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An equatorial coordinate system labels a viewing direction on the celestial sphere by right ascension $\alpha$, increasing eastward around the reference celestial equator, and declination $\delta$, increasing northward from that celestial equator. These two angles locate a direction, without specifying the object's Euclidean distance. In the traditional convention, zero right ascension points toward the reference equinox, the northward crossing of the Sun's annual path through the celestial equator. An hour circle through the direction supplies the north-south arc on which declination is measured. ^[ref].

For a concrete construction, choose Cartesian coordinates with $x$ pointing toward the reference equinox, $z$ toward the north reference celestial pole, and positive $y$ given by the cross product of the positive $z$ and $x$ directions. The direction's unit vector isThis is a spherical coordinate system with polar angle $\pi /2-\delta$ and no radial coordinate. The trigonometric functions require consistent angular units: convert right ascension in hours to radians before evaluating them. The synthetic direction $\alpha =4^{\mathrm{h}}=60^{\circ }$, $\delta =30^{\circ }$ gives $\mathbf{n}=(\sqrt{3}/4,3/4,1/2)$; its squared norm is $3/16+9/16+1/4=1$. A change of reference axes applies a rotation to this unit vector before recovering the two angles. Away from the celestial poles, recover right ascension using the quadrant-aware $atan2(n_y,n_x)$ reduced modulo $2\pi$, and declination using $\arcsin n_z$; a plain arctangent of $n_y/n_x$ loses the quadrant. Adding a common offset to every right ascension describes only the special case of a rotation about the same polar axis.

The reference specification matters. For an equatorial coordinate system tied to the Earth, precession changes the reference celestial equator and reference equinox, and nutation distinguishes their mean and true versions. The date specifying those axes is distinct from the astronomical epoch at which a catalogue's source directions apply: advancing a star's proper motion and changing reference axes are separate operations. A modern catalogue may instead use fixed reference frame axes, aligned approximately with the traditional J2000.0 axes, without a moving reference equinox. Thus a pair of numbers labelled merely "J2000" does not settle every convention. State the reference frame, the astronomical epoch, and whether a direction is catalogued or apparent. ^[ref].

During one night's elementary fixed-direction approximation, a star's right ascension and declination remain constant while its altitude, azimuth and hour angle change because the Earth turns. In angular units, with $\Theta$ the local sidereal time, west-positive hour angle $H$ obeysAt upper meridian transit, $H=0$ and the local sidereal time equals the transiting direction's right ascension. Pair apparent sidereal time with apparent right ascension referred to the true celestial equator and reference equinox of date; mean sidereal time refers to the mean reference equinox. Raw fixed-frame catalogue coordinates require conversion before precision pointing. ^[ref].

This separation makes the equatorial coordinate system useful: a catalogue can identify a target independently of the observer's current astronomical horizon, while local sidereal time tells the observer which hour circle is on the celestial meridian. Precision observations also account for proper motion, stellar parallax, aberration of light and atmospheric refraction; the fixed-direction construction above omits those effects. A reference-axis rotation preserves angles between directions, whereas changing the observer or correcting atmospheric refraction is a different operation. ^[ref].

Write $r=|x|$, $u(x)=\bar{u}(r)$, and let $g(r)$ be the radial profile of the radial derivative $\nabla u(x)\cdot x/|x|$, which may be assumed to be a radial function here. The spherical coordinate system givesSince $1\le r^2\le 4$, both $\bar{u}$ and $g$ belong to the Lp space $L^6(1,2)$. It remains to identify the weak derivative of the radial profile directly.

For $\varphi \in C_c^{\infty }(1,2)$, each component of $\varphi (r)x/r$ is a test function on $A$. Sum the three defining weak derivative identities, using the stated divergence formula:Given an arbitrary test function $\zeta \in C_c^{\infty }(1,2)$, choose $\varphi (r)=\zeta (r)/r^2$. Dividing by $4\pi$ gives $\int g\zeta =-\int \bar{u}{\zeta }^{\prime }$, exactly the definition of the weak derivative. We have provedThis radial Sobolev profile on an annulus estimate follows from the two weighted integrals above and equivalence of the usual Sobolev norms.

In the standard spherical coordinate system, $\theta$ is measured from the positive $z$ axis. The radial unit vector in spherical coordinates isThe coordinate forms of the elliptic paraboloid are given in the labelled parts below.

For the other surface, $r^2\cos 2\theta =r^2({\cos }^2\theta -{\sin }^2\theta )$ converts toFor $a>0$, this is a two-sheeted hyperboloid, with two disconnected sheets starting at $z=\pm \sqrt{a}$ and opening along the $z$ axis. For $a<0$, it is a one-sheet hyperboloid:Its waist is the circle $z=0$, $x^2+y^2=|a|$. Both are surfaces of revolution about the $z$ axis and approach the cone $z^2=x^2+y^2$ far from the origin.

The figure shows $a=1$ and $a=-1$; scaling all coordinates by $\sqrt{|a|}$ gives the other magnitudes.

Assume a time-independent spherical relative potential $\psi (r)$, with acceleration $\nabla \psi$. The relative energy and specific angular momentum areAlong an orbit, $\dot{E}=\nabla \psi \cdot \mathbf{v}-\mathbf{v}\cdot \nabla \psi =0$ and $\dot{\mathbf{\ell }}=\mathbf{r}\times \nabla \psi =0$, since a spherical gradient is radial. Thus both $E$ and $L$ are integrals of motion. Choose the escape level $\psi =0$ and take the bound galactic distribution function to vanish for $E\le 0$.

For an isotropic $f(E)$, integrate over a spherical coordinate system in velocity space:Differentiating gives ${\rho }^{\prime }(\psi )=2\pi \sqrt{2}{\int }_0^{\psi }f(E)/\sqrt{\psi -E}dE$. Convolve once more with the reciprocal-square-root kernel. Reversing the order of integration, under the usual integrability assumptions, usesTherefore ${\int }_0^{E_0}{\rho }^{\prime }(\psi )/\sqrt{E_0-\psi }d\psi =2\sqrt{2}{\pi }^2{\int }_0^{E_0}f(E)dE$. Differentiating proves Eddington inversion:Here ${\pi }^2$ is outside the square root, as in the original PDF; the TeX transcription wrongly places it inside. The assumptions include a locally integrable bound distribution and sufficient density regularity, with no extra unbound or boundary population. An inverted expression must additionally be nonnegative to be physical. Since the velocity measure and $f$ are invariant under all velocity rotations, $⟨v_r^2⟩=⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩=⟨v^2⟩/3$ whenever these moments exist; mixed moments vanish.

For the half-anisotropic distribution function, write the polar velocity angle from the radial direction as $\eta$. Then $L=rv\sin \eta$, andThe apparent $1/L$ singularity is integrable because it cancels the factor $v\sin \eta$ in the velocity measure. Integrating $\eta \in [0,\pi ]$ and $\phi \in [0,2\pi ]$ givesThe density relation requires $r\rho \to 0$ at the escape boundary and $f_E\ge 0$ for a physical distribution. Consider a monotonically decreasing $\psi$. The largest allowed radius at fixed binding energy is the radial-orbit limit $r_E$ with $\psi (r_E)=E$; a nonzero-$L$ orbit normally has a smaller apocenter because of its centrifugal term. Changing from $\psi$ to $r$ givesThe numerator in the original PDF instead uses $\rho +d\rho /d\psi$. That is a genuine printed error: the chain rule for $d(r\rho )/d\psi$ requires $\rho +rd\rho /dr$. It is not legitimate to prove the printed expression as written. A concrete counterexample is the unit-parameter Hernquist model at $r=1$: ${\psi }^{\prime }=-1/4$, ${\rho }^{\prime }=-(5/2)\rho$ and $d\rho /d\psi =10\rho$. The printed quotient is $-44\rho$, whereas the correct $g$ is $6\rho >0$. The former would give a negative distribution.

The angular weight after cancellation is uniform in $\eta$, so the averages of ${\cos }^2\eta$ and ${\sin }^2\eta$ are both $1/2$. The azimuthal angle splits the tangential term equally. Hence $⟨v_r^2⟩=⟨v_{\theta }^2+v_{\varphi }^2⟩$ and each individual tangential moment is half the radial one. The velocity-anisotropy parameter is consequentlyThis is a radially biased constant-anisotropy distribution function, not an isotropic one or purely radial motion. It does not by itself prove dynamical stability; it preferentially weights small angular momentum while remaining integrable.

For the Hernquist model, use the Poisson equation for Newtonian gravity with the relative-potential sign, ${\nabla }^2\psi =-4\pi G\rho$. Since ${\psi }^{\prime }=-GM/(r+a{)}^2$,Its enclosed mass is $M{r}^2/(r+a{)}^2$, which tends to $M$. Expressing the augmented density in the relative potential gives $r\rho =a{\psi }^3/(2\pi G^3{M}^2)$. The preceding derivative therefore giveswith zero bound-model population outside the allowed energy domain. It is nonnegative and reproduces the density by direct integration. The Hernquist model has a central density cusp, so the formulas for the local $1/L$ distribution are interpreted at $r>0$. As a further check, the radial second moment obtained from the same integral is $\psi /4$ and each tangential second moment is $\psi /8$, consistent with $\beta =1/2$.

Write the angular dependence as $f(\varphi )={\sum }_{n\ge 1}{a}_n\sin n\varphi +{\sum }_{n\ge 0}{b}_n\cos n\varphi$, so $u^{\prime }=\sin \theta f(\varphi )e_{\varphi }$. In the spherical coordinate system, $n=e_r$ and $e_r\times e_{\varphi }=-e_{\theta }$, withThus the integral in the torque-free rotation of a spherical squirmer has componentsThe vanishing follows from north-south cancellation, including the potentially relevant first azimuthal harmonics. In the $z$ direction, orthogonality of complex exponentials givesConsequently every rotational component is determined:This assumes an integrable prescribed angular function, with the Fourier series convergent sufficiently to justify the surface integrals; a finite Fourier sum is enough for the construction below. Other harmonics can generate exterior flows even though they contribute no net rotation.

For flow-free rotation of a spherical squirmer, choose $b_0\ne 0$ and set every $a_n$ and every $b_n$ with $n\ge 1$ to zero. Then $u^{\prime }=b_0\sin \theta e_{\varphi }$, while the rigid-rotation velocity isThe total laboratory-frame boundary velocity is therefore zero everywhere. With fluid at rest at infinity, Uniqueness of Stokes flow givesA constant pressure supplies zero resultant force and torque, so this is consistent with free motion. The imposed surface actuation counter-rotates relative to the material sphere; no claim is being made that ordinary no-slip rigid rotation produces zero flow.

In a spherical coordinate system,Since $n_x\mp i{n}_y=\sin \theta e^{\mp i\varphi }$,

Differentiating the spherical coordinate systemgives the orthogonal decompositionwhereThus the scale factors of orthogonal coordinates are

Since the total differential satisfiescomparison of coefficients gives the gradient

In the spherical coordinate system, $|d\mathbf{x}{|}^2=d{r}^2+r^{2}d{\Omega }_2^2$ and $\mathbf{x}\cdot d\mathbf{x}=rdr$. HenceThus the static coordinates on anti-de Sitter spacetime havewhich is positive for every $r\ge 0$.

The radial unit vector in the spherical coordinate system isThe coordinate tangent in the polar direction has length $r$, so its corresponding unit vector isDifferentiating ${\mathbf{e}}_r$ at fixed $\varphi$ therefore givesThe factor is not $r$: the corresponding scale factors of orthogonal coordinates occur in the derivative of the position vector, whereas the question differentiates its unit direction. The formula is understood away from the coordinate singularities where a complete spherical basis is not uniquely specified.

Substitute $U=R(r)F(\theta )$ into the axisymmetric Laplace equation and multiply by $r^2/(RF)$. The radial and angular terms must be separate constants:ThusWith $x=\cos \theta$, the angular equation becomesRequiring boundedness on the entire polar interval, and using the supplied characterization of bounded Legendre polynomial solutions, gives $\lambda =n(n+1)$ and $F\propto P_n(\cos \theta )$ for $n=0,1,2,\dots$. Trying $R=r^s$ gives $s(s+1)=n(n+1)$, hence $s=n$ or $s=-n-1$.

The globally angular-regular separated modes and their superposition are thereforeCoefficients and convergence are determined by the radial domain and boundary data. A domain containing the origin excludes negative radial powers for a regular solution. If the angular domain excludes a polar axis of the spherical coordinate system and singular angular behavior is allowed, additional Legendre-function branches are possible; the displayed family is the one regular over the complete sphere, as intended by the hint.

For the nonrelativistic Coulomb Hamiltonian with reduced electron-proton mass $\mu$, separation in Spherical coordinate system gives ${\psi }_{n\ell m}(r,\theta ,\varphi )=R_{n\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The quantum numbers obey $n\ge 1$, $0\le \ell \le n-1$ and $-\ell \le m\le \ell$. These are joint eigenstates of energy, $L^2$ and $L_z$, with the latter two eigenvalues $\ell (\ell +1){\hslash }^2$ and $m\hslash$.

The Coulomb bound-state energies and characteristic length areFor example, ${\psi }_{100}=(\pi a_{\mu }^3{)}^{-1/2}{e}^{-r/a_{\mu }}$ is spherically symmetric and has no radial node. In general the radial part contains an exponential times a Generalized Laguerre polynomial, with $n-\ell -1$ radial nodes. The spherical harmonic specifies angular structure and azimuthal phase; real linear combinations of degenerate $m$ states give familiar directional orbitals.

The leading Coulomb energy depends on $n$ alone, giving $n^2$ spatial states for each principal shell, or $2n^2$ including electron spin. Fine structure, Lamb shift and Hyperfine structure lift parts of this ideal degeneracy. An atomic orbital is a probability amplitude, not a classical planetary path.^[ref]

A radial profile is a one-variable function $\bar{u}$ representing a radial function $u$ by $u(x)=\bar{u}(|x|)$. On an annulus $a<|x|<b$, with $a>0$, the spherical coordinate system identifies its Lp norm with the weighted profile norm through $\int |u{|}^p=|S^{n-1}|{\int }_a^b|\bar{u}(r){|}^p{r}^{n-1}dr$. Thus equality of two radial functions almost everywhere identifies their profiles almost everywhere in the ordinary one-dimensional sense.

At fixed angular coordinates, differentiating the Cartesian coordinates with respect to radial distance gives this unit vector. It is the local direction of increasing radius in the spherical coordinate system.

A right circular cone with vertex $O$ selects all directions within a half-opening angle $\theta$ of its axis. Here $\theta$ is measured from the axis to a boundary direction, so the full opening in a meridian plane is $2\theta$. For $0\le \theta \le \pi$, interpret the selected directions as a spherical cap; for $\theta >\pi /2$, this is a region larger than a hemisphere.

The spherical coordinate system gives the area element $dA=R^2\sin \alpha d\alpha d\varphi$ on the sphere. Dividing by $R^2$ and integrating over the cap yieldsThe factor $\sin \alpha$ matters: a thin angular strip near a pole contains fewer directions than an equally wide strip near the equator. Treating equal intervals of both spherical coordinates as equal solid angles would overcount the polar directions.

At $\theta =45^{\circ }$, the solid angle is $\pi (2-\sqrt{2})\approx 1.8403$ steradians, or about $14.64\%$ of all directions. At $\theta =90^{\circ }$ it is $2\pi$ steradians, and at $180^{\circ }$ it is $4\pi$ steradians. For a small $\theta$ measured in radians, the Taylor series $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$ gives the small-angle approximation $\Omega \approx \pi {\theta }^2$. Doubling a narrow half-opening angle therefore approximately quadruples its coverage.