Vector potential
# 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 at least as fast as $1/\|\mathbf {x} \|$ for $\|\mathbf {x} \|\to \infty$. 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,

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

Here, $\nabla _{y}\times$ is curl for variable y. Substituting curl[v] for the current density j of the retarded potential, you will get this formula. In other words, v corresponds to the H-field.

You can restrict the integral domain to any single-connected region Ω. That is, A' below is also a vector potential of v;

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

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.

By analogy with Biot-Savart's law, the following ${\boldsymbol {A''}}({\textbf {x}})$ is also qualify as a vector potential for v.

    ${\boldsymbol {A''}}({\textbf {x}})=\int _{\Omega }{\frac {{\boldsymbol {v}}({\boldsymbol {y}})\times ({\boldsymbol {x}}-{\boldsymbol {y}})}{4\pi |{\boldsymbol {x}}-{\boldsymbol {y}}|^{3}}}d^{3}{\boldsymbol {y}}$

Substitute j (current density) for v and H (H-field)for A, we will find the Biot-Savart law.

Let ${\textbf {p}}\in \mathbb {R}$ and let the Ω be a star domain centered on the p then, translating Poincaré's lemma for differential forms into vector fields world, the following ${\boldsymbol {A'''}}({\boldsymbol {x}})$ is also a vector potential for the ${\boldsymbol {v}}$

${\boldsymbol {A'''}}({\boldsymbol {x}})=\int _{0}^{1}s(({\boldsymbol {x}}-{\boldsymbol {p}})\times ({\boldsymbol {v}}(s{\boldsymbol {x}}+(1-s){\boldsymbol {p}}))\ ds$

## 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.

