# Vector potential

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 gradient is a given vector field.

Formally, given a vector field v, a vector potential is a $C^{2}$ vector field A such that

$\mathbf {v} =\nabla \times \mathbf {A} .$ ## Consequence

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.

## Theorem

Let

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

be a solenoidal vector field which is twice continuously differentiable. Assume that v(x) decreases sufficiently fast as ||x|| → ∞. Define
$\mathbf {A} (\mathbf {x} )={\frac {1}{4\pi }}\int _{\mathbb {R} ^{3}}{\frac {\nabla _{y}\times \mathbf {v} (\mathbf {y} )}{\left\|\mathbf {x} -\mathbf {y} \right\|}}\,d^{3}\mathbf {y} .$

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.

## Nonuniqueness

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

$\mathbf {A} +\nabla f,$

where f 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, or gauge freedom, and requires choosing a gauge.