The Jeans Instability Criterion

Derivation

The Jeans Instability Criterion

D-347 Home PU-308 Threads force · waves · matter Depends on euler-equations-fluid, poisson-equation-gravity

Statement

For a uniform, self-gravitating, ideal fluid of density \(\rho_0\) and sound speed \(c_s\), linearising the continuity, Euler and Poisson equations about the static background and inserting a plane-wave perturbation \(\propto e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}\) yields the dispersion relation \(\omega^{2}=c_s^{2}k^{2}-4\pi G\rho_0\). Modes with \(k<k_J\equiv\sqrt{4\pi G\rho_0}/c_s\) have \(\omega^{2}<0\) and grow exponentially, defining the Jeans length \(\lambda_J=2\pi/k_J\) and the Jeans mass \(M_J\) above which a cloud collapses gravitationally.

Why it matters

The Jeans criterion is the fundamental threshold of structure formation: it fixes the smallest mass a gas cloud can have and still collapse under its own gravity against internal pressure. It sets the characteristic scale of star formation in molecular clouds and, in the expanding universe, the mass scale of the first bound structures.

Physically it is a contest between two speeds. Pressure information propagates at \(c_s\); gravity collapses a region on the free-fall time \(\sim(G\rho_0)^{-1/2}\). When a perturbation is too large for pressure to communicate across it before gravity wins, it becomes unstable. The Jeans length is precisely the wavelength at which those two timescales balance.

Assumptions

Ideal, inviscid, barotropic fluid with \(P=c_s^2\rho\) (isothermal closure).If dropped, an extra energy equation is needed; \(c_s\) is replaced by an adiabatic \(\sqrt{\gamma P/\rho}\) and cooling can lower the effective pressure support, shifting \(k_J\).

Small perturbations, so products of perturbed quantities are negligible.If dropped, the equations stay nonlinear; the dispersion relation does not exist and one must integrate the full collapse (runaway, fragmentation).

Uniform, static, infinite background: \(\rho_0,c_s\) constant, \(\mathbf{v}_0=0\).If dropped, plane waves are no longer eigenmodes; gradients or rotation introduce position-dependent coefficients and mode coupling.

The "Jeans swindle": the uniform background is assumed to satisfy \(\nabla\Phi_0=0\) despite \(\nabla^2\Phi_0=4\pi G\rho_0\neq 0\).If dropped honestly, no static uniform self-gravitating medium exists; the clean result is recovered only in an expanding background or a Jeans-mass-conserving comoving analysis, which justifies the swindle a posteriori.

No magnetic fields, rotation, viscosity or radiation pressure.If dropped, magnetic tension and rotation add restoring forces, raising the critical mass (magnetic/rotational support) and making instability anisotropic in \(\mathbf{k}\).

Derivation

\[ \frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})=0,\qquad \frac{\partial\mathbf{v}}{\partial t}+(\mathbf{v}\cdot\nabla)\mathbf{v}=-\frac{1}{\rho}\nabla P-\nabla\Phi,\qquad \nabla^2\Phi=4\pi G\rho,\qquad P=c_s^2\,\rho \]

Governing system: mass continuity, the Euler equation (prior result euler-equations-fluid), Poisson's equation for self-gravity (prior result poisson-equation-gravity), closed by an isothermal equation of state. A

\[ \rho=\rho_0+\delta\rho,\qquad \mathbf{v}=\delta\mathbf{v},\qquad \Phi=\Phi_0+\delta\Phi,\qquad P=P_0+c_s^2\,\delta\rho \]

Split every field into a uniform static background (subscript 0) plus a small perturbation. The pressure perturbation follows from the equation of state: \(\delta P=(dP/d\rho)\,\delta\rho=c_s^2\,\delta\rho\). A

\[ \mathbf{v}_0=0,\qquad \nabla\rho_0=\nabla P_0=0,\qquad \nabla\Phi_0\stackrel{\text{swindle}}{=}0 \]

The background must be a solution. A uniform medium trivially satisfies continuity and the equation of state; the Euler equation forces \(\nabla\Phi_0=0\), which we adopt (the Jeans swindle) even though Poisson gives \(\nabla^2\Phi_0=4\pi G\rho_0\neq0\). This is the one non-rigorous input, cured by an expanding background. C

\[ \frac{\partial\,\delta\rho}{\partial t}+\rho_0\,\nabla\cdot\delta\mathbf{v}=0 \]

Linearised continuity: substitute the split fields, use \(\mathbf{v}_0=0\) and \(\nabla\rho_0=0\), and discard the second-order term \(\nabla\cdot(\delta\rho\,\delta\mathbf{v})\). B

\[ \frac{\partial\,\delta\mathbf{v}}{\partial t}=-\frac{c_s^2}{\rho_0}\nabla\,\delta\rho-\nabla\,\delta\Phi \]

Linearised Euler: the advective term \((\delta\mathbf{v}\cdot\nabla)\delta\mathbf{v}\) is second order and dropped; \(\frac{1}{\rho}\nabla P\to\frac{1}{\rho_0}\nabla(c_s^2\delta\rho)\) to first order; \(\nabla\Phi_0=0\) removes the background force. B

\[ \nabla^2\,\delta\Phi=4\pi G\,\delta\rho \]

Linearised Poisson: subtract the background \(\nabla^2\Phi_0=4\pi G\rho_0\) (formally, under the swindle) to leave the perturbation equation, which is exact since Poisson is already linear. A

\[ \delta\rho,\ \delta\mathbf{v},\ \delta\Phi\ \propto\ e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}\ \Rightarrow\ \partial_t\to-i\omega,\quad \nabla\to i\mathbf{k} \]

The linearised system has constant coefficients, so Fourier plane waves are exact eigenmodes: each derivative becomes an algebraic factor. Only the longitudinal part of \(\delta\mathbf{v}\) (along \(\mathbf{k}\)) couples to \(\delta\rho\); transverse motions decouple and are non-self-gravitating shear. B

\[ -i\omega\,\delta\rho+i\rho_0\,\mathbf{k}\cdot\delta\mathbf{v}=0,\qquad -i\omega\,\delta\mathbf{v}=-\frac{c_s^2}{\rho_0}\,i\mathbf{k}\,\delta\rho-i\mathbf{k}\,\delta\Phi,\qquad -k^2\,\delta\Phi=4\pi G\,\delta\rho \]

Apply the substitutions of Step 7 to Steps 4–6. Continuity and Poisson become scalar; Euler is projected onto \(\hat{\mathbf{k}}\), writing \(\delta v\equiv\hat{\mathbf{k}}\cdot\delta\mathbf{v}\). A

\[ \delta\Phi=-\frac{4\pi G}{k^2}\,\delta\rho,\qquad \delta v=\frac{\omega}{\rho_0 k}\,\delta\rho \]

Solve the Poisson relation for the potential and the continuity relation for the longitudinal velocity, in terms of the density perturbation \(\delta\rho\). A

\[ \omega\,\delta v=\frac{c_s^2}{\rho_0}k\,\delta\rho+k\,\delta\Phi=\frac{c_s^2}{\rho_0}k\,\delta\rho-\frac{4\pi G}{k}\,\delta\rho \]

Take the \(\hat{\mathbf{k}}\)-component of the Euler relation (cancel the common \(i\)) and substitute \(\delta\Phi\) from Step 9. B

\[ \omega\cdot\frac{\omega}{\rho_0 k}\,\delta\rho=\left(\frac{c_s^2 k}{\rho_0}-\frac{4\pi G}{k}\right)\delta\rho \;\Longrightarrow\; \frac{\omega^2}{\rho_0 k}=\frac{c_s^2 k}{\rho_0}-\frac{4\pi G}{k} \]

Insert \(\delta v\) from Step 9 into Step 10 and cancel the non-trivial amplitude \(\delta\rho\neq0\) (a trivial \(\delta\rho=0\) mode carries no perturbation). B

\[ \boxed{\;\omega^2=c_s^2 k^2-4\pi G\rho_0\;} \]

Multiply through by \(\rho_0 k\). This is the Jeans dispersion relation: the free acoustic branch \(\omega^2=c_s^2k^2\) softened by the destabilising self-gravity term \(-4\pi G\rho_0\). A

\[ \omega^2<0 \iff c_s^2 k^2<4\pi G\rho_0 \iff k<k_J\equiv\frac{\sqrt{4\pi G\rho_0}}{c_s} \]

Instability (exponential growth, \(\omega=i|\omega|\)) occurs exactly when self-gravity beats pressure, i.e. for wavenumbers below the critical Jeans wavenumber \(k_J\). B

\[ \lambda_J=\frac{2\pi}{k_J}=c_s\sqrt{\frac{\pi}{G\rho_0}},\qquad M_J=\frac{4}{3}\pi\rho_0\left(\frac{\lambda_J}{2}\right)^3=\frac{\pi}{6}\rho_0\,\lambda_J^{3} \]

Convert the critical wavenumber to the Jeans length, and define the Jeans mass as the gas contained in a sphere of diameter \(\lambda_J\) (a standard convention; factors of order unity vary between texts). A

Result

\[ \omega^2=c_s^2k^2-4\pi G\rho_0,\qquad \lambda_J=c_s\sqrt{\frac{\pi}{G\rho_0}},\qquad M_J=\frac{\pi}{6}\rho_0\lambda_J^{3}=\frac{\pi^{5/2}}{6}\frac{c_s^3}{G^{3/2}\rho_0^{1/2}} \]

Reading. A perturbation behaves like a sound wave shifted downward by gravity. Short wavelengths (\(k>k_J\), \(\lambda<\lambda_J\)) keep \(\omega^2>0\) and merely oscillate: pressure crosses the region faster than gravity can collapse it. Long wavelengths (\(k<k_J\)) have \(\omega^2<0\), so one root grows as \(e^{|\omega|t}\) — gravitational collapse. The Jeans mass \(M_J\propto c_s^3\rho_0^{-1/2}\propto T^{3/2}\rho_0^{-1/2}\) is the least mass that collapses at given \(T\) and \(\rho_0\).

Units check. \([c_s^2k^2]=(\mathrm{m\,s^{-1}})^2(\mathrm{m^{-1}})^2=\mathrm{s^{-2}}\); \([G\rho_0]=(\mathrm{m^3\,kg^{-1}\,s^{-2}})(\mathrm{kg\,m^{-3}})=\mathrm{s^{-2}}\), so \(\omega^2\) is in \(\mathrm{s^{-2}}\). For \(\lambda_J\): \([c_s\sqrt{1/(G\rho_0)}]=\mathrm{m\,s^{-1}}\cdot\mathrm{s}=\mathrm{m}\). For \(M_J\): \([\rho_0\lambda_J^3]=\mathrm{kg\,m^{-3}}\cdot\mathrm{m^3}=\mathrm{kg}\). All consistent.

Limiting cases

\(k\to\infty\) (small scales): \(\omega^2\to c_s^2k^2\), pure sound waves — gravity negligible, always stable.
\(k\to0\) (large scales): \(\omega^2\to-4\pi G\rho_0\), maximal growth rate \(|\omega|=\sqrt{4\pi G\rho_0}\), independent of scale — the pressure-free free-fall limit.
\(k=k_J\) (marginal): \(\omega=0\), a stationary neutral mode marking the collapse threshold.
\(G\to0\) (no self-gravity): \(\omega^2=c_s^2k^2\), recovers ordinary acoustics; \(\lambda_J\to\infty\), nothing collapses.
\(c_s\to0\) (cold/pressureless): \(k_J\to\infty\), \(\lambda_J\to0\): all scales collapse (Zel'dovich/dust limit).

Breaks when

Finite amplitude. Once \(\delta\rho/\rho_0\sim1\) the linearisation of Steps 4–5 fails; collapse becomes nonlinear, non-spherical, and fragments — the single growth rate no longer applies.
Non-uniform or bounded background. Density gradients, boundaries, rotation, or a bulk flow break the constant-coefficient assumption, so plane waves are no longer eigenmodes and \(k_J\) becomes position- and direction-dependent.
Magnetised gas. Magnetic tension supplies an extra restoring force; instability becomes anisotropic and the critical mass rises to the magnetic critical (mass-to-flux) value, which the pure-hydrodynamic \(M_J\) misses entirely.
Rapid heating/cooling. If radiative cooling makes the collapse effectively isothermal or the gas heats adiabatically, the relevant \(c_s\) (isothermal vs adiabatic \(\sqrt{\gamma}\,c_s\)) and hence \(k_J\) change; a strongly cooling gas can fragment far below the naive \(M_J\).

Failure modes

Sign confusion: writing \(\omega^2=c_s^2k^2+4\pi G\rho_0\) — making gravity stabilising. The self-gravity term must be negative for collapse.
Instability on the wrong side of \(k_J\): claiming short wavelengths collapse. It is long wavelengths (\(k<k_J\), \(\lambda>\lambda_J\)) that are unstable.
Adiabatic vs isothermal sound speed: using \(\sqrt{\gamma P/\rho}\) when the cloud is isothermally cooled (or vice versa); this misplaces \(k_J\) by a factor \(\sqrt{\gamma}\).
Forgetting the Poisson sign: taking \(\nabla^2\delta\Phi=-4\pi G\delta\rho\); the attractive convention gives \(\delta\Phi=-4\pi G\delta\rho/k^2\), essential for the destabilising term.
Density confusion in \(M_J\): using number density \(n\) directly instead of mass density \(\rho_0=\mu m_H n\) (include helium via \(\mu\)).
Transverse modes: trying to make shear (\(\delta\mathbf{v}\perp\mathbf{k}\)) unstable; only the longitudinal, compressive mode couples to gravity.

Discussion

The Jeans criterion is best read as a timescale competition. The sound-crossing time of a region of size \(\lambda\) is \(t_s\sim\lambda/c_s\); the gravitational free-fall time is \(t_{\rm ff}\sim(G\rho_0)^{-1/2}\). Setting \(t_s\sim t_{\rm ff}\) gives \(\lambda\sim c_s/\sqrt{G\rho_0}\), which is \(\lambda_J\) up to the \(\sqrt{\pi}\). When the region is larger than \(\lambda_J\), pressure cannot re-establish equilibrium before gravity collapses it, and the perturbation runs away. The dispersion relation makes this exact and quantitative.

The Jeans mass scaling \(M_J\propto T^{3/2}\rho_0^{-1/2}\) has deep consequences. As a cloud collapses isothermally, \(\rho_0\) rises while \(T\) stays roughly fixed, so \(M_J\) falls: sub-regions that were individually stable become unstable, and the cloud fragments hierarchically. This runaway fragmentation is why molecular clouds form clusters of stars rather than one monolithic object, and it terminates only when the gas becomes optically thick and heats, raising \(M_J\) again to set a minimum stellar mass.

In cosmology the same analysis, carried out in an expanding background, replaces the Jeans swindle with a genuine result: perturbations on comoving scales above the (comoving) Jeans length grow, but only as power laws \(\delta\propto t^{2/3}\) in matter domination rather than exponentially, because the Hubble expansion continually dilutes \(\rho_0\) and drains the collapse. Before recombination, photon pressure keeps \(c_s\approx c/\sqrt{3}\) enormous, so the baryonic Jeans mass is huge and baryons cannot collapse; after recombination \(c_s\) plummets and the Jeans mass drops by orders of magnitude, unlocking the formation of the first bound clouds. The pressureless \(c_s\to0\) limit of the dispersion relation is exactly the dark-matter (dust) case that seeds all structure.

Common misconceptions. The Jeans instability is not a resonance or a wave that "breaks"; it is a genuine change in the character of the mode from oscillatory (\(\omega\) real) to exponential (\(\omega\) imaginary) as \(k\) crosses \(k_J\). Also, \(\lambda_J\) is not a fixed length of nature — it depends on the local \(\rho_0\) and \(c_s\) and shrinks as a cloud collapses. Finally, the "swindle" is not a fudge that changes the answer: the identical dispersion relation emerges rigorously from the expanding-universe treatment, so the physics is sound even though the static-uniform starting point is formally inconsistent.

Worked examples

Dense molecular cloud core: \(T=10\ \mathrm{K}\), \(n_{\rm H_2}=10^{4}\ \mathrm{cm^{-3}}\), mean molecular weight \(\mu=2.3\). Find \(c_s\), \(\lambda_J\), \(M_J\).

\[ c_s=\sqrt{\frac{k_B T}{\mu m_H}},\qquad \rho_0=\mu m_H\,n,\qquad \lambda_J=c_s\sqrt{\frac{\pi}{G\rho_0}},\qquad M_J=\frac{\pi}{6}\rho_0\lambda_J^3 \]

Symbols first. Use isothermal \(c_s\) since dense cores cool efficiently. Convert \(n=10^4\ \mathrm{cm^{-3}}=10^{10}\ \mathrm{m^{-3}}\). A

\[ \mu m_H=2.3\times1.67\times10^{-27}=3.84\times10^{-27}\ \mathrm{kg} \]

\[ c_s=\sqrt{\frac{1.38\times10^{-23}\times10}{3.84\times10^{-27}}}=\sqrt{3.59\times10^{4}}\approx1.90\times10^{2}\ \mathrm{m\,s^{-1}} \]

\[ \rho_0=3.84\times10^{-27}\times10^{10}=3.84\times10^{-17}\ \mathrm{kg\,m^{-3}} \]

\[ G\rho_0=6.674\times10^{-11}\times3.84\times10^{-17}=2.56\times10^{-27}\ \mathrm{s^{-2}} \]

\[ \lambda_J=1.90\times10^{2}\sqrt{\frac{3.1416}{2.56\times10^{-27}}}=1.90\times10^{2}\times3.50\times10^{13}\approx6.7\times10^{15}\ \mathrm{m}\approx0.22\ \mathrm{pc} \]

\[ M_J=\frac{\pi}{6}\rho_0\lambda_J^3=0.524\times3.84\times10^{-17}\times(6.7\times10^{15})^3\approx6\times10^{30}\ \mathrm{kg} \]

Numbers substituted after the symbolic forms. \(1\ \mathrm{pc}=3.086\times10^{16}\ \mathrm{m}\), \(M_\odot=1.99\times10^{30}\ \mathrm{kg}\). A

\[ c_s\approx0.19\ \mathrm{km\,s^{-1}},\quad \lambda_J\approx0.22\ \mathrm{pc},\quad M_J\approx3\ M_\odot \]

Reading. A cold dense core roughly a few solar masses in a fifth of a parsec is Jeans-unstable — exactly the scale of low-mass star formation.

Units check. \(\mathrm{m\,s^{-1}}\), \(\mathrm{m}\to\mathrm{pc}\), \(\mathrm{kg}\to M_\odot\) all consistent.

Diffuse giant molecular cloud: \(T=15\ \mathrm{K}\), \(n=100\ \mathrm{cm^{-3}}\), \(\mu=2.3\). Find \(\lambda_J\), \(M_J\), and the growth timescale of the longest-wavelength mode.

\[ c_s=\sqrt{\frac{k_B T}{\mu m_H}},\qquad \rho_0=\mu m_H n,\qquad \lambda_J=c_s\sqrt{\frac{\pi}{G\rho_0}},\qquad \tau=\frac{1}{|\omega|_{\max}}=\frac{1}{\sqrt{4\pi G\rho_0}} \]

Symbols first; \(\tau\) uses the \(k\to0\) maximal growth rate from the dispersion relation. \(n=100\ \mathrm{cm^{-3}}=10^{8}\ \mathrm{m^{-3}}\). B

\[ c_s=\sqrt{\frac{1.38\times10^{-23}\times15}{3.84\times10^{-27}}}=\sqrt{5.39\times10^{4}}\approx2.32\times10^{2}\ \mathrm{m\,s^{-1}} \]

\[ \rho_0=3.84\times10^{-27}\times10^{8}=3.84\times10^{-19}\ \mathrm{kg\,m^{-3}},\qquad G\rho_0=2.56\times10^{-29}\ \mathrm{s^{-2}} \]

\[ \lambda_J=2.32\times10^{2}\sqrt{\frac{3.1416}{2.56\times10^{-29}}}=2.32\times10^{2}\times3.50\times10^{14}\approx8.1\times10^{16}\ \mathrm{m}\approx2.6\ \mathrm{pc} \]

\[ M_J=\frac{\pi}{6}\rho_0\lambda_J^3=0.524\times3.84\times10^{-19}\times(8.1\times10^{16})^3\approx1.1\times10^{32}\ \mathrm{kg}\approx54\ M_\odot \]

\[ \tau=\frac{1}{\sqrt{4\pi\times2.56\times10^{-29}}}=\frac{1}{\sqrt{3.22\times10^{-28}}}=\frac{1}{1.79\times10^{-14}}\approx5.6\times10^{13}\ \mathrm{s}\approx1.8\ \mathrm{Myr} \]

\(1\ \mathrm{Myr}=3.156\times10^{13}\ \mathrm{s}\). B

\[ \lambda_J\approx2.6\ \mathrm{pc},\quad M_J\approx54\ M_\odot,\quad \tau\approx1.8\ \mathrm{Myr} \]

Reading. A lower-density cloud has a much larger Jeans mass (tens of \(M_\odot\)) and collapses on a few-Myr free-fall time — consistent with GMC lifetimes and the formation of stellar groups.

Units check. \(\lambda_J\) in \(\mathrm{m}\to\mathrm{pc}\); \(M_J\) in \(\mathrm{kg}\to M_\odot\); \(\tau=[\mathrm{s^{-2}}]^{-1/2}=\mathrm{s}\to\mathrm{Myr}\).

Problems

A cloud has \(c_s=0.30\ \mathrm{km\,s^{-1}}\) and \(\rho_0=1.0\times10^{-18}\ \mathrm{kg\,m^{-3}}\). Compute the Jeans wavenumber \(k_J\) and Jeans length \(\lambda_J\).
Solution
\(k_J=\sqrt{4\pi G\rho_0}/c_s\). \(4\pi G\rho_0=4\times3.1416\times6.674\times10^{-11}\times10^{-18}=8.39\times10^{-28}\ \mathrm{s^{-2}}\); \(\sqrt{\ }=2.90\times10^{-14}\ \mathrm{s^{-1}}\). \(k_J=2.90\times10^{-14}/300=9.65\times10^{-17}\ \mathrm{m^{-1}}\). \(\lambda_J=2\pi/k_J=6.51\times10^{16}\ \mathrm{m}\approx2.1\ \mathrm{pc}\).
Show from \(\omega^2=c_s^2k^2-4\pi G\rho_0\) that the marginal wavelength is \(\lambda_J=c_s\sqrt{\pi/(G\rho_0)}\), and find the critical wavelength for \(c_s=0.2\ \mathrm{km\,s^{-1}}\), \(\rho_0=4\times10^{-17}\ \mathrm{kg\,m^{-3}}\).
Solution
Marginal mode: \(\omega=0\Rightarrow c_s^2k_J^2=4\pi G\rho_0\Rightarrow k_J=\sqrt{4\pi G\rho_0}/c_s\). Then \(\lambda_J=2\pi/k_J=2\pi c_s/\sqrt{4\pi G\rho_0}=c_s\sqrt{4\pi^2/(4\pi G\rho_0)}=c_s\sqrt{\pi/(G\rho_0)}\). Numerically: \(G\rho_0=6.674\times10^{-11}\times4\times10^{-17}=2.67\times10^{-27}\); \(\pi/(G\rho_0)=1.18\times10^{27}\); \(\sqrt{\ }=3.43\times10^{13}\ \mathrm{s}\); \(\lambda_J=200\times3.43\times10^{13}=6.9\times10^{15}\ \mathrm{m}\approx0.22\ \mathrm{pc}\).
Using \(M_J\propto T^{3/2}\rho_0^{-1/2}\), by what factor does the Jeans mass change if the temperature doubles and the density increases by a factor of 100?
Solution
\(M_J\propto T^{3/2}\rho_0^{-1/2}\). Factor \(=(2)^{3/2}\times(100)^{-1/2}=2.83\times0.10=0.283\). The Jeans mass drops to about \(0.28\times\) its original value — denser, only mildly warmer gas fragments to smaller masses.
For the maximally unstable long-wavelength mode (\(k\to0\)), find the e-folding growth time \(\tau=1/\sqrt{4\pi G\rho_0}\) for \(\rho_0=3.8\times10^{-17}\ \mathrm{kg\,m^{-3}}\), and compare it to the free-fall time \(t_{\rm ff}=\sqrt{3\pi/(32G\rho_0)}\).
Solution
\(4\pi G\rho_0=4\times3.1416\times6.674\times10^{-11}\times3.8\times10^{-17}=3.19\times10^{-26}\ \mathrm{s^{-2}}\); \(\tau=1/\sqrt{3.19\times10^{-26}}=1/(1.79\times10^{-13})=5.6\times10^{12}\ \mathrm{s}\approx0.18\ \mathrm{Myr}\). Free-fall: \(t_{\rm ff}=\sqrt{3\pi/(32G\rho_0)}\); \(G\rho_0=2.54\times10^{-27}\), \(32G\rho_0=8.12\times10^{-26}\), \(3\pi/8.12\times10^{-26}=1.16\times10^{26}\), \(\sqrt{\ }=1.08\times10^{13}\ \mathrm{s}\approx0.34\ \mathrm{Myr}\). So \(t_{\rm ff}\approx1.9\,\tau\): the e-folding and free-fall times agree to a factor of order unity.
A perturbation has wavelength \(\lambda=5\ \mathrm{pc}\) in a medium with \(c_s=0.25\ \mathrm{km\,s^{-1}}\) and \(\rho_0=2\times10^{-18}\ \mathrm{kg\,m^{-3}}\). Is it stable or unstable? Compute \(\omega^2\) and, if unstable, the growth rate.
Solution
\(k=2\pi/\lambda\), \(\lambda=5\times3.086\times10^{16}=1.543\times10^{17}\ \mathrm{m}\Rightarrow k=4.07\times10^{-17}\ \mathrm{m^{-1}}\). \(c_s^2k^2=(250)^2(4.07\times10^{-17})^2=6.25\times10^{4}\times1.66\times10^{-33}=1.04\times10^{-28}\ \mathrm{s^{-2}}\). \(4\pi G\rho_0=4\times3.1416\times6.674\times10^{-11}\times2\times10^{-18}=1.68\times10^{-27}\ \mathrm{s^{-2}}\). \(\omega^2=1.04\times10^{-28}-1.68\times10^{-27}=-1.57\times10^{-27}\ \mathrm{s^{-2}}<0\): unstable. Growth rate \(|\omega|=\sqrt{1.57\times10^{-27}}=3.96\times10^{-14}\ \mathrm{s^{-1}}\), e-folding time \(\approx2.5\times10^{13}\ \mathrm{s}\approx0.8\ \mathrm{Myr}\). (Check: \(\lambda_J=c_s\sqrt{\pi/(G\rho_0)}=250\sqrt{3.1416/(1.33\times10^{-28})}=250\times4.86\times10^{13}=1.2\times10^{16}\ \mathrm{m}\approx0.39\ \mathrm{pc}<5\ \mathrm{pc}\), confirming instability.)