The continuum condition for astrophysical gases: the mean free path, the Knudsen number, and the four systems in which the fluid picture fails outright.
Everything in this book rests on four equations — conservation of mass, of momentum, of energy, and an equation of state — applied to a substance treated as a continuum. Later modules use them to compute the radius of a supernova remnant (Module 8), the mass at which an interstellar cloud must collapse (Module 5), and the rate at which a black hole can swallow gas (Module 9). Before any of that, one question has to be settled, and it is not a formality.
A continuum has a density at every point. Real gas does not: it is a finite number of particles with empty space between them. The fluid equations are an approximation whose validity has to be earned, and in astrophysics the gas is thin enough that earning it is not obvious.
Here is the scale of the problem. A specialist vacuum laboratory can routinely pump a chamber down to about $10^{-13}$ torr. At room temperature that pressure corresponds to a number density
$$ n \;=\; \frac{P}{k_B T} \;=\; \frac{10^{-13}\times 1333.22\ \mathrm{dyn\,cm^{-2}}}{(1.381\times10^{-16}\ \mathrm{erg\,K^{-1}})(300\ \mathrm{K})} \;=\; 3.2\times10^{3}\ \mathrm{cm^{-3}}, \tag{1.1}$$where $P$ is the pressure, $T$ the absolute temperature, and $k_B$ Boltzmann's constant. The warm neutral interstellar medium — the gas that fills most of the volume of the Milky Way's disc — has a number density of roughly $0.5\ \mathrm{cm^{-3}}$. It is therefore about $6.4\times10^{3}$ times more rarefied than that laboratory vacuum. Even against the lowest reported laboratory pressures, near $10^{-15}$ torr and $32\ \mathrm{cm^{-3}}$, the interstellar medium is still 64 times emptier.
Both vacuum figures are inputs, not results of this book. $10^{-13}$ torr is the pressure a dedicated extreme-high-vacuum chamber reaches routinely; $10^{-15}$ torr is at the bottom of the range reported for record measurements. Reported extreme-high-vacuum pressures below about $10^{-12}$ torr are themselves difficult to measure, because the gauge outgasses into the volume it is measuring. Treat both as order-of-magnitude anchors; the argument needs only that they exceed $0.5\ \mathrm{cm^{-3}}$, and the more conservative of them exceeds it by 64.So the object we are about to call a fluid is, by a wide margin, emptier than the emptiest thing anyone has ever built. And yet the fluid equations describe it well. The purpose of this module is to say exactly why, to produce the single number that decides the question, and to identify the systems where the answer comes out the other way.
The logic runs in one direction and each step depends on the last. A fluid description is a claim that the velocity distribution of the particles is fixed by a handful of local numbers (§2). Collisions are what make that claim true (§3). How far a particle travels between collisions is the mean free path, computed for neutral gas in §4 and for ionised gas in §5. Comparing that length to the size of the structure you want to resolve gives the Knudsen number, and the size of the Knudsen number decides which equations you are entitled to write down (§6).
Start with the complete description and then say what the fluid description throws away.
The distribution function carries everything there is to know about the gas short of tracking individual particles. Its evolution is governed by the Boltzmann equation,
$$ \frac{\partial f}{\partial t} \;+\; \mathbf{v}\cdot\nabla_{\mathbf{x}} f \;+\; \mathbf{a}\cdot\nabla_{\mathbf{v}} f \;=\; \left(\frac{\partial f}{\partial t}\right)_{\!\mathrm{coll}}, \tag{2.1}$$in which $\mathbf{a}$ is the acceleration produced by any smooth external force — gravity, a large-scale electric or magnetic field — and the right-hand side collects the effect of short-range encounters between particles. The three terms on the left say that $f$ is conserved along particle trajectories in the six-dimensional space $(\mathbf{x},\mathbf{v})$; the term on the right says that collisions move particles abruptly from one velocity to another and therefore break that conservation.
Equation (2.1) is exact and unusable. It is a partial differential equation for a function of six variables plus time, and the collision term is an integral over the velocities of every possible collision partner. The fluid description is the bet that we never need it.
The saving is enormous: a function of six variables collapses to five functions of three. But nothing so far explains why $f$ should be Maxwellian. That is the work of collisions.
Boltzmann's H-theorem states that the collision term in (2.1) drives $f$ towards the Maxwellian (2.2) and that the Maxwellian is the only distribution the collision term leaves alone. We take that result as given; what we need here is not the theorem but its rate, because a gas is only Maxwellian if the collisions get there before the flow carries the gas somewhere else.
The cheapest honest model of the collision term is the relaxation, or BGK, approximation, due to Bhatnagar, Gross and Krook (1954):
$$ \left(\frac{\partial f}{\partial t}\right)_{\!\mathrm{coll}} \;\simeq\; -\,\frac{f - f_M}{\tau}, \tag{3.1}$$where $\tau$ is the collision time, the mean interval between collisions suffered by one particle. Equation (3.1) says only this: collisions push $f$ towards the local Maxwellian, and they do it on a timescale $\tau$. It is not exact — it replaces a complicated integral operator by a single relaxation rate — but it reproduces the scaling we need, and that scaling is the whole content of this section.
Proposition 1 is the hinge of the entire module, so it is worth saying in words what the algebra did. A particle arriving at a point carries the properties of the place where it last collided, one mean free path away. If the gas changes appreciably over that distance, the arriving particles are a mixture of populations from measurably different conditions, and their combined velocity distribution is not the Maxwellian of the point they have arrived at. The mismatch is proportional to how much the gas changes over one mean free path, which is $\lambda/L$. If $\lambda/L$ is small, every particle arriving at a point comes from somewhere effectively identical, and the local Maxwellian is exact for all practical purposes.
Everything now reduces to computing $\lambda$, which is a different calculation for neutral gas and for ionised gas.
Neutral atoms interact only when they nearly touch. That makes the calculation short.
The photosphere is the layer of the Sun from which visible light escapes — conventionally, the surface at which the optical depth in the visible reaches $2/3$. There the gas pressure is about $1.2\times10^{5}\ \mathrm{dyn\,cm^{-2}}$, and the temperature is close to the Sun's effective temperature, $T_{\rm eff}=5772\ \mathrm{K}$. The second statement is not a coincidence but very nearly a definition: in a grey atmosphere the Eddington approximation gives $T^4(\tau) = \tfrac{3}{4}T_{\rm eff}^4(\tau+\tfrac{2}{3})$, so $T=T_{\rm eff}$ exactly at $\tau=2/3$, which is the surface we have just chosen. We do not need to look up the density; the ideal gas law gives it:
$$ n \;=\; \frac{P}{k_B T} \;=\; \frac{1.2\times10^{5}}{(1.381\times10^{-16})(5772)} \;=\; 1.51\times10^{17}\ \mathrm{cm^{-3}}. \tag{4.2}$$ Two inputs here are quoted rather than derived. The photospheric gas pressure $1.2\times10^{5}\ \mathrm{dyn\,cm^{-2}}$ is a standard model-atmosphere value at $\tau=2/3$ and varies by tens of per cent between published models. The atomic cross-section for neutral hydrogen at photospheric temperatures is taken as $\sigma \approx 10^{-15}\ \mathrm{cm^2}$, an order-of-magnitude value consistent with a hard-sphere radius of a few Bohr radii, and is the weaker of the two: every photospheric number below inherits its uncertainty of roughly a factor of two. Neither affects any conclusion, because the Knudsen numbers that follow are smaller than one by nine orders of magnitude.Then (4.1) gives
$$ \lambda = \frac{1}{(1.51\times10^{17})(10^{-15})} = 6.6\times10^{-3}\ \mathrm{cm} = 66\ \mu\mathrm{m}. \tag{4.3}$$Sixty-six micrometres, in an object $6.957\times10^{10}\ \mathrm{cm}$ across. Section §6 turns that comparison into a number.
Most astrophysical gas is ionised, and the argument of §4 breaks at the first step. The Coulomb force has no range: two charges interact at any separation, so there is no distance $d$ inside which a collision happens and outside which it does not. Worse, the naive cross-section diverges. We have to define what counts as a collision.
Consider two particles of charge $Z_1e$ and $Z_2e$ with reduced mass $\mu = m_1m_2/(m_1+m_2)$, approaching with relative speed $v_{\rm rel}$ and impact parameter $b$ — the perpendicular distance between the incoming asymptote and the target. Classical two-body Coulomb scattering gives the deflection angle $\theta$ through
$$ \tan\frac{\theta}{2} \;=\; \frac{b_{90}}{b}, \qquad b_{90} \;\equiv\; \frac{Z_1Z_2e^2}{\mu\,v_{\rm rel}^2}. \tag{5.1}$$The length $b_{90}$ is therefore the impact parameter that produces a $90^\circ$ deflection, which is why it carries that subscript. For a gas of like particles of charge $e$ and mass $m$ at temperature $T$, we have $\mu=m/2$, and the mean square relative speed of two independently Maxwellian particles is $\langle v_{\rm rel}^2\rangle = 2\langle v^2\rangle = 6k_BT/m$. Hence $\mu\langle v_{\rm rel}^2\rangle = 3k_BT$ and
$$ b_{90} \;=\; \frac{e^2}{3k_BT}. \tag{5.2}$$One step there deserves to be named rather than passed over. Equation (5.1) defines $b_{90}$ for a single pair with a definite relative speed; (5.2) substitutes the mean square relative speed of the whole distribution. Since $b_{90}\propto v_{\rm rel}^{-2}$, and the average of a reciprocal is not the reciprocal of an average, (5.2) is an approximation and not a consequence of (5.1). It is a good enough one to carry the scaling, and §5.3 measures exactly how much it costs: a factor of 2.05 in the final answer, which is why the numerical coefficient there is taken from the properly averaged calculation rather than from this one.
At the temperature of the gas between galaxies, $T=10^8\ \mathrm{K}$, equation (5.2) gives $b_{90}=5.6\times10^{-12}\ \mathrm{cm}$, which is $1.8\times10^{3}$ times smaller than the $10^{-8}\ \mathrm{cm}$ scale of an atom. If only encounters closer than $b_{90}$ counted, the cross-section would be $\pi b_{90}^2 = 9.7\times10^{-23}\ \mathrm{cm^2}$ and ionised gas would essentially never collide. It does collide, and the reason, drawn in Fig. 2, is that the many distant, feeble encounters add up.
For $b \gg b_{90}$, equation (5.1) linearises to $\theta \approx 2b_{90}/b$. Single distant encounters are weak, but they are numerous, because the number of targets at impact parameters between $b$ and $b+db$ grows as $2\pi b\,db$. The deflections have random signs, so they do not add; their squares do. A particle moving at speed $v$ through targets of density $n$ accumulates mean square deflection at the rate
$$ \frac{d\langle\theta^2\rangle}{dt} \;=\; n v\int \theta^2(b)\,2\pi b\,db \;=\; n v\int \left(\frac{2b_{90}}{b}\right)^{\!2} 2\pi b\,db \;=\; 8\pi n v\, b_{90}^2 \int_{b_{\min}}^{b_{\max}}\frac{db}{b}. \tag{5.3}$$The integral is $\ln(b_{\max}/b_{\min})$. Every decade of impact parameter contributes equally: the weakness of distant encounters is exactly cancelled by how many of them there are. This is why the Coulomb result is dominated not by close collisions but by the accumulated effect of many grazing ones, and why the limits of the integral must be supplied by physics outside the two-body problem.
The upper limit is set by screening. In a plasma, a positive charge attracts electrons and repels ions until its field is neutralised beyond the Debye length
$$ \lambda_D \;=\; \left(\frac{k_B T}{4\pi n_e e^2}\right)^{1/2}, \tag{5.4}$$where $n_e$ is the electron number density. Past $\lambda_D$ there is no field left to scatter off, so $b_{\max}=\lambda_D$. The lower limit is the larger of two lengths: the classical $b_{90}$, below which the linearisation fails, and the reduced de Broglie length $\hbar/(m v)$, below which the notion of a classical impact parameter fails outright, because the particle cannot be localised transversely to better than that. (The reduced de Broglie length is not the thermal de Broglie wavelength $h/\sqrt{2\pi m k_B T}$, which at $T=10^8$ K is $7.45\times10^{-10}\ \mathrm{cm}$, larger by a factor of 4.34; the two are easily confused, and the choice shifts $\ln\Lambda$ by 1.5.) Their ratio defines
Evaluate it for the intracluster medium, $n_e=10^{-3}\ \mathrm{cm^{-3}}$ and $T=10^8\ \mathrm{K}$. Equation (5.4) gives $\lambda_D = 2.18\times10^{6}\ \mathrm{cm}$, or 21.8 km. The classical $b_{90}$ is $5.57\times10^{-12}\ \mathrm{cm}$; the electron de Broglie length is $\hbar/(m_ev_e) = 1.72\times10^{-10}\ \mathrm{cm}$, which is 31 times larger and therefore the one that applies. Then
$$ \ln\Lambda = \ln\frac{2.18\times10^{6}}{1.72\times10^{-10}} = \ln\left(1.27\times10^{16}\right) = 37.1. \tag{5.5}$$The mass-independence is worth a moment, because it is easy to misread. Neither $b_{90}$ nor $\lambda$ contains the particle mass: $b_{90}=e^2/3k_BT$ depends on the charge and the temperature alone. So at a common temperature an electron and a proton in the same plasma travel the same distance between large-angle deflections, even though the electron moves $\sqrt{m_p/m_e}=42.9$ times faster. What differs by that factor is the time, not the distance: the electron completes its 22 kpc in $1/43$ of the time the proton takes. For ions of charge $Ze$ the distance shortens as $Z^{-4}$, since $b_{90}\propto Z^2$ enters squared.
Read the scaling before the numbers. The mean free path goes as $T^2/n$: hot, thin plasma is nearly collisionless, and cold, dense plasma is strongly collisional. The $T^2$ is the more surprising half. It comes from $b_{90}\propto T^{-1}$ squared — a fast particle spends less time near any given target and is deflected less — and it is steep enough that a factor of 10 in temperature changes the mean free path by 100.
The second half of that definition is not a technicality, and it is where most errors of judgement enter. The Knudsen number is a property of the question, not of the gas. The photospheric gas of §4.1 has one mean free path, 66 µm, but two Knudsen numbers. Measured against the solar radius it is $9.6\times10^{-14}$; measured against the pressure scale height of about 150 km, which is the scale on which convection organises itself and therefore the scale a model of granulation must resolve, it is $4.4\times10^{-10}$. Both are minute, so here the distinction changes nothing. In §7 we meet a case where it changes everything.
Proposition 1 established that $f - f_M$ is of order $\mathrm{Kn}$. The systematic exploitation of that fact is the Chapman–Enskog expansion, which posits
$$ f \;=\; f_M\left(1 + \mathrm{Kn}\,\phi_1 + \mathrm{Kn}^2\phi_2 + \cdots\right) \tag{6.1}$$and takes moments of the Boltzmann equation order by order. Each order produces a different set of fluid equations.
| Order retained | Equations obtained | What they contain | Truncation error | Usable for |
|---|---|---|---|---|
| $\mathrm{Kn}^0$ | Euler | Pressure only; no viscosity, no conduction. Entropy conserved along streamlines. | $O(\mathrm{Kn})$ | $\mathrm{Kn}\lesssim10^{-2}$ |
| $\mathrm{Kn}^1$ | Navier–Stokes | Viscous stress $\propto\mu\nabla\mathbf{u}$ and heat flux $\propto-\kappa\nabla T$, with $\mu$ and $\kappa$ computed from $\lambda$. | $O(\mathrm{Kn}^2)$ | $10^{-2}\lesssim\mathrm{Kn}\lesssim10^{-1}$ |
| $\mathrm{Kn}^2$ | Burnett | Higher gradient terms. Known to admit unphysical short-wavelength instabilities. | $O(\mathrm{Kn}^3)$, but unstable | rarely usable |
| $\mathrm{Kn}\gtrsim10^{-1}$ | none — the series fails | No closed fluid description exists; the kinetic equation (2.1) must be solved. | — | — |
The thresholds in the last column are conventions, and it is worth saying where they come from rather than leaving them to look like measurements. Truncating (6.1) after the term of order $\mathrm{Kn}^{k}$ leaves a relative error of order $\mathrm{Kn}^{k+1}$, so demanding one per cent accuracy from the Euler equations demands $\mathrm{Kn}\lesssim10^{-2}$, and the same demand on Navier–Stokes admits $\mathrm{Kn}\lesssim10^{-1}$. Engineering practice in rarefied gas dynamics draws the same boundaries and adds a fourth regime above $\mathrm{Kn}\approx10$, where particles cross the whole system without meeting each other. This book does not use that last distinction, because the astrophysical question is only ever whether a fluid description exists at all.
Modules 2 to 11 of this book use the Euler equations. That is the first row, and it is the right choice for most astrophysical problems — but only because the Knudsen numbers come out where they do, which is the subject of the next section. The transport coefficients in the second row are worth deriving now, because their scaling produces one of the more consequential results in the book.
Two consequences are worth stating plainly. First, the density-independence of viscosity is not an artefact of the plasma case: it is Maxwell's result for any gas. He derived it in 1860, did not believe the experiments then available, and tested it himself with Katherine Maxwell in 1865–66, finding the viscosity of air constant over a wide range of pressure. Second, the $T^{5/2}$ law makes hot plasma a very good conductor of heat. Raising the temperature from $10^6$ to $10^8$ K raises $\kappa$ by $(100)^{5/2}=10^5$ from the power alone, or by $8.8\times10^{4}$ once the accompanying rise in $\ln\Lambda$ from 32.5 to 37.1 is included. This is why unsuppressed thermal conduction would erase the temperature structure observed in cluster cores within a fraction of the cluster's age, and it is the reason the question of what suppresses it is a live one. We return to it in Module 12, where magnetic fields provide the answer.
We can now place real systems. Fig. 3 shows the density–temperature plane for ionised gas, with contours of constant Coulomb mean free path from (5.6) and the principal astrophysical plasmas marked. The contours are lines of $T^2/n = \mathrm{constant}$, bent slightly by the variation of $\ln\Lambda$ across the plane.
Fig. 4 collapses the same information onto the axis that actually decides which equations may be used.
The table collects the numbers behind the two figures.
| System | $n$ / cm$^{-3}$ | $T$ / K | $\lambda$ | $L$ | Kn | Model |
|---|---|---|---|---|---|---|
| Solar photosphere | $1.5\times10^{17}$ | 5772 | 66 µm | 150 km | $4.4\times10^{-10}$ | Euler |
| Room air (for scale) | $2.5\times10^{19}$ | 293 | 68 nm | 3 m | $2.3\times10^{-8}$ | Euler |
| Warm neutral ISM | 0.5 | 8000 | 134 AU | 100 pc | $6.5\times10^{-6}$ | Euler |
| Intracluster medium, core | $10^{-2}$ | $3\times10^{7}$ | 0.22 kpc | 100 kpc | $2.2\times10^{-3}$ | Euler |
| Intracluster medium, outskirts | $10^{-4}$ | $10^{8}$ | 222 kpc | 2 Mpc | $1.1\times10^{-1}$ | marginal |
| Solar wind at 1 AU | 5 | $1.2\times10^{5}$ | 1.9 AU | 1 AU | $1.9$ | none |
Three readings of the table matter. First, the size of astrophysical systems buys back most of what their rarefaction costs, but not always all of it, and the two cases are worth separating. Room air has $\mathrm{Kn}=2.3\times10^{-8}$. The solar photosphere is 165 times more rarefied than room air and yet reaches $\mathrm{Kn}=4.4\times10^{-10}$, smaller by a factor of 51, because the scale it must resolve is 150 km rather than 3 m. The warm neutral interstellar medium does worse: its mean free path of 134 AU, set against 100 pc, gives $\mathrm{Kn}=6.5\times10^{-6}$, which is 286 times larger than room air's. There, rarefaction wins. The conclusion survives anyway, and that is the real point — $6.5\times10^{-6}$ still sits more than three decades below the $\mathrm{Kn}\approx10^{-2}$ ceiling of the Euler regime. The fluid description holds for interstellar gas not because that gas resembles air but because it has three decades of margin to spend.
Second, the intracluster medium is where the approximation begins to strain, and it strains at the edges rather than the centre, because density falls outward faster than temperature does. Third, one row is different in kind.
At 1 AU the solar wind has $\mathrm{Kn}\approx1.9$. By the hierarchy of §6 there should be no fluid description of it at all, and the situation is more extreme than $\mathrm{Kn}\approx2$ on its own suggests. Take the radial scalings of problem K1, under which $\lambda(r) = 1.94\,r^{-2/3}$ AU with $r$ measured in AU, and integrate the collision rate along the outflow. The expected number of Coulomb collisions a proton suffers between the corona and Earth's orbit is $$ \int_0^{1}\frac{dr}{\lambda(r)} \;=\; \frac{1}{1.94}\int_0^1 r^{2/3}\,dr \;=\; \frac{3}{5\times1.94} \;=\; 0.31. $$ Fewer than one proton in three arriving at Earth has collided even once since it left the Sun. And yet magnetohydrodynamic models of the solar wind reproduce its large-scale structure — the spiral field, the stream interaction regions, the propagation of coronal mass ejections — with considerable success. Something other than collisions is confining the particles.
Put the solar wind numbers in. The interplanetary field at 1 AU is about 5 nT, or $5\times10^{-5}$ G in Gaussian units. A proton at $T=1.2\times10^5$ K has thermal speed $v_{\rm th}=\sqrt{2k_BT/m_p}=4.454\times10^{6}\ \mathrm{cm\,s^{-1}}$. Equation (8.1) gives
$$ r_g \;=\; \frac{(1.673\times10^{-24})(4.454\times10^{6})(2.998\times10^{10})}{(4.803\times10^{-10})(5\times10^{-5})} \;=\; 9.3\times10^{6}\ \mathrm{cm} \;=\; 93\ \mathrm{km}. \tag{8.2}$$Against $L=1\ \mathrm{AU}=1.5\times10^{8}\ \mathrm{km}$, the ratio is $r_g/L = 6.2\times10^{-7}$. Transverse to the field the particles are confined not by collisions but by the field itself, and confined more tightly than collisions would have managed by the ratio $\lambda/r_g = 3.1\times10^{6}$. Fig. 5 shows what that confinement looks like, and what it leaves untouched. This is the rescue, and it is the reason magnetohydrodynamics works in regimes where ordinary hydrodynamics has no right to.
Two systems are routinely called fluids in casual speech and are not fluids under any reading of Definition 2.
Stars in a galaxy. Take the solar neighbourhood: a stellar number density of about $0.1\ \mathrm{pc^{-3}}$, a velocity dispersion near $20\ \mathrm{km\,s^{-1}}$, and a geometric cross-section for physical contact of $\sigma=\pi(2R_\odot)^2 = 6.1\times10^{22}\ \mathrm{cm^2}$. Proposition 2 then gives a mean free path of $1.6\times10^{6}$ Gpc and a mean time between stellar collisions of $7.7\times10^{19}$ years, which is $5.5\times10^{9}$ Hubble times. Stars do not touch.
They do, however, exchange energy at a distance through gravity, and the relevant question is whether those encounters accumulate into anything. The timescale on which they do is the two-body relaxation time. For a self-gravitating system of $N$ bodies with crossing time $t_{\rm cross}=R/v$, Binney and Tremaine (2008) give
$$ t_{\rm relax} \;\approx\; \frac{0.1\,N}{\ln N}\; t_{\rm cross}. \tag{8.3}$$The $N/\ln N$ carries the same logic as the Coulomb logarithm of §5.2, and for the same reason: gravity, like the Coulomb force, is long-ranged, so many weak encounters dominate and the sum over impact parameters produces a logarithm. Evaluating (8.3):
| System | $N$ | $R$ | $v$ | $t_{\rm cross}$ | $t_{\rm relax}$ | $t_{\rm relax}/t_{\rm Hubble}$ |
|---|---|---|---|---|---|---|
| Globular cluster | $10^{5}$ | 10 pc | 5 km s$^{-1}$ | $2.0\times10^{6}$ yr | $1.7\times10^{9}$ yr | 0.12 |
| Milky Way disc | $10^{11}$ | 10 kpc | 200 km s$^{-1}$ | $4.9\times10^{7}$ yr | $1.9\times10^{16}$ yr | $1.4\times10^{6}$ |
| Galaxy cluster (galaxies) | $10^{3}$ | 1 Mpc | 1000 km s$^{-1}$ | $9.8\times10^{8}$ yr | $1.4\times10^{10}$ yr | 1.0 |
The three rows land in three different places, and the spread is the point. A galaxy is collisionless by a factor of a million: the stellar orbits it has today are the orbits it was born with, and no fluid description of them is admissible. A globular cluster relaxes in about an eighth of a Hubble time, so over its life it is collisional — which is why globular clusters undergo core collapse and mass segregation, processes that have no analogue in galaxies. The galaxies within a cluster sit almost exactly on the boundary.
Dark matter. Here the constraint runs the other way: we do not know the cross-section, and its smallness is an observational result. The merging cluster 1E 0657-56, the Bullet Cluster, separates its collisional component (the X-ray-emitting intracluster gas, which shocked and lagged) from its collisionless component (the galaxies and the lensing mass, which passed through). Randall and collaborators (2008) turned that separation into an upper limit on the dark matter self-interaction cross-section per unit mass, $\sigma/m \lt 0.7\ \mathrm{cm^2\,g^{-1}}$ at 68% confidence. Since $\lambda = 1/(\rho\,\sigma/m)$ for a medium of mass density $\rho$, and a cluster of $10^{15}\,M_\odot$ within 1 Mpc has a mean density $\rho = M/(\tfrac{4}{3}\pi R^3) = 1.6\times10^{-26}\ \mathrm{g\,cm^{-3}}$, which we round to $2\times10^{-26}$, even the extreme allowed value gives
$$ \lambda \;=\; \frac{1}{(2\times10^{-26})(0.7)} \;=\; 7.1\times10^{25}\ \mathrm{cm} \;=\; 23\ \mathrm{Mpc}, \tag{8.4}$$so $\mathrm{Kn}\ge23$ against the megaparsec size of a cluster. The core is denser than that mean by about a factor of six, and taking $\rho=10^{-25}\ \mathrm{g\,cm^{-3}}$ there gives $\lambda\ge4.6$ Mpc and $\mathrm{Kn}\ge4.6$ — smaller, but still greater than one in the region where a collision would have been most likely. Dark matter is collisionless, and the Bullet Cluster measurement is essentially a direct measurement of that fact.
All three systems — galaxies, dark matter, and the collisionless solar wind along $\mathbf{B}$ — are governed by the Boltzmann equation (2.1) with the right-hand side set to zero. That is the collisionless Boltzmann, or Vlasov, equation, and the discipline built on it is stellar dynamics rather than fluid dynamics. This book does not develop it. It is named here so that the boundary of what follows is explicit.
The fluid description we have earned is a single-component, isotropic, local-thermodynamic-equilibrium continuum. Each of those words is an idealisation that later modules either repay or flag.
| Idealisation | What it assumes | Where it is repaid |
|---|---|---|
| One fluid | Electrons and ions share one temperature. | See below — it fails behind cluster shocks. |
| Isotropic pressure | A single scalar $p$, no dependence on direction relative to $\mathbf{B}$. | Module 12 (magnetohydrodynamics; CGL noted in §8.1) |
| Ideal (Euler) flow | $\mu=\kappa=0$: no viscosity or conduction. | §6 gives the coefficients; Module 11 needs them for accretion discs |
| No self-gravity | Gravity, if present, is externally imposed. | Modules 3 and 5 (hydrostatic stars; the Jeans instability) |
| No magnetic field | The gas is unmagnetised or the field is dynamically negligible. | Module 12 |
| No radiation | Photons neither carry momentum nor dominate the energy budget. | Module 13 |
| Non-relativistic | $v \ll c$ and $k_BT \ll mc^2$. | Not repaid in this book; noted where it binds |
The first row deserves more than a pointer, because it fails inside the very system we used to validate §5.3. Electrons and protons at the same temperature carry very different momenta, and energy transfer between the two species is slow. Sarazin (1988) gives the equilibration times for intracluster conditions in the ratio $1 : 43 : 1870$ for electron–electron, proton–proton and proton–electron exchange respectively — the middle ratio being $\sqrt{m_p/m_e}=42.9$ and the last being close to $m_p/m_e = 1836$. At $T_m\approx10^8$ K and $n_e\approx10^{-3}\ \mathrm{cm^{-3}}$ the proton–electron equilibration time is about $6\times10^{8}$ years. A merger shock crossing a cluster takes of order $10^{9}$ years, so the gas behind such a shock is genuinely two-temperature: the protons are heated first and the electrons lag. Since X-ray observations measure the electron temperature, a single-fluid reading of the data underestimates the true thermal energy in exactly the regions where mergers are most violent.
That is a fair summary of the position this module leaves us in. The fluid equations are licensed for almost every system in the table of §7, the licence is quantitative and checkable, and the places where it lapses are identified rather than hidden. Module 2 writes the equations down.
The script below reproduces the two comparisons against published values, the Coulomb logarithm of §5.2 and the intracluster mean free path of §5.3. Every other plasma number in this module comes from the same four functions with different arguments. It has no dependencies beyond NumPy.
import numpy as np
kB = 1.380649e-16 # erg/K
e = 4.80320471e-10 # esu
me = 9.1093837015e-28 # g
hbar = 1.054571817e-27 # erg s
kpc = 3.0856775814913673e21 # cm
SPITZER = 3.0**1.5/(4.0*np.sqrt(np.pi))
def debye(n, T):
return np.sqrt(kB*T/(4.0*np.pi*n*e*e))
def b_min_e(T):
ve = np.sqrt(3.0*kB*T/me)
return max(e*e/(3.0*kB*T), hbar/(me*ve))
def ln_lambda(n, T):
return np.log(debye(n, T)/b_min_e(T))
def lam_coulomb(n, T, lnL=None):
lnL = ln_lambda(n, T) if lnL is None else lnL
return SPITZER*(kB*T)**2/(n*e**4*lnL)
n, T = 1.0e-3, 1.0e8
print("ln Lambda =", round(ln_lambda(n, T), 1), "(Sarazin 1988: 37.8)")
lam = lam_coulomb(n, T, lnL=37.8)/kpc
print("lambda =", round(lam, 1), "kpc (Sarazin 1988: 23 kpc)")
print("ratio =", round(lam/23.0, 3))
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Nothing about the gas changed; $L$ changed. The quoted figure uses $L=100$ pc, appropriate to a large cloud complex. The mean free path is 134 AU, so any structure whose interesting scale approaches that length has a Knudsen number of order unity. A concrete example is the transition layer at the edge of a cold neutral cloud, across which temperature falls from 8000 K to under 100 K; such interfaces are observed to be thin, and a model that must resolve a 100 AU interface has $\mathrm{Kn}\sim1$ and no fluid description of the interface itself. The bulk flow around the cloud remains perfectly fluid. The correct statement is that the Euler equations hold for interstellar problems whose resolution requirement exceeds a few thousand AU.
A hard sphere interacts over one length and one only, so there is a single scale and nothing to integrate over. The Coulomb force acts at every separation, and the strength of an encounter falls as $1/b$ while the number of encounters available at that separation rises as $b$. The two effects cancel in the mean-square deflection, so each logarithmic interval of impact parameter contributes the same amount, and summing over intervals from $b_{\min}$ to $b_{\max}$ produces the logarithm of their ratio.
If $\lambda_D$ were infinite, the integral in (5.3) would diverge at its upper limit and the deflection rate would be unbounded — that is, a plasma would have zero mean free path, which is plainly false. Screening is what makes the Coulomb problem finite. The finiteness is supplied by the collective response of the plasma, not by the two-body problem, which is why $\ln\Lambda$ cannot be derived from Rutherford scattering alone.
It measures a Knudsen number, or equivalently places a lower bound on a mean free path. In the collision the gas, which has $\mathrm{Kn}\ll1$, behaved as a fluid: it shocked, dissipated kinetic energy and was left behind between the two subclusters. The galaxies and the lensing mass, having $\mathrm{Kn}\gg1$, passed through each other without interacting. The separation between the two components is therefore a direct measure of the difference in their collisionality, and converting the observed offset into a limit gives $\sigma/m \lt 0.7\ \mathrm{cm^2\,g^{-1}}$ (Randall et al. 2008), hence $\lambda\ge23$ Mpc and $\mathrm{Kn}\ge23$.
The gas is essential because it is the control. Without a component known in advance to be collisional, the fact that the lensing mass passed through would carry no information: one would not know whether the collision had been violent enough to stop anything. The gas establishes that it was.
Let $\mathbf{v}_1,\mathbf{v}_2$ be independent and Maxwellian with the same $m$ and $T$, and put $\mathbf{v}_{\rm rel}=\mathbf{v}_1-\mathbf{v}_2$. Because the two are independent Gaussians in each Cartesian component, each component of $\mathbf{v}_{\rm rel}$ is Gaussian with variance $2(k_BT/m)$ — variances of independent variables add. So $\mathbf{v}_{\rm rel}$ is itself Maxwellian with the substitution $m \to m/2$, which is the reduced mass, as expected.
The mean speed of a Maxwellian is $\langle v\rangle = \sqrt{8k_BT/\pi m}$. Applying it with $m\to m/2$ gives $\langle v_{\rm rel}\rangle = \sqrt{16 k_BT/\pi m} = \sqrt{2}\,\langle v\rangle$.
The collision rate per particle is $n\sigma\langle v_{\rm rel}\rangle$, but the distance covered per unit time by the particle itself is $\langle v\rangle$, so $$\lambda = \frac{\langle v\rangle}{n\sigma\langle v_{\rm rel}\rangle} = \frac{1}{\sqrt{2}\,n\sigma}.$$ The distinction between the two speeds is the whole content of the factor: use the relative speed for the rate, the particle's own speed for the distance.
For $b\gg b_{90}$ the right side of (5.1) is small, so $\tan(\theta/2)\approx\theta/2$ and $\theta\approx 2b_{90}/b$.
In time $dt$ the particle sweeps an annulus of radius $b$, width $db$ and length $v\,dt$, of volume $2\pi b\,db\,v\,dt$. With target density $n$ the encounter rate in that annulus is $2\pi n v b\,db$.
Deflections from independent encounters have random azimuth, so the mean vector deflection vanishes and the mean square adds: $$\frac{d\langle\theta^2\rangle}{dt} = \int \theta^2(b)\,2\pi n v b\,db = 2\pi n v\int \frac{4b_{90}^2}{b^2}\,b\,db = 8\pi n v b_{90}^2\int\frac{db}{b},$$ which is (5.3). Substituting $b = e^{s}$ gives $db/b = ds$, so the integrand in $s$ is the constant $8\pi n v b_{90}^2$: every $e$-fold in impact parameter contributes equally. This is why the answer depends on the ratio $b_{\max}/b_{\min}$ only through its logarithm, and why an error of a factor of two in either limit changes $\ln\Lambda\approx37$ by less than 2%.
The conduction time over a length $L$ follows from the diffusion of heat: with thermal diffusivity $\chi = \kappa/(n k_B)\simeq \tfrac{1}{3}v_{\rm th}\lambda$, we have $t_{\rm cond}\sim L^2/\chi \sim 3L^2/(v_{\rm th}\lambda)$. The sound-crossing time is $t_{\rm s}\sim L/c_s$ with $c_s\propto v_{\rm th}$. Therefore $$\frac{t_{\rm cond}}{t_{\rm s}} \;\sim\; \frac{3L^2/(v_{\rm th}\lambda)}{L/v_{\rm th}} \;=\; \frac{3L}{\lambda} \;=\; 3\,\mathrm{Kn}^{-1}.$$ So the ratio is just the inverse Knudsen number, and with $\lambda\propto T^2/n$ from (5.6), $$\frac{t_{\rm cond}}{t_{\rm s}} \;\propto\; \frac{nL}{T^{2}}.$$
A cool core has high $n$, low $T$ and small $L$; an outskirt has low $n$, high $T$ and large $L$. Using the Knudsen numbers tabulated in §7, the ratio $3\,\mathrm{Kn}^{-1}$ is $1.4\times10^{3}$ in the core and $27$ in the outskirts. Conduction is therefore far too slow to matter dynamically in a core over a sound-crossing time, but in the outskirts it is within a factor of thirty of being competitive — which is the same statement as $\mathrm{Kn}\approx0.1$ there, read in the time domain instead of the length domain.
Write $r$ in AU. Then $n = 5r^{-2}$ and $T = 1.2\times10^5 r^{-4/3}$. From (5.6), $\lambda\propto T^2/n$, so $$\lambda(r) = \lambda(1)\cdot\frac{(r^{-4/3})^2}{r^{-2}} = \lambda(1)\,r^{-8/3+2} = \lambda(1)\,r^{-2/3},$$ with $\lambda(1)=1.9$ AU. The Knudsen number against $L=r$ is $$\mathrm{Kn}(r) = \frac{1.9\,r^{-2/3}}{r} = 1.9\,r^{-5/3}.$$ Setting this to one gives $r = 1.9^{3/5} = 1.49$ AU.
Two readings. First, $\mathrm{Kn}$ decreases outward here, and the crossing lies just outside Earth's orbit: the wind at 1 AU is already at $\mathrm{Kn}=1.9$, so on this model the fluid description fails everywhere inside about 1.5 AU and recovers beyond it. That is an artefact worth naming — it follows from assuming $L=r$, which makes the available scale grow faster than the mean free path shrinks. Judged instead against a fixed structure, such as the 0.01 AU thickness of a stream interaction region, the wind is collisionless throughout. Second, the exponent $-5/3$ is shallow, so the crossing radius is insensitive to the inputs: doubling $\lambda(1)$ moves it only to 2.3 AU. The conclusion that the solar wind is not collisional at scales a spacecraft can resolve is robust; the precise crossing radius is not, and adiabatic cooling is itself a poor model, since the observed wind cools more slowly than $r^{-4/3}$ because it is heated in transit.
From (5.6) with the Spitzer coefficient, $\lambda = 0.733(k_BT)^2/(n_e e^4\ln\Lambda)$. At $T=5\times10^7$ K, $n_e=10^{-2}\ \mathrm{cm^{-3}}$ and $\ln\Lambda=37$ this evaluates to $\lambda_0 = 0.575$ kpc. Since $\lambda\propto n_e^{-1}$ at fixed temperature, $$\lambda(r) = 0.575\left[1+(r/100)^2\right]\ \mathrm{kpc}, \qquad \mathrm{Kn}(r) = \frac{0.575\left[1+(r/100)^2\right]}{r},$$ with $r$ in kpc. Evaluating: $\mathrm{Kn} = 1.15\times10^{-2}$ at 100 kpc, $2.99\times10^{-2}$ at 500 kpc, $5.81\times10^{-2}$ at 1000 kpc, and $1.00\times10^{-1}$ at 1740 kpc. The threshold is crossed at $r\approx1.73$ Mpc, near the virial radius of a massive cluster.
Note the shape of $\mathrm{Kn}(r)$: it falls from the very centre, reaches a minimum where $r = r_c$, and rises as $r$ beyond that, because $\lambda$ grows as $r^2$ while $L$ grows only as $r$. The outskirts are always the fragile region.
On the resolution question the answer is the opposite of comfortable. At $r=1.73$ Mpc the mean free path is $\lambda = 0.575\times302 = 173$ kpc — seventeen times larger than the 10 kpc cell. The simulation therefore resolves, in detail, a range of scales on which the gas has no fluid description at all, and it solves the Euler equations on every one of them. Its cells at that radius are not small parcels of fluid; they are boxes containing particles that will not collide before they have crossed many boxes. Refining the mesh makes this worse, not better. Nothing is wrong with the arithmetic in such a simulation; what is wrong is the claim that its resolved structure in the outskirts represents gas dynamics.
The requirement $t_{\rm cond}/t_{\rm s}\ge100$ means $3/\mathrm{Kn}\ge100$, that is $\mathrm{Kn}\le0.03$, that is $\lambda \le 0.03\times50\ \mathrm{kpc} = 1.5\ \mathrm{kpc}$.
Invert (5.6) for $T$: $$T = \left(\frac{\lambda\, n_e e^4 \ln\Lambda}{0.733\,k_B^2}\right)^{1/2}.$$ With $\lambda = 1.5\times3.086\times10^{21} = 4.63\times10^{21}$ cm, $n_e=2\times10^{-2}$, $e^4 = 5.32\times10^{-38}$, $\ln\Lambda=36$ and $k_B=1.381\times10^{-16}$: numerator $= 4.63\times10^{21}\times2\times10^{-2}\times5.32\times10^{-38}\times36 = 1.77\times10^{-16}$; denominator $= 0.733\times1.907\times10^{-32} = 1.40\times10^{-32}$; ratio $=1.27\times10^{16}$, so $T = 1.13\times10^{8}$ K.
Since real cool cores are at $2$–$4\times10^{7}$ K, well under this bound, the condition is comfortably satisfied and unsuppressed conduction is slow compared with sound crossing in a core. Note what this does not say: the relevant comparison for the cooling-flow problem is conduction against the radiative cooling time, not against the sound-crossing time, and the cooling time in a dense core is far longer than $t_{\rm s}$. Choosing the comparison timescale is the whole of the physics here; the arithmetic is the easy part.
| Symbol | Meaning | Units (CGS) | Introduced |
|---|---|---|---|
| $f(\mathbf{x},\mathbf{v},t)$ | distribution function | cm$^{-6}$ s$^3$ | Definition 1 |
| $f_M$ | local Maxwellian | cm$^{-6}$ s$^3$ | (2.2) |
| $n$, $n_e$ | number density; electron number density | cm$^{-3}$ | Definition 2 |
| $\rho$ | mass density, $\rho = nm$ | g cm$^{-3}$ | Proposition 4 |
| $\mathbf{u}$ | bulk (fluid) velocity | cm s$^{-1}$ | Definition 2 |
| $v_{\rm th}$ | thermal speed, $\sqrt{2k_BT/m}$ | cm s$^{-1}$ | Definition 3 |
| $T$ | temperature | K | Definition 2 |
| $\tau$ | collision time, $\lambda/v_{\rm th}$ | s | Definition 3 |
| $\lambda$ | mean free path | cm | Definition 3 |
| $\sigma$ | collision cross-section | cm$^2$ | Definition 4 |
| $b$, $b_{90}$ | impact parameter; its $90^\circ$ value | cm | (5.1), (5.2) |
| $\lambda_D$ | Debye screening length | cm | (5.4) |
| $\ln\Lambda$ | Coulomb logarithm | — | Definition 5 |
| $\mathrm{Kn}$ | Knudsen number, $\lambda/L$ | — | Definition 6 |
| $L$ | smallest scale the problem must resolve | cm | Definition 6 |
| $\mu$, $\kappa$ | dynamic viscosity; thermal conductivity | g cm$^{-1}$ s$^{-1}$; erg cm$^{-1}$ s$^{-1}$ K$^{-1}$ | (6.2) |
| $r_g$ | gyroradius | cm | (8.1) |
| $t_{\rm relax}$ | two-body relaxation time | s | (8.3) |
| Book-wide note. In this module and in Module 2, $\mu$ is the dynamic viscosity. From Module 6 on, $\mu$ means the mean molecular weight throughout the book, as it already does in Modules 3, 4 and 5, and the dynamic viscosity is written $\eta$. | |||
Numerical values compared against published results are identified in the text at the point of use. The comparisons made in this module are:
Values used as inputs rather than as checks carry a provenance note where they appear: the laboratory vacuum pressures of §1, the photospheric pressure and neutral-hydrogen cross-section of §4.1, and the representative solar-wind parameters of §7. The cluster mean density used in (8.4) is derived in place from $10^{15}\,M_\odot$ within 1 Mpc rather than quoted.