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Linearize the Collisionless Boltzmann equation and the perturbation Poisson equation about a homogeneous velocity distribution $f_0$. A plane wave gives $f_1={\varphi }_1\mathbf{k}\cdot {\nabla }_v{f}_0/(\mathbf{k}\cdot \mathbf{v}-\omega )$. Integrating this density and using $-k^2{\varphi }_1=4\pi G\int f_1{d}^{3}v$ yields $1+(4\pi G/k^2)\int [\mathbf{k}\cdot {\nabla }_v{f}_0/(\mathbf{k}\cdot \mathbf{v}-\omega )]d^{3}v=0$. Growing modes have $\mathrm{Im}\omega >0$ and no real-velocity pole; real and damped frequencies require the causal continuation prescription. The Newtonian homogeneous background requires the Jeans swindle.

The Jeans swindle is the removal of the unperturbed uniform density's gravitational field when treating an infinite homogeneous Newtonian medium as static. Without such subtraction or an external supporting prescription, a nonzero constant density and zero background gravitational acceleration cannot satisfy Poisson's equation. Linear perturbations then obey ${\nabla }^2\delta \varphi =4\pi G\delta \rho$. This is a specified idealization, not an exact isolated infinite equilibrium.

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.

Using the supplied first critical value ${\xi }_c\simeq 6.5$, the critical radius of a pressure-confined isothermal sphere isNo radius in metres can be specified without $K$ and a central mass density. Here $c_s$ is the isothermal sound speed, consistent with the fixed-temperature variations in part (d). At a given central mass density, equilibria truncated at $\xi <{\xi }_c$ lie on the first stable branch, while increasing the dimensionless outer radius beyond this first threshold gives an unstable configuration. Numerical integration gives ${\xi }_c\simeq 6.451$, consistent with the paper's rounded $6.5$.

For a transparent comparison with the given Jeans length, set the uniform comparison density ${\rho }_0={\rho }_c$ and use the same speed of sound. ThenBoth scales express the competition between pressure support and self-gravity, with characteristic length $c_s/\sqrt{G\rho }$. Their numerical closeness under this density choice does not mean that the models impose the same threshold: $r_c$ is a boundary radius, whereas ${\lambda }_J$ is the wavelength of a perturbation. Comparing $r_c$ with ${\lambda }_J/2$ would instead give a factor about $2.07$. Using the sphere's mean or boundary density instead of its central density changes the coefficient again, so the reference density must always be stated.

The pressure-confined isothermal sphere gives an actual finite, centrally concentrated hydrostatic equilibrium and an explicit confining pressure, making its radial stability threshold more self-consistent for a roughly spherical isothermal cloud. The uniform Jeans instability model gives a simpler local criterion, includes perturbations at different wavelengths, and does not require constructing a spherical equilibrium. Its ideal uniform static infinite background requires the Jeans swindle or an explicitly specified treatment of background gravity. Neither model includes rotation, magnetic fields, turbulent support or a realistic temperature evolution, so neither is a universal collapse criterion.

In the periodic cube, $H_0=|\mathbf{v}{|}^2/2=(2\pi /L{)}^2|\mathbf{J}{|}^2/2$, so $\mathbf{\Omega }=(2\pi /L)\mathbf{v}$. The supplied action-angle variables are canonical: $d^3\theta d^{3}J=d^{3}x{d}^{3}v$. For $\mathbf{k}=2\pi \mathbf{n}/L$,Consequently c becomesFor each nonzero wavenumber, the Poisson equation for Newtonian gravity givesSubstitute the preceding response and add the externally imposed Newtonian gravitational potential:ThereforeThis gravitational dielectric function has the opposite susceptibility sign to the electron dielectric function, reflecting attractive rather than repulsive collective interactions. The homogeneous mean field is explicitly subtracted in this fictitious periodic model: the Poisson equation for Newtonian gravity acts on density fluctuations, and the $\mathbf{k}=0$ mode is omitted. Otherwise a nonzero uniform positive mass density would be incompatible with a periodic potential. This is the convention often called the Jeans swindle.

For analytic continuation, resolve velocity parallel to $\mathbf{k}$: let $u=\hat{\mathbf{k}}\cdot \mathbf{v}$ and $F(u)=\int f_0(u,{\mathbf{v}}_{\perp })d^2{v}_{\perp }$. ThenStart at $\mathrm{Im}\omega >0$ and deform the Landau contour continuously below the moving pole. At real $\omega$, the integral isFor $\mathrm{Im}\omega <0$, this causal continuation equals the undeformed real-axis integral plus $2\pi i{F}^{\prime }(\omega /k)$ wherever $F^{\prime }$ has the necessary analytic extension. It must not be replaced by the undeformed integral alone.

Zeros of ${ϵ}_{\mathbf{k}}$ are self-consistent collective Landau modes: with no external forcing, a nonzero Newtonian gravitational potential is possible at such a root. For time dependence $e^{-i\omega t}$, upper-half-plane roots grow, lower-half-plane roots describe Landau damping, and real roots are neutral. Damped Landau modes are poles of the continued response; they need not be regular discrete eigenfunctions of the original collisionless operator.

Regularity qualification. Continuation throughout the complex plane presupposes suitable analytic $F$. Arbitrary smooth $F$ need not allow it. For example, $F(u)=C\exp [-1/(1-u^2)]$ for $|u|<1$ and zero elsewhere is smooth but not analytic at $u=\pm 1$; its Cauchy integral cannot be holomorphically continued across those endpoints. The Landau contour gives the standard causal continuation on the domains where the equilibrium permits it, with possible singularities retained.

On a length scale $\ell$, pressure communicates support over a sound-crossing time $\ell /v_s$, whereas self-gravity acts over a time of order $(G{\rho }_0{)}^{-1/2}$. When the former is longer, gravity can amplify an overdensity before pressure smooths it. Thus Jeans instability occurs at scales of order $\ell \gtrsim v_s/\sqrt{G{\rho }_0}$; the precise numerical factor depends on whether scale means wavelength or inverse wavenumber.

Take a uniform static background, small perturbations, an inviscid barotropic fluid, and $\delta p=v_s^2\delta \rho$ with $v_s^2=(dp/d\rho {)}_0>0$. An infinite Newtonian homogeneous density cannot itself have zero background gravitational acceleration while satisfying the unmodified Poisson equation for Newtonian gravity. Specify the Jeans swindle: subtract the homogeneous field, or regard an external prescription as balancing it, and apply Poisson's equation only to perturbations. The linearized continuity equation and Euler equations for an inviscid fluid areTaking a time derivative of the first equation and substituting the second yields ${\partial }_t^2\delta \rho =v_s^2{\nabla }^2\delta \rho +4\pi G{\rho }_0\delta \rho$. A plane wave $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ therefore givesFor $0<k<k_J$ there is exponential growth at rate $\sqrt{4\pi G{\rho }_0-v_s^2{k}^2}$. At $k_J$ the density mode is neutral to this linear order, and above it there are acoustic oscillations. The critical wavelength is $2\pi /k_J=\sqrt{\pi }{v}_s/\sqrt{G{\rho }_0}$. Transverse velocity perturbations decouple as zero-frequency vortical modes in this idealization.

For the attractive Yukawa potential, assume a real screening parameter $\alpha \ge 0$. Its Green function obeys $({\nabla }^2-{\alpha }^2)(-Gm{e}^{-\alpha r}/r)=4\pi Gm{\delta }^{(3)}(\mathbf{r})$, including the delta source. Thus $\delta {\varphi }_{\mathbf{k}}=-4\pi G\delta {\rho }_{\mathbf{k}}/(k^2+{\alpha }^2)$. The same fluid elimination gives Jeans instability with Yukawa gravity:There exists a growing nonzero-wavenumber mode precisely whenEquality gives no exponentially growing nonzero mode. If screening exceeds this threshold, the finite-range gravitational response is too weak to overcome pressure at any wavelength. Otherwise all sufficiently long but nonconstant waves remain unstable, with a longer minimum unstable wavelength than in Newtonian gravity. For $\alpha >0$ the long-wave gravitational contribution also scales as $k^2$, so the growth rate tends to zero as $k\to 0$; screening does not produce a second positive lower cutoff in wavenumber. A homogeneous screened background can have the constant gravitational potential $-4\pi G{\rho }_0/{\alpha }^2$, with zero force, so no Newtonian background subtraction is needed for nonzero screening. The limits $k\to 0$ and $\alpha \to 0$ are nonuniform. An exact $k=0$ density change is not one of the growing finite-wavelength modes. In the pressureless edge case $v_s=0$, every finite $k>0$ is unstable for positive density and finite screening.

Use the same specified homogeneous Newtonian background and Jeans swindle, but describe stars by a mass-normalized galactic distribution function. The Collisionless Boltzmann equation is ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{x}f-\nabla \varphi \cdot {\nabla }_{v}f=0$. Let $f=f_0(\mathbf{v})+f_1{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ and similarly perturb the gravitational potential. To first order,Eliminating $f_1$ and the nonzero gravitational potential amplitude gives the collisionless Jeans dispersion relationFor growing modes $\mathrm{Im}\omega >0$ there is no real-velocity pole. Real and damped frequencies require a causal contour prescription or analytic continuation; one must not simply integrate through a pole without specifying it.

Take $b>0$ in the Cauchy velocity distribution. Its normalization is $\int f_0{d}^{3}v={\rho }_0$. Orient $\mathbf{k}$ along the $u=v_z$ axis and integrate over the perpendicular velocities:At the marginal mode $\omega =0$, $F^{\prime }(u)/u=-2{\rho }_{0}b/[\pi (u^2+b^2{)}^2]$, with its finite limiting value at zero. The provided Gamma function integral at power two givesConsequentlyTo check that this really separates growing modes, put $c=\omega /k$ in the upper half-plane. Residue integration, or integration by parts followed by the Cauchy resolvent, gives $\int F^{\prime }(u)/(u-c)du={\rho }_0/(c+ib{)}^2$. The upper-half-plane solution has $\omega =i(\sqrt{4\pi G{\rho }_0}-kb)$, which grows exactly for $0<k<k_J$. At the threshold it approaches the marginal mode continuously. Although $b$ plays the threshold role of a velocity scale, this distribution's second moment diverges: the radial integrand for $\int v^2{f}_0{d}^{3}v$ tends to a nonzero constant at large speed. It is therefore incorrect to identify $b^2$ with a finite velocity dispersion or to infer this result by substituting an rms speed into a fluid formula.

Under the Jeans swindle, longitudinal density disturbances in a static uniform gas have ${\omega }^2=c_s^2{k}^2-4\pi G{\rho }_0$. For $0<k<k_J=\sqrt{4\pi G{\rho }_0}/c_s$, one solution grows as $e^{s(k)t}$ with the displayed rate. It decreases monotonically with wavenumber, so the longest available wavelengths grow fastest and the growth-rate supremum is $\sqrt{4\pi G{\rho }_0}$ as $k\to 0^{+}$. There is no preferred finite fastest wavelength in this model. The exact uniform $k=0$ density change is not an ordinary finite-potential Fourier perturbation of the fixed background, and finite-cloud boundaries restrict the available long modes.