Module 5 — The Jeans Instability

Gravity against pressure: the Jeans dispersion relation, the Jeans swindle stated openly and repaired twice, first by an expanding background and then by a static isothermal sphere, and the Bonnor–Ebert criterion, which calls the dense core Barnard 68 unstable where the Jeans criterion calls it stable. Two checks against published observations refute the reading of the fitted cores as stable equilibria and refute the static isothermal sphere as a description of the outer half of Barnard 68.

1. The problem: a cloud that two criteria judge differently

Modules 2 to 4 treated gravity as a field imposed from outside. Module 2 listed that as an idealisation, because a gas that cannot be the source of its own gravity cannot collapse under it. Module 4 held gravity fixed while it perturbed the gas, and noted that the perturbation of gravity is decisive at long wavelengths. This module removes the restriction. The gas now makes its own gravitational field, and the question is when that field wins against the pressure of the gas.

The question has a concrete astronomical form. Barnard 68 is a small, dark, isolated cloud of cold molecular gas. The fit of Alves, Lada & Lada (2001), as tabulated by Kandori et al. (2005), gives it a mass of $2.10\,M_\odot$ inside a radius of $1.25\times10^{4}$ AU at an assumed distance of $125$ pc, with a temperature of $16$ K. Its mean density is therefore $1.5245\times10^{-19}\ \mathrm{g\,cm^{-3}}$.

Two stability criteria are derived in this module, and both can be applied to these numbers. The first, Jeans's criterion, treats the cloud as a piece of a uniform medium. It gives a critical mass of $2.97\,M_\odot$, larger than the cloud's mass, so it calls the cloud stable. The second, the Bonnor–Ebert criterion, treats the cloud as a pressure-bounded isothermal sphere and uses the shape of its density profile. The measured shape parameter is $\xi_{\max} = 6.9\pm0.2$ against a critical value of $6.4508$, so it calls the cloud unstable.

Both criteria come from the same fluid equations. They disagree because the first is derived about a state that does not satisfy those equations, and §7.2 proves that the Jeans criterion at the mean density calls every isothermal equilibrium sphere stable. §3 derives the Jeans criterion in the standard way. §4 shows what the standard derivation discards, the "Jeans swindle", and repairs it with an expanding background. §5 repairs it a second time with a static equilibrium that does exist. §6 and §7 take the result to Barnard 68 and to thirteen other Bok globules (fourteen other rows of one published table).

2. Self-gravity and the isothermal closure

The gas is described by the continuity equation and the Euler equation of Module 2, (5.1) and (7.3), with Module 2's $p$ written $P$. The acceleration $\mathbf{g}$ is now written as the gradient of a gravitational potential $\Phi$, and $\Phi$ is produced by the gas itself through Poisson's equation:

Here $\rho$ is the mass density, $\mathbf{u}$ the bulk velocity, $P$ the pressure and $G$ the gravitational constant. Equation (2.3) couples the gas back to its own field. It is what Modules 2 to 4 left out.

The system needs a closure. For the cold interstellar clouds of this module the closure is isothermal. Module 2, in the solution to its problem D3, showed that holding $T$ fixed gives $P \propto \rho$ with a constant coefficient, and named the physical assumption: heating and cooling are fast enough to hold each parcel at the ambient temperature however quickly it is compressed.

In cold molecular gas the hydrogen is molecular and the helium atomic. With one helium atom for every ten hydrogen nuclei, each $\mathrm{H_2}$ molecule comes with $0.2$ helium atoms. The mass per $\mathrm{H_2}$ molecule is $(2.016 + 0.2\times4.003)\,m_u = 2.8\,m_u$ and the number of particles is $1.2$, so $\mu = 2.8/1.2 = 7/3$. (Module 2 writes the particle mass as $\mu_{\rm m}m_p$; this module uses $\mu\,m_u$, as Modules 3 and 4 do.) Kandori et al. (2005) round this to $2.33$, and this module uses their value so that its numbers can be compared with their tables. The mass density is $\rho = 2.8\,m_u\,n(\mathrm{H_2})$. At $10$ K, (2.4) gives $c_s = 0.1889\ \mathrm{km\,s^{-1}}$.

3. The Jeans instability

Module 4 perturbed a static equilibrium and found sound waves. The same method is applied here to a gas that makes its own gravity. The equilibrium Jeans chose is the simplest possible one: a medium of uniform density $\rho_0$ and uniform pressure $P_0$, at rest, filling all space. §4 shows that this state is not a solution of (2.1)–(2.3). This section follows Jeans and makes the assumption explicitly, so that its consequences can be traced.

With $G = 0$, (3.1) is the isothermal sound wave of Module 4. Gravity subtracts a constant from $\omega^2$. For short wavelengths the pressure term wins and the perturbation oscillates. For long wavelengths the right-hand side is negative, $\omega$ is imaginary, and one of the two solutions grows exponentially. Fig. 1 shows the relation.

Perturbations with $\lambda\gt\lambda_J$ grow, and those with $\lambda\lt\lambda_J$ oscillate. The Jeans mass is a convention, not a derived threshold: (3.1) is a statement about plane waves, and the choice of a sphere of diameter $\lambda_J$ is one of several in use. A cube of side $\lambda_J$ gives a mass larger by $6/\pi = 1.9099$. The factor matters in §7.2, where the Jeans mass is compared with a measured mass, and it is stated there.

For a molecular cloud core at $10$ K and $n(\mathrm{H_2}) = 10^4\ \mathrm{cm^{-3}}$, so that $\rho_0 = 4.6495\times10^{-20}\ \mathrm{g\,cm^{-3}}$, (3.2) and (3.3) give $\lambda_J = 40177$ AU, or $0.19479$ pc, and $M_J = 2.6584\,M_\odot$.

3.1 Two timescales that differ by a fixed factor

The growth rate of an unstable mode is $s = (4\pi G\rho_0 - c_s^2k^2)^{1/2}$, and it is largest for $k\to0$, where $s\to(4\pi G\rho_0)^{1/2}$. The inverse of that limit is the growth time $t_{\rm grow} \equiv (4\pi G\rho_0)^{-1/2}$. A second timescale is used for the same purpose in the literature: the time a pressureless uniform sphere takes to collapse to a point.

The two timescales are the same order and are not equal. The free-fall time is $1.92$ growth times. For the $10$ K core, $t_{\rm grow} = 1.6047\times10^{5}$ yr and $t_{\rm ff} = 3.0871\times10^{5}$ yr. This module quotes whichever one a source uses and names it.

3.2 The Jeans mass in three regimes

Table 1 applies (3.2) and (3.3) to three gases. The Jeans mass scales as $T^{3/2}\mu^{-3/2}\rho_0^{-1/2}$, so between the first two rows a factor of $800$ in temperature, $4\times10^{4}$ in density and $1.83$ in $\mu$ move it by a factor of $1.1\times10^{7}$.

The warm neutral medium has a Jeans length of $1.49$ kpc and a Jeans mass of $3\times10^{7}\,M_\odot$. A uniform gas in that state cannot fragment into anything smaller. The cold core has a Jeans mass of a few solar masses, which is the scale of a star. The difference is almost entirely temperature and density, and it is why star formation happens in cold molecular gas.

4. The Jeans swindle, and a background that has no swindle

Proposition 1 assumed that $\nabla\Phi_0 = \mathbf{0}$. The next proposition shows that this assumption contradicts the equations the proposition was derived from.

Jeans's derivation keeps the perturbed Poisson equation $\nabla^2\Phi_1 = 4\pi G\rho_1$ and drops the unperturbed one. This step is known as the Jeans swindle. The name is not a judgement that (3.1) is useless; it is a statement that its derivation is inconsistent.

The size of what is dropped can be computed. Two parcels a distance $L$ apart in a uniform medium accelerate towards each other at $g(L) = (4\pi/3)G\rho_0L$, by Gauss's theorem applied about either one. The time over which that acceleration changes a distance $L$ appreciably is $(L/g)^{1/2}$. Its ratio to the growth time is

for every $L$, density and temperature. For the $10$ K core at $L = \lambda_J$, $g = 7.8128\times10^{-9}\ \mathrm{cm\,s^{-2}}$ and $(L/g)^{1/2} = 2.7794\times10^{5}$ yr against $t_{\rm grow} = 1.6047\times10^{5}$ yr. The discarded background acceleration acts on the same timescale as the instability it is used to derive. It is not a small correction that can be neglected in a first approximation. (The time to fall a distance $L$ from rest at constant $g$ is $(2L/g)^{1/2}$, larger by $\sqrt2$; the conclusion is the same.)

There are two ways to remove the inconsistency. The first keeps the medium uniform and lets it move: an expanding or contracting uniform medium is an exact solution. The second keeps it static and lets the density vary: a stratified cloud is an exact solution. This section takes the first; §5 takes the second.

4.1 The expanding background

The background potential is now kept, and it is balanced by the deceleration of the expansion rather than by a pressure gradient. The choice of origin in $\Phi_0$ is arbitrary: moving the origin adds a uniform velocity and a uniform acceleration, which changes no relative motion. The perturbation problem about this background has no swindle in it.

Equation (4.3) has the sign structure of (3.1): the physical wavenumber is $k/a$, and the right-hand side is positive, driving growth, exactly when $k/a\lt k_J$. The Jeans length reappears without any discarded term. The new term $2H\dot\delta$ is a drag produced by the expansion. It changes the growth from exponential to a power law.

To see this, take a background with zero pressure in which the expansion is set by the gas itself, $a\propto t^{2/3}$. Then $H = 2/(3t)$, and (4.2) gives $4\pi G\rho_0 = 2/(3t^2)$. In the limit $k\to0$, (4.3) becomes $\ddot\delta + (4/3t)\dot\delta - (2/3t^2)\delta = 0$. Trying $\delta\propto t^p$ gives $p^2 + p/3 - 2/3 = 0$, so $p = 2/3$ or $p = -1$. The growing mode is $\delta\propto t^{2/3}\propto a$.

A perturbation that starts growing at $z_* = 1089.92$ therefore grows by the factor by which $a$ grows, $1+z_* = 1090.92$, for the growing mode alone (a perturbation that starts with $\dot\delta = 0$ puts $3/5$ of its amplitude in that mode), not by an exponential of the elapsed time over $t_{\rm grow}$. This is the rescue: the instability survives, and its rate is changed by the term that the swindle had discarded.

5. A static equilibrium that exists: the isothermal sphere

The second repair keeps the gas at rest and gives up uniformity. A static isothermal gas that makes its own gravity must be stratified, as the atmospheres of Module 3 were. Module 3 solved the polytropic case, $P\propto\rho^{1+1/n}$, through the Lane–Emden equation. The isothermal closure (2.4) is the limit $n\to\infty$, and it needs its own variables.

Property (d) is the whole difference from Module 3. A polytrope with $n\lt5$ reaches zero density at a finite radius. The isothermal sphere never does; its mass grows without bound as $\xi\to\infty$, since by (b) and (c) the mass approaches $2c_s^2r/G$. A real isothermal cloud must therefore be cut off by something outside it.

The numerical solution used below is a fourth-order Runge–Kutta integration of (5.2) started from the series (a). The series and the integration agree to a ratio of $1.000004$ at $\xi = 0.05$ and $1.0000002$ at $\xi = 0.2$. The difference falls as $\xi$ grows, so it is the start-up error of the integration, not the truncation of the series; the comparison tests the start of the integration. At large $\xi$ the solution oscillates about the singular sphere with a decaying amplitude: the ratio $\rho/\rho_{\rm SIS}$ is $1.1918$ at $\xi = 10$, $0.8998$ at $20$, $0.8425$ at $50$, $0.9242$ at $100$ and $1.0158$ at $200$. Because of the overshoot, the local logarithmic slope $-d\ln\rho/d\ln r = \xi\psi'$ does not rise monotonically to $2$. It reaches a maximum of $2.5176$ at $\xi = 8.993$ and then decays. Every regular isothermal sphere is this one solution rescaled, so this maximum holds for every $\xi_{\max}$; on the integrated range $0\lt\xi\le200$ the later extrema decay towards $2$. §7.3 uses that ceiling. The upper panel of Fig. 2 shows the profile and the singular sphere it oscillates about.

For a fixed temperature, a Bonnor–Ebert sphere has two parameters, $\rho_c$ and $\xi_{\max}$. Its shape, the profile $\rho/\rho_c$ against $r/R$, depends only on $\xi_{\max}$. This is what makes the model testable against a single measured profile: one number fixes the shape.

The function $m(\xi)$ rises from zero, reaches a maximum and falls. The maximum, located on the numerical solution and refined by a parabola through the three bracketing samples, is

Halving the integration step moves $\xi_{\rm crit}$ by $8.7\times10^{-8}$ and $m_{\rm crit}$ by $2.9\times10^{-9}$. The coefficient $1.18$ in the Bonnor–Ebert mass $M_{\rm BE} = 1.18\,c_s^4/(G^{3/2}P_{\rm ext}^{1/2})$ is $m_{\rm crit}$. Kandori et al. (2005) give the critical state as $\xi_{\max} = 6.5$ with a contrast of $14$, which (5.6) reproduces to rounding.

The maximum has a direct consequence that needs no stability theory. For a given external pressure and temperature, no isothermal sphere in hydrostatic equilibrium has a mass above $M_{\rm BE}$. A cloud of larger mass in that environment has no static state to be in. The stability statement is the next proposition, and it needs an assumption.

Two limits of the proposition must be stated. First, it rests on the static criterion, which is an assumption here; a linear analysis of radial pulsations of the sphere is not carried out in this module. Second, it says nothing beyond $\xi_{\max} = 8.9937$, where the radius at fixed mass reaches its minimum, $R_{\min} = GM/(2.5176\,c_s^2)$, and the sign test is no longer used. The test must not be used there: at $\xi_{\max} = 10$ it returns $d\ln P_{\rm ext}/d\ln R = -12.09$, which would read as stable, on a sphere that lies further along the unstable branch. The general tool is the turning-point method, in which stability changes only where the equilibrium sequence has an extremum; it is not developed here. The criterion is applied below only to values of $\xi_{\max}$ on the range where it holds, or to state that a value lies above $\xi_{\rm crit}$.

The only rate that an isothermal sphere contains is $(G\rho_c)^{1/2}$ multiplied by a function of $\xi_{\max}$. The timescale on which an unstable sphere departs from equilibrium is therefore $(G\rho_c)^{-1/2}$ times a factor that depends on $\xi_{\max}$. The factor becomes infinite at $\xi_{\rm crit}$, where the growth rate passes through zero, and it is not computed here.

A short program reproduces (5.6). It integrates (5.2) and locates the maximum of $m(\xi)$ on the grid.

import numpy as np


def rhs(x, y):
    psi, dpsi = y
    return np.array([dpsi, np.exp(-psi) - 2*dpsi/x])


h, x = 1e-3, 1e-3
y = np.array([x*x/6, x/3])
best = (0.0, 0.0)
while x < 12:
    k1 = rhs(x, y)
    k2 = rhs(x + h/2, y + h*k1/2)
    k3 = rhs(x + h/2, y + h*k2/2)
    k4 = rhs(x + h, y + h*k3)
    y, x = y + h*(k1 + 2*k2 + 2*k3 + k4)/6, x + h
    m = x*x*y[1]*np.exp(-y[0]/2)/np.sqrt(4*np.pi)
    best = max(best, (m, x))
m, xc = best
print(f"xi_crit = {xc:.3f}, m_crit = {m:.4f}")

Running it prints $\xi_{\rm crit} = 6.451$ and $m_{\rm crit} = 1.1822$. The grid step of $10^{-3}$ limits $\xi_{\rm crit}$ to three decimals; figs/m05_numbers.py refines it.

6. Barnard 68 and the Bok globules

A Bok globule is a small, isolated dark cloud, seen against the stars behind it. The dust in the cloud dims and reddens those stars, and the reddening of each star measures the column of material in front of it. Mapping many stars gives the column density across the cloud, and averaging it in rings gives a radial profile. A Bonnor–Ebert sphere projected onto the sky predicts that profile with one shape parameter.

Kandori et al. (2005) fitted Bonnor–Ebert spheres to near-infrared extinction profiles of ten globules and tabulated, with them, fits from the literature to four more, including Alves, Lada & Lada (2001) for Barnard 68. Their Table 4 has fifteen rows, because the Coalsack appears twice, from two different papers. Each row gives $\xi_{\max}$ with its fitted error and a centre-to-edge contrast. Their Table 5 gives the temperature, distance, radius, mass and external pressure.

For Barnard 68 the fit is $\xi_{\max} = 6.9\pm0.2$, a precision of $2.9$ per cent. The first Barnard 68 line of Table 5 gives $T = 16$ K, a distance of $125$ pc, $R = 1.25\times10^{4}$ AU, $M = 2.10\,M_\odot$ and $P_{\rm ext}/k_B = 1.8\times10^{5}\ \mathrm{K\,cm^{-3}}$. With $\mu = 2.33$, $c_s = 0.2389\ \mathrm{km\,s^{-1}}$. The mean density from $M$ and $R$ is $1.5245\times10^{-19}\ \mathrm{g\,cm^{-3}}$, against $1.5\times10^{-19}$ quoted by Burkert & Alves (2009), a ratio of $1.0163$. The free-fall time (3.4) at that density is $1.7049\times10^{5}$ yr, against their $0.17\times10^{6}$ yr, a ratio of $1.0029$.

What the fit determines. The profile fixes the shape, $\xi_{\max}$, but not the scale. Kandori et al. (2005, section 4.2.2) show, crediting Lai et al. (2003), that substituting $R\propto d$ and $\rho_c\propto d^{-1}$ for a cloud at distance $d$ into the definition of $\xi_{\max}$ gives $T/d = \text{constant}$. The same fitted shape is therefore consistent with a line of temperature-distance pairs. The second Barnard 68 line of Table 5, from Hotzel, Harju & Juvela (2002), takes the measured temperature $10\pm1.2$ K and reads $d = 85$ pc, $R = 0.85\times10^{4}$ AU and $M = 0.90\,M_\odot$. At fixed $\xi_{\max}$ the mass scales as $c_s^2R\propto Td$, which predicts a mass ratio of $2.3529$ between the two lines; the table gives $2.3333$, a ratio of $0.9917$.

Is the support thermal? Lada et al. (2003) measure a gas temperature of $10.5$ K and line widths of $0.18\pm0.01\ \mathrm{km\,s^{-1}}$ in $\mathrm{C^{18}O}$ and $0.15\pm0.01\ \mathrm{km\,s^{-1}}$ in $\mathrm{C^{34}S}$. The thermal width (full width at half maximum) of a $\mathrm{C^{18}O}$ molecule, mass $30\,m_u$, at $10.5$ K is $0.1270\ \mathrm{km\,s^{-1}}$. Subtracting it in quadrature leaves a non-thermal width of $0.1275\ \mathrm{km\,s^{-1}}$. The thermal width of the mean particle, mass $2.33\,m_u$, is $0.4558\ \mathrm{km\,s^{-1}}$, $3.57$ times larger. Lada et al. state from their own analysis that thermal pressure exceeds non-thermal pressure by a factor of $4$ to $5$ in the central regions. The premise that pressure is mainly thermal is supported; the premise that it is isothermal is tested in §7.3.

A validation, not a check. The contrast column of Kandori's Table 4 was "determined from $\xi_{\max}$ value" (Table 4, note b). It is the authors' own integration of (5.2). Integrating (5.2) here reproduces all fifteen entries, which span $6.63$ to $364$, with a largest residual of $2.2$ per cent (CB 110: $96.38$ computed against $94.30$ printed). For Barnard 68 the computed contrast is $16.56$ against $16.6$. This confirms that the integrator of this module and the integrator behind the table agree. It says nothing about the clouds.

7. The checks

Three statements are tested against measurements. The first is that the fitted globules are stable static equilibria. The second compares the two stability criteria on one cloud. The third is that the gas of Barnard 68 is isothermal.

7.1 Check 1: the fitted cores are not stable equilibria

Fig. 3 places every row of Kandori's Table 4 against $\xi_{\rm crit} = 6.4508$. Twelve of the fifteen rows lie above it. Counting globules rather than rows, and only those without an embedded star, Kandori et al. write: "there are three stable starless globules and eight unstable starless globules." Barnard 68 lies $(6.9 - 6.4508)/0.2 = 2.25$ standard deviations above the critical value. Lynds 694-2, which contains a star, lies $6.18$ standard deviations above. The two rows below by a significant margin are CB 87, at $3.38$ standard deviations, and Lynds 498, at $4.38$. The Coalsack's two rows give opposite verdicts ($5.8$ and $7.0$); Kandori's count of three stable globules includes it as stable.

What this refutes must be stated precisely. It does not refute hydrostatic equilibrium, the isothermal closure or the fits, because these are the instrument by which $\xi_{\max}$ was measured. It refutes the reading that these clouds are long-lived stable equilibria of the kind the fits describe. How fast an unstable sphere departs is not computed here. The rate is $(G\rho_c)^{1/2}$ times a function of $\xi_{\max}$ that passes through zero at $\xi_{\rm crit}$, where the sequence changes stability, so a sphere just above the critical point, as Barnard 68 is at $\xi_{\max}/\xi_{\rm crit} = 1.0696$, can depart much more slowly than the free-fall time, $1.70\times10^{5}$ yr at its mean density.

The tension is sharpest for Barnard 68 because it has been read as long-lived. Lada et al. (2003) interpret its line profiles as oscillations of its outer layers. Burkert & Alves (2009) note that on this reading the cloud should be older than one oscillation time, $2R/V = 3\times10^{6}$ yr for velocities $V\approx0.04\ \mathrm{km\,s^{-1}}$. That is $17.6$ free-fall times at the mean density. They state this in order to argue against it, and propose that a future collision with a small core will trigger its collapse. Kandori et al. offer a different resolution: a sphere that begins to collapse from a nearly critical state "should mimic marginally unstable Bonnor-Ebert spheres for a long time". Neither resolution is tested here.

7.2 The anchor: why the Jeans criterion cannot see the instability

The Jeans criterion can be applied to Barnard 68 at its own mean density and at the temperature its own fit requires. With $\rho_0 = 1.5245\times10^{-19}\ \mathrm{g\,cm^{-3}}$ and $c_s = 0.2389\ \mathrm{km\,s^{-1}}$, (3.2) gives $\lambda_J = 28066$ AU, or $0.1361$ pc, which is $1.1226$ times the cloud's diameter of $25000$ AU. Equation (3.3) gives $M_J = 2.9713\,M_\odot$, against the published $2.10\,M_\odot$, a ratio $M_J/M = 1.4149$. The Jeans criterion calls the cloud stable; the Bonnor–Ebert criterion calls it unstable at $2.25\sigma$. The next proposition shows that the first verdict is not specific to Barnard 68.

At $\xi_{\max} = 6.9$, (7.1) gives $1.5409$; at $\xi_{\rm crit}$ it gives $1.5709$. The published value $1.4149$ differs from $1.5409$ by the factor $(2.10/1.9839)^{3/2}$, the internal inconsistency of the published row found below. The Jeans criterion at the mean density therefore calls every isothermal equilibrium sphere stable, including every sphere on the unstable branch. This is not an observation about Barnard 68. It is what the uniform-background assumption of Proposition 1 costs.

Proposition 9 explains the two lines of Table 5. On the second line ($10$ K, $85$ pc, $0.90\,M_\odot$), $M_J/M = 1.3972$, against $1.4149$ on the first, a ratio of $0.9875$. Both lines carry the same $\xi_{\max}$, so (7.1) requires equal ratios; the residual is the rounding and inconsistency of the tabulated $M$ and $R$. The cube convention of §3 multiplies (7.1) by $1.9099$. A value below $1$ needs a temperature and mass that are not a Bonnor–Ebert sphere: $10.5$ K with $M = 2.10\,M_\odot$ at the $125$ pc density gives $M_J/M = 0.7522$, but at fixed $\xi_{\max}$ Proposition 7 requires $M\propto Td$, so that pair describes no fitted sphere.

The two criteria disagree because they ask different questions. The Jeans criterion asks whether a uniform medium of the cloud's mean density is unstable. The cloud is not uniform: its centre is $16.6$ times denser than its edge. The Bonnor–Ebert criterion is stated in exactly that contrast. The uniform medium that the Jeans criterion assumes is the state that Proposition 3 showed cannot exist.

The mass itself is less certain than either ratio. From $P_{\rm ext}$ and $\xi_{\max} = 6.9$, (5.4) gives $2.2496\,M_\odot$, $1.0712$ times the published value. From $R$ and $\xi_{\max}$, Proposition 7 gives $1.9839\,M_\odot$, a ratio of $0.9447$. The two routes differ by $13.4$ per cent, so the published row is not internally consistent at better than that, and no mass statement about this cloud is better than that. The critical mass at the published pressure is $M_{\rm BE} = 2.2548\,M_\odot$. The mass of a sphere at $\xi_{\max} = 6.9$ is $0.9977$ of the maximum. A mass this close to $M_{\rm BE}$ is compatible with both branches of Fig. 2; only the contrast separates them.

7.3 Check 2: the outer profile of Barnard 68 is not a static isothermal sphere

Nielbock et al. (2012) modelled Herschel far-infrared maps of Barnard 68 with radiative transfer and derived the radial profiles of dust temperature and gas density. Two of their results bear on the closure.

The first is the temperature. The dust temperature falls from $16.7^{+1.3}_{-1.0}$ K at the edge to $8.2^{+2.1}_{-0.7}$ K at the centre, a factor of $2.04$. An isothermal sphere assumes a factor of $1$. Dust temperature is not gas temperature, so this is suggestive rather than decisive.

The second is the density slope, which Nielbock et al. derive from the same radiative-transfer model. Their abstract gives $n_{\rm H}\propto r^{-3.5}$ outside the inner plateau. §5 showed that the local slope $-d\ln\rho/d\ln r$ of an isothermal sphere does not exceed $2.5176$ on the integrated range, for any $\xi_{\max}$. The measured slope is $1.390$ times that ceiling.

A comparison at matched radius is sharper. Their adopted profile is $n_{\rm H}(r) = \Delta n/[1 + (r/r_0)^2]^{\eta/2} + n_{\rm out}$ with $r_0 = 35''$, $\eta = 4.0$, $n_0 = \Delta n + n_{\rm out} = 3.4\times10^{5}\ \mathrm{cm^{-3}}$ and $n_{\rm out} = 4\times10^{2}\ \mathrm{cm^{-3}}$ (their equation 8 and adopted values). The Bonnor–Ebert sphere of Alves et al. has $\xi_{\max} = 6.9$ at $100''$, so its local slope at angular radius $\theta$ is $\xi\psi'$ at $\xi = 6.9\,\theta/100''$. Both profiles are functions of angle, so neither distance enters. Fig. 4 compares them. The measured slope exceeds the isothermal ceiling for $\theta\gt46.2''$, that is over the outer $53.8$ per cent of the fitted radius. At $60''$ it is $2.9309$ against $2.0219$ for the fitted sphere; at $100''$ it is $3.2428$ against $2.4660$, a ratio of $1.315$.

What is refuted is the conjunction of four assumptions for the outer half of Barnard 68: static, spherical, isothermal and self-gravitating. The fluid equations are not refuted. The slope test alone does not say which assumption fails. A gas whose temperature rises outward, as the dust temperature does here, has a pressure that falls more slowly than its density, so it can hold a steeper density profile in hydrostatic balance; departure from spherical symmetry and a non-static outer layer are the other candidates, and this module separates none of them. The result also bears on Check 1: $\xi_{\max} = 6.9$ was fitted with the isothermal sphere that fails over the outer $53.8$ per cent of the radius, so the $2.25\sigma$ of Check 1 holds only inside that model.

8. What this model captures, and what it misses

The module used the fluid equations (2.1)–(2.3) with the isothermal closure (2.4), about a uniform background in §3–§4 and about a static sphere in §5. Table 2 names each idealisation and where it is repaid.

9. Problems

Solutions are given for every problem. Each numerical answer is verified by figs/m05_problems_check.py.

Conceptual ↑ contents

Derivational ↑ contents

Computational ↑ contents

Notation

Sources

Every published value this module compares against, with what was taken from it.

1. The Bok globule fits. R. Kandori, Y. Nakajima, M. Tamura, K. Tatematsu, Y. Aikawa, T. Naoi, K. Sugitani, H. Nakaya, T. Nagayama, T. Nagata, M. Kurita, D. Kato, C. Nagashima and S. Sato, Near-Infrared Imaging Survey of Bok Globules: Density Structure, The Astronomical Journal 130, 2166–2184 (2005), arXiv:astro-ph/0506205. Quoted: Table 4, $\xi_{\max}$, its error and $n_c/n_{\rm edge}$ for fifteen rows, with note b ("determined from $\xi_{\max}$ value"); Table 5, the two Barnard 68 lines; the critical state $\xi_{\max} = 6.5$ and contrast $14$; the count of three stable and eight unstable starless globules; section 4.2.2, the $T/d$ degeneracy; the collapsing-sphere statement "should mimic marginally unstable Bonnor-Ebert spheres for a long time".
2. The original Barnard 68 fit. J. F. Alves, C. J. Lada and E. A. Lada, Nature 409, 159 (2001). Not read; its values enter through Kandori et al. (2005), Tables 4 and 5.
3. Barnard 68 kinematics and temperature. C. J. Lada, E. A. Bergin, J. F. Alves and T. L. Huard, The Dynamical State of Barnard 68: A Thermally Supported, Pulsating Dark Cloud, The Astrophysical Journal 586, 286–295 (2003), arXiv:astro-ph/0211507. Quoted: gas temperature $10.5$ K; line widths $0.18\pm0.01\ \mathrm{km\,s^{-1}}$ ($\mathrm{C^{18}O}$) and $0.15\pm0.01\ \mathrm{km\,s^{-1}}$ ($\mathrm{C^{34}S}$); thermal pressure $4$–$5$ times the non-thermal pressure in the central regions; the pulsation interpretation.
4. Barnard 68 timescales. A. Burkert and J. Alves, The Inevitable Future of the Starless Core Barnard 68, The Astrophysical Journal 695, 1308–1314 (2009), doi:10.1088/0004-637X/695/2/1308, arXiv:0809.1457. Quoted from page 1: mean density $1.5\times10^{-19}\ \mathrm{g\,cm^{-3}}$; collapse time $(3\pi/32G\rho)^{1/2} = 0.17\times10^{6}$ yr; $V\approx0.04\ \mathrm{km\,s^{-1}}$ and $\tau_{\rm dyn} = 2R/V = 3\times10^{6}$ yr; the proposed collision with a smaller core.
5. Barnard 68 dust temperature and density. M. Nielbock, R. Launhardt, J. Steinacker, A. M. Stutz, Z. Balog et al., The Earliest Phases of Star formation (EPoS) observed with Herschel: the dust temperature and density distributions of B68, Astronomy & Astrophysics 547, A11 (2012), arXiv:1208.4512. Quoted: dust temperature $16.7^{+1.3}_{-1.0}$ K at the edge and $8.2^{+2.1}_{-0.7}$ K at the centre; $n_{\rm H}\propto r^{-3.5}$ outside the inner plateau (abstract); the adopted profile, equation 8, with $r_0 = 35''$, $\eta = 4.0$, $n_0 = 3.4\times10^{5}\ \mathrm{cm^{-3}}$, $n_{\rm out} = 4\times10^{2}\ \mathrm{cm^{-3}}$; the second fit with slopes $3.4$–$3.7$ for $r\le190''$.
6. Cosmological parameters. Planck Collaboration (N. Aghanim et al.), Planck 2018 results. VI. Cosmological parameters, Astronomy & Astrophysics 641, A6 (2020), arXiv:1807.06209. Table 2, column TT,TE,EE+lowE+lensing: $\Omega_bh^2 = 0.02237\pm0.00015$, $H_0 = 67.36\pm0.54\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$, $z_* = 1089.92\pm0.25$. Used as inputs only.
7. Cited through Kandori et al., not read. W. B. Bonnor, Monthly Notices of the Royal Astronomical Society 116, 351 (1956); R. Ebert, Zeitschrift für Astrophysik 37, 217 (1955); S. Hotzel, J. Harju and M. Juvela, Astronomy & Astrophysics 395, L5 (2002); S. Lai, T. Velusamy, W. D. Langer and T. B. H. Kuiper, The Astronomical Journal 126, 311 (2003).
8. Constants. CODATA 2018 for $G$, $k_B$ and $m_u$; IAU 2015 Resolution B3 for $GM_\odot$.
9. Earlier modules. Continuity (5.1) and the Euler equation (7.3) of Module 2, and the isothermal closure of its problem D3; the Lane–Emden equation (Proposition 7) of Module 3; the perturbation method (Definition 2), the sign convention and $c_{\rm N}$ (1.1) of Module 4.

Module 5 of Astrophysical Fluid Dynamics. Previous: Module 4 — Sound Waves and Linear Perturbation Theory