Consider a smooth vector field $\mathbf u\colon\Omega\to\mathbb R^3$ defined on an open domain $\Omega\subseteq\mathbb R^3$ such that $\mathbf u$ has zero divergence and zero curl on $\Omega$, that is,
$$\begin{align}\nabla\cdot\mathbf u&=0,\\\nabla\times\mathbf u&=0.\end{align}$$
Is there a specific technical name for such a vector field?
Wikipedia calls it a Laplacian vector field, but
- it does not cite any references, and
- it asserts that any such vector field is the gradient of a harmonic function, but this is only true if $\Omega$ is simply connected (counterexample: $\big(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\big)$ on the region $x^2+y^2>0$),
so I'm disinclined to trust it. Can anyone provide references supporting this or any other name?
Best Answer
In geometric calculus literature (see, for example, Doran and Lasenby), such a function is called monogenic. Monogenic functions are generalizations of complex analytic (or holomorphic) functions. This condition is strictly stronger than being harmonic--all monogenic functions are harmonic, but not all harmonic functions are monogenic.
The term monogenic is not restricted to vector fields, as well; a scalar field with zero gradient would also be referred to as monogenic.
You can also consult this page by Gull, Lasenby, and Doran.
Edit: Phrased in the language of geometric calculus, we define the vector derivative of a vector field $u$ as $\nabla u$, given by
$$\nabla u = \nabla \cdot u + \nabla \wedge u$$
When $\nabla u = 0$, then $\nabla^2 u = \nabla \wedge (\nabla \cdot u) + \nabla \cdot (\nabla \wedge u) = 0$ as well, fulfilling the harmonic condition.