# Calculating the gravitational potential of a plane

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.

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.
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: $$\begin{equation} A\frac{\mathrm{d}\Phi(z)}{\mathrm{d}z} -A\frac{\mathrm{d}\Phi(-z)}{\mathrm{d}z} =4\pi G\Sigma A \Rightarrow \Phi(z)=2\pi G\Sigma|z|. \end{equation}$$
• Then you can use the Heaviside function $\Theta$ with $\Theta'(z)=\delta(z)$. Since you need to integrate twice, notice that $|z|'=\operatorname{sgn}(z)=2\Theta(z)-1$ if $z\neq 0$. Therefore $|z|''=2\delta(z)$, which explains why the $4$ turns into a $2$ for the result. Neither the absolute value nor the Heaviside function is differentiable for $z=0$ and here you need to define the latter as an infinite derivative and (to make it worse) even put into an equation with a distribution. As a mathmatician, I find this derivation extremely ugly. Apr 15, 2022 at 2:34