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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.

In a stationary axisymmetric gravitational potential, conservation of angular momentum gives $L_z=R^2\dot{\varphi }$. Eliminating $\dot{\varphi }$ introduces the effective potentialFor an equatorial circular orbit, radial balance is ${\partial }_R{\Phi }_{\mathrm{eff}}(R_c,0)=0$, givingVertical balance also requires ${\Phi }_{,z}(R_c,0)=0$.

For epicyclic motion, assume a twice differentiable stationary gravitational potential symmetric under $z\mapsto -z$, small displacements $|x|,|z|\ll R_c$, and a stable circular orbit. Choose the epicyclic guiding center using the conserved $L_z$. Reflection symmetry makes ${\Phi }_{,Rz}(R_c,0)=0$, eliminating radial-vertical coupling at first order. Expanding the equations about $(R_c,0)$ givesThe derivatives are evaluated at the guiding centre, and stability requires ${\kappa }^2,{\nu }^2>0$. Without midplane symmetry, a mixed Hessian term can couple the two oscillations. The radial epicyclic frequency and vertical epicyclic frequency are the frequencies of these independent linear oscillations.

Set $\Omega =L_z/R_c^2$. Along the family of equatorial circular orbits, ${\Phi }_{,R}=R{\Omega }^2$. Differentiating this relation and adding the centrifugal contribution givesTo interpret the common frequency range, let $q=d\log v_c/d\log R$. Since $v_c=R\Omega$, one has ${\kappa }^2=2(1+q){\Omega }^2$. A Keplerian disk has $q=-1/2$ and $\kappa =\Omega$, a flat galaxy rotation curve has $q=0$ and $\kappa =\sqrt{2}\Omega$, and solid-body rotation has $q=1$ and $\kappa =2\Omega$. Typical galactic rotation curves lie between these slopes. Thus $\Omega \lesssim \kappa \lesssim 2\Omega$ is a useful galactic range, not a theorem for every possible axisymmetric potential. As a precise sufficient example, epicyclic frequency bounds for monotone spherical density follow from ${\kappa }^2={\Omega }^2+4\pi G\rho$ and $0\le \rho \le 3M/(4\pi R^3)$.

Use a Cartesian frame rotating with the epicyclic guiding center, with $x$ pointing radially outwards and $y=R_c(\varphi -\Omega t)$ in the direction of rotation. To first order, conservation of angular momentum givesThe radial harmonic oscillator solution and its azimuthal integral areA constant in $y$ merely changes the azimuthal origin of the guiding centre. The epicyclic ellipse obeys $x^2/X^2+y^2/Y^2=1$. In the usual frequency range it is elongated azimuthally, and the star travels clockwise when $x$ points right and $y$ up: its small motion relative to the prograde guiding centre is retrograde.

For the Oort constants, subtract the defining expressions to obtain $\Omega =A-B$ and insert $R{\Omega }^{\prime }=-2A$ into the radial epicyclic frequency formula:The solar-neighbourhood values give $\Omega =26.5\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, $\kappa =\sqrt{1272}\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, andThe solar epicyclic ellipse is therefore about $1.49$ times longer azimuthally than radially.

A complete radial oscillation takes $T_r=2\pi /\kappa$. The oscillatory part of $\dot{\varphi }$ has zero average over this interval, so the epicyclic azimuthal advance isThis describes the advance between consecutive radial turning points of the same type; it need not be a full $2\pi$ revolution. For the Sun,The azimuthal advance estimate uses the same linear epicyclic motion approximation as the axis ratio.

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.