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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 sphere of radius $r_0$ and uniform mass density ${\rho }_0$, with no surrounding mass and gravitational potential zero at infinity, the spherical shell theorem gives $V_c^2=4\pi G{\rho }_0{r}^2/3$ inside. Matching to the exterior point-mass potential gives $\Phi (r)=-(2\pi G{\rho }_0/3)(3r_0^2-r^2)$. Since $v_e^2=-2\Phi$, the displayed relation follows for $0<r\le r_0$. Exterior shells leave the interior circular speed unchanged but deepen the potential and change the escape velocity; a uniform core alone is insufficient to determine escape.

An inner NFW profile has $M_i(r_i)=A{r}_i^2$, where $A=2\pi {\rho }_s{r}_s$. If the final total circular speed is $V$, circular-shell adiabatic contraction gives $r_i=(V^2/(GA){)}^{1/3}{r}^{2/3}$ and $M_D^{\prime }=A^{1/3}(V^2/G{)}^{2/3}{r}^{4/3}$. Thus ${\rho }_D^{\prime }=A^{1/3}(V^2/G{)}^{2/3}{r}^{-5/3}/(3\pi )$. The total density producing the flat curve is $V^2/(4\pi G{r}^2)$, so the dark fraction tends to zero as $r^{1/3}$ despite its increase relative to the original cusp.

A spherical stellar model with total mass $M$, scale radius $a$, inner density proportional to $r^{-2}$ and outer density proportional to $r^{-4}$. Integrating its mass density gives $M(r)=Mr/(r+a)$, hence Newtonian gravitational potential $\Phi =(GM/a)\log [r/(r+a)]$, normalized to zero at infinity. Its circular speed satisfies $v_c^2=GM/(r+a)$: it tends to a constant inside the scale radius and becomes Keplerian outside. The central density cusp has vanishing enclosed mass as $r\to 0$ and does not imply a central point mass.

The reflection-symmetric Newtonian gravitational potential $\Phi =v_0^2\log [(\sqrt{R^2+z^2}+|z|)/R_{\ast }]$ is harmonic off the plane. Its normal-derivative jump is $2v_0^2/R$, so the Poisson equation for Newtonian gravity gives surface density of a disk $S=v_0^2/(2\pi GR)$ and volume mass density $\rho =S\delta (z)$. The midplane circular speed is constant. The scale-free model has infinite total mass and does not use a finite-mass boundary condition at infinity.

For an NFW profile, $x=r/r_s$ and $v_c^2=GM(r)/r$. The circular speed rises as $r^{1/2}$ at small radius and eventually decreases as $[\log (r/r_s)/r{]}^{1/2}$. Its maximum occurs at $x\simeq 2.16258$, with $v_{c,\max }^2\simeq 2.71706G{\rho }_s{r}_s^2$.

The exponent of the point-mass weight is $\beta$, with $0<\beta <1$. At the origin the unreferenced relative potential has integrand $GB{b}^{\beta -1}$, whose integral diverges at its upper endpoint. In the origin-referenced potential the two terms cancel pointwise there. More generally, first form their difference at finite cutoffs and then pass to the limit. For any fixed nonzero $(R,z)$, the small-$b$ singular term is $-b^{\beta -1}$, which is integrable because $\beta >0$. At large $b$,so the weighted difference is integrable because $\beta <1$. Hence the referenced potential is well defined, and its value at the origin is zero. It is a potential difference for an infinite astrophysical disk, not a potential with zero at infinity. In fact rescaling all lengths shows ${\psi }_{II}(\lambda R,\lambda z)={\lambda }^{\beta }{\psi }_{II}(R,z)$; its finite values approach zero at the origin even though its force is singular there.

For $R>0$, differentiation under the convergent force integral in the plane givesUse $b=R\sqrt{t/(1-t)}$, for which $db=(R/2)t^{-1/2}(1-t{)}^{-3/2}dt$. The powers of $R$, $t$, and $1-t$ then givewhere the Euler beta function evaluates the constant asThe first Gamma function argument is $(1+\beta )/2$: it follows from adding one to the exponent of $t$. Both Euler beta function arguments are positive in the stated range.

For a circular orbit in the plane, $V^2/R=-{\partial }_R{\psi }_{II}$. The prescribed circular speed therefore fixes $C=K^2$ andIndependently, the Kuzmin reflection method givesHere the power of $b$ is one larger than in the force integral, so its Euler beta function arguments differ. Substituting the normalization fixed by the circular speed yieldsThis scale-free Kuzmin superposition has nonnegative surface density, finite central enclosed mass, and infinite total mass. Its force and surface density approach the Mestel disk values as $\beta \downarrow 0$; that limiting disk requires a different potential reference at the origin.

Put $x=r/r_1$. The NFW profile becomes ${\rho }_D={\rho }_1/[x(1+x{)}^2]$. Integrating spherical shells givesDefine $F(x)=[\log (1+x)-x/(1+x)]/x$. The spherical circular speed is thereforeAt small $x$, the bracket is $x^2/2-2x^3/3+O(x^4)$, givingAt large $x$, the bracket is $\log x-1+O(1/x)$, givingThus the curve rises from zero, reaches one broad maximum, and eventually falls as $[\log r/r{]}^{1/2}$. The untruncated NFW profile has logarithmically divergent mass, but its circular speed still tends to zero.

Differentiating $F$ shows that its stationary point satisfiesThe numerator of $F^{\prime }$ has derivative $x(1-x)/(1+x{)}^3$: it increases up to one and decreases thereafter, crossing zero once at positive $x>1$. Numerical bisection gives $x_{\max }\simeq 2.16258$ and $F(x_{\max })\simeq 0.2162166$. HenceEquivalently $v_{\max }\simeq 0.4650\sqrt{4\pi G{\rho }_1{r}_1^2}$. This is the NFW mass and circular-speed profile.

Observed galaxy rotation curves commonly remain approximately flat across much of the measured outer luminous disk. A pure NFW profile is not an exactly flat curve: it rises in the inner cusp and ultimately declines. Nevertheless, its broad maximum can look nearly flat over a finite radial range, so observing a flat segment does not by itself exclude an NFW halo.

The measured circular speed responds to the total gravitational potential, including stars and gas as well as dark matter. For components in the same plane,For spherical components this reduces to $G{\sum }_j{M}_j(<r)/r$; a disk contribution must instead be computed from its actual disk potential. Baryonic mass can supply much of the inner rotation and modify the halo through slow contraction or more impulsive evolution. Thus the appropriate comparison is with the combined mass model, not an isolated halo-only curve. At arbitrarily large radius a finite isolated galaxy must eventually have a declining circular speed; an observed flat segment is not a claim of flat rotation to infinity.

Use the usual spherical circular-shell adiabatic contraction approximation: initial dark-matter orbits are circular, slow accumulation keeps them circular, and dark shells do not cross. Angular momentum is conserved in a spherical potential, and the vanishing radial action stays zero as an adiabatic invariant. For a circular shell,Here $r_i$ is the initial shell radius and $r$ its final radius. The final flat circular speed $V=2v_{\mathrm{NFW},\mathrm{c},\max }$ gives $M_f(r)=V^{2}r/G$.

For an initial shell in the inner NFW profile, write $A=2\pi {\rho }_1{r}_1$, so $M_i(r_i)\simeq A{r}_i^2$. The invariant givesConservation of dark shell mass yieldsand differentiation givesTo express the requested ratios, set $K=F(x_{\max })\simeq 0.2162166$ and $B=8K\simeq 1.72973$, so $V^2/(GA)=B{r}_1$. Since the original cusp has ${\rho }_D(r)\simeq A/(2\pi r)$,The total density of the final spherical flat-rotation model is ${\rho }_{\mathrm{tot}}=V^2/(4\pi G{r}^2)$. Its dark fraction isSubtracting dark matter from the total gives the added-matter density, ${\rho }_B={\rho }_{\mathrm{tot}}(1-h)$, and thusThese expressions describe the leading inner approximation, not an extension to radii where the denominator becomes negative or the initial shell is outside the NFW cusp. As $r/r_1\to 0$, dark matter is enhanced relative to its original density while its fraction of the even steeper total density tends to zero. This is inner NFW contraction to a flat rotation curve.

The orbital assumptions matter. An initial density profile does not specify the phase-space distribution function. For eccentric orbits, slow evolution conserves their actions, but $rM(r)$ at an instantaneous radius is not generally their invariant. The above ratios are the intended circular-shell estimate; their numerical coefficients cannot be uniquely inferred from the density profile and slow accumulation alone.

Work at $r\gg R_{\mathrm{sch}}$ in the region where the central black hole dominates the gravitational potential. Let $J$ be a star's specific angular momentum. Near its very small pericentre, the centrifugal term balances the deep gravitational potential. For an approximately parabolic passage with capture pericentre $r_c=R_{\mathrm{sch}}$,At radius $r$, a velocity making angle $\vartheta$ with the inward radial direction has $J=rv\sin \vartheta$. Therefore the loss cone half-angle isTaking the characteristic speed to be the circular speed $v^2\sim G{M}_H/r$ giveswith Newtonian circular-speed normalization ${\vartheta }_{\mathrm{lc}}^2\simeq 2R_{\mathrm{sch}}/r$. The inward cap occupies solid angle ${\Omega }_{\mathrm{lc}}\simeq \pi {\vartheta }_{\mathrm{lc}}^2$, or a fraction ${\vartheta }_{\mathrm{lc}}^2/4$ of an isotropic velocity sphere. Counting low-angular-momentum trajectories in both radial directions changes only an order-one factor; each such orbit is removed at its next pericentre.

The capture occurs in the strong-field region, so this Newtonian pericentre estimate is not an exact relativistic capture coefficient. The Schwarzschild capture calculation alters an order-one constant, while retaining the small-angle scaling ${\vartheta }_{\mathrm{lc}}\propto r^{-1/2}$ for these characteristic Keplerian speeds.

The separated disk modes. Away from the plane of a razor-thin disk, there is no volume source, so the Newtonian gravitational potential obeys the Laplace equation in cylindrical coordinates. Separation of variables givesHere $k>0$ is a radial wavenumber, with units of inverse length. Regularity on the axis selects the Bessel function of the first kind $J_0(kR)$ rather than the singular second solution. Decay away from the disk, continuity across it and reflection symmetry select $Z=e^{-k|z|}$. These conditions apply to each nonzero-wavenumber mode; a uniform sheet is a separate zero-mode problem.

Integrate the Poisson equation for Newtonian gravity across the disk to obtain the jump conditionFor the unit-amplitude separated mode, the jump is $-2k{J}_0(kR)$. ThusAn individual basis density is signed; positive physical surface densities are built by superposition.

An arbitrary axisymmetric surface density. Define its order-zero Hankel transform byMatching the coefficients of the mode densities gives the Hankel representation of thin-disk gravity,The boundary conditions exclude added external fields. For sufficiently decaying surface density, this is the isolated-disk potential; for infinite scale-free disks, only potential differences or regularized forces need be finite.

The Mestel disk. Put $C={\Sigma }_0{R}_0$. The Bessel function integral ${\int }_0^{\infty }{J}_0(k{R}^{\prime })d{R}^{\prime }=1/k$, understood with a decaying regulator, gives $\hat{\Sigma }=C/k$. Differentiate the potential before removing the regulator and use $J_0^{\prime }=-J_1$:The last integral is $1/R$, since ${\int }_0^{\infty }{J}_1(u)du=[-J_0(u){]}_0^{\infty }=1$. Therefore the circular speed and enclosed mass areConsequently $v_c^2=GM(R)/R$. This equality follows from the specific Mestel disk calculation; the spherical shell theorem cannot be applied to general disks. Its total mass and absolute potential referenced to infinity diverge. A convenient finite midplane reference is $\Phi (R,0)=v_0^2\log (R/R_0)$.

The rotating distribution function. Use specific energy $E=(v_R^2+v_{\varphi }^2)/2+\Phi$ and $L_z=R{v}_{\varphi }$. For the printed positive exponential, take $\epsilon =-E$ and $s:=\sigma >0$; $s$ has units of squared velocity. If instead $\epsilon$ means ordinary total energy and $\sigma >0$, the velocity integral diverges, so that literal convention cannot define the requested galactic distribution function. Equivalently one can retain ordinary energy and choose the printed $\sigma$ negative.

With the binding-energy convention, write the amplitude as $A$ to distinguish it from the function:Integration over both radial velocities and positive azimuthal velocities givesThe half-line integral follows by setting $u=v_{\varphi }^2/(2s)$ in the defining Gamma function integral and requires $q>-1$. Matching its radial power and normalization to the Mestel disk yields the rotating Mestel disk distribution function parametersSince $v_0^2,s>0$, the required integrability condition is automatic. The amplitude uses the stated potential reference; changing the additive potential constant changes $A$ without changing the physical distribution.

Use $M$ for the mass parameter throughout; the PDF uses $m$ in its formulas and $M$ in the subsequent prose for the same quantity. Set $a=r_j>0$. The Jaffe model density simplifies toIts enclosed mass isIt tends to $M$ as $r\to \infty$, verifying the total mass. It tends to zero at the center, so the cusp carries no extra point mass.

Spherical symmetry gives ${\Phi }^{\prime }(r)=GM(r)/r^2=GM/[r(r+a)]$. Integrating with $\Phi (\infty )=0$ givesThis also directly satisfies the Poisson equation for Newtonian gravity: differentiating $r^2{\Phi }^{\prime }=GMr/(r+a)$ gives $4\pi G{r}^2\rho (r)$. The circular speed isFor $r\ll a$, $v_c=\sqrt{GM/a}[1-r/(2a)+O((r/a{)}^2)]$ is approximately constant. For $r\gg a$, $v_c=\sqrt{GM/r}[1-a/(2r)+O((a/r{)}^2)]\propto r^{-1/2}$. Thus the finite-mass model has an inner nearly flat galaxy rotation curve and an outer Keplerian decline.

The stated relation follows for an isolated uniform sphere ending at $r_0$, with the gravitational potential zero at infinity. Uniform density in a core alone does not fix the escape velocity: exterior shells add a constant interior gravitational potential while leaving the interior force unchanged.

Under the isolated-sphere assumption, the enclosed mass and circular speed areThe boundary value is $\Phi (r_0)=-Gm(r_0)/r_0$. Integrating ${\Phi }^{\prime }=Gm(r)/r^2$ inwards givesThus the escape and circular speeds inside an isolated uniform sphere satisfyThe finite central limit is $v_e^2(0)=4\pi G{\rho }_0{r}_0^2$, although $V_c(0)=0$. Adding an exterior spherical shell of mass $m_s$ at radius $s>r_0$ raises $v_e^2$ by $2G{m}_s/s$ and is a direct counterexample to deducing the boxed relation from a uniform core alone.

Inside the uniform sphere, $V_c^2=3{\sigma }^2{r}^2/r_0^2$. Thus, with $X=V_c/(\sqrt{2}\sigma )$, part (b) givesThe Gaussian integral needed for the lowered Maxwellian velocity distribution isSubstitution $v=\sqrt{2}\sigma y$ therefore yieldsSince $V_c<v_e$ throughout the core, the integrand is positive and so is $I(X)$ for $X>0$. Using $m{n}_0={\rho }_0$, part (a)'s massive-test-body drag magnitude isThe slow drift of a circular orbit is determined by torque, not by equating its changing circular speed to the tangential drag. Here $V_c=\omega r$, with constant $\omega =\sqrt{4\pi G{\rho }_0/3}$. The specific angular momentum $j=r{V}_c=\omega r^2$ obeysHence the positive local decay timescale isThere is a genuine factor-of-two discrepancy with the requested denominator $32{\pi }^2$. That denominator would give $V_c/(2D)$, which is the timescale for a flat galaxy rotation curve, not this uniform-density core. It also follows from the incorrect replacement ${\dot{V}}_c=-D$, which drops the $V_c\dot{r}$ term in $d(r{V}_c)/dt$. The literal signed definition $r/(2\dot{r})$ is negative for an inward drift; it equals $-{\tau }_{+}$. A positive decay time requires the absolute value or a minus sign.

This is the slow circular inspiral under tangential drag calculation. Its assumptions require $D$ small enough for nearly circular adjustment and a meaningful large Coulomb logarithm in stellar dynamics; the leading-log approximation is not uniform at $V_c\to 0$. Also, integrating the truncated distribution all the way to $v_e$ does not give exactly $n_0$. The model uses $n_0$ as the stated core normalization; exact self-consistency of a sharply truncated uniform core with that distribution is not implied.

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.

The divergence theorem states ${\int }_V\nabla \cdot \mathbf{F}dV={\oint }_{\partial V}\mathbf{F}\cdot \mathbf{n}dS$. With the Poisson equation for Newtonian gravity, this givesor gravitational-acceleration flux $\oint \mathbf{g}\cdot \mathbf{n}dS=-4\pi GM(V)$.

Use cylindrical radius $R=r\sin \theta$, so $r(1+|\cos \theta |)=r+|z|$. An arbitrary reference length in the logarithm only adds a constant potential. Off the equatorial plane, the radial part of ${\nabla }^2\log [r(1+|\cos \theta |)]$ is $1/r^2$. Writing $c=\cos \theta$, its angular part is $r^{-2}d[(1-c^2)d\log (1+|c|)/dc]/dc=-1/r^2$ on either open hemisphere. The two cancel: there is no volume density away from $z=0$.

At the plane, ${\partial }_z\varphi {|}_{0^{+}}=v_0^2/R$ and ${\partial }_z\varphi {|}_{0^{-}}=-v_0^2/R$. Applying the gravitational flux law to a thin pillbox gives $4\pi G\Sigma ={\partial }_z\varphi {|}_{0^{+}}-{\partial }_z\varphi {|}_{0^{-}}$. ConsequentlyThis is the Mestel disk potential-density pair. There is no additional point mass at the origin: the enclosed mass tends to zero linearly as the enclosing radius shrinks.

At $z=0$, $\varphi =v_0^2\log R$ up to a constant, so radial circular balance givesOn a sphere of radius $r$, ${\partial }_r\varphi =v_0^2/r$ at every nonsingular angular point. Its flux is $4\pi v_0^{2}r$, and henceDirect integration of the surface density confirms $2\pi {\int }_0^r\Sigma (R)RdR=v_0^{2}r/G$. Thus $v_{\mathrm{circ}}^2=GM(<r)/r$ even though the source is a razor-thin disk. This equality is a special scale-free property of the Mestel disk, not evidence that the source is spherical. In a general disk, matter outside the orbit and nonspherical interior forces prevent this enclosed-mass formula for the local circular speed. The model has infinite total mass, so its potential is defined by differences rather than a zero at infinity.

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.