# Curvature of electrostatic potential is zero

Could you please expound upon this claim? I found such claim on Zangwill's Classical Electrodynamics, which states that constraint coming from Laplacian equation implies

1. electrostatic potential has zero curvature, and

2. is not bounded in at least one direction.

What kind of curvature are we talking about here? Please explain both points for me.

The electric potential $\phi:\mathbb{R}^3\to\mathbb{R}$ is the solution to Laplace's equation and therefore a harmonic function. Harmonic functions enjoy several nice properties, some of them listed on the Wikipedia page.

Concerning OP's second point, let us mention that there is a theorem similar to Liouville's theorem from complex analysis that a bounded harmonic function defined on the whole $\mathbb{R}^3$ is a constant function .

Concerning OP's first point, Zangwill is looking at the graph

$${\rm graph}(\phi)~=~ \{({\bf r}, \phi({\bf r}))~\in~ \mathbb{R}^4\mid {\bf r}\in\mathbb{R}^3 \} ~\subset~ \mathbb{R}^4 .$$

The graph of $\phi$ is a 3-dimensional submanifold with possible curvature embedded in $\mathbb{R}^4$. The metric on the graph is induced from the standard metric on $\mathbb{R}^4$.

• And this induced metric happens to be a flat metric is what he is saying? – Quantization Sep 23 '14 at 21:55
• The induced metric is not necessarily flat, but the mean curvature is zero. – Qmechanic Sep 23 '14 at 22:01

There are several definitions of the term curvature in mathematics, but I think the simplest explanation of the first point is in regards to the second derivative of a function. The Laplacian is really just the second derivative of the electric potential equal to zero. Think of a straight line. Such an object has a finite first derivative but zero second derivative. In contrast, a parabola will have finite first and second derivatives, and thus a finite curvature.

I think this is consistent with @Qmechanic's post, but he/she is much more adept at math than I so I will let them clarify.

Side Note: The Poisson equation for an electrostatic potential, for comparison, can have curvature under the definition I used.

• The issue here is cases like $V(x,y) = x^2 - y^2$. The point $(x,y) = (0,0)$ is a saddle point, and the Gaussian curvature there is negative. However, the mean curvature, which is roughly what the Laplacian corresponds to, at this point is zero. – higgsss Sep 24 '14 at 15:35