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.
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.
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.
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 $\gamma$.
Proposition 1 gives a sign. Its magnitude is a frequency, and the frequency is a measurable property of a star.
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.
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.
| $r/R$ | $\nabla$ | $\nabla_\mu$ | $H$ [cm] | $N$ [$\mu$Hz], $\nabla_{\rm ad}=0.4$ | $N$ [$\mu$Hz], plateau | $N$ [$\mu$Hz], $+$ Ledoux |
|---|---|---|---|---|---|---|
| 0.0500 | 0.3251 | $+0.37196$ | $1.194\times10^{10}$ | 144.84 | 140.59 | 352.04 |
| 0.0999 | 0.3220 | $+0.22043$ | $7.510\times10^{9}$ | 233.70 | 227.12 | 453.84 |
| 0.1496 | 0.3032 | $+0.09531$ | $6.127\times10^{9}$ | 306.06 | 299.15 | 426.24 |
| 0.1996 | 0.2629 | $+0.02629$ | $5.393\times10^{9}$ | 386.01 | 379.87 | 415.78 |
| 0.3016 | 0.2116 | $+0.00256$ | $4.861\times10^{9}$ | 427.47 | 422.53 | 425.46 |
| 0.4002 | 0.1968 | $+0.00134$ | $5.027\times10^{9}$ | 373.34 | 369.35 | 370.59 |
| 0.5018 | 0.1962 | $+0.00155$ | $5.424\times10^{9}$ | 304.51 | 301.26 | 302.43 |
| 0.5979 | 0.2074 | $+0.00166$ | $5.759\times10^{9}$ | 247.82 | 245.02 | 246.09 |
| 0.6804 | 0.2616 | $+0.00906$ | $5.781\times10^{9}$ | 186.66 | 183.72 | 189.83 |
| 0.7099 | 0.3259 | $+0.02423$ | $5.661\times10^{9}$ | 132.67 | 128.73 | 149.43 |
| 0.7500 | 0.3963 | $-0.00307$ | $5.197\times10^{9}$ | 29.34 | imaginary | imaginary |
| 0.8493 | 0.3960 | $-0.00390$ | $3.477\times10^{9}$ | 33.20 | imaginary | imaginary |
| 0.9499 | 0.3870 | $-0.01165$ | $1.199\times10^{9}$ | 91.25 | 74.51 | imaginary |
| 0.9800 | 0.3466 | $-0.02543$ | $4.374\times10^{8}$ | 297.04 | 284.75 | 197.70 |
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.
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 $0.1$.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.
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.
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.
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.
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.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 $P = 1.2\times10^{5}\ \mathrm{dyn\,cm^{-2}}$.
| $T$ [K] | $x$ | $\nabla_{\rm ad}$ | $\nabla_{\rm ad}/0.4$ |
|---|---|---|---|
| 4000 | 0.0000 | 0.39998 | 1.0000 |
| 6000 | 0.0002 | 0.38991 | 0.9748 |
| 8000 | 0.0066 | 0.26086 | 0.6521 |
| 10000 | 0.0623 | 0.12512 | 0.3128 |
| 12000 | 0.2803 | 0.09479 | 0.2370 |
| 15000 | 0.8209 | 0.13080 | 0.3270 |
| 20000 | 0.9916 | 0.35374 | 0.8844 |
| 30000 | 0.9998 | 0.39928 | 0.9982 |
| 100000 | 1.0000 | 0.40000 | 1.0000 |
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.
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.
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.
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.
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.
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.
| $r/R$ | $T$ [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 | $2.083\times10^{6}$ | $5.589\times10^{9}$ | $1.2161\times10^{11}$ | $1.925\times10^{-7}$ | $5.2750\times10^{3}$ | 0.00024 | 2118871 |
| 0.7500 | $1.791\times10^{6}$ | $5.197\times10^{9}$ | $1.1209\times10^{11}$ | $2.478\times10^{-7}$ | $5.5548\times10^{3}$ | 0.00027 | 1871201 |
| 0.8002 | $1.341\times10^{6}$ | $4.391\times10^{9}$ | $9.8460\times10^{10}$ | $4.072\times10^{-7}$ | $6.1562\times10^{3}$ | 0.00035 | 1426538 |
| 0.8493 | $9.504\times10^{5}$ | $3.477\times10^{9}$ | $8.7413\times10^{10}$ | $7.543\times10^{-7}$ | $7.0417\times10^{3}$ | 0.00048 | 987565 |
| 0.9005 | $5.852\times10^{5}$ | $2.385\times10^{9}$ | $7.7748\times10^{10}$ | $1.861\times10^{-6}$ | $8.6531\times10^{3}$ | 0.00075 | 551281 |
| 0.9499 | $2.690\times10^{5}$ | $1.199\times10^{9}$ | $6.9864\times10^{10}$ | $8.469\times10^{-6}$ | $1.2415\times10^{4}$ | 0.00159 | 193192 |
| 0.9800 | $9.740\times10^{4}$ | $4.374\times10^{8}$ | $6.5639\times10^{10}$ | $7.413\times10^{-5}$ | $2.1505\times10^{4}$ | 0.00472 | 40681 |
| 0.9831 | $8.120\times10^{4}$ | $3.617\times10^{8}$ | $6.5235\times10^{10}$ | $1.122\times10^{-4}$ | $2.3984\times10^{4}$ | 0.00580 | 30159 |
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.
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 $F = L_\odot/(4\pi R_\odot^2)$:
| Quantity | Value |
|---|---|
| $H = k_BT/(\mu m_u g)$ | $142.9$ 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 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.
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.
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.
| Denominator | What it is | Value | Departure of $2\ell = 572$ km |
|---|---|---|---|
| $22$ km | the error on the mean diameter | $1050\pm22$ km | $21.7\,\sigma$ |
| $480$ km | the 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.
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.
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:
| Quantity | Value | Source |
|---|---|---|
| $\rho$ 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, $(5/2)k_B/(\mu m_u)$ |
| $\delta T$ | $245.9$ K | measured 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 |
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
$$ v_c = 0.816\ \mathrm{km\,s^{-1}} \quad\text{against the measured } 0.65\ \mathrm{km\,s^{-1}}, \quad\text{a ratio of } 1.25. $$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.
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.
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.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 corrected function, used everywhere below, is
$$ \frac{\Lambda(T)}{\Gamma} \;=\; 10^{7}\exp\!\left(\frac{-1.184\times10^{5}}{T+1000}\right) + 1.4\times10^{-2}\,\sqrt{T}\,\exp\!\left(\frac{-92}{T}\right)\ \mathrm{cm^{3}}, \qquad \Gamma = 2\times10^{-26}\ \mathrm{erg\,s^{-1}}. \tag{8.3}$$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. ∎
| $T$ [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 | $1.2587\times10^{-29}$ | 5.1000 | 1588.9549 | 31779.1 | stable | stable |
| 40 | $1.7755\times10^{-28}$ | 2.8000 | 112.6469 | 4505.9 | stable | stable |
| 80 | $7.9298\times10^{-28}$ | 1.6500 | 25.2212 | 2017.7 | stable | stable |
| 100 | $1.1159\times10^{-27}$ | 1.4200 | 17.9235 | 1792.4 | stable | stable |
| 184 | $2.3037\times10^{-27}$ | 1.0000 | 8.6818 | 1597.5 | stable | stable |
| 300 | $3.5689\times10^{-27}$ | 0.8067 | 5.6040 | 1681.2 | UNSTABLE | stable |
| 600 | $5.8836\times10^{-27}$ | 0.6533 | 3.3993 | 2039.6 | UNSTABLE | stable |
| 1000 | $8.0761\times10^{-27}$ | 0.5920 | 2.4764 | 2476.4 | UNSTABLE | stable |
| 2000 | $1.1959\times10^{-26}$ | 0.5460 | 1.6724 | 3344.8 | UNSTABLE | stable |
| 5039 | $2.0128\times10^{-26}$ | 0.9994 | 0.9936 | 5007.0 | UNSTABLE | stable |
| 6000 | $3.0380\times10^{-26}$ | 4.6673 | 0.6583 | 3950.0 | stable | stable |
| 8000 | $4.1170\times10^{-25}$ | 11.0214 | 0.0486 | 388.6 | stable | stable |
| 10000 | $4.2587\times10^{-24}$ | 9.7247 | 0.0047 | 47.0 | stable | stable |
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.
(9.1) 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.
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.
| Quantity | KI02 fit, this module | W03 full calculation | ratio |
|---|---|---|---|
| $P_{\min}/k$ [K cm$^{-3}$] | 1597.5 | 1960.0 | 0.815 |
| $P_{\max}/k$ [K cm$^{-3}$] | 5007.0 | 4810.0 | 1.041 |
| geometric mean [K cm$^{-3}$] | 2828.2 | 3070.0 | 0.921 |
| $T$ at $P_{\max}$ [K] | 5039.4 | 5040.0 | 0.9999 |
| $T$ at $P_{\min}$ [K] | 184.0 | 258.0 | 0.713 |
| $n$ at $P_{\max}$ [cm$^{-3}$] | 0.994 | 0.860 | 1.155 |
| $n$ at $P_{\min}$ [cm$^{-3}$] | 8.682 | 6.910 | 1.256 |
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.
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.
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.$\varphi_{\rm PAH}$ 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 $\varphi_{\rm PAH} = 0.25$.
| 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 |
|---|---|---|---|---|---|---|---|---|
| Standard | 0.50 | 1960 | 4810 | 3070 | YES | 0.790 | 1.238 | adopted |
| Low $\varphi_{\rm PAH}$ | 0.25 | 1560 | 3150 | 2220 | NO | 1.207 | 1.713 | poorly matched |
| High $\varphi_{\rm PAH}$ | 1.00 | 2270 | 5970 | 3680 | YES | 0.637 | 1.033 | safely ruled out |
| Low $n_{\rm PAH}/n$ | 0.50 | 1580 | 3920 | 2490 | YES | 0.970 | 1.527 | poorly matched |
| Low $G_0$ | 0.50 | 1460 | 3980 | 2410 | YES | 0.955 | 1.578 | safely ruled out |
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.
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 medium | Value | Status |
|---|---|---|
| below $P_{\min}$, static two-phase prediction | $0$ per cent | predicted |
| below $P_{\min}$, their own fitted lognormal | $5.0$ per cent | fitted |
| below $P_{\min}$, measured, all sight lines | $29$ per cent | measured |
| below $P_{\min}$, measured, low-starlight subsample | $23$ per cent | measured |
| inside the window, from the fit | $67.0$ per cent | fitted |
| above $P_{\max}$, from the fit | $28.0$ per cent | fitted |
| above $\log(p/k) = 5.5$, measured | $0.050$ per cent | measured |
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.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.
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 $[1960, 4810]$.
Each idealisation, and where it is repaid if it is repaid at all.
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.
| $\alpha$ | $\nabla-\nabla_{\rm ad}$ | $v_c$ [km s$^{-1}$] | $v_c/c_s$ | $\ell$ [km] |
|---|---|---|---|---|
| 1.0 | 1.5319 | 2.74 | 0.339 | 143 |
| 1.5 | 0.8922 | 3.14 | 0.388 | 214 |
| 2.0 | 0.6079 | 3.45 | 0.427 | 286 |
| 3.0 | 0.3541 | 3.95 | 0.489 | 429 |
| $\nabla_\mu$ | Schwarzschild | Ledoux margin | Ledoux |
|---|---|---|---|
| 0.00 | UNSTABLE | $-0.050$ | UNSTABLE |
| 0.02 | UNSTABLE | $-0.030$ | UNSTABLE |
| 0.05 | UNSTABLE | $+0.000$ | stable (marginal) |
| 0.10 | UNSTABLE | $+0.050$ | stable |
| $n_{\rm eq}$ [cm$^{-3}$] | $P/k$ [K cm$^{-3}$] | $t_{\rm cool}$ | $c_s$ [km s$^{-1}$] | $c_st_{\rm cool}$ | $\lambda_F$ | |
|---|---|---|---|---|---|---|
| CNM, $T = 70$ K | 31.777 | 2224 | $7.248\times10^{11}$ s $= 0.023$ Myr | 0.874 | 0.020 pc | 0.00049 pc $= 101$ au |
| WNM, $T = 6000$ K | 0.658 | 3950 | $6.213\times10^{13}$ s $= 1.969$ Myr | 8.091 | 16.29 pc | 0.09627 pc $= 19857$ au |
| $P$ [dyn cm$^{-2}$] | minimum $\nabla_{\rm ad}$ | at $T$ [K] |
|---|---|---|
| $1.0\times10^{4}$ | 0.08097 | 10580 |
| $1.0\times10^{5}$ | 0.09337 | 12126 |
| $1.2\times10^{5}$ | 0.09448 | 12266 |
| $1.0\times10^{6}$ | 0.10912 | 14129 |
| $1.0\times10^{7}$ | 0.12933 | 16798 |
| $r/R$ | $\ell$ | $v_c$ [km s$^{-1}$] | $\ell/v_c$ |
|---|---|---|---|
| 0.750 | $1.039\times10^{10}$ cm $= 103941$ km | 0.056 | 21.66 d |
| 0.849 | 69541 km | 0.070 | 11.43 d |
| 0.900 | 47703 km | 0.087 | 6.38 d |
| 0.950 | 23985 km | 0.124 | 2.24 d |
| 0.9831 | 7233 km | 0.240 | 8.38 h |
| photosphere, $\tau = 2/3$ | 286 km | 3.45 | 1.38 min |
| Symbol | Meaning | Units (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 module | cm | Proposition 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 coefficient | cm, $\mathrm{g\,cm^{-3}}$ | 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 excess | cm, $\mathrm{cm\,s^{-1}}$, K | Definition 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 book | s, cm | D3, 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-section | cm, 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, $\mathrm{erg\,K^{-1}}$, g | Proposition 1, Proposition 7 |
| $\tau$ | optical depth, in "$\tau = 2/3$" only | — | §6.2 |
Every published value this module compares against, with what was taken from it.
afd/data/bs05_agsop.dat, used here only as a stratification, exactly as Module 3 used it.module04.html:260; the Coulomb logarithm of Module 1, Definition 5.Module 6 of Astrophysical Fluid Dynamics. Previous: Module 5 — The Jeans Instability