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For $u=u_c[1+e\cos \nu (\psi -{\psi }_0)]$ with $\nu >0$ and $e\ge 1$, the positive-radius scattering branch has $|\nu (\psi -{\psi }_0)|<\arccos (-1/e)\le \pi$. Thus the principal complex argument in the displayed expression is continuous throughout that branch, unlike at an apocentre of a bound orbit. Its derivative is $\nu$, so the angle modulo $2\pi$, or the associated unit apsidal vector in the orbital plane, is a stationary integral of motion fixing the periapsis direction. Together with energy and three components of specific angular momentum, it gives five regular scattering labels even for irrational $\nu$. Periodicity is therefore necessary for five orbit labels in recurrent bound motion, not in general for escaping trajectories.

A stellar orbit with sensitive dependence on its initial phase-space point, often diagnosed by a positive Lyapunov exponent. It need not explore the whole energy hypersurface: regular islands and disconnected chaotic regions can coexist. If an orbit is dense in a connected energy-shell region, every continuous integral of motion is constant on its closure; that stronger condition excludes further independent continuous constraints within that region.

On the regular set of two uncoupled oscillators with irrational frequency ratio, every continuous globally defined integral of motion is constant on each fixed-action flat torus, because every orbit is dense there. A differentiable integral is consequently a function of the two actions, so its differential of a smooth map lies in their span. Local angle combinations can still give extra local integrals; they fail to be single-valued on the torus.

An isolating integral is a regular single-valued integral of motion whose fixed value restricts accessible phase space to a lower-dimensional level set. Functional independence of several such integrals means their differentials are linearly independent on the region considered. A central potential has specific orbital energy and the three components of specific angular momentum as four generically independent integrals. A Kepler orbit has a fifth independent integral supplied by its fixed apsidal direction, encoded by the Laplace-Runge-Lenz vector; its vector components obey two constraints with energy and angular momentum. Circular and radial orbits are degenerate cases of this count.

For $\Phi (R,\varphi -\Omega t,z)$ with constant pattern angular velocity $\Omega$, ordinary specific energy satisfies $\dot{E}={\partial }_t\Phi =-\Omega {\partial }_{\varphi }\Phi$ and specific axial angular momentum satisfies ${\dot{L}}_z=-{\partial }_{\varphi }\Phi$. Their combination $E_J$ is conserved. This is the rotating-frame integral of motion; in the circular restricted three-body problem, the usual Jacobi constant is an affine rescaling of it. Thus time dependence in an inertial frame does not alone imply the absence of all integrals.

One integral can occur in an autonomous system with a fully chaotic region and no additional continuous orbit constraints. The specific orbital energy is conserved in a time-independent gravitational potential. If a chaotic stellar orbit is dense in a connected region of its energy hypersurface, every continuous integral of motion is constant on that region. Thus it cannot supply a new independent constraint there, and energy is the only independent isolating integral. A generic nonaxisymmetric potential can contain such regions; not every orbit in such a potential is necessarily chaotic.

An isolating integral is a single-valued integral of motion $I(\mathbf{x},\mathbf{v})$ whose fixed value confines an orbit to a lower-dimensional level set of phase space. Along the flow, $dI/dt=0$; for an autonomous Hamiltonian, this means $\{I,H\}=0$. Several isolating integrals are independent when their gradients are linearly independent on the regular region considered. Counting them requires a specified domain: merely local orbit labels in a flow box do not establish global single-valued integrals. The following cases concern regular globally meaningful orbit constraints.

In the distinction relevant here, an integral of motion is a single-valued function $I(\mathbf{x},\mathbf{v})$ of instantaneous phase space, with no explicit time dependence, constant along the orbit. In an autonomous Hamiltonian system this is $\{I,H\}=0$, where the bracket is the Poisson bracket. An integral must be defined consistently on the phase space region under discussion; an arbitrary multivalued formula obtained by solving one trajectory does not supply a global integral. Some conventions call explicitly time-dependent conserved functions integrals too; the stationary interpretation is needed for the following counts.

One isolating integral occurs when specific orbital energy is the only independent stationary invariant. An autonomous chaotic orbit that densely explores a connected five-dimensional specific orbital energy surface is the standard case. Every continuous integral of motion must then be constant on that surface, so it is locally a function of specific orbital energy rather than an additional independent integral. Irregular motion need not mean zero integrals; specific orbital energy still isolates the orbit from other specific orbital energy surfaces.

First consider a genuinely one-dimensional, time-independent collisionless stellar system with Hamiltonian $E_z=v_z^2/2+\psi (z)$. Choose $\psi (0)=0$. For bound vertical oscillations, a phase-mixed population has equal occupations of the upward and downward branches of each orbit. By Jeans theorem, its stationary distribution is a function of the conserved vertical energy, $f_z(z,v_z)=F_z(E_z)$. Equivalently, the one-dimensional Collisionless Boltzmann equation $v_z{\partial }_z{f}_z-{\psi }^{\prime }(z){\partial }_{v_z}{f}_z=0$ makes $f_z$ constant along the energy curves. Branch-dependent incoming and outgoing populations or time-dependent waves would not satisfy this assumption.

Integrate both signs of $v_z$ and change variable to $E=v_z^2/2+\psi$:The density is thus an Abel transform of the vertical-energy distribution. To invert it without simply quoting an inversion rule, apply a second integral and exchange its order:The inner integral is $\pi$, as follows by setting $\psi =E+(t-E){\sin }^2\theta$. Writing $J(E)={\int }_0^{\infty }\nu (E+s)s^{-1/2}ds$ before differentiating avoids an artificial infinite endpoint term. Under sufficient differentiability and decay,The vertical-energy Abel inversion of a stellar distribution is thereforeThis proves the normalization, including the factor from the two velocity signs. The additive origin of $\psi$ merely changes the origin of energy.

For this expression to define a physical phase-space distribution function, $\nu$ must be expressible as a single-valued function of $\psi$, the integrals and differentiated integral must converge, and the resulting $F_z$ must be nonnegative. Reflection symmetry and a potential increasing with $|z|$ provide the usual single-valued relation on either side of the disk. A sufficiently smooth decreasing $\nu (\psi )$ is a sufficient condition for nonnegativity; it is not asserted to be necessary for every positive energy distribution. If there is a finite escape energy, replace the infinite limits by that energy and impose the appropriate cutoff and boundary conditions. The displayed infinite-limit formula presumes either a confining potential or a distribution extended by zero above its allowed energy, with no omitted boundary population.

A three-dimensional galactic disk does not automatically meet these hypotheses. Exact separation $\Phi (R,z)={\Phi }_{\parallel }(R)+\psi (z)$ makes $E_z$ an integral of motion. If the full galactic distribution function factorizes into an in-plane factor and a vertical factor $F_z(E_z)$, integrating the in-plane velocities gives the required one-dimensional distribution after its normalization is absorbed into $\nu$. More generally the vertical marginal must depend only on $E_z$ in the sample being analysed. Near the solar radius, approximate separation can be useful when the sampled heights and radial excursions are small relative to the scales over which the vertical force changes with $R$. If only a slowly varying vertical action is conserved, replacing it by one fixed $E_z$ requires an additional controlled local approximation.

The population must be sufficiently stationary and phase mixed, with negligible mean vertical flow, and the data must measure a well-defined tracer population with a known selection function. Radial coupling and the tilted velocity ellipsoid must be negligible or modelled. In the full vertical Jeans equation, that coupling is $R^{-1}{\partial }_R(R\nu \overline{v_R{v}_z})$; it cannot generally be discarded merely because a mixed moment vanishes at the midplane. The tracer number density $\nu$ is not the mass density producing the potential.

To infer the local mass, measure the tracer density as a function of height and its vertical velocity distribution near the midplane. With $\psi (0)=0$, the latter directly samples $F_z(v_z^2/2)$. Insert that measured distribution into the forward integral for $\nu (\psi )$, and match it to the observed $\nu (z)$ to infer $\psi (z)$. Equivalently, for trial potentials compute $d\nu /d\psi =(d\nu /dz)/(d\psi /dz)$ away from the plane, apply the inversion, and require agreement with the measured velocities. Density data alone do not determine both an unknown potential and an unknown energy distribution.

Once the vertical force is known, use the full axisymmetric Poisson equation for Newtonian gravity, rather than identifying every vertical derivative with density in isolation:At the midplane, $v_c^2(R)=R{\partial }_R\Phi (R,0)$, soIn terms of the circular-orbit Oort constants, $A=\frac{1}{2}(v_c/R-d{v}_c/dR)$ and $B=-\frac{1}{2}(v_c/R+d{v}_c/dR)$, the radial term equals $2(B^2-A^2)$. It vanishes for a locally flat rotation curve, not for an arbitrary rotating disk. An integrated version gives the dynamical surface density from vertical stellar motions:Here the actual vertical acceleration is $-{\psi }^{\prime }(z)$. The sign of the radial correction follows directly from the Poisson equation for Newtonian gravity.

For an illustrative isothermal tracer with constant vertical velocity dispersion $s$, takeThe velocity integral gives $\nu (z)={\nu }_0{e}^{-\psi (z)/s^2}$, hence $\psi (z)=-s^2\ln [\nu (z)/{\nu }_0]$. For a locally flat rotation curve this yieldsThis example shows explicitly how a measured tracer scale and velocity scale constrain the gravitating density; using the tracer's own density as the mass source would instead impose an unjustified self-gravitating model.

The inferred density includes luminous and faint stars, gas and dark matter. Comparing it with an independently estimated baryonic census tests for extra gravitating mass. An excess need not be a dark disk: a dark matter halo also contributes to the local force. Conversely, agreement with the baryonic density locally does not exclude a halo at other radii or heights. Vertical disequilibrium, radial mixing, errors in selection or tracer populations, and neglected velocity tilt can mimic a mass discrepancy, so the one-dimensional equilibrium assumptions must be tested when interpreting such an inference.

An integral of motion is a function of phase space that is constant along a trajectory. Put $\varphi =-\psi +\mathrm{constant}$, so the stellar acceleration is $\dot{\mathbf{v}}=\nabla \psi$. In a time-independent spherical Newtonian gravitational potential,The second equality uses the radial direction of $\nabla \psi$. Thus both the binding energy and the full specific angular momentum vector are integrals of motion.

The Collisionless Boltzmann equation isFor a stationary galactic distribution function, this says $df/dt=0$ on every characteristic, so $f$ is constant on each stellar orbit. Locally choose a complete set of orbit labels $I_j$, themselves integrals of motion; constancy along the characteristic means $f=F(I_1,\dots ,I_s)$. Conversely, for any such differentiable $F$, the chain rule gives $df/dt={\sum }_j{F}_{,j}{\dot{I}}_j=0$. This proves Jeans theorem in its orbit-invariance form. It does not claim that just two particular integrals classify every orbit in an arbitrary potential. For the specified spherical model, $F(E,L)$ is a rotationally invariant stationary choice.

For the velocity integration, write $v_r=v\cos \eta$, $v_{\perp }=v\sin \eta$, $L=rv\sin \eta$, and set $k=r/r_a$. At fixed position the velocity-dependent factor isUse ${\int }_0^{\infty }{v}^2{e}^{-A{v}^2/(2{\sigma }^2)}dv=\sqrt{\pi /2}{\sigma }^3{A}^{-3/2}$ and integrate over $\chi$. The mass density isSince ${\rho }_c=\rho (0)={\rho }_1{e}^{\psi (0)/{\sigma }^2}$,These integrals use the stated untruncated galactic distribution function over all velocities; an additional escape-energy cutoff would change the result.

In local orthogonal velocity components the same exponent is $-[v_r^2+(1+k^2)(v_{\theta }^2+v_{\varphi }^2)]/(2{\sigma }^2)$. Each component is a centred Gaussian distribution. The one-dimensional identity $\int v^2{e}^{-A{v}^2/(2{\sigma }^2)}dv/\int e^{-A{v}^2/(2{\sigma }^2)}dv={\sigma }^2/A$ yieldsIn particular the velocity-anisotropy parameter is ${\beta }_{\mathrm{an}}=1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)=r^2/(r_a^2+r^2)$. This is Gaussian angular-momentum suppression in a stellar distribution.

The radial laws also require self-consistency through the Poisson equation for Newtonian gravity:For $r\ll r_a$, outside a sufficiently small core, the model is almost isotropic and isothermal. Seek its scale-free envelope with $\rho =A/r^2$. Its distribution-density relation gives ${\psi }^{\prime }=-2{\sigma }^2/r$, and the Poisson equation for Newtonian gravity gives $A={\sigma }^2/(2\pi G)$. Thus the intended intermediate singular isothermal sphere behaviour iswhere a core scale is $r_c=\sigma /\sqrt{4\pi G{\rho }_c}$.

The stated $r^{-2}$ law cannot be the actual central limit of a regular finite-density model. For finite ${\rho }_c$ and $\psi (0)$, regularity and the Poisson equation for Newtonian gravity instead giveand $v_c^2=(4\pi G{\rho }_c/3)r^2+O(r^4)$. A broad intermediate $r^{-2}$ region exists only if $r_c\ll r_a$; the radius $r_a$ alone does not ensure it.

For the outer radially anisotropic isothermal halo asymptotics, put $s=\log (r/r_a)$, $y=[\psi (r)-\psi (0)]/{\sigma }^2$ and $\lambda =4\pi G{\rho }_c{r}_a^2/{\sigma }^2$. At $r\gg r_a$ the density relation and Poisson equation becomeThe slow outer balance is $y_s\sim -\lambda e^y$, so $d(e^{-y})/ds\sim \lambda$, giving $e^y\sim 1/(\lambda s)$. Its consistency check is $y\sim -\log (\lambda s)$, for which $y_{ss}/y_s\sim -1/s\to 0$. ThereforeThe reference radius $r_{\ast }$ changes only subleading logarithmic terms. The circular speed falls as $(\log r{)}^{-1/2}$, rather than remaining exactly flat.

Physically, the model has an almost isotropic central region and, if the scales separate, a flat-rotation isothermal envelope. Beyond $r_a$, suppression of large specific angular momentum leaves predominantly radial orbits: radial velocity dispersion remains $\sigma$, while transverse dispersion falls as $r^{-1}$. The outer density acquires a logarithmic suppression and the galaxy rotation curve declines slowly. The enclosed mass still grows approximately as $r/\log r$, so this untruncated model has infinite total mass.

Write $\mu =G{M}_{\odot }$, and interpret the specified radial force field as acceleration, or force per unit test-particle mass. Since it is a central force, the specific angular momentum $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ is conserved: $\dot{\mathbf{h}}=\mathbf{r}\times \mathbf{F}=0$. For $h=|\mathbf{h}|\ne 0$, motion is confined to the fixed plane perpendicular to $\mathbf{h}$. Choose the orientation of the polar angle so that $h=r^2\dot{\theta }>0$.

To derive the Binet equation here, let $u=1/r$ and differentiate with respect to $\theta$. Then $\dot{r}=-h{u}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{u}^2(u^{\prime \prime }+u)$. Equating this to the radial acceleration givesPut $K=h^2+C$ and $\alpha =\sqrt{K/h^2}$. For $K>0$, the solution isThe radial potential energy per unit mass is $V(r)=-\mu /r+C/(2r^2)$, so the conserved specific orbital energy isSubstitute the displayed $u$ and $\dot{r}=-h{u}^{\prime }$ to obtain $\mathcal{E}={\mu }^2(e^2-1)/(2K)$. Therefore defineFor regular bound motion, $K>0$ and $-{\mu }^2/(2K)\le \mathcal{E}<0$, giving $a>0$ and $0\le e<1$. Since $K/\mu =a(1-e^2)$, choose the angular origin at a pericentre, so ${\theta }_0=0$, and obtainThe turning radii are $r_{\mathrm{p}}=a(1-e)$ and $r_{\mathrm{a}}=a(1+e)$. Thus $a$ is the arithmetic mean of the turning radii; when the orbit precesses it should not be interpreted as the semimajor axis of a fixed ellipse. If $K\le 0$, the centrifugal barrier needed for a regular two-turning-point bound orbit is absent, and this bound-orbit parametrization does not apply.

For $\alpha =1$, one has $C=0$ and the orbit is a closed Kepler orbit: an ellipse of eccentricity $e$ with the Sun at a focus. In addition to $\mathcal{E}$ and the three components of $\mathbf{h}$, the Laplace-Runge-Lenz vectoris conserved. Indeed,so $\dot{\mathbf{A}}=0$. It points toward the fixed pericentre. The constraints $\mathbf{A}\cdot \mathbf{h}=0$ and $A^2={\mu }^2+2\mathcal{E}{h}^2$ leave five generically independent isolating integrals. One may takechoosing a component $A_j$ whose differential is independent of the other four on the local region under consideration. The extra integral fixes the apsidal direction in the already fixed orbital plane.

For $\alpha \ne 1$, successive pericentres are separated by $2\pi /\alpha$ in polar angle. The generic bound orbit is a rosette in that same plane. Its perihelion advance per radial cycle isFor $\alpha <1$ this is prograde; for $\alpha >1$ it is retrograde. If $\alpha$ is rational, the orbit closes after a finite number of radial cycles; if it is irrational and $0<e<1$, its successive radial phases densely fill the accessible annulus. A generic independent set of single-valued isolating integrals isThe apsidal direction is no longer fixed. For irrational frequency ratio, any continuous integral of motion must be constant on the whole two-dimensional orbital torus, because one orbit is dense there; there is no further independent continuous integral specifying a fixed apsidal direction. Exceptional closed resonant orbits do not provide a fifth global integral on an open region of phase space. Circular and radial orbits are degenerate cases of these generic counts.

The plotted precession is enlarged for visibility. Both examples have $e=0.5$ and the same turning radii; the second has $\alpha =\sqrt{0.9}$.

For the precession rate, it is useful to derive the radial period rather than assume an unperturbed angular period. Set $r=a(1-e\cos \chi )$ in the radial energy equation. On a complete radial cycle,Hence $n=2\pi /P_r=\sqrt{\mu /a^3}$. Also $K=\mu a(1-e^2)$ and $h^2=K-C$, so ${\alpha }^{-1}=\sqrt{1-C/K}$. For $|C|\ll K$, inverse-cube perturbation of Kepler apsidal motion givesThere is a dimensional issue with the printed parameter: $C$ has units $L^4/T^2$ and $\mu$ has units $L^3/T^2$, so $\eta =aC/\mu$ has units $L^2$. Retaining that definition literally, the requested expression isA dimensionless perturbation coefficient would instead be $\tilde{\eta }=C/(\mu a)=\eta /a^2$. In that convention the rate is $-n\tilde{\eta }/[2(1-e^2)]$. These two conventions must not be conflated.

Use the stated residual precession and period. The advance in one radial cycle isWith $e=0.206$, the small-perturbation estimate becomesThe negative sign is required: an additional attractive inverse-cube acceleration advances perihelion, whereas the positive-$C$ term would make it regress. If a length-unit estimate of the printed $\eta$ is desired, $P_r=0.24\mathrm{yr}$ and the solar value $\mu \simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ give $a\simeq 5.78\times 10^{10}\mathrm{m}\simeq 0.386\mathrm{au}$ through the radial-period law. ThusThe dimensionless value is the estimate obtainable directly from the supplied period, eccentricity and advance; the dimensional value additionally uses the solar gravitational parameter to set the length scale.

Take the separate oscillator energiesThey depend on disjoint canonical coordinate pairs, soThus they are first integrals in involution. Their differentialsare nonzero precisely when $(x_i,y_i)\ne (0,0)$ and have disjoint supports. Hence they are functionally independent on the specified open set.

To understand all further global integrals of motion, use the actionsDirect differentiation gives $d{x}_i\wedge d{y}_i=d{\theta }_i\wedge d{R}_i$, so these are action-angle variables withWhen ${\alpha }_2/{\alpha }_1$ is irrational, every orbit is dense in its fixed-action two-dimensional flat torus. One elementary proof samples the orbit at times $2\pi j/{\alpha }_1$: the first angle returns, and the second undergoes an irrational rotation. These samples are dense in the second circle. Allowing a fixed additional time supplies any desired first angle, proving density in the whole flat torus.

A continuous, globally defined integral of motion must have the same value on an orbit and its closure. Therefore it is constant on every fixed-action flat torus. A $C^1$ such integral has the form $F=\varphi (R_1,R_2)$, with $\varphi$ $C^1$, by evaluating $F$ at a fixed choice of angles. HenceIt cannot be independent of $E_1,E_2$. This proves there are at most two globally independent differentiable integrals. If a third global integral were independent at a point on an excluded coordinate plane, independence would persist in a neighbourhood and hence at nearby points of the regular open set, again a contradiction.

The word global matters: on an angular chart, ${\alpha }_2{\theta }_1-{\alpha }_1{\theta }_2$ is a local integral of motion independent of the actions. It is not single-valued on the flat torus when the frequency ratio is irrational. Thus the assertion would be false for arbitrary local, rather than global, integrals.

A function $F$ is an integral of motion of $H$ when it is constant along every trajectory of the Hamiltonian flow. Infinitesimally,The antisymmetry of a Lie bracket of the Poisson bracket gives $X_{F}H=\{H,F\}=0$. Hence $H$ is also an integral of motion of $F$.

For the commuting assertion, ${\mathcal{L}}_{X_F}\omega =d({\iota }_{X_F}\omega )=d^{2}F=0$, because $\omega$ is closed. The contraction identity for a Lie bracket of vector fields then givesSince the symplectic form is a nondegenerate bilinear form at every point, the Hamiltonian vector fields commute:With the chosen convention the general identity is $[X_F,X_H]=X_{\{H,F\}}$. Equivalently, it follows by applying the Jacobi identity for the Poisson bracket to an arbitrary test function. Local flows commute whenever both compositions are defined; no global completeness assumption is needed.

Put $V=|v_0|>0$, take $q>0$ without loss of generality, and write $s=R^2+z^2/q^2$. The Newtonian gravitational potential is singular at the origin; all pointwise claims below concern $s>0$. The zero of the Newtonian gravitational potential is the one in the paper, with an implicit fixed length unit inside the logarithm.

The Poisson equation for Newtonian gravity in cylindrical coordinates givesThe two contributions to the Laplacian areConsequently the mass density of the axisymmetric logarithmic gravitational potential isThe coefficient of $R^2$ is positive. The coefficient of $z^2$ determines the density positivity for an axisymmetric logarithmic potential: strict positivity at every noncentral point requires $q^2>1/2$ and $v_0\ne 0$. Nonnegativity permits $q^2=1/2$, where the mass density vanishes on the noncentral symmetry axis. For $q^2<1/2$ it is negative near that axis. Either sign of a nonzero real $q$ gives the same model, and $q=0$ is undefined. Setting $v_0=0$ yields vacuum away from the singular origin, not strictly positive mass density. The $r^{-2}$ central cusp is locally integrable and contains no hidden point mass: the Newtonian gravitational field flux through a shrinking sphere is of order its radius.

An equatorial circular orbit requires inward gravitational acceleration equal to $v_c^2/R$. Thus the circular speed isThis is a flat galaxy rotation curve. The singular center does not define an additional circular orbit at $R=0$.

The specific orbital energy $E=\frac{1}{2}(v_R^2+v_{\varphi }^2+v_z^2)+\Phi$ is conserved because the Newtonian gravitational potential is time independent. The axial specific angular momentum $L_z=R{v}_{\varphi }$ is conserved because the Newtonian gravitational potential is axisymmetric. Hence the Jeans theorem, or directly $dF(E,L_z^2)/dt=0$ in the Collisionless Boltzmann equation, permits a stationary two-integral galactic distribution function. Using $L_z^2$ makes this galactic distribution function even under reversal of the azimuthal velocity, so it describes a nonstreaming population. A general axisymmetric population can have an odd part in $L_z$, or dependence on a third integral of motion; neither is required for this construction.

At each position define $\rho ⟨g⟩=\int gF{d}^{3}v$. The two-integral galactic distribution function is invariant under swapping $v_R$ and $v_z$, since only their sum of squares occurs in $E$. It is also even in each velocity component. Swapping integration variables proves meridional velocity isotropy of a two-integral distribution, while reflecting one variable makes each mixed integrand odd. ThereforeAll mean velocities vanish, so these are also the corresponding statements for the velocity ellipsoid and its centered covariance tensor. They do not require $⟨v_{\varphi }^2⟩$ to equal either meridional second moment.

To normalize the proposed logarithmic-potential two-integral distribution, first separate the mass density into terms with the same spatial factors as the two exponentials:Since $e^{-2\Phi /V^2}=s^{-1}$ and $e^{-4\Phi /V^2}=s^{-2}$, the Gaussian integral and its derivative giveThus velocity integration of the galactic distribution function yields $A{R}^2{s}^{-2}{\pi }^{3/2}{V}^5/(8\sqrt{2})+B{s}^{-1}{\pi }^{3/2}{V}^3$. Matching the independent spatial factors determinesThis verifies both the stationary Collisionless Boltzmann equation and the required Poisson equation for Newtonian gravity. A shift $\Phi \mapsto \Phi +C$ changes the normalizations to $A{e}^{4C/V^2}$ and $B{e}^{2C/V^2}$; the physical mass density is unchanged.

There is an important distinction between positive mass density and a nonnegative galactic distribution function. For $1/2\le q^2\le 1$, both coefficients above are nonnegative. For a prolate model $q^2>1$, $A<0$ and one must test the sum rather than reject the model just because one term is negative. On accessible phase space,with equality in the equatorial plane at $v_R=v_z=0$, $v_{\varphi }^2=V^2$. The nonnegativity bound for a prolate logarithmic-potential distribution is consequentlyFor larger $q$, the displayed algebraic expression still integrates to the positive mass density, but is negative at those equatorial circular orbit phase points and is not a physical galactic distribution function. This extra restriction is distinct from the earlier density-only answer.

A phase-space distribution function describes the mass in a small cell of positions and velocities: $F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$ is that mass. Its velocity integral is the spatial mass density, and $F/\rho$ at a fixed position is a normalized velocity probability density. A number-normalized galactic distribution function differs by the stellar mass convention.

Take $a>0$ and $v_0>0$. For the cusped logarithmic spherical galaxy model, differentiate the gravitational potential and use the spherical Poisson equation for Newtonian gravity:ThusA circular orbit requires $v_c^2/r={\Phi }^{\prime }$. Its galaxy rotation curve is thereforeIt rises as $v_0\sqrt{r/a}$ at small radius and approaches $v_0$ from below at large radius.

The outer flat galaxy rotation curve is qualitatively useful for galaxies. However, the density has an inner $1/r$ cusp and an outer $1/r^2$ tail, while $M(r)\sim v_0^{2}r/G$ diverges. The model therefore requires an outer truncation for a finite galaxy, and does not describe a central constant-density core. Its gravitational potential cannot be normalized to zero at infinity.

For a star, the specific energy is $E=v^2/2+\Phi$ and the specific angular momentum is $\mathbf{L}=\mathbf{x}\times \mathbf{v}$. Since the gravitational potential is stationary and spherical,Thus both the specific energy and the entire angular-momentum vector are conserved.

For Jeans theorem, follow a stellar trajectory in phase space. The Collisionless Boltzmann equation is preciselyIn a steady system ${\partial }_{t}F=0$, so the value of the galactic distribution function is constant along every trajectory: it is itself an integral of motion. Conversely, if $I_1,\dots ,I_m$ are time-independent integrals of motion, then $F=h(I_1,\dots ,I_m)$ satisfies $dF/dt=\sum h_{,a}{\dot{I}}_a=0$. Locally, a complete set of invariant orbit labels expresses any steady galactic distribution function this way. This is the weak form of Jeans theorem; a stronger assertion about a small set of isolating integrals needs additional orbital regularity or mixing assumptions. In particular, the theorem does not say that every steady galaxy must have a galactic distribution function depending on only $E$ and $L$.

For the proposed constant-anisotropy distribution function, write $v_T=(v_{\theta }^2+v_{\varphi }^2{)}^{1/2}$ and $L=r{v}_T$. Introduce a polar angle $\psi$ in the tangential velocity plane. Then $d^{3}v=d{v}_r{v}_{T}d{v}_{T}d\psi$, so the apparent $1/L$ singularity cancels against the velocity Jacobian. This inverse-angular-momentum velocity integration makes the density finite for every $r>0$.

Let $t=e^{-\Phi /v_0^2}=a/(a+r)$, $c_1=1$, $c_2=2$, and ${\alpha }_k=k/(2v_0^2)$. Each exponential component contributesHere the two Gaussian integrals and the angular integral have product ${\pi }^2/{\alpha }_k$. Consequently,The galactic distribution function is nonnegative, gives the required density, and depends on conserved $E,L$; by Jeans theorem it is a self-consistent steady solution.

Differentiating the Gaussian integral with respect to ${\alpha }_k$ shows that each component has $⟨v_r^2{⟩}_k=1/(2{\alpha }_k)=v_0^2/k$, and the same value for $⟨v_T^2{⟩}_k$. Angular averaging gives $⟨v_{\theta }^2{⟩}_k=⟨v_{\varphi }^2{⟩}_k=v_0^2/(2k)$. Weighting these by ${\rho }_1,{\rho }_2$ yields the velocity dispersionsThe radial numerator includes $2r$, as in the original PDF. All mean velocities vanish by parity. The same parity, or the zero angular average of $\sin \psi \cos \psi$, gives $⟨v_r{v}_{\theta }⟩=⟨v_r{v}_{\varphi }⟩=⟨v_{\theta }{v}_{\varphi }⟩=0$.