12. Magnetohydrodynamics

One extra term in the momentum equation and one extra equation from Maxwell, and four measurements that the same pair reaches: a wound field, a surface the wind crosses once, a cloud a field can hold, and a transport coefficient that is wrong by twenty-three decades.

Contents

Ten modules of this book have written the same sentence in ten different ways: there is a magnetic field here, and it matters, and we are not carrying it. Module 1 ended with a plasma whose mean free path exceeds the system size and could say only that a field must be doing the confining. Module 2 left the intracluster medium with a Reynolds number of about $600$ and a thermal conduction problem it handed on without a number. Module 5 could not say what holds a subcritical cloud up. Module 7 needed a statement accurate to one degree about a field its source never measured. Module 10 confirmed Kolmogorov's exponent most cleanly on a magnetic spectrum in a plasma its own argument does not describe. Twenty such promises are on the record, each with the file and line it was printed at, and this module exists to pay them.

What it costs is small, and that is the surprise. One term is added to Module 2's momentum equation — the Lorentz force — and one equation is added to close it, which is Ohm's law fed into Faraday's. Nothing else changes. Module 2 §4 proved its moment hierarchy under the hypothesis $\nabla_{\mathbf v}\cdot\mathbf a = 0$, and observed in the same breath that the Lorentz force $\mathbf a = (q/mc)\,\mathbf v\times\mathbf B$ satisfies it, so that "Magnetohydrodynamics, in Module 12, therefore inherits (4.1) unchanged". That sentence is honoured here and not reproved. So is Module 2 §11's: "Proposition 3 was proved in a form that admits the Lorentz force, so nothing here needs redoing."

This module has no terrestrial measurement, and §4 says so where a reader will first miss it. Every other module in this book puts a laboratory number between the derivation and the sky — Module 4's speed of sound in air, Module 6's dry adiabat, Module 10's Kolmogorov constant. All four checks here are astronomical. What stands in the terrestrial slot is a reproduction of another chapter of this book, and it is labelled as one.

1. When is a fluid magnetised?

"Magnetised" is not one question. It is two, and the same gas can answer them differently: can the field push the gas? — which is the plasma beta — and does a particle remember the field between collisions? — which is the magnetisation $\omega_{\rm c}\tau$. The first is settled for all five census rows in this section; the second is a number this module does not need until §8, where it is computed for the intracluster medium and comes out at $10^{12.71}$, and it is defined here so that §8's $(\omega_{\rm c}\tau)^{-2}$ arrives with its symbol already known.

Definition 1 (magnetic pressure, the plasma beta, and the Alfvén speed). For a field of strength $B$ in a gas of number density $n$ at temperature $T$ and mass density $\rho$, define $$ p_{\rm mag} \equiv \frac{B^2}{8\pi}, \qquad \beta \equiv \frac{p_{\rm gas}}{p_{\rm mag}} = \frac{8\pi nk_{\rm B}T}{B^2}, \qquad v_{\rm A} \equiv \frac{B}{\sqrt{4\pi\rho}}. \tag{1.1}$$ $\beta$ asks whether the field can push the gas. $v_{\rm A}$ is a combination of $B$ and $\rho$ with the dimensions of a speed; that it is the speed of anything is not assumed here and is proved in Proposition 6. Both $p_{\rm mag}$ and the tension that accompanies it are derived, not posited, in §3.
Definition 2 (the magnetisation). $\omega_{\rm c}\tau$, where $\omega_{\rm c} = ZeB/mc$ is the gyrofrequency and $\tau$ the collision time, is the gyro-phase in radians that a particle turns between collisions. The collision time meant throughout this module is Braginskii's — his equations (2.5e) for electrons and (2.5i) for ions — because this module's §8 defines its transport coefficients on it. For the intracluster medium of §8 it comes out at $4.3\times10^{12}$, which is why §8's suppression factor is $10^{-26}$ and not $10^{-2}$. §8.1 prints the one other collision time this book uses, says exactly how the two differ, and accounts for the difference exactly, as $\sqrt{3/2}$ times a ratio of Coulomb logarithms.
Proposition 1 (the two speeds are one ratio). For an ideal gas with ratio of specific heats $\gamma$, $$ \frac{v_{\rm A}}{c_s} = \sqrt{\frac{2}{\gamma\beta}} \tag{1.2}$$ identically, where $c_s = \sqrt{\gamma p_{\rm gas}/\rho}$. The mass per particle cancels.
$v_{\rm A}^2/c_s^2 = (B^2/4\pi\rho)\big/(\gamma p_{\rm gas}/\rho) = B^2/(4\pi\gamma p_{\rm gas})$. The density has cancelled, and $B^2/4\pi = 2p_{\rm mag}$, so the ratio is $2p_{\rm mag}/(\gamma p_{\rm gas}) = 2/(\gamma\beta)$. ∎

This is why $\beta$ is the census variable and $v_{\rm A}$ is not. The Alfvén speed needs a mean mass per particle and therefore a composition; $\beta$ needs neither. The table below prints both, and prints (1.2) beside the ratio of two separately computed speeds, so that if a $\mu$ is ever changed in one place and not the other the last column stops reading $1.0000$.

system$n$ / cm⁻³$T$ / K$B$ / G$\mu$$\beta$$v_{\rm A}$ / km s⁻¹$c_s$ / km s⁻¹$v_{\rm A}/c_s$(1.2)
solar photosphere$1.51\times10^{17}$$5772$$5$$1.30$$1.21\times10^{5}$$0.02474$$7.844$$0.003154$$0.003154$
solar corona$10^{9}$$10^{6}$$10$$0.61$$0.0347$$886.4$$150.7$$5.881$$5.881$
solar wind, 1 au$5$$1.2\times10^{5}$$5\times10^{-5}$$0.61$$0.833$$62.67$$52.21$$1.200$$1.200$
intracluster medium$10^{-3}$$10^{8}$$10^{-6}$$0.61$$347$$88.64$$1507$$0.05881$$0.05881$
molecular cloud$10^{2}$$10$$10^{-5}$$2.33$$0.0347$$1.434$$0.2439$$5.881$$5.881$
The field column is an order-of-magnitude census and nothing in this module is checked against any entry in it. Row by row: the photospheric $5$ G is a quiet-Sun mean and has no source in this project — the NASA NSSDC Sun Fact Sheet, which was read, gives a polar field of $1$–$2$ G, sunspots at $3000$ G, prominences at $10$–$100$ G, plages at $200$ G and bright network at $25$ G, and none of those is this number. The coronal $10$ G is copied from Module 7 §9.5, which prints "an active-region corona of $10$ gauss" and adds "the number is this module's", so it is illustrative there too. The solar wind's $5$ nT is Module 1's census value, reused unchanged by Module 10 §6.3. The intracluster $1\ \mu$G is Module 10 §6.1's and its Problem D1's. The cloud's $10\ \mu$G has no source at this density: Troland & Crutcher (2008) measure a total mean $|\mathbf B| = 16.4\ \mu$G, but at $n({\rm H_2})$ of a few $\times10^{3}\ \mathrm{cm^{-3}}$, thirty times this row. $\beta$ spans five decades down that column and no statement in this module turns on a factor of two in any of the five. Every number in this module is produced by afd/figs/m12_numbers.py.
10-14 10-10 10-6 10-2 102 106 10-15 10-11 10-7 10-3 101 105 β = 1 β = 100 β = 0.01 solar photosphere (β = 1.21e+05) solar corona (β = 0.0347) solar wind, 1 au (β = 0.833) intracluster medium (β = 347) molecular cloud (β = 0.0347) gas pressure nkBT / dyn cm-2 magnetic pressure B2/8π / dyn cm-2 The corona and the cloud have the SAME β = 0.0347 and lie 12.0 decades apart in pressure. Orange: the field wins. Teal: the gas wins. The solar wind sits 0.079 of a decade from β = 1.
The two pressures of Definition 1, for the five census rows, on logarithmic axes spanning twenty decades. The dashed diagonal is $\beta = 1$; the two faint diagonals are $\beta = 100$ and $\beta = 0.01$. The active-region corona and the molecular cloud have the same $\beta$ to four figures — both $0.0347$, because $B^2/nT$ happens to agree — and therefore sit on the same diagonal while lying $12.0$ decades apart in absolute pressure. A figure that plotted $\beta$ alone could not show that. The solar wind sits $0.079$ of a decade from $\beta = 1$, which is the physical reason §6 has a surface to find.

Two things in that figure are worth stating in words. First, $\beta$ is a ratio and not a field strength: the corona at $10$ G and the cloud at $10\ \mu$G — six decades apart in $B$ — are equally magnetised in this sense. Second, as Fig. 1 shows, four of the five rows are far from $\beta = 1$ and the fifth is not. Where $\beta \gg 1$ the gas drags the field; where $\beta \ll 1$ the field steers the gas; the solar wind does both, at different radii, and §6 finds the radius where the balance tips.

1.1 Every speed in this module, and how many components each one has

A one-dimensional quantity compared with a three-dimensional one has been the defect found in four consecutive modules of this book. Module 10 §6.4 adopted the remedy that works, which is not a resolution to be careful but a printed table. This module inherits it, and it is load-bearing here: the factor $\sqrt{3/2}$ that appears in §8.1 is nothing but a component count.

quantityformulacomponentswhere it is used, and why that count is right
$v_\perp$$\sqrt{2k_{\rm B}T/m}$the two components perpendicular to $\mathbf B$the gyroradius $r_g = v_\perp/\omega_{\rm c}$. A gyro-orbit is a motion in the plane perpendicular to the field, so the speed that sets its radius averages over two components and not three. Module 1's convention, its Proposition 5 and its (8.2)
$v_{\rm rms}$$\sqrt{3k_{\rm B}T/m}$3-D root mean squarea Spitzer mean free path is $v_{\rm rms}\tau$. This is why (8.3) below has a $\sqrt{3/2}$ in it and not a $1$
$\bar v$$\sqrt{8k_{\rm B}T/\pi m}$3-D mean of $|\mathbf v|$Module 10's kinetic viscosity $\nu = \lambda\bar v/3$. It appears in this module only where Module 10's number is quoted, never in a new calculation
$c_s$$\sqrt{\gamma k_{\rm B}T/\mu m_{\rm u}}$a bulk wave speed, not a particle speed(1.2), and the magnetosonic speeds of §4
$v_{\rm A}$$B/\sqrt{4\pi\rho}$a bulk wave speedDefinition 1; shown to be a wave speed at Proposition 6

No two of the three thermal speeds are compared anywhere in this module before a reader has been given this table. The two bulk speeds, $c_s$ and $v_{\rm A}$, are compared once already — at Proposition 1 and in the census table above — and that comparison is (1.2), which is an identity and needs no component count. The three thermal speeds stand in the fixed ratios $v_\perp : v_{\rm rms} : \bar v = 1 : 1.2247 : 1.1284$, and each is correct for its own job.

2. The induction equation, and why flux freezing is a number

The field obeys Maxwell's equations; the gas obeys Module 2's. What couples them is Ohm's law, and what comes out is a single equation for $\mathbf B$ with exactly two terms.

Proposition 2 (the induction equation). For a fluid of uniform electrical conductivity $\sigma_{\rm e}$ moving non-relativistically, $$ \frac{\partial\mathbf B}{\partial t} \;=\; \nabla\times(\mathbf u\times\mathbf B) \;+\; \eta_{\rm m}\nabla^2\mathbf B, \qquad \eta_{\rm m}\equiv\frac{c^2}{4\pi\sigma_{\rm e}}, \tag{2.1}$$ with $\nabla\cdot\mathbf B = 0$ preserved for all time if it holds once.
Ohm's law in the frame of the moving fluid is $\mathbf j = \sigma_{\rm e}(\mathbf E + \mathbf u\times\mathbf B/c)$, so $\mathbf E = \mathbf j/\sigma_{\rm e} - \mathbf u\times\mathbf B/c$. Ampère's law without the displacement current gives $\mathbf j = (c/4\pi)\nabla\times\mathbf B$. Substituting both into Faraday's law $\partial\mathbf B/\partial t = -c\,\nabla\times\mathbf E$, $$ \frac{\partial\mathbf B}{\partial t} = \nabla\times(\mathbf u\times\mathbf B) - \frac{c^2}{4\pi\sigma_{\rm e}}\nabla\times(\nabla\times\mathbf B), $$ and $\nabla\times(\nabla\times\mathbf B) = \nabla(\nabla\cdot\mathbf B) - \nabla^2\mathbf B = -\nabla^2\mathbf B$. Taking the divergence of (2.1) gives $\partial(\nabla\cdot\mathbf B)/\partial t = \eta_{\rm m}\nabla^2(\nabla\cdot\mathbf B)$, a diffusion equation for $\nabla\cdot\mathbf B$; if it starts at zero it stays there. ∎
Dropping the displacement current is an ordering, and here is the ordering. $|\partial_t\mathbf E/c|$ against $|\nabla\times\mathbf B|$ is of order $(U/c)^2$ once $E \sim UB/c$ is used, so the term is smaller than the one kept by $(U/c)^2$. For the fastest flow in this module's census, the solar wind at $435.6\ \mathrm{km\,s^{-1}}$, that is $2.1\times10^{-6}$. It is an ordering and not an identity, and it fails for any relativistic flow — which is why this module's equations do not apply to a pulsar magnetosphere or a jet, and §11 says so.
Proposition 3 (the magnetic Reynolds number, and why size is the variable). The ratio of the two terms of (2.1) on a flow of speed $U$ over a scale $L$ is $$ \mathrm{R_m} \;\equiv\; \frac{UL}{\eta_{\rm m}}. \tag{2.2}$$ With Braginskii's $\sigma_\parallel = 1.96\,n_{\rm e}e^2\tau_{\rm e}/m_{\rm e}$ and his $\tau_{\rm e}\propto T^{3/2}/(n\ln\Lambda)$, the density cancels and $$ \eta_{\rm m} \;\propto\; \frac{\ln\Lambda}{T^{3/2}} \tag{2.3}$$ depends on the temperature and the Coulomb logarithm alone. Moreover $$ \frac{L^2/\eta_{\rm m}}{L/U} \;=\; \frac{UL}{\eta_{\rm m}} \;=\; \mathrm{R_m} \tag{2.4}$$ identically: the ohmic decay time measured in the system's own dynamical times is the magnetic Reynolds number.
The inductive term is of order $UB/L$ and the resistive term of order $\eta_{\rm m}B/L^2$; their ratio is $UL/\eta_{\rm m}$. For (2.3), $\sigma_\parallel \propto n\tau_{\rm e}$ and $\tau_{\rm e}\propto T^{3/2}/(n\ln\Lambda)$, so $\sigma_\parallel\propto T^{3/2}/\ln\Lambda$ and $\eta_{\rm m} = c^2/4\pi\sigma_\parallel \propto \ln\Lambda/T^{3/2}$. Equation (2.4) is arithmetic: $(L^2/\eta_{\rm m})(U/L) = UL/\eta_{\rm m}$. ∎

Equation (2.3) is the reason astrophysical plasmas are good conductors, and it is not the reason a reader expects. A large plasma is a good conductor through its size, not through its density. Two gases at the same temperature have the same resistivity whether one is a million times denser than the other; what differs is $L$, and $L$ enters $\mathrm{R_m}$ linearly. The table makes the point, and its last column is (2.4) used as a self-check on the column above it.

system$T$ / K$\ln\Lambda$$\eta_{\rm m}$ / cm² s⁻¹$L$ / cm$\mathrm{R_m}$$L^2/\eta_{\rm m}$ / yr$L/U$ / yrratio
solar corona$10^{6}$$18.66$$9.768\times10^{3}$$1.000\times10^{10}$$1.024\times10^{12}$$3.244\times10^{8}$$3.169\times10^{-4}$$1.024\times10^{12}$
solar wind, 1 au$1.2\times10^{5}$$26.10$$3.286\times10^{5}$$1.496\times10^{13}$$1.821\times10^{15}$$2.158\times10^{13}$$1.185\times10^{-2}$$1.821\times10^{15}$
intracluster medium$10^{8}$$37.08$$1.941\times10^{1}$$1.851\times10^{23}$$1.564\times10^{29}$$5.596\times10^{37}$$3.577\times10^{8}$$1.564\times10^{29}$
molecular cloud$10$$10.58$$1.751\times10^{11}$$3.086\times10^{19}$$4.653\times10^{13}$$1.723\times10^{20}$$3.704\times10^{6}$$4.653\times10^{13}$
The molecular cloud row is wrong, it is printed anyway, and the reason it is wrong is the physics. A $10$ K cloud is not fully ionised. Its resistivity is set by ion–neutral collisions — ambipolar diffusion — and not by electron–ion ones, so the Spitzer conductivity behind (2.3) is a formula evaluated outside its own domain, in exactly the sense Module 10 Proposition 6 names for $\mathrm{Re}$ at $\mathrm{Kn}\sim1$. This module does not derive ambipolar diffusion and refuses the row in print. Nothing is checked against $\mathrm{R_m} = 4.653\times10^{13}$.

The refusal is made once, here, and pointed back to twice. §9's protoplanetary disc is neutral for the same reason. And §7's mass-to-flux criterion is an equilibrium statement whose missing mechanism is precisely this one: a subcritical cloud leaks flux through ambipolar diffusion and becomes supercritical, and that is how a cloud the field can hold today collapses tomorrow. Naming the mechanism is all this module does with it; §11 records that nobody in this book derives it.
Proposition 4 (Alfvén's theorem). If $\eta_{\rm m} = 0$, the magnetic flux $$ \Phi(t) \;=\; \int_{S(t)}\mathbf B\cdot\mathrm{d}\mathbf S \tag{2.5}$$ through any surface $S(t)$ carried with the fluid is constant in time.
A surface moving with the fluid changes the flux through it in two ways: the field changes in place, and the boundary sweeps through the field. The transport theorem for a surface integral gives $$ \frac{\mathrm{d}\Phi}{\mathrm{d}t} = \int_{S}\frac{\partial\mathbf B}{\partial t}\cdot\mathrm{d}\mathbf S \;-\; \oint_{\partial S}(\mathbf u\times\mathbf B)\cdot\mathrm{d}\boldsymbol\ell, $$ where the second term is the sweeping, written with the sign that follows from $\mathrm{d}\mathbf S = \mathrm{d}\boldsymbol\ell\times\mathbf u\,\mathrm{d}t$ on the swept strip. Stokes's theorem turns the line integral into $\int_S\nabla\times(\mathbf u\times\mathbf B)\cdot\mathrm{d}\mathbf S$, so $$ \frac{\mathrm{d}\Phi}{\mathrm{d}t} = \int_{S}\Big[\frac{\partial\mathbf B}{\partial t} - \nabla\times(\mathbf u\times\mathbf B)\Big]\cdot\mathrm{d}\mathbf S = \int_{S}\eta_{\rm m}\nabla^2\mathbf B\cdot\mathrm{d}\mathbf S $$ by (2.1), which vanishes when $\eta_{\rm m} = 0$. The surface was arbitrary. ∎
Flux freezing is a measurement, not an idealisation. The smallest $\mathrm{R_m}$ in the table is the solar corona's, $1.024\times10^{12}$, which is $\mathbf{12.0}$ decades. By (2.4) that is the number of dynamical times a coronal loop would need in order to lose its field by ohmic decay. The word "decades" is printed beside the exponent that produced it, because a phrase like "many orders of magnitude" cannot be checked and would have been wrong: this book's own build sessions have three times printed an order-of-magnitude phrase that no saved run supports, and a phrase, unlike an exponent, is read by no mechanical check.

The result of Proposition 4 is Alfvén's, in a one-page note of 1942. That paper was not fetched for this book and is cited for the result only, in the same way Module 1 cites Maxwell (1867); its volume, pages and date were confirmed at Crossref and nothing else about it is claimed here. The proof above is this module's own.

3. Magnetic pressure and magnetic tension

The field acts on the gas through the Lorentz force on the current it carries. What that force is, mechanically, is the content of this section, and it is what Module 2 §11, Module 3 §11, Module 4 §8, Module 5 §8 and Module 7 §5 have each been promised.

Proposition 5 (the Lorentz force splits into a pressure and a tension). The force per unit volume on a conducting fluid is $$ \mathbf f = \frac{1}{c}\,\mathbf j\times\mathbf B = \frac{1}{4\pi}(\nabla\times\mathbf B)\times\mathbf B = -\nabla\!\left(\frac{B^2}{8\pi}\right) + \frac{(\mathbf B\cdot\nabla)\mathbf B}{4\pi}. \tag{3.1}$$ The first term is an isotropic pressure $p_{\rm mag} = B^2/8\pi$. The second is a tension $B^2/4\pi$ along the field; where a field line has radius of curvature $R_c$ it contributes a force per volume $B^2/(4\pi R_c)$ directed toward the centre of curvature.
Use the identity $\nabla(\mathbf B\cdot\mathbf B) = 2(\mathbf B\cdot\nabla)\mathbf B + 2\mathbf B\times(\nabla\times\mathbf B)$, so that $$ (\nabla\times\mathbf B)\times\mathbf B = -\mathbf B\times(\nabla\times\mathbf B) = (\mathbf B\cdot\nabla)\mathbf B - \tfrac12\nabla B^2. $$ Dividing by $4\pi$ gives (3.1). For the curvature, write $\mathbf B = B\hat{\mathbf b}$ with $\hat{\mathbf b}$ a unit vector. Then $$ \frac{(\mathbf B\cdot\nabla)\mathbf B}{4\pi} = \frac{B}{4\pi}\frac{\partial}{\partial\ell}\big(B\hat{\mathbf b}\big) = \hat{\mathbf b}\,\frac{\partial}{\partial\ell}\!\left(\frac{B^2}{8\pi}\right) + \frac{B^2}{4\pi}\frac{\partial\hat{\mathbf b}}{\partial\ell}, $$ where $\partial/\partial\ell$ is the derivative along the field line. The first piece is a pressure gradient along $\hat{\mathbf b}$, which exactly cancels the field-parallel part of $-\nabla(B^2/8\pi)$ — so the magnetic pressure acts only across the field, never along it. The second piece uses the Frenet relation $\partial\hat{\mathbf b}/\partial\ell = -\hat{\mathbf n}/R_c$ with $\hat{\mathbf n}$ pointing away from the centre of curvature, giving a force $B^2/(4\pi R_c)$ toward it. ∎
The cancellation in that proof is the whole physical content, and it is easy to read past. A magnetic field is not a gas with pressure $B^2/8\pi$. Along its own direction the pressure term is cancelled exactly, and a straight uniform field exerts no force at all. What a field resists is being squeezed (the pressure, acting across the lines) and being bent (the tension, acting along them). Every result in §4 to §9 is one of these two, or both.

system$p_{\rm gas}$ / dyn cm⁻²$p_{\rm mag}$ / dyn cm⁻²$p_{\rm mag}/p_{\rm gas}$
solar photosphere$1.2000\times10^{5}$$9.9472\times10^{-1}$$8.2893\times10^{-6}$
solar corona$1.3806\times10^{-1}$$3.9789\times10^{0}$$2.8819\times10^{1}$
solar wind, 1 au$8.2839\times10^{-11}$$9.9472\times10^{-11}$$1.2008\times10^{0}$
intracluster medium$1.3806\times10^{-11}$$3.9789\times10^{-14}$$2.8819\times10^{-3}$
molecular cloud$1.3806\times10^{-13}$$3.9789\times10^{-12}$$2.8819\times10^{1}$

The last column is $1/\beta$ by construction and carries no new information; it is printed so that a reader can see the two pressures as pressures. Note that the photosphere's $p_{\rm mag}\approx1\ \mathrm{dyn\,cm^{-2}}$ is the dynamic pressure $\tfrac12\rho v^2$ of sea-level air moving at $0.41\ \mathrm{m\,s^{-1}}$ — a draught, not a wind —, against a gas pressure comparable to the Earth's atmosphere — which is the sense in which a $5$ G field is, for the photosphere, negligible.

4. Three wave speeds, and what one of them costs Module 7

Module 4 derived sound from a gas whose only restoring force was its own pressure, and its captures-and-misses table names magnetic pressure and tension as the restoring forces it left out. Adding them turns one wave into three.

Proposition 6 (the three MHD wave speeds). Linearising the ideal MHD equations about a uniform state with field $\mathbf B_0$, for a plane wave with wavevector at angle $\theta$ to $\mathbf B_0$, the phase speeds are $$ v_{\rm Alfv\acute en} = v_{\rm A}\cos\theta, \qquad v_{\pm}^2 = \tfrac12\left(v_{\rm A}^2 + c_s^2\right) \pm \tfrac12\sqrt{\left(v_{\rm A}^2+c_s^2\right)^2 - 4v_{\rm A}^2c_s^2\cos^2\theta}, \tag{4.1}$$ with $v_+$ the fast and $v_-$ the slow magnetosonic speed. At $\theta = 0$, $v_+ = \max(v_{\rm A},c_s)$ and $v_- = \min(v_{\rm A},c_s)$. At $\theta = 90^\circ$ the slow and Alfvén speeds both vanish and only the fast wave survives, at $\sqrt{v_{\rm A}^2+c_s^2}$.
Write $\rho = \rho_0+\rho_1$, $\mathbf u = \mathbf u_1$, $\mathbf B = \mathbf B_0 + \mathbf B_1$ and keep first order. Continuity and the adiabatic closure give $p_1 = c_s^2\rho_1$; the induction equation (2.1) at $\eta_{\rm m}=0$ gives $\partial_t\mathbf B_1 = \nabla\times(\mathbf u_1\times\mathbf B_0)$; and the momentum equation with (3.1) gives $$ \rho_0\,\partial_t\mathbf u_1 = -\nabla p_1 - \nabla\!\left(\frac{\mathbf B_0\cdot\mathbf B_1}{4\pi}\right) + \frac{(\mathbf B_0\cdot\nabla)\mathbf B_1}{4\pi}. $$ Substituting $\propto\exp[i(\mathbf k\cdot\mathbf x-\omega t)]$ turns continuity into $\rho_1 = \rho_0(\mathbf k\cdot\mathbf u_1)/\omega$ and the induction equation into $$ \mathbf B_1 = \frac{\mathbf B_0(\mathbf k\cdot\mathbf u_1) - \mathbf u_1(\mathbf k\cdot\mathbf B_0)}{\omega}. $$ Putting both into the momentum equation and writing $\mathbf v_{\rm A} \equiv \mathbf B_0/\sqrt{4\pi\rho_0}$, so that $|\mathbf v_{\rm A}| = v_{\rm A}$, gives one vector equation in $\mathbf u_1$ alone: $$ \omega^2\mathbf u_1 = \left(c_s^2+v_{\rm A}^2\right)(\mathbf k\cdot\mathbf u_1)\,\mathbf k \;+\; (\mathbf k\cdot\mathbf v_{\rm A})\Big[(\mathbf k\cdot\mathbf v_{\rm A})\mathbf u_1 - (\mathbf v_{\rm A}\cdot\mathbf u_1)\mathbf k - (\mathbf k\cdot\mathbf u_1)\mathbf v_{\rm A}\Big]. \tag{4.1a}$$ Now choose axes: $\mathbf B_0 = B_0\hat{\mathbf z}$ and $\mathbf k = k(\sin\theta,\,0,\,\cos\theta)$, so $\mathbf k\cdot\mathbf v_{\rm A} = kv_{\rm A}\cos\theta$.

The $y$ component decouples. $u_y$ appears in neither $\mathbf k\cdot\mathbf u_1$ nor $\mathbf v_{\rm A}\cdot\mathbf u_1$, and $\mathbf k$ and $\mathbf v_{\rm A}$ have no $y$ component, so every term on the right of (4.1a) except the first inside the bracket vanishes for it: $\omega^2u_y = k^2v_{\rm A}^2\cos^2\theta\,u_y$. This motion is perpendicular to both $\mathbf k$ and $\mathbf B_0$, so it compresses nothing — $\mathbf k\cdot\mathbf u_1 = 0$ means $\rho_1 = 0$ and $p_1 = 0$ — and the only restoring force left is the tension. That is the Alfvén wave, and the proof shows what it is: a transverse wave on a string under tension $B^2/4\pi$ with mass per unit length $\rho$, whose speed $\sqrt{\rm tension/\rho}$ is exactly $v_{\rm A}$ — which is the promised justification for calling $v_{\rm A}$ a speed.

The $x$ and $z$ components remain, and they lie in the plane of $\mathbf k$ and $\mathbf B_0$. Write $v \equiv \omega/k$ and $\Delta \equiv \sin\theta\,u_x + \cos\theta\,u_z$, so $\mathbf k\cdot\mathbf u_1 = k\Delta$. The $z$ component of (4.1a) is $$ v^2u_z = (c_s^2+v_{\rm A}^2)\Delta\cos\theta + v_{\rm A}^2\cos^2\theta\,u_z - v_{\rm A}^2\cos^2\theta\,u_z - v_{\rm A}^2\cos\theta\,\Delta = c_s^2\cos\theta\,\Delta, $$ the two tension terms in $u_z$ cancelling exactly — a field does nothing to motion along itself, which is §3's cancellation seen again. The $x$ component is $$ v^2u_x = (c_s^2+v_{\rm A}^2)\Delta\sin\theta + v_{\rm A}^2\cos^2\theta\,u_x - v_{\rm A}^2\cos\theta\sin\theta\,u_z. $$ Collecting, and using $(c_s^2+v_{\rm A}^2)\sin^2\theta + v_{\rm A}^2\cos^2\theta = c_s^2\sin^2\theta + v_{\rm A}^2$, $$ \begin{pmatrix} v^2 - c_s^2\sin^2\theta - v_{\rm A}^2 & -c_s^2\sin\theta\cos\theta \\ -c_s^2\sin\theta\cos\theta & v^2 - c_s^2\cos^2\theta \end{pmatrix}\begin{pmatrix}u_x\\ u_z\end{pmatrix} = 0, $$ and a non-trivial solution needs the determinant to vanish: $$ \big(v^2 - c_s^2\sin^2\theta - v_{\rm A}^2\big)\big(v^2 - c_s^2\cos^2\theta\big) - c_s^4\sin^2\theta\cos^2\theta = 0. $$ The two $c_s^4\sin^2\theta\cos^2\theta$ terms cancel, leaving $$ v^4 - v^2\left(v_{\rm A}^2+c_s^2\right) + v_{\rm A}^2c_s^2\cos^2\theta = 0, $$ a quadratic in $v^2$ whose two roots are (4.1). At $\cos\theta = 1$ it factorises as $(v^2-v_{\rm A}^2)(v^2-c_s^2)=0$; at $\cos\theta = 0$ the constant term vanishes, so one root is $0$ and the other $v_{\rm A}^2+c_s^2$. ∎
system$c_s$$v_{\rm A}$fast, $\theta=90^\circ$fast, $\theta=0$slow, $\theta=0$
solar photosphere$7.844$$0.02474$$7.844$$7.844$$0.02474$
solar corona$150.7$$886.4$$899.1$$886.4$$150.7$
solar wind, 1 au$52.21$$62.67$$81.57$$62.67$$52.21$
intracluster medium$1507$$88.64$$1510$$1507$$88.64$
molecular cloud$0.2439$$1.434$$1.455$$1.434$$0.2439$

All speeds in km s⁻¹. Note the photosphere row: with $v_{\rm A}$ three hundred times smaller than $c_s$, the field changes the wave speeds of Module 4 by three parts in a million, which is why Module 4 was entitled to ignore it and says so.

4.1 Module 7's one-degree bound, recomputed — a validation, not a check

Module 7 §9.5 states the debt in one line: "A purely hydrodynamic calculation that needs a one-degree statement about a field it never mentions is a calculation waiting for magnetohydrodynamics." Its Proposition 9 says that Kelvin–Helmholtz growth at the observed rate requires $$ B\cos\theta \;\lt \; \Delta U\sqrt{\frac{2\pi\rho_h\rho_l}{\rho_h+\rho_l}} \tag{4.2}$$ for equal fields on both sides of the interface, with $4\pi$ in place of $2\pi$ if the field is one-sided. That inequality is recomputed here from Proposition 6's dispersion relation rather than copied, so that the comparison below is able to fail.

A validation of the machinery, and not a check against nature. With $\rho_h = 5.8607\times10^{-16}$ and $\rho_l = 2.6210\times10^{-16}\ \mathrm{g\,cm^{-3}}$ read out of Module 7 §9.1's provenance note, and $\Delta U = 20\ \mathrm{km\,s^{-1}}$ read from the sentence that calls it "the shear measured on the flank of a coronal mass ejection", (4.2) gives $\mathbf{0.06747}$ G against Module 7's printed $0.0675$, ratio $0.9995$, and $\mathbf{0.09541}$ G against its $0.0954$ on the one-sided hypothesis, ratio $1.0001$. The whole angle table agrees: $3.87^\circ$, $1.93^\circ$, $0.77^\circ$, $0.39^\circ$, $0.08^\circ$ at $B = 1, 2, 5, 10, 50$ G. Nothing independent is being tested. This is labelled the way Module 5 labelled Kandori's contrast column: a check that two parts of this book agree, which is worth having and is not evidence about the Sun.
The angle is measured from perpendicular, and the first draft of this module's generator got it backwards. (4.2) bounds $B\cos\theta$, so at fixed $B$ it bounds $\cos\theta$, and the permitted departure from perpendicular is $\arcsin(\cos\theta)$ and not $\arccos(\cos\theta)$. Writing $\arccos$ printed $86.13^\circ$ where Module 7 prints $3.87^\circ$ — a ninety-degree error that no mechanical check in this project reads, and which was caught only because Module 7 had printed its table.

And here is the gap this module cannot fill. The book's shape is: derive the terrestrial result, change the regime, check against a measured astronomical number. Module 4 had the speed of sound in air, Module 6 the dry adiabat, Module 10 the Kolmogorov constant. Magnetohydrodynamics has terrestrial measurements — Alfvén waves were first seen in liquid mercury, and every tokamak is an MHD experiment — and none of them was read for this book. What occupies the terrestrial slot above is a reproduction of Module 7, which is a check against this book and not against a laboratory. The four checks of §§5–8 are all astronomical, and a reader should weigh them knowing that no bench-top number stands behind them.

5. The anchor: the Parker spiral

Module 9 built a wind that leaves the Sun radially and is still accelerating at 1 au. The Sun rotates. A field line is frozen to the gas by Proposition 4 and its foot is carried round by the rotation while its body streams outward — so the line is drawn into a spiral, and the shape of that spiral is fixed by two numbers this book already has: the solar rotation rate and the wind speed. Nothing is fitted.

Proposition 7 (the Parker spiral). For a steady, spherically symmetric, radial wind of speed $v(r)$ from a source surface at $r_0$, on a star rotating at angular rate $\Omega$, the frozen-in field has $$ B_r(r) = B_r(r_0)\left(\frac{r_0}{r}\right)^{2}, \qquad \frac{B_\phi}{B_r} = -\,\frac{\Omega\,(r-r_0)\sin\theta}{v}, \tag{5.1}$$ with $\theta$ the colatitude. The first follows from $\nabla\cdot\mathbf B = 0$ alone and is exact; the second is the kinematics of a line whose foot rotates.
The radial part. $\nabla\cdot\mathbf B = 0$ integrated over a sphere of radius $r$ gives $\oint B_r\,\mathrm{d}S = 4\pi r^2\langle B_r\rangle = $ constant, since no flux is created between two spheres. Hence $B_r r^2$ is constant. No dynamics entered, which is why this half is exact and is the half the control in Check 1 tests.

The azimuthal part. Work in the frame rotating with the star, in which the flow is steady. There the velocity is $\mathbf v' = v\hat{\mathbf r} - \Omega(r-r_0)\sin\theta\,\hat{\boldsymbol\phi}$ — Parker's equation (24), with $r_0$ his $b$ — and in a steady state the field must be parallel to the flow, for otherwise the flow would sweep the field into a configuration that changes with time. Hence $\mathbf B \parallel \mathbf v'$, so $$ \frac{B_\phi}{B_r} = \frac{v'_\phi}{v'_r} = -\frac{\Omega(r-r_0)\sin\theta}{v}, $$ which is Parker's equation (26). ∎
Parker's $(r-r_0)$ is carried, and the module prints what dropping it would cost. His (26) reads $B_\phi = B(\theta,\phi_0)(\omega/v_m)(r-b)(b/r)^2\sin\theta$. The familiar $\Omega r\sin\theta/v$ is his large-$r$ limit, stated as a limit, and his (27) places the $45^\circ$ surface at $r\simeq(v_m/\omega)\sin\theta$ only after $b$ has been dropped. Printing $\Omega r\sin\theta/v$ while citing (26) is the one thing that is not defensible, so this module does neither. At 1 au with $r_0 = 2.5\,R_\odot$, carrying $r_0$ lowers $-B_\phi/B_r$ from $0.9840$ to $0.9726$, by $1.16$ per cent; dropping it raises the ratio by $1.18$ per cent. The two figures are one term seen from its two ends, and $r_0/1\,\mathrm{au} = 0.01163$ is why. The source surface is written $r_0$ throughout, because $b$ is Module 1's impact parameter and Module 10's forcing parameter.

The observable consequence is the total field's radial falloff. Writing $x \equiv \Omega(r-r_0)\sin\theta/v$, $$ |\mathbf B| = B_r\sqrt{1+x^2}, \qquad \frac{\mathrm{d}\ln|\mathbf B|}{\mathrm{d}\ln r} = -2 + \frac{x^2}{1+x^2}\left(\frac{r}{r-r_0} - \frac{\mathrm{d}\ln v}{\mathrm{d}\ln r}\right). \tag{5.2}$$ Fig. 2 draws both halves of this. At small $r$ the field is nearly radial, $x\ll1$, and the index is $-2$. At large $r$ it is nearly azimuthal, $x\gg1$, and the index tends to $-1$. Between the two, over the range a spacecraft actually flew, the index is something in between — and that is a measurement.

1 au spiral angle 44.20° at 1 au, Ω = 2π/25.38 d v = 435.6 km s-1 0.3 0.5 0.7 10 20 50 Parker, fitted index -1.7751 measured -1.5460 ± 0.110 unwound radial field, −2 heliocentric distance / au |B| / nT Mean fits. The band is the YEAR-TO-YEAR scatter 0.110, not the formal fit error 0.018.
Left: four field lines from the source surface at $2.5\,R_\odot$ out to 1 au in the equatorial plane, with the highlighted line's tangent drawn at 1 au against the radial direction. The angle between them is $44.20^\circ$ on the mean-fit wind speed. Right: the total field over the Helios range $0.29$–$0.98$ au: the Parker prediction of (5.2), whose best-fit power law over that window is $-1.7751$; the measured power law $-1.5460$ with a band showing the year-to-year scatter $\pm0.110$ of Venzmer & Bothmer's Fig. 9, not the formal fit error $0.018$; and a straight $r^{-2}$ for an unwound field. The prediction and the measurement are close enough that the eye does not separate them over this window, while the unwound line plainly separates — but this is the mean fit set, on which the gap is $2.08$ year-to-year scatters and $12.7$ formal errors; the median set, where the gap is $0.92$ scatters, is closer still. That is the visual form of the verdict below, and the verdict is stated so that it survives either set.

5.1 Check 1 — the anchor

Venzmer & Bothmer (2018) fitted power laws $x(r) = d\,r^{e}$ to combined Helios 1 and 2 data over $0.29$–$0.98$ au and published both mean and median fits, with errors on every coefficient. Module 9 reads the mean fits; Module 2 reads the median row. This module prints both, as equals, and never mixes them inside one comparison — which is the rule an earlier draft of the generator broke by quoting the median field against the mean density, two denominators from one table.

median fitsmean fits
Parker prediction, fitted over $0.29$–$0.98$ au$-1.7557$$-1.7751$
measured field index, Table 3$-1.6550\pm0.017$$-1.5460\pm0.018$
gap, in year-to-year scatters ($0.110$)$0.92$$2.08$
gap, in formal fit errors$5.9$$12.7$
an unwound $-2.000$, in scatters$3.14$$4.13$
prediction nearer the measurement than $-2$ is, by a factor of$3.43$$1.98$
CONTROL: density index against $-2-\alpha_v$, in density scatters ($0.072$)$0.49$$0.54$
Check 1 — CONFIRMED that the field is wound; NOT RESOLVED as to how much. On both fit sets the Parker prediction lies nearer the measured index than an unwound radial field does — nearer by a factor of $3.43$ on the median fits and $1.98$ on the mean — the ratio of the two gaps, in which the scatter cancels. On neither is it inside the formal fit error: $5.9$ and $12.7$ errors away. The verdict is stated in the form that survives the choice of fit set and the choice of error bar, because a verdict that depended on either would be a choice of denominator presented as a result.

The control is what makes this a measurement and not a coincidence, and it is Module 2's preferred shape: one exact conservation law confirmed and one modelling assumption tested, from the same table. Mass conservation through a steady spherical flow forces $\rho\propto r^{-2-\alpha_v}$, where $\alpha_v$ is the measured velocity exponent of the same fit set — which predicts $-2.0580$ (median) and $-2.0490$ (mean). The measured density indices are $-2.093\pm0.046$ and $-2.010\pm0.038$, gaps of $0.0350$ and $0.0390$. These are divided by the density exponent's own year-to-year scatter, $0.072$, and not by the field's $0.110$ — Venzmer & Bothmer's Fig. 9 gives one scatter per fitted quantity, and using the field's here would be the two-denominators error this section refuses two paragraphs above. The gaps are therefore $0.49$ and $0.54$ scatter units. So this data set can see a near-inverse-square falloff when there is one. The field does not show one, and (5.1)'s second half says why: the field is wound, and the density is not.

Which error bar, and why the scatter. The formal errors $0.017$ and $0.018$ describe how well a power law fits one combined data set. The year-to-year scatter $0.110$ of their Fig. 9 describes how much the exponent itself moves from one year of solar wind to the next. Module 9 made the same choice on the same table for the same reason: the formal error describes a regression and the scatter describes the solar wind. Both are printed above, and both verdicts are stated, because the choice is a modelling decision and not a result.

5.2 Sensitivity, priced rather than hidden

variantfitted indexshift
baseline (median fits)$-1.7557$$0.0000$
equatorial rotation rate, $24.47$ d$-1.7428$$+0.0129$
source surface $2.0\,R_\odot$$-1.7553$$+0.0004$
source surface $3.0\,R_\odot$$-1.7561$$-0.0004$
$r_0 = 0$, the textbook form$-1.7536$$+0.0020$
constant wind speed$-1.7527$$+0.0030$
the whole mean-fit wind law: $435.6\ \mathrm{km\,s^{-1}}$ at 1 au, $\alpha_v = +0.049$$-1.7751$$-0.0194$

The largest shift in that table is $0.0194$, smaller than the year-to-year scatter $0.110$ by a factor of $5.7$. The verdict turns on none of them. That sentence is the point of the table.

The rotation rate is two choices, not one, and the module prices them separately. $25.38$ d is the NASA NSSDC sidereal period, $609.12$ hours, adopted at $16^\circ$ latitude; the fact sheet's own law $14.37 - 2.33\sin^2L - 1.56\sin^4L$ deg/day reproduces it there to four digits. $24.47$ d is Snodgrass & Ulrich (1990), $360/14.71$, from Doppler cross-correlation of the supergranulation network — a different tracer, not just a different latitude. The $3.71$ per cent between them factorises exactly: $1.0236$ for the tracer at either latitude, $1.0132$ for the latitude on either tracer, product $1.0371$, against $25.3808/24.4731 = 1.0371$. Only the latitude is a choice this module makes; the tracer is a disagreement between two ways of watching the Sun.

6. The Alfvén surface, bracketed by two measurements

Close to the Sun the field is strong and the wind slow, so $v \lt v_{\rm A}$ and the field controls the flow: it can enforce corotation, and it carries away angular momentum. Far out the wind is fast and the field weak, so $v \gt v_{\rm A}$ and the flow drags the field, which is what §5 just measured. In between there is a surface where the two are equal, and it is the boundary between those two regimes.

Proposition 8 (the wind crosses its own radial Alfvén speed exactly once). For the wind of Proposition 7, mass conservation gives $\rho \propto r^{-2}v^{-1}$ and $\nabla\cdot\mathbf B = 0$ gives $B_r\propto r^{-2}$, so the radial Alfvén speed obeys $$ v_{\rm A,r} \;=\; \frac{B_r}{\sqrt{4\pi\rho}} \;\propto\; \frac{r^{-2}}{\sqrt{r^{-2}v^{-1}}} \;=\; r^{-1}v^{1/2}. \tag{6.1}$$ If $v$ is non-decreasing outward, $v/v_{\rm A,r}\propto r\,v^{1/2}$ is strictly increasing, so the equation $v = v_{\rm A,r}$ has exactly one root $r_{\rm A}$.
$\dot M = 4\pi r^2\rho v$ constant gives $\rho\propto r^{-2}v^{-1}$; substituting into $B_r/\sqrt{4\pi\rho}$ with $B_r\propto r^{-2}$ gives (6.1). Then $v/v_{\rm A,r}\propto v\cdot r\,v^{-1/2} = r\,v^{1/2}$, a product of two non-decreasing positive factors one of which is strictly increasing. A strictly increasing continuous function crosses the value $1$ at most once, and does so at least once given that it is $\ll1$ at the base and $\gg1$ at 1 au. ∎
The Alfvén speed in (6.1) takes $B_r$ and not $|\mathbf B|$, and the first draft of this module's generator used $|\mathbf B|$. Two reasons, either of which settles it. (i) §5 has just proved that $|\mathbf B|$ does not scale as $r^{-2}$; $B_r$ does, by $\nabla\cdot\mathbf B = 0$, and it is the $r^{-2}$ scaling the extrapolation inward uses. (ii) The measured quantity is the Weber–Davis Alfvén radius, which Verscharen, Bale & Velli define on the radial Alfvén speed, because it is the radial components that enter the flow deflection they fit. Using $|\mathbf B|$ inflated every radius below by $\sqrt{1+x_1^2}$, a factor near $1.40$, and changed the module's headline.
5 10 20 50 100 200 20 50 100 200 500 Verscharen FLS1 12.08 R☉ PSP I1, a LOWER bound ≥ 19.8 R☉ eight model variants, 15.76–19.35 R☉ wind speed 435.6 km s-1 radial Alfvén speed, ∝ 1/r heliocentric distance / R☉ speed / km s-1 the two differ by 1.64, and are not the same quantity every variant exceeds the first by 1.305–1.602 and falls short of the second by 1.023–1.256
The wind speed and the radial Alfvén speed of (6.1) against heliocentric distance, on the mean fits with protons only, and the crossing they define. The shaded vertical band spans all eight model variants — two fit sets, constant or power-law $v$, protons or $4$ per cent helium — which run from $15.76$ to $19.35\,R_\odot$. The two published values are marked: Verscharen, Bale & Velli's $12.080\pm0.236\,R_\odot$, and Parker Solar Probe interval I1's $19.8\,R_\odot$, drawn with an arrow because it is a lower bound and not a value.

6.1 Check 2 — and the result is a bracket

quantityvalue
all eight extrapolated $r_{\rm A}$$15.76$–$19.35\,R_\odot$
Verscharen, Bale & Velli (2021) Table 3, first fast-latitude scan$12.080\pm0.236\,R_\odot$
Parker Solar Probe interval I1, a lower bound$\ge19.8\,R_\odot$
the band against Verscharen$1.305$ to $1.602$ times
the band short of the PSP boundby $1.023$ to $1.256$
the two published values against each other$1.639$
Check 2 — BRACKETED, and it is a measurement rather than a verdict. Every one of the eight variants lies above Verscharen, Bale & Velli's surface and below Parker Solar Probe's lower bound. A single spherically symmetric extrapolation is therefore refuted by both, in opposite directions — and it could not have satisfied both, because the two published numbers are $1.639$ apart and are not the same quantity. Verscharen, Bale & Velli fit a steady axisymmetric Weber–Davis surface to Ulysses fast-latitude scans. Kasper et al. report local crossings of a corrugated boundary along one trajectory, on two days. No spherically symmetric model can be both.
The PSP number is a bound and it has a direction. Kasper et al.'s Table I lists three sub-Alfvénic intervals in encounter 8, at $19.8\to18.4$, $16.0\to16.0$ and $17.7\to18.0\,R_\odot$, with median Alfvén Mach numbers $0.79$, $0.49$ and $0.88$. Every one is below $1$, so each radius is a point at which the spacecraft was already inside the surface — never the surface itself. The outermost, $19.8\,R_\odot$, is therefore a lower bound on $r_{\rm A}$ along that ray on that day, and a model value below it violates it rather than agreeing with it.

And "inside their $16$–$20\,R_\odot$ range" is a sentence this module does not write. It would be true and would mean nothing: that range is a union over three longitudes and two days, not a surface anything can be inside of.

Fig. 3 shows why the bracket is a result rather than an accident of the headline choice: the band of eight variants is narrower than the gap between the two published surfaces. Two extrapolations are priced here and neither is hidden. First, $v\propto r^{+0.058}$ is a fit over $0.29$–$0.98$ au used at about $0.08$ au, a factor $3.5$ outside its own range; it moves $r_{\rm A}$ by $6.9$ per cent against a constant speed, which is small, and this book prices that kind of thing rather than assuming it. Second, the real wind is slower close in than any power law fitted beyond $0.29$ au says — which is the physical reason a Weber–Davis radius comes out smaller than this extrapolation does, and it is the direction of the discrepancy with Verscharen.

A cross-module number, settled here. Module 9 §9 prints the solar wind's Alfvén surface as "near $12\,R_\odot$", sourced in its own sources list to Verscharen's first fast-latitude scan. Its generator afd/figs/m09_numbers.py prints, in an aside in its own run and not on the shipped page, "the Alfven surface near 19 R_sun" — the Parker Solar Probe crossing, a different method, a different definition of the surface, and not in Module 9's sources list. Module 9's page is right and its generator's aside is not. This module prints both numbers, names both methods and states that they are different quantities, so no two pages of this book contradict each other. No edit to Module 9 is proposed.

7. Magnetic support against collapse

Module 5 §8 is owed this: magnetic pressure and tension add support against collapse, and the statement that results is not an extra pressure term but a critical mass-to-flux ratio. Below it no field strength is enough; above it the cloud is held for as long as the flux stays with the gas.

Definition 3 (the mass-to-flux ratio). For a cloud of mass $M$ threaded by magnetic flux $\Phi$, $$ \lambda_\Phi \;\equiv\; \frac{M/\Phi}{(M/\Phi)_{\rm crit}}, \qquad \left(\frac{M}{\Phi}\right)_{\rm crit} = \frac{c_\Phi}{\sqrt G}. \tag{7.1}$$ $\lambda_\Phi \gt 1$ is supercritical: the field cannot hold the cloud. The symbol carries its subscript because bare $\lambda$ is Module 1's mean free path book-wide, and both appear in this module.
Proposition 9 (the critical ratio is a pure number). For a flux-frozen, self-gravitating cloud, the gravitational and magnetic energies both scale as $1/R$ at fixed $M$ and $\Phi$, so their ratio is independent of the radius and the criterion for collapse is a condition on $M/\Phi$ alone. The coefficient is $$ c_\Phi = \frac{c_1\sqrt5}{3\pi} = 0.125745 \quad\text{(uniform sphere)}, \qquad c_\Phi = \frac{1}{2\pi} = 0.159155 \quad\text{(flattened cloud)}. \tag{7.2}$$
The form is derived. For a uniform sphere of mass $M$ and radius $R$ the gravitational energy is $W = -3GM^2/5R$. Flux freezing fixes $\Phi = \pi R^2\langle B\rangle$, so $\langle B\rangle = \Phi/\pi R^2$ and the magnetic energy inside the cloud is $\mathcal M = (B^2/8\pi)(4\pi R^3/3) = \Phi^2/(6\pi^2R)$. Both go as $1/R$, so their sum is either positive or negative for all $R$ at once: compressing the cloud cannot change the sign. The virial condition $\mathcal M + W = 0$ therefore gives a critical ratio and not a critical radius, $$ \frac{\Phi^2}{6\pi^2R} = \frac{3GM^2}{5R} \;\Longrightarrow\; \frac{M}{\Phi} = \frac{1}{3\pi}\sqrt{\frac{5}{2G}} = \frac{0.16777}{\sqrt G}. $$ That is the estimate this argument produces, and it is not the normalisation (7.2) uses. Mouschovias & Spitzer's equation (2) is written against $(1/3\pi)\sqrt{5/G} = 0.23726/\sqrt G$, which is $\sqrt2$ larger — it counts a magnetic energy twice the energy of the field inside the cloud, and where that second factor of two comes from is not derived here and is not derived anywhere in this book. Their $c_1$ is measured against their normalisation and not against the estimate above, so $c_1 = 0.53$ is not the size of the correction to this line: the ratio of their coefficient to the estimate above is $0.125745/0.16777 = 0.7495$. Every number in this module uses their formula with their $c_1$, so none of them turns on this.

The coefficients are stated, not derived. Mouschovias & Spitzer (1976) show that the virial estimate is an approximation and fit a correction factor to exact equilibrium models: their equation (2) reads $M_{c\Phi} = (c_1/3\pi)(5/G)^{1/2}\Phi$ and their equation (4) gives $c_1 = 0.53$, whence $c_\Phi = 0.53\sqrt5/3\pi = 0.125745$. Nakano & Nakamura (1978) solve the flattened case and find the disc stable when $\sigma_0/B_0 \lt (4\pi^2G)^{-1/2}$ — a column density over a field — so $c_\Phi = 1/2\pi$ exactly. ∎
This module never prints $0.13$. The number that circulates for Mouschovias & Spitzer's coefficient is $0.13$; their two-page note prints $c_1 = 0.53$ and never prints $0.13$ at all. The correct value is $0.125745$, and $0.13$ is $1.0338$ times it. Two further traps from the same paper, because both are ways of being wrong by a lot: their $c_2 = 0.60$ applies to the maximum external pressure and not to the mass, and their footnote 1 warns that what Spitzer (1968) writes $(M_c/M)^{2/3}$ is here $(M_{c\Phi}/M)^2$, because $M_{c\Phi}$ is the critical mass at fixed flux and $M_c$ the critical mass at fixed initial density, with $M_{c\Phi}/M = (M_c/M)^{1/3}$. A module that quotes one symbol for the other is wrong by a cube.

Applied to this module's $10$ pc cloud — $n_{\rm tot} = 100\ \mathrm{cm^{-3}}$ at $\mu = 2.33$, $R = 5$ pc, $B = 10\ \mu$G — the two geometries give:

geometry$c_\Phi$$(M/\Phi)_{\rm crit}$ / g Mx⁻¹$M_\Phi$ / $M_\odot$$\lambda_\Phi$
uniform sphere (Mouschovias & Spitzer)$0.125745$$4.8673\times10^{2}$$1.831\times10^{3}$$1.635$
flattened cloud (Nakano & Nakamura)$0.159155$$6.1605\times10^{2}$$2.317\times10^{3}$$1.292$

The cloud's mass is $2.993\times10^{3}\,M_\odot$. Both answers are supercritical: on this census field the cloud is not held. The two differ by the coefficient ratio $1.2657$.

7.1 Check 3 — the coefficient, and the sign

Troland & Crutcher (2008) observed 34 dark cloud cores for about $500$ hours with Arecibo in OH Zeeman, and converted each measurement to a mass-to-flux ratio with their equation (2), $$ \lambda \;=\; 7.6\times10^{-21}\,\frac{N({\rm H_2})}{B_{\rm los}}, \qquad N \text{ in cm}^{-2},\ B \text{ in }\mu\text{G}, \tag{7.3}$$ stating in the line above it that their critical ratio is Nakano & Nakamura's — the same constant (7.2) computes. So their coefficient and this module's can be checked against each other rather than merely quoted together.

Check 3 — CONFIRMED on the coefficient and on the sign; REFUTED that a mass-to-flux ratio can be quoted without its geometry.
(a) The coefficient. Rebuilding (7.3) from $c_\Phi = 1/2\pi$ and $\mu = 2.8$ per H₂ molecule gives $\mathbf{7.6022\times10^{-21}}$ with the hydrogen mass and $7.5473\times10^{-21}$ with the atomic mass unit, against their printed $7.6\times10^{-21}$. Their coefficient carries two significant figures, so the discriminating statement is a rounding and not a ratio: $7.6022$ rounds to $7.6$ and $7.5473$ rounds to $7.5$. The atomic mass unit is excluded and the hydrogen mass is not, and the $0.7$ per cent between them is $m_{\rm p}/m_{\rm u} = 1.00728$.
(b) The sign. Their geometry-corrected values are $\lambda_c = 1.4$–$2.1$ and $\lambda_{1/2,c} = 1.7$–$2.6$, and their own sentence is the one to quote: the cores "appear to be slightly supercritical by about a factor of $2$".
(c) What is refuted. The sphere-to-sheet coefficient ratio $1.2657$ spans $\mathbf{0.381}$ of the log width of that entire measured range. A $\lambda_\Phi$ quoted without its assumed geometry is not a measurement, and this module's own table above is the demonstration: the same cloud, the same field, $1.635$ or $1.292$.
This is not a check on this module's cloud, and the density is why. Troland & Crutcher's cores sit at "a typical density of a few times $10^{3}\ \mathrm{cm^{-3}}$". This module's row is a $10$ pc cloud at $n_{\rm tot} = 100\ \mathrm{cm^{-3}}$, thirty times less dense. The check is on the critical coefficient and on the sign of the answer; it is not a check on the value of $\lambda_\Phi$ for this particular cloud, and their raw uncorrected $\lambda\approx4.2$ (mean) and $5.2$ (median) are not the numbers to quote either.
Two mass conventions, and a module that mixes them is wrong by 20 per cent. Troland & Crutcher's $\mu = 2.8$ is per H₂ molecule; this module's $\mu = 2.33$ is per particle. They are the same mass density counted two ways: with He/H $= 0.1$ by number, $\rho/(n_{\rm H_2}m_{\rm H}) = 1.4/0.5 = 2.8$ and $\rho/(n_{\rm tot}m_{\rm H}) = 1.4/0.6 = 2.333$. So this module's $n_{\rm tot} = 100\ \mathrm{cm^{-3}}$ row is $n({\rm H_2}) = 83.3\ \mathrm{cm^{-3}}$, and nothing in this project's check battery reads that distinction. It is the same shape of defect as Module 10's one-dimensional against three-dimensional dispersion and Module 9's $\mu = 1/2$, one layer down.
Crutcher (2012), ARA&A 50, 29, was NOT read. It is paywalled and neither OpenAlex nor Semantic Scholar knows of an open-access copy; both were queried by DOI. It is listed in Sources because a reader looking for the full Zeeman compilation should be sent to it, and it carries "not read" in its entry. Nothing in this module rests on it. Troland & Crutcher (2008) is the primary survey behind the part of that review this module's §7 needed, and it gives a better check than the review would have, because it names the critical ratio it divides by.

8. Transport across a field, and the answer that is wrong by twenty-three decades

Module 2 §9 put two problems together and pointed at this module for both: the intracluster medium's Reynolds number of about $600$, which is not large enough for the turbulence that is observed, and the thermal-conduction problem it names in the same sentence and attributes to Module 1. Neither module puts a number on the conduction half: Module 1's Problem K3 compares conduction with the sound-crossing time and finds it slow, and says in the same solution that the comparison the cooling-flow problem needs is against the radiative cooling time, which this book has not made. So the second problem arrives here as a scaling and a suspicion, not as a measured discrepancy. It said the two "have the same suspected resolution — magnetic suppression of transport across field lines — and Module 12 treats both together." This section is that treatment, and it ends by refuting the classical form of the resolution.

Proposition 10 (perpendicular transport is suppressed by $(\omega_{\rm c}\tau)^{-2}$). In an unmagnetised gas a random walk of step $\lambda$ per collision time $\tau$ gives a diffusivity $\sim\lambda^2/\tau$. Across a field, a particle cannot travel $\lambda$ between collisions: it returns to within a gyroradius of where it started. The step is therefore $r_g$ rather than $\lambda$, and $$ \frac{D_\perp}{D_\parallel} \;\sim\; \frac{r_g^2}{\lambda^2} \;=\; \frac{1}{(\omega_{\rm c}\tau)^2}, \tag{8.1}$$ since $\lambda = v\tau$ and $r_g = v/\omega_{\rm c}$. The same factor governs the conductivity and the viscosity, with different numerical coefficients.
Between collisions the guiding centre is displaced along $\mathbf B$ by $\lambda$ and across it by at most $2r_g$; the perpendicular displacement per collision is therefore $O(r_g)$ and uncorrelated from one collision to the next. A random walk of $N$ steps of size $s$ in time $N\tau$ has diffusivity $s^2/\tau$, so $D_\perp/D_\parallel = r_g^2/\lambda^2$. With $\lambda = v\tau$ and $r_g = v/\omega_{\rm c}$ the speed cancels and the ratio is $(\omega_{\rm c}\tau)^{-2}$. ∎

Braginskii (1965) does the calculation properly and supplies the coefficients. For $Z = 1$ his equations (2.12) and (2.13) on p. 217 give the electron conduction, $$ \kappa_\parallel = 3.16\,\frac{n_{\rm e}T_{\rm e}\tau_{\rm e}}{m_{\rm e}}, \qquad \kappa_\perp = 4.66\,\frac{n_{\rm e}T_{\rm e}}{m_{\rm e}\omega_{\rm e}^2\tau_{\rm e}}, \qquad \frac{\kappa_\perp}{\kappa_\parallel} = \frac{4.66}{3.16}\frac{1}{(\omega_{\rm e}\tau_{\rm e})^2}, \tag{8.2}$$ and his equations (2.22) and (2.23) on p. 218 give the ion viscosity, which Module 10 §6.2 already uses, with $\eta_0^{\rm B} = 0.96\,n_iT_i\tau_i$ and $\eta_1^{\rm B} = (3/10)\,n_iT_i/(\omega_i^2\tau_i)$.

Every Braginskii coefficient in this module is for $Z = 1$, and every plasma in it is hydrogenic, so that is the right column. His Table 1 on p. 216 tabulates the $Z$-dependence and it is worth printing, because it prices the assumption rather than hiding it: $\kappa_\parallel^{\rm e}$ runs $3.16$, $4.9$, $6.1$, $6.9$, $12.5$ for $Z = 1, 2, 3, 4, \infty$, and $\kappa_\perp^{\rm e}$ runs $4.66$, $4.0$, $3.7$, $3.6$, $3.2$. Note also that his collision-time denominators are $Z^2n_i$ and $Z^4n_i$, not $n_{\rm e}$; at $Z = 1$ they are the same number. All seven coefficients used here were read from the Consultants Bureau scan, whose pages carry two article pages each — article page $= 2\times$(PDF page)$+200$ for the left half, confirmed against the printed $214$ and $215$.

8.1 Two magnetisations for one plasma, and the exact account of their ratio

The intracluster medium yields two different numbers for $\omega_i\tau_i$, and a module that printed both without explanation would leave a reader with two magnetisations for one gas.

definitionvaluewhat it is
$\lambda/r_g$, equivalently $\omega_i\tau_i$ with $\tau = \lambda/v_\perp$$5.1767\times10^{12}$a count: the number of gyro-orbits a proton completes between collisions, using Module 1's Coulomb mean free path at $\ln\Lambda = 37.8$, $22.504$ kpc, which that module prints rounded to $22.5$
$\omega_i\tau_i$ with Braginskii's $\tau_i$$4.3087\times10^{12}$the magnetisation of Definition 2, and the one that belongs in (8.1) and (8.2)
The module's $\omega_{\rm c}\tau$ is Braginskii's, because his transport coefficients are defined on his $\tau$. Comparing a coefficient computed with one collision time against a suppression computed with another is a convention error dressed as a result. The other number is kept, as a count and never as a magnetisation: an intracluster proton completes $10^{12.71}$ gyro-orbits between collisions, which is the intracluster counterpart of the $\lambda/r_g = 3.1\times10^{6}$ that Module 1 §8 printed for the solar wind.

The ratio of the two is $1.2015$, and it decomposes exactly: $$ \frac{\lambda/r_g}{\omega_i\tau_i^{\rm B}} \;=\; \sqrt{\tfrac32}\times\frac{\ln\Lambda_{\rm e}({\rm derived})}{\ln\Lambda({\rm census})} \;=\; 1.224745 \times \frac{37.081}{37.8} \;=\; 1.224745\times0.980985 \;=\; 1.2015. \tag{8.3}$$ The $\sqrt{3/2}$ is exact and it is a component count. A Spitzer mean free path is $v_{\rm rms}\tau$ with the three-dimensional rms speed $\sqrt{3k_{\rm B}T/m}$, while the first row of the table divides it by $\sqrt{2k_{\rm B}T/m}$, the rms of the two components perpendicular to $\mathbf B$ — which is the speed a gyroradius takes. Read §1.1's table and the factor is not a discrepancy but an identity. The second factor is the electron-branch Coulomb logarithm derived here against Module 1's census value; the difference is $1.9$ per cent and is not a correction to either module.

quantity, intracluster medium at $n = 10^{-3}$, $T = 10^{8}$ K, $B = 1\ \mu$Gvalue
$\ln\Lambda$, electron branch, derived$37.08$
Braginskii $\tau_{\rm e}$$7.4227\times10^{12}$ s
Braginskii $\tau_i$$4.4981\times10^{14}$ s
$\tau_i/\tau_{\rm e}$, against $\sqrt{2m_{\rm p}/m_{\rm e}} = 60.5995$$60.5995$, ratio $1.000000$
$\omega_{\rm e}\tau_{\rm e}$$1.3055\times10^{14}$
$\omega_i\tau_i$$4.3087\times10^{12}$
$\kappa_\perp/\kappa_\parallel$$8.6523\times10^{-29}$
$\nu_\perp/\nu_\parallel$$1.6833\times10^{-26}$

The $\tau_i/\tau_{\rm e}$ row is a self-check: Braginskii's two collision times differ by $\sqrt{2m_{\rm p}/m_{\rm e}}$ exactly, so if the row ever stops reading $1.000000$ one of the two has been edited alone.

10-30 10-25 10-20 10-15 10-10 10-5 100 Zhuravleva et al. (2019): suppressed by 10–1000, a ONE-SIDED bound unsuppressed Braginskii ν⊥/ν∥ = 1.68e-26 Braginskii κ⊥/κ∥ = 8.65e-29 22.8 to 24.8 decades ratio of perpendicular to parallel transport The measurement lies far nearer the UNSUPPRESSED value than the classical one, so the step is not the gyroradius. What it IS, this book does not compute.
The transport ladder on one logarithmic axis spanning thirty decades. At the right, unsuppressed transport. The shaded band is Zhuravleva et al.'s measured suppression of the Coma cluster's effective viscosity, between $10$ and $1000$ — drawn open at its left-hand end, because it is a one-sided bound and not an interval. Far to the left, Braginskii's classical perpendicular viscosity and conduction. The arrow spans the gap. Zhuravleva measure a viscosity, so the comparison that means anything is with $\nu_\perp/\nu_\parallel$; the conduction mark is the analogous quantity, not a second measurement.

8.2 Check 4 — the classical answer is refuted, and the direction is named

Check 4 — CONFIRMED that the intracluster medium is magnetised; REFUTED that classical perpendicular transport describes it.
Confirmed. $\omega_i\tau_i = 4.3087\times10^{12}$, and $\log_{10}(\lambda/r_g) = \mathbf{12.71}$. A proton completes $10^{12.71}$ gyro-orbits between collisions, so the gas is magnetised by $12.71$ decades, and the ratio Module 1 §8 printed for the solar wind, $\lambda/r_g = 3.1\times10^{6}$, is $6.2$ decades smaller.
Refuted. Braginskii's classical viscous suppression factor is $\mathbf{5.941\times10^{25}}$. Zhuravleva et al. (2019) bound the measured suppression of the Coma cluster's effective viscosity at $\sim10$ to $\sim1000$. The classical factor is larger than the measured one by between $\mathbf{22.8}$ and $\mathbf{24.8}$ decades. Both decade figures are printed rather than rounded into a phrase, because a phrase cannot be checked and the phrase this one replaced read "twenty-odd".

The direction of the failure is the physical content, and Fig. 4 is where to read it. The measurement lies far nearer the unsuppressed Coulomb value than the classical perpendicular one. That is not a small correction to (8.1); it says the premise of (8.1) is wrong for a real cluster. The field in a cluster is tangled, so heat and momentum follow field lines that wander, and a particle that has travelled one mean free path along a line has also moved a long way across the mean field direction. The step is therefore neither $\lambda$ nor $r_g$, and it is set by the field-line random walk and by pressure-anisotropy microinstabilities.

What the effective transport is, this book does not compute. Module 10 printed Zhuravleva's bound and refused to state a value; this module does the same. Two caveats travel with that bound and both are the authors': the factor depends on the Prandtl number, which they leave free, and "it is not at present possible to prove that the observed fluctuations are associated with the turbulence in clusters as we do not have direct velocity measurements." §11 records that nobody in this book repays it, rather than pointing at a module that will not deliver.

9. The magnetorotational instability

Module 8 §6.3 names the magnetorotational instability in its list of what this module builds, so it is not optional. It is done locally: the shear is assumed, not derived from a disc model, because disc structure — the $\alpha$ prescription, the thin-disc solution — is Module 11.

The problem it solves is stated in one line. A Keplerian disc has angular velocity $\Omega\propto R^{-3/2}$, so its specific angular momentum $\ell = \Omega R^2 \propto R^{1/2}$ increases outward. By the Rayleigh criterion that is hydrodynamically stable: displace a ring outward and it has too little angular momentum to stay there, and it falls back. A purely hydrodynamic Keplerian disc does not transport angular momentum, and therefore does not accrete. Adding the weakest imaginable magnetic field destroys that stability.

Proposition 11 (the local MRI dispersion relation). Let $q \equiv -\mathrm{d}\ln\Omega/\mathrm{d}\ln R$, so $q = 3/2$ for a Keplerian disc, and let the epicyclic frequency be $\kappa_{\rm ep}^2 = 2(2-q)\Omega^2$. For an axisymmetric, incompressible perturbation with wavevector $\mathbf k\parallel\hat{\mathbf z}$ along a weak vertical field, writing $\sigma \equiv \gamma/\Omega$ and $x \equiv kv_{\rm A}/\Omega$, $$ \sigma^4 + \sigma^2\!\left(\frac{\kappa_{\rm ep}^2}{\Omega^2} + 2x^2\right) + x^2\!\left(x^2 - 4 + \frac{\kappa_{\rm ep}^2}{\Omega^2}\right) = 0. \tag{9.1}$$ There is a growing root whenever $x^2 \lt 2q$. At $q = 3/2$ this gives instability for $kv_{\rm A} \lt \sqrt3\,\Omega$, a maximum growth rate $$ \gamma_{\max} = \frac{q}{2}\,\Omega = 0.75\,\Omega \quad\text{at}\quad kv_{\rm A} = \frac{\sqrt{15}}{4}\,\Omega = 0.96825\,\Omega, \tag{9.2}$$ and marginal stability at $kv_{\rm A} = \sqrt3\,\Omega = 1.73205\,\Omega$.
Work in a local frame corotating at $R_0$ with coordinates $(x_1,x_2,x_3)$ along $(\hat{\mathbf R},\hat{\boldsymbol\phi},\hat{\mathbf z})$, and linearise the ideal MHD equations for displacements $\boldsymbol\xi\propto\exp(\gamma t + ikx_3)$. In the shearing sheet the equations of motion for a displacement are $\ddot\xi_1 = 2\Omega\dot\xi_2 + 2q\Omega^2\xi_1 + f_1$ and $\ddot\xi_2 = -2\Omega\dot\xi_1 + f_2$: the Coriolis coupling is $2\Omega$ in both, and the background shear appears once, as the tidal term $2q\Omega^2\xi_1$. A vertical field contributes a tension $f_i = -(kv_{\rm A})^2\xi_i$ to each in-plane component — the field lines act as springs joining the fluid element to its unperturbed radius. So $$ (\gamma^2 + k^2v_{\rm A}^2 - 2q\Omega^2)\,\xi_1 - 2\Omega\gamma\,\xi_2 = 0, \qquad 2\Omega\gamma\,\xi_1 + (\gamma^2 + k^2v_{\rm A}^2)\,\xi_2 = 0, $$ and the determinant condition is $$ (\gamma^2 + k^2v_{\rm A}^2 - 2q\Omega^2)(\gamma^2 + k^2v_{\rm A}^2) + 4\Omega^2\gamma^2 = 0, $$ which is (9.1) after dividing by $\Omega^4$ and using $\kappa_{\rm ep}^2/\Omega^2 = 2(2-q) = 4-2q$.

The field-free limit is the check that this pair is the right one. At $k = 0$ it reduces to $\sigma^2(\sigma^2 + 4 - 2q) = 0$, so $\sigma^2 = 2q-4 = -\kappa_{\rm ep}^2/\Omega^2$, which is $-1$ for a Keplerian disc: an oscillation at the epicyclic frequency and no growth. A Keplerian disc with no field is stable, as the opening of this section says, and any linearisation that does not reproduce that is wrong.

The instability condition. Equation (9.1) is a quadratic in $\sigma^2$ with positive coefficient on $\sigma^2$, so it has a positive root for $\sigma^2$ if and only if its constant term is negative, that is if and only if $x^2 - 4 + \kappa_{\rm ep}^2/\Omega^2 \lt 0$. Substituting $\kappa_{\rm ep}^2/\Omega^2 = 2(2-q)$ gives $x^2 \lt 2q$, so $x \lt \sqrt3$ at $q = 3/2$.

The maximum. Differentiate (9.1) implicitly with respect to $x^2$ and set $\mathrm{d}\sigma^2/\mathrm{d}x^2 = 0$: the surviving condition is $2\sigma^2 + 2x^2 - 4 + \kappa_{\rm ep}^2/\Omega^2 = 0$. With $q = 3/2$, $\kappa_{\rm ep}^2/\Omega^2 = 1$, this is $\sigma^2 = (3 - 2x^2)/2$. Substituting into (9.1) and solving gives $x^2 = 15/16$ and $\sigma^2 = 9/16$, hence $\sigma = 3/4$ and $x = \sqrt{15}/4$. ∎
Two things about this result that the algebra hides.
(i) There is no threshold field. The unstable band is $kv_{\rm A} \lt \sqrt3\,\Omega$, so a weaker field moves the band to larger $k$ — shorter wavelengths — without closing it. The instability disappears only when the longest unstable wavelength no longer fits inside the disc. That is why the condition that matters in practice is $\lambda_{\rm marg} \lt H$ and not a condition on $B$ alone.
(ii) The growth rate does not depend on the field at all — Fig. 5 is flat-topped at $0.75\,\Omega$ whatever the disc. $\gamma_{\max} = q\Omega/2$ contains no $B$, no $n$ and no $T$. A weaker field changes which wavelength grows fastest, not how fast it grows. In e-folds per orbit that is $2\pi(q/2) = 3\pi/2 = 4.712$ for every Keplerian disc — an identity, not a property of any particular one.
0.0 0.5 1.0 1.5 2.0 0.00 0.25 0.50 0.75 √15/4 = 0.96825, γ = 0.75Ω marginal, √3 = 1.73205 no growth beyond here λ = H at 1 au, where β = 104 λ = H at 10 au, where β = 1.73e+03 kvA/Ω γ/Ω The peak height is q/2 and the marginal wavenumber is √3 for EVERY Keplerian disc. A disc is stabilised when its dashed line passes to the RIGHT of √3.
The growing root of (9.1) at $q = 3/2$. The curve rises from zero, peaks at $\gamma = 0.75\,\Omega$ where $kv_{\rm A} = \sqrt{15}/4\,\Omega$, and reaches zero at $\sqrt3\,\Omega$, beyond which there is no growth at all — a hard upper edge in wavenumber, which is a hard lower edge in wavelength. The two dotted lines mark, for the worked discs below, the wavenumber whose wavelength is exactly one scale height; both lie inside the unstable band, so both discs have unstable modes that fit. A disc is stabilised when its dotted line passes to the right of $\sqrt3$.
quantity$R = 1$ au, $B = 1$ G, $n = 10^{14}$, $T = 300$ K$R = 10$ au, $B = 0.01$ G, $n = 10^{12}$, $T = 50$ K
$\Omega_{\rm K}$$1.9910\times10^{-7}$ rad s⁻¹ ($1.00$ yr)$6.2960\times10^{-9}$ rad s⁻¹ ($31.62$ yr)
$\gamma_{\max}$$1.4932\times10^{-7}$ s⁻¹$4.7220\times10^{-9}$ s⁻¹
e-folds per orbit$4.712 = 3\pi/2$$4.712 = 3\pi/2$
$v_{\rm A}$ / km s⁻¹$0.1434$$0.0143$
$c_s$ / km s⁻¹$1.3357$$0.5453$
$H = c_s/\Omega$$0.04485$ au $= 0.0448\,R$$0.57897$ au $= 0.0579\,R$
$\lambda$ at fastest growth$0.03125$ au $= 0.6967\,H$$0.09881$ au $= 0.1707\,H$
$\lambda$ at marginal stability$0.01747$ au $= 0.3895\,H$$0.05524$ au $= 0.0954\,H$
$\beta$$1.041\times10^{2}$$1.735\times10^{3}$
$\lambda_{\rm marg} \lt H$?YES, unstableYES, unstable
These two discs are a configuration and not an object, in the sense Module 7 §7.1 uses the phrase: nothing is checked against them, no observation of a real disc enters, and they exist to make the algebra of (9.2) numerical. The Keplerian $q = 3/2$ is assumed and not derived; deriving it needs a disc model, which is Module 11.

The result of Proposition 11 is Balbus & Hawley's (1991), and that paper was not fetched for this book. It is cited for the result only, with its volume, pages and year confirmed at Crossref — exactly as Alfvén (1942) is cited in §2. The derivation above is this module's own, so no unread paper supplies a number here.
And the condition this book cannot check is the ionisation. The MRI needs the gas coupled to the field, and a $300$ K protoplanetary disc at $n = 10^{14}\ \mathrm{cm^{-3}}$ is very nearly neutral — the same defect as §2's molecular cloud row, and refused there once for both. The $\lambda_{\rm marg} \lt H$ verdicts above are therefore statements about an ideal MHD disc, and whether either of these discs is ideal enough is a question about its ionisation fraction that this module does not compute. Real discs are believed to have "dead zones" for exactly this reason.

10. Anisotropy, critical balance, and Module 10's debt

Module 10 §7.2 confirmed Kolmogorov's $-5/3$ most cleanly on the magnetic spectrum of the solar wind, and said plainly what was uncomfortable about that: the confirmed exponent belongs to a magnetised plasma, which is not the fluid its Proposition 2 describes, and the same exponent follows from a quite different argument about magnetised turbulence. Its own sentence is the one to keep: "agreement of an exponent does not select between two derivations that give the same exponent."

Proposition 12 (critical balance gives the same exponent, and an anisotropy K41 does not have). Suppose an Alfvénic cascade keeps the linear Alfvén crossing time and the nonlinear turnover time equal at every scale: $$ k_\parallel v_{\rm A} \;\sim\; k_\perp\,\delta v_\perp. \tag{10.1}$$ If the perpendicular cascade is Kolmogorov, $\delta v_\perp \sim (\varepsilon/k_\perp)^{1/3}$, then $$ k_\parallel \;\sim\; k_\perp^{2/3}L^{-1/3}, \qquad E(k_\perp) \;\propto\; k_\perp^{-5/3}. \tag{10.2}$$
Substituting $\delta v_\perp \sim \varepsilon^{1/3}k_\perp^{-1/3}$ into (10.1) gives $k_\parallel \sim k_\perp^{2/3}\varepsilon^{1/3}/v_{\rm A}$; writing $\varepsilon \sim v_{\rm A}^3/L$ at the outer scale, where the balance is imposed, gives $\varepsilon^{1/3}/v_{\rm A} = L^{-1/3}$ and hence the first of (10.2). For the second, the perpendicular energy spectrum follows from $\delta v_\perp^2 \sim k_\perp E(k_\perp)$ with $\delta v_\perp \propto k_\perp^{-1/3}$, so $E \propto k_\perp^{-5/3}$. ∎
The exponent is not the discriminant. The anisotropy is, and this book does not measure it. K41 and critical balance predict the same $-5/3$ from different premises, so a measured $-5/3$ selects neither. What distinguishes them is the first half of (10.2): K41 predicts isotropic eddies, and critical balance predicts eddies that grow more elongated along the field at every smaller scale, with $k_\parallel \propto k_\perp^{2/3}$. No anisotropy measurement was read for this book, and this module therefore declares the discriminating measurement outside it. Module 10's four indices — $1.66$, $1.72$, $1.66$, $1.58$, mean $1.6550$ against $5/3$, ratio $0.9930$ — are reprinted here so that the two modules quote one set, and no new check is built on them.
Goldreich & Sridhar's own restriction travels in the same sentence as their result, every time this module states it. The third and fourth sentences of their abstract read: "Here, as in Paper I, we restrict attention to the symmetric case where the oppositely directed waves carry equal energy fluxes. This precludes application to the solar wind in which the outward flux significantly exceeds the ingoing one." Proposition 12 above is therefore a statement about a symmetric Alfvénic cascade, and it may not be presented as a prediction for the solar-wind indices reprinted beside it — which are precisely the measurement its authors excluded.

This matters for a shipped page. Module 10 §7.2 and its captures-and-misses table both attribute the $5/3$ of the solar wind's magnetic spectrum to Goldreich & Sridhar as an alternative prediction. No number depends on it — Module 10's sources list already records that they are "not used numerically anywhere" — so it is a prose attribution and not a computed claim, and the correction belongs to Module 10's author and not to this module. Module 12 carries the caveat whether or not that edit is made.

Note also that the bare exponent $2/3$ is not dimensional. Their relation is $k_\parallel \approx k_\perp^{2/3}L^{-1/3}$, and the outer scale $L$ is what makes the two sides agree in units. A module that printed $k_\parallel \sim k_\perp^{2/3}$ alone would be printing a relation between quantities of different dimensions.

11. What this theory captures, and what it misses

Every row below is an idealisation this module makes, what it costs, and where — if anywhere — it is repaid. Where nobody repays it, the row says so, rather than pointing at a module that will not deliver.

idealisation made herewhat it costswhere it is repaid
One fluid, one temperatureIon and electron temperatures equilibrate in the intracluster medium on $t_{\rm eq}(p,e)\approx6\times10^{8}$ yr (Module 2 §11), which is not short compared with anything about a clusterNobody. Single-fluid MHD has one temperature by construction, and this book contains no two-fluid module. Module 2 §11's and Module 9 §9's debts are declared unpaid here
Isotropic pressure — no CGL, no firehose, no mirrorA magnetised collisionless plasma has $p_\parallel$ and $p_\perp$, and the anisotropy drives the instabilities that are, in Module 1 §8's words, "the plasma's substitute for collisions". They are also the microinstabilities that §8 could not computeReturned to, not derived. Module 1 §8 promises that "Module 12 develops magnetohydrodynamics and returns to this point", and that is exactly what §11.1 does. (Gate D placed this return in §10; it was moved here on drafting, because a debt about what the module leaves out belongs beside the table that lists what it leaves out.) No threshold is printed, because no CGL source was read for this book
Ideal Ohm's law — no ambipolar diffusion, no Hall termThe $10$ K cloud of §2 and the $300$ K disc of §9 are both nearly neutral, so the Spitzer resistivity is outside its domain in both; and §7's criterion is an equilibrium with no mechanism for leaving it, when ambipolar leakage is exactly that mechanismNobody in this book. Named once in §2 and cross-referenced from §7 and §9
Classical (Braginskii) perpendicular transportIt is wrong by $22.8$ to $24.8$ decades against the only measured bound. The field is tangled and the step is neither $\lambda$ nor $r_g$Nobody. What the effective transport is, this book does not compute. This row does not point forward, because there is no module that would deliver it
A steady, spherically symmetric, single-latitude wind in §§5–6$\Omega$ and the source-surface radius are held fixed, and the real Alfvén surface is corrugated — which is why Kasper's three crossings in two days span $16.0$–$19.8\,R_\odot$Not repaid; priced in §5.2's table and named in §6.1's verdict
The anisotropy of the cascade is not measured$-5/3$ does not separate K41 from critical balance; the discriminant is $k_\parallel\propto k_\perp^{2/3}L^{-1/3}$Nobody in this book. §10 says the discriminating measurement is outside it
Non-relativistic flow, in the ordering that drops the displacement current$(U/c)^2 = 2.1\times10^{-6}$ for the fastest flow here, and $O(1)$ for a pulsar magnetosphere or a relativistic jetNot repaid; named in §2, and relativistic MHD is not in this book
Oblique MHD shock jump conditions are not derivedModule 8 §9 calls its refusal of a Voyager comparison "permanent rather than provisional" on the ground that Module 12 does not derive themDeliberately absent. Deriving them here would make a shipped module false
Disc structure, the $\alpha$ prescription, the thin-disc solution§9 assumes $q = 3/2$ and derives nothing about the disc that produces itModule 11 (accretion discs)
Magnetic reconnectionThe ohmic term of (2.1) is negligible everywhere in §2's table, yet solar flares change the field's topology in minutes. Reconnection is how a plasma with $\mathrm{R_m} = 10^{12}$ nevertheless does that, by making $L$ locally tinyNobody. No shipped module promises it, and this row names it as absent rather than as deferred
No terrestrial measurement in this moduleAll four checks are astronomical; the one reproduction is of another chapter of this bookNot repaid; named in §4

11.1 The point Module 1 asked this module to return to

Module 1 §8 observed that the magnetic rescue of the fluid description is partial, and partial in a measurable way. A field constrains motion across itself and does nothing to motion along itself, so a magnetised collisionless plasma is not an ordinary fluid with a scalar pressure: it has $p_\parallel$ along the field and $p_\perp$ across it, and the two relax toward each other only through collisions the plasma is not having. The resulting description is the Chew–Goldberger–Low or double-adiabatic theory. Pressure anisotropy is routinely measured in the solar wind, and it drives the firehose and mirror instabilities, which scatter particles and so act as the plasma's substitute for collisions.

That is the return, and it is all of it. Single-fluid MHD as built in §2 and §3 assumes a scalar pressure and therefore cannot contain any of it. No instability threshold is printed here, because no CGL source was read for this book, and a threshold quoted from memory is the failure this project keeps finding in its own drafts. What can be said without a source is the structural point: the same anisotropy that Module 1 named is the leading candidate for the microinstabilities that §8 needed and could not compute, so two of this book's unpaid debts are probably one debt.

12. Problems

Every modelling input a solution uses is stated in its own problem. Where an input is a census value with no source, or an assumption rather than a measurement, the statement says so — because a problem that hides its assumptions teaches the wrong habit. All nine are rebuilt from their statements alone by afd/figs/m12_problems_check.py, which imports nothing from the numbers script.

C1. An active-region corona has $n = 10^{9}\ \mathrm{cm^{-3}}$, $T = 10^{6}$ K and $B = 10$ G — the field taken from Module 7 §9.5, which states that the number is illustrative and not a measurement. Take $\mu = 0.61$ and $\gamma = 5/3$. Find $p_{\rm gas}$, $p_{\rm mag}$, $\beta$, $v_{\rm A}$ and $c_s$, and then obtain $v_{\rm A}/c_s$ twice: once from the two speeds and once from (1.2). Why must the two agree exactly rather than approximately?
Solution

$\rho = n\mu m_{\rm u} = 10^{9}\times0.61\times1.66054\times10^{-24} = 1.0129\times10^{-15}\ \mathrm{g\,cm^{-3}}$.

$p_{\rm gas} = nk_{\rm B}T = 10^{9}\times1.380649\times10^{-16}\times10^{6} = 1.3806\times10^{-1}\ \mathrm{dyn\,cm^{-2}}$, and $p_{\rm mag} = B^2/8\pi = 100/(8\pi) = 3.9789\ \mathrm{dyn\,cm^{-2}}$. Hence $\beta = 0.1381/3.979 = \mathbf{0.0347}$: the field carries $28.8$ times the gas pressure.

$v_{\rm A} = B/\sqrt{4\pi\rho} = 10/\sqrt{4\pi\times1.0129\times10^{-15}} = 8.864\times10^{7}\ \mathrm{cm\,s^{-1}} = \mathbf{886.4\ km\,s^{-1}}$. $c_s = \sqrt{\gamma k_{\rm B}T/\mu m_{\rm u}} = \sqrt{(5/3)(1.380649\times10^{-16})(10^{6})/(1.0129\times10^{-24})} = 1.507\times10^{7}\ \mathrm{cm\,s^{-1}} = \mathbf{150.7\ km\,s^{-1}}$.

The ratio of the two speeds is $8.8635\times10^{7}/1.5072\times10^{7} = \mathbf{5.8807}$ — take them unrounded, because $886.4/150.7$ as printed gives $5.8826$ and the point of the next line is an exact agreement. From (1.2), $\sqrt{2/(\gamma\beta)} = \sqrt{2/((5/3)(0.034699))} = \mathbf{5.8807}$. The two agree to six figures.

They must agree exactly because (1.2) is an identity, not an approximation. The proof of Proposition 1 cancels $\rho$ between the two speeds, so the same $\mu$ enters both and leaves. If the two columns ever disagreed it would not mean the physics had changed; it would mean a $\mu$ had been altered in one place and not the other. That is why the table in §1 prints both columns and their ratio.

C2. A coronal loop $100$ Mm long ($L = 10^{10}$ cm) has $n = 10^{9}\ \mathrm{cm^{-3}}$, $T = 10^{6}$ K and a flow along it of $U = 10\ \mathrm{km\,s^{-1}}$ — a loop flow speed, not a wind speed. Using Braginskii's $\sigma_\parallel = 1.96\,n e^2\tau_{\rm e}/m_{\rm e}$ with his $\tau_{\rm e}$ of equation (2.5e) and $\ln\Lambda = 18.66$, find $\eta_{\rm m}$, $\mathrm{R_m}$, the ohmic decay time $L^2/\eta_{\rm m}$, and the dynamical time $L/U$. Then compute the ratio of the last two and say why you did not need to.
Solution

$\eta_{\rm m} = c^2/4\pi\sigma_\parallel = \mathbf{9.768\times10^{3}\ cm^2\,s^{-1}}$. For comparison, that is about sixty-five thousand times the kinematic viscosity of air; a coronal plasma is a poor conductor by laboratory standards and an extraordinary one by astrophysical ones, and the next line is why.

$\mathrm{R_m} = UL/\eta_{\rm m} = (10^{6})(10^{10})/9.768\times10^{3} = \mathbf{1.024\times10^{12}}$.

$L^2/\eta_{\rm m} = 10^{20}/9.768\times10^{3} = 1.024\times10^{16}\ \mathrm{s} = \mathbf{3.244\times10^{8}\ yr}$, and $L/U = 10^{10}/10^{6} = 10^{4}\ \mathrm{s} = \mathbf{3.169\times10^{-4}\ yr}$.

Their ratio is $1.024\times10^{12}$ — the magnetic Reynolds number again, to six figures. You did not need to compute it, because (2.4) says $(L^2/\eta_{\rm m})/(L/U) = UL/\eta_{\rm m}$ identically. The point of doing it anyway is that the identity turns the ohmic-time column into a check on the $\mathrm{R_m}$ column beside it: if one is ever edited alone, the ratio stops closing.

And the comparison is with the loop's own dynamical time, deliberately. Comparing $3.244\times10^{8}$ yr with the age of the universe would be meaningless — the loop does not live that long. Comparing it with the $10^{4}$ s the loop takes to do anything at all is the comparison that says the field is frozen in.

C3. Take $\Omega$ from the sidereal rotation period $25.38$ d — the NSSDC value, adopted at $16^\circ$ latitude — a source surface $r_0 = 2.5\,R_\odot$, $\theta = 90^\circ$, and a wind speed held constant at $v = 435.6\ \mathrm{km\,s^{-1}}$ — Venzmer & Bothmer's mean-fit 1-au value, the same one Problem D2 uses — which the Helios fits do not support and which is assumed here only to keep the arithmetic in one line. Using (5.1), find $-B_\phi/B_r$ and the spiral angle at $1$ au and at $5$ au, and the local index (5.2) at each. Then say what dropping $r_0$ would do at $1$ au.
Solution

$\Omega = 2\pi/(25.38\times86400) = \mathbf{2.86533\times10^{-6}\ rad\,s^{-1}}$ and $r_0 = 2.5\times6.957\times10^{10} = 1.739\times10^{11}$ cm $= 0.011626$ au.

At 1 au: $-B_\phi/B_r = \Omega(r-r_0)/v = (2.86533\times10^{-6})(1.4960\times10^{13} - 1.739\times10^{11})/(4.356\times10^{7}) = \mathbf{0.9726}$. The spiral angle is $\arctan(0.9726) = \mathbf{44.20^\circ}$, and (5.2) with $\mathrm{d}\ln v/\mathrm{d}\ln r = 0$ gives $-2 + (0.9726^2/1.9459)(1.0118) = \mathbf{-1.5082}$.

At 5 au: $-B_\phi/B_r = \mathbf{4.9088}$, angle $\mathbf{78.49^\circ}$, index $\mathbf{-1.0376}$.

The spiral tightens outward, and the index climbs from $-1.51$ toward the asymptotic $-1$: by 5 au the field is almost entirely azimuthal, and the total field is falling off nearly as $1/r$ rather than as $1/r^2$.

Dropping $r_0$ replaces $(r-r_0)$ by $r$ and raises $-B_\phi/B_r$ at 1 au from $0.9726$ to $0.9840$, by $\mathbf{1.18}$ per cent; equivalently, carrying $r_0$ lowers it by $1.16$ per cent. The two percentages are the same term seen from its two ends, and $r_0/1\ \mathrm{au} = 0.011626$ is why they are nearly equal. At 5 au the effect is five times smaller still, which is exactly why Parker states $\Omega r\sin\theta/v$ as a large-$r$ limit.

D1. A spherical cloud $10$ pc across has $n_{\rm tot} = 100\ \mathrm{cm^{-3}}$ per particle at $\mu = 2.33$, so that $n({\rm H_2}) = 83.3\ \mathrm{cm^{-3}}$; take $R = 5$ pc and $B = 10\ \mu$G, an order-of-magnitude census value for which this book has no source at this density. Find $M$, and then $M_\Phi$ and $\lambda_\Phi$ for both geometries of (7.2). Which of the two answers should be quoted, and why is that the wrong question?
Solution

$\rho = n\mu m_{\rm u} = 100\times2.33\times1.66054\times10^{-24} = 3.8691\times10^{-22}\ \mathrm{g\,cm^{-3}}$, and with $R = 5\ \mathrm{pc} = 1.5428\times10^{19}$ cm, $M = \tfrac43\pi R^3\rho = 5.952\times10^{36}\ \mathrm{g} = \mathbf{2.993\times10^{3}\,M_\odot}$.

The flux is $\Phi = \pi R^2B = \pi(1.5428\times10^{19})^2(10^{-5}) = 7.478\times10^{33}$ Mx, and $M_\Phi = (c_\Phi/\sqrt G)\Phi$.

Sphere, $c_\Phi = 0.53\sqrt5/3\pi = 0.125745$: $(M/\Phi)_{\rm crit} = \mathbf{4.8673\times10^{2}}$ g Mx⁻¹, $M_\Phi = \mathbf{1.831\times10^{3}\,M_\odot}$, $\lambda_\Phi = 2993/1831 = \mathbf{1.635}$.

Sheet, $c_\Phi = 1/2\pi = 0.159155$: $(M/\Phi)_{\rm crit} = \mathbf{6.1605\times10^{2}}$ g Mx⁻¹, $M_\Phi = \mathbf{2.317\times10^{3}\,M_\odot}$, $\lambda_\Phi = \mathbf{1.292}$.

Neither should be quoted alone, and that is the answer. Both are supercritical, so the conclusion — the field cannot hold this cloud — is the same either way and is robust. But the two numbers differ by the coefficient ratio $1.2657$, and Troland & Crutcher's entire measured range for real cores is $1.4$ to $2.6$. The geometric factor alone spans $\mathbf{0.381}$ of the log width of that range. A $\lambda_\Phi$ quoted without its assumed geometry is not a measurement, and quoting $1.635$ as though it were one would be claiming a precision the modelling does not have.

Note also what has been assumed about the flux: $\Phi = \pi R^2B$ takes a uniform field threading the cloud's mid-plane. A tangled field of the same $|B|$ threads less net flux and so makes the cloud more supercritical still.

D2. At $1$ au take $n = 7.57\ \mathrm{cm^{-3}}$, $|\mathbf B| = 6.05$ nT and $v = 435.6\ \mathrm{km\,s^{-1}}$ — Venzmer & Bothmer's mean fits — with protons only, $-B_\phi/B_r = 0.9726$ from Problem C3, $v$ held constant, $B_r\propto r^{-2}$ and $\rho\propto r^{-2}v^{-1}$. Find $B_r$ at 1 au, the radial Alfvén speed there, and $r_{\rm A}$. Then compute what you would have got using $|\mathbf B|$, and say why that answer is wrong.
Solution

$\sqrt{1+x_1^2} = \sqrt{1+0.9726^2} = \mathbf{1.3950}$, so $B_r = 6.05/1.3950 = \mathbf{4.3370}$ nT $= 4.337\times10^{-5}$ G.

$\rho = n m_{\rm p} = 7.57\times1.6726\times10^{-24} = 1.2662\times10^{-23}\ \mathrm{g\,cm^{-3}}$, so $v_{\rm A,r} = B_r/\sqrt{4\pi\rho} = \mathbf{34.38\ km\,s^{-1}}$ and $v/v_{\rm A,r} = 435.6/34.38 = \mathbf{12.669}$.

At constant $v$, (6.1) gives $v_{\rm A,r}\propto r^{-1}$, so the crossing is at $r_{\rm A} = (1\ \mathrm{au})\,v_{\rm A,r}(1\ \mathrm{au})/v = 1/12.669$ au $= 0.07893$ au $= \mathbf{16.97\,R_\odot}$.

With $|\mathbf B|$ in place of $B_r$ the Alfvén speed at 1 au would be $1.3950$ times larger and so would $r_{\rm A}$, giving $\mathbf{23.68\,R_\odot}$ on these inputs.

That answer is wrong twice over. First, the extrapolation inward uses $B\propto r^{-2}$, and §5 proved that $|\mathbf B|$ does not scale that way — $B_r$ does, by $\nabla\cdot\mathbf B = 0$. Using $|\mathbf B|$ at 1 au and then propagating it as $r^{-2}$ mixes two different quantities. Second, the surface being compared against is the Weber–Davis one, which Verscharen, Bale & Velli define on the radial Alfvén speed because it is the radial components that enter the flow deflection they fit. Either reason alone settles it.

A caution about which number this is. $16.97\,R_\odot$ is one of eight variants; §6.1 uses the whole band $15.76$–$19.35\,R_\odot$, precisely so that the verdict cannot depend on which row is called the headline.

D3. An intracluster medium has $n = 10^{-3}\ \mathrm{cm^{-3}}$, $T = 10^{8}$ K and $B = 1\ \mu$G. Using Braginskii's $\tau_{\rm e}$ with its $4\sqrt{2\pi}$, the derived electron-branch $\ln\Lambda = 37.08$ (Module 1's census value is $37.8$), and his $\kappa_\parallel = 3.16$, $\kappa_\perp = 4.66$ for $Z = 1$, find $\omega_{\rm e}\tau_{\rm e}$ and $\kappa_\perp/\kappa_\parallel$. Before quoting the suppression as the answer, check it against Zhuravleva et al.'s measured $10$–$1000$.
Solution

$\tau_{\rm e} = \mathbf{7.4227\times10^{12}}$ s and $\omega_{\rm e} = eB/m_{\rm e}c = \mathbf{17.588\ rad\,s^{-1}}$, so $\omega_{\rm e}\tau_{\rm e} = \mathbf{1.3055\times10^{14}}$.

By (8.2), $\kappa_\perp/\kappa_\parallel = (4.66/3.16)(\omega_{\rm e}\tau_{\rm e})^{-2} = 1.4747/(1.7043\times10^{28}) = \mathbf{8.6523\times10^{-29}}$.

That answer must be checked before it is quoted, and it fails the check. A perpendicular conductivity $8.65\times10^{-29}$ of the parallel value would leave a cluster core with essentially no conduction across the field at all — and cluster cores are not observed to be thermally isolated across the field. The classical coefficient does not describe a real cluster.

And the comparison must be made on the right quantity. Zhuravleva et al. measure a viscosity, so the number to compare is $\nu_\perp/\nu_\parallel$. The ion magnetisation is $\omega_i\tau_i = 4.3087\times10^{12}$, giving $\nu_\perp/\nu_\parallel = (0.3/0.96)(\omega_i\tau_i)^{-2} = \mathbf{1.6833\times10^{-26}}$, a classical suppression factor of $5.941\times10^{25}$. Against their measured $10$ to $1000$, the classical value is too small by between $\mathbf{22.8}$ and $\mathbf{24.8}$ decades. Comparing $\kappa_\perp/\kappa_\parallel$ with a viscosity bound would be two different quantities in one sentence, and this problem exists to stop that.

Why the two $\ln\Lambda$ values. $37.08$ is the electron-branch logarithm derived from this $n$ and $T$; $37.8$ is Module 1's census value, reused by Module 10. The $1.9$ per cent between them is one of the two factors in (8.3), and it is not a correction to either module.

K1. A protoplanetary disc at $R = 1$ au around $1\,M_\odot$ has $B = 1$ G, $n = 10^{14}\ \mathrm{cm^{-3}}$, $T = 300$ K and $\mu = 2.33$. Take $q = -\mathrm{d}\ln\Omega/\mathrm{d}\ln R = 3/2$, assumed and not derived. Find $\Omega_{\rm K}$, $\gamma_{\max}$, the number of e-folds per orbit, $v_{\rm A}$, $c_s$, $H = c_s/\Omega$, and the marginal wavelength $\lambda_{\rm marg} = 2\pi v_{\rm A}/(\sqrt3\,\Omega)$. Is the disc unstable? Then explain why your e-folds-per-orbit answer contains no $B$, no $n$ and no $T$.
Solution

$\Omega_{\rm K} = \sqrt{GM/R^3} = \mathbf{1.9910\times10^{-7}\ rad\,s^{-1}}$, a period of $\mathbf{1.00}$ yr, as it must be at 1 au around a solar mass.

$\gamma_{\max} = (q/2)\Omega = 0.75\times1.9910\times10^{-7} = \mathbf{1.4932\times10^{-7}\ s^{-1}}$, so the e-folding time is $6.70\times10^{6}$ s $= 77.5$ days.

E-folds per orbit $= 2\pi\gamma_{\max}/\Omega = 2\pi\times0.75 = \mathbf{4.712}$.

$\rho = 10^{14}\times2.33\times1.66054\times10^{-24} = 3.869\times10^{-10}\ \mathrm{g\,cm^{-3}}$, so $v_{\rm A} = \mathbf{0.1434\ km\,s^{-1}}$ and $c_s = \mathbf{1.3357\ km\,s^{-1}}$; $\beta = 104.1$. Then $H = c_s/\Omega = 6.709\times10^{11}\ \mathrm{cm} = \mathbf{0.04485}$ au, and $\lambda_{\rm marg} = 2\pi v_{\rm A}/(\sqrt3\Omega) = 2.613\times10^{11}\ \mathrm{cm} = \mathbf{0.01747}$ au $= \mathbf{0.3895\,H}$.

Yes, unstable. $\lambda_{\rm marg} \lt H$: the longest unstable wavelength fits inside the disc with room to spare, so the MRI operates.

Why $4.712$ contains none of the inputs. (9.2) gives $\gamma_{\max} = q\Omega/2$ with no dependence on $v_{\rm A}$ whatever, so e-folds per orbit is $2\pi(q/2) = 3\pi/2$ for every Keplerian disc. It is an identity about the shear, not a property of this disc. What the field does control is which wavelength grows fastest, and therefore whether any unstable wavelength fits — which is the calculation of $\lambda_{\rm marg}/H$ above, and the only place $B$, $n$ and $T$ enter the verdict.

What this does not establish. The MRI needs the gas coupled to the field, and a $300$ K disc at $n = 10^{14}\ \mathrm{cm^{-3}}$ is very nearly neutral. The verdict above is a statement about an ideal MHD disc; whether this disc is ideal enough is a question about its ionisation fraction that this module does not compute.

K2. Module 7 §9.1 gives $\rho_h = 5.8607\times10^{-16}$ and $\rho_l = 2.6210\times10^{-16}\ \mathrm{g\,cm^{-3}}$ for the two sides of a coronal-mass-ejection flank, and calls $\Delta U = 20\ \mathrm{km\,s^{-1}}$ the measured shear — an upper limit. Using (4.2), find the bound on $B\cos\theta$ for equal fields and for a one-sided field, and then the permitted angle at $B = 1$ G and $B = 10$ G. The angle is measured from perpendicular to $\mathbf B$; which inverse trigonometric function do you need, and what does the other one give?
Solution

The prefactor is $\sqrt{2\pi\rho_h\rho_l/(\rho_h+\rho_l)} = \sqrt{2\pi(5.8607)(2.6210)\times10^{-32}/(8.4820\times10^{-16})} = 3.3734\times10^{-8}$ in CGS. With $\Delta U = 2\times10^{6}\ \mathrm{cm\,s^{-1}}$,

$B\cos\theta \lt \mathbf{0.06747}$ G for equal fields, and $\sqrt2$ times that, $\mathbf{0.09541}$ G, for a one-sided field. Module 7 prints $0.0675$ and $0.0954$; the ratios are $0.9995$ and $1.0001$.

At $B = 1$ G, $\cos\theta \lt 0.06747$, so $\theta$ must lie within $\arcsin(0.06747) = \mathbf{3.87^\circ}$ of perpendicular. At $B = 10$ G, $\cos\theta \lt 0.006747$ and the angle is $\mathbf{0.39^\circ}$. Both match Module 7 §9.5's table exactly.

You need $\arcsin$. The bound is on $\cos\theta$, where $\theta$ is measured from the field; the departure from perpendicular is $90^\circ - \theta$, whose sine is $\cos\theta$. Using $\arccos$ instead gives $\arccos(0.06747) = \mathbf{86.13^\circ}$ — the complement, and a ninety-degree error. The two densities are Module 7's own printed values and are quoted, not recomputed: they follow there from $n_e = 3\times10^{8}\ \mathrm{cm^{-3}}$ carried from Module 1's census and $\mu_e = 2/(1+X) = 1.1765$ at $X = 0.70$, and the paper itself gives only their ratio, $\sqrt5$.

It is worth noticing that $86.13^\circ$ looks perfectly plausible on its own: it is a legal angle, it varies sensibly with $B$, and nothing about it announces that it is wrong. The only thing that catches it is a second source printing the same table, which is why Module 7 printing its angle column was worth more than it looked.

K3. For the intracluster medium of Problem D3, compute the magnetisation two ways: as $\lambda/r_g$, with $\lambda$ Module 1's Coulomb mean free path at $\ln\Lambda = 37.8$ and $r_g = v_\perp/\omega_{\rm c}$ with $v_\perp = \sqrt{2k_{\rm B}T/m_{\rm p}}$; and as $\omega_i\tau_i$ with Braginskii's $\tau_i$ at the derived $\ln\Lambda = 37.081$. Account for the ratio exactly, and say which of the two belongs in $(\omega_{\rm c}\tau)^{-2}$.
Solution

$\lambda = \mathbf{22.504}$ kpc $= 6.9440\times10^{22}$ cm — the Spitzer mean free path at this $n$, $T$ and $\ln\Lambda = 37.8$, which Module 1 prints rounded to $22.5$ kpc. Carry the unrounded value: $22.5$ kpc exactly would give $5.1759\times10^{12}$ below, and this module prints five figures. $v_\perp = \sqrt{2k_{\rm B}T/m_{\rm p}} = \mathbf{1285\ km\,s^{-1}}$ and $\omega_{\rm c} = eB/m_{\rm p}c = 9.579\times10^{-3}\ \mathrm{rad\,s^{-1}}$, so $r_g = v_\perp/\omega_{\rm c} = 1.3414\times10^{10}\ \mathrm{cm} = \mathbf{1.3414\times10^{5}}$ km. Hence $\lambda/r_g = \mathbf{5.1767\times10^{12}}$.

Braginskii's $\tau_i$ at $\ln\Lambda = 37.081$ is $4.4981\times10^{14}$ s, so $\omega_i\tau_i = \mathbf{4.3087\times10^{12}}$.

The ratio is $5.1767/4.3087 = \mathbf{1.2015}$, and it is two factors and nothing else:

$\sqrt{3/2} = \mathbf{1.224745}$, times $\ln\Lambda_{\rm e}/\ln\Lambda = 37.081/37.8 = \mathbf{0.980985}$, giving $1.224745\times0.980985 = \mathbf{1.2015}$.

The $\sqrt{3/2}$ is a component count and is exact. A Spitzer mean free path is $v_{\rm rms}\tau$ with the three-dimensional rms speed $\sqrt{3k_{\rm B}T/m}$; the first calculation divided it by $\sqrt{2k_{\rm B}T/m}$, the rms of the two components perpendicular to $\mathbf B$, because that is the speed a gyroradius takes. Dividing by a speed $\sqrt{2/3}$ times too small makes the answer $\sqrt{3/2}$ times too large. The second factor is not a convention at all but two different evaluations of the Coulomb logarithm, differing by $1.9$ per cent.

Braginskii's belongs in $(\omega_{\rm c}\tau)^{-2}$, because his transport coefficients $1.96$, $3.16$, $4.66$, $3.9$ and $2$ are defined on his $\tau$. Using the other one there would suppress the transport by a further factor $1.2015^2 = 1.44$ for no physical reason. The first number is still worth having, but as a count: a proton turns $10^{12.71}$ gyro-orbits between collisions.

Notation

Every symbol used in this module, with its units and where it is defined. Every glyph this module gives a meaning that an earlier module gave a different meaning is resolved here — $\eta$, $\lambda$, $\beta$, $\sigma$, $\kappa$, $\Omega$ and $\mu$ — rather than left to the reader, and one cross-module constant differs from an earlier module's by a stated amount.

symbolmeaningunitsdefined
$\eta_{\rm m}$the magnetic diffusivity $c^2/4\pi\sigma_{\rm e}$, always subscripted. Bare $\eta$ is the dynamic viscosity book-wide from Module 6 on, by the notation notes in Module 1 and Module 2 — Module 6 itself never writes the glyph, and the quantity is Module 1's Proposition 4, eq. (6.2), where it is $\mu$, the Kolmogorov dissipation scale in Module 10, an interface displacement and a density ratio in Module 7, and a density-profile exponent in Module 5; Module 9 wrote $\eta_{\rm rad}$ to keep clear of it. A fourth bare meaning would carry no meaning at all, so this module never uses the bare glyph. Braginskii's dynamic viscosities appear only as the ratio $\nu_\perp/\nu_\parallel$, in this module's §8cm² s⁻¹Proposition 2, (2.1)
$\lambda$, $\lambda_\Phi$, $\lambda_{\rm MRI}$, $\lambda_{\rm marg}$Bare $\lambda$ is Module 1's mean free path. It is used here in §1 and §8. $\lambda_\Phi$ is the mass-to-flux ratio of Definition 3, Troland & Crutcher's own symbol with a subscript, and occurs only in this module's §7. $\lambda_{\rm MRI}$ and $\lambda_{\rm marg}$ are the fastest-growing and marginal wavelengths of §9 and occur only there. Braginskii writes the Coulomb logarithm $\lambda$; this book writes $\ln\Lambda$, as Modules 1 and 10 do, and his glyph is named once and never used. No section uses two of the fourcm; —; cmModule 1; Definition 3; (9.2)
$\beta$the plasma beta $8\pi nk_{\rm B}T/B^2$. Module 3's $\beta$ is the gas-pressure fraction $P_{\rm gas}/P$ whose radiation share is $1-\beta$; Module 2's and Module 9's is a fitted velocity exponent. Venzmer & Bothmer's fitted exponents are written $\alpha_n$ and $\alpha_v$ throughout this module, so no bare $\beta$ is ever a fit exponent here, and Module 3's does not appear—Definition 1, (1.1)
$\sigma_{\rm e}$the electrical conductivity, always subscripted, used only inside the proof of Proposition 2 and in (2.3). Bare $\sigma$ is Module 1's collision cross-section book-wide from Module 6, and $\sigma_{ij}$ is Module 2's viscous stress; neither appears here. Nakano & Nakamura's $\sigma_0$ is a column density and is quoted, inside their own sentence, only in this module's §7s⁻¹Proposition 2
$\kappa_\parallel$, $\kappa_\perp$, $\kappa_{\rm ep}$the thermal conductivities along and across the field, always with $\parallel$ or $\perp$. Module 4's $\kappa$ is an evanescent decay rate; different module. The epicyclic frequency, which the disc literature writes $\kappa$, is written $\kappa_{\rm ep}$ here and appears only inside the proof of Proposition 11 — the statement of that proposition and every number from it use $q$ and $\Omega$ alone. The table says so because a reader who knows that literature will look for iterg cm⁻¹ s⁻¹ K⁻¹; s⁻¹(8.2); Proposition 11
$\Omega_\odot$, $\Omega_{\rm K}$angular velocities, both subscripted at first use: the solar rotation rate in §5 and the Keplerian rate in §9. Bare $\Omega$ is used only inside §9's algebra, where no solar rotation occurs. Module 3's $\Omega$ is a gravitational potential energy and does not appearrad s⁻¹(5.1); (9.2)
$\mu$mean mass per particle measured against $m_{\rm u}$, this book's convention from Module 3 onward: $2.33$ for the cloud, $0.61$ for ionised gas, $1.30$ for the photosphere. Troland & Crutcher's $\mu = 2.8$ is per H₂ molecule and against the hydrogen mass, and the two are the same mass density counted differently: $\rho/(n_{\rm H_2}m_{\rm H}) = 1.4/0.5 = 2.8$ against $\rho/(n_{\rm tot}m_{\rm H}) = 1.4/0.6 = 2.333$. So $n_{\rm tot} = 100\ \mathrm{cm^{-3}}$ is $n({\rm H_2}) = 83.3\ \mathrm{cm^{-3}}$, and the mass unit in (7.3) is fixed to a hydrogen mass and no finer, because their coefficient carries two significant figures—§1; §7.1
Seven further rows, which are not collisions but are declared because they are new to this book or carry a convention.
$v_{\rm A}$the Alfvén speed $B/\sqrt{4\pi\rho}$. New to this book. Declared at Definition 1 as a combination with the dimensions of a speed and shown to be a wave speed in the proof of Proposition 6cm s⁻¹Definition 1
$\Phi$, $M_\Phi$magnetic flux and the magnetic critical mass. No gravitational potential appears in this module; only $GM/R$ inside the virial argumentMx; gDefinition 3, (7.1)
$\mathrm{R_m}$the magnetic Reynolds number $UL/\eta_{\rm m}$. New to this book, and declared beside $\mathrm{Kn}$ (Module 1), $\mathrm{Re}$ (Module 10) and $\mathrm{Ma_{th}}$ (Module 2), none of which appears here—Proposition 3, (2.2)
$q$the shear parameter $-\mathrm{d}\ln\Omega/\mathrm{d}\ln R$, taken as $3/2$ and assumed, not derived. Module-local to §9; it is not a heat flux anywhere in this module—Proposition 11
$r_0$the source-surface radius, $2.5\,R_\odot$. Parker writes $b$, and this module does not, because $b$ is Module 1's impact parameter and Module 10's forcing parametercm(5.1)
$\alpha_n$, $\alpha_v$Venzmer & Bothmer's fitted density and velocity exponents in $x(r) = d\,r^{e}$. The same meaning as Module 9 §7 and Module 2 §10, deliberately, because all three modules read the same table. Module 5's $\alpha$ is a length scale, Module 6's a mixing-length parameter, Module 10's $P_{\rm turb}/P_{\rm th}$, and Module 11's will be the disc viscosity parameter; none appears here—§5.1
$\tau_{\rm e}$, $\tau_i$Braginskii's collision times, his (2.5e) and (2.5i), both for $Z = 1$. $\tau_i/\tau_{\rm e} = \sqrt{2m_{\rm p}/m_{\rm e}} = 60.5995$ is printed against the computed value, so the two cannot drift apart. His denominators are $Z^2n_i$ and $Z^4n_i$, not $n_{\rm e}$; at $Z = 1$ they are the same number and every plasma here is hydrogenicsDefinition 2; §8.1

One cross-module constant, and the size of the difference. §8 uses the derived electron-branch Coulomb logarithm $\ln\Lambda = 37.081$ where Module 1's census value is $37.8$ and Module 10 reuses that census value. The difference is $1.9$ per cent, it is not a correction to either module, and it is one of the two factors in (8.3). §1's table of speeds is the companion device for the other factor.

Sources

Every published value this module compares against, with what was read and how. Where a paper was not read, the entry says so, and nothing in this module rests on it.