# 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 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.

## 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

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

${\mathbf {A} }+\nabla m$ 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, or gauge freedom, and requires choosing a gauge.