Calculating the gravitational potential of a plane

Question

I'm trying to prove that the potential generated by a infinitesimally thin surface density with constant $\Sigma$ at $z = 0$ is equal to $$\Phi = 2 \pi G \Sigma |z|$$ using Poisson's equation $$\nabla^2 \Phi = 4 \pi G \rho$$ where $\rho$ is the volume density.

So given that the mass is distributed essentially all over the plane $z=0$ the next relation can be drawn $$\rho = \delta(z) \Sigma$$

Now replacing this in the Poisson equation we get $$\frac{\mathrm{d}^2\Phi}{\mathrm{d}z^2} = 4 \pi G \Sigma \delta(z)$$

By integrating over $z$, given the definition of the Dirac Delta we have $$\frac{\mathrm{d}\Phi}{\mathrm{d}z} = 4\pi G \Sigma - C$$ Now this is where I have my doubts about my proceedings. If I take the derivative with respect to $z$ of the last equality, the right hand side becomes zero, which is not consistent with the Poisson equation. What am I doing wrong here.

I know I can get the right answer easily using the Gauss theorem, but I need to solve it using this method.

Answer 1

The antiderivative of the delta-function isn't a constant, it's the Heaviside function. So you should have $$ \frac{d\Phi}{dz} = 4 \pi G \Sigma \Theta(z) - C = \begin{cases} 4 \pi G \Sigma - C & z > 0 \\ -C & z < 0 \end{cases} $$ where $C$ is a constant of integration. Requiring that the solution be symmetric about $z = 0$ means that $\Phi(z) = \Phi(-z)$, or equivalently $\Phi'(z) = - \Phi'(-z)$. This then sets the value of the integration constant $C$.

There are also plenty of non-symmetric solutions to Laplace's equation in this situation — just substitute $\Phi \to \Phi + \alpha z$, where $\alpha$ is a constant. (Or more generally, add in any functions with $\nabla^2 \Phi = 0$). This is why the constant $C$ is not determined until you demand that the solution be symmetric.

Answer 2

You can also use the flux law of Gauss. It's easier and bypasses the Dirac $\delta$ distribution, which you shouldn't use like it's a function. Take a volume $V$, then using Gauss integral law we have: \begin{equation} \oint_{\partial V}\nabla\Phi\cdot\mathrm{d}\mathbf{A} =\int_V\Delta\Phi\mathrm{d}V =4\pi G\int_V\rho\mathrm{d}V =4\pi Gm(V), \end{equation} where $m(V)$ denoted the mass inside the volume $V$. Take an area $A$ on the $xy$-plane and extend it perpenticular to the $xy$-plane in both positive and negative direction using a distance $z$ to get a volume $V$. Because of symmetry, we have $\Phi(x,y,z)=\Phi(z)=\Phi(-z)$ and therefore no flux of $\nabla\Phi$ through the walls. Using the upper equation we have: \begin{equation} A\frac{\mathrm{d}\Phi(z)}{\mathrm{d}z} -A\frac{\mathrm{d}\Phi(-z)}{\mathrm{d}z} =4\pi G\Sigma A \Rightarrow \Phi(z)=2\pi G\Sigma|z|. \end{equation}