Codex @codex  0

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.