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A shearing-wave ansatz in a uniform inviscid isothermal shearing sheet removes background advection when the radial wavevector grows as $k_x(t)=k_x(0)+S{k}_{y}t$. The razor-thin disk Poisson kernel reduces the pressure-plus-gravity amplitude to $[c_s^2-2\pi G{\Sigma }_0/k]{\tilde{\Sigma }}^{\prime }/{\Sigma }_0$. Continuity couples it to $i{k}_x{v}_x+i{k}_y{v}_y$, while the two momentum equations have rotational couplings $-2\Omega v_y$ and $(2\Omega -S)v_x$. The linearized vortensity amplitude remains constant, because the barotropic pressure and gravitational forces have zero curl and the background vortensity is uniform.

The local axisymmetric density/velocity system of a pressured dust layer with linear gas drag and a fixed Keplerian shearing sheet gas flow has this cubic dispersion relation. The $\Omega /2$ azimuthal velocity coefficient includes advection of the Keplerian shear. The razor-thin disk Poisson kernel supplies the $|k|$ self-gravity term. Keeping the full amplitude determinant retains the zero-frequency branch and avoids dividing out the secular mode.

Evaluating the potential of a razor-thin disk at height $ϵ$ gives a softened horizontal force. Its Fourier multiplier is $e^{-kϵ}$ times the razor-thin disk Poisson kernel. This suppresses short wavelengths while retaining the long-wave potential. A real finite-thickness profile generally has a different full reduction factor.

Adopt the planar Fourier transform conventionExtend the disk potential to three dimensions. The Poisson equation for Newtonian gravity becomesFor $k>0$, decay away from an isolated disk and continuity across it give $\tilde{\varphi }=B(\mathbf{k})e^{-k|z|}$. Integration through the plane yields $-2kB=4\pi G\tilde{\Sigma }$, hence the razor-thin disk Poisson kernelThe given transform of $K(\mathbf{x})=1/|\mathbf{x}|$ is $\tilde{K}=2\pi /k$. The convolution theorem therefore gives the forward potential relationFor the reverse relation, $\widetilde{{\Delta }_2\varphi }=-k^2\tilde{\varphi }$, soInvert the transform to obtain the thin-disk potential-density inversionAn additive constant in $\varphi$ is annihilated by ${\Delta }_2$. These formulas hold for suitable isolated finite disks, and in a distributional or regularized sense for less regular profiles. The uniform zero-wavenumber sheet needs a separate vertical solution; the infinite Mestel disk requires potential differences or an infrared regularization, since its absolute forward integral diverges.

Apply the shearing-wave ansatz to each disturbance. Acting on its phase givesThus a spatially uniform amplitude system exists whenFor a nonzero horizontal wavenumber $k=|\mathbf{k}|$, the razor-thin disk Poisson kernel follows from $({\partial }_z^2-k^2){\tilde{\Phi }}_d^{\prime }=4\pi G{\tilde{\Sigma }}^{\prime }\delta (z)$. The potential decays above and below the sheet as ${\tilde{\Phi }}_{d,m}^{\prime }{e}^{-k|z|}$. Integrating across the midplane gives the jump $-2k{\tilde{\Phi }}_{d,m}^{\prime }=4\pi G{\tilde{\Sigma }}^{\prime }$, henceSubstitute this potential and the phase evolution into the linearized equations. The barotropic shearing-wave gravitational amplitude system isThe self-gravity sign is attractive and opposes the pressure response. The uniform $k=0$ mass density shift requires its own gravitational reference and is not obtained by dividing by $k$.

Write the perturbation velocity as $(u,v)$, the density amplitude as ${\sigma }_1$, and ${\gamma }_d=ϵ\Omega$. Axisymmetry removes advection by the background azimuthal flow, but the radial perturbation advects the Keplerian shear: $u{\partial }_x{U}_{g,y}=-3\Omega u/2$. Combining this with the Coriolis force gives the linearized shearing sheet equationsThe razor-thin disk Poisson kernel supplies the self-gravity term. The $\Omega /2$ coefficient, rather than $2\Omega$, is essential: it includes the perturbed advection of the background velocity.

Put $A=c^2{k}^2-2\pi G{\sigma }_0|k|$ and ${\omega }^2={\Omega }^2+A$. The determinant of the three amplitude equations isExpanding it gives the dust gravitational dispersion relation with gas dragUsing a determinant avoids division by $s$ and retains the neutral/secular branch. The radial epicyclic frequency of this Keplerian shearing sheet is $\Omega$, so the three terms in ${\omega }^2$ represent rotational support, self-gravity and dust pressure.

Use a local, homogeneous razor-thin disk approximation in a frame rotating with constant $\Omega$. The unperturbed planar velocity is zero in this frame, and the large-scale gravitational and centrifugal forces balance. Neglect viscosity, magnetic fields, thickness and background gradients across a wavelength. Take small planar disturbances, with wavelength short compared with the galaxy's background scale but long enough for a fluid description. The barotropic closure of a razor-thin disk is $P=K{\Sigma }^{\gamma }$, with $K$ fixed and a positive derivativeSolid-body rotation has no shear and has radial epicyclic frequency $\kappa =2|\Omega |$. By rotational symmetry of the local model choose a wavevector along $x$, and write perturbations proportional to $e^{i(kx-\omega t)}$.

Let ${\Sigma }_1,u,v,{\Phi }_1$ be the surface density, $x$-velocity, $y$-velocity and Newtonian gravitational potential amplitudes. The linearized continuity equation and Euler equations with the Coriolis force areThe perturbing Newtonian gravitational potential solves the Poisson equation for Newtonian gravityIts decaying solution is proportional to $e^{-|k||z|}$. The jump in its derivative is $-2|k|{\Phi }_1(0)=4\pi G{\Sigma }_1$, giving the razor-thin disk Poisson kernelEliminating $v$ from the two momentum equations, and using ${\Sigma }_1=k{\Sigma }_{0}u/\omega$, gives the density-wave branchThe complete linear system also has a zero-frequency balanced mode; it is not the growing density-wave branch, and division by $\omega$ in this elimination excludes it. The result is the uniformly rotating gas-sheet dispersion relation: pressure opposes compression at large wavenumber, self-gravity promotes compression, and rotation provides epicyclic support.

A mode is exponentially unstable when ${\omega }^2<0$. With $\Omega =0$ and $c>0$, this occurs atThe uniform $k=0$ perturbation is marginal rather than growing in this local calculation. With $c=0$, rotation fails to stabilize sufficiently short waves:These limits show why both pressure and rotation are needed for stability at all wavelengths.

For nonzero pressure and rotation, put $q=|k|$. Complete the square:Its minimum lies at $q_{\ast }=\pi G{\Sigma }_0/c^2$. Thus a growing mode exists precisely whenThe printed inequality has the opposite physical meaning: it is the condition for no exponentially growing density wave. Equality is marginal. In the usual gas Toomre stability criterion, $Q=\kappa c/(\pi G{\Sigma }_0)=2|\Omega |c/(\pi G{\Sigma }_0)$, so instability is $Q<1$ and stability is $Q\ge 1$. The unstable wavenumber band of a rotating gas sheet iswhen $Q<1$.

If ${\Sigma }_0$ and $\Omega \ne 0$ remain fixed while $c^2$ decreases slowly, first reach marginality atThe first wavelength to become unstable just below this threshold is the marginal fragmentation wavelength of a rotating sheet:Density maxima are separated by approximately this wavelength. The expected fragment size is of this order; an overdense half-wave has width about ${\ell }_{\mathrm{crit}}/2$. Linear theory fixes a preferred wavelength, not an exact nonlinear clump radius or shape. Further cooling shifts the fastest-growing wavelength to $2c^2/(G{\Sigma }_0)$. A rough fragment mass is consequently of order ${\Sigma }_0{\ell }_{\mathrm{crit}}^2$, with a geometrical factor depending on the nonlinear fragmentation pattern.

For nonzero real wavenumber $k$, seek a Fourier mode of the Newtonian gravitational potential in the form ${\Phi }^{\prime }=f(z)e^{ikx}$. Away from the razor-thin sheet, the Poisson equation for Newtonian gravity becomes $f^{\prime \prime }-k^{2}f=0$. Requiring the perturbation to decay on both sides and remain continuous gives $f=A{e}^{-|k||z|}$.

Integrating the Poisson equation for Newtonian gravity through $z=0$ fixes the derivative jump:Thus the perturbing gravitational potential isAt the midplane this reduces to the razor-thin disk Poisson kernel. The observable real disturbance is the real part. Writing the perturbation amplitude as ${\Sigma }_a$ instead of ${\Sigma }_0$ gives the same formula with ${\Sigma }_a$; linearity makes it independent of the background density. The $k=0$ disturbance is a uniformly changed sheet and has potential proportional to $|z|$, rather than a decaying nonzero-wavenumber solution.

The three terms represent competing physical effects:

In the razor-thin disk approximation, the three-dimensional Poisson equation isFor a horizontal Fourier mode with wavevector $\mathbf{k}$ and $k=|\mathbf{k}|>0$, the Fourier transform of this equation isThe solution that decays away from the disk isConsequently the required razor-thin disk Poisson kernel in the midplane isThe spatially uniform $k=0$ background is excluded from this local perturbation formula.

Write the perturbations as $(\sigma ,u,v)e^{i{k}_{x}x+\lambda t}$, set $k=|k_x|$, and evaluate all unmarked background quantities at ${\Sigma }_0$. DefineThe linearized mass conservation equation isThe radial and azimuthal momentum equations arewhere the razor-thin disk Poisson kernel gives $\varphi =-2\pi G\sigma /k$. The term proportional to $\beta$ comes from perturbing the density-dependent background shear stress $T_{xy}=-\nu \Sigma S$.

Eliminating $\sigma ,u,v$ and using $S=3\Omega /2$, so that the radial epicyclic frequency obeys ${\kappa }_r^2=2\Omega (2\Omega -S)={\Omega }^2$, gives the dispersion relation

For an isolated razor-thin disk, define the two-dimensional Fourier transform by $\tilde{g}(\mathbf{k})=\int e^{-i\mathbf{k}\cdot \mathbf{x}}g(\mathbf{x})d^{2}x$. At $k>0$, the razor-thin disk Poisson kernel gives the displayed inversion. Since the transform of $1/|\mathbf{x}|$ is $2\pi /k$, inversion also gives $\Sigma =(4{\pi }^{2}G{)}^{-1}(1/|\mathbf{x}|)\ast {\Delta }_2\Phi$. The forward relation is $\Phi =-G(1/|\mathbf{x}|)\ast \Sigma$. An additive potential constant is irrelevant; a uniform infinite sheet and scale-free divergent potentials require separate zero-mode or potential-difference treatment rather than an arbitrary decaying boundary condition.

For a homogeneous inviscid razor-thin disk approximation with solid-body rotation and barotropic sound speed $c$, the compressive wave branch obeys ${\omega }^2=c^2{k}^2-2\pi G{\Sigma }_0|k|+4{\Omega }^2$. The razor-thin disk Poisson kernel supplies the self-gravity term, and the radial epicyclic frequency is $2|\Omega |$. Growing modes have ${\omega }^2<0$; a separate zero-frequency balanced mode can also exist.