10. Turbulence

One dimensional argument, and the four fluids that test it: a laboratory, the solar wind, a cluster of galaxies and a molecular cloud.

Contents

1. The nonlinear term, and the number that decides

Module 2 derived the Navier–Stokes equation and then spent the rest of the book throwing its viscous term away. The licence to do that came from one number, and Module 2 §9 printed it for four gases. This module asks what the equation does when the viscous term is small but not zero — which is the ordinary condition of almost every fluid in the sky, and which is not the same thing as the viscous term being absent.

The distinction is the whole subject. Set $\nu = 0$ exactly and the Euler equations conserve energy; there is nowhere for energy to go. Set $\nu$ small and the flow develops structure on smaller and smaller scales until the scales are small enough for viscosity to act, and energy is destroyed at a rate that does not go to zero as $\nu$ does. That is the anomaly at the centre of turbulence, and everything in this module is a consequence of taking it seriously.

Definition 1 (Reynolds number, and the turbulent regime). For a flow of characteristic speed $U$ over a characteristic scale $L$ in a fluid of kinematic viscosity $\nu = \mu/\rho$, $$ \mathrm{Re} \;\equiv\; \frac{UL}{\nu}. \tag{1.1}$$ It is the ratio of the inertial term $(\mathbf{u}\cdot\nabla)\mathbf{u} \sim U^2/L$ of Module 2's equation (7.5) to the viscous term $\nu\nabla^2\mathbf{u} \sim \nu U/L^2$. The flow is called turbulent when $\mathrm{Re}$ is large enough for the nonlinearity to dominate at every scale between $L$ and the scale at which viscosity finally acts.

"Large enough" is not a number this module derives. Pipe flow goes turbulent near $\mathrm{Re}\approx 2300$ and a flat-plate boundary layer near $\mathrm{Re}\approx 5\times10^{5}$: two thresholds differing by more than two orders of magnitude for the same fluid, because the threshold depends on the geometry and on how the flow is disturbed, not on $\nu$ alone. What matters below is not the threshold but the enormous margin by which astrophysical flows clear it. Both numbers are quoted here as inputs and are checked nowhere in this module.

Proposition 1 (Module 2's identity, re-derived). If the viscosity is the kinetic-theory value $\mu \simeq \tfrac13\rho\,\bar v\,\lambda$ of Module 1, with $\lambda$ the mean free path and $\bar v$ a thermal speed, then $$ \mathrm{Re} \;=\; \frac{\rho UL}{\tfrac13\rho\,\bar v\,\lambda} \;=\; 3\,\frac{U}{\bar v}\,\frac{L}{\lambda} \;=\; \frac{3\,\mathrm{Ma_{th}}}{\mathrm{Kn}}, \qquad \mathrm{Ma_{th}}\equiv\frac{U}{\bar v},\quad \mathrm{Kn}\equiv\frac{\lambda}{L}. \tag{1.2}$$ The identity holds for any thermal speed, provided the same one is used in the viscosity and in $\mathrm{Ma_{th}}$: $\bar v$ appears once in each, to the same power and on opposite sides of the ratio, so the two occurrences cancel and the $\mathrm{Re}$ they produce is unchanged. What is not unchanged is $\mathrm{Ma_{th}}$ itself, which is why a module must say which speed it means.
Substitute $\mu = \tfrac13\rho\bar v\lambda$ into (1.1) written as $\mathrm{Re}=\rho UL/\mu$. The density cancels, leaving $3(U/\bar v)(L/\lambda)$, which is the right-hand side. The clause about the speed is the part worth care. Nothing above fixes what $\bar v$ means; it enters the viscosity and the Mach number to the same power, so replacing it everywhere by any other thermal speed leaves (1.2) true and leaves $\mathrm{Re}$ where it was, while moving $\mathrm{Ma_{th}}$. It is therefore not enough to say which identity is being used; a module must say which speed. §6.4 prints every speed this module uses and how many components each one averages over. ∎

This module's choice, stated once. Throughout, $\bar v \equiv \sqrt{8k_{\rm B}T/\pi m}$, the mean thermal speed — the speed that enters the kinetic-theory viscosity $\nu = \lambda\bar v/3$. Module 2 made the other consistent choice, $v_{\rm th}=\sqrt{2k_{\rm B}T/m}$, which is the speed Module 1 uses for a gyroradius. The two differ by $\sqrt{\pi/4} = 0.8862$, so a Reynolds number computed here is smaller than the same gas's Reynolds number in Module 2 by $\sqrt{4/\pi} = 1.1284$ — and the reconciliation is printed in §6 rather than left for a reader to notice.

Equation (1.2) is the reason this module can be written at all. It converts a Reynolds number, which needs a viscosity, into a Knudsen number, which needs only a mean free path — and Module 1 tabulated mean free paths for every gas in this book. Two examples, both computed from Module 1's own inputs:

System$\mathrm{Kn}=\lambda/L$$\mathrm{Ma_{th}}=U/\bar v$$3\,\mathrm{Ma_{th}}/\mathrm{Kn}$
laboratory air, $L = 1$ m, $U = 10\ \mathrm{m\,s^{-1}}$$6.800\times10^{-8}$$0.0218$$9.613\times10^{5}$
solar photosphere, $L = H = 150$ km, $U = 2\ \mathrm{km\,s^{-1}}$$4.427\times10^{-10}$$0.2063$$1.398\times10^{9}$
The photosphere row uses Module 1's photospheric number density $n = P/k_{\rm B}T = 1.506\times10^{17}\ \mathrm{cm^{-3}}$ at $P = 1.2\times10^{5}\ \mathrm{dyn\,cm^{-2}}$ and $T = 5772$ K, giving $\lambda = 66\ \mu\mathrm{m}$ against a $150$ km scale height, and $\mu = 1.30$ in $\bar v = \sqrt{8k_{\rm B}T/\pi\mu m_{\rm u}}$, which is what fixes the $\mathrm{Ma_{th}}$ column. The hard-sphere cross-section $\sigma_{\rm H}=10^{-15}\ \mathrm{cm^2}$ it inherits is good to a factor of about two, so read $1.398\times10^{9}$ to one significant figure. The air row's $\lambda = 68$ nm is Module 1's, and its $\mu = 28.96$ is printed in §6. Every number in this module is produced by afd/figs/m10_numbers.py.

A Reynolds number of $10^{9}$ in the layer of the Sun one can photograph is the fact this module exists to explain. It is not a number any laboratory reaches, and it is at the small end of the astrophysical range. What the rest of this section, and §3, will show is that because it is so large, almost nothing about the flow can be computed — and that a surprising amount about it can nevertheless be predicted, with no calculation at all, from dimensions alone.

2. Kolmogorov 1941, by dimensional analysis alone

The argument in this section is due to A. N. Kolmogorov, in a three-page note of 1941, and it is the most economical piece of physics in this book. It solves no equation. It makes one assumption about what a turbulent flow forgets, and then counts dimensions.

Definition 2 (the dissipation rate and the inertial range). Let $\varepsilon$ be the energy dissipated per unit mass per unit time, with dimensions $\mathrm{cm^2\,s^{-3}}$. In a statistically steady flow stirred at the scale $L$, $\varepsilon$ is also the rate at which energy is supplied, since none accumulates. The inertial range is the band of scales $$ \eta \;\ll\; \ell \;\ll\; L \tag{2.1}$$ on which neither the stirring scale $L$ nor the viscosity $\nu$ appears in the statistics, so that $\varepsilon$ and the scale $\ell$ — equivalently the wavenumber $k = 1/\ell$ — are the only quantities available. Here $\eta$ is the scale at which viscosity acts, constructed in §3.
That is an assumption, and it is the only one. Equation (2.1) asserts that a scale in the middle of the range has forgotten both how the energy was put in and how it will be taken out, and remembers only how fast it is passing through. Nothing proves it. Everything in §5 and §7 is a test of it, and §11 names the respect in which it is known to be false.
Proposition 2 (the K41 spectrum). Let $E(k)$ be the energy spectrum per unit mass, defined so that $\int_0^\infty E(k)\,\mathrm{d}k$ is the kinetic energy per unit mass and $E$ therefore has dimensions $\mathrm{cm^3\,s^{-2}}$. If in the inertial range $E$ depends only on $\varepsilon$ and $k$, then $$ E(k) \;=\; C\,\varepsilon^{2/3}k^{-5/3} \tag{2.2}$$ with $C$ a dimensionless constant, and the exponents are unique. The corresponding velocity increment across a separation $\ell$ is $$ \delta v(\ell) \;=\; (\varepsilon\ell)^{1/3}. \tag{2.3}$$
Write the most general such dependence, $E = C\,\varepsilon^{a}k^{b}$, and match dimensions. With $[\varepsilon] = \mathrm{cm^2\,s^{-3}}$ and $[k] = \mathrm{cm^{-1}}$, $$ \mathrm{cm^3\,s^{-2}} \;=\; (\mathrm{cm^2\,s^{-3}})^{a}\,(\mathrm{cm^{-1}})^{b}. $$ The two units give two equations in two unknowns: $$ \text{seconds:}\quad -2 = -3a \;\Rightarrow\; a = \tfrac23; \qquad \text{centimetres:}\quad 3 = 2a - b \;\Rightarrow\; b = 2a-3 = -\tfrac53. \tag{2.4}$$ The system is $2\times2$ with a non-singular matrix, so the solution is unique: no other pair of exponents is dimensionally possible. That is the force of the result — nothing was fitted, and the exponent $-5/3$ is not a best-fit slope but the only slope two units permit. For (2.3), the same two quantities give exactly one speed at the scale $\ell$: $[\varepsilon\ell] = \mathrm{cm^3\,s^{-3}}$, whose cube root is a speed. Consistency with (2.2) is a check rather than a second assumption — the energy in a band of wavenumbers of width $\sim k$ around $k$ is $kE(k) \sim C(\varepsilon/k)^{2/3}$, and its square root is $(\varepsilon\ell)^{1/3}$ up to the constant. ∎

The units line in Definition 2 fixes the exponent, and changing it changes the subject. If $E(k)$ had been defined as an energy rather than an energy per unit wavenumber — that is, if the $\mathrm{cm}$ count in $\mathrm{cm^3\,s^{-2}}$ were $2$ rather than $3$ — the second equation of (2.4) would give $b = -2/3$ and the whole subject would have a different exponent. The $5/3$ is a statement about a spectral density.

What dimensional analysis cannot supply is $C$. It is a pure number, the same for every flow if the assumption of Definition 2 is right, and different from flow to flow if it is wrong. §5 measures it in the laboratory, before this module builds anything astrophysical on it. Fig. 1 draws what (2.2) asserts, and draws beside it the exponent $-2$ that §9 will derive for a field of shocks, so that the two can be compared by eye.

100 101 102 103 104 105 10-9 10-7 10-5 10-3 10-1 101 k L, wavenumber in units of the outer scale E(k), arbitrary units energy in at L energy out at η inertial range: 4 decades E ∝ k−5/3, Kolmogorov E ∝ k−2, Burgers ×21.5 here, ×2.15 after one decade
The inertial range is a band, bounded at both ends. Energy enters at the stirring scale $L$ on the left and leaves at the dissipation scale $\eta$ on the right; between them (2.2) applies and the slope is $-5/3$. The dashed line is the $k^{-2}$ of a field of shocks, Proposition 8, anchored at the same point on the left so that only the slopes are being compared: after one decade the two differ by a factor $10^{1/3} = 2.15$, and across the four decades drawn, by $21.5$. The vertical normalisation is schematic and the band width is a choice: no measured spectrum is plotted anywhere in this module, and four decades of inertial range is what $\mathrm{Re} = (10^{4})^{4/3} = 2.15\times10^{5}$ would give by Proposition 4. The curvature at the right-hand end is an exponential cut-off drawn to show that the range ends; its shape is not predicted by anything in §2.

3. The dissipation scale, and what a simulation would cost

The inertial range of Definition 2 has a lower end, and the same counting that fixed the spectrum fixes where it is. This section builds that scale, and then uses it to price a direct numerical simulation — which is the calculation that decides, once and for all, that astrophysical turbulence is not going to be computed from first principles.

Proposition 3 (the Kolmogorov microscales). At the small-scale end the available quantities are $\nu$ and $\varepsilon$, and they form exactly one length, one speed and one time: $$ \eta \;=\; \left(\frac{\nu^3}{\varepsilon}\right)^{1/4},\qquad u_\eta \;=\; (\nu\varepsilon)^{1/4},\qquad \tau_\eta \;=\; \left(\frac{\nu}{\varepsilon}\right)^{1/2}. \tag{3.1}$$ These are not three statements but one, because $$ \frac{u_\eta\,\eta}{\nu} \;=\; 1 \tag{3.2}$$ identically: the Reynolds number of the smallest eddy is exactly unity, which is what it means for viscosity to act there.
With $[\nu] = \mathrm{cm^2\,s^{-1}}$ and $[\varepsilon] = \mathrm{cm^2\,s^{-3}}$, seek $\nu^{a}\varepsilon^{b}$ with the dimensions of a length: $2a+2b = 1$ and $-a-3b = 0$, so $b = -a/3$ and $a = 3/4$, giving $\eta = (\nu^3/\varepsilon)^{1/4}$. The same two equations with the dimensions of a speed give $a = b = 1/4$, and with those of a time, $a = -b = 1/2$. Each system is $2\times2$ and non-singular, so each answer is the only one. For (3.2), substitute: $u_\eta\eta/\nu = (\nu\varepsilon)^{1/4}(\nu^3/\varepsilon)^{1/4}/\nu = (\nu^4)^{1/4}/\nu = 1$. The prep script evaluates this numerically and prints $1.000000$, which is the intended reading: it is an algebraic identity, and a numerical value differing from one would mean a typing error and nothing else. ∎

Put numbers in. For air near the ground, stirred at the rate $\varepsilon = 10^{-3}\ \mathrm{W\,kg^{-1}} = 10\ \mathrm{erg\,g^{-1}\,s^{-1}}$ typical of the atmospheric boundary layer, with $\nu = 0.15\ \mathrm{cm^2\,s^{-1}}$:

$$ \eta = 1.355\ \mathrm{mm},\qquad u_\eta = 1.11\ \mathrm{cm\,s^{-1}},\qquad \tau_\eta = 122.47\ \mathrm{ms}. \tag{3.3}$$
Precision, stated once here and inherited everywhere below. The $0.15\ \mathrm{cm^2\,s^{-1}}$ is a two-significant-figure value for which no primary source was obtained: the standard reference, Kadoya, Matsunaga & Nagashima (1985), is listed in Sources and was not served by any archive tried. Every quantity in (3.3) therefore carries two significant figures whatever width it is printed at, and so does every number derived from it, including Problem C1. The extra digits are carried only so that the identity (3.2) closes to $1.000000$. This statement is not repeated at each later use.
Proposition 4 (the number of degrees of freedom, and the cost of resolving them). With $U$ the outer-scale speed and $\varepsilon \sim U^3/L$, $$ \frac{L}{\eta} \;=\; \mathrm{Re}^{3/4}, \qquad N \;=\; \left(\frac{L}{\eta}\right)^{3} \;=\; \mathrm{Re}^{9/4}. \tag{3.4}$$ A time-explicit code is limited by the Courant condition to a step $\delta t \sim \eta/U$ — the largest speed on the smallest cell — so one outer turnover $L/U$ costs $L/\eta = \mathrm{Re}^{3/4}$ steps and $$ \mathrm{Re}^{9/4}\times\mathrm{Re}^{3/4} \;=\; \mathrm{Re}^{3} \tag{3.5}$$ point-updates in total.
Take $\varepsilon = U^3/L$, which is (2.3) at $\ell = L$ rearranged, and substitute into $\eta = (\nu^3/\varepsilon)^{1/4}$: $$ \frac{L}{\eta} \;=\; L\left(\frac{U^3}{L\nu^3}\right)^{1/4} \;=\; \left(\frac{UL}{\nu}\right)^{3/4} \;=\; \mathrm{Re}^{3/4}. $$ A cube of side $L$ resolved to $\eta$ needs $(L/\eta)^3$ cells, which is (3.4). For the step count, $(L/U)/(\eta/U) = L/\eta$, and multiplying by the cell count gives $\mathrm{Re}^{9/4+3/4} = \mathrm{Re}^{3}$. ∎

A softer count exists and it is wrong for this purpose, by a factor that can be named. Dividing the turnover time $L/U$ by the Kolmogorov time $\tau_\eta$ rather than by $\eta/U$ gives $\mathrm{Re}^{1/2}$ steps and $\mathrm{Re}^{11/4}$ updates. That is the number of turnovers of the smallest eddy, not the number of steps a code takes, and it understates the cost by exactly $$ \frac{U}{u_\eta} \;=\; \mathrm{Re}^{1/4}. \tag{3.6}$$ Both columns are printed below so the difference is visible rather than argued about. $\mathrm{Re}^{3}$ is the standard result; Pope, Turbulent Flows, their Sect. 9.1.2 states it — a section number this book has not confirmed against a copy, as Sources records.

$\mathrm{Re}$$L/\eta = \mathrm{Re}^{3/4}$$N = \mathrm{Re}^{9/4}$one snapshotCFL steps per turnoverwork, $\mathrm{Re}^{3}$the softer count, $\mathrm{Re}^{11/4}$
$10^{3}$$177.8$$5.623\times10^{6}$$1.35\times10^{8}$ B$178$$10^{9}$$1.78\times10^{8}$
$10^{4}$$1000$$10^{9}$$2.4\times10^{10}$ B$10^{3}$$10^{12}$$10^{11}$
$10^{6}$$3.162\times10^{4}$$3.162\times10^{13}$$7.589\times10^{14}$ B$3.16\times10^{4}$$10^{18}$$3.16\times10^{16}$
$10^{8}$$10^{6}$$10^{18}$$2.4\times10^{19}$ B$10^{6}$$10^{24}$$10^{22}$
$10^{10}$$3.162\times10^{7}$$3.162\times10^{22}$$7.589\times10^{23}$ B$3.16\times10^{7}$$10^{30}$$3.16\times10^{27}$
$10^{12}$$10^{9}$$10^{27}$$2.4\times10^{28}$ B$10^{9}$$10^{36}$$10^{33}$

The snapshot column stores three velocity components at eight bytes each, so 24 bytes per grid point, and nothing else — no density, no pressure, no field, and no second copy for the time integrator. It is a floor.

The number that makes Module 14 necessary. At $\mathrm{Re} = 10^{6}$ — a large wind tunnel, and three orders of magnitude below the photosphere of §1 — a single velocity snapshot is 759 TB and one outer turnover costs $10^{18}$ point-updates. The molecular cloud of §6, at $\mathrm{Re} = 8.1\times10^{7}$, needs $6.24\times10^{17}$ grid points. Direct numerical simulation of an astrophysical flow is not expensive; at $10^{18}$ point-updates per outer turnover it is beyond any machine that has been built or proposed, and every astrophysical code therefore solves a different problem from the one written down here.

Two more evaluations of (3.1), both of which will be used later, and one of which refutes its own premise:

Flow$\varepsilon = U^3/L$$\eta$$L/\eta$
molecular cloud, $\sigma_L = 2.64\ \mathrm{km\,s^{-1}}$ over $10$ pc$5.963\times10^{-4}\ \mathrm{erg\,g^{-1}\,s^{-1}}$$2.4$ au$8.544\times10^{5}$
intracluster medium, $164\ \mathrm{km\,s^{-1}}$ over $60$ kpc, unmagnetised $\nu$$2.382\times10^{-2}\ \mathrm{erg\,g^{-1}\,s^{-1}}$$64.67$ kpcnone: $\eta/L = 1.078$
Read the second row as a refutation, not as a dissipation scale. It gives $\eta/L = 1.078$: the scale at which viscosity is supposed to take over is larger than the system, so there is no inertial range at all and (2.1) is empty. The line also uses $\varepsilon = U^3/L$, which is itself a K41 relation, so it has used a cascade to locate the end of a cascade it has just shown cannot exist. It is the $\mathrm{Kn} = 0.375$ of §6 seen from the other end, and §6 is where the way out is. With the magnetised viscosity of that section the same formula gives $\eta = 3.887$ au, which is $3.1\times10^{-10}$ of the system size.

4. The one exact result, and its minus sign

Everything in §2 and §3 is dimensional analysis: correct up to a constant that dimensions cannot fix. There is exactly one result in the theory of turbulence that is exact, that follows from the Navier–Stokes equation rather than from counting units, and that has no adjustable constant at all.

Proposition 5 (Kolmogorov's four-fifths law). For homogeneous, isotropic, statistically stationary turbulence at high Reynolds number, the third-order longitudinal structure function $$ S_3(\ell) \;\equiv\; \big\langle\,\delta u_\parallel(\ell)^3\,\big\rangle, \qquad \delta u_\parallel(\ell) \;\equiv\; \big[\,\mathbf{u}(\mathbf{x}+\boldsymbol{\ell}) - \mathbf{u}(\mathbf{x})\,\big]\cdot\hat{\boldsymbol{\ell}}, $$ satisfies, for $\eta \ll \ell \ll L$, $$ S_3(\ell) \;=\; -\tfrac45\,\varepsilon\,\ell. \tag{4.1}$$ Stated, not derived here. The derivation is von Kármán and Howarth's, from the two-point correlation equation; it is a page of tensor algebra that would displace two of this module's four checks, and it is quoted in the same way Module 9 quotes the Eddington luminosity. Its four hypotheses — homogeneous, isotropic, stationary, $\mathrm{Re}\to\infty$ — are named in this sentence because each of them fails somewhere in this module.
Not given; see the statement. What can be checked here without the derivation is consistency with (2.3): $S_3 \propto \varepsilon\ell$ has the dimensions of a speed cubed, and $(\varepsilon\ell)^{1/3}$ is that speed, so the exact law and the dimensional estimate agree on the scaling and differ only by the coefficient $4/5$. The ratio of the two is $(4/5)^{1/3} = 0.9283$, which is the number Problem C3 asks a reader to recognise. ∎

The minus sign is the physics. A symmetric distribution of velocity increments has zero third moment, so (4.1) says the distribution is skewed: at any separation in the inertial range, large negative longitudinal increments are more common than large positive ones. Fluid elements approach each other along the line joining them more violently than they separate. That asymmetry is the direction of the cascade written as a number — energy flows from large scales to small, and the sign of $S_3$ is what says so. Reverse it and the energy would flow the other way, which is what happens in two-dimensional turbulence and is one of the reasons that subject is different.

The second-order structure function $S_2(\ell) = \langle\delta u_\parallel^2\rangle$ has no such exact law; it is $C_2(\varepsilon\ell)^{2/3}$, with $C_2$ related to the spectral constant of (2.2) by an isotropy factor. §5 quotes both.

5. The terrestrial check: is there one Kolmogorov constant?

Proposition 2 predicts an exponent and leaves a number free. The exponent is what most accounts of K41 quote; the number is where the assumption of Definition 2 is actually tested, and it is tested in a laboratory. This section comes before any astrophysical number in this module, and the ordering is deliberate: §9 and Problem K2 use $C$ to evaluate a spectrum in a molecular cloud, and a constant used in the sky should be pinned on the ground first. Modules 3 and 4 put their terrestrial measurement early for the same reason.

The convention matters, because three different constants are called "the Kolmogorov constant" in the literature and they differ by factors of two and more. Sreenivasan (1995) defines his through the one-dimensional longitudinal spectrum $E_{11}(k_1)$, the thing a single hot wire in a single mean flow actually measures:

$$ E_{11}(k_1) \;=\; C_K\,\varepsilon^{2/3}k_1^{-5/3}. \tag{5.1}$$

The three-dimensional constant $C$ of (2.2) is related to it by an isotropy factor, and the second-order structure function by another:

$$ C \;=\; \tfrac{55}{18}\,C_K, \qquad C_2 \;=\; 4.02\,C_K. \tag{5.2}$$

Both factors are Sreenivasan's, from his Section I, and both assume local isotropy — the same assumption Definition 2 makes. They are quoted, not derived.

A second Reynolds number, and it is not Definition 1's. Experimenters report the Taylor-microscale Reynolds number $R_\lambda \equiv u'\lambda_T/\nu$, where $u'$ is the root-mean-square velocity fluctuation of one component and $\lambda_T$ is the Taylor microscale $u'/\langle(\partial u/\partial x)^2\rangle^{1/2}$. They report it because $\lambda_T$ can be measured from a single probe where $L$ often cannot. For homogeneous turbulence $R_\lambda \propto \mathrm{Re}^{1/2}$, so a range in $R_\lambda$ is half as many decades as the same range in $\mathrm{Re}$. This relation is standard and is quoted here, not derived. $R_\lambda$ is new to this book, is not numbered as a definition because nothing later depends on it, and appears only in this section.

Sreenivasan compiled more than a hundred measurements of $C_K$ — grid turbulence, jets, wakes, boundary layers, a tidal channel — and found:

Check 2 — CONFIRMED, and note exactly what is confirmed. $C_K = 0.53 \pm 0.055$ across more than $100$ flows, a fractional scatter of $10.4$ per cent, over $R_\lambda$ from $50$ to about $10^{4}$ — which is $\mathbf{2.30}$ decades in $R_\lambda$ and, since $\mathrm{Re}\sim R_\lambda^{2}$, $\mathbf{4.60}$ decades in $\mathrm{Re}$. K41 does not predict the value $0.53$. It predicts that there is a value: that one number describes flows of entirely different geometry across four and a half decades of Reynolds number. A ten per cent scatter is that prediction, met.
The $\pm0.055$ is the scatter of the experiments, not the error on their mean; with $N \gt 100$ the error on the mean would be smaller by $\sqrt{N} \gt 10$, and quoting it would turn a statement about universality into a statement about arithmetic. The span is stated in both variables deliberately: "three decades", the phrase one might reach for, is neither of them.

In the three-dimensional form (5.2) the compiled value is $C = 1.619 \pm 0.168$, and the second-order structure constant is $C_2 = 2.131 \pm 0.221$. Sreenivasan then revises his own number: allowing that the local-isotropy estimate of $\varepsilon$ undercounts by about ten per cent moves $C_K$ to about $0.50$ and the three-dimensional constant to $1.53$. This module therefore quotes the range $1.53$–$1.62$ wherever a three-dimensional constant is needed, including in Problem K2, rather than a single digit that would hide a correction the author himself applied.

One entry of the compilation deserves separate treatment, because it is the measurement the subject is usually said to rest on. Grant, Stewart & Moilliet (1962) measured spectra in a tidal channel at $R_\lambda$ of $3000$–$18000$ and obtained $C_K = 0.47\pm0.02$, three-dimensional form $C = 1.436\pm0.061$: a departure of $-1.09$ population standard deviations from the compiled mean, which is to say, an ordinary member of the population.

That number is quoted at second hand and the module says so. The 1962 paper is paywalled and was not read. The value above is read from Sreenivasan's Table IV, whose own footnote records that the tabulated constants are averages over seventeen data sets read off log–log plots with a straight edge, and that Kraichnan judged a different reading of the same plots would raise the value by more than ten per cent. The $R_\lambda$ range is Sreenivasan's estimate, not Grant et al.'s measurement. Nothing in this module depends on the entry; it is included because a reader who knows the subject will look for it.
And a gap in this module, stated here rather than in a footnote. This module has no terrestrial fitted slope with a published error bar. Every measured exponent it quotes is a solar-wind exponent, from §7. What the laboratory data contribute is the constant and its universality — a different claim, and a weaker one, than a measured $-5/3$. A reader who wants the exponent measured must take it from a spacecraft, which is the next-but-one section.

6. Four fluids, and the Knudsen number that disqualifies two

Definition 1 needs a kinematic viscosity. Module 1 showed that in a dilute gas $\nu \simeq \lambda\bar v/3$, and Module 2 showed that this expression is the first-order term of an expansion in $\mathrm{Kn}$ — so it is meaningful only when $\mathrm{Kn}$ is small. This section evaluates $\mathrm{Re}$ for four fluids and finds that in two of the four, the expression has no right to be evaluated at all.

Fluidinputs$\lambda$$\nu$ (cm² s⁻¹)$U$, $L$$\mathrm{Re}$$\mathrm{Kn}$
laboratory air$288.15$ K, $\mu = 28.96$$68$ nm$0.15$ measured$10\ \mathrm{m\,s^{-1}}$, $1$ m$6.667\times10^{5}$$6.800\times10^{-8}$
molecular cloud$10$ K, $\mu = 2.33$, $n = 10^{2}\ \mathrm{cm^{-3}}$$10^{13}$ cm $=0.67$ au$1.0048\times10^{17}$$2.64\ \mathrm{km\,s^{-1}}$, $10$ pc$8.107\times10^{7}$$3.241\times10^{-7}$
intracluster medium$10^{8}$ K, $n = 10^{-3}\ \mathrm{cm^{-3}}$, $\ln\Lambda = 37.8$$22.5$ kpc$3.3557\times10^{30}$$164\ \mathrm{km\,s^{-1}}$, $60$ kpc$0.90$$0.375$
solar wind at 1 au$1.2\times10^{5}$ K, $n = 5\ \mathrm{cm^{-3}}$, $\ln\Lambda = 26.10$$1.936$ au$4.8492\times10^{19}$$400\ \mathrm{km\,s^{-1}}$, $1$ au$12.34$$1.936$
The air row uses the measured viscosity $0.15\ \mathrm{cm^2\,s^{-1}}$; kinetic theory's $\lambda\bar v/3$ gives $0.1040$, a ratio of $0.69$. That $30$ per cent is the honest accuracy of a coefficient of $1/3$ in a real gas, and this module uses a measured viscosity wherever one exists and the kinetic-theory value where none does. The ICM's $22.5$ kpc is Module 1's mean free path against Sarazin's published $23$ kpc; the solar wind's four inputs are Module 1's census values, and Module 1 §8 prints $\lambda = 1.9$ AU and $r_g = 93$ km from exactly them. Mean free paths in ionised rows are Coulomb; in neutral rows, hard-sphere at $\sigma_{\rm H} = 10^{-15}\ \mathrm{cm^2}$, good to a factor of about two.

Two rows agree with the identity (1.2) to every digit, and one is deliberately excluded from that test:

Fluid$\mathrm{Kn}$$\mathrm{Ma_{th}}$$3\,\mathrm{Ma_{th}}/\mathrm{Kn}$$UL/\nu$ratio
molecular cloud, 10 pc$3.241\times10^{-7}$$8.758$$8.107\times10^{7}$$8.107\times10^{7}$$1.0000$
intracluster medium, 60 kpc$3.751\times10^{-1}$$0.113$$9.048\times10^{-1}$$9.048\times10^{-1}$$1.0000$
solar wind, 1 au$1.936$$7.964$$1.234\times10^{1}$$1.234\times10^{1}$$1.0000$

The air row is absent because it uses the measured viscosity rather than $\lambda\bar v/3$, so the identity cannot hold there by construction: the discrepancy would be exactly the $0.69$ above. A ratio of $1.0000$ in the other three rows is not a physical result; it is a guard. If one of the mean free paths above were ever changed and another not, this column would stop reading $1.0000$.

Proposition 6 (when there is no viscosity to put in). If $\mathrm{Kn} \gtrsim 1/3$, the Chapman–Enskog expansion of Module 1 is failing, the Navier–Stokes viscosity does not exist as a property of the gas, and $UL/\nu$ is a formula evaluated outside its own domain. By (1.2), such a gas has $$ \mathrm{Re} \;\lt\; 1 \quad\text{whenever}\quad \mathrm{Kn} \;\gt\; 3\,\mathrm{Ma_{th}}, \tag{6.1}$$ so "the Reynolds number is small" and "the fluid description has failed" are one statement in such a gas, not two. Where a magnetic field is present, the transport step across the field is not $\lambda$ but the gyroradius $r_g$, so the perpendicular viscosity is bounded above and the perpendicular Reynolds number below: $$ \nu_\perp \;\le\; \frac{r_g\,\bar v}{3}, \qquad \mathrm{Re}_\perp \;\ge\; \frac{UL}{r_g\bar v/3}. \tag{6.2}$$
The first part is (1.2) rearranged: $\mathrm{Re} = 3\,\mathrm{Ma_{th}}/\mathrm{Kn} \lt 1$ exactly when $\mathrm{Kn} \gt 3\,\mathrm{Ma_{th}}$. Nothing further is needed, and that is the point — one identity turns a statement about a transport coefficient into a statement about a mean free path, which is why Fig. 2 can draw both on one plane. For (6.2), Module 1 §8 showed that across $\mathbf{B}$ a charged particle's excursion between scatterings is bounded by $r_g$ however long it goes without colliding, whereas along $\mathbf{B}$ it is free to travel $\lambda$. A diffusivity is a step squared over a time, so replacing the step $\lambda$ by $r_g$ in $\nu = \lambda\bar v/3$ gives the perpendicular value, and it is an upper bound because scattering can only be more effective than the gyro-orbit alone, never less. Dividing $UL$ by an upper bound on $\nu$ gives a lower bound on $\mathrm{Re}_\perp$, which is (6.2). It is printed as a bound in both directions throughout this module and never as a value. ∎
10-8 10-6 10-4 10-2 100 102 100 102 104 106 108 Knudsen number Kn = λ/L Reynolds number Re laboratory air, 1 m molecular cloud, 10 pc intracluster medium, 60 kpc solar wind, 1 au Math = 0.01 Math = 0.1 Math = 1 Math = 10 grey lines: Module 2’s identity Re = 3 Math/Kn Kn ≥ 1/3: no Navier–Stokes viscosity exists the Chapman–Enskog expansion of Module 1 is failing here open marker: Re from the MEASURED ν, not from λv̄/3
The formula, and the region it has left. The grey lines are the identity (1.2), $\mathrm{Re} = 3\,\mathrm{Ma_{th}}/\mathrm{Kn}$, one for each thermal Mach number; a fluid moves along its own line as the system size changes. The shaded band on the right is $\mathrm{Kn} \ge 1/3$, where the Chapman–Enskog expansion of Module 1 is failing and there is no Navier–Stokes viscosity to put into (1.1). Laboratory air and the molecular cloud sit $6.7$ and $6.0$ decades clear of that band; the intracluster medium and the solar wind sit inside it. The air point is drawn open because its $\mathrm{Re}$ uses the measured viscosity rather than $\lambda\bar v/3$, which is why it is the one point that does not lie on its own grey line — it sits a factor $9.613\times10^{5}/6.667\times10^{5} = 1.44$ below it, which is the reciprocal of the $0.69$ above, and that factor is the accuracy of kinetic theory's coefficient rather than an error.

6.1 The intracluster medium, and the field that rescues it

The Coulomb mean free path in a cluster core is $22.5$ kpc — a third of the region Hitomi observed. The ion mean thermal speed is $\bar v_i = \sqrt{8k_{\rm B}T/\pi m_{\rm p}} = 1450\ \mathrm{km\,s^{-1}}$, so $\nu = \lambda\bar v_i/3 = 3.3557\times10^{30}\ \mathrm{cm^2\,s^{-1}}$ and

$$ \mathrm{Re} \;=\; \frac{(1.64\times10^{7})(60\ \mathrm{kpc})}{3.3557\times10^{30}} \;=\; 0.90, \qquad \mathrm{Kn} \;=\; \frac{22.5}{60} \;=\; 0.375. \tag{6.3}$$
Read those two together, not separately. $\mathrm{Re} \lt 1$ because $\mathrm{Kn} \gt 1/3$, and by (1.2) they are one statement. At $\mathrm{Kn} = 0.375$ the expansion of Module 1 is failing, so $0.90$ is a formula evaluated outside its own domain of validity. It does not say the intracluster medium is a barely turbulent fluid. It says the collisional fluid description does not apply to the intracluster medium — and the observations of §8 show gas that is moving, which no laminar reading of $\mathrm{Re} = 0.90$ could accommodate.
The speed in $\lambda\bar v/3$ here is the ion mean thermal speed at $m = m_{\rm p}$, not the mean speed of an average particle at $\mu_{\rm ICM} = 0.61$. A viscosity is carried by whatever carries the momentum, and in a hydrogen plasma that is the ions; $\mu_{\rm ICM} = 0.61$ is a mass per particle with the electrons counted in, and it belongs in the sound speed and in the pressure ratio of §8, not here. Using it here would inflate $\nu$ by $\sqrt{1.00728/0.61} = 1.285$.

The way out is Module 1's, and (6.2) is what it buys. At $B = 1\ \mu$G the proton gyroradius is $r_g = 1.341\times10^{10}\ \mathrm{cm} = 8.97\times10^{-4}$ au, so

$$ \frac{\lambda}{r_g} \;=\; 5.177\times10^{12}, \qquad \nu_\perp \;\le\; \frac{r_g\bar v_i}{3} \;=\; 6.4824\times10^{17}\ \mathrm{cm^2\,s^{-1}}, \qquad \mathrm{Re}_\perp \;\ge\; 4.684\times10^{12}. \tag{6.4}$$

That is the cascade §8 measures. It exists because the field, not the collisions, sets the transport step across $\mathbf{B}$ — the same one-dimensional rescue Module 1 §8 described, here turned into a number.

The opposite bound is worth printing beside it. Braginskii's classical perpendicular viscosity, $0.3125\,\nu_\parallel (r_g/\lambda)^2$ — note the square — is $4.0775\times10^{4}\ \mathrm{cm^2\,s^{-1}}$, which would give $\mathrm{Re}_\perp = 7.446\times10^{25}$.

The two bounds differ by $1.59\times10^{13}$, and neither is the answer. That factor is $3.07$ times $\lambda/r_g = 5.18\times10^{12}$ and not $\lambda/r_g$ itself, because the classical value is built on Braginskii's $\nu_\parallel$, which carries his coefficient $0.96$ and his root-mean-square speed rather than this module's $\lambda\bar v/3$. Neither bound is the effective viscosity of the intracluster medium, which is set by pressure-anisotropy microinstabilities that scatter ions far more often than Coulomb collisions do. This module bounds $\mathrm{Re}_\perp$. It does not state it.

And the bound has been measured once. Zhuravleva et al. (2019) compared a twelve-day Chandra image of an offset region of the Coma cluster — where the lower density raises $\lambda$ to about $30$ kpc — against direct numerical simulation, and found no viscous cut-off where a Coulomb-collisional plasma must have one. Their conclusion is that "the effective viscosity must be suppressed by at least a factor of $\sim10$ to $\sim1000$ for $\mathrm{Pr} \le 1$". So the collisional $\mathrm{Re} = 0.90$ of (6.3) is a floor that one observation lifts by one to three orders of magnitude.

Two caveats, and both are theirs. The suppression factor depends on the Prandtl number, which they leave as a free parameter — which is why the bound is a range of two orders of magnitude and not a number. 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." $\mathrm{Pr}$ is quoted here inside their sentence and is computed nowhere in this module.

What the two intracluster rows do and do not license

Evaluating (1.1) at $L = 1$ Mpc and $V = 300\ \mathrm{km\,s^{-1}}$ with the same $\nu$ gives $\mathrm{Re} = 27.6$. Read that row narrowly. It carries the core mean free path, $22.5$ kpc at $n = 10^{-3}\ \mathrm{cm^{-3}}$ and $T = 10^{8}$ K, out to a larger $L$. It is not a cluster outskirts: $\lambda \propto T^2/(n\ln\Lambda)$, and a real cluster at $1$ Mpc has a lower density, which raises $\lambda$ and lowers $\mathrm{Re}$ again. Module 1's census makes the direction concrete: its intracluster outskirts row, at $n = 10^{-4}\ \mathrm{cm^{-3}}$ and $T = 10^{8}$ K over $L = 2$ Mpc, has $\lambda = 222$ kpc and $\mathrm{Kn} = 0.11$ — ten times the mean free path of the core row, at a radius twice as large. Module 3 §3 put an intracluster medium at $r = 1$ Mpc in its scale-height table but gave no density there, so this book has no sourced $(n,T)$ pair at that radius and the row may not be read as one.

What the evidence in this book actually is, stated at its exact width. One sourced intracluster Knudsen number: Hitomi's, at $30$–$60$ kpc, equal to $0.375$. One paper ruling the collisional description out in a second cluster: Zhuravleva's, whose offset Coma region has $\lambda \approx 30$ kpc — but this book has no scale $L$ for her fluctuation spectrum, so no Knudsen number may be computed from it, only her own conclusion quoted. That is the whole of the evidence, and the prose may not widen it into a claim about every scale at which the intracluster medium has been measured.

6.2 Braginskii's coefficient, and a convention that is not a coefficient

Kinetic theory's $\lambda\bar v/3$ carries a coefficient of $1/3$ that was obtained by hand-waving over directions. Braginskii (1965) computed the transport coefficients of a plasma properly, and comparing the two is the natural check. It has to be done carefully, because there are two different comparisons here and only one of them is about a coefficient.

First: what Braginskii's $\tau_i$ is. He does not leave it free. His equation (2.5i) defines it, $$ \tau_i \;=\; \frac{3\sqrt{m_i}\,T_i^{3/2}}{4\sqrt{\pi}\,\ln\Lambda\,e^4Z^4 n_i} \;=\; 4.4126\times10^{14}\ \mathrm{s}, \tag{6.5}$$ and with the Spitzer mean free path of Module 1 this is exactly $\lambda/v_{\rm rms}$, because $3^{3/2}/(4\sqrt{\pi})$ divided by $\sqrt{3}$ is $3/(4\sqrt\pi)$, which is his coefficient. Exactly is meant in the algebra; in the arithmetic it holds to four figures, because the $22.5$ kpc printed above is the unrounded Coulomb mean free path rounded for display. That identity validates nothing about Module 1's mean-free-path coefficient, because a Spitzer mean free path is $v_{\rm rms}\tau$; it fixes which convention this book is using, and the answer is Braginskii's.

Second: the coefficient check. His ion viscosity is $\eta_0^{\rm B} = 0.96\,n_iT_i\tau_i$, so $\nu_{\rm B} = 0.96\,k_{\rm B}T\tau_i/m_i = 3.4966\times10^{30}\ \mathrm{cm^2\,s^{-1}}$, and since $k_{\rm B}T/m = v_{\rm rms}^2/3$,

$$ \frac{\nu_{\rm B}}{\lambda v_{\rm rms}/3} \;=\; 0.9600 \qquad\text{exactly.} \tag{6.6}$$
Check on the coefficient, and on nothing else. Braginskii's proper kinetic calculation gives $0.96$ where the elementary estimate of Module 1 gives $1$, at the same speed. Four per cent. That is what (6.6) says, and it is the only coefficient statement in this module.

Third: a number that looks like a second coefficient and is not one. Against this module's own $\lambda\bar v/3$, which uses the mean speed, the same viscosity gives

$$ \frac{\nu_{\rm B}}{\lambda\bar v/3} \;=\; 0.96\sqrt{\frac{3\pi}{8}} \;=\; 1.0420, \tag{6.7}$$

whatever the temperature, the density or the mean free path. It is printed only in that algebraic form, because writing it as a decimal invites a reader to treat $1.042$ as a rival to $0.96$. The extra $8.5$ per cent is the mean speed against the root-mean-square speed:

$$ \frac{\bar v}{v_{\rm rms}} \;=\; \sqrt{\frac{8}{3\pi}} \;=\; 0.9213, \qquad \frac{0.96}{0.9213} \;=\; 1.0420. \tag{6.8}$$

Equation (6.8) is the whole of the gap between (6.6) and (6.7). It is a component-count and averaging statement, not a physical one: both speeds are three-dimensional, and they differ only in whether the average is taken of $v$ or of $v^2$. A reader who took $0.96$ and $1.042$ to be two estimates of one coefficient would conclude that two calculations of the plasma viscosity disagree, when in fact one calculation has been divided by two different speeds.

Notation. Braginskii's dynamic viscosities are written $\eta_0^{\rm B}$ and $\eta_1^{\rm B}$ in this module, always with both marks, and appear only in this subsection; bare $\eta$ is the Kolmogorov dissipation scale of Proposition 3 throughout. He gives $\eta_1^{\rm B} = (3/10)n_iT_i/(\omega_i^2\tau_i)$ with $\eta_2^{\rm B} = 4\eta_1^{\rm B}$, and it is $\eta_1^{\rm B}/\eta_0^{\rm B} = 0.3125\,(r_g/\lambda)^2$ that produced the classical perpendicular viscosity of §6.1.

6.3 The solar wind, and which solar wind it is

At 1 au the Coulomb mean free path on Module 1's census values is $1.936$ au, with $\ln\Lambda = 26.10$ derived rather than assumed. So

$$ \mathrm{Kn} \;=\; 1.936, \qquad \mathrm{Re} \;=\; 12.34, \qquad r_g \;=\; 93\ \mathrm{km}, \qquad \frac{\lambda}{r_g} \;=\; 3.117\times10^{6}, \qquad \mathrm{Re}_\perp \;\ge\; 3.846\times10^{7}. \tag{6.9}$$

The story is the intracluster medium's, and worse: on these values the mean free path is twice the system size, so there is no Navier–Stokes viscosity to put into $UL/\nu$ at all. The same Bohm-like upper bound on $\nu_\perp$ gives the same kind of lower bound on $\mathrm{Re}_\perp$.

Which solar wind is this, and why it is not Module 9's. The four inputs above — $n = 5\ \mathrm{cm^{-3}}$, $T = 1.2\times10^{5}$ K, $B = 5$ nT, $U = 400\ \mathrm{km\,s^{-1}}$ — are Module 1's census values, and Module 1 §8 prints $\lambda = 1.9$ AU, $r_g = 93$ km and $\lambda/r_g = 3.1\times10^{6}$ from exactly them. Module 9 uses a different set — Venzmer & Bothmer (2018) Table 3 median fits — because its job is an acceleration along a fitted radial profile, not a census. Changing them here alone would make this module disagree with the module that owns the census in order to agree with one that is doing something else. So the difference is priced instead of hidden:
quantityModule 1 censusVenzmer & Bothmer median fits (Module 9)
inputs$n = 5\ \mathrm{cm^{-3}}$, $T = 1.2\times10^{5}$ K, $B = 5$ nT, $U = 400\ \mathrm{km\,s^{-1}}$$n = 7.57\ \mathrm{cm^{-3}}$, $T = 9.67\times10^{4}$ K, $B = 6.05$ nT, $U = 435.6\ \mathrm{km\,s^{-1}}$
$\lambda$$1.9363$ au$0.8456$ au
$\mathrm{Kn}$ at $L = 1$ au$1.9363$$0.8456$
$\mathrm{Re}$$12.34$$34.28$
$r_g$$93$ km$69$ km

The conclusion does not move. $\mathrm{Kn}$ is of order one on both sets, so the collisional viscosity is outside its domain on both, and the magnetic rescue is needed on both. What does move is one phrase. "The mean free path is twice the system size" belongs to Module 1's values alone; on Venzmer & Bothmer's it is $0.85$ of the system size. One further reason not to mix the two: their $T$ is the proton temperature, and $\ln\Lambda$ is evaluated at the same $T$ for both sets in the table above, whereas an electron temperature is what belongs in $\ln\Lambda$.

6.4 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 three consecutive modules of this book. The remedy adopted here is not a resolution to be careful but a printed table: every speed, dispersion and width this module uses, with the number of components each one averages over and the reason that count is right for its job. No two of them are compared anywhere in this module without a reader having been given this table first.

quantityformulacomponentswhere it is used, and why that count
mean thermal speed $\bar v$$\sqrt{8k_{\rm B}T/\pi m}$3-D mean of the speed$\nu = \lambda\bar v/3$ in kinetic theory, and $\mathrm{Ma_{th}} = U/\bar v$ in (1.2). It is the average of $|\mathbf{v}|$, not of $v_x$
rms thermal speed $v_{\rm rms}$$\sqrt{3k_{\rm B}T/m}$3-D rmsBraginskii's $\tau_i = \lambda/v_{\rm rms}$, his (2.5i). The ratio $\bar v/v_{\rm rms} = \sqrt{8/3\pi} = 0.9213$ is the whole of (6.8)
perpendicular thermal speed $v_\perp$$\sqrt{2k_{\rm B}T/m}$2-D rms, across $\mathbf{B}$$r_g = mv_\perp c/(ZeB)$, Module 1 Proposition 5. A gyroradius is set by the motion perpendicular to the field and by nothing else, so a two-component rms is the correct speed and not a mistake for the three-component one
Larson's $\sigma_L$fitted, $1.10\,L^{0.38}\ \mathrm{km\,s^{-1}}$3-Dhe adds it into a three-dimensional $\sigma^2$. It is the outer-scale $U$ of the cloud row in §6
Larson's thermal term$0.32\ \mathrm{km\,s^{-1}}$ at 10 K3-Dsame reason. The matching one-dimensional speed is $0.185\ \mathrm{km\,s^{-1}}$, and that is the one a one-dimensional linewidth must be compared with
Solomon's $\sigma_v$fitted, $1.0\,S^{0.5}\ \mathrm{km\,s^{-1}}$1-Done linewidth per cloud
Hitomi's $\sigma_v$measured, $164\ \mathrm{km\,s^{-1}}$1-D, line of sighta Doppler width is one component. The three-dimensional Mach number is $\sqrt3\,\sigma_v/c_s = 0.273$, against the one-dimensional $0.158$, and §8 prints both
$\sigma_s$$\sqrt{\ln(1+b^2\mathcal{M}^2)}$not a speedthe width of $s = \ln(\rho/\langle\rho\rangle)$, dimensionless. Definition 3 and Proposition 9, in §9

6.5 Reconciling this section with Module 2's table

Module 2 §9 printed $\mathrm{Kn} \approx 1.9$, $\mathrm{Ma_{th}} = 9.2$ and $\mathrm{Re} \approx 14$ for the solar wind at 1 au. This module gets $\mathrm{Re} = 12.34$ for the same gas. A reader is entitled to ask which is right, and the answer is that both are, in different conventions, and the difference factorises exactly:

The Knudsen numbers agree exactly, at $1.9363$ against its printed $1.9$, because a mean free path involves no thermal-speed convention at all. That is the practical argument for stating a Knudsen number wherever a plasma Reynolds number is quoted: it is the convention-free half of (1.2).

Module 2's intracluster row is not comparable in the same way, and no reconciliation is attempted for it: it uses $n = 10^{-2}\ \mathrm{cm^{-3}}$, $T = 3\times10^{7}$ K, $L = 100$ kpc and $u = 300\ \mathrm{km\,s^{-1}}$, which is a different gas at a different radius from the $(10^{-3}, 10^{8})$ core of §6.1. Its $\mathrm{Re}\approx580$ and this module's $0.90$ are answers to different questions.

7. The anchor: the solar wind at 1 au

This is the module's anchor, and it earns that name for a reason no other check in it can claim: one table, from one instrument, over one fit window, confirms one prediction and refutes another. A check that can only ever confirm is a weak instrument, because it cannot tell a law from a coincidence. This one is sharp enough to separate two fields of the same plasma.

Podesta, Roberts & Goldstein (2007) took Wind spacecraft data in the ecliptic at 1 au and fitted power laws to the magnetic, velocity, kinetic-energy and total-energy frequency spectra over four intervals: 1995 May, 1997 December, 2000 November and 2003 June — one near solar minimum, one near maximum, one on the ascending phase and one on the descending phase, each 54 to 81 days long. Every exponent was fitted independently of every other, with no fluid theory imposed.

Taylor's hypothesis is doing work here, and it is an assumption. A spacecraft measures a time series at a fixed point, so what is fitted is a frequency spectrum $P(f)$. Converting it to a wavenumber spectrum assumes the structures are swept past the spacecraft faster than they evolve or propagate — that the wind speed greatly exceeds the Alfvén and sound speeds. On Module 1's census values the Alfvén speed is $v_A = B/\sqrt{4\pi\rho} = 48.8\ \mathrm{km\,s^{-1}}$ with $\rho = n m_{\rm p}$, so $U/v_A = 8.2$: the wind sweeps a structure past the spacecraft about eight times faster than the structure can propagate. That is a ratio of order ten and not of order a hundred, so the assumption is good rather than exact. It is not repaid anywhere in this book. Frequencies in this section are written $f$, in hertz, and never $\nu$, which is the kinematic viscosity throughout this module.

The fit window is $10^{-3} \le f \le 10^{-2}$ Hz. Their Sect. 3 bounds the window's own effect: moving the endpoints over $7\times10^{-4}$ to $1.5\times10^{-3}$ Hz and $7\times10^{-3}$ to $1.5\times10^{-2}$ Hz changes an exponent "typically by 1%", always less than the tabulated error. The tabulated errors are $99$ per cent confidence limits, not standard deviations: divide by $2.576$ for one $\sigma$. Fig. 3 plots all eight fitted indices with their bars, against $5/3$ and against $3/2$; the table below gives the same eight numbers.

intervalmagnetic indexdeparture from $5/3$, in its own 99 % half-widthsin $\sigma$velocity indexdeparture from $5/3$, in half-widthsin $\sigma$
1 — 1995 May 23, 54 d$1.66\pm0.02$$-0.33$$-0.9$$1.50\pm0.02$$-8.33$$-21.5$
2 — 1997 Dec 14, 54 d$1.72\pm0.02$$+2.67$$+6.9$$1.59\pm0.02$$-3.83$$-9.9$
3 — 2000 Nov 15, 81 d$1.66\pm0.01$$-0.67$$-1.7$$1.52\pm0.02$$-7.33$$-18.9$
4 — 2003 Jun 11, 53.854 d$1.58\pm0.02$$-4.33$$-11.2$$1.50\pm0.02$$-8.33$$-21.5$
1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80 5/3 3/2 interval 1 interval 2 interval 3 interval 4 magnetic mean 1.6550 velocity mean 1.5275 spectral index of P(f) magnetic velocity bars are 99 per cent confidence limits; divide by 2.576 for one σ
Eight fitted exponents, and the two predictions they are being asked about. Orange is magnetic, teal is velocity; the two dashed grey lines are $5/3$, Kolmogorov's prediction from Proposition 2, and $3/2$, the Iroshnikov–Kraichnan value for a magnetised cascade, drawn in neutral grey because neither belongs to either colour. The dotted coloured lines are the two means. Error bars are 99 per cent confidence limits; divide by $2.576$ for one standard deviation. The magnetic points straddle $5/3$ — one clearly above, one clearly below — while all four velocity points lie below it, on the same side, which is the shape of the result that needs no statistics at all.

7.1 The refutation, which needs no statistics

Read the velocity column of the table before reading any mean. Every one of the four velocity indices lies below $5/3$, by between $3.8$ and $8.3$ of its own $99$ per cent half-width — that is, between $9.9$ and $21.5\sigma$ — and all four lie on the same side. Four independent intervals, spanning eight years and the whole solar cycle, each excluding the prediction by its own fit error, in the same direction.

Check 1(a) — REFUTED. One Kolmogorov exponent does not describe both fields of the solar wind. The velocity spectrum is not $k^{-5/3}$ in any of the four intervals, and the statement requires no averaging over intervals, no $n = 4$ statistics, and no assumption about how the four intervals are distributed.

Measured against $3/2$ instead, the same four velocity indices depart by $0.0$, $4.5$, $1.0$ and $0.0$ half-widths: three of the four are consistent with $3/2$ and interval 2 is not. That is weaker than "the velocity spectrum is $k^{-3/2}$", and it is what the table supports. What has been established is that $5/3$ is excluded for the velocity in all four intervals, and that $3/2$ fits three of them.

7.2 The confirmation, which is of a mean and of nothing finer

The magnetic column behaves differently, and the difference is the point of the check. The four values are $1.66$, $1.72$, $1.66$, $1.58$: their departures from $5/3$ in their own half-widths are $-0.3$, $+2.7$, $-0.7$ and $-4.3$, so one sits above the prediction and three below, and only two of the four agree with it within their own fit errors. The largest single departure is interval 4 at $4.3$ half-widths, which is $11.2\sigma$. The interval-to-interval scatter is real, and it is larger than the fit errors. Nothing in this table confirms $5/3$ interval by interval, and the prose does not say it does.

What is consistent with $5/3$ is the mean of the four, tested against the scatter of the four rather than against the fit errors:

fieldmeans.d. of the fours.e.m.vs $5/3$: $t$ (3 dof), $p$, equivalent $\sigma$vs $3/2$: $t$, $p$, equivalent $\sigma$
magnetic$1.6550$$0.0574$$0.0287$$-0.41$, $p = 0.7118$, $0.37\sigma$$+5.40$, $p = 0.0125$, $2.50\sigma$
velocity$1.5275$$0.0427$$0.0214$$-6.52$, $p = 0.0073$, $2.68\sigma$$+1.29$, $p = 0.2883$, $1.06\sigma$
Those are not sigmas, and the table says what they are. With $n = 4$ the statistic is Student's $t$ on three degrees of freedom, and a $t$ of $6.52$ on 3 dof earns $p = 0.0073$ — the equivalent of $2.7$ Gaussian $\sigma$, not $6.5$. The same correction turns the magnetic result against $3/2$, nominally "$5.4$ standard errors", into $2.5\sigma$. Any mean of a handful of measurements needs its $t$, its degrees of freedom and its $p$ printed, never a bare count of standard errors. This is why §7.1 leads with the per-interval statement, which needs none of this machinery.
Check 1(b) — CONFIRMED, of the mean. The mean magnetic index is $1.6550$ against the predicted $1.6667$: a ratio of $\mathbf{0.9930}$, and $t = -0.41$ on 3 dof, $p = 0.71$. Kolmogorov's exponent, derived in Proposition 2 from two units and no fitting, describes the mean magnetic spectrum of the solar wind to seven parts in a thousand. It is a statement about the mean of four intervals and not about any one of them, and against $3/2$ the roles swap, weakly: the magnetic mean departs at $2.50\sigma$ and the velocity mean not at all.

Two further entries of the same table, for completeness and for the caution each carries. The total-energy indices are $1.60$, $1.68$, $1.63$, $1.55$, mean $1.6150$: the total sits between the two fields, nearer the magnetic one. And the paper's Alfvén ratio $E_{\rm kin}/E_{\rm mag}$ over the fit band runs $0.50$–$1.10$, $0.40$–$0.85$, $0.50$–$0.80$ and $0.60$–$0.85$ across the four intervals. It reaches $1.10$ in interval 1, so the magnetic energy is comparable to or larger than the kinetic across these intervals, and this module does not write that it dominates them.

What is confirmed is an exponent, not a mechanism. The confirmed $5/3$ is a magnetic spectrum in an Alfvénic plasma, which is not the fluid Proposition 2 describes — and $5/3$ for the magnetic field is also what Goldreich & Sridhar (1995) predict, from a quite different argument about anisotropic critically balanced cascades. Agreement of an exponent does not select between two derivations that give the same exponent. Module 12 is where the magnetic case is done properly, and §11 lists this as the single most important entry in its table.

8. The cluster check: what "static" costs, measured

Module 3 built its hydrostatic equilibria on the idealisation it lists in its own §11 as “Static: $\mathbf{u}=\mathbf{0}$”, applied it to an intracluster medium, and repaid that idealisation through Module 4. This section prices it: in one cluster core, with one measurement, how much of the mass does a hydrostatic analysis miss because the gas is moving?

One thing Module 3 did not do, stated because it would be easy to assume otherwise. Module 3 did not name turbulence as the debt behind that row; the word does not occur in that module at all. What it did was list “Static: $\mathbf{u}=\mathbf{0}$” among its idealisations, put an intracluster medium at $r = 1$ Mpc in its §3 scale-height table, and use that row in its Problem C3. This section supplies a price for the idealisation, measured, and claims nothing about what Module 3 anticipated.

In 2016 the Hitomi satellite's calorimeter resolved the iron K line in the core of the Perseus cluster, for a few weeks before the spacecraft was lost. In the region $30$–$60$ kpc from the nucleus it measured a line-of-sight velocity dispersion

$$ \sigma_v \;=\; 164 \pm 10\ \mathrm{km\,s^{-1}} \quad (90\%\ \text{confidence}), \tag{8.1}$$

which as a one-standard-deviation error is $\pm 6.1\ \mathrm{km\,s^{-1}}$, together with $k_{\rm B}T = 4.1 \pm 0.1$ keV for the same region. An independent determination from the hydrogen-like Lyman-$\alpha$ line gives $160 \pm 16\ \mathrm{km\,s^{-1}}$, consistent with (8.1).

At $\mu_{\rm ICM} = 0.61$ and $\gamma = 5/3$ the sound speed of that gas is $1040\ \mathrm{km\,s^{-1}}$, so

$$ \mathcal{M}_{\rm 1D} \;=\; \frac{\sigma_v}{c_s} \;=\; 0.158, \qquad \mathcal{M}_{\rm 3D} \;=\; \frac{\sqrt3\,\sigma_v}{c_s} \;=\; 0.273. \tag{8.2}$$

Both are printed because both are used, and the factor $\sqrt3$ between them is the component count of §6.4: a Doppler width is one component and a Mach number of a flow is three. The gas is subsonic on either reading, which is what makes the incompressible theory of §2 the right one to apply here and the wrong one to apply in §10.

8.1 The turbulent pressure, and what the density does not do

For isotropic turbulence the turbulent pressure is $\tfrac13\rho\langle v^2\rangle = \tfrac13\rho(3\sigma_v^2) = \rho\sigma_v^2$, and the thermal pressure of an ideal gas is $\rho k_{\rm B}T/(\mu m_{\rm u})$. The ratio is

$$ \alpha \;\equiv\; \frac{P_{\rm turb}}{P_{\rm th}} \;=\; \frac{\mu m_{\rm u}\sigma_v^2}{k_{\rm B}T} \;=\; 4.15\ \text{per cent}. \tag{8.3}$$
The density cancels; the mean mass per particle does not. That is why a line width and a temperature suffice, with no density measurement anywhere — and it is why $\mu = 0.61$ must be named every time (8.3) is used, including in every problem statement that uses it, because the ratio is linear in it.
What (8.3) is, and three things it is not — kept apart deliberately.
  1. It is a validation of the arithmetic, not a check against nature. The paper's own headline is "turbulent pressure support in the gas is 4% or less of the thermodynamic pressure". The $4.15$ per cent above reproduces that from the paper's own dispersion and the paper's own temperature. Nothing independent is being tested — the same treatment Module 5 gave Kandori's contrast column, where the label reads in full “A validation, not a check.”
  2. The "or less" is inherited, not established here. Both numbers come from the same $164\ \mathrm{km\,s^{-1}}$, which the paper says its pixel self-calibration and its angular resolution each push up. So $4.15$ per cent is an upper-ish value for exactly the reason $4$ per cent is, and this module does not treat "or less" as a bound it has independently obtained.
  3. No ratio may be printed to three decimals. "4%" carries one significant figure and spans $3.5$ to $4.5$ per cent, against which $4.15$ gives a ratio anywhere from $0.92$ to $1.18$. A ratio quoted to three decimals would claim that a discrepancy had been measured when none has been.

8.2 The mass a hydrostatic analysis misses

Proposition 7 (the hydrostatic mass bias). Suppose the gas is supported against gravity by $P_{\rm tot} = P_{\rm th} + P_{\rm turb}$, and suppose $\alpha = P_{\rm turb}/P_{\rm th}$ is constant with radius. Then an observer who infers a mass from the thermal pressure gradient alone obtains $M_{\rm th}$, whereas the true mass is $M = (1+\alpha)M_{\rm th}$, and the fraction of the mass missed is $$ \frac{M - M_{\rm th}}{M} \;=\; \frac{\alpha}{1+\alpha}. \tag{8.4}$$ The constant-$\alpha$ assumption is stated before the number, not after it.
Hydrostatic balance for the total pressure is $\mathrm{d}P_{\rm tot}/\mathrm{d}r = -GM\rho/r^2$, which is Module 3's equation with the turbulent term added. If $\alpha$ does not depend on $r$ then $P_{\rm tot} = (1+\alpha)P_{\rm th}$ and $\mathrm{d}P_{\rm tot}/\mathrm{d}r = (1+\alpha)\,\mathrm{d}P_{\rm th}/\mathrm{d}r$, so $$ -\frac{GM\rho}{r^2} \;=\; (1+\alpha)\frac{\mathrm{d}P_{\rm th}}{\mathrm{d}r} \;=\; -(1+\alpha)\frac{GM_{\rm th}\rho}{r^2}, $$ giving $M = (1+\alpha)M_{\rm th}$ and hence (8.4). Drop the constant-$\alpha$ assumption and the result changes in kind, not in size: $\mathrm{d}P_{\rm tot}/\mathrm{d}r = (1+\alpha)\mathrm{d}P_{\rm th}/\mathrm{d}r + P_{\rm th}\,\mathrm{d}\alpha/\mathrm{d}r$, so the bias acquires a term in the difference between the two pressures' logarithmic scale lengths, and a single dispersion measured in a single annulus cannot supply it. ∎
Check 3 — CONFIRMED: the price of one of Module 3's idealisations. At $\alpha = 4.15$ per cent, (8.4) gives a hydrostatic mass bias of $\mathbf{3.98}$ per cent — and $\mathbf{7.66}$ per cent if the large-scale shear the paper says could "at most double" the estimate does so. Treating this cluster core as static costs about four per cent of its mass, or about eight in the worst case the observers themselves allow. The idealisation is a good one, and now it is a good one with a number attached.
Scope, stated: one pointing, one annulus, one cluster. Hitomi observed $30$–$60$ kpc in one cluster core, and the paper itself restricts the claim to "the central regions". Cluster masses used for cosmology are measured near $r_{500}$, far outside this radius, where no comparable measurement exists. Nothing in this section licenses a four-per-cent correction to a cluster mass function.

Three further numbers from the same observation, each of which cuts against reading (8.3) too confidently. The central region — which contains the active galactic nucleus — gives $187 \pm 13\ \mathrm{km\,s^{-1}}$ at $3.8$ keV, so $\alpha = 5.82$ per cent there: higher, and not cleanly attributable to a cascade. A velocity gradient of $150 \pm 70\ \mathrm{km\,s^{-1}}$ runs across the $60$ kpc image, which is bulk motion rather than turbulence and is the shear the "at most doubling" refers to. And the systematic budget on the $164$ is energy resolution $\pm6$, temperature $\pm2$ and calibration scatter $-3\ \mathrm{km\,s^{-1}}$ — all smaller than the $\pm10$ statistical limit, and all quoted by the collaboration.

9. Supersonic turbulence, and the density it makes

Everything so far has assumed the flow is incompressible, and this is the first place the assumption is named. Proposition 2's dimensional argument has no sound speed in it, the $4/5$ law is derived for an incompressible fluid, and the intracluster medium of §8 is subsonic at $\mathcal{M}_{\rm 3D} = 0.273$, which is why it was the right place to apply the theory. A molecular cloud is not subsonic. Its turbulent speeds exceed its sound speed by a factor of ten, and this section works out what changes.

Definition 3 (sonic Mach number, and the log density). $\mathcal{M} \equiv U/c_s$, with $c_s$ the adiabatic sound speed of Module 4 and $U$ the root-mean-square turbulent speed — not a bulk speed and not a one-component dispersion. And $$ s \;\equiv\; \ln\!\left(\frac{\rho}{\langle\rho\rangle}\right), \tag{9.1}$$ the logarithm of the density in units of its mean, which is the variable in which the statistics below are simple.

For the cloud row of §6 — $10$ K, $\mu = 2.33$, $\gamma = 5/3$ — the sound speed is $c_s = 0.2439\ \mathrm{km\,s^{-1}}$, against Larson's three-dimensional $\sigma_L = 2.639\ \mathrm{km\,s^{-1}}$ at $10$ pc. So $\mathcal{M} = 10.82$: ten times the sound speed, where the intracluster medium was a quarter of it. The flow is a field of shocks, and the first thing to ask is what a field of shocks does to a spectrum.

Proposition 8 (a field of shocks gives $k^{-2}$, and $h = 1/2$). A velocity field consisting of isolated discontinuities has $$ E(k) \;\propto\; k^{-2}, \qquad \delta v(\ell) \;\propto\; \ell^{1/2}. \tag{9.2}$$
Take a single step of height $\Delta u$ at $x_0$: $u(x) = \Delta u\,H(x-x_0)$. Its Fourier transform is $$ \hat u(k) \;=\; \Delta u\,\frac{\mathrm{e}^{-\mathrm{i}kx_0}}{\mathrm{i}k}, \qquad |\hat u(k)|^2 \;=\; \frac{(\Delta u)^2}{k^2}, $$ so the energy per unit wavenumber falls as $k^{-2}$. That $|\hat u(k)|^2$ is a one-dimensional spectral density, and the $E(k)$ of (2.2) is the three-dimensional one; for a power law the two carry the same exponent and differ only by a constant factor — the isotropy factor $55/18$ of (5.2) is that constant for the Kolmogorov case — so the exponent $-2$ transfers and the prefactor does not. A superposition of randomly placed steps adds these spectra incoherently and leaves the exponent unchanged; only the prefactor counts the shocks. For the increment, consider two points a distance $\ell$ apart in a field with $n_s$ shocks per unit length. The probability that a shock lies between them is $n_s\ell$ for $n_s\ell \ll 1$, and when one does, the increment is $\Delta u$; otherwise it is the small smooth variation, which may be neglected. Then $$ \big\langle\delta v(\ell)^2\big\rangle \;\simeq\; n_s\ell\,(\Delta u)^2 \;\propto\; \ell, $$ so the root-mean-square increment scales as $\ell^{1/2}$ and $h = 1/2$, against Kolmogorov's $h = 1/3$ from (2.3). The two statements in (9.2) are consistent with each other by Proposition 10 below, which is the general bridge between an increment exponent and a spectral one. ∎

Two exponents are now in play — $5/3$ from a smooth cascade and $2$ from a field of shocks — and Fig. 1 drew them together at the start of this module so that the difference would be visible: a factor $2.15$ per decade. §10 measures which one a molecular cloud obeys.

9.1 What supersonic turbulence does to the density

Incompressible turbulence has no density statistics to speak of. Supersonic turbulence has nothing else: shocks compress gas, and the distribution of densities they produce is what makes stars.

Definition 4 (the forcing parameter). $b$ is the ratio of the fractional density dispersion to the Mach number, $\sigma_\rho/\langle\rho\rangle = b\mathcal{M}$. It runs from about $1/3$ for purely solenoidal (divergence-free) forcing in three dimensions to about $1$ for purely compressive (curl-free) forcing. Both endpoints are stated, from Federrath et al. (2010), their Sect. 5 and Fig. 8; this module quotes them and computes nothing about the forcing itself. The value $1/3$ is a three-dimensional statement: solenoidal forcing drives one of the three components compressively by a cascade, and the fraction is $1/3$ for that reason.
Proposition 9 (the log-normal density distribution). In isothermal supersonic turbulence $s$ of (9.1) is normally distributed with variance $$ \sigma_s^2 \;=\; \ln\!\left(1 + b^2\mathcal{M}^2\right) \tag{9.3}$$ — stated, Federrath et al. (2010) equations (18) and (19) — and with mean $$ s_0 \;=\; -\tfrac12\sigma_s^2, \tag{9.4}$$ which is forced, not fitted. The fraction of the volume with $\rho/\langle\rho\rangle$ above a threshold $x$ is then $$ f(\gt x) \;=\; \tfrac12\,\mathrm{erfc}\!\left[\frac{\ln x - s_0}{\sigma_s\sqrt2}\right]. \tag{9.5}$$
Equation (9.3) is quoted; it is a fit to simulations, not a theorem, and the module marks it as such. Equations (9.4) and (9.5) follow from it. For (9.4): if $s$ is normal with mean $s_0$ and variance $\sigma_s^2$, then $\langle\mathrm{e}^{s}\rangle = \exp(s_0 + \tfrac12\sigma_s^2)$, the standard Gaussian moment-generating result. But $\mathrm{e}^{s} = \rho/\langle\rho\rangle$ by (9.1), whose volume average is $1$ by construction. Setting $\exp(s_0+\tfrac12\sigma_s^2) = 1$ gives $s_0 = -\tfrac12\sigma_s^2$. The mean of $s$ is therefore not an independent parameter: the most probable density is below the mean density, and by an amount fixed entirely by the width. For (9.5): $f(\gt x) = \int_{\ln x}^{\infty} (2\pi\sigma_s^2)^{-1/2}\exp[-(s-s_0)^2/2\sigma_s^2]\,\mathrm{d}s$. Substituting $t = (s-s_0)/(\sigma_s\sqrt2)$ turns this into $\pi^{-1/2}\int_{t_0}^\infty \mathrm{e}^{-t^2}\mathrm{d}t = \tfrac12\mathrm{erfc}(t_0)$ with $t_0 = (\ln x - s_0)/(\sigma_s\sqrt2)$. ∎
$\mathcal{M}$$b$$\sigma_s$$\sigma_\rho/\langle\rho\rangle = b\mathcal{M}$$s_0$most probable $\rho/\langle\rho\rangle$
$1$$0.333$$0.3246$$0.33$$-0.053$$0.9487$
$1$$1.000$$0.8326$$1.00$$-0.347$$0.7071$
$5$$0.333$$1.1529$$1.67$$-0.665$$0.5145$
$5$$1.000$$1.8050$$5.00$$-1.629$$0.1961$
$10$$0.333$$1.5793$$3.33$$-1.247$$0.2873$
$10$$1.000$$2.1483$$10.00$$-2.308$$0.0995$
$20$$0.333$$1.9536$$6.67$$-1.908$$0.1483$
$20$$1.000$$2.4483$$20.00$$-2.997$$0.0499$

At $\mathcal{M} = 10$ the width of $s$ is $1.5793$ for purely solenoidal forcing and $2.1483$ for purely compressive — a ratio of only $1.36$. The consequence is not in the width. It is in the tail, and Fig. 4 is drawn with a logarithmic lower panel for that reason.

-8 -6 -4 -2 0 2 4 6 0.0 0.1 0.2 10-6 10-4 10-2 b = 1/3, solenoidal, σs = 1.5793 b = 1, compressive, σs = 2.1483 same two curves, logarithmic axis: the tail is where the difference lives ρ > 100 ⟨ρ⟩ volume beyond the line: 6.46×10−4 against 1.05×10−4, a factor 6.1 s = ln(ρ/⟨ρ⟩), at Mach 10 p(s) p(s)
A width ratio of $1.36$ becomes a tail ratio of $6.1$. The probability density of $s = \ln(\rho/\langle\rho\rangle)$ at $\mathcal{M} = 10$, for solenoidal forcing ($b = 1/3$, teal) and compressive forcing ($b = 1$, orange). Both are centred below $s = 0$ by (9.4), which is why the most probable density is less than the mean. The upper panel is linear and shows the widths; the lower panel plots the same two curves on a logarithmic axis, with the same colour key, because on a linear axis the region that matters is indistinguishable from zero. The shaded band is $\rho \gt 100\langle\rho\rangle$, and the fraction of the volume inside it is $6.46\times10^{-4}$ for compressive forcing against $1.05\times10^{-4}$ for solenoidal.
threshold $x = \rho/\langle\rho\rangle$$f(\gt x)$ at $b = 1/3$$f(\gt x)$ at $b = 1$ratio
$10$$1.230\times10^{-2}$$1.594\times10^{-2}$$1.3$
$100$$1.054\times10^{-4}$$6.459\times10^{-4}$$6.1$
$1000$$1.211\times10^{-7}$$8.949\times10^{-6}$$73.9$
Why the forcing matters, in one line. At a hundred times the mean density, compressive forcing puts $6.1$ times as much volume above the threshold as solenoidal forcing does — and at a thousand times, $73.9$ times as much — from a distribution that is only $1.36$ times wider. Star formation happens in that tail. A quantity that is insensitive to the forcing in the bulk can be enormously sensitive to it in the tail, and that is a property of the log-normal rather than of any particular cloud.

10. The interstellar check: the size–linewidth exponent

The last check asks whether the smooth cascade of §2 describes a molecular cloud — specifically, whether its size–linewidth exponent is Kolmogorov's $h = 1/3$ — and the answer is no. It is the only refutation in this module that goes against K41 itself rather than against one of its hypotheses, and the way it is obtained needs a bridge between two different kinds of statistic, which is the proposition that follows.

Proposition 10 (from an increment exponent to a spectral exponent). Suppose $E(k)$ is a power law $Ak^{-p}$ with $p \gt 1$ over a range of wavenumbers, and let $\delta v(\ell)$ be the root-mean-square increment across the corresponding separations. Then $\delta v(\ell) \propto \ell^{h}$ with $h = (p-1)/2$, equivalently $$ E(k) \;\propto\; k^{-(1+2h)}. \tag{10.1}$$ In particular $h = 1/3$ gives $5/3$ and $h = 1/2$ gives $2$. The power-law hypothesis is the one that cannot be dropped: a measured increment exponent fixes a spectral exponent only if there is a spectral power law to fix, and §10 measures the increment exponent, not the spectrum.
The velocity increment across a separation $\ell$ samples all wavenumbers above $k = 1/\ell$, so $$ \delta v(\ell)^2 \;\simeq\; \int_{1/\ell}^{\infty} E(k)\,\mathrm{d}k. $$ Put $E(k) = Ak^{-p}$ with $p \gt 1$, which is required for the integral to converge at the upper limit. Then $$ \int_{1/\ell}^{\infty}Ak^{-p}\,\mathrm{d}k \;=\; \frac{A}{p-1}\,\ell^{\,p-1}, $$ so $\delta v \propto \ell^{(p-1)/2}$. Matching to $\ell^{h}$ gives $h = (p-1)/2$, that is $p = 1+2h$, which is (10.1). Substituting $h = 1/3$ gives $p = 5/3$, recovering Proposition 2, and $h = 1/2$ gives $p = 2$, recovering Proposition 8. ∎
This module measures an exponent, and infers a spectrum. What §10 compares against the sky is $h$, a size–linewidth exponent. The corresponding $E(k) \propto k^{-2}$ follows through (10.1) and through the assumption stated below about two-point versus ensemble statistics. No interstellar power spectrum is measured anywhere in this module, and the only measured spectral exponents in it are Podesta's eight, in §7.

10.1 Two fits, forty years apart

Larson (1981) collected linewidths and sizes from the literature and drew a line through them:

$$ \sigma_L \;=\; 1.10\,L^{0.38}\ \mathrm{km\,s^{-1}}, \qquad 0.1 \lesssim L \lesssim 100\ \mathrm{pc}, \tag{10.2}$$

with $\sigma_L$ three-dimensional and no error bar of any kind: the paper calls it an eye-fitted line, and its only stated scatter is $\mathrm{rms}(\log\sigma) = 0.14$, a factor of $1.38$ in $\sigma$. The exponent $0.38$ lies $+0.0467$ from Kolmogorov's $1/3$ and $-0.1200$ from Burgers' $1/2$. Larson drew the obvious conclusion himself, on his p. 816: "it is perhaps remarkable that the observed relation $\sigma \propto L^{0.38}$ is so close to the Kolmogoroff law $\sigma \propto L^{0.33}$."

Solomon, Rivolo, Barrett & Yahil (1987) did the measurement properly, with one survey, one distance scale and one cloud-finding algorithm applied to $273$ Galactic molecular clouds:

$$ \sigma_v \;=\; (1.0 \pm 0.1)\,S^{\,0.5\,\pm\,0.05}\ \mathrm{km\,s^{-1}}, \tag{10.3}$$

a least-squares fit, with $\sigma_v$ one-dimensional, $S$ the cloud size in parsecs, and a dispersion in $\log\sigma_v$ of $\pm0.11$ dex.

What the $\pm0.05$ is, established before it is used for anything. Their p. 737: "The quoted error in the exponent is larger than the formal fit error and represents an estimate of the systematic error based on several fits to the data with varying definitions of cloud boundaries." It is a spread over where a cloud ends. It is not a standard deviation, and no $\sigma$ is quoted from it anywhere in this module. What is printed instead is the gap measured in whole envelopes.

10.2 The comparison, and what it does and does not decide

The assumption, stated before the verdict and not after it. $\delta v(\ell) \sim \ell^{1/3}$ is a two-point statistic inside one flow: the typical velocity difference between two points a distance $\ell$ apart. Solomon's exponent comes from a regression across 273 different clouds, each contributing one linewidth and one size. Reading the second as the first assumes that a cloud's internal dispersion samples the same cascade a structure function would, and that the cloud-to-cloud ensemble behaves statistically as one object. Larson made that identification explicitly and it is standard practice. It is still an assumption, and the refutation below depends on it.
Check 4 — REFUTED. The measured exponent is $0.50$. Kolmogorov's $1/3$ lies $0.1667$ below it, which is $3.33$ times the whole systematic envelope of $\pm0.05$; Burgers' $1/2$ lies on it, a gap of $0.0000$ envelopes. Incompressible Kolmogorov scaling does not describe a molecular cloud. A gap of three envelopes is a refutation rather than a demolition, and the module prints the gap rather than a confidence level it cannot compute.

The statistical case is stronger than the number printed, not weaker. The formal fit error is smaller than the systematic envelope — the authors say so — so quoting the envelope is the conservative choice. And the authors reach the same verdict in their own words, on the same page: "Larson (1981), from a literature search, suggested that the observed exponent was close to 1/3 and reflected a Kolmogorov turbulent spectrum. This hypothesis is ruled out by the data presented here."

Why Solomon rather than Larson, when the two exponents differ. Larson's $0.38$ is much closer to $1/3$, and if it were the number in hand the verdict would be different. It is shown here beside Solomon's $0.50$ so that a reader sees both, and the reason this module weights Solomon's is stated rather than assumed: Larson's line is eye-fitted from a literature compilation with no error bar of any kind, its only stated scatter being a factor of $1.38$ in $\sigma$; Solomon's is a least-squares fit to $273$ clouds from one survey with a stated error envelope. Larson's exponent is the reason the question was asked. Solomon's is the reason it can be answered.

The exponent refutes Kolmogorov. It does not select Burgers. $\sigma_v \propto S^{1/2}$ is also exactly what virial equilibrium at constant surface density gives: $\sigma^2 \sim GM/S$ with $M \sim \Sigma S^2$ yields $\sigma \sim (G\Sigma)^{1/2}S^{1/2}$, independent of any cascade. Solomon et al. read their own result that way, their constant mean surface density being $170\ M_\odot\,\mathrm{pc^{-2}}$. Two quite different physical pictures predict the same exponent, and this module does not claim the measurement distinguishes them.

An order-of-magnitude cross-check of that reading, and its limit. Putting their normalisation $\sigma_v(1\ \mathrm{pc}) = 1.0\ \mathrm{km\,s^{-1}}$ into $\sigma_{\rm 1D}^2 = \pi G\Sigma S/3$ gives $\Sigma = 222\ M_\odot\,\mathrm{pc^{-2}}$ against the $170$ they quote, a ratio of $1.31$. The factor $\pi/3$ is this book's choice of a uniform sphere, not Solomon et al.'s virial coefficient, so a ratio of order one is all the comparison can show, and it is quoted for no more than that.

10.3 One sensitivity, and the Mach numbers that follow

Solomon et al. also state, on the same p. 737, what their result would be without the $2\ \mathrm{km\,s^{-1}}$ velocity extrapolation they apply to their line profiles: the constant of (10.3) becomes $0.83$ rather than $1.0$, a reduction of $17$ per cent. That moves every Mach number below by $17$ per cent and leaves the exponent — and therefore Check 4 — untouched.

cloud size $S$$\sigma_v$ from (10.3), 1-D$\mathcal{M}_{\rm 1D} = \sigma_v/c_s$ at $c_s = 0.244\ \mathrm{km\,s^{-1}}$
$1$ pc$1.00\ \mathrm{km\,s^{-1}}$$4.10$
$10$ pc$3.16\ \mathrm{km\,s^{-1}}$$12.97$
$100$ pc$10.00\ \mathrm{km\,s^{-1}}$$41.00$

Every entry is supersonic, which is the physical reason the incompressible theory should not have been expected to hold. And the comparison is done in matched components: these are one-dimensional linewidths against a one-dimensional sound-speed ratio. Larson's own thermal term, $0.32\ \mathrm{km\,s^{-1}}$, is three-dimensional because he adds it into a three-dimensional $\sigma^2$; the matching one-dimensional speed is $0.185\ \mathrm{km\,s^{-1}}$, and that is the one a one-dimensional linewidth must be compared with. Against either, every row above is supersonic by a wide margin.

11. What this theory captures, and what it misses

The theory of this module is Kolmogorov's 1941 argument, which assumes an incompressible, homogeneous, isotropic, unmagnetised fluid at infinite Reynolds number with scale-invariant statistics. Not one of those adjectives describes any of the four fluids it was applied to. The table names each idealisation, what it costs, and where — or whether — it is repaid.

Idealisation made hereWhat it costsWhere it is repaid
Incompressible, homogeneous, isotropic turbulence (Propositions 2–5)It excludes exactly the regime §9 and §10 are about. A molecular cloud runs at $\mathcal{M} = 10.82$§9 in part, through Burgers; not fully repaid in this book
No magnetic field in the K41 argumentLarger than any other row. Check 1's confirmed exponent is a magnetic spectrum in an Alfvénic plasma, and $5/3$ for $\mathbf{B}$ is also what Goldreich & Sridhar (1995) predict from a quite different argument. What is confirmed is the exponent, not the mechanismModule 12 (magnetohydrodynamics)
Taylor's frozen-flow hypothesis, in Check 1The spacecraft measures a frequency spectrum; converting it to a wavenumber spectrum assumes the wind speed greatly exceeds the wave speeds. At 1 au it exceeds the Alfvén speed by $8.2$, which is good and not exactnot repaid; named in §7 where it is used
A Navier–Stokes viscosity, in $\mathrm{Re} = UL/\nu$At $\mathrm{Kn}$ of order one there is none. The collisional Reynolds numbers of the intracluster medium and the solar wind are formulae outside their own domain, and Proposition 6 says so in one lineProposition 6 inside this module, and Module 12 for the field that rescues it
The effective viscosity of the intracluster mediumIt is set by pressure-anisotropy microinstabilities, not by Coulomb collisions. Zhuravleva et al. (2019) bound the suppression at $10$ to $1000$ and leave $\mathrm{Pr}$ freeNobody repays it. This book contains no module that computes it, and this row says so rather than pointing at one that will not deliver
$\mathrm{Re}^{3}$ as the cost of a simulationIt makes direct numerical simulation of any astrophysical flow impossible. The $10$ pc cloud alone needs $6.24\times10^{17}$ grid pointsModule 14 (numerical methods)
One exponent, no intermittencyK41 assumes the statistics are scale-invariant. They are not: the higher-order structure functions do not follow $S_n \propto \ell^{n/3}$, which is why K62 and the multifractal models existNot in this book. Named here and named as absent
Constant $P_{\rm turb}/P_{\rm th}$ with radius, in Proposition 7Drop it and the mass bias depends on the two pressures' scale lengths, which one annulus cannot supplynot repaid; named where the number is printed, in §8.2
A single $\nu$ for air, quoted to two figuresEvery quantity in (3.3) and in Problem C1 inherits two significant figures, because no primary source for $0.15\ \mathrm{cm^2\,s^{-1}}$ was obtainednot repaid; stated once in §3 and in Sources

Read as a whole, the table says something that is worth saying plainly. Of the four checks in this module, the one that confirms K41 most cleanly is the one whose physical setting K41 least describes. The magnetic spectrum of a collisionless magnetised plasma at $\mathrm{Kn} \approx 1.9$ has no business obeying an incompressible hydrodynamic argument, and its mean exponent matches to seven parts in a thousand. The molecular cloud, which at least is a fluid in the sense of Module 1, does not obey it at all. A reader who wants one sentence to carry out of this module should take that one, and Module 12 is where the first half of it is explained.

12. Problems

Nine problems in the house set: three conceptual (C), three derivational (D), three computational (K). Every numerical answer below is produced by afd/figs/m10_numbers.py and independently verified by afd/figs/m10_problems_check.py, which imports nothing from it and rebuilds every modelling input from the problem statement.

C1. Take air near the ground with kinematic viscosity $\nu = 0.15\ \mathrm{cm^2\,s^{-1}}$ — a two-significant-figure value for which this book obtained no primary source — stirred at $\varepsilon = 10^{-3}\ \mathrm{W\,kg^{-1}}$. First convert $\varepsilon$ to CGS, then find $\eta$, $u_\eta$ and $\tau_\eta$, and verify (3.2). To how many figures may the three be quoted?
Solution

$1\ \mathrm{W\,kg^{-1}} = 10^{7}\ \mathrm{erg\,s^{-1}}/10^{3}\ \mathrm{g} = 10^{4}\ \mathrm{erg\,g^{-1}\,s^{-1}}$, so $\varepsilon = 10^{-3}\ \mathrm{W\,kg^{-1}} = 10\ \mathrm{erg\,g^{-1}\,s^{-1}} = 10\ \mathrm{cm^2\,s^{-3}}$. The conversion is worth doing explicitly: a factor $10^4$ mislaid here moves $\eta$ by $10$.

Then $\eta = (\nu^3/\varepsilon)^{1/4} = (0.15^3/10)^{1/4} = 0.1355\ \mathrm{cm} = 1.355$ mm; $u_\eta = (\nu\varepsilon)^{1/4} = (1.5)^{1/4} = 1.11\ \mathrm{cm\,s^{-1}}$; $\tau_\eta = (\nu/\varepsilon)^{1/2} = (0.015)^{1/2} = 0.12247\ \mathrm{s} = 122.47$ ms. Check: $u_\eta\eta/\nu = 1.11\times0.1355/0.15 = 1.000000$, as (3.2) requires identically.

Two figures. $\nu$ carries two and every one of the three is a power of it, so $\eta \approx 1.4$ mm, $u_\eta \approx 1.1\ \mathrm{cm\,s^{-1}}$, $\tau_\eta \approx 0.12$ s is the honest statement. The extra digits above are carried only so the identity closes to $1.000000$; they are not accuracy.

C2. A molecular cloud $10$ pc across has Larson's three-dimensional velocity dispersion $\sigma_L = 2.64\ \mathrm{km\,s^{-1}}$ and, at $T = 10$ K with $\mu = 2.33$ and $\gamma = 5/3$, a sound speed $c_s = 0.244\ \mathrm{km\,s^{-1}}$. Find the eddy turnover time at the outer scale, and the time sound would take to cross the cloud. Which is the relevant dynamical time, and why?
Solution

$10\ \mathrm{pc} = 3.0857\times10^{19}$ cm. Turnover: $L/\sigma_L = 3.0857\times10^{19}/2.64\times10^{5} = 1.169\times10^{14}\ \mathrm{s} = 3.70$ Myr. Sound crossing: $L/c_s = 3.0857\times10^{19}/2.439\times10^{4} = 1.265\times10^{15}\ \mathrm{s} = 40.1$ Myr.

The ratio of the two is the Mach number, $10.82$, which is the cleanest way to see that $\mathcal{M}$ is a ratio of times as much as of speeds. The turnover time is the relevant one: it is the time in which the outer eddy delivers its energy to the cascade, and therefore the time in which the cloud's structure is rearranged. Sound is ten times too slow to mediate anything.

The three-dimensionality of $\sigma_L$ matters here. Larson adds his dispersion into a three-dimensional $\sigma^2$, so $2.64\ \mathrm{km\,s^{-1}}$ is the full speed and not one component. Using a one-dimensional linewidth in its place would give a turnover time $\sqrt3$ too long.

C3. For the same cloud, take $\varepsilon = U^3/L$ with $U = \sigma_L$ — note that this is a K41 relation and not a measurement — and evaluate the four-fifths law (4.1) at $\ell = 1$ pc. Then compare $|S_3|^{1/3}$ with $(\varepsilon\ell)^{1/3}$, and say what the ratio is without computing it a second time.
Solution

$\varepsilon = (2.64\times10^{5})^3/3.0857\times10^{19} = 5.963\times10^{-4}\ \mathrm{erg\,g^{-1}\,s^{-1}}$. At $\ell = 1\ \mathrm{pc} = 3.0857\times10^{18}$ cm,

$S_3 = -\tfrac45\varepsilon\ell = -0.8\times5.963\times10^{-4}\times3.0857\times10^{18} = -1.472\times10^{15}\ \mathrm{cm^3\,s^{-3}}$, so $|S_3|^{1/3} = 1.138\ \mathrm{km\,s^{-1}}$, against $(\varepsilon\ell)^{1/3} = 1.225\ \mathrm{km\,s^{-1}}$.

The ratio is $(4/5)^{1/3} = 0.9283$, and it could have been written down before any arithmetic, because both quantities are the cube root of the same product and differ only by the factor $4/5$. The dimensional estimate (2.3) and the exact law (4.1) agree on the scaling and differ by eight per cent in the speed — which is the sense in which "$\delta v \sim (\varepsilon\ell)^{1/3}$" is a good estimate rather than a loose one.

The minus sign says the distribution of longitudinal increments is skewed toward approach, which is the direction of the cascade. Note also what has been assumed to get here: $\varepsilon = U^3/L$ is (2.3) at the outer scale, so this problem has used K41 to evaluate the one result in the subject that does not depend on K41.

D1. An intracluster medium has $n = 10^{-3}\ \mathrm{cm^{-3}}$, $T = 10^{8}$ K and $\ln\Lambda = 37.8$, giving a Coulomb mean free path of $22.5$ kpc and an ion mean thermal speed of $1450\ \mathrm{km\,s^{-1}}$. Turbulence of $U = 164\ \mathrm{km\,s^{-1}}$ fills a region $L = 60$ kpc across. Compute $\varepsilon$ and the dissipation scale $\eta$ using the unmagnetised viscosity $\nu = \lambda\bar v_i/3$. Then repeat with the magnetised bound $\nu_\perp \le r_g\bar v_i/3$ at $B = 1\ \mu$G. Before quoting either answer, check whether the first one is admissible.
Solution

$\nu = \lambda\bar v_i/3 = (6.943\times10^{22})(1.450\times10^{8})/3 = 3.3557\times10^{30}\ \mathrm{cm^2\,s^{-1}}$ and $\varepsilon = U^3/L = (1.64\times10^{7})^3/(1.8514\times10^{23}) = 2.382\times10^{-2}\ \mathrm{erg\,g^{-1}\,s^{-1}}$. Then $\eta = (\nu^3/\varepsilon)^{1/4} = 1.995\times10^{23}\ \mathrm{cm} = 64.67$ kpc.

That answer must be rejected, and rejecting it is the point of the problem. $\eta/L = 64.67/60 = 1.078 \gt 1$: the scale at which viscosity is supposed to take over is larger than the system. There is no inertial range, so (2.1) is empty and $\eta$ is not a dissipation scale but a demonstration that the premise is wrong. The same conclusion arrives from the other direction as $\mathrm{Kn} = 22.5/60 = 0.375$, and the two are one statement by (1.2). Note also that $\varepsilon = U^3/L$ is itself a K41 relation, so the calculation used a cascade to locate the end of a cascade it has just shown cannot exist.

With the field: $r_g = 1.341\times10^{10}$ cm at $1\ \mu$G, so $\nu_\perp \le r_g\bar v_i/3 = 6.4824\times10^{17}\ \mathrm{cm^2\,s^{-1}}$, and $\eta \ge (\nu_\perp^3/\varepsilon)^{1/4} = 5.815\times10^{13}\ \mathrm{cm} = 3.887$ au. That is $3.1\times10^{-10}$ of the system size, so the inertial range now spans nine and a half orders of magnitude. Both signs must be carried: $\nu_\perp$ is an upper bound, so $\eta$ is a lower bound and $\mathrm{Re}_\perp \ge 4.684\times10^{12}$ is a lower bound too. Neither is a value.

D2. In isothermal supersonic turbulence at root-mean-square sonic Mach number $\mathcal{M} = 10$, compute $\sigma_s$ from (9.3) and the fraction of the volume with $\rho \gt 100\langle\rho\rangle$ from (9.5), for the two forcing extremes: $b = 1/3$ (purely solenoidal in three dimensions) and $b = 1$ (purely compressive), both from Federrath et al. (2010). Comment on the ratio of the widths against the ratio of the tails.
Solution

$\sigma_s = \sqrt{\ln(1+b^2\mathcal{M}^2)}$. At $b = 1/3$: $b\mathcal{M} = 3.333$, $1+b^2\mathcal{M}^2 = 12.11$, $\sigma_s = 1.5793$. At $b = 1$: $b\mathcal{M} = 10$, $1+100 = 101$, $\sigma_s = 2.1483$. The widths differ by a factor $1.36$.

$s_0 = -\sigma_s^2/2$ is $-1.247$ and $-2.308$. With $\ln 100 = 4.605$, equation (9.5) gives $t_0 = (4.605+1.247)/(1.5793\sqrt2) = 2.620$ and $(4.605+2.308)/(2.1483\sqrt2) = 2.275$, so $f(\gt100) = \tfrac12\mathrm{erfc}(2.620) = 1.054\times10^{-4}$ and $\tfrac12\mathrm{erfc}(2.275) = 6.459\times10^{-4}$.

A factor $1.36$ in the width becomes a factor $6.1$ in the tail — and at $\rho \gt 1000\langle\rho\rangle$, a factor $73.9$. The reason is that the tail of a Gaussian is controlled by the exponent $-t_0^2$, and $t_0$ is inversely proportional to $\sigma_s$; a small change in the denominator is a large change after exponentiation. Since star formation is a threshold process in the far tail, the forcing that a cloud experiences matters far more for its star-formation rate than for any bulk property of its density field.

Note that $\sigma_s$ is dimensionless and is not a speed; it is the width of $\ln(\rho/\langle\rho\rangle)$, and comparing it with any of the speeds of §6.4 would be meaningless.

D3. A molecular cloud has $n = 10^{2}\ \mathrm{cm^{-3}}$ of molecular hydrogen at $T = 10$ K with $\mu = 2.33$, a neutral cross-section $\sigma_{\rm H} = 10^{-15}\ \mathrm{cm^2}$, a size $L = 10$ pc and Larson's three-dimensional dispersion $\sigma_L = 2.64\ \mathrm{km\,s^{-1}}$ as its outer-scale speed. Compute $\mathrm{Re}$, then $L/\eta$, two ways.
Solution

$\lambda = 1/(n\sigma_{\rm H}) = 10^{13}\ \mathrm{cm} = 0.67$ au and $\bar v = \sqrt{8k_{\rm B}T/\pi\mu m_{\rm u}} = 0.3014\ \mathrm{km\,s^{-1}}$, so $\nu = \lambda\bar v/3 = 1.0048\times10^{17}\ \mathrm{cm^2\,s^{-1}}$ and $$\mathrm{Re} = \frac{(2.64\times10^{5})(3.0857\times10^{19})}{1.0048\times10^{17}} = 8.107\times10^{7}.$$

By (3.4), $L/\eta = \mathrm{Re}^{3/4} = (8.107\times10^{7})^{3/4} = 8.54\times10^{5}$. Directly: $\varepsilon = 5.963\times10^{-4}$ from C3 gives $\eta = (\nu^3/\varepsilon)^{1/4} = 3.6\times10^{13}\ \mathrm{cm} = 2.4$ au, and $3.0857\times10^{19}/3.6\times10^{13} = 8.5\times10^{5}$. The two agree, as they must: (3.4) was derived by putting $\varepsilon = U^3/L$ into the definition of $\eta$.

Check the licence before trusting the number. $\mathrm{Kn} = 10^{13}/3.0857\times10^{19} = 3.2\times10^{-7}$, six orders of magnitude inside the Chapman–Enskog domain — the $6.0$ decades Fig. 2 draws for this row, measured against its $\mathrm{Kn} = 1/3$ boundary — so unlike D1 this Reynolds number is a real one. The cross-section is good to about a factor of two, so read $\mathrm{Re}$ to one significant figure; nothing in this module depends on more.

K1. Price a direct numerical simulation at $\mathrm{Re} = 10^{6}$. Give the grid side, the total number of points, and the size of one velocity snapshot storing three components at eight bytes each, so 24 bytes per point, and nothing else. Then give the number of point-updates in one outer turnover on the CFL count $\mathrm{Re}^{3}$, and say how much the softer $\mathrm{Re}^{11/4}$ count would understate it.
Solution

Side: $\mathrm{Re}^{3/4} = (10^{6})^{3/4} = 31623$ points. Total: $\mathrm{Re}^{9/4} = 3.162\times10^{13}$ points. Snapshot: $24\times3.162\times10^{13} = 7.589\times10^{14}\ \mathrm{B} = 759$ TB.

Work on the CFL count: $\mathrm{Re}^{3} = 10^{18}$ point-updates per outer turnover. On the Kolmogorov-time count: $\mathrm{Re}^{11/4} = 3.16\times10^{16}$, smaller by $\mathrm{Re}^{1/4} = 31.6$ — which is (3.6), $U/u_\eta$. The softer count is the number of turnovers of the smallest eddy; it is not the number of steps a time-explicit code takes, because the code is limited by the largest speed on the smallest cell and not by the local eddy time.

And $759$ TB is one snapshot of one field at a Reynolds number three orders of magnitude below the solar photosphere of §1. A time integrator needs at least two copies, and a magnetohydrodynamic code needs three more fields.

K2. (a) How large a line-of-sight velocity dispersion would the Perseus core need for a hydrostatic mass analysis to miss 20 per cent of the mass, at $k_{\rm B}T = 4.1$ keV and $\mu = 0.61$? Compare with Hitomi's measured $164\ \mathrm{km\,s^{-1}}$. (b) Evaluate $E(k)$ at $k = 1/(1\ \mathrm{pc})$ in the $10$ pc cloud of D3, taking the three-dimensional Kolmogorov constant $C = 1.62$ — and state why the answer should be read as a range.
Solution

(a) Invert (8.4): a bias of $0.20 = \alpha/(1+\alpha)$ needs $\alpha = 0.25$. Then (8.3) rearranges to $\sigma_v = \sqrt{\alpha\,k_{\rm B}T/(\mu m_{\rm u})}$. With $k_{\rm B}T = 4.1\ \mathrm{keV} = 6.569\times10^{-9}$ erg and $\mu m_{\rm u} = 0.61\times1.6605\times10^{-24} = 1.013\times10^{-24}$ g, $k_{\rm B}T/(\mu m_{\rm u}) = 6.485\times10^{15}\ \mathrm{cm^2\,s^{-2}}$, so $\sigma_v = \sqrt{1.621\times10^{15}} = 4.03\times10^{7}\ \mathrm{cm\,s^{-1}} = 403\ \mathrm{km\,s^{-1}}$.

That is $2.46$ times the measured $164\ \mathrm{km\,s^{-1}}$ — and, since $\alpha \propto \sigma_v^2$, a factor $2.46$ in the dispersion is a factor $6.0$ in the pressure ratio. A twenty-per-cent hydrostatic mass bias in this core is not marginally excluded; it is excluded by a factor of six in the quantity that matters. The $\mu = 0.61$ is carried explicitly because the answer scales as $\mu^{-1/2}$ and the density never enters.

(b) $E(k) = C\varepsilon^{2/3}k^{-5/3}$ with $\varepsilon = 5.963\times10^{-4}\ \mathrm{cm^2\,s^{-3}}$ and $k = 1/(3.0857\times10^{18}\ \mathrm{cm})$: $\varepsilon^{2/3} = 7.08\times10^{-3}$, $k^{-5/3} = 6.55\times10^{30}$, so $E = 1.62\times7.08\times10^{-3}\times6.55\times10^{30} = 7.503\times10^{28}\ \mathrm{cm^3\,s^{-2}}$.

Read it as a range. §5 gives the three-dimensional constant as $1.53$–$1.62$: $1.62$ from Sreenivasan's compiled $C_K = 0.53$, and $1.53$ from his own revision to $0.50$ after allowing that the local-isotropy estimate of $\varepsilon$ undercounts by about ten per cent. So $E$ at this wavenumber is $7.1$ to $7.5\times10^{28}\ \mathrm{cm^3\,s^{-2}}$, and quoting a single value to four figures would claim a precision the constant does not have. The units are worth checking: $\mathrm{cm^3\,s^{-2}}$, an energy per unit mass per unit wavenumber, as Definition 2 requires.

K3. (a) Give the Reynolds number of the intracluster medium of D1 without a magnetic field, printing the Knudsen number beside it, and then with one. (b) At what Reynolds number does a direct numerical simulation need $N = 10^{9}$ grid points, and what does that say about the largest flow anyone has simulated?
Solution

(a) Unmagnetised: $\mathrm{Re} = UL/\nu = 0.90$ at $\mathrm{Kn} = 0.375$. The two must be quoted together, because $\mathrm{Kn} = 0.375$ is outside the Chapman–Enskog domain and the $0.90$ is therefore a formula evaluated where it does not apply — not a statement that the gas is nearly laminar. By (1.2) the two facts are one fact: $\mathrm{Re} \lt 1$ because $\mathrm{Kn} \gt 3\,\mathrm{Ma_{th}} = 0.34$.

Magnetised: $\nu_\perp \le r_g\bar v_i/3 = 6.4824\times10^{17}\ \mathrm{cm^2\,s^{-1}}$, so $\mathrm{Re}_\perp \ge 4.68\times10^{12}$. It is a lower bound and must never be written as a value, because (6.2) bounds $\nu_\perp$ from above: scattering can only be more effective than the bare gyro-orbit. The true effective viscosity is set by microinstabilities, and Zhuravleva et al. (2019) bound its suppression below the Coulomb value at a factor of $10$ to $1000$.

(b) $N = \mathrm{Re}^{9/4} = 10^{9}$ gives $\mathrm{Re} = (10^{9})^{4/9} = 1.000\times10^{4}$. A $1000^3$ grid — a large calculation — resolves a Reynolds number of ten thousand. The photosphere of §1 is at $1.4\times10^{9}$, five orders of magnitude higher, and by (3.4) it would need $\mathrm{Re}^{9/4} = 3.8\times10^{20}$ points. No increase in computer power closes a gap of eleven and a half orders of magnitude in $N$, which is why astrophysical codes model the small scales instead of resolving them.

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 — $\eta$, $\nu$, $b$, $L$, $E$ and $\alpha$ — rather than left to the reader, and one cross-module constant differs from an earlier module's by a stated amount.

symbolmeaningunitsdefined
$\eta$the Kolmogorov dissipation scale. 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 $\mu$; Module 7 borrowed it for an interface displacement and Module 9 wrote $\eta_{\rm rad}$ to keep clear of it. Here the bare glyph is the dissipation scale, declared at Proposition 3. Braginskii's dynamic viscosities are written $\eta_0^{\rm B}$ and $\eta_1^{\rm B}$, always with both marks, and appear only in §6.2; everywhere else this module uses the kinematic $\nu = \eta/\rho$ and needs no bare dynamic viscosity at allcmProposition 3, (3.1)
$\nu$kinematic viscosity, the Module 7 precedent. $\nu$ is a cyclic frequency in Module 3 and Module 4 ($\nu_{\rm ac}$, $\nu_{\max}$). Podesta's spectra are frequency spectra and this module writes their frequency $f$, in hertz — in the fit window $10^{-3} \le f \le 10^{-2}$ Hz, in $P(f)$, and on the figure axis$\mathrm{cm^2\,s^{-1}}$Definition 1
$\sigma$bare $\sigma$ stays Module 1's collision cross-section, used in the cloud's $\lambda = 1/(n\sigma_{\rm H})$. Everything else is subscripted: $\sigma_v$ a one-dimensional linewidth (Solomon, Hitomi), $\sigma_L$ Larson's three-dimensional dispersion, $\sigma_s$ the width of $s$, $\sigma_\rho$ the density dispersion. The "$N\sigma$" idiom for a departure appears in §7 only, as it does in Modules 3, 5, 6 and 7$\mathrm{cm^2}$; $\mathrm{cm\,s^{-1}}$; —§6.4
$S_3(\ell)$, $S$$S_3$ is the third-order structure function and always carries its subscript, including in Problem C3. Solomon et al. write their cloud size $S$, and this module keeps their symbol inside §10, where no structure function occurs; their surface density is $\Sigma = 170\ M_\odot\,\mathrm{pc^{-2}}$$\mathrm{cm^3\,s^{-3}}$; pcProposition 5; (10.3)
$b$the forcing parameter, $1/3$ solenoidal to $1$ compressive. Module-local: Module 1's $b$ and $b_{90}$ are impact parameters and Module 7's is the bubble depth-to-wavelength ratio $h_B/\langle\lambda\rangle$; neither appears here, and $b$ is used only in §9 and Problem D2—Definition 4
$\varepsilon$dissipation rate per unit mass. Module 9's $\epsilon$ is Marrone et al.'s equipartition fraction, confined to its §8.2; different module, no overlap here$\mathrm{cm^2\,s^{-3}}$Definition 2
$L$, $\ell$the outer (stirring) scale and a separation inside the inertial range. Module 1's $L$ is the smallest scale to resolve and Module 3's is a luminosity; neither appears here. $\ell$ is never a mean free path, which stays $\lambda$cmDefinition 2
$\lambda$mean free path, Module 1 Definition 3, unchanged. No wavelength occurs in this module: wavenumbers are written $k = 1/\ell$ and never $k = 2\pi/\lambda$. $\lambda_T$, the Taylor microscale, occurs only in §5 and carries its subscriptcmModule 1
$\mathcal{M}$the root-mean-square sonic Mach number $U/c_s$. Module 7's $M = \Delta U/c_s$ is the same construction on a shear. No stellar mass appears in this module except $M_\odot$ inside $\Sigma = 170\ M_\odot\,\mathrm{pc^{-2}}$, and $M$ without a subscript in §8.2 is a cluster mass, declared there—Definition 3
$E(k)$, $C$, $C_K$, $C_2$the energy spectrum per unit mass and per unit wavenumber, and the three constants of (2.2), (5.1) and (5.2). Module 8's $E$ is an explosion energy in erg; different module. The units of $E(k)$ are printed because they are what fixes the exponent $-5/3$$\mathrm{cm^3\,s^{-2}}$; —Proposition 2; §5
$\mathrm{Kn}$, $\mathrm{Re}$, $R_\lambda$, $\mathrm{Ma_{th}}$, $\mathrm{Pr}$Knudsen (Module 1), Reynolds (Definition 1), Taylor-microscale Reynolds (§5, new to this book), thermal Mach (Module 2, with this module's own speed convention — see §6.5), Prandtl. $\mathrm{Pr}$ is quoted inside Zhuravleva's own sentence and is computed nowhere—as listed
$\bar v$, $v_{\rm rms}$, $v_\perp$the mean, root-mean-square and perpendicular thermal speeds, $\sqrt{8k_{\rm B}T/\pi m}$, $\sqrt{3k_{\rm B}T/m}$ and $\sqrt{2k_{\rm B}T/m}$. All three are live in this module and each is correct for its own job; the table in §6.4 gives the component count of every one$\mathrm{cm\,s^{-1}}$§6.4
$\alpha$$P_{\rm turb}/P_{\rm th}$, confined to §8 and Problem K2. Module 2's and Module 9's $\alpha$ is a fitted density exponent, Module 5's a length scale and Module 6's a mixing-length parameter; none appears here—(8.3)

One cross-module constant, and the size of the difference. Module 3 §3 used $\mu = 0.6$ for its intracluster medium; this module uses $\mu_{\rm ICM} = 0.61$, the standard fully-ionised value, in the sound speed of (8.2) and in the pressure ratio (8.3). The difference is $1.7$ per cent, it is not a correction to Module 3, and no quantity in either module changes sign or order because of it. The ion viscosity of §6.1 uses neither: it uses $m_{\rm p}$, for the reason given there.

Sources

Every published value this module compares against, with what was taken from it and from where. Eight papers were listed for source verification and all eight were fetched as PDFs and read locally; where the scan had no usable text layer the page was rendered and read as an image. No web summariser was used anywhere. Volume, page and year for every DOI below were confirmed at Crossref rather than recalled. Three entries were not read and say so in place. Two more carry no read status and say so: Sarazin, inherited unchanged from Module 1, which owns it; and Pope, quoted for a standard result. One convention, carried from Module 8: a § on its own is a section of this book and is a link; a section of a cited paper is written "their Sect. N", with no § and no link.

  1. The anchor. J. J. Podesta, D. A. Roberts and M. L. Goldstein, Spectral exponents of kinetic and magnetic energy spectra in solar wind turbulence, ApJ 664, 543–548 (2007), doi:10.1086/519211.
    Their Table 2: the magnetic, velocity, kinetic-energy and total-energy exponents of four intervals with 99 per cent confidence limits, reproduced in §7. Their Table 1: the intervals — 1995 May 23 to Jul 16 (54 d), 1997 Dec 14 to 1998 Feb 06 (54 d), 2000 Nov 15 to 2001 Feb 04 (81 d), 2003 Jun 11 to Aug 03 (53.854 d). Wind spacecraft, ecliptic plane, 1 au.
    Their Sect. 3 bounds the fit window: moving the endpoints over $7\times10^{-4}$–$1.5\times10^{-3}$ Hz and $7\times10^{-3}$–$1.5\times10^{-2}$ Hz changes an exponent "typically by 1%", always less than the tabulated error. The Alfvén ratios quoted in §7.2 are theirs.
  2. The Kolmogorov constant. K. R. Sreenivasan, On the universality of the Kolmogorov constant, Physics of Fluids 7(11), 2778–2784 (1995), doi:10.1063/1.868656. Author's copy at users.ictp.it/~krs/pdf/1995_003.pdf.
    His eq. (2) defines $C_K$ through the one-dimensional longitudinal spectrum; his Sect. I gives the isotropy factors $55/18$ and $4.02$; his Fig. 3 and Sect. V give $0.53\pm0.055$ over more than 100 spectra with $R_\lambda$ from 50 to about $10^{4}$, and the revision to about $0.50$.
    His Table IV, row "Tidal channel", gives $0.47\pm0.02$ at $R_\lambda$ 3000–18000, with the footnote about straight-edge readings of seventeen data sets and Kraichnan's judgement that a different reading would raise it by more than ten per cent.
  3. The first clean $-5/3$ measurement. H. L. Grant, R. W. Stewart and A. Moilliet, Turbulence spectra from a tidal channel, J. Fluid Mech. 12, 241 (1962). NOT READ: the publisher's copy is paywalled. Its $C_K$ is quoted in §5 only as tabulated by Sreenivasan (1995), and the module says so where the number is used. The $R_\lambda$ range quoted is Sreenivasan's estimate, not Grant et al.'s measurement.
  4. The cluster. Hitomi Collaboration, The quiescent intracluster medium in the core of the Perseus cluster, Nature 535, 117–121 (2016), arXiv:1607.04487.
    Abstract and their p. 3: line-of-sight velocity dispersion $164\pm10\ \mathrm{km\,s^{-1}}$ at 30–60 kpc, all errors at 90 per cent confidence; the Lyman-$\alpha$ value $160\pm16$; the gradient $150\pm70\ \mathrm{km\,s^{-1}}$ across the core; the systematic budget $\pm6$, $\pm2$, $-3\ \mathrm{km\,s^{-1}}$.
    Their p. 4: $k_{\rm B}T = 4.1\pm0.1$ keV in that region and $3.8\pm0.1$ keV in the centre, with $187\pm13\ \mathrm{km\,s^{-1}}$ there.
    Abstract, in full: "Turbulent pressure support in the gas is 4% or less of the thermodynamic pressure, with large scale shear at most doubling that estimate." The Methods and Extended Data were checked and nothing tighter exists.
  5. The size–linewidth relation, first. R. B. Larson, Turbulence and star formation in molecular clouds, MNRAS 194, 809–826 (1981), read from the ADS scan.
    His eq. (1), p. 814: $\sigma = 1.10\,L^{0.38}\ \mathrm{km\,s^{-1}}$, an eye-fitted line with $\mathrm{rms}(\log\sigma) = 0.14$, valid for $0.1 \lesssim L \lesssim 100$ pc. His $\sigma$ and his thermal term $0.32\ \mathrm{km\,s^{-1}}$ are both three-dimensional (his p. 814).
    His p. 816: "it is perhaps remarkable that the observed relation $\sigma \propto L^{0.38}$ is so close to the Kolmogoroff law $\sigma \propto L^{0.33}$."
  6. The size–linewidth relation, properly. P. M. Solomon, A. R. Rivolo, J. Barrett and A. Yahil, Mass, luminosity, and line width relations of Galactic molecular clouds, ApJ 319, 730–741 (1987), ADS scan; their eq. (1) was verified against a rendered image of the page, not the OCR layer.
    Their eq. (1), p. 731: $\sigma_v = (1.0\pm0.1)\,S^{0.5\pm0.05}\ \mathrm{km\,s^{-1}}$, 273 clouds, dispersion in $\log\sigma_v$ of $\pm0.11$. Abstract: constant mean surface density $170\ M_\odot\,\mathrm{pc^{-2}}$.
    Their p. 737 settles three things this module depends on: what the $\pm0.05$ is ("larger than the formal fit error … an estimate of the systematic error based on several fits to the data with varying definitions of cloud boundaries", so not a $\sigma$); the ruling on Larson ("This hypothesis is ruled out by the data presented here"); and the sensitivity (without the $2\ \mathrm{km\,s^{-1}}$ extrapolation the constant of eq. (1) is 0.83, a reduction of 17 per cent).
  7. The forcing parameter and the log-normal. C. Federrath, J. Roman-Duval, R. S. Klessen, W. Schmidt and M.-M. Mac Low, Comparing the statistics of interstellar turbulence in simulations and observations, A&A 512, A81 (2010), arXiv:0905.1060.
    Their eqs. (18) and (19): $\sigma_\rho/\langle\rho\rangle = b\mathcal{M}$ and $\sigma_s^2 = \ln(1+b^2\mathcal{M}^2)$, citing Padoan et al. (1997) and Federrath et al. (2008b). Their Fig. 8 and eq. (20): $b$ falls from about 1 for purely compressive forcing to about $1/3$ in three dimensions for purely solenoidal forcing.
  8. The transport coefficients of a plasma. S. I. Braginskii, Transport processes in a plasma, in Reviews of Plasma Physics, Vol. 1, ed. M. A. Leontovich, Consultants Bureau, New York (1965), pp. 205–311, translated by Herbert Lashinsky. The original translation was read, not a review restating it; the scan has no text layer, so every equation below was read from a rendered page.
    His p. 215, eq. (2.5i): the definition of $\tau_i$. His p. 217, eq. (2.19): $\pi_{ab} = -\eta_0W_{ab}$. His p. 218: eq. (2.22) $\eta_0^i = 0.96\,n_iT_i\tau_i$ and eq. (2.23) $\eta_1^i = (3/10)n_iT_i/(\omega_i^2\tau_i)$ with $\eta_2 = 4\eta_1$. The electron values on the same page are 0.73 and 0.51 at $Z = 1$.
  9. The contested intracluster viscosity. I. Zhuravleva, E. Churazov, A. A. Schekochihin, S. W. Allen, A. Vikhlinin and N. Werner, Suppressed effective viscosity in the bulk intergalactic plasma, Nature Astronomy 3, 832–837 (2019), doi:10.1038/s41550-019-0794-z, arXiv:1906.06346.
    A twelve-day Chandra observation of an offset region of the Coma cluster against direct numerical simulation: "the effective viscosity must be suppressed by at least a factor of $\sim$10 to $\sim$1000 for Pr $\le$ 1", and "the assumption of a Coulomb-collision-dominated plasma is ruled out for the Coma Cluster". Both of their caveats are quoted in §6.1.
  10. The Coulomb mean free path. C. L. Sarazin, X-ray Emission from Clusters of Galaxies (1988), his Sect. 5.4 — the mean free path and $\ln\Lambda = 37.8$, both reused unchanged from Module 1, whose anchor is $22.5$ kpc against his published 23 kpc.
  11. The viscosity of air. K. Kadoya, N. Matsunaga and A. Nagashima, Viscosity and thermal conductivity of dry air in the gaseous phase, J. Phys. Chem. Ref. Data 14, 947–970 (1985), doi:10.1063/1.555744, confirmed at Crossref. NOT READ: no copy was served by NIST, ADS or AIP. The value $\nu = 0.15\ \mathrm{cm^2\,s^{-1}}$ therefore carries two significant figures throughout this module and is labelled as having no primary source wherever it is used.
  12. The original. A. N. Kolmogorov, Dokl. Akad. Nauk SSSR 30, 299 (1941). NOT READ; cited for the result only, as Module 1 does for Maxwell (1867). The derivation reproduced in Proposition 2 is the standard dimensional argument and is not a translation of his.
  13. Quoted for a standard result, not compared against. S. B. Pope, Turbulent Flows, Cambridge (2000), their Sect. 9.1.2 — the $\mathrm{Re}^{3}$ cost of a direct numerical simulation. NOT READ: no copy was obtained, and the section number is recalled rather than confirmed against a copy. The result is derived in full in Proposition 4 and depends on nothing quoted. And P. Goldreich and S. Sridhar (1995) — named in §11 as a second derivation of a $5/3$ magnetic spectrum, and not used numerically anywhere.

Reproducing the numbers

Every number in this module comes from one script, which takes no arguments and prints its whole argument in order:

cd afd/figs
python m10_numbers.py          # every number in the prose and the problems
python m10_problems_check.py   # rebuilds the nine problems independently
python m10_build_figs.py       # writes the four SVGs
python splice.py ../module10.html

The check script imports nothing from the numbers script: it re-derives each problem's answer from the inputs printed in that problem's statement, which is the only way to catch a modelling choice that lives in the generator and not in the prose. That failure mode cost Module 9 a problem statement, and the rule it earned is applied to all nine slots here.