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Given a continuously differentiable vector field $\bf a$, demonstrate the equivalence (iff) between the requirement that it satisfies ${\bf a}\cdot(\nabla \times {\bf a})=0$ and that it has the representation ${\bf a}=\lambda \nabla \phi$, where $\lambda, \phi$ are scalar functions. Stated in another way, why is ${\bf a}\cdot(\nabla \times {\bf a})=0$ a necessary and sufficent condition for vector fields to have normal congruences.

Many reference cites Lord Kelvin as the source of this theorem, but his proof, though classical, is too magnetism-oriented to readily understand. Thanks!

Boyangcy
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1 Answers1

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Necessary follows from standard vector identities: \begin{eqnarray} \nabla\times (\lambda\nabla\phi) &=& \lambda \nabla \times \nabla \phi + \nabla \lambda \times \nabla \phi = \nabla \lambda \times \nabla \phi \\ \lambda \nabla \phi \cdot \nabla \lambda \times \nabla \phi &=& \lambda \nabla \lambda \cdot \nabla \phi \times \nabla \phi = 0. \end{eqnarray}

I doubt that you can obtain sufficiency without considering boundary conditions or the topology of the domain on which you look for solutions. In the case of the more standard vector calculus for instance, you have that a curl-free vector field has a potential with matching boundary conditions if the domain is simply connected.

Guido Kanschat
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