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Expand a unit-mass particle's Lagrangian about an axisymmetric circular orbit of radius $r_0$ and frequency ${\Omega }_0$, using $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Circular-orbit balance removes the linear radial term; total-time-derivative invariance of a Lagrangian removes $r_0{\Omega }_0\dot{y}$. To second order,Here $q$ is the orbital shear parameter and ${\Omega }_z$ is the vertical epicyclic frequency. The Euler-Lagrange equations retain the local Coriolis acceleration and tidal gravity.

Put $r_S=2R_g$. On the midplane, circular orbit balance gives $r{\Omega }^2={\partial }_r\Phi =GM/(r-r_S{)}^2$. The specific angular momentum, specific orbital energy and orbital shear parameter of a Paczyński-Wiita circular orbit are thereforeFor a small radial displacement at fixed specific angular momentum, the effective potential stability criterion gives the radial epicyclic frequencyHere ${\kappa }_r$ is the question's epicyclic $\kappa$. Its square is negative for $2R_g<r<6R_g$, so a radial displacement grows instead of undergoing epicyclic motion. The orbit at $r_{\mathrm{in}}=6R_g$ is marginal, and the outer orbits are stable to small radial displacements. At this inner edge,This efficiency belongs to the Paczyński-Wiita potential; it is an approximation to the relativistic black-hole result.

For unit mass, the Lagrangian in cylindrical coordinates isIntroduce shearing sheet coordinates by $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Evaluate all derivatives of $\Phi$ at $(r_0,0)$. The reference circular orbit obeys ${\Phi }_r=r_0{\Omega }_0^2$, while midplane symmetry gives ${\Phi }_z={\Phi }_{rz}=0$. Its Taylor expansion isExpanding the kinetic energy to the same order givesThe terms linear in $x$ cancel by circular-orbit balance. Discard the constant and the term $r_0{\Omega }_0\dot{y}$ by total-time-derivative invariance of a Lagrangian; these do not change the Euler-Lagrange equations. The particle Lagrangian in a shearing sheet is thereforewith shearing-sheet tidal potentialThe second form follows from ${\Phi }_r=r{\Omega }^2$, the orbital shear parameter $q_0=-d\log \Omega /d\log r$ and the vertical epicyclic frequency ${\Omega }_z^2={\Phi }_{zz}$.

For inward accretion rate $\dot{M}>0$, the steady conservation of angular momentum equation is $\mathcal{G}=\dot{M}(h-h_{\mathrm{in}})$ when the inner torque vanishes. The viscous torque in an accretion disk is $\mathcal{G}=2\pi qh\bar{\nu }\Sigma$, where $h=r^2\Omega$ and $q$ is the orbital shear parameter. ThusThe familiar Keplerian accretion disk follows when $q=3/2$ and $h\propto r^{1/2}$. The general form also applies to the non-Keplerian Paczyński-Wiita circular orbit rotation law.