Potential of an axisymmetric disc with constant rotation velocity I am having trouble understanding why the form of the 3D potential for a disc with a constant rotation velocity for circular orbits of stars within the disc
\begin{equation}
v(R) = v_0, \tag{1}
\end{equation}
must be of the form
\begin{equation}
\Phi(r,z)=v_0^2 \ln{(r+|z|)},\tag{2}
\end{equation}
where $(r,\theta,\phi)$ are spherical co-ordinates and $(R,\theta,z)$ are cylindrical co-ordinates.
The definition of the potential $\Phi$ by Green (in terms of the point-mass Green's function) is
\begin{equation}
\Phi(\mathbf{x}) = -G \int{\frac{\rho({\mathbf{y}})}{|\mathbf{x}-\mathbf{y}|} \; d^3 \mathbf{y}}.\tag{3}
\end{equation}
And I have already worked out that the surface density is
\begin{equation}
\Sigma(R) = \frac{v_0^2}{2\pi G} \frac{1}{R} \delta(z),\tag{4}
\end{equation}
that is, the disc is infinitesimally thin.
Mathematically, I cannot see how this can possibly give a $z$-dependence, since the $\delta(z)$ knocks it out immediately! I can however see physically that the potential must depend on $z$ independently of $r$, since it should be axisymmetric, not spherically symmetric.
I would be grateful for some advice on this apparent discrepancy between the physics of the problem and its mathematical description.
 A: Hints: 


*

*Note that the derivative of the sign function
$$ {\rm sgn}^{\prime}(z)~=~2\delta(z) \tag{A}$$
is twice the Dirac delta distribution. This fact seems to be at the heart of OP's question.

*Repeated differentiations of the Mestel disk potential 
$$\Phi~:=~ v_0^2 \ln(r+|z|), \qquad r~:=~\sqrt{R^2+z^2}, \tag{B}$$
leads to 
$$\frac{\partial \Phi}{\partial z}~=~v_0^2\frac{{\rm sgn}(z)}{r},\tag{C}$$
$$\frac{\partial ^2\Phi}{\partial z^2}~=~-v_0^2\frac{|z|}{r^3}+\frac{2v_0^2}{R}\delta(z),\tag{D}$$
$$\frac{1}{R}\frac{\partial}{\partial R}R\frac{\partial\Phi}{\partial R}~=~\frac{v_0^2|z|}{r^3},\tag{E}$$
$$4\pi G \rho~=~\nabla^2\Phi~=~\frac{2v_0^2}{R}\delta(z).\tag{F}$$
The above calculations can be given rigorous meaning in distribution theory, i.e. with the help of test functions. 

*For a thin 2D disk, the mass density is
$$\rho~=~\Sigma \delta(z),\tag{G}$$
so that the surface density is 
$$ \Sigma~\stackrel{(F)+(G)}{=}~\frac{v_0^2}{2\pi G R}.\tag{H}$$
References:


*

*J. Binney & S. Tremaine, Galactic Dynamics, 2nd edition (2008); p. 99.

