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The midplane galaxy rotation curve fixes the radial derivative of the Newtonian gravitational potential, not every exterior boundary condition. Adding $a|z|$ leaves that radial derivative unchanged but adds a uniform sheet of surface density of a disk $a/(2\pi G)$. The pure scale-free Mestel disk excludes such extra sources.

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.

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 planar Mestel disk has $F(E)\propto e^{-E/v_0^2}$, giving independent centered Gaussian distributions for $v_R,v_{\varphi }$ with velocity dispersions ${\sigma }_R={\sigma }_{\varphi }=v_0$. Its streaming velocity is zero. A maximally prograde stellar distribution instead has mean $\sqrt{2/\pi }{v}_0$ and azimuthal variance $v_0^2(1-2/\pi )$, while all second raw moments remain unchanged.