= Solution
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 $R_*$ 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> gives
$$
\Phi(R,0)=v_0^2\log(R/R_*).
$$
A reflection-symmetric <harmonic function> with this midplane boundary value is
$$
\boxed{\Phi(R,z)=v_0^2\log\frac{\sqrt{R^2+z^2}+|z|}{R_*}.}
$$
Verify this continuation using the <Poisson equation for Newtonian gravity>. For $z>0$, set $r=\sqrt{R^2+z^2}$; then
$$
\Phi_R=\frac{v_0^2}{R}\left(1-\frac zr\right),\qquad
\Phi_z=\frac{v_0^2}{r},\qquad
\frac1R\partial_R(R\Phi_R)=\frac{v_0^2z}{r^3},\qquad
\Phi_{zz}=-\frac{v_0^2z}{r^3}.
$$
Hence $\nabla^2\Phi=0$ 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 gives
$$
\boxed{S(R)=\frac{v_0^2}{2\pi GR},\qquad
\rho(R,z)=\frac{v_0^2}{2\pi GR}\delta(z).}
$$
At 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 $F(\boldsymbol x,\boldsymbol v)d^2x\,d^2v$ 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 $u=|\boldsymbol v|^2/2$ and write $F=f(\boldsymbol x,u)$. The stationary <Collisionless Boltzmann equation> becomes
$$
\boldsymbol v\cdot\left(\nabla_x f-f_u\nabla_x\Phi\right)=0.
$$
Since this holds in every <velocity> direction, $\nabla_x f=f_u\nabla_x\Phi$. In coordinates $(\boldsymbol x,E)$ with $E=u+\Phi$, this is exactly $\nabla_x f|_E=0$. Therefore \b[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>:
$$
S(R)=\int_{\mathbb R^2}F\!\left(\Phi+\frac{v^2}{2}\right)d^2v
=2\pi\int_0^\infty F(\Phi+v^2/2)\,v\,dv
=2\pi\int_\Phi^\infty F(E)\,dE.
$$
On the <astrophysical disk> $R=R_*e^{\Phi/v_0^2}$, so $S(\Phi)=v_0^2e^{-\Phi/v_0^2}/(2\pi GR_*)$. Differentiate the integral with respect to its lower limit:
$$
F(\Phi)=-\frac1{2\pi}\frac{dS}{d\Phi},\qquad
\boxed{F(E)=\frac1{4\pi^2GR_*}e^{-E/v_0^2}.}
$$
The boundary value $S\to0$ as $\Phi\to\infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_*=1$ recovers the printed normalization. Changing the additive <energy> zero changes this prefactor accordingly.
At a fixed radius, the normalized <velocity> density is
$$
\frac{F(E)}{S(R)}=
\frac1{2\pi v_0^2}\exp\left[-\frac{v_R^2+v_\phi^2}{2v_0^2}\right].
$$
It 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> are
$$
\boxed{\sigma_R^2=\sigma_\phi^2=v_0^2,\qquad
\langle v_Rv_\phi\rangle=0,\qquad \langle v_\phi\rangle=0.}
$$
An 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 $L_z=Rv_\phi$ are integrals of the motion. Its density and even <velocity> moments are unchanged. Its <streaming velocity> is
$$
\boxed{\langle v_\phi\rangle_+=\langle|v_\phi|\rangle
=\frac{2}{\sqrt{2\pi}v_0}\int_0^\infty v_\phi e^{-v_\phi^2/(2v_0^2)}dv_\phi
=\sqrt{\frac2\pi}\,v_0.}
$$
This <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 $\langle v_\phi^2\rangle_+=v_0^2$ while $\sigma_{\phi,+}^2=v_0^2(1-2/\pi)$. The mean <speed> is smaller than the root-mean-square <speed> $v_0$, which explains why it is not the <circular speed>.
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