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 ${\displaystyle \mathbf {v} }$, a vector potential is a ${\displaystyle C^{2}}$ vector field ${\displaystyle \mathbf {A} }$ such that $${\displaystyle \mathbf {v} =\nabla \times \mathbf {A} .}$$

If a vector field ${\displaystyle \mathbf {v} }$ admits a vector potential ${\displaystyle \mathbf {A} }$, then from the equality $${\displaystyle \nabla \cdot (\nabla \times \mathbf {A} )=0}$$ (divergence of the curl is zero) one obtains $${\displaystyle \nabla \cdot \mathbf {v} =\nabla \cdot (\nabla \times \mathbf {A} )=0,}$$ which implies that ${\displaystyle \mathbf {v} }$ must be a solenoidal vector field.

Let $${\displaystyle \mathbf {v} :\mathbb {R} ^{3}\to \mathbb {R} ^{3}}$$ be a solenoidal vector field which is twice continuously differentiable. Assume that ${\displaystyle \mathbf {v} (\mathbf {x} )}$ decreases at least as fast as ${\displaystyle 1/\|\mathbf {x} \|}$ for ${\displaystyle \|\mathbf {x} \|\to \infty }$. Define $${\displaystyle \mathbf {A} (\mathbf {x} )={\frac {1}{4\pi }}\int _{\mathbb {R} ^{3}}{\frac {\nabla _{\mathbf {s} }\times \mathbf {v} (\mathbf {s} )}{\left\|\mathbf {x} -\mathbf {s} \right\|}}\,d^{3}\mathbf {s} }$$ where ${\displaystyle \nabla _{\mathbf {s} }\times }$ denotes curl with respect to variable ${\displaystyle \mathbf {s} }$. Then ${\displaystyle \mathbf {A} }$ is a vector potential for ${\displaystyle \mathbf {v} }$. That is, $${\displaystyle \nabla \times \mathbf {A} =\mathbf {v} .}$$

The integral domain can be restricted to any simply connected region ${\displaystyle \Omega }$. That is, ${\displaystyle \mathbf {A'} }$ also is a vector potential of ${\displaystyle \mathbf {v} }$, where $${\displaystyle \mathbf {A'} (\mathbf {x} )={\frac {1}{4\pi }}\int _{\Omega }{\frac {\nabla _{\mathbf {s} }\times \mathbf {v} (\mathbf {s} )}{\left\|\mathbf {x} -\mathbf {s} \right\|}}\,d^{3}\mathbf {s} .}$$

A generalization of this theorem is the Helmholtz decomposition theorem, 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 the Biot–Savart law, ${\displaystyle \mathbf {A''} (\mathbf {x} )}$ also qualifies as a vector potential for ${\displaystyle \mathbf {v} }$, where

$${\displaystyle \mathbf {A''} (\mathbf {x} )=\int _{\Omega }{\frac {\mathbf {v} (\mathbf {s} )\times (\mathbf {x} -\mathbf {s} )}{4\pi \left|\mathbf {x} -\mathbf {s} \right|^{3}}}d^{3}\mathbf {s} }$$

Substituting ${\displaystyle \mathbf {j} }$ (current density) for ${\displaystyle \mathbf {v} }$ and ${\displaystyle \mathbf {H} }$ (H-field) for ${\displaystyle \mathbf {A} }$, yields the Biot–Savart law.

Let ${\displaystyle \Omega }$ be a star domain centered at the point ${\displaystyle \mathbf {p} }$, where ${\displaystyle \mathbf {p} \in \mathbb {R} ^{3}}$. Applying Poincaré's lemma for differential forms to vector fields, then ${\displaystyle \mathbf {A'''} (\mathbf {x} )}$ also is a vector potential for ${\displaystyle \mathbf {v} }$, where

$${\displaystyle \mathbf {A'''} (\mathbf {x} )=\int _{0}^{1}s\left[(\mathbf {x} -\mathbf {p} )\times \mathbf {v} (s\mathbf {x} +(1{-}s)\mathbf {p} )\right]ds}$$

The vector potential admitted by a solenoidal field is not unique. If ${\displaystyle \mathbf {A} }$ is a vector potential for ${\displaystyle \mathbf {v} }$, then so is $${\displaystyle \mathbf {A} +\nabla f,}$$ where ${\displaystyle f}$ 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.