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 $$𝐯=\mathrm{\nabla }\times 𝐀.$$

Consequence

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

Theorem

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

The integral domain can be restricted to any simply connected region $\mathrm{\Omega }$. That is, ${𝐀}^{\prime }$ also is a vector potential of $𝐯$, where $${𝐀}^{\prime }(𝐱)=\frac{1}{4\pi }\underset{\mathrm{\Omega }}{\int }\frac{{\mathrm{\nabla }}_y\times 𝐯(𝐲)}{\Vert 𝐱-𝐲\Vert }{d}^3𝐲.$$

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, ${𝐀}^{″}(𝐱)$ also qualifies as a vector potential for $𝐯$, where

${𝐀}^{″}(𝐱)=\underset{\mathrm{\Omega }}{\int }\frac{𝐯(𝐲)\times (𝐱-𝐲)}{4\pi |𝐱-𝐲{|}^3}{d}^3𝐲$.

Substituting $𝐣$ (current density) for $𝐯$ and $𝐇$ (H-field) for $𝐀$, yields the Biot-Savart law.

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

${𝐀}^{‴}(𝐱)=\underset{0}{\overset{1}{\int }}s((𝐱-𝐩)\times (𝐯(s𝐱+(1-s)𝐩))ds$

Nonuniqueness

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

See also

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

References

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