Laplacian, Poisson & Laplace's Equation

Chapter 2 — Vector Calculus

The Laplacian is the natural second-derivative operator acting on a scalar field. Combining it with Gauss's law gives two of the most important partial differential equations in electrostatics: Poisson's equation (where there's charge) and Laplace's equation (where there isn't).

The Laplacian

The Laplacian of a scalar field \(f\) is the divergence of its gradient:

\[\nabla^2 f \equiv \vec{\nabla}\cdot(\vec{\nabla} f) = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}.\]

The operator \(\nabla^2\) (sometimes written \(\Delta\)) is a scalar: it takes a scalar field in and returns a scalar field. Intuitively, \(\nabla^2 f\) at a point measures how much \(f\) differs from its average over a small surrounding sphere — positive Laplacian means the point is a local dip, negative means a local bump.

Poisson's equation

Write Gauss's law in differential form and substitute \(\vec{E} = -\vec{\nabla} V\):

\[\vec{\nabla}\cdot\vec{E} = -\vec{\nabla}\cdot(\vec{\nabla} V) = -\nabla^2 V = \rho/\epsilon_0.\]

Rearranging:

\[\boxed{\ \nabla^2 V = -\rho/\epsilon_0\ }\qquad \text{(Poisson's equation)}.\]

This is the master PDE of electrostatics. Every potential \(V(\vec{r})\) for every charge distribution \(\rho(\vec{r})\) satisfies it. The "general" solution is the Coulomb-integral form you already know:

\[V(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}\,dV',\]

but Poisson's equation lets you find \(V\) via other methods too (boundary-value problems, separation of variables, numerical relaxation).

Laplace's equation

In a region where \(\rho = 0\) (charge-free but not necessarily the whole universe), Poisson's equation degenerates to

\[\boxed{\ \nabla^2 V = 0\ }\qquad \text{(Laplace's equation)}.\]

Solutions are called harmonic functions. Any physical electrostatic potential is harmonic in every charge-free region.

Mean-value property

If \(V\) is harmonic in a region containing a sphere of radius \(r\) centered at a point \(P\), then the average of \(V\) over that sphere equals \(V(P)\):

\[V(P) = \frac{1}{4\pi r^2}\oint_{\text{sphere}} V\,dA.\]

This is surprisingly strong: the value of \(V\) at the center is exactly the spherical average on any surrounding sphere (as long as no charge lives inside). It's often easier to argue physically: the potential energy of a thin spherical shell of charge \(q\) in an external field \(\phi\) is \(q\bar{\phi}\); equating this two different ways (shell-in-external-field vs charge-built-outside-shell) forces \(\bar{\phi} = \phi(P)\).

Consequence: no interior max or min

A harmonic function can have no strict local maximum or minimum in the interior of its domain. If \(V\) dipped below the surrounding values at some point, the average over a surrounding sphere would be higher than \(V(P)\), contradicting the mean-value property. Extrema live on the boundary.

Earnshaw's theorem

An immediate physics corollary: a charged particle cannot be held in stable electrostatic equilibrium in a region of empty space. Stable equilibrium requires the particle's potential energy \(U = qV\) to have a local minimum. But in empty space, \(V\) is harmonic, so \(U\) is harmonic, so \(U\) has no interior minimum. At best you'll get saddle points — which means any perturbation can be amplified along some direction.

This is why ion traps need oscillating fields (Paul trap) or magnetic fields (Penning trap) to actually confine charged particles.

Worked example

Verify that \(V(x,y) = x^2 - y^2\) is harmonic, and that it has no interior extrema.

\(\nabla^2 V = 2 + (-2) + 0 = 0\). Yes, harmonic. Now look at critical points: \(\vec{\nabla} V = (2x, -2y, 0)\), zero at the origin. But the Hessian is \(\begin{pmatrix}2&0\\0&-2\end{pmatrix}\) — saddle, not extremum. Consistent.

Practice Problems