Vector potential

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In vector calculus, a vector potential is a vector field whose curl is a given vector field. This is analogous to a scalar potential, which is a scalar field whose negative gradient is a given vector field.

Formally, given a vector field v, a vector potential is a vector field A such that

\mathbf{v} = \nabla \times \mathbf{A}.

If a vector field v admits a vector potential A, then from the equality

\nabla \cdot (\nabla \times \mathbf{A}) = 0

(divergence of the curl is zero) one obtains

\nabla \cdot \mathbf{v} = \nabla \cdot (\nabla \times \mathbf{A}) = 0,

which implies that v must be a solenoidal vector field.

An interesting question is then if any solenoidal vector field admits a vector potential. The answer is affirmative, if the vector potential satisfies certain conditions.

Contents

Let

\mathbf{v} : \mathbb R^3 \to \mathbb R^3

be solenoidal vector field which is twice continuously differentiable. Assume that v(x) decreases sufficiently fast as ||x||→∞. Define

\mathbf{A} (\mathbf{x}) = \frac{1}{4 \pi} \nabla \times \int_{\mathbb R^3} \frac{ \mathbf{v} (\mathbf{y})}{\left\|\mathbf{x} -\mathbf{y} \right\|} \, d\mathbf{y}.

Then, A is a vector potential for v, that is,

\nabla \times \mathbf{A} =\mathbf{v}.

A generalization of this theorem is the Helmholtz decomposition which states that any vector field can be decomposed as a sum of a solenoidal vector field and an irrotational vector field.

The vector potential admitted by a solenoidal field is not unique. If A is a vector potential for v, then so is

\mathbf{A} + \nabla m

where m is any continuously differentiable scalar function. This follows from the fact that the curl of the gradient is zero.

This nonuniqueness leads to a degree of freedom in the formulation of electrodynamics, and is referred to as choosing a gauge.

  • Fundamentals of Engineering Electromagnetics by David K. Cheng, Addison-Wesley, 1993.
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