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Use a steady, isotropic, self-gravitating dark matter halo, with $V$ its constant one-dimensional velocity dispersion. The Spherical Jeans equation and spherical mass conservation readFor the scale-free outer region, put $\rho =A{r}^{-p}$. Its enclosed mass scales as $r^{3-p}$, so $GM(r)/r$ is constant only if $p=2$; the Spherical Jeans equation then gives $2V^2=GM(r)/r=4\pi GA$. Thus the singular isothermal sphere hasIts circular speed is $v_c=\sqrt{2}V$. If a speed denoted $V$ were instead the circular speed, the coefficient would be $V^2/(4\pi G)$, but the printed variable is a velocity dispersion.

The assumptions matter. A constant velocity dispersion and spherical shape alone do not specify a unique central mass density or orbital distribution. Even with isotropy, the Spherical Jeans equation integrates to $\rho ={\rho }_0\exp [-(\Phi -{\Phi }_0)/V^2]$, and the self-gravitating Poisson equation for Newtonian gravity givesThere are finite-central-density isothermal solutions as well as the singular one; the $r^{-2}$ expression is the intended outer scale-free halo approximation. A finite halo must also end or steepen at large radius, since the untruncated singular isothermal sphere has infinite total mass.

The stellar population has appreciable random velocity dispersion, and its declining radial density creates an outward stress gradient that supplies some support against gravity; cold gas has much less random-motion support and follows nearly the local circular speed. Neglecting velocity-ellipsoid tilt, the radial Jeans equation gives $v_c^2-{\overline{v}}_{\varphi }^2={\sigma }_R^2[-d\ln (n_{\ast }{\sigma }_R^2)/d\ln R-1+{\sigma }_{\varphi }^2/{\sigma }_R^2]$, generally positive for the stellar disk. The slower mean stellar rotation is asymmetric drift.

Let $R$ be a characteristic radius enclosing mass $M$, and let $\sigma$ denote a one-dimensional velocity dispersion. Gravity and random-motion support give a relation $GM/R=\eta {\sigma }^2$, where the dimensionless structure coefficient $\eta$ depends on the chosen radius and density profile. A crossing-based galactic dynamical time is thereforeThe scaling follows by equating acceleration scales ${\sigma }^2/R$ and $GM/R^2$, not by identifying a complete orbital period with a crossing time. An explicit scale-free example checks the coefficient: constant circular speed and spherical symmetry give $M(<r)=v_c^{2}r/G$ and $\rho =v_c^2/(4\pi G{r}^2)$. With isotropic constant velocity dispersion, the Spherical Jeans equation becomesso $v_c^2=2{\sigma }^2$, $M(<R)=2{\sigma }^{2}R/G$, and $R/\sigma =GM/(2{\sigma }^3)$. This is the singular isothermal sphere scaling, valid over an interior radial range; a literal infinite model has no finite total mass, and a hard finite boundary needs a pressure term or an outer transition. A free-fall time of a uniform sphere or orbital period changes the numerical factor. The subsequent estimates use the nominal scale $GM/{\sigma }^3$, keeping these order-unity structural uncertainties explicit.

Older disc stellar populations generally have larger velocity dispersions, partly through accumulated galactic disk heating, so random-motion stresses supply more of their support against gravity. The radial Jeans equation then requires a smaller mean rotation than the circular speed, producing larger stellar asymmetric drift than in dynamically colder young stars.

The neutral-hydrogen 21-centimeter line traces cold gas far beyond the bright stellar galactic disk, where a dark matter halo contributes strongly to the galaxy rotation curve. With relatively small random velocities, this gas approximately follows circular orbits, allowing its Doppler velocities to measure the circular speed after inclination and pressure corrections, whereas stellar dispersions require an orbital-anisotropy model. Its extended radial reach and comparatively simple force tracing make neutral gas especially useful for outer-halo constraints, although warps and noncircular motions can invalidate that simplicity.

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.

Let $M_h,R_h$ describe the halo and let $v=v_{\mathrm{vir}}$. The cosmic baryon fraction is $f_b={\Omega }_b/{\Omega }_m=0.20$, so the specified settled fraction gives $M_d=0.10M_h$. The specific angular momentum constraint at the disk edge isTo infer an actual circular speed one must specify the disk's radial mass distribution: enclosed mass does not determine a disc rotation curve. Introduce a finite geometry coefficient $\kappa$ by $v_d^2=\kappa G{M}_d/R_d$. A rounded, centrally concentrated disk permits a monopole estimate $\kappa \simeq 1$ at its outer edge; it is an explicit approximation, not the spherical shell theorem applied exactly to a razor-thin disk. Ignore the force of the unsettled baryons as well as the dark matter within the disk. Using $v^2=G{M}_h/R_h$, the angular-momentum estimate of a self-gravitating galactic disk givesAn exact disk answer cannot be fixed by total disk mass alone; the dependence on $\kappa$ records that missing input.

For the numerical virial radius of a dark-matter halo, adopt mean density ${\Delta }_c=200$ times the critical density at the formation epoch. This conventional definition givesThe supplied expansion law gives $H(3)=286.49\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, and hence $R_h=v/(10H)=97.74\mathrm{kpc}$. Taking $\kappa =1$ yieldsThe corresponding virial mass of a dark-matter halo is $M_h\simeq 1.78\times 10^{12}{M}_{\odot }$, and $M_d\simeq 1.78\times 10^{11}{M}_{\odot }$. Other overdensity conventions give $R_d\propto {\Delta }_c^{-1/2}$ at fixed $H,v,\kappa$.

Interpret the wavelength separation as an observed-frame local mean near the redshift in question. Since Lyman-alpha absorption appears at ${\lambda }_{\mathrm{obs}}={\lambda }_{\alpha }(1+z)$, the incidence isAssume one counted absorption system per intercepted disk, no unrelated forest systems or missed absorbers, unity neutral covering fraction, and a locally slowly varying population. Let ${\sigma }_{\mathrm{abs}}$ be the proper interception cross-section and $n_c$ the comoving number density. The proper density is $n_p=n_c(1+z{)}^3$, and the proper line element along the light path is $d{l}_p=cdz/[H(z)(1+z)]$. Thus the absorber incidence and comoving number density relation isFor a thin circular disk, the projected area is $\pi R_d^2|\cos i|$. Isotropically oriented normals have $⟨|\cos i|⟩=1/2$, so the random-orientation absorbing-disk cross-section is ${\sigma }_{\mathrm{abs}}=\pi R_d^2/2$. ConsequentlyIf all disks are taken face-on instead, $n_c\simeq 0.398{\mathrm{Mpc}}^{-3}$. Generally the random-orientation result scales as ${\kappa }^2({\Delta }_c/200)$ and is divided by the neutral covering fraction. The finite mean redshift spacing is sizable, so a precision inference would integrate the incidence over the actual redshift interval rather than identify it with one local value.

There is also a halo abundance mass-budget bound on this formal result. The present mean matter density for these parameters is ${\bar{\rho }}_{m,0}={\Omega }_{m}3H_0^2/(8\pi G)\simeq 3.40\times 10^{10}{M}_{\odot }{\mathrm{Mpc}}^{-3}$. Distinct haloes of this mass cannot have $n_c{M}_h>{\bar{\rho }}_{m,0}$, even if every matter particle belonged to them. Yet the inferred random-orientation population hasEven the face-on estimate exceeds this bound by about twenty-one. The numerical incidence result is conditional; the supplied population assumptions are not cosmologically consistent under this standard virial and compact-disk estimate. A larger neutral-gas absorption radius, a different absorber population, or different physical inputs are needed. For example, random orientations would require an absorbing radius at least about $63\mathrm{kpc}$ merely to reach the all-matter upper bound, much larger than the calculated centrifugal radius. This check does not change the algebraic answer, but prevents treating it as a realizable population.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

Use a stationary, non-streaming tracer population with spherical symmetry in its full velocity distribution. In particular, the mean velocities vanish, the mixed covariances vanish, and the two tangential components have equal velocity dispersions. WriteHere ${\sigma }_t^2$ is the variance of one tangential component. In the alternative convention using the sum of both tangential variances, the same velocity-anisotropy parameter is $1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)$. These definitions agree; they must not be mixed by inserting an extra factor of two.

The time derivative and mean-velocity terms in the supplied Jeans equation vanish, as do its angular mixed-moment terms. Its remaining geometric term is $2\nu ({\sigma }_r^2-{\sigma }_t^2)/r$. Thus the Spherical Jeans equation becomesThe no-streaming assumption excludes a steady radial flow; stationarity of the density alone would not remove its advective terms.

By the shell theorem, the gravitational potential of a spherical mass distribution satisfies $d\Phi /dr=GM(<r)/r^2$. Expanding the derivative above, dividing by $\nu$, and using logarithmic derivatives givesThe tracer density $\nu$ need not be the density of the gravitating matter. The circular speed is $V_c^2=rd\Phi /dr=GM(<r)/r$, whether or not the tracers themselves follow circular orbits.

Now let $\beta$ be constant, $\nu ={\nu }_0(r/r_0{)}^{-\gamma }$, and $V_c^2=V_0^2(r/r_0{)}^{2\alpha }$. With $y={\sigma }_r^2$, the Spherical Jeans equation readsFor a power-law ansatz $y=K{V}_c^2$, $y^{\prime }=2\alpha y/r$, so $(\gamma -2\beta -2\alpha )y=V_c^2$. Therefore the scale-free spherical Jeans solution isThe positivity condition is necessary for this particular solution to be a physical velocity dispersion.

There is a boundary condition implicit in selecting this solution. The integrating factor $r^{2\beta -\gamma }$ gives, when $D=\gamma -2\beta -2\alpha \ne 0$,The homogeneous term represents an additional radial-pressure boundary contribution. It is removed, for example, by requiring $r^{2\beta }\nu {\sigma }_r^2\to 0$ at infinity. Under that condition,and convergence requires $D>0$, giving the boxed result. If $D=0$, the solution instead has the logarithmic formso the quoted constant proportionality does not apply. A finite outer boundary can retain a homogeneous term even when $D\ne 0$.

To derive the heliocentric projection of a spherical velocity dispersion, put the Sun at vector ${\mathbf{R}}_{\odot }$ of length $R_{\odot }$, and a tracer at $\mathbf{r}=r\hat{\mathbf{r}}$. Its heliocentric line of sight isIf $\chi$ is the angle between $\hat{\mathbf{\ell }}$ and the Galactocentric radial direction, thenAfter correction for the Sun's motion, the intrinsic covariance matrix has eigenvalues ${\sigma }_r^2,{\sigma }_t^2,{\sigma }_t^2$ in the local spherical basis. Projection onto the line of sight therefore givesAt fixed $r$, ${\sigma }_r$ and $\beta$ are constant across the spherical shell. Averaging its squared line of sight velocity, with zero mean, givesFor a uniformly sampled Galactocentric shell, the solid angle measure makes $u$ uniform on $[-1,1]$, giving the geometric correction for heliocentric velocity dispersionTo evaluate it, set $a=r^2+R_{\odot }^2$ and $b=2r{R}_{\odot }$, and divide the integrand:The odd term integrates to zero, while ${\int }_{-1}^{1}du/(a-bu)=b^{-1}\ln [(a+b)/(a-b)]$. For $r>0$, $R_{\odot }>0$, and $r\ne R_{\odot }$, this givesThe removable limit at $r=R_{\odot }$ is $A=1/2$. As $r/R_{\odot }\to \infty$, $A\simeq 2R_{\odot }^2/(3r^2)$, while as $r/R_{\odot }\to 0$, $A\to 2/3$. If $R_{\odot }=0$, $A=0$ and the observed velocity dispersion is exactly radial. For isotropic velocities, $\beta =0$, the correction vanishes at every radius.

The averaging measure matters. If directions are instead uniform in heliocentric solid angle at a fixed Galactocentric radius $r>R_{\odot }$, let $\vartheta$ be the angle between the line of sight and ${\mathbf{R}}_{\odot }$. Geometry gives ${\sin }^2\chi =(R_{\odot }/r{)}^2{\sin }^2\vartheta$, so that particular weighting gives $A=2R_{\odot }^2/(3r^2)$ exactly. The logarithmic expression above uses uniform tracer weighting on a Galactocentric shell. For an observational sample, use its actual angular selection weights in $A=⟨{\sin }^2\chi ⟩$; the boxed dispersion relation remains valid.

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.

SetThe spherical Poisson equation givesThe numerator identitysplits this into two scale-free density-potential components:Thus

At small radius, $D\sim 2b$ and the first term dominates:The cusp has finite enclosed mass because $r^2\rho \sim r^{1/2}$. Moreover,so the circular speed tends to zero as $r^{1/4}$. At large radius, $D\sim r$ andSince $r^2\rho \sim r^{-3/2}$ is integrable at infinity and $\Phi \sim -GM/r$, the total mass is finite and equals $M$.