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For a radial function, the Laplacian is $\Delta \varphi ={\varphi }_{rr}+2{\varphi }_r/r$. Introduce the retarded and advanced null coordinates and the radial reduction of the three-dimensional wave equation:The chain rule gives ${\partial }_t={\partial }_u+{\partial }_v$ and ${\partial }_r=-{\partial }_u+{\partial }_v$. Consequently the inhomogeneous wave equation isMultiplying the radial wave equation by $r$ removes its first-order radial derivative:Thus the particularly useful retarded and advanced null coordinates form isThese formulas hold for $r>0$. At the axis, smooth spherical symmetry imposes ${\varphi }_r(t,0)=0$ and $w(t,0)=0$; equivalently $w$ has a smooth odd extension across $r=0$.

A spherically symmetric stellar model has no preferred direction in the tangent plane at a fixed radius. Rotations about the radial axis exchange its two transverse velocity directions, so $⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩$. This uses spherical symmetry of the velocity distribution, not just a spherical mass density.

The augmented density $\rho (r,\psi )$ treats radius and relative potential as independent variables, agreeing with the physical density only along $\psi =\psi (r)$. For any regular bound spherical galactic distribution function $F(E,L)$, write $P_r=\int v_r^{2}F{d}^{3}v$ and $P_t=\int v_{\theta }^{2}F{d}^{3}v$. At fixed $r$, $F_{\psi }=F_E$ and ${\partial }_{v_r}F=-v_r{F}_E$. Integration by parts therefore givesFor $v_t^2=v_{\theta }^2+v_{\varphi }^2$, the velocity measure is $2\pi v_{t}d{v}_{t}d{v}_r$, and $r{F}_L={\partial }_{v_t}F+v_t{F}_E$. Integrating the transverse derivative gives $-2P_r$, while integrating the radial derivative in $\int v_r^2{v}_t^2{F}_E{d}^{3}v$ gives $2P_t$. Consequentlywhere $\psi$ is fixed. These identities show that the augmented representation holds for any such spherical distribution, not only a monomial ansatz. With radial pressure vanishing at the zero-binding boundary, set Along the physical curve, the chain rule gives $dA/dr={\int }_0^{\psi }{\partial }_r\rho (r,q)dq+\rho (r,\psi ){\psi }^{\prime }(r)$. The Spherical Jeans equation then requiresThus the Jeans moments from an augmented density areThe partial derivative holds $q$ fixed. Substitution cancels the integral of ${\partial }_r\rho$ and leaves precisely $\rho {\psi }^{\prime }$ in the Spherical Jeans equation. This is a representation by a chosen augmented density, not a unique solution determined by the one-variable density: different extensions off the physical curve encode different anisotropies, and not every formal extension necessarily gives a nonnegative galactic distribution function.

For the hypervirial density-potential family, differentiate the potential and use the spherical shell theorem:Differentiating $M(<r)$ and dividing by $4\pi r^2$ givesThe enclosed mass tends to $M$ at infinity and to zero at the centre for every $p>0$. The same result follows from ${\nabla }^2\psi =-4\pi G\rho$, with the stated sign convention.

From now on set $G=M=a=1$. The separable augmented density is $\rho (r,q)=(p+1)r^{p-2}{q}^{2p+1}/(4\pi )$. Integrating it givesand henceWith $\psi =(1+r^p{)}^{-1/p}$ these are the requested radial expressions. The velocity-anisotropy parameter is $\beta =1-p/2$, and the total mean square speed is $⟨v^2⟩=\psi /2$.

The local kinetic-energy density is $t=\rho ⟨v^2⟩/2=\rho \psi /4$. With the potential zero at infinity, the gravitational energy density is $w=\rho \Phi /2=-\rho \psi /2$, where $\Phi =-\psi$. ThereforeThis is the local virial relation of the hypervirial model, a special property stronger than the global virial theorem. To compute the global energies as well, set $I=\int \rho \psi d^{3}x$. The substitution $u=r^p$ yields the global binding integral of the hypervirial modelThe beta function integral converges for all $p>0$. ThusRestoring dimensions multiplies $I,T,W$ by $G{M}^2/a$.

To derive the hypervirial distribution function, let $E=\psi -v^2/2$ be the positive relative energy and $L=rv\sin \vartheta$ the magnitude of the specific angular momentum. Seek $F=C{L}^{p-2}{E}^s$ for $E>0$ and zero otherwise. Its density isThe angular integral is $\sqrt{\pi }\Gamma (p/2)/\Gamma ((p+1)/2)$. Substitution $z=v^2/(2\psi )$ makes the radial integralMatching the augmented-density power ${\psi }^{2p+1}$ gives $s=(3p+1)/2$. Matching its coefficient gives the normalization of the hypervirial distribution functionAll integrals are finite in velocity for $p>0$, and $C>0$. Because $E$ and $L$ are orbital integrals, this is a steady solution by the Jeans theorem. As checks, $p=1$ gives $C=3/(4{\pi }^3)$ for the Hernquist model, and $p=2$ gives $C=24\sqrt{2}/(7{\pi }^3)$ for the isotropic Plummer model.

The density and potential do not uniquely fix the stellar distribution. The preceding coefficient is fixed within the chosen separable power-law model, but nonuniqueness of spherical galactic distribution functions remains without that extra restriction. An explicit positive counterexample uses the same $p=2$ density. Define the Osipkov-Merritt distribution function variable $Q=E-L^2/2$ and takeRescale the transverse velocity by $\sqrt{1+r^2}$. Then $Q=\psi -[v_r^2+(1+r^2)v_t^2]/2$ and the velocity Jacobian contributes $(1+r^2{)}^{-1}$. Direct integration givesexactly the same Plummer model density. This anisotropic Plummer distribution with unit anisotropy radius has $\beta =r^2/(1+r^2)$, so it is distinct from the isotropic $E^{7/2}$ distribution while remaining positive and spherical. Its second moments differ, so it does not share the extra augmented-density and local-virial restrictions of the chosen hypervirial model.

False. Adiabatic invariance of an orbital action preserves actions under sufficiently slow regular evolution, not the ratios of their frequencies. With action-angle variables, ${\Omega }_i=\partial H/\partial J_i$ can change even when $J_i$ does not. A closed bound orbit requires commensurate radial and angular frequencies; rational commensurability is not an adiabatic invariant.

For an explicit counterexample, slowly change the physical potential from Kepler form towith a small, slowly varying positive $\alpha$. The radial epicyclic frequency and circular angular frequency at a nearly circular orbit areInitially their ratio is one. For generic nonzero $\alpha$, $\Omega /\kappa$ is irrational, producing the Kepler-harmonic epicyclic precession. An initially slightly eccentric ellipse with nonzero radial action therefore becomes a precessing nonclosed orbit, while its radial action remains approximately fixed and its angular momentum remains exactly fixed by spherical symmetry.

Adiabatic actions do not preserve orbital closure. Exceptional circular or resonantly commensurate final orbits do not rescue a claim about every initially closed orbit.

In electrostatics, Gauss's law is ${\oint }_{\partial \Omega }\mathbf{E}\cdot \mathbf{n}dS=Q_{\mathrm{enclosed}}/{ϵ}_0$, or locally $\nabla \cdot \mathbf{E}={\rho }_e/{ϵ}_0$, where ${\rho }_e$ is charge density and ${ϵ}_0$ is vacuum permittivity. Spherical symmetry makes the electric field radial, and Gauss's law givesChoose the additive constant of the electrostatic potential by $\varphi (\infty )=0$. Using $\mathbf{E}=-\nabla \varphi$ and continuity of potential across each charged shell givesThe field jumps at the shell surfaces, so the displayed formulas are for the requested open regions. The potential difference of the spherical capacitor is $\Delta \varphi =Q(1/a-1/b)/(4\pi {ϵ}_0)$. Thus its capacitance and electrostatic energy areThe energy follows either by charging work ${\int }_0^{Q}q/Cdq$, or directly from $W=({ϵ}_0/2)\int |\mathbf{E}{|}^2{d}^{3}x=Q^2(1/a-1/b)/(8\pi {ϵ}_0)$.

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.

Assume a Newtonian, spherically symmetric, nonrotating body in hydrostatic equilibrium, supported by a polytropic equation of state $P=K{\rho }^{1+1/n}$ with constant $K>0$ and $n>0$. The enclosed mass and hydrostatic pressure support equation areEliminate $m$ by first writing $r^2{\rho }^{-1}dP/dr=-Gm$ and then differentiating:Let ${\rho }_c$ be central mass density, and introduce the Lane-Emden variables for a stellar polytropewhereThen $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Substitution cancels the dimensional factors and yields the Lane-Emden equationThe central conditions enforce the chosen central mass density and regular spherical symmetry. The local regular expansion is $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. If a finite first zero ${\xi }_1$ exists and surface pressure is negligible, the physical radius is $R=a{\xi }_1$, and the Lane-Emden mass formula is $M=4\pi a^3{\rho }_c[-{\xi }_1^2{\theta }^{\prime }({\xi }_1)]$. For $0<n<5$ the standard isolated solution has a finite surface; $n=5$ has infinite extent, so a finite surface must not be assumed for all indices. Irradiation, composition stratification and non-polytropic equations of state require more general structure equations.

By spherical symmetry, rotate a geodesic into the equatorial plane $\theta =\pi /2$. The geodesic conserved quantities from Killing vectors arewhere the dot denotes a suitable affine parameter. Normalize the tangent by $g(\dot{x},\dot{x})=-ϵ$, with $ϵ=1$ for timelike proper time, $ϵ=-1$ for spacelike proper length, and $ϵ=0$ for null geodesics. ThenTherefore the geodesic potential of the magnetic dilaton black hole isTogether with the two conserved first integrals, this reduces the radial motion to ${\dot{r}}^2/2+\tilde{V}=0$. To cover turning points as well, the radial geodesic equation itself givesThus a constant-radius geodesic must satisfy both $\tilde{V}=0$ and ${\tilde{V}}^{\prime }=0$. The intermediate static-coordinate formulas apply where $V\ne 0$; the radial equation extends through a regular horizon in regular coordinates.

For an affine parameter $\lambda$, the geodesic Lagrangian isVarying $S=\int \mathcal{L}d\lambda$ gives the geodesic equations

The rotational Killing vector fields of the spherical symmetry conserve the angular-momentum vector. Its direction is fixed, so every nonradial orbit lies in the plane through the symmetry centre orthogonal to that vector. A spatial rotation can make this the equatorial plane. Equivalently, the initial conditions $\theta =\pi /2$ and $\dot{\theta }=0$ solve the $\theta$ equation uniquely. This is the planarity of geodesics in spherical symmetry; a radial geodesic has zero angular momentum and may be assigned any plane.

On $\theta =\pi /2$, the cyclic coordinates $t$ and $\varphi$ produce geodesic conserved quantities from Killing vectors,For a timelike geodesic parametrized by proper time, the proper-time normalization is $g_{ab}{\dot{x}}^a{\dot{x}}^b=-1$. Substitution of $E$ and $L$ gives the first-order radial equationThis is the timelike geodesic effective potential for static Anti-de Sitter spacetime with curvature radius one.

A geodesic passing through $r=0$ must have $L=0$. If it moves away from the origin, $E>1$, andTaking proper time $s=0$ at departure givesIt turns around at $s=\pi /2$ and first returns to the origin atThis energy-independent refocusing is the characteristic radial timelike geodesic in anti-de Sitter spacetime result; for curvature radius $a$, the answer is $\pi a$.