# 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

$$v(R) = v_0, \tag{1}$$

must be of the form

$$\Phi(r,z)=v_0^2 \ln{(r+|z|)},\tag{2}$$

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

$$\Phi(\mathbf{x}) = -G \int{\frac{\rho({\mathbf{y}})}{|\mathbf{x}-\mathbf{y}|} \; d^3 \mathbf{y}}.\tag{3}$$

And I have already worked out that the surface density is

$$\Sigma(R) = \frac{v_0^2}{2\pi G} \frac{1}{R} \delta(z),\tag{4}$$

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.

• The $z$ inside the $\delta$ is, when put into equation $(3)$, the z-component of $\mathbf{y}$. The $z$ you see in $(2)$ is the z-component of $\mathbf{x}$, which is alive and well since it's not being integrated over and is not affected by the delta function. – Javier May 25 '16 at 20:49

1. 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.
2. 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.
3. For a thin 2D disk, the mass density is $$\rho~=~\Sigma \delta(z),\tag{G}$$ so that the surface density is $$\Sigma~\stackrel{(F)+(G)}{=}~\frac{v_0^2}{2\pi G R}.\tag{H}$$