place to another. Readers coming from engineering courses may associate heat transfer chiefly with convection. Here, convection is only one of the three modes of energy transport; the other two are radiation and conduction.

The mathematical situation shows the same gap. If we take the mass coordinate $m$ as the independent variable, the unknowns describing the structure of the star are the radius $r$ , the pressure $P$ , the temperature $T$ , and the local luminosity $L$ , which we will define shortly. The density $\rho$ is not a separate unknown; it follows from $P$ and $T$ through the equation of state. Four unknowns require four equations. Chapter 2 gave us two of them. The two missing ones are the subject of this chapter: the energy conservation equation, which keeps account of where energy is produced, and the energy transport equation, which tells us how energy flows.

In this chapter we will do the following, in order. First we will set up the energy conservation equation so that it includes nuclear production, neutrino losses, and gravitational energy release. Then we will treat the journey of radiation through the star as a random walk and derive the radiative transport equation; along the way we will also obtain the relation $a = 4\sigma/c$ , which we used without derivation in Chapter 2. Next we will examine the obstacle in front of the radiation, namely the opacity, and its correct average, the Rosseland mean. We will then build, in turn, the Schwarzschild and Ledoux criteria, which tell us when convection sets in once radiation can no longer keep up; mixing length theory, which computes how much energy convection carries; and conduction, which becomes dominant in degenerate matter. Finally we will bring the four equations together with their boundary conditions.

The goal is this: by the end of the chapter, we will have all the equations needed to compute the internal structure of a star, and the problem will be mathematically closed. The only missing piece is the question of where the energy comes from; we leave that to Chapter 4.

3.1 The Energy Conservation Equation

First, a definition. Consider an imaginary sphere of radius $r$ , concentric with the star. Let $L(r)$ denote the net energy passing outward through the surface of this sphere per unit time, and call it the local luminosity. At the stellar surface this quantity is exactly the luminosity we defined in Chapter 1. At the centre, since nothing remains inside the sphere, $L(0) = 0$ .

[!note] Derivation (Energy conservation in a stationary star) Assume the star is in thermal equilibrium everywhere, i.e. no layer heats up or cools down with time. The mass of the thin spherical shell between radii $r$ and $r + dr$ is, as we saw in Section 2.1, $dm = 4\pi r^2 \rho, dr$ . Let $\varepsilon$ denote the energy produced per unit mass per unit time; its unit is $\mathrm{W,kg^{-1}}$ .

$L(r)$ enters the shell from inside, $L(r+dr)$ leaves from outside, and $\varepsilon, dm$ is produced within the shell. If no energy accumulates in the shell, the outgoing energy equals the sum of the incoming and the produced energy: $L(r + dr) = L(r) + \varepsilon, dm.$ Calling the difference on the left $dL$ : $\frac{dL}{dm} = \varepsilon \qquad \text{or} \qquad \frac{dL}{dr} = 4\pi r^2 \rho, \varepsilon.$

3.1.1 Sources and sinks

Here $\varepsilon$ is the net rate of energy input per unit mass into the shell, and it consists of three separate contributions.

Nuclear energy generation $\varepsilon_\text{nuc}$ . The energy released by nuclear reactions. How it is calculated is the subject of Chapter 4.
Neutrino losses $\varepsilon_\nu$ . Neutrinos interact so weakly with matter that they leave the star without stopping. The energy they carry does not contribute to the local luminosity; it is lost directly. The energy of neutrinos emitted in nuclear reactions is usually subtracted from the nuclear generation at the outset; $\varepsilon_\nu$ instead represents the neutrinos produced by thermal processes in hot, dense plasma. We will return to these processes in Section 4.7.
Gravitational energy release $\varepsilon_\text{grav}$ . This arises if the star is not in thermal equilibrium, i.e. if it is contracting or expanding. To see what it is, we need to relax the assumption of stationarity.

[!note] Derivation (General energy equation) Let $u$ be the internal energy per unit mass and $v = 1/\rho$ the specific volume. By the first law of thermodynamics, the heat $dq$ added to a unit mass is the sum of the increase in internal energy and the work of expansion: $dq = du + P, dv.$ The net heat entering a unit mass in the shell per unit time is the energy produced minus the losses minus the energy flowing out: $\frac{\partial q}{\partial t} = \varepsilon_\text{nuc} - \varepsilon_\nu - \frac{\partial L}{\partial m}.$ The time derivatives are taken at fixed mass coordinate $m$ , i.e. following the same shell (the Lagrangian viewpoint of Section 2.1). Equating the two expressions and isolating $\partial L/\partial m$ : $\frac{\partial L}{\partial m} = \varepsilon_\text{nuc} - \varepsilon_\nu - \left(\frac{\partial u}{\partial t} + P\frac{\partial v}{\partial t}\right).$ The expression in parentheses is the same as $T,\partial s/\partial t$ in terms of the specific entropy $s$ , because $T,ds = du + P,dv$ . Hence, with the definition $\varepsilon_\text{grav} \equiv -T\frac{\partial s}{\partial t},$ the energy conservation equation takes its final form: $\boxed{\frac{\partial L}{\partial m} = \varepsilon_\text{nuc} - \varepsilon_\nu + \varepsilon_\text{grav}}$

[!tip] Physical Interpretation In a contracting star, matter loses heat ( $T,ds < 0$ ), so $\varepsilon_\text{grav} > 0$ : contraction behaves like an energy source. There is a small puzzle here. In Section 2.6 we saw that a contracting star heats up; how can entropy decrease while temperature rises? The answer is hidden in the density. The entropy per unit mass of a monatomic ideal gas is, up to an additive constant, proportional to $\ln(T^{3/2}/\rho)$ . During contraction, as long as the density rises faster than the $3/2$ power of the temperature, $T^{3/2}/\rho$ decreases and the entropy drops. The virial theorem's statement that "half of the released gravitational energy goes into heat and half into radiation" is in fact nothing other than the $\varepsilon_\text{grav}$ term summed over the whole star. The Kelvin–Helmholtz timescale of Section 2.7 is precisely the time for which this term alone can power the star.

In a star in thermal equilibrium, $\varepsilon_\text{grav} \approx 0$ , and integrating gives $L = \int_0^M (\varepsilon_\text{nuc} - \varepsilon_\nu), dm.$ A main-sequence star is in this state: it radiates what it produces.

[!example] Example 3.1.1 The Sun's average energy generation per kilogram is $\bar\varepsilon_\odot = \frac{L_\odot}{M_\odot} \approx \frac{3.83 \times 10^{26}\ \mathrm{W}}{1.99 \times 10^{30}\ \mathrm{kg}} \approx 1.9 \times 10^{-4}\ \mathrm{W,kg^{-1}}.$ A resting human body produces about 100 W of heat; for a 70-kilogram person, that is 1.4 W per kilogram. Per kilogram, the Sun produces about seven thousand times less energy than a human. What makes the Sun bright is not efficiency, but its mass of $2 \times 10^{30}$ kilograms.

3.2 Radiative Transport

Photons inside a star bear no resemblance to the photons we imagined travelling freely through space in Chapter 1. Stellar matter is so dense that a photon, after travelling a very short distance, scatters off an electron or is absorbed and re-emitted by an atom. Energy leaks out through the sum of these countless small steps.

[!abstract] Definition 3.2.1 (Opacity and mean free path) The opacity $\kappa$ of a material is the effective absorption and scattering cross section per unit mass; its unit is $\mathrm{m^2,kg^{-1}}$ . In a medium of density $\rho$ , the photon mean free path is given by $\ell = \frac{1}{\kappa\rho}.$

The second expression follows directly from the definition. Over a path $dx$ a photon interacts with probability $\kappa\rho,dx$ ; a beam of $N$ photons is attenuated as $dN = -N\kappa\rho,dx$ , giving $N(x) = N_0 e^{-\kappa\rho x}$ . The mean distance a photon travels before interacting is $\langle x\rangle = \int_0^\infty x, e^{-\kappa\rho x}, \kappa\rho, dx = \frac{1}{\kappa\rho}.$ We will examine the opacity itself in detail in Section 3.3; for now let us use it as a measure of the obstacle in the photon's way.

3.2.1 Random walk

[!note] Derivation (Random walk) Let each step of the photon have length $\ell$ and a random direction. After $N$ steps in total, its displacement from the starting point is the sum of the step vectors: $\mathbf d = \boldsymbol\ell_1 + \boldsymbol\ell_2 + \cdots + \boldsymbol\ell_N$ . Let us take the mean of the square of this vector: $\langle d^2\rangle = \sum_i\sum_j \langle \boldsymbol\ell_i\cdot\boldsymbol\ell_j\rangle = \sum_i \langle \ell_i^2\rangle + \sum_{i\neq j}\langle \boldsymbol\ell_i\cdot\boldsymbol\ell_j\rangle.$ Since the steps are independent and randomly oriented, the terms with $i \neq j$ average to zero. What remains is $\langle d^2\rangle = N\ell^2$ . So in $N$ steps the photon gets only $\sqrt N,\ell$ away.

To escape from a star of radius $R$ requires $N \approx R^2/\ell^2$ steps. The total path travelled is $N\ell = R^2/\ell$ , and the time this takes is $t_\text{diff} \approx \frac{R^2}{\ell c}.$

[!example] Example 3.2.1 The mean density of the Sun is $\bar\rho \approx 1.4 \times 10^3\ \mathrm{kg,m^{-3}}$ . For the opacity, let us take $\kappa \approx 0.1\ \mathrm{m^2,kg^{-1}}$ , a value suitable for the interior (we will estimate this value ourselves in Section 3.3). Then $\ell \approx \frac{1}{(0.1)(1.4\times10^3)}\ \mathrm m \approx 7\times10^{-3}\ \mathrm m,$ i.e. one step of the photon is about 7 millimetres. The time: $t_\text{diff} \approx \frac{(7.0\times10^8)^2}{(7\times10^{-3})(3.0\times10^8)}\ \mathrm s \approx 2\times10^{11}\ \mathrm s \approx 7\times10^3\ \text{yr}.$ Our rough estimate gives something of order ten thousand years. Because the opacity rises rapidly towards the outer layers, the real time is longer; detailed solar models give values between tens of thousands and hundreds of thousands of years.

[!info] Aside By "photon" here we really mean energy. Along the way, photons are absorbed and re-emitted many times; radiation that starts at the centre with X-ray energies has turned into visible light by the time it reaches the surface. What walks outward is not a single photon, but the energy itself.

3.2.2 The radiative diffusion equation

The random walk gave us the time. Now we want how much energy passes through a point in the star per unit time, i.e. the flux.

[!note] Derivation (Radiative flux) Treat a small piece of the star as locally plane-parallel and call the radial direction $z$ . Let $u(z)$ be the energy density of the radiation. Each photon crossing the plane at $z = 0$ made its last interaction roughly a distance $\ell$ behind: a photon whose direction makes an angle $\theta$ with the vertical comes from height $z = -\ell\cos\theta$ and carries the energy density there. Since the radiation is distributed equally in all directions, the energy density per unit solid angle is $u/(4\pi)$ ; this energy flows at speed $c$ , and the component normal to the plane brings a factor $\cos\theta$ . The net flux through the plane is: $F = \int_{4\pi} \frac{c, u(-\ell\cos\theta)}{4\pi}\cos\theta, d\Omega.$ Since the free path $\ell$ is very small compared to the lengths over which the energy density varies, the first two terms of a Taylor expansion suffice: $u(-\ell\cos\theta) \approx u(0) - \ell\cos\theta,(du/dz)$ . Substituting: $F = \frac{c}{4\pi}\left[u(0)\int_{4\pi}\cos\theta,d\Omega - \ell\frac{du}{dz}\int_{4\pi}\cos^2\theta,d\Omega\right].$ The first integral vanishes: upward- and downward-going photons cancel. The second is computed in spherical coordinates: $\int_{4\pi}\cos^2\theta,d\Omega = 2\pi\int_0^\pi \cos^2\theta\sin\theta,d\theta = \frac{4\pi}{3}.$ Thus we find $F = -\frac{c\ell}{3}\frac{du}{dz}.$ Now write $r$ in place of $z$ , substitute $\ell = 1/(\kappa\rho)$ , and use the value of the energy density in local thermodynamic equilibrium: $u = aT^4$ (Section 2.3). Since the chain rule gives $du/dr = 4aT^3, dT/dr$ : $\boxed{F = -\frac{4acT^3}{3\kappa\rho}\frac{dT}{dr}}$ To convert the flux to the local luminosity, multiply by the area of the sphere: $L = 4\pi r^2 F$ . Isolating the temperature gradient, $\frac{dT}{dr} = -\frac{3\kappa\rho L}{16\pi ac, r^2 T^3},$ and multiplying this by $dr/dm = 1/(4\pi r^2\rho)$ we reach the Lagrangian form: $\boxed{\frac{dT}{dm} = -\frac{3\kappa L}{64\pi^2 ac, r^4 T^3}}$

[!tip] Physical Interpretation The isotropic part carries no net flux; what creates the flux is the variation of the energy density from place to place. The language of the equation is the language of heat conduction: the flux is proportional to the temperature gradient and inversely proportional to the opacity. Since the temperature decreases outward ( $dT/dr < 0$ ), the flux points outward.

The same equation has another reading. We write the radiation pressure as $P_\text{rad} = aT^4/3$ , following the standard notation of the international literature. Taking its derivative and comparing with the flux equation gives $\frac{dP_\text{rad}}{dr} = \frac{4aT^3}{3}\frac{dT}{dr} = -\frac{\kappa\rho}{c}F.$ Outflowing radiation transfers momentum to the matter that absorbs it and pushes it outward. When this push equals gravity, the star reaches the maximum luminosity it can carry; in Chapter 6 we will derive this limit, the Eddington limit, from here.

3.2.3 The relation between the radiation constant and the Stefan–Boltzmann constant

When writing the radiation pressure in Section 2.3, we used the relation $a = 4\sigma/c$ between the radiation constant $a$ and the Stefan–Boltzmann constant $\sigma$ without derivation. In the flux equation above, $a$ and $c$ also appear side by side; it is time to establish their connection to $\sigma$ . We now pay off the debt.

[!note] Derivation ( $a = 4\sigma/c$ ) The most fundamental way to describe a radiation field is the specific intensity: $I$ is the energy passing per unit time, per unit area, within a unit solid angle. The Planck function $B_\nu$ of Section 1.4 is the specific intensity per unit frequency of a blackbody; let us call its integral over frequency $B = \int B_\nu, d\nu$ .

Energy density. Consider photons travelling normal to a surface $dA$ , within a solid angle $d\Omega$ . In a short time $dt$ , an energy $I,dA,dt,d\Omega$ crosses the surface; this energy fills a cylinder of length $c,dt$ and volume $c,dt,dA$ . The contribution of photons in that direction to the energy density is therefore $I,d\Omega/c$ . Summing over all directions: $u = \frac{1}{c}\int_{4\pi} I, d\Omega.$ Blackbody radiation is the same in all directions ( $I = B$ ); hence $u = 4\pi B/c$ .

Flux. We computed the flux emerging from a blackbody surface in Section 1.4: only the outward-facing hemisphere is counted, with a $\cos\theta$ weight, giving $F = \pi B = \sigma T^4$ . Hence $B = \sigma T^4/\pi$ .

Combining. Putting the two together: $u = \frac{4\pi}{c}\cdot\frac{\sigma T^4}{\pi} = \frac{4\sigma}{c}T^4.$ Comparing with $u = aT^4$ : $\boxed{a = \frac{4\sigma}{c}}$ Numerical value: $a = 4(5.670\times10^{-8})/(2.998\times10^8) \approx 7.566\times10^{-16}\ \mathrm{J,m^{-3},K^{-4}}$ . This is the same value we used in Section 2.3.

[!tip] Physical Interpretation Where does the factor of four come from? The energy density counts radiation going in every direction and sweeps the full solid angle, $4\pi$ . The flux leaving a surface counts only the outward-facing hemisphere, moreover with a $\cos\theta$ weight, and gives $\pi$ . Four is the ratio of these two numbers; $c$ is the speed that converts energy density into energy flow. With this relation, the product $4ac$ in the flux equation equals $16\sigma$ ; we can also write the radiative flux as $F = -(16\sigma T^3/3\kappa\rho), dT/dr$ .

3.2.4 Why is the diffusion approximation so good?

Our derivation rested on two assumptions: the radiation is locally blackbody radiation ( $u = aT^4$ ), and the energy density changes very little over one free path. Let us test these for the Sun. The average temperature gradient between the centre and the surface is $\frac{T_c}{R_\odot} \approx \frac{1.6\times10^7\ \mathrm K}{7.0\times10^8\ \mathrm m} \approx 2\times10^{-2}\ \mathrm{K,m^{-1}}.$ Over a free path of about 7 millimetres the temperature changes by only $1.6\times10^{-4}$ K. Since the temperature is millions of kelvin, the relative change is of order $10^{-11}$ . A photon sees almost the same temperature at the two ends of its free path.

This is why local thermodynamic equilibrium holds almost perfectly in the stellar interior and the diffusion approximation is extremely reliable. The condition of local thermodynamic equilibrium we mentioned for the photosphere in Section 1.4 holds far more robustly deep inside. The approximation breaks down only very close to the surface, where a photon can escape into space without waiting for its next interaction. We will return to this boundary region in Section 3.7.

3.2.5 The radiative gradient

In stellar structure, a dimensionless gradient that tells how temperature varies together with pressure is very useful.

[!abstract] Definition 3.2.2 (Temperature gradient) The ratio of the relative changes in temperature and pressure when moving from one point in the star to a neighbouring one is called the temperature gradient: $\nabla \equiv \frac{d\ln T}{d\ln P}.$

In hydrostatic equilibrium the pressure decreases outward; so does the temperature. Since both change in the same direction, $\nabla$ is a positive number.

[!note] Derivation (Radiative gradient) Let $\nabla_\text{rad}$ denote the gradient that would arise if all the energy were carried by radiation. Divide the Lagrangian forms of the temperature and pressure equations by each other (the pressure equation is from Section 2.2): $\frac{dT/dm}{dP/dm} = \frac{-3\kappa L/(64\pi^2 ac, r^4T^3)}{-Gm/(4\pi r^4)} = \frac{3\kappa L}{16\pi ac, Gm, T^3}.$ The factors of $r^4$ in the denominators cancelled. The left-hand side is $dT/dP$ ; to reach the definition of $\nabla$ , multiply both sides by $P/T$ : $\boxed{\nabla_\text{rad} = \frac{3\kappa L P}{16\pi ac, Gm, T^4}}$

[!tip] Physical Interpretation The radiative gradient measures how heavy a load is placed on radiation's back. If the opacity is large (the photon's path is blocked) or $L/m$ is large (a lot of energy must pass per unit mass), radiation needs a steep temperature drop to carry the energy. Near the centre, $L \approx \varepsilon_c m$ (where $\varepsilon_c$ is the energy generation rate at the centre), so the ratio $L/m$ approaches the central energy generation rate; in stars whose energy generation is very sensitive to temperature, the central $\nabla_\text{rad}$ is therefore very large. But there is a limit to how steep a temperature drop can be. Once that limit is exceeded, stellar matter begins to boil. We will derive the limit in Section 3.4.

3.3 Opacity

The $\kappa$ in the radiative transport equation is a measure of the resistance stellar matter offers to radiation. The analogy with resistance in an electric circuit is apt: the flux is proportional to the temperature gradient divided by the opacity, just as the current is proportional to the voltage divided by the resistance. What determines the opacity are the microscopic interactions between photons and matter. The type of these interactions, in turn, depends on temperature, density, and composition.

Real opacity also depends strongly on frequency. So we will first find the correct way to reduce the frequency-dependent opacity to a single number, and then move on to the sources of opacity.

3.3.1 The Rosseland mean

[!note] Derivation (Rosseland mean) Let $\kappa_\nu$ denote the frequency-dependent opacity and write the diffusion equation separately for each frequency. Let the energy density of radiation with frequency between $\nu$ and $\nu + d\nu$ be $u_\nu, d\nu$ . In local thermodynamic equilibrium, the relation whose frequency-integrated form we found in the previous section also holds per frequency: $u_\nu = 4\pi B_\nu(T)/c$ . The flux per frequency is: $F_\nu = -\frac{c}{3\kappa_\nu\rho}\frac{du_\nu}{dr} = -\frac{4\pi}{3\kappa_\nu\rho}\frac{\partial B_\nu}{\partial T}\frac{dT}{dr}.$ Here we used the chain rule: $B_\nu$ depends on $r$ only through $T$ . For the total flux, integrate over frequency: $F = -\frac{4\pi}{3\rho}\frac{dT}{dr}\int_0^\infty \frac{1}{\kappa_\nu}\frac{\partial B_\nu}{\partial T}, d\nu.$ We want to bring this expression into the same form as $F = -(4acT^3/3\kappa\rho), dT/dr$ , valid when the opacity is frequency-independent. To do so, compute the integral of the weighting function: $\int_0^\infty \frac{\partial B_\nu}{\partial T}, d\nu = \frac{d}{dT}\int_0^\infty B_\nu, d\nu = \frac{d}{dT}\left(\frac{\sigma T^4}{\pi}\right) = \frac{4\sigma T^3}{\pi} = \frac{acT^3}{\pi}.$ In the last step we used $a = 4\sigma/c$ . Now if we define the mean opacity as $\frac{1}{\kappa_R} \equiv \frac{\displaystyle\int_0^\infty \frac{1}{\kappa_\nu}\frac{\partial B_\nu}{\partial T}, d\nu}{\displaystyle\int_0^\infty \frac{\partial B_\nu}{\partial T}, d\nu},$ the flux becomes $F = -\frac{4\pi}{3\kappa_R\rho}\cdot\frac{acT^3}{\pi}\frac{dT}{dr} = -\frac{4acT^3}{3\kappa_R\rho}\frac{dT}{dr}.$ This is identical to the equation of Section 3.2; only $\kappa$ has been replaced by $\kappa_R$ . This mean was defined by Svein Rosseland in 1924, and it has kept the name Rosseland mean opacity.

[!tip] Physical Interpretation The Rosseland mean is a harmonic mean: what is averaged is not the opacity itself but its inverse. Energy flows through the frequency windows where the opacity is small, just as traffic flows through the open lanes. A narrow absorption line with very large opacity contributes almost nothing to the mean. A broad transparent window, by contrast, pulls the mean down. The weighting function $\partial B_\nu/\partial T$ peaks around $x = h\nu/kT \approx 3.8$ ; the mean is most sensitive to frequencies where the photon energy is a few $kT$ .

[!info] Aside The harmonic mean has another consequence: the opacities of different sources cannot be added after averaging. The correct procedure is to add the opacities per frequency first and then take the mean. Computing the Rosseland means of the sources separately and adding them underestimates the true opacity, because a transparent window left by one source may be closed by another.

3.3.2 Sources of opacity

There are four basic processes a photon encounters in stellar matter: one scattering process and three absorption processes. To these we will add a special process dominant in the outer layers of cool stars: the negative hydrogen ion.

Electron scattering. A free electron oscillates under the influence of an incident electromagnetic wave and re-radiates the energy it receives in other directions. The photon is not absorbed; only its direction changes.

[!note] Derivation (Thomson cross section) A wave with electric field $E = E_0\cos\omega t$ gives the electron an acceleration $a_e = eE/m_e$ . An accelerating charge radiates, according to the Larmor formula known from electrodynamics, a power $P = \frac{e^2 a_e^2}{6\pi\varepsilon_0 c^3}.$ Since $\cos^2\omega t$ averages to $1/2$ over time, the mean radiated power is $\langle P\rangle = \frac{e^4 E_0^2}{12\pi\varepsilon_0 m_e^2 c^3}.$ The flux carried by the incident wave is $S = \tfrac12\varepsilon_0 c E_0^2$ . The cross section is the ratio of scattered power to incident flux: $\sigma_T = \frac{\langle P\rangle}{S} = \frac{e^4}{6\pi\varepsilon_0^2 m_e^2 c^4} = \frac{8\pi}{3}r_e^2 \approx 6.652\times10^{-29}\ \mathrm{m^2},$ where $r_e = e^2/(4\pi\varepsilon_0 m_e c^2) \approx 2.818\times10^{-15}$ m is the classical electron radius. The result is frequency-independent; it holds as long as the photon energy is much smaller than the electron rest-mass energy ( $h\nu \ll m_e c^2$ ).

[!note] Derivation (Electron scattering opacity) The opacity per unit mass is the product of the number of electrons per unit mass and $\sigma_T$ : $\kappa_\text{es} = \sigma_T n_e/\rho$ . In fully ionised matter, each hydrogen atom contributes one electron and has a mass of about one atomic mass unit. Helium and heavier elements contribute about half an electron per mass unit (helium brings two electrons with its four units of mass). Thus $\frac{n_e}{\rho} = \frac{1}{m_u}\left(X + \frac Y2 + \frac Z2\right) = \frac{1+X}{2m_u},$ since $Y + Z = 1 - X$ . In terms of $\mu_e$ from Section 2.3, this means $\mu_e = 2/(1+X)$ . The result: $\boxed{\kappa_\text{es} = \frac{\sigma_T(1+X)}{2m_u} \approx 0.020,(1+X)\ \mathrm{m^2,kg^{-1}}}$

Electron scattering opacity depends on neither temperature nor density. As long as matter is fully ionised, it is close to the smallest value the opacity can take, and it is the dominant source in hot, low-density regions, especially in the interiors of massive stars. For a solar-like composition ( $X \approx 0.7$ ) its value is about $0.034\ \mathrm{m^2,kg^{-1}}$ .

Free–free absorption. An electron passing near an ion can absorb a photon and continue as a faster free electron. A free electron on its own cannot do this, because energy and momentum cannot be conserved simultaneously; the ion is the third body that takes up the momentum. This process is the inverse of bremsstrahlung (braking radiation).

[!note] Derivation (Kramers' law) In 1923 Hendrik Kramers found the absorption coefficient of this process with a semiclassical calculation. The absorption per unit path length scales with the product of the electron and ion densities, with the $-1/2$ power of temperature, and with the $-3$ power of frequency: $\kappa_\nu\rho \propto n_e n_i, T^{-1/2}\nu^{-3}.$ Deriving the cross section itself extends into quantum electrodynamics; we will take the scaling behaviour and show how the temperature dependence arises. For the opacity per unit mass, divide by $\rho$ . Since both the electron and ion densities are proportional to $\rho$ , $\kappa_\nu \propto \rho, T^{-1/2}\nu^{-3}.$ Now take the Rosseland mean. Switching to the dimensionless variable $x = h\nu/kT$ , we have $\nu^3 = (kT/h)^3 x^3$ and $\frac{1}{\kappa_\nu} \propto \frac{T^{1/2}}{\rho}\left(\frac{kT}{h}\right)^3 x^3.$ In the Rosseland definition, the weighting function appears in both the numerator and the denominator of the fraction; its temperature-dependent factors cancel, leaving only a numerical integral over $x$ . Therefore $1/\kappa_R \propto T^{1/2}T^3/\rho$ , i.e. $\boxed{\kappa_\text{ff} \propto \rho, T^{-7/2}}$

With the numerical coefficient, one writes approximately $\kappa_\text{ff} \approx 3.7\times10^{18},(1-Z)(1+X),\rho, T^{-7/2}\ \mathrm{m^2,kg^{-1}},$ where $\rho$ is in $\mathrm{kg,m^{-3}}$ and $T$ in kelvin. The factor $(1+X)$ in the formula comes from the number of electrons, and the factor $(1-Z) = X + Y$ from the squared charges of the hydrogen and helium ions.

Bound–free absorption. A sufficiently energetic photon can eject an electron bound to an atom or ion. This photoionisation process is possible only when the photon energy exceeds the ionisation energy; above the threshold the cross section again falls roughly as $\nu^{-3}$ . The Rosseland mean therefore gives the same $\rho,T^{-7/2}$ behaviour. The difference is this: the process works only if there are ions still holding some of their electrons. At stellar interior temperatures hydrogen and helium are fully ionised, so bound–free absorption is supplied mostly by heavy elements, i.e. metals. The coefficient is therefore proportional to $Z$ : $\kappa_\text{bf} \approx 4\times10^{21},\frac{Z(1+X)}{t},\rho,T^{-7/2}\ \mathrm{m^2,kg^{-1}}.$ Here $t$ is a correction called the "guillotine factor", typically taking values between 1 and 100. The common $\rho,T^{-7/2}$ behaviour of free–free and bound–free absorption is called Kramers opacity.

Bound–bound absorption. A photon whose energy exactly equals the difference between two bound levels can lift an electron from one level to the other. This is the process that creates spectral lines in Section 1.6. Although individual lines are narrow, inside a star they are broadened by pressure and the Doppler effect; moreover, there are millions of them. Many-electron elements such as iron in particular create a pronounced bump in the opacity around $2\times10^5$ K. We will meet this bump again when studying pulsating stars in Section 10.5. The bound–bound contribution does not fit into a simple formula; it requires tables computed from detailed atomic data.

The negative hydrogen ion. In the atmospheres and outer layers of relatively cool stars like the Sun, the dominant source of opacity is a surprising particle: the negative hydrogen ion, $\mathrm H^-$ , in which two electrons are bound to a proton. The second electron is bound by only 0.754 eV; so any photon with wavelength shorter than 1.64 micrometres can detach it. For this ion to form, free electrons must be present in the medium. These electrons are supplied by easily ionised metals (such as sodium, magnesium, calcium). As the temperature rises, the number of free electrons increases, and the $\mathrm H^-$ abundance, and with it the opacity, rises rapidly. Between about 3000 and 6000 K one can write $\kappa_{\mathrm H^-} \approx 8\times10^{-34}\left(\frac{Z}{0.02}\right)\rho^{1/2},T^9\ \mathrm{m^2,kg^{-1}}.$ The ninth power of temperature! This steep increase will play a key role in Chapter 6, when we explain why fully convective stars lie on an almost vertical line (the Hayashi track) in the HR diagram.

Process	Approximate behaviour	Where dominant
Electron scattering	$0.020(1+X)$ , independent of $\rho$ and $T$	Hot, fully ionised, low-density regions
Free–free	$\propto (1-Z)(1+X),\rho,T^{-7/2}$	Fully ionised, dense regions
Bound–free	$\propto Z(1+X),\rho,T^{-7/2}$	Regions where metals are partially ionised
Bound–bound	From tables	Iron bump ( $T \approx 2\times10^5$ K)
$\mathrm H^-$	$\propto Z,\rho^{1/2},T^9$	Cool atmospheres and outer layers
Conduction	Section 3.6	Degenerate regions

Table 1: The main sources of opacity in stellar matter

[!info] Aside Modern stellar models compute opacity not with formulae but with detailed tables. The most widely used are those of the OPAL project at Lawrence Livermore laboratory (Iglesias and Rogers, 1996) and of the international Opacity Project; at low temperatures, molecules and dust grains are also taken into account. Kramers' formulae remain indispensable, however, for building intuition and for scaling analysis.

[!example] Example 3.3.1 At the centre of the Sun, $\rho \approx 1.5\times10^5\ \mathrm{kg,m^{-3}}$ and $T \approx 1.57\times10^7$ K. Because part of the hydrogen has been converted into helium, $X \approx 0.35$ at the centre. Electron scattering: $\kappa_\text{es} \approx 0.020\times1.35 \approx 0.027\ \mathrm{m^2,kg^{-1}}.$ For free–free absorption, first compute $T^{7/2} \approx 1.5\times10^{25}$ ; then $\kappa_\text{ff} \approx \frac{3.7\times10^{18}\times0.98\times1.35\times1.5\times10^5}{1.5\times10^{25}} \approx 0.05\ \mathrm{m^2,kg^{-1}}.$ Together with the bound–free and bound–bound contributions of metals, detailed tables give a total opacity of order $0.1\ \mathrm{m^2,kg^{-1}}$ . With this value the free path at the centre is $\ell = \frac{1}{\kappa\rho} \approx \frac{1}{(0.1)(1.5\times10^5)}\ \mathrm m \approx 7\times10^{-5}\ \mathrm m.$ At the centre of the Sun, a photon interacts again after travelling less than a tenth of a millimetre.

3.4 Convection: When Radiation Cannot Keep Up

When water heats up in a pot on the stove, the hot water at the bottom rises and the cold water at the surface sinks; heat is carried by the matter itself. The same thing happens somewhere in most stars. The question is: under what condition does radiative transport become unstable, so that matter begins to boil?

[!note] Derivation (Stability criterion) Consider a small parcel of matter inside the star; from now on we will simply call it an element (or blob). Push the element upward by $\Delta r$ from its position and make two assumptions:

The element is always in pressure equilibrium with its surroundings. Pressure differences are equalised at the speed of sound, i.e. very quickly.

The element exchanges no heat with its surroundings along the way, i.e. it expands adiabatically. Heat exchange is a much slower process than pressure equalisation.

If the element remains denser than its surroundings at its new position, it falls back and the layer is stable. If it is less dense, buoyancy keeps pushing it upward; the layer is unstable and convection begins.

Let us write how the density changes with pressure, temperature, and mean molecular weight for a general equation of state: $d\ln\rho = \alpha, d\ln P - \delta, d\ln T + \varphi, d\ln\mu,$ $\alpha \equiv \left(\frac{\partial\ln\rho}{\partial\ln P}\right)_{T,\mu},\quad \delta \equiv -\left(\frac{\partial\ln\rho}{\partial\ln T}\right)_{P,\mu},\quad \varphi \equiv \left(\frac{\partial\ln\rho}{\partial\ln\mu}\right)_{P,T}.$ For an ideal gas, $\rho \propto P\mu/T$ , so $\alpha = \delta = \varphi = 1$ .

Compare the element (e) with its surroundings (s). In both, the pressure changes in the same way. The element's composition does not change along the way, while its temperature changes adiabatically. Both the temperature and the composition of the surroundings change according to their own gradients: $\left(\frac{d\ln\rho}{dr}\right)_e = \alpha\frac{d\ln P}{dr} - \delta\left(\frac{d\ln T}{dr}\right)_\text{ad},$ $\left(\frac{d\ln\rho}{dr}\right)_s = \alpha\frac{d\ln P}{dr} - \delta\left(\frac{d\ln T}{dr}\right)_s + \varphi\left(\frac{d\ln\mu}{dr}\right)_s.$ Since the densities are initially equal, if the element remains denser than its surroundings after rising (stability), the element's density decreases more slowly than that of its surroundings: $(d\ln\rho/dr)_e > (d\ln\rho/dr)_s$ . Taking the difference: $-\delta\left(\frac{d\ln T}{dr}\right)_\text{ad} + \delta\left(\frac{d\ln T}{dr}\right)_s - \varphi\left(\frac{d\ln\mu}{dr}\right)_s > 0.$ Write all gradients in terms of pressure. By definition $d\ln T/dr = \nabla, d\ln P/dr$ ; for adiabatic change define $\nabla_\text{ad} \equiv (d\ln T/d\ln P)_\text{ad}$ , and for composition $\nabla_\mu \equiv d\ln\mu/d\ln P$ : $\frac{d\ln P}{dr}\left[\delta\nabla - \delta\nabla_\text{ad} - \varphi\nabla_\mu\right] > 0.$ In hydrostatic equilibrium the pressure decreases outward, i.e. $d\ln P/dr < 0$ . For the inequality to hold, the square bracket must be negative: $\nabla < \nabla_\text{ad} + \frac{\varphi}{\delta}\nabla_\mu.$ In a radiative layer the actual gradient is $\nabla_\text{rad}$ . The condition for the layer to remain radiative is therefore $\boxed{\nabla_\text{rad} < \nabla_\text{ad} + \frac{\varphi}{\delta}\nabla_\mu} \qquad \text{(Ledoux criterion)}$ In a layer of uniform composition ( $\nabla_\mu = 0$ ) the condition simplifies: $\boxed{\nabla_\text{rad} < \nabla_\text{ad}} \qquad \text{(Schwarzschild criterion)}$

[!tip] Physical Interpretation The language of the Schwarzschild criterion is very simple: if the temperature drop imposed by radiation is steeper than the rate at which a rising element can cool by its own expansion, the element stays hotter than its surroundings as it rises. What is hotter is less dense, and it keeps rising. This criterion was found in 1906 by Karl Schwarzschild, working on the solar atmosphere.

The composition term was added by Paul Ledoux in 1947. When the composition gradient is positive ( $\nabla_\mu > 0$ ), i.e. the mean molecular weight increases inward, a rising element carries upward a composition heavier than its surroundings, and this weight pulls it back. The region just above a core that has burned its hydrogen into helium is such a place. A composition gradient is stabilising. Layers that are unstable according to the Schwarzschild criterion but stable according to the Ledoux criterion host slow and complicated mixing; this situation is called semiconvection. In Chapters 7 and 9 we will see the importance of this subtle distinction in the evolution of massive stars.

3.4.1 The adiabatic gradient

On one side of the criterion is $\nabla_\text{rad}$ ; we derived it in Section 3.2. The $\nabla_\text{ad}$ on the other side depends on the thermodynamics of the matter.

[!note] Derivation ( $\nabla_\text{ad}$ for an ideal gas) The internal energy per unit mass of a monatomic ideal gas is $u = \tfrac32 kT/(\mu m_u)$ . With the ideal gas law $P = \rho kT/(\mu m_u)$ , this means $u = \tfrac32 Pv$ ( $v = 1/\rho$ ). In an adiabatic process $du + P,dv = 0$ : $\frac32(P,dv + v,dP) + P,dv = 0 \implies \frac52 P,dv + \frac32 v,dP = 0 \implies \frac{dP}{P} = -\frac53\frac{dv}{v}.$ So $P \propto v^{-5/3} \propto \rho^{5/3}$ . The temperature varies as $T \propto P/\rho \propto P^{1-3/5} = P^{2/5}$ : $\boxed{\nabla_\text{ad} = \frac25 = 0.4}$

[!note] Derivation ( $\nabla_\text{ad}$ for pure radiation) The energy of a radiation gas of volume $V$ is $aT^4V$ , and its pressure is $aT^4/3$ . In an adiabatic process $d(aT^4V) + (aT^4/3),dV = 0$ : $4aT^3V,dT + aT^4,dV + \frac{aT^4}{3},dV = 0 \implies \frac{dT}{T} = -\frac13\frac{dV}{V}.$ So $TV^{1/3}$ is constant. Since the pressure is proportional to $T^4$ , $P \propto V^{-4/3} \propto \rho^{4/3}$ and $\boxed{\nabla_\text{ad} = \frac14}$

Note the exponent $4/3$ in the second result. In Section 2.3, the pressure of ultra-relativistic electrons also grew as the $4/3$ power of density. This is no coincidence: every gas made of particles moving at or very close to the speed of light has an adiabatic exponent of $4/3$ . In Chapter 13 we will see why this exponent means a stability limit for white dwarfs.

[!note] Derivation (Mixture of gas and radiation) In real stellar matter, gas and radiation coexist. The internal energy per unit mass and the total pressure are: $u = \frac32\frac{P_\text{gas}}{\rho} + \frac{aT^4}{\rho},\qquad P = P_\text{gas} + \frac{aT^4}{3},\qquad P_\text{gas} = \frac{\rho kT}{\mu m_u}.$ Use the parameter $\beta = P_\text{gas}/P$ from Section 2.4: $P_\text{gas} = \beta P$ and $aT^4 = 3(1-\beta)P$ .

At fixed composition the adiabatic condition is $du + P,d(1/\rho) = 0$ . Let us write the terms one by one. The first term is $\tfrac32 kT/(\mu m_u)$ , so its differential is $\tfrac32(P_\text{gas}/\rho),d\ln T$ . The differential of the second term is $(aT^4/\rho)(4,d\ln T - d\ln\rho)$ , and the work term is $P,d(1/\rho) = -(P/\rho),d\ln\rho$ . Setting the sum to zero, multiplying by $\rho/P$ , and writing in terms of $\beta$ : $\frac32\beta,d\ln T + 3(1-\beta)(4,d\ln T - d\ln\rho) - d\ln\rho = 0,$ that is, $\left[\frac{3\beta}{2} + 12(1-\beta)\right]d\ln T = (4 - 3\beta),d\ln\rho.$ Let us also write the change in pressure in the same variables. Since the gas pressure is proportional to $\rho T$ and the radiation pressure to $T^4$ , $dP = P_\text{gas}(d\ln\rho + d\ln T) + 4P_\text{rad},d\ln T$ ; hence $d\ln P = \beta,d\ln\rho + (4 - 3\beta),d\ln T.$ From the first equation, $d\ln\rho = \dfrac{24 - 21\beta}{2(4-3\beta)},d\ln T$ ; substituting into the second: $d\ln P = \frac{\beta(24 - 21\beta) + 2(4-3\beta)^2}{2(4-3\beta)},d\ln T = \frac{32 - 24\beta - 3\beta^2}{2(4-3\beta)},d\ln T.$ The adiabatic gradient is the inverse of this: $\boxed{\nabla_\text{ad} = \frac{2(4-3\beta)}{32 - 24\beta - 3\beta^2}}$ Let us check: for $\beta = 1$ (pure gas) we get $2/5$ , and for $\beta = 0$ (pure radiation) $8/32 = 1/4$ . It reproduces both limiting cases correctly.

[!tip] Physical Interpretation As the share of radiation pressure increases, the adiabatic gradient decreases; matter boils more easily. In Section 2.4 we saw that $\beta$ decreases in massive stars. This is one of the reasons their centres are prone to convection.

Another place where the adiabatic gradient decreases is in partial ionisation zones. Part of the energy supplied to compressed gas goes into ionising atoms instead of raising the temperature (the Saha equilibrium of Section 2.5). The temperature rises less, and in the hydrogen ionisation zone $\nabla_\text{ad}$ can drop to of order 0.1. This is one reason why the outer layers of the Sun are convective.

3.4.2 Where does convection set in?

The Schwarzschild criterion can be violated in two ways: $\nabla_\text{rad}$ grows or $\nabla_\text{ad}$ shrinks. Looking at the expression derived in Section 3.2, $\nabla_\text{rad} = \frac{3\kappa LP}{16\pi ac,Gm,T^4},$ three typical situations stand out.

Where the opacity is large. In the outer layers of cool stars, $\mathrm H^-$ and partial ionisation make the opacity very large; together with the reduced adiabatic gradient, this makes the outer envelope convective. In the Sun, the convection zone begins at about 0.71 of the radius from the centre and extends to the surface.
Where $L/m$ is large. If energy generation is very sensitive to temperature, almost all the energy is produced in a small mass at the centre, and $L/m$ becomes very large there. In Chapter 4 we will see that the CNO cycle has exactly such a sensitivity. Stars more massive than about 1.2 solar masses therefore have a convective core.
Very low-mass stars. Stars below about 0.35 solar masses are fully convective from centre to surface.

The Sun sits in the middle of this picture: its centre transports energy by radiation, and its outer envelope by convection.

3.5 Mixing Length Theory

The Schwarzschild criterion told us where convection sets in. Now there are two more questions: How much energy does convection carry, and what is the actual temperature gradient $\nabla$ in a convection zone? The exact answer rests on the physics of turbulent flows, which is still not fully solved. Stellar physics overcomes this difficulty with a semi-empirical model. The idea proposed by Ludwig Prandtl in 1925 for laboratory turbulence was applied to stars by Erika Böhm-Vitense in 1958. The model is called mixing length theory (MLT).

The idea is this: convective elements travel on average a distance $\ell_m$ from where they were born, then mix with their surroundings, lose their identity, and deposit their excess heat there. This mixing length is written in terms of the local pressure scale height: $\ell_m = \alpha_\text{MLT} H_P$ . Here $\alpha_\text{MLT}$ is a dimensionless free parameter.

[!note] Derivation (Pressure scale height) Define the length over which the pressure decreases by a factor of $e$ : $H_P \equiv -dr/d\ln P = -P,(dr/dP)$ . Since hydrostatic equilibrium gives $dP/dr = -\rho g$ with $g = Gm/r^2$ , $\boxed{H_P = \frac{P}{\rho g}}$

[!note] Derivation (Convective velocity and flux) Let $\nabla_e$ denote the gradient the element shows along its own path. If the element lost no heat at all along the way, we would have $\nabla_e = \nabla_\text{ad}$ . The gradient of the surroundings is $\nabla$ .

Temperature excess. Since $d\ln T/dr = \nabla,d\ln P/dr = -\nabla/H_P$ , in the surroundings $dT/dr = -T\nabla/H_P$ , and in the element $-T\nabla_e/H_P$ . How much hotter the element is than its surroundings after rising by $\Delta r$ : $\frac{\Delta T}{T} = (\nabla - \nabla_e)\frac{\Delta r}{H_P}.$ The elements we observe at a given moment are, on average, halfway through their journey. So taking $\Delta r = \ell_m/2$ : $\frac{\Delta T}{T} = \frac12(\nabla - \nabla_e)\frac{\ell_m}{H_P}.$ Velocity. For an element in pressure equilibrium, the density difference is $\Delta\rho/\rho = -\delta,\Delta T/T$ . The buoyancy force per unit mass is $f = -g,\Delta\rho/\rho = g\delta,\Delta T/T$ , and it grows linearly as the element rises: $f = g\delta(\nabla - \nabla_e),\Delta r/H_P$ . The work done by this force over $\Delta r$ is $\int_0^{\Delta r} f,dr' = \frac{g\delta(\nabla - \nabla_e)}{2H_P}(\Delta r)^2.$ Assume that half of the work goes into the element's motion and the other half into pushing aside the surrounding matter. Then $v^2/2$ equals half of this work; substituting $\Delta r = \ell_m/2$ : $v^2 = \frac{g\delta(\nabla - \nabla_e),\ell_m^2}{8H_P}.$ Flux. Each element carries an excess heat of $c_P,\Delta T$ per unit mass; $c_P$ is the specific heat at constant pressure. Rising hot elements and sinking cool elements contribute to the flux with the same sign. The convective flux is: $F_\text{conv} = \rho,c_P,v,\Delta T.$ Substituting the velocity and the temperature excess: $F_\text{conv} = \rho c_P T\cdot\frac{(\nabla - \nabla_e),\ell_m}{2H_P}\cdot\ell_m\sqrt{\frac{g\delta(\nabla - \nabla_e)}{8H_P}}.$ Finally, since $2\sqrt8 = 4\sqrt2$ : $\boxed{F_\text{conv} = \frac{\rho c_P T\sqrt{g\delta},\ell_m^2}{4\sqrt2,H_P^{3/2}}(\nabla - \nabla_e)^{3/2}}$

[!info] Aside The numerical factors (half the path, half the work) are arbitrary, and different sources make different choices. This uncertainty is absorbed into $\alpha_\text{MLT}$ . The parameter is calibrated by requiring the solar model at the present age to reproduce the Sun's observed radius and luminosity, yielding a value between 1.5 and 2. This is the weak point of the theory; yet it works surprisingly well.

3.5.1 Heat loss of the element and closure of the theory

We have two unknowns: the gradient of the surroundings $\nabla$ and the gradient of the element $\nabla_e$ . Two more relations are needed to determine them. The first comes from how much heat the element loses by radiation along the way, and the second from the sharing of the total flux between radiation and convection.

[!note] Derivation (Radiative heat loss of the element) Let the element have size $d$ , surface area $S$ , and volume $V$ . The element is hotter than its surroundings by $\Delta T$ , and this difference is bridged across a boundary layer of thickness about $d/2$ . Using the flux formula of Section 3.2, the energy the element loses by radiation per unit time is: $\lambda = \frac{4acT^3}{3\kappa\rho}\frac{\Delta T}{d/2}S.$ This loss pulls the element's temperature below the adiabatic value. Since the element moves with speed $v$ , the temperature lost per unit path length is $\lambda/(\rho V c_P v)$ ; this is the deviation of the element's gradient from the adiabatic gradient: $\frac{T}{H_P}(\nabla_e - \nabla_\text{ad}) = \frac{\lambda}{\rho V c_P v}.$ Substituting the loss rate $\lambda$ and the expression $\Delta T = \tfrac12 T(\nabla - \nabla_e),\ell_m/H_P$ : $\nabla_e - \nabla_\text{ad} = \frac{4acT^3\ell_m}{3\kappa\rho^2 c_P v}\frac{S}{Vd}(\nabla - \nabla_e).$ The ratio $S/(Vd)$ here depends on the shape of the element; for a sphere of diameter $d$ it would be $6/d^2$ . Standard mixing length theory takes $d = \ell_m$ and uses the value $9/(2\ell_m^2)$ for this ratio; the difference is again absorbed into $\alpha_\text{MLT}$ . With this choice, $\frac{\nabla_e - \nabla_\text{ad}}{\nabla - \nabla_e} = \frac{6acT^3}{\kappa\rho^2 c_P,\ell_m v}.$ The inverse of the ratio on the left is the ratio of the heat the element carries to the heat it loses along the way; we may call it the efficiency of convection. Now substitute the expression above for $v$ and collect all local quantities into a single parameter: $U \equiv \frac{3acT^3}{c_P\rho^2\kappa,\ell_m^2}\sqrt{\frac{8H_P}{g\delta}}.$ Substituting $v = \ell_m\sqrt{g\delta(\nabla - \nabla_e)/8H_P}$ for the velocity, the right-hand side becomes $2U/\sqrt{\nabla - \nabla_e}$ , and the relation takes this simple form: $\nabla_e - \nabla_\text{ad} = 2U\sqrt{\nabla - \nabla_e}.$

[!note] Derivation (Sharing of the flux) We defined the total flux as $F = L/(4\pi r^2)$ , and $\nabla_\text{rad}$ as the gradient that would arise if radiation carried this entire flux. Solving the radiative gradient expression for $L$ and dividing by $4\pi r^2$ : $F = \frac{4acG}{3}\frac{T^4 m}{\kappa P r^2}\nabla_\text{rad}.$ In a convection zone, radiation still carries part of the flux, but with the actual gradient $\nabla$ : $F_\text{rad} = (4acG/3)(T^4m/\kappa Pr^2),\nabla$ . Convection carries the rest: $F_\text{conv} = F - F_\text{rad} = \frac{4acG}{3}\frac{T^4m}{\kappa Pr^2}(\nabla_\text{rad} - \nabla).$ On the other hand, since $g = Gm/r^2$ and $H_P = P/(\rho g)$ , we have $Gm/(Pr^2) = 1/(\rho H_P)$ . Equating this to the boxed $F_\text{conv}$ expression above and isolating $(\nabla - \nabla_e)^{3/2}$ : $(\nabla - \nabla_e)^{3/2} = \frac{16\sqrt2}{3}\frac{acT^3}{\kappa\rho^2 c_P,\ell_m^2}\sqrt{\frac{H_P}{g\delta}},(\nabla_\text{rad} - \nabla).$ Compare the factor on the right with $U$ . By definition $U = 6\sqrt2,(acT^3/\kappa\rho^2c_P\ell_m^2)\sqrt{H_P/g\delta}$ , so the factor is $(16\sqrt2/3)/(6\sqrt2) = 8/9$ times $U$ : $(\nabla - \nabla_e)^{3/2} = \frac89 U(\nabla_\text{rad} - \nabla).$

[!note] Derivation (The cubic equation) Let us combine the two relations. For brevity let $y \equiv \sqrt{\nabla - \nabla_e}$ . From the heat-loss relation, $\nabla_e = \nabla_\text{ad} + 2Uy$ , so $\nabla - \nabla_\text{ad} = (\nabla - \nabla_e) + (\nabla_e - \nabla_\text{ad}) = y^2 + 2Uy = (y+U)^2 - U^2.$ Also define $\xi \equiv y + U$ and $W \equiv \nabla_\text{rad} - \nabla_\text{ad}$ . Then $\nabla - \nabla_\text{ad} = \xi^2 - U^2$ and $\nabla_\text{rad} - \nabla = W - (\xi^2 - U^2)$ . The flux relation $y^3 = \tfrac89 U(\nabla_\text{rad} - \nabla)$ thus becomes an equation in a single unknown: $\boxed{(\xi - U)^3 + \frac{8U}{9}\left(\xi^2 - U^2 - W\right) = 0}$ At a given point, $U$ and $W$ are computed from local quantities. Solving the equation gives the physical root $\xi$ , and the actual gradient follows from $\nabla = \nabla_\text{ad} + \xi^2 - U^2$ . This is exactly what stellar evolution codes do in convection zones.

[!tip] Physical Interpretation The parameter $U$ we defined is a measure of the element's tendency to lose heat by radiation. Note that it is inversely proportional to the square of the density: in low-density matter the element loses its heat quickly.

Efficient convection ( $U \ll 1$ ). When $U$ is small in the cubic equation, $\xi$ also stays small and the equation becomes approximately $\xi^3 \approx \tfrac89 UW$ . Hence $\nabla - \nabla_\text{ad} \approx \xi^2 \approx (8UW/9)^{2/3}$ , i.e. as $U \to 0$ , $\nabla \to \nabla_\text{ad}$ . Convection can carry any flux while keeping the gradient just above the adiabatic value.

Inefficient convection ( $U \gg 1$ ). This time, writing $\xi = U + \eta$ with $\eta$ small, the dominant terms of the equation give $\tfrac{16}{9}U^2\eta \approx \tfrac89 UW$ ; hence $\eta \approx W/(2U)$ and $\nabla - \nabla_\text{ad} = \xi^2 - U^2 \approx 2U\eta \approx W$ . That is, $\nabla \to \nabla_\text{rad}$ : because the element loses its heat along the way, it can carry almost nothing, and radiation does the job once again.

[!example] Example 3.5.1 Consider a point in the lower part of the Sun's convection zone, near its base: $r \approx 4.9\times10^8$ m (0.71 of the radius), $T \approx 2.2\times10^6$ K, $\rho \approx 190\ \mathrm{kg,m^{-3}}$ . The flux through this radius is $F = \frac{L_\odot}{4\pi r^2} \approx \frac{3.83\times10^{26}}{4\pi(4.9\times10^8)^2}\ \mathrm{W,m^{-2}} \approx 1.3\times10^8\ \mathrm{W,m^{-2}}.$ Writing the temperature excess in terms of the velocity gives $\Delta T/T = 4v^2/(g\delta\ell_m)$ ; substituting into the flux gives $F_\text{conv} = 4\rho c_P T v^3/(g\delta\ell_m)$ . For a monatomic ideal gas, $c_V = \tfrac32 k/(\mu m_u)$ and $c_P = c_V + k/(\mu m_u) = \tfrac52 k/(\mu m_u)$ , so $c_P T = \tfrac52 P/\rho = \tfrac52 gH_P$ , and the flux takes a very simple form: $F_\text{conv} = \frac{10,\rho v^3}{\delta,\alpha_\text{MLT}}.$ For a rough estimate, assume convection carries the entire flux, with $\delta = 1$ and $\alpha_\text{MLT} = 1.8$ : $v = \left(\frac{\delta,\alpha_\text{MLT}F}{10\rho}\right)^{1/3} \approx \left(\frac{1.8\times1.3\times10^8}{1900}\right)^{1/3}\ \mathrm{m,s^{-1}} \approx 50\ \mathrm{m,s^{-1}}.$ That is about fifty metres per second; the speed of sound at the same location is about 220 kilometres per second. Let us extract the gradient difference from the velocity expression. For this, $gH_P = P/\rho = kT/(\mu m_u) \approx 3.0\times10^{10}\ \mathrm{m^2,s^{-2}}$ (with $\mu \approx 0.6$ ), so $\nabla - \nabla_e = \frac{8v^2}{\delta,\alpha_\text{MLT}^2,gH_P} \approx \frac{8\times2.5\times10^3}{3.2\times3.0\times10^{10}} \approx 2\times10^{-7}.$ The effect of the heat the element loses along the way is much smaller still: taking the opacity at this point to be about $2\ \mathrm{m^2,kg^{-1}}$ , the efficiency of convection comes out of order a million. That is, the element's own gradient deviates from the adiabatic gradient by only a millionth of the difference $\nabla - \nabla_e$ .

Conclusion: In the deep convection zones of stars, the actual gradient differs from the adiabatic gradient by less than one part in a million. In structure calculations one can write directly $\nabla = \nabla_\text{ad}$ there, and the value of $\alpha_\text{MLT}$ hardly matters in these regions.

[!info] Aside As we approach the surface, the density drops, $U$ grows, and convection loses its efficiency. In a thin layer just below the photosphere, the actual gradient clearly exceeds the adiabatic gradient; this layer is called the superadiabatic layer. The structure here is sensitive to $\alpha_\text{MLT}$ , and it determines the radius of stars with convective envelopes. We will return to this sensitivity when deriving the Hayashi track in Chapter 6. The granulation we see on the surface of the Sun, with cells about a thousand kilometres across, is a direct image of convection. There the velocities rise to the order of kilometres per second, because much less dense matter must carry the same flux: since the flux is proportional to $\rho v^3$ , the velocity must increase as the density falls.

3.6 Conductive Transport

The third mode is the transfer of heat through collisions of matter particles: conduction. In daily life this is the most familiar mode; the hand holding a hot glass of tea receives heat by conduction. In most stars, however, the contribution of conduction is negligibly small. To see why, and when this reverses, let us write the two transport modes in the same language.

[!note] Derivation (Thermal conductivity from kinetic theory) Kinetic theory gives the same form for heat conduction, whatever the carriers. Let the carriers have speed $v$ , mean free path $\ell$ , and heat capacity per unit volume $C$ . Let us repeat the flux calculation of Section 3.2 exactly. There we had the gradient of the energy density; here the change in energy density is $du = C,dT$ : $F = -\frac{v\ell}{3}\frac{du}{dr} = -\frac13 Cv\ell\frac{dT}{dr} \equiv -k\frac{dT}{dr},\qquad k = \frac13 Cv\ell.$ Apply this formula to photons: $C = d(aT^4)/dT = 4aT^3$ , $v = c$ , and $\ell = 1/(\kappa\rho)$ . The result: $k_\text{rad} = \frac13(4aT^3)\frac{c}{\kappa\rho} = \frac{4acT^3}{3\kappa\rho}.$ This is exactly the radiative flux of Section 3.2. Radiative transport is, in fact, conduction by the photon gas.

[!note] Derivation (Conductive opacity) Let $k_\text{cond}$ denote the thermal conductivity of the electrons, and to cast it in the same mould as radiation, define a conductive opacity: $k_\text{cond} \equiv \frac{4acT^3}{3\kappa_\text{cond},\rho}.$ The two modes operate side by side under the same temperature gradient, and their fluxes add: $F = -(k_\text{rad} + k_\text{cond})\frac{dT}{dr} = -\frac{4acT^3}{3\rho}\left(\frac{1}{\kappa_\text{rad}} + \frac{1}{\kappa_\text{cond}}\right)\frac{dT}{dr}.$ So without changing the radiative transport equation at all, it suffices to replace $\kappa$ with $\boxed{\frac1\kappa = \frac{1}{\kappa_\text{rad}} + \frac{1}{\kappa_\text{cond}}}$

[!tip] Physical Interpretation Opacities combine like resistors in parallel: current goes through whichever channel is easiest, and the total resistance is smaller than even the smaller one. Conduction becomes important only when $\kappa_\text{cond}$ becomes small enough to compete with the radiative opacity.

This does not happen in an ordinary, non-degenerate plasma. Although the heat capacity per unit volume of the electrons is larger than that of the photons, electrons are much slower than light; moreover, because of Coulomb collisions with ions, their free paths are many times shorter than those of photons. At the centre of the Sun, these factors combine to keep the contribution of conduction negligibly small.

In degenerate matter the picture reverses. As we saw in Section 2.3, the electrons fill the Fermi sphere. For an electron to scatter, the state it goes into must be empty; but all states below the Fermi surface are occupied. The Pauli principle forbids most collisions, and the electron's free path becomes enormously long. Electrons at the Fermi surface are also very fast. Although the heat capacity decreases (only electrons within about $kT$ of the Fermi surface can carry heat), the long free path and high speed more than compensate. As a result, $\kappa_\text{cond}$ falls far below the radiative opacity and conduction becomes dominant.

[!example] Example 3.6.1 This is why the interior of a white dwarf is almost isothermal: conduction rapidly erases temperature differences. How fast a white dwarf cools is determined not by its conducting core but by the thin, non-degenerate envelope surrounding it; heat leaks out through this envelope as if through a blanket. In Chapter 13 we will build the cooling theory of white dwarfs on this picture. The degenerate helium cores of red giants are, for the same reason, nearly isothermal throughout. We will see the consequences in the helium flash of Chapter 8.

3.7 The Four Equations of Stellar Structure

No piece is missing any more. Taking the mass coordinate $m$ as the independent variable, let us write the four equations together:

\boxed{ \begin{aligned} \frac{\partial r}{\partial m} &= \frac{1}{4\pi r^2\rho}, \[4pt] \frac{\partial P}{\partial m} &= -\frac{Gm}{4\pi r^4}, \[4pt] \frac{\partial L}{\partial m} &= \varepsilon_\text{nuc} - \varepsilon_\nu + \varepsilon_\text{grav}, \[4pt] \frac{\partial T}{\partial m} &= -\frac{GmT}{4\pi r^4 P}\nabla. \end{aligned}}

The first two come from Chapter 2, the third from Section 3.1. The fourth is the combination of the definition of $\nabla$ with hydrostatic equilibrium: $\partial T/\partial m = (dT/dP)(\partial P/\partial m) = (T/P),\nabla,(\partial P/\partial m)$ . Which $\nabla$ to use is decided by the stability criterion:

If $\nabla_\text{rad} \le \nabla_\text{ad}$ (by the Ledoux criterion if there is a composition gradient), the layer is radiative and $\nabla = \nabla_\text{rad}$ is taken. In this case the fourth equation reduces to the radiative equation of Section 3.2.
Otherwise the layer is convective and $\nabla$ is found from mixing length theory; deep inside, in practice, $\nabla = \nabla_\text{ad}$ .

[!note] Derivation (A convenient form for $\varepsilon_\text{grav}$ ) In codes it is easier to work with temperature and pressure rather than entropy. Write the entropy as a function of $T$ and $P$ : $ds = (\partial s/\partial T)_P,dT + (\partial s/\partial P)_T,dP$ . The first coefficient is $c_P/T$ by definition. For the second, a Maxwell relation is used: $(\partial s/\partial P)_T = -(\partial v/\partial T)_P$ . On the other hand, from $v = 1/\rho$ and the definition of $\delta$ , $(\partial v/\partial T)_P = \delta/(\rho T)$ . Thus $T,ds = c_P,dT - \frac\delta\rho,dP \implies \boxed{\varepsilon_\text{grav} = -c_P\frac{\partial T}{\partial t} + \frac\delta\rho\frac{\partial P}{\partial t}}$

The right-hand sides of the four equations require three relations describing the properties of the matter: $\rho = \rho(P, T, X_i),\qquad \kappa = \kappa(\rho, T, X_i),\qquad \varepsilon = \varepsilon(\rho, T, X_i).$ The first is the equation of state (Section 2.4); $\nabla_\text{ad}$ , $c_P$ , and $\delta$ are also derived from it. The second is the opacity (Sections 3.3 and 3.6). The third is energy generation, the subject of Chapter 4. Here $X_i$ is the mass fraction of each element making up the chemical composition.

Time enters the equations in two ways: through the $\varepsilon_\text{grav}$ term and through changes in composition. Nuclear reactions change the composition slowly; convection, by mixing the region it covers, homogenises the composition there. The evolution of a star is a sequence of equilibrium models whose composition changes step by step. The large separation of timescales we saw in Section 2.7 is what makes this approach possible.

3.7.1 Boundary conditions

Four first-order differential equations require four boundary conditions. Two come from the centre and two from the surface.

Centre. At the centre, i.e. at $m = 0$ , we must have $r = 0$ and $L = 0$ . The central pressure and temperature are unknown; finding them is part of the problem.

[!note] Derivation (Behaviour near the centre) The $r^2$ and $r^4$ denominators in the equations go to zero at the centre; this is why numerical solutions are started not exactly at the centre but just outside it. Let us see how the solution behaves there. For very small $m$ , the density can be taken equal to its central value: $\rho \approx \rho_c$ . Then $m \approx \tfrac{4\pi}{3}r^3\rho_c$ , that is, $r \approx \left(\frac{3m}{4\pi\rho_c}\right)^{1/3} \qquad\text{and}\qquad L \approx \varepsilon_c m.$ Here $\varepsilon_c$ is the central value of $\varepsilon$ . For the pressure, substitute $r^4$ : $\frac{dP}{dm} = -\frac{Gm}{4\pi}\left(\frac{4\pi\rho_c}{3m}\right)^{4/3} = -\frac{G}{4\pi}\left(\frac{4\pi\rho_c}{3}\right)^{4/3}m^{-1/3}.$ Integrating gives $P \approx P_c - \frac{3G}{8\pi}\left(\frac{4\pi\rho_c}{3}\right)^{4/3}m^{2/3}.$ The pressure starts from a finite value at the centre and decreases as $m^{2/3}$ , i.e. as $r^2$ .

Surface. The simplest choice is to take $P = 0$ and $T = 0$ at the surface. This is good enough for most of the internal structure, because the pressure and temperature inside are orders of magnitude larger than at the surface. But the surface of a star is not at zero temperature; the surface we observe is the photosphere. In Section 1.4 we said that the photosphere lies at an optical depth of about $2/3$ and that its temperature there is close to the effective temperature. Now we can see where this statement comes from.

[!abstract] Definition 3.7.1 (Optical depth) The optical depth $\tau$ , counted inward from the outside of the star, is defined by $d\tau = -\kappa\rho,dr$ and is zero outside the star. Optical depth is a measure of how many free paths deep we are, as seen from outside.

[!note] Derivation (Eddington's grey atmosphere) Assume the atmosphere is thin (its thickness small compared to the radius), plane-parallel, and grey (opacity independent of frequency). No energy is produced in the atmosphere, i.e. the flux is the same at every depth: $F = \sigma T_\text{eff}^4$ . Assume the diffusion relation holds approximately throughout the atmosphere. The expression $F = -(c/3\kappa\rho),du/dr$ of Section 3.2 becomes, with $d\tau = -\kappa\rho,dr$ , $F = \frac c3\frac{du}{d\tau}.$ Since the flux is constant, the integral is immediate: $u(\tau) = u(0) + \frac{3F}{c}\tau.$ What remains is the surface value $u(0)$ . At the surface ( $\tau = 0$ ) there is no incoming radiation from outside; radiation goes only into the outward-facing hemisphere. Assume the specific intensity in this hemisphere is the same in every direction, $I_0$ (the Eddington approximation). Since the solid angle of the hemisphere is $2\pi$ , the energy density is $u(0) = 2\pi I_0/c$ ; and the flux, by the Lambert calculation of Section 1.4, is $F = \pi I_0$ . So $u(0) = 2F/c$ and $u(\tau) = \frac{3F}{c}\left(\tau + \frac23\right).$ In local thermodynamic equilibrium $u = aT^4 = 4\sigma T^4/c$ ; and the flux is $\sigma T_\text{eff}^4$ . Substituting: $\frac{4\sigma T^4}{c} = \frac{3\sigma T_\text{eff}^4}{c}\left(\tau + \frac23\right) \implies \boxed{T^4(\tau) = \frac34 T_\text{eff}^4\left(\tau + \frac23\right)}$

[!tip] Physical Interpretation Setting $\tau = 2/3$ in the equation gives $T = T_\text{eff}$ . That the photosphere lies at $\tau \approx 2/3$ is not a convention but a result of this simple atmosphere model: the effective temperature of a star is the temperature at the depth we can see from outside, i.e. at an average optical depth of $2/3$ . At the surface ( $\tau = 0$ ) the temperature is not zero; $T^4 = T_\text{eff}^4/2$ , i.e. $T \approx 0.84,T_\text{eff}$ .

[!info] Aside The Eddington approximation is not perfect at the surface. The exact solution of the radiative transfer equation for a grey atmosphere has the form $T^4 = \tfrac34 T_\text{eff}^4,[\tau + q(\tau)]$ . The so-called Hopf function $q(\tau)$ rises slowly from about 0.58 at the surface to 0.71 at depth. Eddington's $2/3$ is a good average sitting between these two values.

[!note] Derivation (Photospheric pressure) Hydrostatic equilibrium also holds in the atmosphere: $dP/dr = -\rho g$ . Writing this in terms of optical depth: $\frac{dP}{d\tau} = \frac{dP/dr}{d\tau/dr} = \frac{-\rho g}{-\kappa\rho} = \frac g\kappa.$ In a thin atmosphere $g$ is constant; taking the opacity as constant too and integrating from $\tau = 0$ (where $P = 0$ ) to $\tau = 2/3$ : $\boxed{P_\text{ph} = \frac23\frac g\kappa}$

The surface boundary conditions are thus determined: at $m = M$ , $T = T_\text{eff}$ and $P = \tfrac23,GM/(R^2\kappa)$ . The effective temperature is in turn related to the luminosity and radius by $L = 4\pi R^2\sigma T_\text{eff}^4$ (Section 1.4). These two conditions connect the interior solution to the atmosphere. Detailed solar atmosphere models give a photospheric pressure of about $10^4$ Pa; this is about a tenth of the atmospheric pressure at sea level on Earth.

3.7.2 The Vogt–Russell theorem

Let us sum up: four differential equations, three material relations, and four boundary conditions. For a given total mass $M$ and a given distribution of composition within the star, $X_i(m)$ , this system generally has a unique solution. This observation is called the Vogt–Russell theorem: the structure of a star is determined by its mass and the distribution of its chemical composition within the star. The name "theorem" is somewhat generous; it has not been mathematically proven, and in some special cases more than one solution can exist. But the practice of stellar physics rests on this principle. At the beginning of Chapter 1 we said, "where there is mass, there is destiny." The mathematics of that destiny is these four equations.

[!info] Aside This system can be solved analytically only in very special cases; polytropic models are the most important of these (Appendix B). Real stellar models are solved numerically, mostly with the Henyey method. The star is divided into hundreds or thousands of shells, the equations are turned into difference equations between these shells, and all shells are solved simultaneously, starting from a guess and correcting iteratively. Open-source codes such as MESA today do this on a laptop in minutes.

[!quote] Next Chapter We close Chapter 3. We have completed the work begun in Chapter 2: the four equations of stellar structure, together with their boundary conditions, are in our hands. We have seen how radiation carries energy through a random walk, how opacity slows this walk, how convection takes over when radiation cannot keep up, and why conduction dominates in degenerate matter. We paid off the $a = 4\sigma/c$ debt left over from Chapter 2; we also showed where the optical depth of $2/3$ mentioned in Chapter 1 comes from.

We deliberately left a gap in the equations: $\varepsilon_\text{nuc}$ . We have written how energy is conserved and how it is transported, but we have not yet said where and how it is produced. In Chapter 4 we descend to the energy source of the star, nuclear reactions: mass defect and binding energy, quantum tunnelling through the Coulomb barrier, the Gamow peak, the pp chain, the CNO cycle, and helium burning. In Chapter 1 we said we would compute the value $Q \approx 26.73$ MeV from the mass defect; we will keep that promise there. Once we find how sensitive energy generation is to temperature, we will also see in numbers the reason for the convective cores that we mentioned only qualitatively in this chapter.