Vector_potential Knowpia

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} .

ConsequenceEdit

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.

TheoremEdit

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 followng ${\boldsymbol {A'''}}({\boldsymbol {x}})$ is also a vector potential for the ${\boldsymbol {v}}$

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

NonuniquenessEdit

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.

ReferencesEdit

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

Summary

ConsequenceEdit

TheoremEdit

NonuniquenessEdit

See alsoEdit

ReferencesEdit