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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.

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.

For a spherical orbit,so the galaxy rotation curve is flat. The isotropic spherical Jeans equation with constant dispersion isSince $\rho \propto r^{-2}$,

WriteFor $X\ll 1$,so Chandrasekhar dynamical friction is linear in $\mathbf{v}$ at low speed. At $X\gg 1$, $\mathcal{F}(X)\to 1$, so its acceleration magnitude decays as $v^{-2}$. At zero speed the wake is symmetric and the drag vanishes; at high speed the subhalo spends too little time deflecting each background particle efficiently.

For a circular orbit $v=v_0$, so $X=1$ and $\mathcal{F}(1)=0.428$. The tangential acceleration isThe specific angular momentum is $L=r{v}_0$, hence $v_0\dot{r}=r{a}_{\mathrm{df}}$ andIntegration from $r_i$ to zero gives

For a circular orbit of radius $r$,Solving for $r$ givesSince $\eta =L/L_{\mathrm{circ}}(E)$,

The friction acceleration is antiparallel to velocity, so locally $\dot{L}/L=(1/v)\dot{v}$ and $\dot{E}=v\dot{v}$. Applying the chain rule $\dot{e}=(de/d\eta )\dot{\eta }$ givesBecause $d\eta /de<0$ and $\dot{v}<0$, at pericentre $v>v_0$ gives $\dot{e}<0$: friction circularizes. At apocentre $v<v_0$ gives $\dot{e}>0$: it makes the orbit more eccentric. Orbit averaging produces substantial cancellation; the denser pericentre region generally gives modest net circularization, but the eccentricity changes much less dramatically than the orbital energy and radius.

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.

Put $x=r/r_s$. The enclosed mass of the Navarro--Frenk--White profile isSince $y/(1+y{)}^2=(1+y{)}^{-1}-(1+y{)}^{-2}$,The circular-orbit condition $v_c^2/r=GM(r)/r^2$ then gives the galaxy rotation curve

For $x\ll 1$,and thereforeFor $x\gg 1$,soThe NFW curve thus rises from the center, peaks, and eventually declines slowly. Typical spiral-galaxy rotation curves remain approximately flat over the observed outer disk, although an NFW halo combined with baryons can approximate such a plateau over a finite interval.

At the maximum, $x_m=2.16$, so the scale radius of a Navarro--Frenk--White profile isWriting $f(x)=\log (1+x)-x/(1+x)$, the characteristic density of a Navarro--Frenk--White profile isWith $G=4.3009\times 10^{-6}\mathrm{kpc}(\mathrm{km}{\mathrm{s}}^{-1}{)}^2{M}_{\odot }^{-1}$ and $f(2.16)=0.4670$, this is

An NFW cusp has $\rho \propto r^{-1}$ and hence $v_c\propto r^{1/2}$. Many dwarf and low-surface-brightness galaxies instead favor a nearly constant-density core, for which $M(r)\propto r^3$ and $v_c\propto r$; after matching the outer speed, this gives the more slowly rising observed inner curve. This is the cusp--core problem. Repeated burst-driven gas outflows can fluctuate the central potential and transfer orbital energy to dark matter, while self-interacting dark matter provides another possible core-forming mechanism. Beam smearing and noncircular motions are observational systematics but do not explain every case.

For an exponential galactic disk,soSince $7\mathrm{kpc}\simeq 2.3R_d$ is near the disk maximum, assigning the full $220\mathrm{km}{\mathrm{s}}^{-1}$ there givesIf the disk mass is increased while the observed total curve is fixed, the halo contribution must decrease in the inner galaxy. An NFW fit therefore needs a lower characteristic density, a larger scale radius and lower concentration, or a correlated combination of these changes.

A maximum disk model raises the stellar mass-to-light ratio as far as the rotation curve permits, conventionally making the disk contribute about $85\%$ of the speed near $2.2R_d$. High stellar surface densities, population-synthesis mass-to-light ratios, microlensing optical depths, and fast bars that have suffered little dynamical-friction braking can support a large disk contribution. Directly measured vertical stellar dispersions often favor submaximal disks, especially in low-surface-brightness galaxies, while stability and cosmological halo constraints may also disfavor an excessively massive disk.

The disk--halo degeneracy follows fromOver a finite radial range, raising the stellar mass-to-light ratio and lowering or broadening the halo can leave the same sum. The degeneracy can be reduced by measuring the disk surface density dynamically from vertical velocity dispersion and scale height, or by constraining the stellar mass-to-light ratio with resolved populations, colors, and a specified initial mass function. Strong or weak gravitational lensing, gas-layer flaring, stellar streams, and rotation data far beyond the optical disk provide further independent halo constraints.

At an abruptly truncated razor-thin disk with nonzero edge surface density of a disk, the midplane radial force diverges logarithmically. Locally, mass occupies a half-plane and the inward force contains a nonzero angular factor multiplying $G{\Sigma }_{\mathrm{edge}}{\int }_{ϵ}^{\ell }ds/s$. The opposite exterior half-plane is absent, so its cancellation does not occur. Finite thickness or a taper taking the edge density to zero regularizes the force. Therefore a finite-edge galaxy rotation curve requires more than the ideal scale-free Mestel disk normalization.

For a slowly evolving circular orbit with tangential deceleration $D>0$, the specific angular momentum is $j=r{V}_c(r)$ and its torque is $\dot{j}=-rD$. Differentiating $j$ gives the displayed drift when $1+q\ne 0$. The positive local decay time $r/(2|\dot{r}|)$ is $(1+q)V_c/(2D)$. In a uniform-density core $q=1$, so it is $V_c/D$; for a flat galaxy rotation curve $q=0$, it is $V_c/(2D)$. Simply setting ${\dot{V}}_c=-D$ neglects the radial contribution to angular momentum change.

In solid-body rotation the angular velocity is independent of distance from the rotation axis, so tangential speed is proportional to cylindrical radius. A galactic region with this galaxy rotation curve has radial epicyclic frequency $\kappa =2\Omega$.

In a spherically dominated Newtonian gravitational field, the circular orbit speed satisfies $v_c^2=GM(<r)/r$. A flat galaxy rotation curve therefore implies $M(<r)=v_0^{2}r/G$. Differentiating the enclosed mass gives $\rho =(4\pi r^2{)}^{-1}dM/dr$, yielding the inverse-square density profile. This is a halo-region approximation, not a mass inversion for a flattened stellar disk. A flat curve extending to infinite radius would imply infinite mass, so a real halo profile must eventually change or truncate.