= Solution
\b[The background and the Jeans prescription.] For a resting fluid, $\mathbf u_0=0$ and continuity is automatic. The remaining <hydrostatic equilibrium> equations are
$$
\boxed{\nabla p_0=-\rho_0\nabla\psi_0,\qquad
\nabla^2\psi_0=4\pi G\rho_0.}
$$
An 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.
\b[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 gives
$$
\boxed{\partial_t\delta\rho+\rho_0\nabla\cdot\mathbf v=0.}
$$
The advective acceleration is quadratic, and the background pressure gradient vanishes, so the Euler equation becomes
$$
\boxed{\partial_t\mathbf v=-\frac1{\rho_0}\nabla\delta p-\nabla\delta\psi.}
$$
With the Jeans prescription and constant <sound speed>, the gravitational and pressure closures are
$$
\boxed{\nabla^2\delta\psi=4\pi G\delta\rho,\qquad
\delta p=c_s^2\delta\rho.}
$$
The density correction to $1/\rho$ multiplying $\nabla\delta p$ is second order and is consistently omitted.
\b[The longitudinal dispersion relation.] Differentiate continuity in time and substitute the divergence of the linearized Euler equation:
$$
\partial_t^2\delta\rho
=\nabla^2\delta p+\rho_0\nabla^2\delta\psi
=c_s^2\nabla^2\delta\rho+4\pi G\rho_0\delta\rho.
$$
For a <plane wave> proportional to $e^{i(\mathbf k\cdot\mathbf x-\omega t)}$, $\partial_t^2\mapsto-\omega^2$ and $\nabla^2\mapsto-k^2$. Therefore
$$
\boxed{\omega^2=c_s^2k^2-4\pi G\rho_0
=c_s^2(k^2-k_J^2),\qquad
k_J=\frac{\sqrt{4\pi G\rho_0}}{c_s}.}
$$
The physical <Jeans length> is
$$
\boxed{\ell_J=\frac{2\pi}{k_J}
=c_s\sqrt{\frac{\pi}{G\rho_0}}.}
$$
Since $G\rho_0$ 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.
\b[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> is
$$
\boxed{s(k)=\sqrt{4\pi G\rho_0-c_s^2k^2},\qquad
\delta\rho\propto e^{s(k)t}.}
$$
Its 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 $k\to0^+$: 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.
\Image[/past-exam-of-the-astrophysical-fluid-dynamics-course-of-the-university-of-cambridge/2018-b4-jeans-spectrum.png]
{title=Original calculated Jeans dispersion and growth-rate curves; only wavelengths longer than the Jeans length grow, with faster growth toward the longest available scales}
\b[Comparing acoustic and gravitational times.] A sound-crossing estimate for a region of size $\ell$ is $t_{\rm sound}\sim\ell/c_s$. Self-gravitational acceleration scales as $G\rho_0\ell$, so collapse over a distance $\ell$ takes $t_{\rm grav}\sim(G\rho_0)^{-1/2}$. Equating these scales gives
$$
\ell\sim\frac{c_s}{\sqrt{G\rho_0}},
$$
which 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 $\dot r^2=2GM_a(1/r-1/a)$. Thus
$$
t_{\rm ff}=\int_0^a\frac{dr}{\sqrt{2GM_a(1/r-1/a)}}
=\frac{\pi}{2\sqrt2}\sqrt{\frac{a^3}{GM_a}}
=\sqrt{\frac{3\pi}{32G\rho_0}},
$$
where $M_a=4\pi\rho_0a^3/3$. Consequently the <Jeans length and gravitational time conventions> satisfy
$$
\boxed{\frac{\ell_J}{c_s}=\sqrt{\frac{\pi}{G\rho_0}}
=\sqrt{\frac{32}{3}}\,t_{\rm ff}\simeq3.27\,t_{\rm ff}.}
$$
They agree as characteristic scales, not as identical exact times. Alternatively, using the length $k_J^{-1}$ gives $1/(c_sk_J)=1/\sqrt{4\pi G\rho_0}$, 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.
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