One dimensional argument, and the four fluids that test it: a laboratory, the solar wind, a cluster of galaxies and a molecular cloud.
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.
"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.
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, $U = 2\ \mathrm{km\,s^{-1}}$ | $4.427\times10^{-10}$ | $0.2063$ | $1.398\times10^{9}$ |
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.
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.
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.
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.
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 $\nu = 0.15\ \mathrm{cm^2\,s^{-1}}$:
$$ \eta = 1.355\ \mathrm{mm},\qquad u_\eta = 1.11\ \mathrm{cm\,s^{-1}},\qquad \tau_\eta = 122.47\ \mathrm{ms}. \tag{3.3}$$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 snapshot | CFL steps per turnover | work, $\mathrm{Re}^{3}$ | 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.
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 $\nu$ | $2.382\times10^{-2}\ \mathrm{erg\,g^{-1}\,s^{-1}}$ | $64.67$ kpc | none: $\eta/L = 1.078$ |
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.
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.
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.
Sreenivasan compiled more than a hundred measurements of $C_K$ — grid turbulence, jets, wakes, boundary layers, a tidal channel — and found:
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.
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.
| Fluid | inputs | $\lambda$ | $\nu$ (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 | $12.34$ | $1.936$ |
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 $1.0000$.
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
$$ \mathrm{Re} \;=\; \frac{(1.64\times10^{7})(60\ \mathrm{kpc})}{3.3557\times10^{30}} \;=\; 0.90, \qquad \mathrm{Kn} \;=\; \frac{22.5}{60} \;=\; 0.375. \tag{6.3}$$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 $\mathrm{Re}_\perp = 7.446\times10^{25}$.
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.
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.
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 $k_{\rm B}T/m = v_{\rm rms}^2/3$,
$$ \frac{\nu_{\rm B}}{\lambda v_{\rm rms}/3} \;=\; 0.9600 \qquad\text{exactly.} \tag{6.6}$$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.
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 $\mathrm{Re}_\perp$.
| quantity | Module 1 census | Venzmer & 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 $\ln\Lambda$.
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.
| quantity | formula | components | where 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 rms | Braginskii'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-D | he 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 K | 3-D | same 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-D | one linewidth per cloud |
| Hitomi's $\sigma_v$ | measured, $164\ \mathrm{km\,s^{-1}}$ | 1-D, line of sight | a 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 speed | the width of $s = \ln(\rho/\langle\rho\rangle)$, dimensionless. Definition 3 and Proposition 9, in §9 |
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.
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.
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.
| interval | magnetic index | departure from $5/3$, in its own 99 % half-widths | in $\sigma$ | velocity index | departure from $5/3$, in half-widths | in $\sigma$ |
|---|---|---|---|---|---|---|
| 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 | $1.58\pm0.02$ | $-4.33$ | $-11.2$ | $1.50\pm0.02$ | $-8.33$ | $-21.5$ |
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.
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.
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:
| field | mean | s.d. of the four | s.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 | $1.5275$ | $0.0427$ | $0.0214$ | $-6.52$, $p = 0.0073$, $2.68\sigma$ | $+1.29$, $p = 0.2883$, $1.06\sigma$ |
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.
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?
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.
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
$$ \alpha \;\equiv\; \frac{P_{\rm turb}}{P_{\rm th}} \;=\; \frac{\mu m_{\rm u}\sigma_v^2}{k_{\rm B}T} \;=\; 4.15\ \text{per cent}. \tag{8.3}$$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.
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.
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.
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.
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.
| $\mathcal{M}$ | $b$ | $\sigma_s$ | $\sigma_\rho/\langle\rho\rangle = b\mathcal{M}$ | $s_0$ | 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.
| threshold $x = \rho/\langle\rho\rangle$ | $f(\gt x)$ at $b = 1/3$ | $f(\gt x)$ at $b = 1$ | ratio |
|---|---|---|---|
| $10$ | $1.230\times10^{-2}$ | $1.594\times10^{-2}$ | $1.3$ |
| $100$ | $1.054\times10^{-4}$ | $6.459\times10^{-4}$ | $6.1$ |
| $1000$ | $1.211\times10^{-7}$ | $8.949\times10^{-6}$ | $73.9$ |
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.
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.
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.
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.
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.
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 here | What it costs | Where 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 argument | Larger 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 mechanism | Module 12 (magnetohydrodynamics) |
| Taylor's frozen-flow hypothesis, in Check 1 | The 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 exact | not 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 line | Proposition 6 inside this module, and Module 12 for the field that rescues it |
| The effective viscosity of the intracluster medium | It is set by pressure-anisotropy microinstabilities, not by Coulomb collisions. Zhuravleva et al. (2019) bound the suppression at $10$ to $1000$ and leave $\mathrm{Pr}$ free | Nobody 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 simulation | It makes direct numerical simulation of any astrophysical flow impossible. The $10$ pc cloud alone needs $6.24\times10^{17}$ grid points | Module 14 (numerical methods) |
| One exponent, no intermittency | K41 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 exist | Not in this book. Named here and named as absent |
| Constant $P_{\rm turb}/P_{\rm th}$ with radius, in Proposition 7 | Drop it and the mass bias depends on the two pressures' scale lengths, which one annulus cannot supply | not repaid; named where the number is printed, in §8.2 |
| A single $\nu$ for air, quoted to two figures | Every 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 obtained | not 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.
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.
$1\ \mathrm{W\,kg^{-1}} = 10^{7}\ \mathrm{erg\,s^{-1}}/10^{3}\ \mathrm{g} = 10^{4}\ \mathrm{erg\,g^{-1}\,s^{-1}}$, 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.
$10\ \mathrm{pc} = 3.0857\times10^{19}$ 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.
$\varepsilon = (2.64\times10^{5})^3/3.0857\times10^{19} = 5.963\times10^{-4}\ \mathrm{erg\,g^{-1}\,s^{-1}}$. 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.
$\nu = \lambda\bar v_i/3 = (6.943\times10^{22})(1.450\times10^{8})/3 = 3.3557\times10^{30}\ \mathrm{cm^2\,s^{-1}}$ 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.
$\sigma_s = \sqrt{\ln(1+b^2\mathcal{M}^2)}$. 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.
$\lambda = 1/(n\sigma_{\rm H}) = 10^{13}\ \mathrm{cm} = 0.67$ 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.
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.
(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.
(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.
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.
| symbol | meaning | units | defined |
|---|---|---|---|
| $\eta$ | 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 all | cm | Proposition 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 $\Sigma = 170\ M_\odot\,\mathrm{pc^{-2}}$ | $\mathrm{cm^3\,s^{-3}}$; pc | Proposition 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 $\lambda$ | cm | Definition 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 subscript | cm | Module 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.
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.
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.