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The physical Jeans length is ${\ell }_J=c_s\sqrt{\pi /(G{\rho }_0)}$. A pressureless uniform sphere of initial density ${\rho }_0$ has free-fall time of a uniform sphere $t_{\mathrm{ff}}=\sqrt{3\pi /(32G{\rho }_0)}$, giving the title relation. Thus a full-wavelength sound-crossing time and this exact free-fall convention agree in scaling, not numerical value. Using the scale $k_J^{-1}$ instead gives $1/(c_s{k}_J)=1/\sqrt{4\pi G{\rho }_0}$, the inverse long-wave static Jeans growth-rate spectrum limit. At the threshold itself pressure balances gravity and the net growth rate vanishes; the pressureless free-fall time must not be confused with that marginal mode's infinite growth time.

For small isentropic perturbations of a fixed-composition gas, linearizing its equation of state gives $p_1=(\partial p/\partial \rho {)}_s{\rho }_1=c_s^2{\rho }_1$. The adiabatic sound speed is evaluated at the uniform background, so it is spatially constant at this order. If fast heat exchange instead enforces constant temperature, use the isothermal sound speed consistently. A completely arbitrary entropy perturbation would require an additional thermodynamic variable.

The uniform static infinite medium requires the Jeans swindle: its background gravitational field is subtracted or externally balanced, and only the perturbation sources the Poisson equation for Newtonian gravity. Linearizing the continuity equation, Euler equation and gravitational equation then givesTaking one more time derivative of the first equation and substituting the other two yieldsA Fourier mode proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ therefore obeys the Jeans instability relationFor $k>k_J$, the mode oscillates as a sound wave. For $0<k<k_J$, $\omega =\pm i\sqrt{4\pi G{\rho }_0-c_s^2{k}^2}$ gives a growing and a decaying mode; the static Jeans growth-rate spectrum has no preferred finite fastest-growing wavelength in the infinite-medium approximation. The threshold $k=k_J$ is marginal at linear order. The Jeans length is ${\lambda }_J=2\pi /k_J=c_s\sqrt{\pi /(G{\rho }_0)}$. Defining the Jeans mass as a sphere whose diameter is this wavelength givesOther radius conventions alter its numerical coefficient; the mass scale compares self-gravity with pressure communication.

For the pressureless finite cloud, the shell theorem gives an edge acceleration $\ddot{r}=-GM/r^2$, with fixed enclosed $M=4\pi r_0^3{\rho }_0/3$. Multiplication by $\dot{r}$ and integration from rest givesThe negative root describes collapse. Substituting $r=r_0{\sin }^2\theta$ gives the free-fall time of a uniform sphereThe surface and inner shells reach the origin together in pressureless homologous spherical collapse; pressure, rotation or shell crossing would change this ideal solution.

Equating this time to the specified radius-based sound-crossing time $r_0/c_s$ gives $r_0=c_s\sqrt{3\pi /(32G{\rho }_0)}$. The resulting crossing-time cloud support mass isThis is the same dimensional mass scale as $M_J$, with a different coefficient because the adopted length and time criteria differ. In particular, $t_{\mathrm{ff}}=r_0/c_s$ is not the exact marginal dispersion-relation condition.

With the requested isotropic support approximation, replace $c_s$ by the Alfvén speed $v_A=B/\sqrt{{\mu }_0{\rho }_0}$. In the crossing-time convention the Magnetic Jeans mass isUsing the wavelength-based Jeans mass convention instead replaces $C_{\mathrm{cross}}$ by ${\pi }^{5/2}/6$. The scaling, rather than a convention-independent coefficient, is the shared prediction.

For a fixed cloud mass undergoing isothermal homologous contraction, $\rho /{\rho }_0=(r_0/r{)}^3$ and $c_s$ is constant. Its number of thermal Jeans masses therefore grows asA cloud initially close to the support threshold has $N_{J,0}$ of order one, so smaller pieces can subsequently become independently unstable: this is isothermal fragmentation.

For approximately isotropic contraction with magnetic flux freezing, $B{r}^2$ stays constant, so $B\propto {\rho }^{2/3}$. Hence $M_B\propto B^3/{\rho }^2$ stays constant andThis ideal magnetic support therefore does not create progressively more unstable masses in the way thermal isothermal support does. The same conclusion follows from the conserved mass-to-flux ratio. A magnetically subcritical cloud cannot initiate unrestricted cross-field collapse under ideal flux freezing alone; flux loss or initially supercritical conditions are needed. A real magnetic field also gives anisotropic support: motion and fragmentation along its field lines are not excluded by this scalar-speed estimate.

The background and the Jeans prescription. For a resting fluid, ${\mathbf{u}}_0=0$ and continuity is automatic. The remaining hydrostatic equilibrium equations areAn infinite uniform static medium with constant $p_0$ and ${\rho }_0>0$ would require $\nabla {\psi }_0=0$ by force balance, but then its Laplacian would vanish, contradicting the Poisson equation. An isolated static uniform background is therefore not an exact equilibrium of this system.

The Jeans swindle suppresses the gravitational field of the uniform background and retains only the field sourced by perturbations. Formally it replaces the source by $\rho -{\rho }_0$ in the perturbation problem. This can be regarded as a specified compensating background or a local idealization; it does not prove that the original infinite isolated medium can be static.

Linearized equations. Put $\delta \rho =\Delta \rho$, $\delta p=\Delta p$, $\delta \psi =\Delta \psi$, and $\mathbf{v}=\Delta \mathbf{u}$. For uniform ${\rho }_0,p_0$, neglect products of perturbations. Continuity givesThe advective acceleration is quadratic, and the background pressure gradient vanishes, so the Euler equation becomesWith the Jeans prescription and constant sound speed, the gravitational and pressure closures areThe density correction to $1/\rho$ multiplying $\nabla \delta p$ is second order and is consistently omitted.

The longitudinal dispersion relation. Differentiate continuity in time and substitute the divergence of the linearized Euler equation:For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$, ${\partial }_t^2\mapsto -{\omega }^2$ and ${\nabla }^2\mapsto -k^2$. ThereforeThe physical Jeans length isSince $G{\rho }_0$ has units inverse square seconds, $k_J$ has units ${\mathrm{m}}^{-1}$ and ${\ell }_J$ metres. This is the compressive density branch of the Jeans instability; transverse velocity disturbances have no pressure or gravity restoring force in this static nonrotating model and are neutral at linear order.

Which scales grow. For $0<k<k_J$, or wavelength $\ell >{\ell }_J$, ${\omega }^2<0$ and one temporal solution grows exponentially. The static Jeans growth-rate spectrum isIts derivative is negative for $0<k<k_J$, so longer wavelengths grow faster. The supremum is $s\to \sqrt{4\pi G{\rho }_0}$ as $k\to 0^{+}$: there is no selected finite fastest wavelength in this infinite-medium calculation. The exactly uniform $k=0$ density change is not an ordinary finite-potential Fourier perturbation with the fixed background subtraction. In a finite cloud, boundaries set the longest allowed scale.

For $k>k_J$, the density perturbation oscillates as a stable acoustic wave. At $k=k_J$ the linear equation is ${\partial }_t^2\delta \rho =0$, so the mode is marginal rather than exponentially growing.

Comparing acoustic and gravitational times. A sound-crossing estimate for a region of size $\ell$ is $t_{\mathrm{sound}}\sim \ell /c_s$. Self-gravitational acceleration scales as $G{\rho }_0\ell$, so collapse over a distance $\ell$ takes $t_{\mathrm{grav}}\sim (G{\rho }_0{)}^{-1/2}$. Equating these scales giveswhich agrees with the Jeans length up to its numerical factor $\sqrt{\pi }$. This is the physical competition between pressure communication and gravitational contraction.

The constants can be checked using the free-fall time of a uniform sphere. For a pressureless shell initially at radius $a$, enclosing fixed mass $M_a$, energy conservation gives ${\dot{r}}^2=2G{M}_a(1/r-1/a)$. Thuswhere $M_a=4\pi {\rho }_0{a}^3/3$. Consequently the Jeans length and gravitational time conventions satisfyThey agree as characteristic scales, not as identical exact times. Alternatively, using the length $k_J^{-1}$ gives $1/(c_s{k}_J)=1/\sqrt{4\pi G{\rho }_0}$, exactly the gravitational inverse frequency entering the dispersion relation. The actual growth time of a marginal Jeans mode is infinite because pressure and gravity cancel there; it should not be confused with a pressureless free-fall time.