Vector potential

If a vector field v admits a vector potential 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 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 v(x) decreases at least as fast as ${\displaystyle 1/\|\mathbf {x} \|}$ for ${\displaystyle \|\mathbf {x} \|\to \infty }$. Define

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

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

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

Here, ${\displaystyle \nabla _{y}\times }$ 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;

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

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 ${\displaystyle {\boldsymbol {A''}}({\textbf {x}})}$ is also qualify as a vector potential for v.

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

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

Let ${\displaystyle {\textbf {p}}\in \mathbb {R} }$ 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 ${\displaystyle {\boldsymbol {A'''}}({\boldsymbol {x}})}$ is also a vector potential for the ${\displaystyle {\boldsymbol {v}}}$

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

The vector potential admitted by a solenoidal field is not unique. If A is a vector potential for 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.

• Fundamental theorem of vector calculus
• Magnetic vector potential
• Solenoid
• Closed and Exact Differential Forms

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