How the Jeans scale separates oscillation and growth

Define the background and the disturbance

Start with uniform density $\rho_0$ and a wave along $x$. Positive velocity $u$ points toward increasing $x$. The relative density disturbance is $\delta=(\rho-\rho_0)/\rho_0$; it is dimensionless and much smaller than one. The sound speed $c_s$ has velocity units.

For isothermal gas, pressure obeys $p=c_s^2\rho$. Holding temperature fixed supplies the pressure law; we are not solving another equation for heating and cooling. The gravitational potential $\phi$ has units of velocity squared. Its negative spatial derivative gives gravitational acceleration.

Sound speed squared here is the pressure derivative with respect to density at fixed temperature. The official-input chapter also encounters an adiabatic derivative. Specify the temperature response during compression before assigning a numerical sound speed.

A uniform, static, infinite self-gravitating background needs a convention for its mean gravity. Here a periodic box responds only to density minus the mean, and its uniform component produces no force. This defines the perturbation problem. An isolated sphere requires its own finite-boundary gravity problem.

Reading: Chandrasekhar & Fermi 1953, Problems of Gravitational Stability in the Presence of a Magnetic Field · Molecular-cloud Jeans standalone scientific reproduction

Join mass conservation to the forces

Mass conservation says that density decreases where more gas leaves than enters. Keep only first-order disturbances: products of a density disturbance and a velocity are second order and are omitted for now. The momentum equation adds the pressure and gravitational accelerations.

A derivative measures change over a small interval. When $\partial_xu>0$, gas on the right moves faster than gas on the left, stretching a small region. The minus sign in continuity makes its density contrast fall. Converging gas has a negative velocity gradient and an increasing density.

Differentiate continuity once more in time. Substitute the momentum equation for the time derivative of velocity, then substitute Poisson’s equation for the second derivative of $\phi$. The result describes the acceleration of the density contrast. Pressure acts through spatial curvature; the gravity term has the same sign as the density contrast.

Track the signs explicitly: $\partial_t^2\delta=-\partial_x(\partial_tu)$. Substitution produces two pairs of minus signs, yielding $c_s^2\partial_x^2\delta+\partial_x^2\phi$. Poisson’s equation then replaces the potential curvature with density contrast. Its positive contribution is the mathematical expression of gravity strengthening an overdensity.

The density amplitude is not a particle moving along a trajectory, but its evolution can still obey an oscillator equation. A second time derivative describes how the rate of change itself accelerates.

Reading: Chandrasekhar & Fermi 1953, Problems of Gravitational Stability in the Presence of a Magnetic Field · Molecular-cloud Jeans standalone scientific reproduction

A cosine wave gives the critical wavelength

Separate position and time by writing $\delta(x,t)=D(t)\cos(kx)$. The dimensionless amplitude is $D$, the angular wavenumber is $k=2\pi/\lambda$, and $\lambda$ is wavelength. Two spatial derivatives of a cosine supply $-k^2$. Short wavelengths therefore have a stronger pressure-restoring term.

When $\omega^2$ is positive, the amplitude oscillates with angular frequency $\omega$. When it is negative, define $\gamma=\sqrt{-\omega^2}$; the solution contains a growing exponential and a decaying exponential. The threshold occurs at $\omega^2=0$.

Both oscillation and growth become slow near the threshold. A nearly unchanged finite-time image can mean that the rate is small. Calculate the condition before interpreting an apparently quiet patch as stable.

Reading: Chandrasekhar & Fermi 1953, Problems of Gravitational Stability in the Presence of a Magnetic Field

Starting at rest selects a different curve

Use box length $L$ as the length unit and $t_0=(4\pi G\rho_0)^{-1/2}$ as the time unit. Define the Jeans ratio $J=L/\lambda_J$. For a fundamental wave spanning the box, $\omega^2t_0^2=J^{-2}-1$. With $J=1.5$, the growth rate is $\gamma t_0=0.745356$.

Choose initial amplitude $D_0=10^{-4}$. If the initial velocity has exactly the inward growing-mode pattern, the amplitude follows $D_0e^{\gamma t}$. At $t=2t_0$ it reaches about $4.4403\times10^{-4}$. Initial velocity decides how much of the growing and decaying solutions is present.

For gas initially at rest, $\dot D(0)=0$. Equal growing and decaying contributions cancel the initial slope. Their average is a hyperbolic cosine, giving amplitude $2.3327\times10^{-4}$ at the same time. The growing contribution dominates only later.

The small-amplitude numerical growth test deliberately prepares the growing velocity. The finite-amplitude gas patch starts at rest. Comparing their early slopes directly would confuse different initial conditions with numerical error.

Reading: Molecular-cloud Jeans standalone scientific reproduction

State a geometry before converting length to mass

A common three-dimensional Jeans mass encloses the background density in a sphere of radius $\lambda_J/2$. Stating the radius matters: a different convention changes the geometric coefficient. A critical wavelength does not assign a unique mass to every shape.

For an illustrative physical scaling, specify a total thermal-particle density of one thousand per cubic centimeter, mean particle mass $2.33$ times an adopted hydrogen-atom mass, box length $1\,\mathrm{pc}$, and $J=1.5$. These choices give sound speed about $187\,\mathrm{m\,s^{-1}}$, temperature $9.90\,\mathrm K$, $t_0=0.554\,\mathrm{Myr}$, and $\lambda_J=0.667\,\mathrm{pc}$. The particle density here is not specifically a count of hydrogen molecules.

The stated spherical definition gives about $8.94M_\odot$. The two-dimensional density picture is invariant in the third direction and integrates mass per unit depth. Assigning the spherical number to its bright central region would combine two different geometries. An isolated homogeneous sphere’s free-fall time is another separately defined reference.

Try a scaling example with no new constants. Double sound speed at fixed density: the critical length doubles, its reference sphere has eight times the volume, and the Jeans mass rises eightfold. Instead quadruple background density at fixed sound speed: the length halves, the sphere volume becomes one eighth, and four times the mass per volume makes the reference mass one half.

The comparison separates a critical length from a critical mass. A smaller reference mass means that a smaller amount of gas can become unstable under the stated assumptions. The total mass of a particular evolving region still follows its own geometry and boundary conditions.

Reading: Molecular-cloud Jeans standalone scientific reproduction · Guszejnov et al. 2020, Can magnetized turbulence set the mass scale of stars? arXiv:2002.01421v1

Follow the approximation as the wave grows

The derivation discarded second-order terms. Once the density contrast becomes large, density and velocity affect each other, and a cosine can develop additional spatial structure. The linear dispersion relation still tests early evolution; later motion requires the actual mass and momentum equations.

A numerical grid introduces another length. Isothermal compression shortens the local Jeans length as density rises, leaving fewer cells across it. The L3 chapter recalculates the same case with a finer grid and an independent method so that a change in the gas can be distinguished from a change caused by discretization.

Reading: Molecular-cloud Jeans standalone scientific reproduction