Physics: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 $$\mathbf{𝐯}=\mathrm{\nabla }\times \mathbf{𝐀}.$$

Consequence

If a vector field v admits a vector potential A, then from the equality $$\mathrm{\nabla }\cdot (\mathrm{\nabla }\times \mathbf{𝐀})=0$$ (divergence of the curl is zero) one obtains $$\mathrm{\nabla }\cdot \mathbf{𝐯}=\mathrm{\nabla }\cdot (\mathrm{\nabla }\times \mathbf{𝐀})=0,$$ 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 $1/\Vert \mathbf{𝐱}\Vert$ for $\Vert \mathbf{𝐱}\Vert \to \mathrm{\infty }$. Define $$\mathbf{𝐀}(\mathbf{𝐱})=\frac{1}{4\pi }\underset{{\mathbb{ℝ}}^3}{\int }\frac{{\mathrm{\nabla }}_y\times \mathbf{𝐯}(\mathbf{𝐲})}{\Vert \mathbf{𝐱}-\mathbf{𝐲}\Vert }{d}^3\mathbf{𝐲}.$$

Then, A is a vector potential for v, that is,　$$\mathrm{\nabla }\times \mathbf{𝐀}=\mathbf{𝐯}.$$ Here, ${\mathrm{\nabla }}_y\times$ 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; $${\mathbf{𝐀}}^{\prime }(\mathbf{𝐱})=\frac{1}{4\pi }\underset{\mathrm{\Omega }}{\int }\frac{{\mathrm{\nabla }}_y\times \mathbf{𝐯}(\mathbf{𝐲})}{\Vert \mathbf{𝐱}-\mathbf{𝐲}\Vert }{d}^3\mathbf{𝐲}.$$

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 ${\mathit{A}}^{″}(\text{x})$ is also qualify as a vector potential for v.

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

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

Let $\text{p}\in \mathbb{ℝ}$ 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 ${\mathit{A}}^{‴}(\mathit{x})$ is also a vector potential for the $\mathit{v}$

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

Nonuniqueness

The vector potential admitted by a solenoidal field is not unique. If A is a vector potential for v, then so is $$\mathbf{𝐀}+\mathrm{\nabla }f,$$ 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
• Solenoid
• Closed and Exact Differential Forms

References

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