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Assume a thin disk has height-independent opacity ${\kappa }_{\mathrm{op}}$ and radiation-to-gas pressure ratio $\beta$, with vertical gravity $-{\Omega }_z^{2}z$. Hydrostatic equilibrium and radiative diffusion implyBalancing its derivative against viscous heating $\rho \nu q^2{\Omega }^2$ fixes the dynamic viscosityIt is independent of height under these assumptions. For a spherically symmetric potential, ${\Omega }_z=\Omega$. Integrating $\rho \nu$ through the full surface-to-surface thickness $2H$ gives $\bar{\nu }\Sigma =2H\rho \nu$, which determines the thickness from the radial accretion rate solution.

For a spherically symmetric potential, the Poisson equation for Newtonian gravity becomesIntegrate from the center, excluding any additional point mass there. The boundary condition is $r^2{\varphi }^{\prime }(r)\to 0$ as $r\to 0$, soThis derives the enclosed-mass form of the spherical shell theorem from the differential equation. Outside the mass, integrating ${\varphi }^{\prime }=GM/r^2$ gives $\varphi =-GM/r+C$; the conventional zero at infinity sets $C=0$.

For the dyadic shells, multiply their constant mass density by their spherical volumes. The $n$th shell contributesSumming the geometric series yieldsAt radius $r=2^{-N}$, the enclosed shells are exactly those with $n\ge N+1$; a boundary itself contributes no mass. ThusDividing by $r^2=4^{-N}$ givesThese dyadic spherical shells with inverse-radius density scaling have the same field magnitude at each shell boundary, although the field varies inside an individual shell. Finally, outside the planet,with zero potential at infinity; without that convention an arbitrary additive constant remains.

In the classical experiment, particles whose impact parameter lies between $b$ and $b+db$ occupy an incident annulus of areaFor a spherically symmetric potential, azimuthal symmetry scatters this annulus into the solid-angle stripThus the central-potential classical differential cross-section isIf the relation between $b$ and $\theta$ has several branches, one sums this expression over every branch. The corresponding diagram consists of an incoming line offset by $b$ from the centre, an outgoing asymptote making angle $\theta$ with the incident direction, and the annulus and solid-angle strip just described.

For quantum scattering by a central potential, the far-field wavefunction iswhere $f(\theta )$ is the scattering amplitude. The incident plane wave has probability currentTo leading order in $1/r$, the scattered spherical wave has radial currentThe scattered rate through area $r^{2}d\Omega$ divided by the incident flux is therefore the quantum differential cross-section from a scattering amplitude:

For a real short-range spherically symmetric potential, the stationary wavefunction must solve the Schrödinger equation, be regular at the origin whenever the potential is finite there, and satisfy the scattering boundary conditionHere $e^{ikz}$ is the incident plane wave, the second term is an outgoing spherical wave, and its coefficient $f(\theta )$ is the scattering amplitude.

Comparing the stated asymptotic expression with the partial-wave expansion of a scattering amplitude giveswhere $P_l$ is a Legendre polynomial. The outgoing-to-incoming coefficient in channel $l$ isPartial-wave unitarity requiresor equivalentlyThe real scattering phase shift ${\delta }_l$ is defined modulo $\pi$ by

For a circular orbit of cylindrical radius $r$ in an axisymmetric gravitational potential,A small vertical displacement satisfiesso the vertical epicyclic frequency isFor a spherically symmetric potential $\Phi (R)$, where $R=(r^2+z^2{)}^{1/2}$,Hence $\boxed{{\Omega }_z=\Omega }$. Geometrically, a slightly tilted circular orbit remains a circular orbit in a different plane, and its height completes one oscillation per revolution.