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An axisymmetric logarithmic Newtonian gravitational potential has spheroidal equipotential surfaces and a flat galaxy rotation curve: $R{\Phi }_R(R,0)=V^2$. The singular model is defined away from the origin; a fixed reference length makes the logarithm dimensionless. Its central $r^{-2}$ mass density cusp is locally integrable, but its total mass diverges at large radius.

This spherical gravitational potential, with $a,v_0>0$, has density $\rho =v_0^2(2a+r)/(4\pi Gr(a+r{)}^2)$ and circular speed $v_c=v_0\sqrt{r/(a+r)}$. It combines a central $1/r$ density cusp with an outer flat galaxy rotation curve, but its total mass is infinite. It differs from a logarithmic potential with a constant-density core.

For a singular isothermal sphere, spherical integration of the mass density givesThe spherical shell theorem makes the radial attraction $GM(<R)/R^2$. Balancing it with circular centripetal acceleration yieldsThus the model has a flat galaxy rotation curve, resembling the outer galaxy rotation curves of many spiral galaxies. Central rotation curves usually rise, and not all real curves are flat throughout. A pressure-supported elliptical galaxy does not itself rotate at this test-particle circular speed.

The model has divergent central mass density, although finite enclosed mass at each finite radius, and an infinite total mass because $M(<R)\propto R$. Its potential is logarithmic, requiring an outer cutoff for a finite escape energy. A globally constant isotropic velocity dispersion, spherical symmetry, and one common profile for all gravitating components are further idealizations. Stellar light commonly has a quite different radial profile from the dark matter halo; real disks are flattened, and halos are neither exact isothermal spheres nor infinite. Tidal truncation also changes the distribution near the satellite boundary.

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.

For an axisymmetric Newtonian gravitational potential, the vacuum Laplace equation in cylindrical coordinates is $R^{-1}{\partial }_R(R{\partial }_R\Phi )+{\partial }_z^2\Phi =0$. Use separation of variables, $\Phi =X(R)Z(z)$. Dividing by $XZ$ and setting the separated constant to $k^2$, with $k>0$, givesThe first equation has solutions $e^{\pm kz}$; with $u=kR$, the second is the order-zero Bessel differential equation. Regularity at the axis selects the Bessel function of the first kind $J_0(kR)$, rather than the singular Bessel function of the second kind $Y_0$. ThusThere are also zero-separation-constant modes, and other boundary conditions can select other separated families. For a localized reflection-symmetric thin disk, decay away from the disk selects $e^{-k|z|}$ on the respective half-spaces.

Apply the divergence theorem to the Poisson equation for Newtonian gravity across a narrow pillbox enclosing disk area. Writing the volume mass density as $\Sigma (R)\delta (z)$ givesReflection symmetry makes ${\Phi }_z(R,0^{+})=2\pi G\Sigma (R)$. A superposition of the regular vacuum modes therefore satisfiesThe order-zero Hankel transform and its inversion now determine the Hankel representation of thin-disk gravity:The sign is fixed by the positive derivative of an attractive Newtonian gravitational potential above the disk. For ordinary finite disks the representation uses the isolated boundary condition, fixing any extra vacuum field or additive constant. Infinite scale-free disks require suitable regularization of the potential itself.

Equatorial circular orbit balance is $v_c^2=R{\Phi }_R(R,0)$. Using $J_0^{\prime }(u)=-J_1(u)$ in the Hankel representation of thin-disk gravity givesDifferentiation under the integral is justified by convergence for regular models, or by first working off the disk and taking the limiting radial field.

Write $C={\Sigma }_0{R}_0$. For the Mestel disk, the supplied Bessel function integral gives $S(k)=-2\pi GC/k$ for $k>0$. The circular speed is thenMeanwhile the interior mass is $M(R)=2\pi {\int }_0^R(C/s)sds=2\pi CR$. ConsequentlyThe Mestel disk has a flat galaxy rotation curve, despite its nonspherical geometry.

Its infinite total mass prevents setting the Newtonian gravitational potential to zero at infinity. Indeed the integral for $\Phi$ diverges at $k=0$ because $S(k)\sim 1/k$. Subtracting the potential at a reference point removes this additive divergence. As an explicit check, the Mestel disk potential-density pair isIt is harmonic for $z\ne 0$, its vertical derivative jump is $4\pi GC/R$, and $R{\Phi }_R(R,0)=2\pi GC$. Equivalently, an Abel regularization of an oscillatory integral with damping factor $e^{-ak}$ in the radial-field integral gives $R{\int }_0^{\infty }{e}^{-ak}{J}_1(kR)dk=1-a/\sqrt{a^2+R^2}$, which tends to one as $a\downarrow 0$. The force is well-defined although the unreferenced potential is not.

The equality with $GM(R)/R$ is special to the Mestel disk, not a thin-disk shell theorem. The spherical shell theorem normally makes that formula possible, but does not apply to an astrophysical disk. Exterior annuli exert outward radial gravitational acceleration inside their radii, while interior annuli do not act exactly as point masses at the origin. For this particular scale-free surface density, their combined contributions happen to produce the same result as the interior mass expression. Changing the radial surface density or truncating the disk generally destroys the equality: enclosed mass does not determine a disc rotation curve.

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 $⟨v_r{v}_{\theta }⟩=⟨v_r{v}_{\varphi }⟩=⟨v_{\theta }{v}_{\varphi }⟩=0$.

The velocity-anisotropy parameter isThe constant-anisotropy distribution function therefore favors radial motion at every radius: the radial variance is twice either individual tangential variance. It still has a substantial population with nonzero angular momentum, rather than exclusively radial orbits. The $L^{-1}$ factor enhances low-angular-momentum stars. The radial velocity dispersion approaches $3v_0^2/4$ at the centre and $v_0^2$ far out; each individual tangential variance is half as large. There is no mean rotation, despite the well-defined galaxy rotation curve for hypothetical circular orbits.

In a stationary axisymmetric gravitational potential, conservation of angular momentum gives $L_z=R^2\dot{\varphi }$. Eliminating $\dot{\varphi }$ introduces the effective potentialFor an equatorial circular orbit, radial balance is ${\partial }_R{\Phi }_{\mathrm{eff}}(R_c,0)=0$, givingVertical balance also requires ${\Phi }_{,z}(R_c,0)=0$.

For epicyclic motion, assume a twice differentiable stationary gravitational potential symmetric under $z\mapsto -z$, small displacements $|x|,|z|\ll R_c$, and a stable circular orbit. Choose the epicyclic guiding center using the conserved $L_z$. Reflection symmetry makes ${\Phi }_{,Rz}(R_c,0)=0$, eliminating radial-vertical coupling at first order. Expanding the equations about $(R_c,0)$ givesThe derivatives are evaluated at the guiding centre, and stability requires ${\kappa }^2,{\nu }^2>0$. Without midplane symmetry, a mixed Hessian term can couple the two oscillations. The radial epicyclic frequency and vertical epicyclic frequency are the frequencies of these independent linear oscillations.

Set $\Omega =L_z/R_c^2$. Along the family of equatorial circular orbits, ${\Phi }_{,R}=R{\Omega }^2$. Differentiating this relation and adding the centrifugal contribution givesTo interpret the common frequency range, let $q=d\log v_c/d\log R$. Since $v_c=R\Omega$, one has ${\kappa }^2=2(1+q){\Omega }^2$. A Keplerian disk has $q=-1/2$ and $\kappa =\Omega$, a flat galaxy rotation curve has $q=0$ and $\kappa =\sqrt{2}\Omega$, and solid-body rotation has $q=1$ and $\kappa =2\Omega$. Typical galactic rotation curves lie between these slopes. Thus $\Omega \lesssim \kappa \lesssim 2\Omega$ is a useful galactic range, not a theorem for every possible axisymmetric potential. As a precise sufficient example, epicyclic frequency bounds for monotone spherical density follow from ${\kappa }^2={\Omega }^2+4\pi G\rho$ and $0\le \rho \le 3M/(4\pi R^3)$.

Use a Cartesian frame rotating with the epicyclic guiding center, with $x$ pointing radially outwards and $y=R_c(\varphi -\Omega t)$ in the direction of rotation. To first order, conservation of angular momentum givesThe radial harmonic oscillator solution and its azimuthal integral areA constant in $y$ merely changes the azimuthal origin of the guiding centre. The epicyclic ellipse obeys $x^2/X^2+y^2/Y^2=1$. In the usual frequency range it is elongated azimuthally, and the star travels clockwise when $x$ points right and $y$ up: its small motion relative to the prograde guiding centre is retrograde.

For the Oort constants, subtract the defining expressions to obtain $\Omega =A-B$ and insert $R{\Omega }^{\prime }=-2A$ into the radial epicyclic frequency formula:The solar-neighbourhood values give $\Omega =26.5\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, $\kappa =\sqrt{1272}\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, andThe solar epicyclic ellipse is therefore about $1.49$ times longer azimuthally than radially.

A complete radial oscillation takes $T_r=2\pi /\kappa$. The oscillatory part of $\dot{\varphi }$ has zero average over this interval, so the epicyclic azimuthal advance isThis describes the advance between consecutive radial turning points of the same type; it need not be a full $2\pi$ revolution. For the Sun,The azimuthal advance estimate uses the same linear epicyclic motion approximation as the axis ratio.

For a spherical system, use the spherical shell theorem with shell mass $dm=4\pi \rho (s)s^{2}ds$. Inner shells contribute $-Gdm/r$ to the gravitational potential, while an outer shell contributes its constant interior potential $-Gdm/s$. With $\Phi (\infty )=0$, assuming the required integrals converge,On differentiation, the two terms containing $\rho (r)r$ cancel. Hence ${\Phi }^{\prime }=GM(r)/r^2$, where $M(r)=4\pi {\int }_0^r\rho (s)s^{2}ds$. Radial balance for a circular orbit gives $v_c^2=r{\Phi }^{\prime }=GM(r)/r$.

For a razor-thin axisymmetric astrophysical disk, the element of mass is $dm=\Sigma (s)sdsd{\varphi }^{\prime }$, where $\Sigma$ is the surface density. Superposing Newtonian gravitational potentials therefore givesAxisymmetry permits $\varphi =0$. Unlike the spherical case, an exterior annulus generally exerts a radial force: enclosed mass does not determine a disc rotation curve.

For the Legendre expansion of thin-disk gravity, split the radial integral at $s=R$. In the inner part the kernel expands in $s/R$, while in the outer part it expands in $R/s$. The angular integrals of odd Legendre polynomials vanish, and those of even degree $n$ equal $2{\alpha }_n$, whereIntroduce $I_n(R)={\int }_0^R\Sigma (s)s^{n+1}ds$ and $J_n(R)={\int }_R^{\infty }\Sigma (s)s^{-n}ds$. The gravitational potential becomesWhen differentiating each bracket, the moving-limit terms cancel: $R^{-(n+1)}{I}_n^{\prime }=\Sigma (R)$ and $R^n{J}_n^{\prime }=-\Sigma (R)$. Since $v_c^2=R{\Phi }^{\prime }$, this givesThe zeroth term has ${\alpha }_0=\pi$ and $M(R)=2\pi I_0$. Separating it provesThe inner correction is inward, while the exterior correction is outward. For a smooth surface density, the original potential singularity at coincident points is integrable. Its radial force is understood through a symmetric Cauchy principal value or a vanishing-thickness regularization; the paired interior and exterior terms above retain the cancellation at $s=R$. This avoids treating the two singular local force contributions separately.

For an exponential galactic disk, write $\Sigma (s)={\Sigma }_0{e}^{-s/R_d}$, so $M_{\infty }=2\pi {\Sigma }_0{R}_d^2$ and $M(R)=M_{\infty }[1-(1+R/R_d)e^{-R/R_d}]$. At large $R$, the missing mass and exterior-ring terms are exponentially small. The leading nonspherical interior term is $k=1$, with ${\alpha }_2=\pi /4$ and $I_2(\infty )=6{\Sigma }_0{R}_d^4$. Consequently the exponential-disk Keplerian asymptotic isThe positive leading correction shows that the rotation curve approaches the Keplerian limit from above. The finite-order large-radius expansion is sufficient here; an infinite moment expansion need not converge for a disk extending to infinity.

A useful special example is the Mestel disk, with surface density $\Sigma (s)=C/s$ for $s>0$, $C>0$. It has $M(R)=2\pi CR$. For every positive even $n$,All the nonspherical corrections cancel, givingThus the Mestel disk has a perfectly flat galaxy rotation curve.

This example has infinite total mass and a singular central surface density. The absolute gravitational potential cannot be set to zero at infinity, but the radial force exists as a limit of disks with increasing outer cutoff. Potential differences are $\Phi (R)-\Phi (R_0)=v_M^2\log (R/R_0)$. The earlier potential integral is therefore interpreted up to a radius-independent divergent constant for this example; the force calculation remains valid. Truncating the Mestel disk gives a more physical finite system but changes the exact flat curve near its edges.

The definition of the virial radius of a dark-matter halo immediately gives the virial mass of a dark-matter haloIn an Einstein-de Sitter universe, spherical collapse gives $v=18{\pi }^2\simeq 178$, conventionally rounded to about $200$.

For the Navarro--Frenk--White profile, put $x=r/r_s=cs$. Direct integration givesUsing $r_s=r_v/c$ and ${\delta }_{\mathrm{char}}=v{c}^{3}g(c)/3$ therefore yieldsAt small radius the bracket is $(cs{)}^2/2+O((cs{)}^3)$, so $M(r)\propto r^2$, consistent with the central $\rho \propto r^{-1}$ cusp. At large radius it is $\log (cs)-1+o(1)$, so the mass diverges logarithmically unless the halo is truncated.

The primordial free streaming of warm dark matter erases small-scale density fluctuations and lowers the central phase-space density, tending to replace the smallest, earliest cold-dark-matter cusps by shallower central profiles or cores.

Since $V_v^2=G{M}_v/r_v$, the spherical potential that vanishes at infinity isIndeed, $rd\Phi /dr=GM(r)/r$, and normalization at $s=1$ requiresThe circular speed is consequentlyIt rises as $s^{1/2}$ near the centre, peaks at $r\simeq 2.16r_s$, and then declines approximately as $\sqrt{\log r/r}$. Increasing the concentration of a dark-matter halo moves the peak inward in units of $r_v$ and raises it relative to $V_v$. Thus an NFW curve can be fairly broad but is not exactly a flat galaxy rotation curve; stellar and gas contributions matter when comparing with an observed galaxy rotation curve.

Lower-mass haloes typically collapse earlier, when the cosmic background density is larger. Their characteristic inner densities are consequently larger relative to the present virial density, producing the mass-concentration relation of dark-matter haloes in which concentration decreases weakly with mass.

For a mass smaller by $1024=2^{10}$,and $c=101024^{0.1}=20$. HenceA halo of roughly $10^9{M}_{\odot }$ and kiloparsec scale naturally hosts a dwarf galaxy, possibly an extremely faint one if star formation is inefficient.

Chandrasekhar dynamical friction is the drag exerted by the overdense gravitational wake that a massive body creates in a background of lighter particles. It transfers orbital energy and angular momentum to the host, causing satellites, star clusters, and massive black holes to spiral inward and promoting galaxy mergers. In a homogeneous isotropic Maxwellian background its force iswhere $\log \Lambda$ is the Coulomb logarithm in stellar dynamics. Its scaling can be reconstructed from the strong-deflection impact parameter $b_{\mathrm{cr}}\sim GM/v^2$: the encountered mass rate is $\sim \rho v\pi b_{\mathrm{cr}}^2$, and multiplying by momentum change $\sim v$ gives $F_{\mathrm{df}}\sim G^2{M}^2\rho /v^2$.

For a spherical host with a flat galaxy rotation curve,This is a singular isothermal sphere with one-dimensional dispersion $\sigma =v_c/\sqrt{2}$, so $X=1$. DefineFor a constant-mass satellite on a circular orbit, the drag magnitude and its torque areIt follows thatThe quadratic radius dependence and inverse mass dependence explain why massive nearby satellites merge much faster than light or distant ones.

Let the satellite also have a flat internal rotation curve of speed $v_s$. Equating its edge density to the host density gives the tidal radiusBecause $M_s(<R_t)=v_s^2{R}_t/G$, the bound mass decreases linearly:Under the question's literal closure that the derivative of the remaining satellite's total orbital angular momentum equals the frictional torque,and henceIf stripped material is explicitly assigned the satellite's instantaneous specific orbital angular momentum, the balance for the bound remnant is instead $M_{s}d(v_{c}r)/dt=-r{F}_{\mathrm{df}}$; that convention gives $t_{\mathrm{df},\mathrm{strip}}=2t_{\mathrm{df},0}$. Both treatments show the robust point: tidal stripping weakens the drag as the orbit shrinks and substantially delays coalescence. In less idealized profiles the mass can fall faster than linearly, producing dynamical-friction stalling by tidal stripping.