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For a smooth horizontal circular hoop of radius $a$, rotated at constant angular velocity $\omega$ about a vertical axis through a rim point, measure $\theta$ at that pivot from the diameter toward the bead. In rotating Cartesian coordinate system coordinates, $\mathbf{r}=a(1+\cos 2\theta ,\sin 2\theta ,0)$. The tangential component of the centrifugal acceleration gives the displayed equation; the Coriolis acceleration is normal to the hoop. The farthest point $\theta =0$ is stable for $\omega \ne 0$, with small oscillation angular frequency $|\omega |$. The pivot itself is unstable. This geometry differs from a vertical rotating hoop and has no tangential gravitational term.

A Cartesian basis is an orthonormal basis associated with the perpendicular axes of a Cartesian coordinate system.

The original PDF measures $\theta$ at the rim point $O$, from the positive $x^{\prime }$-axis to $OP$; it is not the angle at the centre of the circle. The converted TeX omits this entire subpart. In rotating Cartesian coordinate system, the centre is $(a,0)$, and the nonzero intersection of the ray $\mathbf{r}=\rho (\cos \theta ,\sin \theta )$ with $(x^{\prime }-a{)}^2+y^{\prime 2}=a^2$ has $\rho =2a\cos \theta$. ThereforeThe second form stays regular even at the rim point $O$, where the polar radius is zero. Define the outward unit normal from the centre and the increasing-angle unit tangent vector byThe first two derivatives in the rotating reference frame are ${\mathbf{v}}^{\prime }=2a\dot{\theta }\mathbf{\tau }$ and ${\mathbf{a}}^{\prime }=2a\ddot{\theta }\mathbf{\tau }-4a{\dot{\theta }}^2\mathbf{n}$. These describe a bead on a horizontal hoop rotating about its rim.

Let the free surface be at $r=R_1$ and take the Newtonian gravitational potential to vanish at infinity. Mass conservation fixes $R_1=(3M_1/(4\pi {\rho }_1){)}^{1/3}$. The enclosed mass is $m(r)=4\pi {\rho }_1{r}^3/3$, and hydrostatic equilibrium requires $d{p}_0/dr=-{\rho }_{1}Gm(r)/r^2$. With vacuum outside, $p_0(R_1)=0$, givingThe pressure is zero outside. A prescribed constant external pressure would simply add that constant to $p_0$.

The bulk Eulerian and Lagrangian fluid perturbations obey $\delta \rho =0$ because the mass density is uniform and the Lagrangian displacement has zero divergence. The linearized Euler momentum equation in the interior isTaking the curl gives ${\omega }^2\nabla \times \mathbf{\xi }=0$. Thus every nonzero-frequency normal mode has an irrotational flow displacement; the simply connected space formed by the interior admits $\mathbf{\xi }=\nabla U$ and incompressibility gives the Laplace equation ${\nabla }^{2}U=0$. For a regular center, one spherical harmonic component is $U=A_l{r}^l{Y}_l^m$, with $|m|\le l$.

It is essential to retain the surface gravity perturbation of a uniform-density sphere. Although the bulk mass density perturbation is zero, the displaced density discontinuity givesThe Poisson equation implies that $\delta \Phi$ is continuous and its outward radial derivative has the jumpUsing regularity at the center and decay at infinity, writeThe derivative jump is $-(2l+1)C_l{R}_1^{l-1}{Y}_l^m$, so $C_l=-4\pi G{\rho }_{1}l{A}_l/(2l+1)$.

Integrating the bulk Euler momentum equation gives $\delta p/{\rho }_1={\omega }^{2}U-\delta \Phi$ for $l\ge 1$. At the free surface the Lagrangian pressure perturbation vanishes:Combining these relations yields the Kelvin stellar mode frequencyThe $2l+1$ choices of $m$ are degenerate because of spherical symmetry. For $l=1$, $r{Y}_1^m$ is a linear combination of coordinates in a Cartesian coordinate system, so $\nabla U$ is a constant displacement: the zero frequency is rigid translation of the whole isolated body. Such a translation has no restoring force. The constant $l=0$ potential generates no displacement; a radial breathing motion is excluded by incompressibility and a regular center.

The printed irrotational assertion needs the nonzero-frequency qualification, or restriction to the potential incompressible stellar surface modes. It is false for all neutral displacements: $\mathbf{\xi }=\mathbf{\Omega }\times \mathbf{r}$ has zero divergence, zero normal displacement at the surface, and $\delta p=\delta \Phi =0$, so it is a zero-frequency displacement, but $\nabla \times \mathbf{\xi }=2\mathbf{\Omega }\ne 0$. This does not change the Kelvin stellar mode spectrum just derived.

Use the gnomonic projection associated with an undistorted focal plane. Write $\Delta =\alpha -A$ and rotate the celestial Cartesian coordinate system so that the pointing meridian has longitude zero. The stellar direction and an orthonormal basis adapted to the optical axis areHere $\mathbf{e}$ points east and $\mathbf{n}$ points north. Intersect the ray $t\mathbf{s}$ with the plane $\mathbf{x}\cdot \mathbf{b}=f$. It gives $t=f/(\mathbf{s}\cdot \mathbf{b})$, so the detector coordinates areThese equations also avoid spurious singularities caused by writing individual tangents or cotangents.

To put this gnomonic projection in the desired form, let $\rho =({\cos }^2\delta {\cos }^2\Delta +{\sin }^2\delta {)}^{1/2}$ and choose $q$ locally byThus $\cot q=\cot \delta \cos \Delta$. The denominator becomes $\rho \cos (q-D)$, the numerator for $\eta$ becomes $\rho \sin (q-D)$, and $\cos \delta \sin \Delta =\rho \cos q\tan \Delta$. ConsequentlyUse the local meridian chart $\cos \Delta >0$, with $q\in [-\pi /2,\pi /2]$ chosen continuously near $D$ and $q=D$ at the image centre. A visible gnomonic projection requires $\mathbf{s}\cdot \mathbf{b}>0$, but this front-hemisphere condition alone does not select that meridian chart. A narrow field near a celestial pole can cross the opposite meridian; for such fields use the Cartesian expressions above and distinguish the oriented great circle parameter from ordinary declination. In particular, the small-field limit is $\xi \simeq f\cos D(\alpha -A)$ and $\eta \simeq f(\delta -D)$, with angles in radians. The factor $\cos D$ is the shrinking angular distance per unit right ascension near the pole.

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.