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Use the cylindrical coordinate system basis ${\mathbf{e}}_r,{\mathbf{e}}_{\phi },{\mathbf{e}}_z$. The meridional unit tangent vector and outward unit normal vector consistent with the given geometry areAn surface of revolution has $\theta$ independent of $\phi$. ConsequentlyApply the product rule to the vector force balance:Equating the three independent components to zero, and using $r_s=\cos \theta$, givesThus, away from the rotation axis where $r>0$, the scalar mechanical equilibrium equations areThe two geometric factors ${\theta }_s$ and $\sin \theta /r$ are the signed meridional and azimuthal principal curvatures in this convention. The normal equation is the Young–Laplace equation generalized to unequal principal membrane tensions.

In an axisymmetric cylindrical coordinate system, the steady Collisionless Boltzmann equation isLet $\nu =\int f{d}^{3}v$ and $\overline{A}={\nu }^{-1}\int Af{d}^{3}v$. Multiply by $v_R$ and integrate over velocities. The spatial terms give ${\partial }_R(\nu \overline{v_R^2})+{\partial }_z(\nu \overline{v_R{v}_z})$. Integration by parts in $v_R$ supplies $\nu {\Phi }_R-\nu \overline{v_{\varphi }^2}/R$, and the $v_{\varphi }$ derivative supplies $\nu \overline{v_R^2}/R$. The $v_z$ force term vanishes because $v_R$ does not depend on $v_z$. This assumes sufficient decay of the galactic distribution function to discard all velocity boundary terms. The radial Jeans equation is consequentlyReflection symmetry about $z=0$ makes $\nu$ even and $\overline{v_R{v}_z}$ odd. At the midplane the latter vanishes, and ${\nu }^{-1}{\partial }_z(\nu \overline{v_R{v}_z})={\partial }_z\overline{v_R{v}_z}$. Multiplying by $R/\nu$ provesAll quadratic velocities here are population moments, not individual stellar velocities. In particular, symmetry does not make the midplane tilt contribution to asymmetric drift disappear: an odd mixed moment can have a nonzero derivative at zero.

For application to the Milky Way, assume negligible mean radial and vertical flows, and define the circular speed by $v_c^2=R{\Phi }_R(R,0)$. Write $\overline{v_R^2}={\sigma }_R^2$ and $\overline{v_{\varphi }^2}={\overline{v}}_{\varphi }^2+{\sigma }_{\varphi }^2$. Rearranging gives the exact stress-support relationThus the stellar asymmetric drift $v_a=v_c-{\overline{v}}_{\varphi }$ is the right-hand side divided by $v_c+{\overline{v}}_{\varphi }$. For $v_a\ll v_c$, this denominator is approximately $2v_c$. If $\nu \propto e^{-R/h_{\nu }}$ and ${\sigma }_R^2\propto e^{-R/h_{\sigma }}$ locally, the bracket becomes $R/h_{\nu }+R/h_{\sigma }-1+{\sigma }_{\varphi }^2/{\sigma }_R^2$, making the density and dispersion-gradient contributions explicit.

A radially declining random-motion stress supplies some of the force needed to support a galactic disk, so its stars need less ordered azimuthal motion than a population on circular orbits. Hotter, often older, stellar populations therefore generally rotate more slowly than cold populations in the same gravitational potential; this is the observed stellar asymmetric drift. It is a stress-support effect, not a frictional slowing of each orbit. For a velocity ellipsoid aligned approximately with spherical coordinates, $\overline{v_R{v}_z}\simeq ({\sigma }_R^2-{\sigma }_z^2)z/R$, and the midplane tilt term contributes $-({\sigma }_R^2-{\sigma }_z^2)$ rather than zero.

Local measurements of mean rotation and all components of the velocity dispersion, combined with tracer-density and dispersion gradients, can therefore constrain $v_c$ and the radial Newtonian gravitational field. A hotter sample's mean rotation must not be identified directly with the circular speed. The tracer number density need not be the gravitating mass density. Nonaxisymmetric streaming, a nonstationary disk, poorly measured radial gradients, or a neglected tilt term can bias this inference; these are assumptions of the derivation rather than extra free corrections to the circular force.

In the cylindrical coordinate system, $r=\sqrt{x^2+y^2}=5$ and $z=-1$. The point is in the third quadrant, so a representative azimuth in $[0,2\pi )$ is $\pi +\arctan (4/3)$. ThereforeThe angle is defined modulo $2\pi$.

The radial coordinate in the cylindrical coordinate system is the Euclidean distance from the $z$-axis:

Write the position vector as $\mathbf{r}(\rho ,\varphi ,z)=(\rho \cos \varphi ,\rho \sin \varphi ,z)$. Its coordinate tangent vectors areTheir pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, areThus the cylindrical coordinate system is orthogonal wherever it is a nonsingular coordinate chart. At $\rho =0$, the angular tangent vector vanishes and the azimuth is undefined; this is the usual coordinate singularity on the axis, not a failure of the Euclidean geometry.

The method of images solves certain electrostatics problems by replacing a conductor by fictitious electric charges outside the region where the solution is required. The replacement must satisfy the actual boundary values of the electric potential as well as the prescribed sources. The uniqueness of Poisson equation then establishes that the constructed electric potential equals the physical one in that region. The image locations are mathematical devices; they are not real charges inside the metal.

For a point charge $q$ at $(0,0,a)$ above an infinite grounded conducting plane $z=0$, put an image charge $-q$ at $(0,0,-a)$. In a cylindrical coordinate system with coordinates $(s,z)$ the potential in the physical region isThis vanishes at $z=0$, has the required singularity at the real charge, and tends to zero at infinity. With electric field $\mathbf{E}=-\nabla \varphi$, the induced surface charge density follows from the normal field immediately above the conductor:The force on $q$ is the field of the image evaluated at the real charge:The electrostatic energy relative to infinite separation is $U=\frac{1}{2}q{\varphi }_{\mathrm{image}}=-q^2/(16\pi {ϵ}_{0}a)$. The factor $1/2$ accounts for the fact that the induced charge changes as the real charge is assembled; treating the image as an independently fixed physical charge would give the wrong energy.

A second standard geometry is a grounded conducting sphere of radius $b$ with a charge $q$ a distance $d>b$ from its centre. Its exterior field is generated by an image $q^{\prime }=-qb/d$ at distance $b^2/d$ from the centre on the line towards the real charge. On the sphere the distances to the real and image charges are in the ratio $d/b$, making their potentials cancel. For an isolated neutral sphere, an additional central image $+qb/d$ makes the total induced charge zero and leaves the surface an equipotential. The construction is powerful for suitable symmetric boundaries, but arbitrary boundaries do not in general admit a finite image system.

Use separate moving orthonormal bases in the reference and current cylindrical coordinate systems. For pure torsion of an incompressible cylinder, the deformation map isThe unit vectors depend on the angle: ${\partial }_{\theta }{\mathbf{e}}_r={\mathbf{e}}_{\theta }$ and ${\partial }_{\theta }{\mathbf{e}}_{\theta }=-{\mathbf{e}}_r$. ConsequentlyIn particular, the last derivative includes the rotation of ${\mathbf{e}}_r$ along the cylinder. The material gradient has the reference coordinate scale factor $1/R$:The outer products specify the images of the three reference unit vectors. Thus, with current basis vectors indexing rows and reference basis vectors indexing columns,Its determinant is one, checking incompressibility. The radius is unchanged, so the current outer radius in part (d) is $a=A$. The formula at the coordinate axis is understood by continuity in Cartesian coordinates.

Multiplying the deformation gradient by its matrix transpose gives the left Cauchy-Green deformation tensor and its trace:The derivative of the strain energy density isFor brevity put $g(r)={\mu }_1+{\mu }_2(3+{\tau }^2{r}^2{)}^n$. Part (a) gives the Cauchy stress tensor, in the current cylindrical coordinate system orthonormal basis, asIn particular, the end-face shear stress is $T_{\theta z}=2g\tau r$. The undetermined pressure $p$ affects the normal stresses but does not affect this shear stress. No normal traction conditions have been specified to determine $p$.

A deformation map twisting every cross-section of a cylinder through an angle proportional to its reference axial position, without radial or axial extension. In the local current and reference cylindrical coordinate systems, the deformation gradient is $F=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& \tau r\\ 0& 0& 1\end{array}\right)$. Its determinant is one. The twist per unit reference length is $\tau$; an end separation $L$ gives end rotation $\phi =\tau L$. The physical torque is $N=2\pi {\int }_0^A{T}_{\theta z}{r}^{2}dr$.