Module 6 — Convection and Thermal Instability

Hydrostatic balance is not stability. A stratified fluid can fail in two independent ways: buoyancy can stop restoring a displaced parcel, which is convection, and cooling can run away under compression, which is thermal instability. The Schwarzschild and Ledoux criteria, the collapse of the adiabatic gradient in an ionisation zone, mixing-length theory and its price, then Field's criterion and the two-phase interstellar medium. Two checks against measured data: mixing-length theory against solar granulation, and the predicted two-phase pressure window against 89 measured lines of sight.

Contents

1. Two fluids that will not sit still

Module 3 solved the hydrostatic equation and built stars out of it. Nothing in that module asked whether the solutions it found would survive being disturbed. A stratified fluid in hydrostatic balance has every force balanced at every point, and that is a statement about the fluid at rest; it says nothing about what happens to a parcel of it that is nudged. Balance is not stability.

This module asks the stability question twice, of two different restoring mechanisms, and gets two different instabilities.

The first is buoyancy. Displace a parcel upward a little and it expands to the lower ambient pressure. If it arrives denser than its new surroundings it sinks back and the layer is stable. If it arrives lighter it keeps rising, and the layer convects. §2 turns that sentence into an inequality between two logarithmic derivatives, one belonging to the star and one to the gas, and §3 turns the inequality into a frequency. §4 asks what a composition gradient does to it and §5 what partial ionisation does. §6 then asks the quantitative question the criterion cannot answer — how much heat a convecting layer carries — and §7 takes the answer to two measurements of the solar surface.

The second is cooling. Compress a parcel of interstellar gas a little, at the pressure of its surroundings. If the compression makes it radiate away energy faster than it gains it, the parcel cools, and at fixed pressure a cooler parcel is denser, so it contracts further. No gravity is involved at all. §8 derives Field's criterion for that runaway, §9 shows that it carves the interstellar medium into two stable phases separated by a narrow band of pressures, and §10 compares the predicted band with 89 measured sight lines. That comparison is the anchor of this module, and it confirms one prediction and refutes another from the same measurement.

A terrestrial case sets the difficulty of the first question immediately. The International Standard Atmosphere fixes the tropospheric lapse rate at $6.50\ \mathrm{K\,km^{-1}}$, against a dry adiabatic value $g/c_p = 9.761\ \mathrm{K\,km^{-1}}$ for dry air. The ratio is $0.6659$, which makes the standard troposphere a polytrope of index $n = 4.2558$, with $\nabla = 1/(n+1) = 0.19026$ against the diatomic adiabatic value $2/7 = 0.28571$. By the criterion of §2 the troposphere is stable, with $33$ per cent to spare. It nevertheless convects every afternoon. The resolution is that air is not dry: a saturated parcel releases about $2.5\times10^{10}\ \mathrm{erg\,g^{-1}}$ of latent heat as it rises and condenses, which warms it on the way up and lowers its effective lapse rate to $5$–$6\ \mathrm{K\,km^{-1}}$, below the ambient $6.5$. Moist convection is unstable where dry convection is not. Problem C1 works this through.

The Sun has no latent-heat term of that kind, so the solar criterion is the clean one, and it is derived first.

2. The displaced parcel and the Schwarzschild criterion

The whole of this section follows from the Euler equation of Module 2 and one assumption about the parcel. No transport theory is needed, and none is used.

Definition 1 (the two logarithmic gradients). For a stratified fluid in hydrostatic balance, the stratification gradient is the logarithmic temperature gradient of the background, taken along the radius, $$ \nabla \;\equiv\; \frac{d\ln T}{d\ln P}, \tag{2.1}$$ which is Module 3, Definition 5. The adiabatic gradient is the corresponding derivative for a parcel of the gas at fixed specific entropy $s$, $$ \nabla_{\rm ad} \;\equiv\; \left(\frac{\partial\ln T}{\partial\ln P}\right)_{\!s}. \tag{2.2}$$ The first is a property of the star; the second is a property of the gas. Fig. 1 sets out the comparison the criterion will make. Notation. In this module the symbol $\nabla$ with no argument always means (2.1). The vector operator is written only with an argument attached, as in $\nabla\!\cdot\mathbf{u}$ or $\nabla\Phi$.
r P P 0 1 δr the parcel: adiabatic the surroundings T p T vs hotter ⇒ lighter ⇒ rises on cooler ⇒ heavier ⇒ falls back both at P 0 both at P 1
Setup for Proposition 1. A parcel at the level where the ambient pressure is $P_0$ is displaced upward by $\delta r$ to the level $P_1$. It stays in pressure balance with its new surroundings throughout, and changes state adiabatically, so it arrives at $P_1$ with its own temperature $T_{\rm p}$. Only the comparison of $T_{\rm p}$ with the ambient $T$ at the same pressure decides the sign of the buoyancy force, because at equal pressure a hotter ideal gas of the same composition is the lighter one.
Proposition 1 (the Schwarzschild criterion). Consider a plane-parallel stratification of an ideal gas in hydrostatic balance, with (i) uniform chemical composition, so that the mean molecular weight $\mu$ is the same in the parcel and its surroundings, and (ii) $\rho = \mu m_u P/(k_BT)$. Displace a parcel from the level $r$ by $\delta r$ with $|\delta r| \ll H$, where $H \equiv P/(\rho g)$ is the pressure scale height, subject to (iii) instantaneous pressure balance with the surroundings, and (iv) adiabatic change of state. Then to first order in $\delta r$ the density contrast the parcel acquires is $$ \frac{\Delta\rho}{\rho} \;=\; -\,(\nabla-\nabla_{\rm ad})\,\frac{\delta r}{H}, \tag{2.3}$$ and the buoyancy force per unit mass on it is $+g\,(\nabla-\nabla_{\rm ad})\,\delta r/H$, directed along $\delta r$. The force therefore restores the parcel to its starting level if and only if $$ \nabla \;\lt \; \nabla_{\rm ad}. \tag{2.4}$$

Write $\Delta$ for the difference between the parcel and its surroundings at the same height, and $\delta$ for the change along the displacement. Both the parcel and the surroundings are ideal gases of the same $\mu$ by (i) and (ii), so $\ln\rho = \ln P - \ln T + \mathrm{const}$ for each, and

$$ \delta\ln\rho \;=\; \delta\ln P - \delta\ln T .$$

By (iii) the parcel's pressure is the ambient pressure at every instant, so $\delta\ln P$ is the same number for both. By (iv) the parcel's temperature follows its own adiabat, $\delta\ln T_{\rm p} = \nabla_{\rm ad}\,\delta\ln P$, while the surroundings supply $\delta\ln T = \nabla\,\delta\ln P$ by (2.1). Subtracting,

$$ \frac{\Delta\rho}{\rho} \;=\; \delta\ln\rho_{\rm p} - \delta\ln\rho \;=\; -(\nabla_{\rm ad}-\nabla)\,\delta\ln P \;=\;(\nabla-\nabla_{\rm ad})\,\delta\ln P .$$

Hydrostatic balance gives $dP/dr = -\rho g$, so $d\ln P/dr = -\rho g/P = -1/H$ and $\delta\ln P = -\delta r/H$ to first order. Substituting gives (2.3).

Archimedes' principle in the Euler equation is the statement that a parcel of density $\rho+\Delta\rho$ in surroundings of density $\rho$ feels a net force per unit mass $-g\,\Delta\rho/\rho$ in the direction of increasing $r$. With (2.3) that force is $+g(\nabla-\nabla_{\rm ad})\delta r/H$. Since $g$, $H$ and $\rho$ are positive, the force has the sign of $(\nabla-\nabla_{\rm ad})\,\delta r$: it points along $\delta r$, and so amplifies the displacement, when $\nabla \gt \nabla_{\rm ad}$, and against $\delta r$, restoring it, when $\nabla \lt \nabla_{\rm ad}$. ∎

Four premises were used and each will be paid for later. Premise (i), uniform composition, is lifted in §4. Premise (iii), instantaneous pressure balance, requires the parcel to move slowly compared with the sound speed, and §6 finds $v_c/c_s = 0.4271$ at the solar photosphere, where the premise is therefore spent. Premise (iv), adiabatic change, requires the parcel to lose no appreciable heat while it travels, and §6 and §7 find that at the photosphere radiation carries $98.7$ per cent of the flux, so it is spent there too. Only premise (ii) survives everywhere this module uses the criterion.

The criterion has one immediate consequence. For an ideal gas with a constant ratio of specific heats $\gamma$, an adiabatic change obeys $P\propto\rho^\gamma$ and the ideal gas law gives $P\propto\rho T$, so $T\propto P^{(\gamma-1)/\gamma}$ and

$$ \nabla_{\rm ad} \;=\; \frac{\gamma-1}{\gamma}, \tag{2.5}$$

which is $2/5 = 0.400000$ for a monatomic gas and $2/7 = 0.285714$ for a diatomic one. These are the two numbers §3 and Problem C1 compare against. Everything later in the module either changes the left-hand side of (2.4), which §3 does by reading $\nabla$ off a solar model, or the right-hand side, which §5 does by breaking the constancy of $\gamma$.

3. The buoyancy frequency, and where the Sun's convection zone starts

Proposition 1 gives a sign. Its magnitude is a frequency, and the frequency is a measurable property of a star.

Proposition 2 (the Brunt–Väisälä frequency). Under the hypotheses of Proposition 1, and neglecting any drag on the parcel, the displacement obeys $$ \ddot{\delta r} \;=\; -N^2\,\delta r, \qquad N^2 \;\equiv\; \frac{g}{H}\,(\nabla_{\rm ad}-\nabla). \tag{3.1}$$ The layer is stable if and only if $N^2\gt 0$, and a displaced parcel then oscillates about its starting level with angular frequency $N$.

The parcel has unit mass and the force per unit mass computed in Proposition 1 is $+g(\nabla-\nabla_{\rm ad})\delta r/H$. Newton's second law gives $\ddot{\delta r} = g(\nabla-\nabla_{\rm ad})\delta r/H = -(g/H)(\nabla_{\rm ad}-\nabla)\,\delta r$, which is (3.1). When $N^2\gt 0$ this is the equation of a harmonic oscillator of angular frequency $N$; when $N^2\lt 0$ its general solution is $A e^{|N|t}+Be^{-|N|t}$ and the growing branch is convection. The coefficients $g$, $H$, $\nabla$ and $\nabla_{\rm ad}$ are evaluated at the starting level and held fixed, which is the same first-order approximation used in Proposition 1. ∎

Equation (3.1) is the Schwarzschild criterion again, with a rate attached. It also identifies what the criterion costs when it is close: $N^2$ is a small difference of two numbers of order $0.4$, so a model table has to resolve $\nabla$ well to give $N^2$ at all inside a convection zone. That limit is stated below and respected.

3.1 Reading the criterion off a solar model

The stratification is the tabulated standard solar model BS2005-AGS,OP of Bahcall, Serenelli & Basu (2005), the same table Module 3 used. From its columns $\nabla$ is a sliding least-squares slope of $\ln T$ against $\ln P$ over $\pm25$ rows, $g = Gm(r)/r^2$ from the tabulated mass fraction, and $H = P/(\rho g)$.

The table is an input, not a checked quantity. It is used here only as a stratification: a set of $(P,\rho,T,m)$ values to differentiate. Module 3 measured its adiabatic plateau, $\nabla = 0.3957$ over $0.80$–$0.92\,R$, and located the base of its convection zone at $0.7269\,R$; this module's script recovers both numbers with the same locator. That agreement demonstrates that the two scripts read the table identically. It is not a check, because both numbers come from the same file.

Table 1 gives the result and Fig. 2, upper panel, plots it. $N$ is computed twice on the right: once with the ideal monatomic $\nabla_{\rm ad} = 0.4$, and once with the model's own measured plateau $0.3957$, which is what its equation of state — including Coulomb corrections and a little radiation pressure — actually delivers. The last column adds the composition term of §4.

Table 1. The buoyancy frequency through BS2005-AGS,OP. The $\mu$Hz columns are $N/2\pi$. "imaginary" means $N^2\lt 0$ at the quoted precision, which inside the convection zone is not a measurement; see the limit stated below.
$r/R$$\nabla$$\nabla_\mu$$H$ [cm]$N$ [$\mu$Hz], $\nabla_{\rm ad}=0.4$$N$ [$\mu$Hz], plateau$N$ [$\mu$Hz], $+$ Ledoux
0.05000.3251$+0.37196$$1.194\times10^{10}$144.84140.59352.04
0.09990.3220$+0.22043$$7.510\times10^{9}$233.70227.12453.84
0.14960.3032$+0.09531$$6.127\times10^{9}$306.06299.15426.24
0.19960.2629$+0.02629$$5.393\times10^{9}$386.01379.87415.78
0.30160.2116$+0.00256$$4.861\times10^{9}$427.47422.53425.46
0.40020.1968$+0.00134$$5.027\times10^{9}$373.34369.35370.59
0.50180.1962$+0.00155$$5.424\times10^{9}$304.51301.26302.43
0.59790.2074$+0.00166$$5.759\times10^{9}$247.82245.02246.09
0.68040.2616$+0.00906$$5.781\times10^{9}$186.66183.72189.83
0.70990.3259$+0.02423$$5.661\times10^{9}$132.67128.73149.43
0.75000.3963$-0.00307$$5.197\times10^{9}$29.34imaginaryimaginary
0.84930.3960$-0.00390$$3.477\times10^{9}$33.20imaginaryimaginary
0.94990.3870$-0.01165$$1.199\times10^{9}$91.2574.51imaginary
0.98000.3466$-0.02543$$4.374\times10^{8}$297.04284.75197.70

The maximum of $N$ in the radiative interior, over $0.05$–$0.65\,R$, computed with the plateau $\nabla_{\rm ad} = 0.3957$ rather than $0.4$, is

$$ N_{\max} = 2.7022\times10^{-3}\ \mathrm{rad\,s^{-1}} = 430.06\ \mu\mathrm{Hz} \quad\text{at}\quad r = 0.2670\,R, $$

and the corresponding buoyancy period $2\pi/N_{\max}$ is $38.8$ minutes. The maximum lies inside that window because $N\to0$ at the centre, where $g\to0$, and again at the base of the convection zone. Since a gravity mode of the star is a standing pattern of the oscillation of Proposition 2, and no such oscillation can be faster than the local $N$, every solar g mode must have a period longer than $38.8$ minutes. The Sun's famous five-minute oscillations are therefore acoustic, as Module 4 treated them, and the two mode families cannot be confused by their periods. This is a prediction of the model read through Proposition 2, not a check: no measured g-mode period is compared against it here.

3.2 Why the convection zone reaches down to $0.727\,R$

Read the $\nabla$ column of Table 1 outward, from the centre to the surface. It falls through $0.2116$ at $0.3016\,R$ to a shallow minimum — $0.1968$ at $0.4002\,R$ and $0.1962$ at $0.5018\,R$ — and then rises: $0.2074$ at $0.5979\,R$, $0.2616$ at $0.6804\,R$, $0.3259$ at $0.7099\,R$, and $0.3963$ at $0.7500\,R$. Across the whole interval $0.4002\,R$ to $0.7500\,R$, $\nabla$ rises by a factor $2.014$. The interval is not monotone, and the sampling matters: the rows printed above dip to $0.1962$ at $0.5018\,R$, while the full profile reaches its minimum of $0.1937$ at $0.4531\,R$, a radius Table 1 does not sample. Measured from that minimum the factor is $2.046$. Either way the rise carries $\nabla$ straight across the flat adiabatic plateau $0.3957$. The base of the convection zone is where the crossing happens, at $0.7269\,R$, which is $T = 2.02\times10^{6}$ K in the same table.

It is worth being exact about which side moves. In this star the convection zone opens because $\nabla$ rises to meet $\nabla_{\rm ad}$, not because $\nabla_{\rm ad}$ falls to meet $\nabla$: over the whole radiative interior $\nabla_{\rm ad}$ is the flat plateau value. A radiative layer carries its flux with $\nabla = \nabla_{\rm rad}\propto \kappa L P/(m\,T^4)$, so a rising opacity $\kappa$ and a falling temperature both push $\nabla$ up, and both are happening here. This module computes neither. It reads the rise off the model and does not apportion it between opacity and temperature; doing so would need the opacity tables the model was built with, which are not part of this module. §5 shows the other mechanism, the one that moves $\nabla_{\rm ad}$, and shows that it operates somewhere else entirely.

Honest limit on the sign of $N^2$ inside the convection zone. At $r = 0.90\,R$ the table gives $\nabla = 0.394599$ against the plateau $0.395673$, a difference of $-1.07\times10^{-3}$. §6 shows that the true superadiabatic excess there is about $1.9\times10^{-6}$, three orders of magnitude smaller. The table quotes $P$, $\rho$ and $T$ to four significant figures and cannot resolve a difference that small. So $N^2$ in the convection zone is reported here as zero to within the table's precision, and the "imaginary" entries of Table 1 are not a measurement of a negative $N^2$. The radiative-interior maximum is robust by contrast, because there $\nabla_{\rm ad}-\nabla$ is of order $0.1$.
Buoyancy frequency through the Sun 0.0 0.2 0.4 0.6 0.8 1.0 0 100 200 300 400 500 r / R N / 2π (µHz) maximum 430 µHz at 0.27 R so every g mode has period > 38.8 min with the Ledoux composition term convection zone N² = 0 to the table's precision base 0.727 R How superadiabatic the convection zone must be to carry the Sun's luminosity 0.70 0.75 0.80 0.85 0.90 0.95 1.00 10⁻⁷ 10⁻⁶ 10⁻⁵ 10⁻⁴ 10⁻³ 10⁻² 10⁻¹ 1 r / R ∇ − ∇ad 1.9×10⁻⁶ at 0.90 R: adiabatic to one part in 5×10⁵ table ends 0.983 R photosphere, carrying all of L: 0.61 carrying the measured 1.3% of L: 0.034 mixing-length theory, α = 2, L(r) from the table Stratification: BS2005-AGS,OP (Bahcall, Serenelli & Basu 2005). N² uses the measured adiabatic plateau ∇ = 0.3957, not 0.4.
Convection in the Sun. Upper panel: the buoyancy frequency $N(r)$ of Proposition 2 through BS2005-AGS,OP, computed with the model's measured adiabatic plateau $\nabla_{\rm ad} = 0.3957$ rather than $0.4$. It rises from the centre to $430.06\ \mu$Hz at $0.2670\,R$ and falls to zero at the base of the convection zone, $0.7269\,R$; the shaded band beyond it is where $N^2$ is zero to the table's precision. The dashed curve adds the Ledoux composition term of §4, which roughly doubles $N$ inside $0.1\,R$ and makes no difference outside $0.3\,R$. Lower panel: the superadiabatic excess $\nabla-\nabla_{\rm ad}$ that mixing-length theory needs to carry the solar luminosity (§6), on a logarithmic axis, from $1.9\times10^{-6}$ at $0.90\,R$ to order unity at the photosphere. The photospheric point is drawn twice: once carrying all of $L$, and once carrying only the convective flux measured in §7.

4. Ledoux: what a composition gradient does

Premise (i) of Proposition 1 was uniform composition. The Sun has burned hydrogen in its core for $4.6$ Gyr, so its $\mu$ is not uniform, and the premise has to be paid for.

Proposition 3 (the Ledoux criterion). Keep every hypothesis of Proposition 1 except (i), and let the composition of the surroundings vary with depth, with $\nabla_\mu \equiv d\ln\mu/d\ln P$, while the displaced parcel carries its own composition unchanged. Then (2.3) becomes $$ \frac{\Delta\rho}{\rho} \;=\; -\left[\nabla-\nabla_{\rm ad}-\nabla_\mu\right]\frac{\Delta r}{H}, $$ the buoyancy frequency becomes $$ N^2 \;=\; \frac{g}{H}\left[\nabla_{\rm ad}-\nabla+\nabla_\mu\right], \tag{4.1}$$ and the layer is stable if and only if $\nabla \lt \nabla_{\rm ad} + \nabla_\mu$. Here $\Delta r$ is the displacement written $\delta r$ in Proposition 1; it is renamed for this proof only, so that no symbol has to serve both as a variation and as a coefficient.

For an ideal gas $\rho = \mu m_uP/(k_BT)$, so along the displacement $\Delta\ln\rho = \Delta\ln P - \Delta\ln T + \Delta\ln\mu$ for the surroundings, and the same without the last term for the parcel, which carries its composition unchanged. The parcel and its surroundings share $\Delta\ln P$ by premise (iii), so those terms cancel in the difference exactly as in Proposition 1. The temperature terms give $-(\nabla_{\rm ad}-\nabla)\Delta\ln P$ as before, and the composition term gives $-\nabla_\mu\,\Delta\ln P$. Subtracting the surroundings from the parcel,

$$ \frac{\Delta\rho}{\rho} = \left[(\nabla-\nabla_{\rm ad}) - \nabla_\mu\right]\Delta\ln P,$$

and $\Delta\ln P = -\Delta r/H$ as before gives the displayed contrast. Newton's second law with the buoyancy force $-g\Delta\rho/\rho$ then gives (4.1), and stability is $N^2\gt 0$, which is the stated inequality. Setting $\nabla_\mu=0$ recovers (3.1). ∎

For a general equation of state the two composition coefficients $\delta \equiv -(\partial\ln\rho/\partial\ln T)_{P,\mu}$ and $\varphi \equiv (\partial\ln\rho/\partial\ln\mu)_{P,T}$ do not both equal $1$, and (4.1) carries a factor $\delta$ outside the bracket and $\varphi/\delta$ on $\nabla_\mu$ (Kippenhahn & Weigert, section 6.1). Every gas this module treats is an ideal gas, so that generality is quoted and not used.

The sign is the physical content. A $\mu$ that increases inward means $\nabla_\mu \gt 0$, since $P$ also increases inward; a parcel carried up from such a layer keeps its heavy composition and arrives denser than its new surroundings, which stabilises the stratification even where $\nabla \gt \nabla_{\rm ad}$.

Table 1's $\nabla_\mu$ column shows the Sun's. The $\mu$ used there is the table's own, $\mu = \rho k_BT/(P\,m_u)$, so no ionisation assumption enters. The share of $N^2$ contributed by the Ledoux term is $+84.05$ per cent at $0.050\,R$ ($\mu = 0.77684$, $\nabla_\mu = +0.37196$), $+74.96$ per cent at $0.100\,R$ ($\mu = 0.69264$, $\nabla_\mu = +0.22043$), $+50.74$ per cent at $0.150\,R$, $+16.53$ per cent at $0.200\,R$, $+3.07$ per cent at $0.251\,R$, $+1.37$ per cent at $0.302\,R$, $+0.67$ per cent at $0.400\,R$ and $+0.87$ per cent at $0.598\,R$. The gradient left by hydrogen burning is large where it exists and dies away outside the burning core.

Does the Ledoux term move any boundary in the Sun? No, and the measured size of the "no" is worth printing. The Ledoux term matters only where the two criteria could disagree, which is where $\nabla$ is close to $\nabla_{\rm ad}$. Where $\nabla_\mu$ is large, inside $0.2\,R$, the Schwarzschild margin $\nabla_{\rm ad}-\nabla$ runs from $0.0706$ at $0.050\,R$ to $0.1328$ at $0.200\,R$, so those layers are stable by a wide margin under either criterion. Through the radiative zone outside the burning core, over $0.30$–$0.60\,R$, $|\nabla_\mu| \le 0.00256$. Locating the base of the convection zone again with $\nabla_{\rm ad}+\nabla_\mu$ in place of $\nabla_{\rm ad}$, using the same locator, moves it from $0.7269\,R$ to $0.7306\,R$: a shift of $0.0037\,R$, or $0.37$ per cent of the solar radius. And that is an upper bound, because the $\nabla_\mu$ the table gives near the base is partly the effective-$\mu$ artefact described next.
One caution on the outer rows of Table 1, and it is not a footnote. Beyond about $0.72\,R$ the $\mu$ printed there is an effective $\mu = \rho k_BT/(P\,m_u)$, which absorbs the changing ionisation state of hydrogen and helium as well as the composition. Its gradient in those rows is an ionisation effect, not a composition gradient, and it must not be fed into Proposition 3: the parcel does not carry its ionisation state unchanged, it re-ionises. The percentages quoted above therefore stop at $0.60\,R$.

The case where the choice of criterion does matter is a star with a receding convective core, which leaves a large $\nabla_\mu$ and a near-neutral layer in the same place. Such a layer is Schwarzschild-unstable and Ledoux-stable at once, and mixes on a thermal rather than a dynamical timescale — semiconvection. Problem D2 puts numbers on it and shows why the Sun escapes the ambiguity. This module does not develop semiconvection further.

The right-hand side of (2.4) has been treated so far as the constant $0.4$. It is not a constant.

5. The collapse of $\nabla_{\rm ad}$ in a partially ionised gas

Equation (2.5) assumed a constant $\gamma$. In a gas that is partly ionised, some of the energy a compressed parcel receives goes into pulling electrons off atoms instead of into raising the temperature, and the parcel therefore heats less for the same compression. That lowers $\nabla_{\rm ad}$, and it lowers it a long way.

Definition 2 (Saha equilibrium for pure hydrogen). Let $x \equiv n_{\rm II}/n_{\rm H}$ be the ionisation fraction, with $n_{\rm H}$ the total density of hydrogen nuclei, so that $n_e = n_{\rm II} = x\,n_{\rm H}$ and $n_{\rm I} = (1-x)\,n_{\rm H}$. Saha's equation with the statistical weight ratio $g_{\rm II}/g_{\rm I} = 1/2$ for hydrogen is $$ \frac{n_e\,n_{\rm II}}{n_{\rm I}} \;=\; \left(\frac{2\pi m_ek_BT}{h^2}\right)^{3/2}\exp\!\left(-\frac{\chi_{\rm H}}{k_BT}\right) \;\equiv\; A(T), \tag{5.1}$$ with $\chi_{\rm H} = 13.5984340$ eV the ionisation energy of hydrogen.

The total particle density is $n_{\rm H}(1+x)$, so the gas pressure is $P = n_{\rm H}(1+x)k_BT$. Eliminating $n_{\rm H}$ between this and (5.1) gives a closed form with no iteration:

$$ \frac{x^2}{1-x^2} \;=\; \frac{A(T)\,k_BT}{P} \;\equiv\; B(T,P), \qquad x \;=\; \sqrt{\frac{B}{1+B}}. \tag{5.2}$$

For a partially ionised pure-hydrogen gas, the first law with the ionisation energy included in the internal energy gives $$ \nabla_{\rm ad} \;=\; \frac{2+x(1-x)\Phi}{5+x(1-x)\Phi^2}, \qquad \Phi \;\equiv\; \frac{5}{2}+\frac{\chi_{\rm H}}{k_BT}. \tag{5.3}$$

(5.3) is quoted from Kippenhahn & Weigert, section 14.3, and is not derived in this module. It is the only result in Module 6 taken on authority rather than proved. Its two limits are checkable without the derivation, and Proposition 4 checks them: at $x=0$ and $x=1$ it must return the monatomic $2/5$ of (2.5), and it does.
Proposition 4 (the ionisation zone lowers $\nabla_{\rm ad}$). Assume (5.3), with $x$ from (5.2) and $\chi_{\rm H} \gt 0$. Then $\nabla_{\rm ad} \lt 2/5$ at every $0\lt x\lt 1$, and $\nabla_{\rm ad} = 2/5$ exactly at $x=0$ and $x=1$.

Write $q \equiv x(1-x)$, which is positive for $0\lt x\lt 1$ and zero at both ends. Comparing (5.3) with $2/5$ by cross-multiplication, and noting that both denominators are positive,

$$ \nabla_{\rm ad} \lt \tfrac25 \iff 5\,(2+q\Phi) \lt 2\,(5+q\Phi^2) \iff 5q\Phi \lt 2q\Phi^2 \iff q\,\Phi\,(2\Phi-5) \gt 0 .$$

Now $\Phi = 5/2 + \chi_{\rm H}/(k_BT) \gt 5/2$ for every finite $T$, so $2\Phi-5 \gt 0$ and $\Phi\gt 0$. The inequality therefore holds exactly when $q\gt 0$, that is for $0\lt x\lt 1$. At $x=0$ or $x=1$, $q=0$ and (5.3) returns $2/5$, the monatomic value of (2.5). ∎

The size of the effect is set by $\Phi$, and $\Phi$ is large. At the temperature where the depression is deepest on the photospheric isobar, $T = 12266$ K, $\chi_{\rm H}/k_BT = 12.86$ and so $\Phi = 15.36$; at the hotter temperature where the gas is half ionised, $T = 13136$ K, it is $12.01$. Either way the ionisation reservoir a parcel can pour heat into is about twelve times the thermal reservoir of one particle, which is why the depression is not a small correction. Table 2 evaluates (5.2) and (5.3) along an isobar at the photospheric pressure $P = 1.2\times10^{5}\ \mathrm{dyn\,cm^{-2}}$.

Table 2. Ionisation fraction and adiabatic gradient of pure hydrogen along the isobar $P = 1.2\times10^{5}\ \mathrm{dyn\,cm^{-2}}$, from (5.2) and (5.3).
$T$ [K]$x$$\nabla_{\rm ad}$$\nabla_{\rm ad}/0.4$
40000.00000.399981.0000
60000.00020.389910.9748
80000.00660.260860.6521
100000.06230.125120.3128
120000.28030.094790.2370
150000.82090.130800.3270
200000.99160.353740.8844
300000.99980.399280.9982
1000001.00000.400001.0000
The hydrogen ionisation zone drops $\nabla_{\rm ad}$ from $0.4000$ to $0.09448$, a factor $4.23$. The minimum along this isobar is at $T = 12266$ K, where $x = 0.3271$.

The minimum does not sit at $x = 1/2$, and the reason is worth one sentence. At fixed $\Phi$ the numerator and denominator of (5.3) both depend on $x$ only through $q = x(1-x)$, which peaks at $x = 1/2$; but $\Phi$ falls as $T$ rises, so moving along the isobar changes $q$ and $\Phi$ together, and the two effects put the minimum at $x = 0.3271$, on the cool side of half ionisation. Problem K1 follows the minimum across five decades of pressure and finds it drifts from $0.08097$ at $10580$ K to $0.12933$ at $16798$ K — Saha equilibrium needs a hotter gas to ionise a denser one.

5.1 What this explains, and what it does not

It would be natural to say that the collapse of $\nabla_{\rm ad}$ is what makes the outer part of the Sun convective. The model table says otherwise, and the numbers are unambiguous.

So the hydrogen ionisation zone lies outside $0.983\,R$, in the outermost $1.7$ per cent of the radius. It does not open the convection zone, whose outer $27.3$ per cent by radius, from $0.7269\,R$ to the surface, is set the other way round by $\nabla$ rising, as §3.2 showed. What Proposition 4 explains is why convection reaches all the way up to the photosphere instead of stopping below it, and why the strongly superadiabatic layer of §6 sits where it does.

Two mechanisms, two depths. $\nabla$ rising to meet a flat $\nabla_{\rm ad}$ sets the base of the solar convection zone at $0.7269\,R$. $\nabla_{\rm ad}$ collapsing by the factor $4.23$ in the partial-ionisation zone sets its top, above $0.983\,R$, where $\nabla-\nabla_{\rm ad}$ runs from $1.9\times10^{-6}$ to $0.61$.

The criterion has now been derived, corrected for composition, and corrected for ionisation. It still says only whether convection happens. It says nothing about how much heat it carries.

6. Mixing-length theory, and how superadiabatic the Sun must be

A stability criterion cannot give a flux, because it contains no statement about how far a parcel travels before it merges with its surroundings. Mixing-length theory supplies exactly one such statement and derives everything else from it.

Definition 3 (the mixing length). A convecting parcel is assumed to rise a distance $\ell$, keeping its identity, and then to dissolve into its surroundings and deliver its excess heat. The mixing length is taken proportional to the local pressure scale height, $$ \ell \;\equiv\; \alpha\,H, \qquad H = \frac{P}{\rho g}, \tag{6.1}$$ with $\alpha$ a dimensionless parameter of order unity. This module uses $\alpha = 2.0$.
Notation. Within this module $\alpha$ is the mixing-length parameter and nothing else. Module 5's $\alpha$, the length scale of the isothermal sphere, is not meant here.
$\alpha$ is not derived. It is calibrated: a value near $2$ is what makes a solar evolution model reach the observed solar radius at the solar age. Every prediction in §7 therefore rests on a parameter that was fitted to the Sun, and the checks say so where they are made. Problem C3 measures how much freedom that leaves.
Proposition 5 (mixing-length flux and velocity, efficient limit). Let a layer have $\nabla \gt \nabla_{\rm ad}$ and write the superadiabatic excess $\Delta\nabla \equiv \nabla-\nabla_{\rm ad}$. Assume the parcel loses no heat while it travels, so that it follows its own adiabat over the whole mixing length, and take the ideal-gas value $\delta = 1$. Adopt the two standard averaging conventions: the parcel's mean displacement over its life is $\ell/2$, and half the buoyancy work done over that displacement becomes the parcel's kinetic energy. Then the mean temperature excess a parcel carries, its velocity, and the convective flux are $$ \delta T \;=\; T\,\Delta\nabla\,\frac{\ell}{2H}, \qquad v_c \;=\; \frac{\ell}{H}\sqrt{\frac{gH\,\Delta\nabla}{8}}, \tag{6.2}$$ $$ F \;=\; \rho\,v_c\,c_P\,\delta T \;=\; \rho\,c_P\,T\,\alpha^2\sqrt{\frac{gH}{32}}\;\Delta\nabla^{3/2}. \tag{6.3}$$ Consequently, at fixed $F$, $\rho$, $c_P$, $T$, $g$ and $H$, $\Delta\nabla\propto\alpha^{-4/3}$ and $v_c\propto\alpha^{1/3}$.

Proposition 1 gives the temperature difference a parcel acquires after rising a distance $\Delta r$: subtracting the two logarithmic temperature changes at equal pressure, $\Delta T/T = \Delta\nabla\,\Delta r/H$. Averaging over the parcel's life, which is taken to be the mean displacement $\ell/2$, gives the first of (6.2).

For the velocity, the buoyancy force per unit mass at displacement $\Delta r$ is $-g\,\Delta\rho/\rho = g\,\Delta T/T = g\,\Delta\nabla\,\Delta r/H$, using $\Delta\rho/\rho = -\Delta T/T$ for an ideal gas at fixed pressure. The work done over the mean displacement is $$ W = \int_0^{\ell/2} \frac{g\,\Delta\nabla}{H}\,\Delta r\;d(\Delta r) = \frac{g\,\Delta\nabla\,\ell^2}{8H}. $$ Half of that work is assumed to go into the kinetic energy of the parcel and half into pushing the surroundings aside, so $\tfrac12 v_c^2 = \tfrac12 W$, that is $v_c^2 = g\Delta\nabla\,\ell^2/(8H)$, which is the second of (6.2).

The enthalpy a parcel carries per unit mass is $c_P\,\delta T$, and the mass flux carried upward by the convecting fluid is $\rho v_c$, so $F = \rho v_c c_P \delta T$. Substituting (6.2) and $\ell = \alpha H$, $$ F = \rho c_P\,T\,\Delta\nabla\frac{\alpha}{2}\cdot\alpha\sqrt{\frac{gH\Delta\nabla}{8}} = \rho c_P T\,\alpha^2\sqrt{\frac{gH}{32}}\,\Delta\nabla^{3/2},$$ since $\tfrac12\sqrt{1/8} = \sqrt{1/32}$. Holding $F$, $\rho$, $c_P$, $T$, $g$ and $H$ fixed and varying $\alpha$ in (6.3) gives $\Delta\nabla\propto\alpha^{-4/3}$, and putting that into (6.2) gives $v_c\propto\alpha\cdot\Delta\nabla^{1/2}\propto\alpha^{1/3}$. ∎

The factors $1/2$ in $\delta T$ and $1/8$ in $v_c^2$ are the two averaging conventions of the standard theory. A different convention would move both by a numerical factor, which is absorbed into a different calibrated $\alpha$; it changes no physics and no scaling. What the scalings say is that $\alpha$ is a weak handle on the velocity and a strong one on the excess, and Problem C3 exploits exactly that.

6.1 The excess through the solar convection zone

Inverting (6.3) for $\Delta\nabla$ at the flux the star must actually carry, $F = L(r)/(4\pi r^2)$ from the model's own luminosity column, gives Table 3. The local $L(r)$ is used rather than $L_\odot$; inside the convection zone the two agree to five figures.

Table 3. Mixing-length theory through the solar convection zone at $\alpha = 2.0$, from (6.2) and (6.3) with the BS2005-AGS,OP stratification. $c_s$ is the adiabatic sound speed $\sqrt{(5/3)P/\rho}$.
$r/R$$T$ [K]$H$ [cm]$F$ [erg cm$^{-2}$ s$^{-1}$]$\nabla-\nabla_{\rm ad}$$v_c$ [cm s$^{-1}$]$v_c/c_s$$\ell/v_c$ [s]
0.7200$2.083\times10^{6}$$5.589\times10^{9}$$1.2161\times10^{11}$$1.925\times10^{-7}$$5.2750\times10^{3}$0.000242118871
0.7500$1.791\times10^{6}$$5.197\times10^{9}$$1.1209\times10^{11}$$2.478\times10^{-7}$$5.5548\times10^{3}$0.000271871201
0.8002$1.341\times10^{6}$$4.391\times10^{9}$$9.8460\times10^{10}$$4.072\times10^{-7}$$6.1562\times10^{3}$0.000351426538
0.8493$9.504\times10^{5}$$3.477\times10^{9}$$8.7413\times10^{10}$$7.543\times10^{-7}$$7.0417\times10^{3}$0.00048987565
0.9005$5.852\times10^{5}$$2.385\times10^{9}$$7.7748\times10^{10}$$1.861\times10^{-6}$$8.6531\times10^{3}$0.00075551281
0.9499$2.690\times10^{5}$$1.199\times10^{9}$$6.9864\times10^{10}$$8.469\times10^{-6}$$1.2415\times10^{4}$0.00159193192
0.9800$9.740\times10^{4}$$4.374\times10^{8}$$6.5639\times10^{10}$$7.413\times10^{-5}$$2.1505\times10^{4}$0.0047240681
0.9831$8.120\times10^{4}$$3.617\times10^{8}$$6.5235\times10^{10}$$1.122\times10^{-4}$$2.3984\times10^{4}$0.0058030159
At $r = 0.90\,R$ the superadiabatic excess needed to carry the whole solar luminosity is $\nabla-\nabla_{\rm ad} = 1.861\times10^{-6}$, which is $4.7\times10^{-4}$ per cent of $\nabla_{\rm ad}$ itself. The stratification there is adiabatic to one part in $5\times10^{5}$.

That number repays a debt to Module 3. Module 3 fitted the solar convection zone with a polytrope of index $n = 3/2$ — the index of an adiabatic monatomic gas — and found $n_{\rm eff} = 1.5273$, agreeing to $1.8$ per cent. Table 3 says why such a fit can work at all: convection deep in the envelope is so efficient that the departure from an adiabat is six orders of magnitude below the adiabat itself. The polytrope is not an approximation that happens to be good; it is good because the transport is fast.

6.2 The photosphere, where the same theory is strained

The BS2005-AGS,OP table stops at $r = 0.98308\,R$, well below the photosphere, so the surface layer is computed from independent photospheric inputs: the IAU 2015 nominal effective temperature $T = 5772$ K, a gas pressure $P = 1.2\times10^{5}\ \mathrm{dyn\,cm^{-2}}$ at optical depth $\tau = 2/3$, and $\mu = 1.2250$ from the model's surface hydrogen and helium fractions, as Module 4 eq. (4.7) computes it. The gravity is the IAU nominal $g = GM_\odot/R_\odot^2 = 2.7420\times10^{4}\ \mathrm{cm\,s^{-2}}$, and $c_P = (5/2)k_B/(\mu m_u)$ is the monatomic ideal-gas value.

All four photospheric inputs are inputs, not checked quantities, and two of them are model-dependent: $P$ at $\tau = 2/3$ comes from a solar atmosphere calculation, and $\mu = 1.2250$ from the surface hydrogen and helium fractions of the BS2005-AGS,OP model, through Module 4 eq. (4.7). Module 3 used a round $\mu = 1.3$ for the same layer. One convention differs from Module 4: the gravity here uses the IAU nominal radius, as Module 3 does and as every other photospheric constant in this module does, while Module 4 used the model table's radius $6.9598\times10^{10}$ cm. The consequence is $H = 142.9$ km here against the $143.0$ km Module 4 prints. That $0.1$ km is the choice of radius and nothing else; the composition is the same in both.

With those inputs, and requiring convection to carry the entire emergent flux $F = L_\odot/(4\pi R_\odot^2)$:

QuantityValue
$H = k_BT/(\mu m_u g)$$142.9$ km
$\rho = P\mu m_u/(k_BT)$$3.0631\times10^{-7}\ \mathrm{g\,cm^{-3}}$
$c_P$$1.6968\times10^{8}\ \mathrm{erg\,g^{-1}\,K^{-1}}$
$F = L_\odot/(4\pi R_\odot^2)$$6.2939\times10^{10}\ \mathrm{erg\,cm^{-2}\,s^{-1}}$
$\nabla-\nabla_{\rm ad}$ (upper bound)$0.6079$
$v_c$$3.451\ \mathrm{km\,s^{-1}}$
$c_s$$8.080\ \mathrm{km\,s^{-1}}$
$v_c/c_s$$0.4271$
$\ell = \alpha H$$286$ km
turnover $\ell/v_c$$1.38$ min
The superadiabatic excess runs from $1.9\times10^{-6}$ at $0.90\,R$ to $0.61$ at the photosphere, a factor $3\times10^{5}$. Fig. 2, lower panel, draws the whole range on one logarithmic axis. Convection is efficient in the deep envelope and inefficient at the surface, where a parcel is optically thin and radiates its heat excess away before it has travelled a mixing length.
Three premises are spent here, and they do not all push the same way. (a) The photospheric numbers above were obtained by demanding that convection carry all of $L_\odot$ at $\tau = 2/3$. Radiation already carries most of that flux there; §7 measures the split and finds convection carries $1.3$ per cent. Taken alone, that assumption makes $\nabla-\nabla_{\rm ad} = 0.6079$ and $v_c = 3.451\ \mathrm{km\,s^{-1}}$ overestimates. (b) Proposition 5's own hypothesis — the parcel loses no heat over the whole mixing length — is the one the key result above says fails: a photospheric parcel is optically thin and radiates its excess away before it has travelled $\ell$. That failure works the other way on the excess, because a leaky parcel delivers less $\delta T$ for the same $\nabla-\nabla_{\rm ad}$ and so needs a larger gradient to carry a given flux. The two therefore bracket nothing: $0.6079$ is an upper bound with respect to (a) alone, and this module does not compute the correction for (b). (c) $v_c/c_s = 0.4271$, so premise (iii) of Proposition 1 — a slow parcel in instantaneous pressure balance — has failed at the photosphere by a wide margin. All three are read off the table above and none needs a measurement.

The range of turnover times is a separate result. Table 3 and Problem K2 give $\ell/v_c = 21.66$ d at $0.750\,R$, $8.38$ h at the last tabulated row $0.9831\,R$, and $1.38$ min at the photosphere: a factor $2\times10^{4}$ between the two ends of one convection zone. No single mixing length describes both.

7. Check A: mixing-length theory against solar granulation

Mixing-length theory makes two predictions about the solar photosphere that can be compared with measurements taken there. Both measurements are of the quiet Sun at disk centre, and neither used a stellar model to produce the quantity compared.

7.1 A1: the size of a granule

The overturning eddy of Definition 3 is one mixing length across. Seen from above, a convection cell has upflow in the middle and downflow at its edges, one mixing length each way, so the natural prediction for the diameter of a granule is $2\ell$. With $H = 143$ km and $\alpha = 2$, that is $\ell = 286$ km and $2\ell = 572$ km.

Abramenko et al. (2012) measured granule sizes with the New Solar Telescope and fitted the regular-granule population with a Gaussian, obtaining, in their section 3, a mean diameter $d_0 = 1050\pm22$ km and a width $\sigma_d = 480\pm11$ km.

Those two numbers are different quantities and this module prints both, with each denominator named, because they answer different questions.

DenominatorWhat it isValueDeparture of $2\ell = 572$ km
$22$ kmthe error on the mean diameter$1050\pm22$ km$21.7\,\sigma$
$480$ kmthe width of the granule population$\sigma_d = 480\pm11$ km$1.00\,\sigma$

The ratio of measured to predicted is $1050/572 = 1.84$; against $\ell$ alone it is $3.67$. Read the two departures together: $21.7\,\sigma$ says the mean granule is not $572$ km across, and $1.00\,\sigma$ says a $572$ km cell is an entirely ordinary member of the granule population. Abramenko's own Conclusions print "$1050\pm480$ km", which reads as an error bar and is not one; a reader who opens the paper will meet that line, and it is the population width.

A1: the scale is confirmed, the number disagrees. The pressure scale height sets the granule size — $H = 143$ km against a $1050$ km granule, with $\alpha$ of order unity, is the content of the agreement — and the identification of the cell with $2\ell$ is itself ambiguous, since $2\ell$ gives a ratio $1.84$ and $\ell$ alone gives $3.67$. The specific prediction $2\ell = 572$ km is $21.7\,\sigma$ from the measured mean of $1050\pm22$ km. Module 3 reported its $29.4\,\sigma$ as a disagreement rather than a refutation, and this module follows that precedent. Note also that $\alpha = 2$ was calibrated on the solar radius, so this is not a parameter-free prediction; what is parameter-free is that the scale is set by $H$ at all.

7.2 A2: the speed of the upflow

Oba, Iida & Shimizu (2017) measured photospheric velocities with the Hinode spectropolarimeter, using a bisector analysis of Fe I $630.15$ nm with a subsonic filter applied to remove the five-minute oscillation. They find an upward speed falling from $0.65\ \mathrm{km\,s^{-1}}$ at a geometrical height $h = 40$ km to $0.40\ \mathrm{km\,s^{-1}}$ at $h = 163$ km, and an rms filtered convective velocity running from $0.6\ \mathrm{km\,s^{-1}}$ at the deepest level sampled to $0.3\ \mathrm{km\,s^{-1}}$ at the highest.

Two properties of that measurement must travel with it. Their stated velocity error, $0.18\ \mathrm{km\,s^{-1}}$, is a zero-point systematic used as a threshold for calling a pixel up or down; it is not a random error on the quoted speeds, and it is not used as a $\sigma$ anywhere in this module. Their heights are derived from the VAL81 atmosphere model at each intensity level, not measured directly.

Against the $3.451\ \mathrm{km\,s^{-1}}$ of §6.2, the ratios are $5.3$ against the $0.65\ \mathrm{km\,s^{-1}}$ upflow and $5.8$ against the $0.6\ \mathrm{km\,s^{-1}}$ rms. Too fast by a factor $5.3$, against the factor $1.25$ the same theory returns in §7.3 once it is given the measured flux.

The right response is not to stop here. The calculation of §6.2 contains an assumption that is no part of mixing-length theory at all: that convection carries the whole luminosity at $\tau = 2/3$. That assumption can be tested against the same two measurements, and the independent warning $v_c/c_s = 0.43$ already says the theory has been pushed past its own premise.

7.3 A3, worked: what flux does the measured granulation actually carry?

This subsection uses no stellar model beyond the photospheric inputs already in use, and it rests on three assumptions that must be named before the numbers. The density $\rho$ at $\tau = 2/3$ is an input carried from Module 3, not a measurement. The observed intensity contrast is converted to a temperature contrast through a Planck source function, as if the contrast formed at a single depth. And the convective flux is formed as the product of a mean upflow speed with an rms temperature contrast, which assumes the two are perfectly correlated across the granulation pattern. A fourth mismatch is not an assumption but a fact worth stating: Oba et al.'s $0.65\ \mathrm{km\,s^{-1}}$ is measured at $h = 40$ km, $\tau_{500} = 0.55$, while $\rho$ and $c_P$ are taken at $\tau = 2/3$.

Step 1: from intensity contrast to temperature contrast. Abramenko et al. report an rms intensity contrast of $15.5\pm0.6$ per cent at $705.7$ nm. For a Planck function, $\delta I/I = (d\ln B_\lambda/d\ln T)\,(\delta T/T)$ with

$$ \frac{d\ln B_\lambda}{d\ln T} \;=\; \frac{x\,e^x}{e^x-1}, \qquad x \;\equiv\; \frac{hc}{\lambda k_BT}. \tag{7.1}$$

At $\lambda = 705.7$ nm and $T = 5772$ K, $x = 3.5322$ and $d\ln B/d\ln T = 3.6386$. So $\delta T/T = 0.155/3.6386 = 0.04260$, that is $\delta T = 245.9$ K. The bandpass amplifies the contrast by a factor $3.64$: a $15.5$ per cent intensity contrast is only a $4.3$ per cent temperature contrast, and reading the intensity contrast as a temperature contrast directly would overstate the flux by that same factor.

Step 2: the convective flux the measurements imply. Feed the measured $\delta T$ and the measured velocity into the mixing-length flux expression $F = \rho\,v\,c_P\,\delta T$ of Proposition 5, which needs no $\alpha$ once $v$ and $\delta T$ are given:

QuantityValueSource
$\rho$ at $\tau = 2/3$$3.0631\times10^{-7}\ \mathrm{g\,cm^{-3}}$input, from the photospheric $P$, $T$, $\mu$
$v$$0.65\ \mathrm{km\,s^{-1}}$measured, Oba et al. (2017)
$c_P$$1.6968\times10^{8}\ \mathrm{erg\,g^{-1}\,K^{-1}}$input, $(5/2)k_B/(\mu m_u)$
$\delta T$$245.9$ Kmeasured contrast through (7.1)
$F_{\rm conv}$$8.3067\times10^{8}\ \mathrm{erg\,cm^{-2}\,s^{-1}}$the product
$F_{\rm total} = L_\odot/(4\pi R_\odot^2)$$6.2939\times10^{10}\ \mathrm{erg\,cm^{-2}\,s^{-1}}$input
$F_{\rm conv}/F_{\rm total} = 0.0132$. Convection carries $1.3$ per cent of the flux at $\tau = 2/3$; the other $98.7$ per cent is radiation.

Step 3: run the theory again, on the flux it should have been given. The calculation of §6.2 over-drove mixing-length theory by a factor $76$ in flux. Since $v_c\propto F^{1/3}$ at fixed $\alpha$ by (6.3), that is a factor $76^{1/3} = 4.2$ in velocity, which is most of the discrepancy of $5.3$ found in §7.2. Inverting (6.3) at $F = F_{\rm conv}$ instead gives $\nabla-\nabla_{\rm ad} = 0.0340$ and

$$ v_c = 0.816\ \mathrm{km\,s^{-1}} \quad\text{against the measured } 0.65\ \mathrm{km\,s^{-1}}, \quad\text{a ratio of } 1.25. $$
The verdict, in this order. REFUTED: the proposition that convection carries the solar flux at $\tau = 2/3$, which fails by a factor of $76$. SURVIVES: the mixing-length velocity law itself, to $25$ per cent, once it is asked a question it can answer.

What remains genuinely outside mixing-length theory is the split of the flux between radiation and convection, which needs an atmosphere model with radiative transfer that this module does not build. And the rest cannot be checked the same way: deep in the convection zone, where all four premises of Proposition 1 do hold, the excess is $10^{-6}$ and there is nothing to measure directly. The theory is checkable against data only in the one layer where it works worst.

Buoyancy is one way a stratified fluid loses its footing. The other needs no gravity at all.

8. Thermal instability: the loss function and Field's criteria

The question is the same one Proposition 1 asked — a parcel is perturbed, does the perturbation grow? — with a different restoring mechanism. Compress a parcel of interstellar gas slightly at the pressure of its surroundings. Its density rises, so it radiates more; if it also loses the ability to be reheated fast enough, it cools, and at fixed pressure a cooler parcel is denser still. Nothing in this argument involves gravity, buoyancy or a stratification.

Definition 4 (Field's net loss function). Let $\mathcal{L}(\rho,T)$ be the energy losses minus the energy gains of the gas, per gram per second, exclusive of thermal conduction (Field 1965, p. 532). Thermal equilibrium is $\mathcal{L} = 0$.
Notation. The loss function is always written $\mathcal{L}$ in this module. The plain letter $L$ is the solar luminosity, as in §6. The cooling coefficient below is always written $\Lambda(T)$ with its argument shown, to keep it distinct from Module 1's Coulomb logarithm $\ln\Lambda$, which reappears in Problem K3.
Proposition 6 (Field's criteria, 1965, eqs. 4a and 4b). Perturb a parcel in thermal equilibrium by $\delta T$. Two idealised paths are compared. A parcel held at constant density runs away if and only if $$ \left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\!\rho} \lt 0 \qquad\text{(isochoric, Field 4a)}, \tag{8.1}$$ and a parcel that stays in mechanical equilibrium with its surroundings throughout — which requires the perturbation to evolve slowly compared with its own sound-crossing time, and so is perturbed at constant pressure — runs away if and only if $$ \left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\!P} \;=\; \left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\!\rho} - \frac{\rho_0}{T_0}\left(\frac{\partial\mathcal{L}}{\partial\rho}\right)_{\!T} \;\lt \;0 \qquad\text{(isobaric, Field 4b)}. \tag{8.2}$$ If $(\partial\mathcal{L}/\partial\rho)_T \gt 0$, as it is for any cooling rate that increases with density, then (8.1) implies (8.2), and there exist states satisfying (8.2) but not (8.1): the isobaric criterion is the weaker condition, and is met over a wider range of states.

Let the gas sit at $(\rho_0,T_0)$ with $\mathcal{L}(\rho_0,T_0) = 0$, and perturb the temperature by $\delta T$. The energy equation gives $c\,\dot{\delta T} = -\delta\mathcal{L}$ for the appropriate specific heat $c\gt 0$, so the perturbation grows exactly when $\delta\mathcal{L}$ and $\delta T$ have opposite signs, that is when the derivative of $\mathcal{L}$ with respect to $T$ along the path of the perturbation is negative. At constant density that derivative is $(\partial\mathcal{L}/\partial T)_\rho$, which is (8.1).

At constant pressure the density is not free: for an ideal gas of fixed composition $\rho \propto P/T$, so $(\partial\rho/\partial T)_P = -\rho_0/T_0$. The chain rule then gives $$ \left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\!P} = \left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\!\rho} + \left(\frac{\partial\mathcal{L}}{\partial\rho}\right)_{\!T}\left(\frac{\partial\rho}{\partial T}\right)_{\!P} = \left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\!\rho} - \frac{\rho_0}{T_0}\left(\frac{\partial\mathcal{L}}{\partial\rho}\right)_{\!T},$$ which is (8.2). The subtracted term is positive when $(\partial\mathcal{L}/\partial\rho)_T\gt 0$, so $(\partial\mathcal{L}/\partial T)_P \lt (\partial\mathcal{L}/\partial T)_\rho$, and a state satisfying (8.1) satisfies (8.2) while the converse fails whenever the two derivatives straddle zero. ∎

Which criterion applies is not a matter of taste. A parcel held at fixed density but given a different temperature is at the wrong pressure for its surroundings, and the force equation will not let it stay there: it expands or contracts on a sound-crossing time, which in the interstellar medium is far shorter than the cooling time of the warm gas. The isobaric criterion is the one compatible with the momentum equation, and that was Field's point in 1965 against Parker's earlier isochoric treatment. Because (8.2) is the weaker condition, far more of the interstellar medium is unstable than the isochoric criterion would suggest.

A form $(\partial(\mathcal{L}/T)/\partial T)_P \lt 0$ appears in some later texts. It coincides with (8.2) only on the equilibrium curve, where $\mathcal{L} = 0$. This module uses (8.2) as Field printed it.

8.1 The cooling function, tested before it is used

The interstellar gas of this section is heated by photoelectric emission from small grains and polycyclic aromatic hydrocarbons at a rate $\Gamma$ per hydrogen nucleus, taken independent of density and temperature, and cooled at a rate $n^2\Lambda(T)$ per unit volume. Koyama & Inutsuka (2002) fitted $\Lambda(T)$ over the range of the neutral medium with two terms: Lyman-$\alpha$ cooling, which switches on above about $8000$ K, and the [C II] $158\ \mu$m fine-structure line, whose $92$ K excitation temperature appears explicitly in the exponent and which is what cools the cold neutral medium.

Their equation (4) as printed contains two typographical errors, and the printed form misses the measured interstellar pressures by three orders of magnitude. As printed, with coefficient $14$ on the second term and $114800$ in the first exponent, the equilibrium relation of (8.3) below gives $P/k = 1.7924\ \mathrm{K\,cm^{-3}}$ at $100$ K and $6.3632$ at $8000$ K. Corrected, the same relation gives $1792.4$ and $388.6$. Measured interstellar pressures are near $3802\ \mathrm{K\,cm^{-3}}$ (§10). The printed form misses by three orders of magnitude.
The correction is not a guess. Vázquez-Semadeni et al. (2007) print the corrected equations (3) and (4), and their footnote 5 reads: "Note that eq. (4) in Koyama & Inutsuka (2002) contains two typographical errors. The form used here incorporates the necesary corrections, kindly provided by H. Koyama." The change $114800 \to 1.184\times10^{5}$ is a transposition of digits, since $114800 = 1.148\times10^{5}$, not a change of notation.

The corrected function, used everywhere below, is

$$ \frac{\Lambda(T)}{\Gamma} \;=\; 10^{7}\exp\!\left(\frac{-1.184\times10^{5}}{T+1000}\right) + 1.4\times10^{-2}\,\sqrt{T}\,\exp\!\left(\frac{-92}{T}\right)\ \mathrm{cm^{3}}, \qquad \Gamma = 2\times10^{-26}\ \mathrm{erg\,s^{-1}}. \tag{8.3}$$
Proposition 7 (the criterion for this heating law). Let the volumetric heating be $n\Gamma$ and the volumetric cooling $n^2\Lambda(T)$, with $\Gamma$ constant, $\Lambda(T)\gt 0$ and differentiable, and $n = \rho/m_{\rm H}$. Then thermal equilibrium is $$ n_{\rm eq}(T) \;=\; \frac{\Gamma}{\Lambda(T)}, \tag{8.4}$$ and Field's isobaric runaway condition (8.2) reduces at every equilibrium state to $d\ln\Lambda/d\ln T \lt 1$, so that the gas is stable against an isobaric perturbation exactly when $$ \frac{d\ln\Lambda}{d\ln T} \;\gt \; 1 \quad\text{for stability}. \tag{8.5}$$

Per gram, the losses are $n^2\Lambda/\rho = \rho\Lambda/m_{\rm H}^2$ and the gains are $n\Gamma/\rho = \Gamma/m_{\rm H}$, so $$ \mathcal{L}(\rho,T) = \frac{\rho\,\Lambda(T)}{m_{\rm H}^2} - \frac{\Gamma}{m_{\rm H}} .$$ Setting $\mathcal{L}=0$ gives $\rho = m_{\rm H}\Gamma/\Lambda(T)$, that is (8.4). Differentiating, $$ \left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\!\rho} = \frac{\rho}{m_{\rm H}^2}\frac{d\Lambda}{dT}, \qquad \left(\frac{\partial\mathcal{L}}{\partial\rho}\right)_{\!T} = \frac{\Lambda}{m_{\rm H}^2},$$ so (8.2) reads $$ \left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\!P} = \frac{\rho}{m_{\rm H}^2}\left[\frac{d\Lambda}{dT} - \frac{\Lambda}{T}\right] \lt 0 \iff \frac{T}{\Lambda}\frac{d\Lambda}{dT} \lt 1,$$ after dividing by the positive factor $\rho\Lambda/(m_{\rm H}^2T)$. The layer is therefore stable exactly when $d\ln\Lambda/d\ln T \gt 1$, which is (8.5). Note also that $(\partial\mathcal{L}/\partial\rho)_T = \Lambda/m_{\rm H}^2\gt 0$, so Proposition 6's ordering of the two criteria applies: the isochoric runaway condition here is $d\ln\Lambda/d\ln T \lt 0$, which is strictly harder to meet than $d\ln\Lambda/d\ln T \lt 1$, so every isochorically unstable state on this curve is isobarically unstable and not conversely. ∎

Table 4. The corrected Koyama–Inutsuka cooling function, its logarithmic slope, the equilibrium state from (8.4), and the two Field criteria. "UNSTABLE" in the isobaric column means (8.5) is violated.
$T$ [K]$\Lambda$ [erg cm$^3$ s$^{-1}$]$d\ln\Lambda/d\ln T$$n_{\rm eq}$ [cm$^{-3}$]$P/k$ [K cm$^{-3}$]isobaric (8.2)isochoric (8.1)
20$1.2587\times10^{-29}$5.10001588.954931779.1stablestable
40$1.7755\times10^{-28}$2.8000112.64694505.9stablestable
80$7.9298\times10^{-28}$1.650025.22122017.7stablestable
100$1.1159\times10^{-27}$1.420017.92351792.4stablestable
184$2.3037\times10^{-27}$1.00008.68181597.5stablestable
300$3.5689\times10^{-27}$0.80675.60401681.2UNSTABLEstable
600$5.8836\times10^{-27}$0.65333.39932039.6UNSTABLEstable
1000$8.0761\times10^{-27}$0.59202.47642476.4UNSTABLEstable
2000$1.1959\times10^{-26}$0.54601.67243344.8UNSTABLEstable
5039$2.0128\times10^{-26}$0.99940.99365007.0UNSTABLEstable
6000$3.0380\times10^{-26}$4.66730.65833950.0stablestable
8000$4.1170\times10^{-25}$11.02140.0486388.6stablestable
10000$4.2587\times10^{-24}$9.72470.004747.0stablestable

Read the two right-hand columns together, remembering that (8.1) and (8.2) are conditions for runaway while (8.5) is the condition for stability. The isochoric runaway fires nowhere in this table: $\Lambda$ increases with $T$ throughout, so $d\Lambda/dT\gt 0$ everywhere and (8.1) is never met. The isobaric runaway fires over a wide band, wherever the slope falls below $1$. That is the whole content of Field's 1965 correction, in one table.

The slope crosses $1$ twice: at $T = 184.0$ K, where $n_{\rm eq} = 8.682\ \mathrm{cm^{-3}}$ and $P/k = 1597.5\ \mathrm{K\,cm^{-3}}$, and again at $T = 5039.4$ K, where $n_{\rm eq} = 0.994\ \mathrm{cm^{-3}}$ and $P/k = 5007.0\ \mathrm{K\,cm^{-3}}$. Those two states bound the unstable branch, and §9 shows they are also the turning points of the equilibrium pressure.

9. The two-phase window

Proposition 8 (the pressure window). On the thermal-equilibrium curve (8.4), assume further that $d\ln\Lambda/d\ln T - 1$ changes sign exactly twice on the range considered, which Table 4 verifies for (8.3). Then the equilibrium pressure satisfies $$ \frac{P}{k_B} \;=\; n_{\rm eq}(T)\,T \;=\; \frac{\Gamma\,T}{\Lambda(T)}, \qquad \frac{d\ln P}{d\ln T} \;=\; 1 - \frac{d\ln\Lambda}{d\ln T}. \tag{9.1}$$ Hence $P$ has a turning point exactly where $d\ln\Lambda/d\ln T = 1$, the boundary case of (8.5), and $dP/dn\lt 0$ exactly on the unstable branch. Between the local minimum $P_{\min}$ and the local maximum $P_{\max}$ of the curve, a single pressure is attained at three densities, of which the two extreme ones are stable and the middle one is not: two stable phases coexist at the same pressure.

(9.1) is (8.4) multiplied by $T$, and its logarithmic derivative. A turning point of $P$ along the curve is $d\ln P/d\ln T = 0$, which by (9.1) is $d\ln\Lambda/d\ln T = 1$, which is the boundary case of (8.5). For the density, (8.4) gives $d\ln n/d\ln T = -\,d\ln\Lambda/d\ln T \equiv -s$, so on any part of the curve with $s\neq0$, $$ \frac{d\ln P}{d\ln n} = \frac{1-s}{-s} = 1 - \frac{1}{s}.$$ For $s\gt 1$, which is the stable cold branch of Table 4, this is positive and less than $1$; for $0\lt s\lt 1$, the unstable branch, $1/s\gt 1$ and the expression is negative, so $dP/dn\lt 0$ there. A continuous curve that rises, falls and rises again in $P$ as $n$ increases takes every value of $P$ strictly between the local minimum and the local maximum at three densities, one on each monotone piece. The outer two lie on branches with $s\gt 1$, which satisfy (8.5) and are stable; the middle one lies on the falling branch, which does not. ∎

From the corrected fit (8.3), the turning points are

$$ \frac{P_{\min}}{k_B} = 1597.5\ \mathrm{K\,cm^{-3}} \ \ (T = 184.0\ \mathrm{K},\ n = 8.682\ \mathrm{cm^{-3}}), \qquad \frac{P_{\max}}{k_B} = 5007.0\ \mathrm{K\,cm^{-3}}\ \ (T = 5039.4\ \mathrm{K},\ n = 0.994\ \mathrm{cm^{-3}}), $$

with geometric mean $2828.2\ \mathrm{K\,cm^{-3}}$ and width $P_{\max}/P_{\min} = 3.134$. Fig. 3 draws the curve, with the unstable branch between those two points marked separately.

Thermal equilibrium of neutral interstellar gas, and where the cold gas is measured 10⁻² 10⁻¹ 1 10 10² 10³ 300 1000 3000 10000 30000 hydrogen density n (cm⁻³) P / k (K cm⁻³) Pmax = 5007 Pmin = 1597 warm neutral medium cold neutral medium thermally unstable branch: dP/dn < 0, Field (1965) eq. 4b two-phase window, Wolfire et al. (2003) 1960 to 4810 measured CNM pressure, median 3802 ± 0.175 dex Jenkins & Tripp (2011)
The thermal-equilibrium curve of the neutral interstellar medium. Pressure against density along (8.4), on logarithmic axes, with the thermally unstable branch — where $dP/dn\lt 0$ and Field's isobaric criterion (8.5) is violated — drawn separately from the two stable branches. $P_{\min}$ and $P_{\max}$ mark the turning points of Proposition 8; the band is the two-phase window computed by Wolfire et al. (2003), and the bar is the pressure distribution measured by Jenkins & Tripp (2011) that §10 compares against. Curve: Koyama & Inutsuka (2002) eq. 4 with the two typographical corrections printed by Vázquez-Semadeni et al. (2007), $\Lambda/\Gamma = 10^{7}\exp[-1.184\times10^{5}/(T+1000)] + 1.4\times10^{-2}\sqrt{T}\exp(-92/T)\ \mathrm{cm^3}$, $\Gamma = 2\times10^{-26}\ \mathrm{erg\,s^{-1}}$, equilibrium $n = \Gamma/\Lambda(T)$.

9.1 The same window from a full calculation

Wolfire, McKee, Hollenbach & Tielens (2003) computed the same curve without a two-term fit: real gas-phase abundances of carbon and oxygen, grain and PAH photoelectric heating, [C II] and [O I] line cooling, and a radiation field appropriate to the solar neighbourhood. Their Table 3, at galactocentric radius $R = 8.5$ kpc with a cloud column $N_{\rm cl} = 10^{19}\ \mathrm{cm^{-2}}$, gives the row this module uses throughout.

Table 5. The two-term fit against the full calculation. Both columns are calculations. This is a check on the arithmetic of §8 and §9, not a check against nature.
QuantityKI02 fit, this moduleW03 full calculationratio
$P_{\min}/k$ [K cm$^{-3}$]1597.51960.00.815
$P_{\max}/k$ [K cm$^{-3}$]5007.04810.01.041
geometric mean [K cm$^{-3}$]2828.23070.00.921
$T$ at $P_{\max}$ [K]5039.45040.00.9999
$T$ at $P_{\min}$ [K]184.0258.00.713
$n$ at $P_{\max}$ [cm$^{-3}$]0.9940.8601.155
$n$ at $P_{\min}$ [cm$^{-3}$]8.6826.9101.256

A two-term analytic fit reproduces both edges of the window to better than $20$ per cent, and the coldest temperature the warm phase can reach to $0.01$ per cent. Wolfire et al. also quote the general result of Field, Goldsmith & Habing (1969) that $P_{\max}$ is at most about $3P_{\min}$: here the fit gives $3.13$ and the full calculation $2.45$. From §10 onward the full calculation's window $[1960, 4810]$ is the prediction under test, because it is the one whose inputs are traceable to independent measurements.

One property of the source has to be said rather than hidden: Wolfire's $P_{\rm th,ave}$ is the geometric mean of his own two edges to four figures, in all five variants of Table 6 — $\sqrt{P_{\min}P_{\max}}$ gives $3070.4$, $2216.8$, $3681.3$, $2488.7$ and $2410.6$ against the quoted $3070$, $2220$, $3680$, $2490$ and $2410$. So the "geometric mean" row of Table 5 restates the two rows above it rather than testing anything new, and the two readings of the anchor in §10.1 — the ratio to $P_{\rm ave}$ and the position in dex — are one statement written twice.

A comparison that is refused. Heiles & Troland (2003) call $500$–$5000$ K the thermally unstable range for the warm neutral medium; the curve here gives $184$–$5039$ K. The upper edges match to $0.8$ per cent, and that is the edge that matters for their measurement, because a warm parcel becomes unstable by cooling down through it. The lower edges are not the same quantity: $500$ K is a round number they adopt to classify 21-cm absorption components, and $184$ K is where this particular cooling fit turns over. The two lower numbers are not reported here as a check.

The window is narrow in pressure and wide in everything else: a factor $3.13$ in pressure, $8.7$ in density and $27.4$ in temperature (Problem C2). That asymmetry is what makes a pressure survey worth doing. Jenkins & Tripp had already made one before Wolfire et al. published: their 2001 survey of C I fine-structure excitation gave a mean pressure of $2240\ \mathrm{K\,cm^{-3}}$, which Wolfire et al. compare their own window against in their section 8.1. In 2011 the same two authors measured again, on 89 sight lines and with revised C I atomic data, and that second measurement is what §10 puts against the window.

10. The anchor: Jenkins & Tripp (2011) against the window

Jenkins & Tripp (2011) measured thermal pressures in the diffuse cold neutral medium along 89 sight lines, from the fine-structure excitation of neutral carbon observed with the STIS echelle spectrograph on the Hubble Space Telescope. Their mass-weighted lognormal fit, their equation (3), has mean $\log(p/k) = 3.58$ dex and an rms dispersion $\sigma_{\log P} = 0.175$ dex, which they state is a lower limit. The median is $3801.9\ \mathrm{K\,cm^{-3}}$ and the $\pm1\sigma$ range is $2541$ to $5689\ \mathrm{K\,cm^{-3}}$.

This module takes two predictions to that measurement, in the shape Module 2 used: one confirmed and one refuted, from the same source table.

10.1 Confirmed: the measured centre lies inside the predicted window

CONFIRMED. The predicted window is $[1960, 4810]\ \mathrm{K\,cm^{-3}}$ with predicted average pressure $3070$. The measured median $3801.9\ \mathrm{K\,cm^{-3}}$ lies inside it, at a ratio $1.238$ to the predicted average, and $0.738$ of the way from $P_{\min}$ to $P_{\max}$ measured in dex.

The content of that confirmation is that a factor-$2.5$ window exists at all, and sits where it does. Neither of its edges was fitted to a pressure value. Wolfire et al. computed $P_{\min}$ and $P_{\max}$ from grain photoelectric heating and [C II]/[O I] cooling, with gas-phase carbon and oxygen abundances fixed by independent Hubble absorption-line studies and a PAH abundance fixed by ISO infrared emission data. One heating parameter, $\varphi_{\rm PAH} = 0.5$, was calibrated to an observed C I/C II column-density ratio; Wolfire's own detailed working credits that ratio to the C I and H I data compiled by Welty & Hobbs (2001), although an earlier passage cites Jenkins & Tripp (2001) alongside it. Jenkins & Tripp (2001) is the predecessor of, and distinct from, the 2011 pressure measurement compared here.

Two things this module does not claim. It does not claim that the $\varphi_{\rm PAH}$ calibration and the 2011 pressures share a technique or a set of sight lines: a search of the 2001 paper's own text finds no C I/C II ratio, no total carbon column and no N(C II) reported in it. And it does not claim that the two lineages are wholly separate: Wolfire et al. also compared their predicted average pressure against Jenkins & Tripp's (2001) derived value of $2240\ \mathrm{K\,cm^{-3}}$, after the fact and without adjusting any parameter, reporting their own value higher by a factor $3070/2240 = 1.37$, which is $0.14$ dex. The 2011 revision from $2240$ to $3801.9$ is attributed by its authors to three reasons of their own — updated C I atomic rates, revised optical pumping, and a corrected temperature — none of which depends on Wolfire et al.

10.2 How far the confirmation survives its own parameters

$\varphi_{\rm PAH}$ is a real lever on the window, not a formality. Wolfire's Table 4 tests five model variants, three of which vary $\varphi_{\rm PAH}$ discretely while two hold it at $0.5$ and vary other parameters. Raising $\varphi_{\rm PAH}$ raises all three pressures monotonically, and the authors state the size of the effect themselves: over $\varphi_{\rm PAH} = 0.25$ to $1$, $P_{\min}$ changes by a factor $1.5$ and $P_{\max}$ by $1.9$, with $P_{\max}$ decreasing by about $35$ per cent at $\varphi_{\rm PAH} = 0.25$.

Table 6. The measured median $3801.9\ \mathrm{K\,cm^{-3}}$ against all five variants of Wolfire et al. Table 4. The last column is Wolfire's own treatment of each row, on evidence that is not a pressure.
Model$\varphi_{\rm PAH}$$P_{\min}/k$$P_{\max}/k$$P_{\rm ave}/k$median inside?med/$P_{\max}$med/$P_{\rm ave}$Wolfire's treatment
Standard0.50196048103070YES0.7901.238adopted
Low $\varphi_{\rm PAH}$0.25156031502220NO1.2071.713poorly matched
High $\varphi_{\rm PAH}$1.00227059703680YES0.6371.033safely ruled out
Low $n_{\rm PAH}/n$0.50158039202490YES0.9701.527poorly matched
Low $G_0$0.50146039802410YES0.9551.578safely ruled out
The confirmation is not robust across the variants its own authors tested. The median is inside four of the five windows, and in two of those only narrowly, at $0.970$ and $0.955$ of $P_{\max}$ — margins of $118$ and $178\ \mathrm{K\,cm^{-3}}$. In "Low $\varphi_{\rm PAH}$" it lies above $P_{\max} = 3150$, by a factor $1.207$.

What mitigates that must be stated as precisely as the defect, and it is stronger than one row. Wolfire et al. disfavour every non-standard variant, on evidence that is not a pressure. Their four-item summary reads, in their order: "Low $\varphi_{\rm PAH}$" models poorly match the C I*/C$_{\rm tot}$ ratio and the C$^+$ cooling rate, with values lower than observed; "High $\varphi_{\rm PAH}$" models produce temperatures and C I/C II ratios higher than observed; "Low PAH" — Table 4's "Low $n_{\rm PAH}/n$" — gives low C I*/C$_{\rm tot}$ ratios and C$^+$ cooling rate; "Low $G_0$" gives low C I*/C$_{\rm tot}$ and high C I/C II. Separately, they write that only "High $\varphi_{\rm PAH}$" and "Low $G_0$" "can be safely ruled out by the required C I/C II ratio".

The two tiers must be kept apart. Two rows are ruled out in those words. Two more, including the one that breaks the confirmation, are called a worse fit on two independent diagnostics without being ruled out. The only variant Wolfire et al. do not argue against is the standard one, and the standard one contains the measured median.

In the same paragraph as their own "$P_{\max}$ decreases by $\sim35$ per cent", Wolfire et al. call the phases "very robust against variations in the PAH physical and chemical characteristics". This module prints the $35$ per cent and the five rows of Table 6, and does not print the adjective.

One number is stable across all five variants: the measured median exceeds every predicted average pressure, from $1.033$ times the "High $\varphi_{\rm PAH}$" value to $1.713$ times the "Low $\varphi_{\rm PAH}$" one. The measurement runs high against the central model value whatever the parameters. Only the window verdict moves.

10.3 Refuted: a static two-phase medium cannot put cold gas below $P_{\min}$

Below $P_{\min}$ the equilibrium curve has no cold branch: by Proposition 8 the only equilibrium state at such a pressure is the warm one. A strict, static two-phase medium therefore allows none of the cold neutral medium to sit below $P_{\min}$. Jenkins & Tripp measured the cold medium, and they report in their section 10.1.2 that $29$ per cent of it is there.

Fraction of the cold neutral mediumValueStatus
below $P_{\min}$, static two-phase prediction$0$ per centpredicted
below $P_{\min}$, their own fitted lognormal$5.0$ per centfitted
below $P_{\min}$, measured, all sight lines$29$ per centmeasured
below $P_{\min}$, measured, low-starlight subsample$23$ per centmeasured
inside the window, from the fit$67.0$ per centfitted
above $P_{\max}$, from the fit$28.0$ per centfitted
above $\log(p/k) = 5.5$, measured$0.050$ per centmeasured
REFUTED. A static two-phase medium allows $0$ per cent of the cold mass below $P_{\min}$. Jenkins & Tripp's own fitted lognormal already gives $5.0$ per cent. The measurement gives $29$ per cent, an excess over the fit of $5.8$ times. The prediction fails twice over: the fit violates it, and the data violate the fit.

Fig. 4 puts both halves of the anchor on one page: the windows and the measured distribution in its upper panel, and in its lower panel the quantity that carries the disagreement, the fraction of cold gas below $P_{\min}$.

Both $29$ per cent and $23$ per cent are read off the observed histograms of their Fig. 9, not produced by a fit: integrating their eq. (3) lognormal below $P_{\min}$ gives $5.0$ per cent and their eq. (4) polynomial gives $17.4$ per cent, so no fit of theirs can produce $29$. The two numbers describe different populations — $29$ per cent is the complete sample and $23$ per cent the low-starlight subsample — and are not two estimates of one quantity.

What is not refuted. Gas above $P_{\max}$ is allowed. Only the warm phase is forbidden there, and this survey measures the cold phase, so the $28$ per cent above $P_{\max}$ violates nothing and is not counted. The refutation rests on the low side alone, where a static cold medium is genuinely impossible.

The refutation survives the parameter sweep that the confirmation does not. Only the fitted fraction can be re-evaluated at another $P_{\min}$; the measured $29$ per cent is tied to $P_{\min} = 1960\ \mathrm{K\,cm^{-3}}$, the value Jenkins & Tripp themselves quote, and cannot be moved without their data. Across the five variants of Table 6, the fitted lognormal below $P_{\min}$ gives $5.0$ per cent (Standard, $P_{\min}=1960$), $1.4$ (Low $\varphi_{\rm PAH}$, $1560$), $10.0$ (High $\varphi_{\rm PAH}$, $2270$), $1.5$ (Low $n_{\rm PAH}/n$, $1580$) and $0.9$ (Low $G_0$, $1460$) — a range of $0.9$ to $10.0$ per cent, against the $0$ per cent a static medium allows, in every variant.

Two limits on that sweep. First, only the fit-against-static half of the refutation was re-evaluated; the measured excess over the fit, $29$ per cent against $5.0$, exists only at the standard $P_{\min}$. Second, Jenkins & Tripp state that their lognormal understates the observed material outside $3.2 \lt \log(p/k) \lt 4.0$, that is below $1585\ \mathrm{K\,cm^{-3}}$. Three of the five variants have $P_{\min}$ below that line — $1560$, $1580$ and $1460$ — so their $1.4$, $1.5$ and $0.9$ per cent are extrapolations of a fit its own authors say runs low there, and the true fractions are larger. The direction favours the refutation; the rows are labelled extrapolated anyway.

10.4 What fails, precisely: the word "static", not the criterion

Jenkins & Tripp offer two readings of their own excess and prefer the second: either the Wolfire curve does not apply to the gas they observe, or, in their words, "rarefactions caused by turbulence create momentary excursions below the curve". Their dispersion of $0.175$ dex is itself read as turbulence at Mach $1$ to $4$. On that reading the thermal-equilibrium curve is an attractor, and the gas is pushed off it faster than it relaxes back. Problem D3 makes that quantitative by putting the cooling time against the crossing time of a turbulent eddy, and finds the two phases differ by a factor $794$ in cooling length — which is why a cold excursion below $P_{\min}$ can only be momentary while a warm departure can be sustained.

The same failure seen from the warm side. Heiles & Troland (2003), from 21-cm absorption, find that at least $48$ per cent of the warm neutral medium by column density lies at $500$–$5000$ K, inside the thermally unstable band; a strict two-phase medium allows only the small fraction in transit between phases. Their other numbers: the warm medium is $61$ per cent of the H I column at $|b|\gt 10^\circ$, and the cold-medium spin-temperature histogram peaks at $40$ K with a column-weighted median of $70$ K. Both the $0.61$ and the $0.48$ are column-density weighted and both are lower bounds.

The mean state is right; the spread around it is not there. Wolfire et al. compute a cold-medium temperature of $85.0$ K at the average pressure, a ratio of $1.21$ to Heiles & Troland's measured column-weighted median of $70$ K. The two-phase calculation gets the typical temperature and the typical pressure of the neutral interstellar medium right, and gets the width of both distributions wrong, in the same direction, from both ends.
CONFIRMED: the measured centre lies inside the predicted window 1000 2000 3000 5000 10000 15000 P / k (K cm⁻³), logarithmic Wolfire et al. (2003) predicted window: 1960–4810 Koyama–Inutsuka fit window, this module: 1597–5007 Jenkins & Tripp (2011) measured: 3802, ±0.175 dex median / predicted mean = 3802 / 3070 = 1.238 notch = geometric mean REFUTED: cold gas below Pmin, which a static medium forbids 0 5 10 15 20 25 30 35 40 per cent of CNM mass below Pmin = 1960 K cm⁻³ static two-phase medium prediction 0.0% lognormal fit, their eq. 3 same survey, fitted 5.0% measured, all sight lines Jenkins & Tripp, section 10.1.2 29% measured, low starlight their abstract 23% 5.8 × the fit What fails is the word static: Jenkins & Tripp read the excess as turbulent rarefaction, Mach 1–4. Gas above Pmax is allowed for the cold phase and is not counted. Window: Wolfire et al. (2003) Table 3, R = 8.5 kpc, Ncl = 10¹⁹ cm⁻².
The anchor. Upper panel: on a logarithmic pressure axis, the two-phase window computed by Wolfire et al. (2003) from $1960$ to $4810\ \mathrm{K\,cm^{-3}}$ with its predicted average $3070$, the window from the analytic cooling fit of §9 from $1597$ to $5007$, and the pressure distribution measured by Jenkins & Tripp (2011), median $3801.9$ with a dispersion of $0.175$ dex. The measured centre lies inside both windows. Lower panel: the fraction of cold gas below $P_{\min}$, the quantity that carries the disagreement. A static two-phase medium allows none; the fitted lognormal gives $5.0$ per cent; the measurement gives $29$ per cent for the complete sample and $23$ per cent for the low-starlight subsample.

For reference, against the analytic-fit window $[1597, 5007]$ the same lognormal puts $73.7$ per cent of the mass inside, rather than the $67.0$ per cent quoted above for $[1960, 4810]$.

11. What this model captures, and what it misses

Each idealisation, and where it is repaid if it is repaid at all.

12. Problems

Nine problems in the house set: three conceptual (C), three derivational (D), three computational (K). Every printed answer is reproduced independently by afd/figs/m06_problems_check.py.

Conceptual ↑ contents

C1. The International Standard Atmosphere fixes the tropospheric lapse rate at $6.50\ \mathrm{K\,km^{-1}}$. Show that the standard troposphere is Schwarzschild-stable, give its polytropic index, and then explain why thunderstorms exist.
Solution
For a plane-parallel ideal-gas atmosphere with lapse rate $\Gamma_{\rm lapse} \equiv -dT/dz$, hydrostatic balance gives $d\ln P/dz = -1/H$ with $H = k_BT/(\mu m_u g)$, so $$ \nabla = \frac{d\ln T}{d\ln P} = \frac{(1/T)(dT/dz)}{-1/H} = \frac{\Gamma_{\rm lapse}\,H}{T} = \frac{\Gamma_{\rm lapse}}{g/c_p}\cdot\frac{k_B}{\mu m_u c_p} = \frac{\Gamma_{\rm lapse}}{g/c_p}\,\nabla_{\rm ad},$$ using $c_p - k_B/(\mu m_u) = c_v$ and $\nabla_{\rm ad} = (\gamma-1)/\gamma = k_B/(\mu m_u c_p)$ from (2.5). So $\nabla/\nabla_{\rm ad}$ is just the ratio of the actual lapse rate to the dry adiabatic one. With $c_p = 1.0047\times10^{7}\ \mathrm{erg\,g^{-1}\,K^{-1}}$ for dry air and $g = 980.665\ \mathrm{cm\,s^{-2}}$, the dry adiabatic lapse rate is $g/c_p = 9.761\ \mathrm{K\,km^{-1}}$ and the ratio is $6.50/9.761 = 0.6659$. With $\gamma = 7/5$, $\nabla_{\rm ad} = 2/7 = 0.28571$ and $\nabla = 0.19026$. By Definition 5 of Module 3, $n_{\rm eff} = 1/\nabla - 1 = 4.2558$.
So $\nabla \lt \nabla_{\rm ad}$: the standard troposphere is stable by Proposition 1, with $33$ per cent of margin, and yet it convects on any warm afternoon. The premise that fails is (iv), adiabatic change of state. A rising parcel of moist air reaches saturation and then releases latent heat as water condenses, about $2.5\times10^{10}\ \mathrm{erg\,g^{-1}}$, which warms it on the way up. Its effective lapse rate is then $5$–$6\ \mathrm{K\,km^{-1}}$, below the ambient $6.5$, so the moist parcel arrives warmer than its surroundings and keeps rising. Moist convection is unstable where dry convection is not. The Sun has no comparable latent-heat term in its convection zone, which is why §2 derives the criterion there.
C2. Express the thermally unstable branch of §8 as a range in each of temperature, density and pressure. Why does the asymmetry between those three ranges make a pressure survey the right measurement to make?
Solution
From the turning points of Proposition 8 on the corrected fit: temperature $184.0$ to $5039.4$ K, a factor $27.4$; density $0.994$ to $8.682\ \mathrm{cm^{-3}}$, a factor $8.7$; pressure $1597$ to $5007\ \mathrm{K\,cm^{-3}}$, a factor $3.13$. The pressure range is by far the narrowest, and it is narrow for the reason Proposition 8 gives: $P = n_{\rm eq}T$ runs along a curve on which $n$ and $T$ move in opposite directions, so the product barely changes while each factor changes by an order of magnitude.
That is exactly what makes pressure the useful observable. A measured temperature or a measured density of one cloud says little, because both vary by a large factor within the allowed states; a measured pressure pins the state of the medium to within a factor of three, and any excursion outside that factor is immediately a violation of the static two-phase picture rather than a normal fluctuation. §10 is the measurement, and $29$ per cent of its mass falls outside.
C3. The mixing-length parameter $\alpha$ was fitted, not derived. Determine how much of the disagreement in §7.2 could be removed by choosing a different $\alpha$, and state what the answer implies about the freedom in the theory.
Solution
By Proposition 5, at a fixed flux $F$ the superadiabatic excess scales as $\alpha^{-4/3}$ and the velocity only as $\alpha^{1/3}$. Repeating the photospheric calculation of §6.2 at four values:
$\alpha$$\nabla-\nabla_{\rm ad}$$v_c$ [km s$^{-1}$]$v_c/c_s$$\ell$ [km]
1.01.53192.740.339143
1.50.89223.140.388214
2.00.60793.450.427286
3.00.35413.950.489429
Changing $\alpha$ by a factor of $3$ changes the predicted velocity by only $1.44$, in the wrong direction to help: the smallest $\alpha$ gives the slowest parcel, $2.74\ \mathrm{km\,s^{-1}}$, still $4.2$ times the measured $0.65$. The factor $5.3$ of §7.2 cannot be tuned away with $\alpha$. It was removed instead by fixing the flux the theory was asked to carry, in §7.3, which is a statement about the physics and not about the parameter. Note also that the excess moves by a factor $4.3$ across the same range, so $\alpha$ is a strong handle on the stratification and a weak one on the velocity — which is why calibrating $\alpha$ on a stellar radius works, and why the radius calibration says little about the surface flow.

Derivational ↑ contents

D1. Show that for a plane-parallel ideal-gas atmosphere the buoyancy frequency of Proposition 2 can be written $N^2 = (g/T)\,(dT/dz + g/c_p)$, and evaluate it for the isothermal lower stratosphere at $T = 216.65$ K and for the standard troposphere at $T = 288.15$ K.
Solution
From (3.1), $N^2 = (g/H)(\nabla_{\rm ad}-\nabla)$ with $H = k_BT/(\mu m_u g)$. As in C1, $\nabla = -(H/T)\,dT/dz$ and $\nabla_{\rm ad} = k_B/(\mu m_u c_p)$, so $$ N^2 = \frac{g}{H}\left[\frac{k_B}{\mu m_u c_p} + \frac{H}{T}\frac{dT}{dz}\right] = \frac{g\,k_B}{\mu m_u c_p H} + \frac{g}{T}\frac{dT}{dz}.$$ Substituting $H = k_BT/(\mu m_u g)$ in the first term gives $g^2/(c_pT)$, so $N^2 = (g/T)(g/c_p + dT/dz)$ as required. The bracket is the difference between the dry adiabatic lapse rate and the actual one, with signs arranged so that a shallower actual lapse means a stiffer atmosphere.
Isothermal lower stratosphere, $dT/dz = 0$, $T = 216.65$ K: $N^2 = (980.665/216.65)(9.7608\times10^{-5}) = 4.4182\times10^{-4}\ \mathrm{s^{-2}}$, $N = 0.02102\ \mathrm{rad\,s^{-1}}$, period $2\pi/N = 4.98$ min. Troposphere at the ISA lapse rate, $dT/dz = -6.5\times10^{-5}\ \mathrm{K\,cm^{-1}}$, $T = 288.15$ K: $N^2 = 1.1098\times10^{-4}\ \mathrm{s^{-2}}$, $N = 0.01053\ \mathrm{rad\,s^{-1}}$, period $9.94$ min. Both are positive, confirming C1's verdict, and both are of order ten minutes: these are the periods of the lee waves that make the stacked wave clouds downwind of a mountain ridge.
D2. A layer has $\nabla = 0.45$ and $\nabla_{\rm ad} = 0.4$. Find the composition gradient that just holds it together, tabulate the Ledoux verdict for $\nabla_\mu = 0$, $0.02$, $0.05$, $0.10$, and explain why the Sun never presents this ambiguity although its $\nabla_\mu$ reaches four times the critical value.
Solution
Proposition 3 gives stability when $\nabla \lt \nabla_{\rm ad} + (\varphi/\delta)\nabla_\mu$, and for an ideal gas $\varphi/\delta = 1$. The layer is marginal when $\nabla_\mu = \nabla - \nabla_{\rm ad} = 0.050$. Writing the Ledoux margin as $\nabla_{\rm ad} + \nabla_\mu - \nabla$:
$\nabla_\mu$SchwarzschildLedoux marginLedoux
0.00UNSTABLE$-0.050$UNSTABLE
0.02UNSTABLE$-0.030$UNSTABLE
0.05UNSTABLE$+0.000$stable (marginal)
0.10UNSTABLE$+0.050$stable
A layer that is Schwarzschild-unstable and Ledoux-stable is semiconvective: the temperature structure drives overturning, the composition structure resists it, and the layer mixes on a thermal rather than a dynamical timescale, because the parcel must first exchange heat with its surroundings before it can move.
The Sun escapes the ambiguity, and not because its $\nabla_\mu$ is small. At $0.10\,R$ it reaches $\nabla_\mu = 0.220$, more than four times the $0.050$ above. The reason is that the Sun's large $\nabla_\mu$ and its near-neutral layer are in different places. Inside $0.2\,R$, where $\nabla_\mu$ is large, $\nabla_{\rm ad}-\nabla$ runs from $0.0706$ at $0.050\,R$ to $0.1328$ at $0.200\,R$, so the layer is stable under either criterion by $1.41$ to $2.66$ times the $0.050$ this problem needed. Where the criterion is delicate, at $0.72$–$0.75\,R$, $|\nabla_\mu| \le 0.0270$. A star with a receding convective core puts the two in the same place, and then the choice of criterion changes the star's lifetime.
D3. Define the cooling length $c_st_{\rm cool}$ and the Field length $\lambda_F$, evaluate both for the cold and the warm phases on the equilibrium curve, and use the result to explain why the warm medium sustains a large departure from thermal equilibrium while the cold medium's excursions below $P_{\min}$ must be transient.
Solution
Take the cooling time as the thermal energy divided by the net radiated power, $t_{\rm cool} = \tfrac32 nk_BT/(n^2\Lambda(T))$, and the Field length as $\lambda_F = \sqrt{\kappa T/(n^2\Lambda)}$, the wavelength at which conduction erases a temperature contrast as fast as cooling builds it, with the neutral-gas conductivity $\kappa = 2.5\times10^{3}\sqrt{T}$ adopted by Koyama & Inutsuka after Parker (1953). The sound speed is the adiabatic $c_s = \sqrt{(5/3)k_BT/(\mu m_u)}$ with $\mu = 1.27$ for the neutral medium. Both phases are taken at the states the observations report: $T = 70$ K for the cold medium, Heiles & Troland's column-weighted median spin temperature, and $T = 6000$ K for the warm medium, a point on the warm branch near $P_{\max}$.
$n_{\rm eq}$ [cm$^{-3}$]$P/k$ [K cm$^{-3}$]$t_{\rm cool}$$c_s$ [km s$^{-1}$]$c_st_{\rm cool}$$\lambda_F$
CNM, $T = 70$ K31.7772224$7.248\times10^{11}$ s $= 0.023$ Myr0.8740.020 pc0.00049 pc $= 101$ au
WNM, $T = 6000$ K0.6583950$6.213\times10^{13}$ s $= 1.969$ Myr8.09116.29 pc0.09627 pc $= 19857$ au
A disturbance smaller than $c_st_{\rm cool}$ is erased by cooling before sound can cross it, so the gas relaxes back onto the equilibrium curve; a disturbance larger than $c_st_{\rm cool}$ outruns the cooling and carries the gas off the curve. The two cooling lengths differ by a factor $794$. For the cold medium $c_st_{\rm cool} = 0.020$ pc, far below any scale a survey resolves or turbulence drives, so cold gas relaxes back onto the curve almost at once and its excursions below $P_{\min}$ must be transient rarefactions rather than a sustained state — which is exactly how Jenkins & Tripp read their own $29$ per cent of cold mass below $P_{\min}$. For the warm medium $c_st_{\rm cool} = 16.29$ pc, the size of a whole molecular cloud, so turbulence driven on tens of parsecs can hold a large part of the warm volume permanently away from equilibrium. That is consistent with Heiles & Troland finding at least $48$ per cent of the warm column at $500$–$5000$ K, although the two published fractions are not the same quantity and are not compared here as one: $48$ per cent is a warm-medium column fraction inside an unstable temperature band, where gas may sit exactly on the unstable branch of the curve, and $29$ per cent is a cold-medium mass fraction below a pressure edge, where the curve has no cold branch at all. What the two cooling lengths explain is the direction and the size of the asymmetry, not a ratio between those two numbers.
The Field lengths set something different: the resolution a simulation needs, since a grid coarser than $\lambda_F$ cannot represent the smallest condensation the instability makes. Koyama & Inutsuka used $140$ au cells on a $2048\times512$ grid over $1.44\times0.36$ pc for exactly this reason.

Computational ↑ contents

K1. Follow the minimum of $\nabla_{\rm ad}$ from (5.3) across five decades of pressure, from $10^{4}$ to $10^{7}\ \mathrm{dyn\,cm^{-2}}$, and say what the trend implies about the depth of the solar convection zone.
Solution
Solving (5.2) and minimising (5.3) along each isobar:
$P$ [dyn cm$^{-2}$]minimum $\nabla_{\rm ad}$at $T$ [K]
$1.0\times10^{4}$0.0809710580
$1.0\times10^{5}$0.0933712126
$1.2\times10^{5}$0.0944812266
$1.0\times10^{6}$0.1091214129
$1.0\times10^{7}$0.1293316798
The zone moves to higher temperature as the pressure rises, because Saha equilibrium needs a hotter gas to ionise a denser one: raising $P$ at fixed $T$ in (5.2) lowers $B$ and therefore lowers $x$. The depression also becomes shallower, from a factor $4.94$ below $2/5$ at $10^{4}$ to a factor $3.09$ at $10^{7}$. The zone is therefore not a thin surface skin fixed at one temperature — but the whole band, $1.06\times10^{4}$ to $1.68\times10^{4}$ K, still lies far above the $2.02\times10^{6}$ K at the base of the solar convection zone. The trend deepens the ionisation zone; it does not deepen it anywhere near $0.727\,R$, which §5.1 shows is set by $\nabla$ instead.
K2. Compute the convective turnover time $\ell/v_c$ from the base of the solar convection zone to the photosphere, and state the problem the answer poses for a theory of the solar dynamo.
Solution
From Table 3 and §6.2, with $\ell = 2H$ at each radius:
$r/R$$\ell$$v_c$ [km s$^{-1}$]$\ell/v_c$
0.750$1.039\times10^{10}$ cm $= 103941$ km0.05621.66 d
0.84969541 km0.07011.43 d
0.90047703 km0.0876.38 d
0.95023985 km0.1242.24 d
0.98317233 km0.2408.38 h
photosphere, $\tau = 2/3$286 km3.451.38 min
The turnover time falls by a factor $2\times10^{4}$ across one convection zone, from $21.66$ days at the base to $1.38$ minutes at the photosphere. Both the length and the velocity contribute: $\ell$ falls by $364$ and $v_c$ rises by $62$. The dynamo problem is that a single fluid carries eddies whose turnover times differ by four orders of magnitude, so no single mixing length, and no single ratio of turnover time to rotation period, describes the whole zone. The slow deep eddies and the fast surface ones couple to the rotation entirely differently.
K3. Is the hot gas of a galaxy cluster thermally unstable? Take $T = 10^{8}$ K, $n_{\rm H} = 10^{-3}\ \mathrm{cm^{-3}}$, $\mu = 0.6$, bremsstrahlung cooling $\Lambda = 2.1\times10^{-27}\sqrt{T}\ \mathrm{erg\,cm^{3}\,s^{-1}}$, a cluster radius of $1$ Mpc, and $n_e = 1.2\,n_{\rm H}$, $n_{\rm tot} = 2.3\,n_{\rm H}$ for a fully ionised cosmic mixture. Compare the cooling time with the sound-crossing time and the Field length with the cluster radius, then repeat for a cool core at $n_{\rm H} = 10^{-2}\ \mathrm{cm^{-3}}$, $T = 3\times10^{7}$ K, $R = 50$ kpc.
Solution
$\Lambda = 2.1\times10^{-27}\sqrt{10^{8}} = 2.1000\times10^{-23}\ \mathrm{erg\,cm^{3}\,s^{-1}}$. The thermal energy density is $\tfrac32 n_{\rm tot}k_BT = 4.7632\times10^{-11}\ \mathrm{erg\,cm^{-3}}$ and the radiated power is $n_en_{\rm H}\Lambda = 2.5200\times10^{-29}\ \mathrm{erg\,cm^{-3}\,s^{-1}}$, so $t_{\rm cool} = 1.8902\times10^{18}\ \mathrm{s} = 59.90$ Gyr. The sound speed $\sqrt{(5/3)k_BT/(\mu m_u)} = 1520\ \mathrm{km\,s^{-1}}$ crosses $1$ Mpc in $2.0304\times10^{16}\ \mathrm{s} = 0.643$ Gyr, so $t_{\rm cool}/t_{\rm sound} = 93.1$.
For conduction, the Spitzer electron conductivity through Sarazin eq. (5.37), $\kappa = 1.84\times10^{-5}T^{5/2}/\ln\Lambda$ with Module 1's $\ln\Lambda = 37.8$ for the intracluster medium, gives $\kappa = 4.8677\times10^{13}\ \mathrm{erg\,s^{-1}\,cm^{-1}\,K^{-1}}$ and a Field length $\sqrt{\kappa T/(n_en_{\rm H}\Lambda)} = 4504$ kpc, which is $4.50$ times the cluster radius. Both tests agree: conduction smooths any thermally unstable mode before it can grow, and the gas would not have cooled appreciably in a Hubble time even if it could.
The cool core is a different matter, and the reason is geometry rather than microphysics. At fixed $T$, $t_{\rm cool}\propto 1/n$ and $\lambda_F\propto1/n$, so both shrink together and the dimensionless ratio $\lambda_F/(c_st_{\rm cool})$ does not change with density at all. What changes is the size of the region they are compared with. At $n_{\rm H} = 10^{-2}$ and $T = 3\times10^{7}$ K: $t_{\rm cool} = 3.28$ Gyr, $\lambda_F = 74.0$ kpc, $\lambda_F/R = 1.48$ and $t_{\rm cool}/t_{\rm sound} = 55.9$. The core cools well inside a Hubble time and its Field length is a sizeable fraction of its radius rather than several times it. Thermal instability in clusters is a cool-core phenomenon.

Notation

SymbolMeaningUnits (CGS)First used
$\nabla$the stratification's logarithmic temperature gradient $d\ln T/d\ln P$. With no argument attached it is never the vector operator in this module—(2.1)
$\nabla_{\rm ad}$, $\nabla_\mu$adiabatic gradient $(\partial\ln T/\partial\ln P)_s$; composition gradient $d\ln\mu/d\ln P$—(2.2), Proposition 3
$\Delta\nabla$the superadiabatic excess $\nabla-\nabla_{\rm ad}$—Proposition 5
$H$pressure scale height $P/(\rho g)$. Book-wide from this modulecmProposition 1
$N$Brunt–Väisälä (buoyancy) frequency. Module-local: Module 1's $N$ is a particle count and does not appear here$\mathrm{rad\,s^{-1}}$(3.1)
$\Delta r$, $\Delta\rho$the displacement and the density contrast it produces, written $\delta r$ in Proposition 1 and renamed in Proposition 3 so that no symbol is both a variation and a coefficientcm, $\mathrm{g\,cm^{-3}}$Propositions 1, 3
$\delta$, $\varphi$quoted, not used: $-(\partial\ln\rho/\partial\ln T)_{P,\mu}$ and $(\partial\ln\rho/\partial\ln\mu)_{P,T}$, both $1$ for an ideal gas. $\delta T$ in (6.2) is a temperature excess, not this $\delta$—Proposition 3
$\gamma$ratio of specific heats; $5/3$ monatomic, $7/5$ diatomic. Module 4's $\Gamma_1$ is the same quantity for a general gas—(2.5)
$x$, $\chi_{\rm H}$, $\Phi$hydrogen ionisation fraction; ionisation energy $13.5984340$ eV; the combination $5/2+\chi_{\rm H}/k_BT$—, erg, —Definition 2, (5.3)
$\alpha$module-local: the mixing-length parameter $\ell/H$, here $2.0$. Module 5's $\alpha$, the length scale of the isothermal sphere, is not meant—Definition 3
$\ell$, $v_c$, $\delta T$mixing length $\alpha H$; convective velocity; parcel temperature excesscm, $\mathrm{cm\,s^{-1}}$, KDefinition 3, (6.2)
$c_P$, $c_p$, $c_s$specific heat at constant pressure, $(5/2)k_B/(\mu m_u)$ for a monatomic gas; the same for dry air in C1 and D1; adiabatic sound speed $\sqrt{(5/3)P/\rho}$$\mathrm{erg\,g^{-1}\,K^{-1}}$, $\mathrm{cm\,s^{-1}}$(6.3), Table 3
$L$, $F$the solar luminosity and the flux $L/(4\pi r^2)$. Module 1's $L$, the smallest scale to resolve, does not appear in this module$\mathrm{erg\,s^{-1}}$, $\mathrm{erg\,cm^{-2}\,s^{-1}}$§6.1
$\mathcal{L}(\rho,T)$Field's net loss function: losses minus gains, per gram per second, exclusive of conduction. Never written $L$$\mathrm{erg\,g^{-1}\,s^{-1}}$Definition 4
$\Lambda(T)$, $\Gamma$cooling coefficient, always written with its argument; photoelectric heating rate per hydrogen nucleus, $2\times10^{-26}\ \mathrm{erg\,s^{-1}}$. Module-local: Module 3's $\Gamma$ is a lapse rate, written $\Gamma_{\rm lapse}$ in C1; Module 1's $\ln\Lambda$ is the Coulomb logarithm and keeps that form in K3$\mathrm{erg\,cm^{3}\,s^{-1}}$, $\mathrm{erg\,s^{-1}}$(8.3)
$n$number density. In C1 only, $n$ is also a polytropic index, written $n_{\rm eff}$ after Module 3$\mathrm{cm^{-3}}$(8.4), C1
$P_{\min}$, $P_{\max}$, $P_{\rm ave}$the two turning points of the equilibrium curve and the model's predicted average pressure$\mathrm{dyn\,cm^{-2}}$, quoted as $P/k_B$ in $\mathrm{K\,cm^{-3}}$Proposition 8
$t_{\rm cool}$, $\lambda_F$cooling time $\tfrac32 nk_BT/(n^2\Lambda)$; Field length $\sqrt{\kappa T/(n^2\Lambda)}$. Never written $\tau_{\rm cool}$: $\tau$ already carries other meanings in this books, cmD3, K3
$\sigma_d$, $\sigma_{\log P}$the width of the granule-size distribution, $480$ km; the dispersion of $\log(p/k)$, $0.175$ dex. Both statistical widths are subscripted, because from this module a bare $\sigma$ is a collision cross-sectioncm, dex§7.1, §10
$\varphi_{\rm PAH}$Wolfire et al.'s scale factor on electron–PAH collision rates; $0.5$ in their standard model—§10.2
$\mu$, $m_u$, $k_B$, $m_{\rm H}$mean molecular weight per particle, measured against $m_u$; atomic mass unit; Boltzmann's constant; hydrogen mass—, g, $\mathrm{erg\,K^{-1}}$, gProposition 1, Proposition 7
$\tau$optical depth, in "$\tau = 2/3$" only—§6.2

Sources

Every published value this module compares against, with what was taken from it.

  1. Thermal instability. G. B. Field, Thermal Instability, The Astrophysical Journal 142, 531 (1965), ADS scan. Quoted from p. 532, equations (1)–(4b): $\mathcal{L}(\rho,T)$ defined as "energy losses minus energy gains, per gram of material per second exclusive of thermal conduction"; (4a) the isochoric criterion $(\partial\mathcal{L}/\partial T)_\rho\lt 0$; (4b) the isobaric criterion $(\partial\mathcal{L}/\partial T)_P = (\partial\mathcal{L}/\partial T)_\rho - (\rho_0/T_0)(\partial\mathcal{L}/\partial\rho)_T\lt 0$, and Field's statement that condensations are governed by (4b) rather than by Parker's (4a). The later form $(\partial(\mathcal{L}/T)/\partial T)_P\lt 0$ coincides with (4b) only on the equilibrium curve; this module uses (4b) as printed.
  2. The two-phase window (the prediction under test). M. G. Wolfire, C. F. McKee, D. Hollenbach and A. G. G. M. Tielens, Neutral Atomic Phases of the Interstellar Medium in the Galaxy, The Astrophysical Journal 587, 278–311 (2003), arXiv:astro-ph/0207098. Quoted: Table 3, row $R = 8.5$ kpc, $N_{\rm cl} = 10^{19}\ \mathrm{cm^{-2}}$ — $P_{\min}/k = 1960$, $P_{\max}/k = 4810$, $P_{\rm th,ave}/k = 3070\ \mathrm{K\,cm^{-3}}$, warm-phase $T$ from $8310$ to $5040$ K and $n$ from $0.209$ to $0.860\ \mathrm{cm^{-3}}$, cold-phase $T$ from $258$ to $61.6$ K and $n$ from $6.91$ to $71.0\ \mathrm{cm^{-3}}$, cold-phase $T_{\rm ave} = 85.0$ K; Table 4, all five model variants of §10.2; the sentences ruling out "High $\varphi_{\rm PAH}$" and "Low $G_0$" and the four-item summary disfavouring all four non-standard variants; the $\varphi_{\rm PAH} = 0.5$ calibration on a C I/C II ratio, credited in their section 8.1 working to Welty & Hobbs (2001); their statements that $P_{\min}$ changes by a factor $1.5$ and $P_{\max}$ by $1.9$ over $\varphi_{\rm PAH} = 0.25$ to $1$, with $P_{\max}$ falling by about $35$ per cent at $0.25$; their section 8.1 post-hoc comparison against Jenkins & Tripp's (2001) mean pressure of $2240\ \mathrm{K\,cm^{-3}}$. Their gas-phase abundances come from HST absorption studies (Sofia et al. 1997; Cardelli et al. 1996; Meyer et al. 1998) and their PAH abundance from ISO emission (Tielens 1999); those papers were not opened here. Their photoelectric-heating formula was not re-derived, and Bakes & Tielens (1994) and Draine (1978, 1987) were not opened, so whether the heating physics carries a further pressure dependence beyond $\varphi_{\rm PAH}$ is an open question, not a closed one.
  3. The measured pressures (the anchor). E. B. Jenkins and T. M. Tripp, The Distribution of Thermal Pressures in the Diffuse, Cold Neutral Medium of our Galaxy. II. An Expanded Survey of Interstellar C I Fine-Structure Excitations, The Astrophysical Journal 734, 65–86 (2011), arXiv:1104.2323. Quoted: abstract and eq. (3) — mass-weighted lognormal with mean $\log(p/k) = 3.58$, rms dispersion at least $0.175$ dex, normalisation $2.30\times10^{23}\ \mathrm{cm^{-2}}$ per dex, and the statement that outside $3.2\lt \log(p/k)\lt 4.0$ the fit understates the observed material; section 10.1.2 — "A substantial fraction of the material (29%) … is detected at pressures below those permissible for a static CNM, $(p/k)_{\min} = 1960\ \mathrm{cm^{-3}\,K}$, as defined by the standard model … presented by Wolfire et al. (2003)", read by them as turbulent rarefaction; abstract — $23$ per cent for the low-intensity subsample, about $0.05$ per cent above $\log(p/k) = 5.5$, turbulent Mach number $1\lt M\lt 4$; and their attribution of the rise from the 2001 value to updated C I atomic rates, revised optical pumping and a corrected temperature.
  4. Paper I, cited but not compared against. E. B. Jenkins and T. M. Tripp, The Distribution of Thermal Pressures in the Interstellar Medium from a Survey of C I Fine-Structure Excitation, The Astrophysical Journal Supplement Series 137, 297–340 (2001), arXiv:astro-ph/0107177. Its mean pressure of $2240\ \mathrm{K\,cm^{-3}}$ enters only through Wolfire et al.'s section 8.1. Its text was searched for a C I/C II column-density ratio, a total carbon column and an N(C II): none appears, which is why §10.1 does not claim a shared technique between the $\varphi_{\rm PAH}$ calibration and the 2011 pressures.
  5. The warm-side measurement. C. Heiles and T. H. Troland, The Millennium Arecibo 21 Centimeter Absorption-Line Survey. II., The Astrophysical Journal 586, 1067–1093 (2003), arXiv:astro-ph/0207105. Quoted from the abstract and sections 2.3.2 and 9.2.1: at least $48$ per cent of the warm neutral medium by column density lies in the thermally unstable range $500$–$5000$ K; $61$ per cent of H I is warm at $|b|\gt 10^\circ$; the cold-medium spin-temperature histogram peaks at $40$ K with a column-weighted median of $70$ K. Both fractions are column-density weighted lower bounds. Their $500$ K is a classification boundary, not a computed turning point, and §9 declines to compare it with the $184$ K of the fit.
  6. The cooling function, and its two typographical errors. H. Koyama and S. Inutsuka, An Origin of Supersonic Motions in Interstellar Clouds, The Astrophysical Journal 564, L97 (2002), arXiv:astro-ph/0112420, equations (4) and (5), and their $140$ au grid cells on $2048\times512$ over $1.44\times0.36$ pc. The corrected form is printed as equations (3) and (4) of E. Vázquez-Semadeni, G. C. Gómez, A. K. Jappsen, J. Ballesteros-Paredes, R. F. González and R. S. Klessen, Molecular Cloud Evolution. II., The Astrophysical Journal 657, 870–883 (2007), arXiv:astro-ph/0608375, whose footnote 5 reads: "Note that eq. (4) in Koyama & Inutsuka (2002) contains two typographical errors. The form used here incorporates the necesary corrections, kindly provided by H. Koyama."
  7. Granule velocities. T. Oba, Y. Iida and T. Shimizu, Height Dependence of the Velocity Structure of Photospheric Convection in Granules and Intergranular Lanes with Hinode/SOT, The Astrophysical Journal 836, 40 (2017), arXiv:1612.06175. Quoted from the abstract and section 4.4: upward speed falling from $0.65$ to $0.40\ \mathrm{km\,s^{-1}}$ with height and downward speed rising from $0.30$ to $0.50\ \mathrm{km\,s^{-1}}$ with depth; rms filtered convective velocity $0.6\ \mathrm{km\,s^{-1}}$ at $I/I_0 = 0.75$ falling to $0.3$ at $I/I_0 = 0.40$; stated velocity error $0.18\ \mathrm{km\,s^{-1}}$; Table 1, which places $I/I_0 = 0.75$ at $h = 40$ km, $\tau_{500} = 0.55$ and $I/I_0 = 0.40$ at $h = 163$ km, $\tau_{500} = 0.10$. Those heights are derived from the VAL81 model atmosphere, not measured.
  8. Granule sizes and contrast. V. I. Abramenko, V. B. Yurchyshyn, P. R. Goode, I. N. Kitiashvili and A. G. Kosovichev, Detection of Small-scale Granular Structures in the Quiet Sun with the New Solar Telescope, The Astrophysical Journal Letters 756, L27 (2012), arXiv:1208.4337. Quoted from their section 3: the regular granules are Gaussian with $d_0 = 1050\pm22$ km and $\sigma_d = 480\pm11$ km; the mini-granules below $600$ km follow a power law of index $-1.82\pm0.12$; the area-contribution function peaks at $1080$–$1300$ km; and the rms intensity contrast $15.5\pm0.6$ per cent at $705.7$ nm.
  9. The solar stratification. J. N. Bahcall, A. M. Serenelli and S. Basu, The Astrophysical Journal 621, L85–L88 (2005). The tabulated model BS2005-AGS,OP, at afd/data/bs05_agsop.dat, used here only as a stratification, exactly as Module 3 used it.
  10. The textbook results. R. Kippenhahn and A. Weigert, Stellar Structure and Evolution (Springer, 1990), sections 6.1, 7.2 and 14.3: the Schwarzschild and Ledoux criteria, the mixing-length flux and velocity in the efficient limit with their $1/2$ and $1/8$ conventions, and $\nabla_{\rm ad}$ for a partially ionised gas, equation (5.3).
  11. Cited through Wolfire et al., not read. D. E. Welty and L. M. Hobbs, The Astrophysical Journal Supplement Series 133, 345 (2001), for the C I and H I data behind the $\varphi_{\rm PAH}$ calibration; G. B. Field, D. W. Goldsmith and H. J. Habing, The Astrophysical Journal 155, L149 (1969), for the general result that $P_{\max}\lesssim 3P_{\min}$.
  12. Transport coefficients. The electron conductivity $\kappa = 1.84\times10^{-5}T^{5/2}/\ln\Lambda$ was read from C. L. Sarazin, X-ray Emissions from Clusters of Galaxies (1988), eq. (5.37), which attributes it to L. Spitzer, Physics of Fully Ionized Gases; Spitzer's book was not opened, and its 1956 and 1962 printings are its first and second editions. Sarazin section 5.4 also supplies $\ln\Lambda = 37.8$ for the intracluster medium, carried over from Module 1. The neutral-gas conductivity $K = 2.5\times10^{3}\sqrt{T}$ is E. N. Parker (1953) as adopted by Koyama & Inutsuka below their eq. (3); Parker's paper was not opened.
  13. Constants and standards. CODATA 2018 for $k_B$, $m_u$, $m_e$, $h$, $G$ and $\chi_{\rm H}$; IAU 2015 Resolution B3 for $GM_\odot$, $R_\odot$, $L_\odot$ and $T_{\rm eff,\odot} = 5772$ K; ISO 2533:1975 and the US Standard Atmosphere 1976 for the ISA lapse rate, $T_0 = 288.15$ K and the tropopause temperature $216.65$ K.
  14. Earlier modules. The Euler equation (7.3) of Module 2; the hydrostatic equation, the temperature gradient $\nabla$ of Definition 5, the $n = 3/2$ polytrope and the photospheric inputs of Module 3; the composition $\mu = 1.2250$ of Module 4 eq. (4.7) and its photospheric scale height at module04.html:260; the Coulomb logarithm of Module 1, Definition 5.

Module 6 of Astrophysical Fluid Dynamics. Previous: Module 5 — The Jeans Instability