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 [math]\displaystyle{ C^2 }[/math] vector field A such that [math]\displaystyle{ \mathbf{v} = \nabla \times \mathbf{A}. }[/math]

Consequence

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

Theorem

Let [math]\displaystyle{ \mathbf{v} : \R^3 \to \R^3 }[/math] be a solenoidal vector field which is twice continuously differentiable. Assume that v(x) decreases at least as fast as [math]\displaystyle{ 1/\|\mathbf{x}\| }[/math] for [math]\displaystyle{ \| \mathbf{x}\| \to \infty }[/math]. Define [math]\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}. }[/math]

Then, A is a vector potential for v, that is,　[math]\displaystyle{ \nabla \times \mathbf{A} =\mathbf{v}. }[/math] Here, [math]\displaystyle{ \nabla_y \times }[/math] 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; [math]\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}. }[/math]

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

[math]\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} }[/math]

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

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

[math]\displaystyle{ \boldsymbol{A'''}(\boldsymbol{x}) =\int_0^1 s ((\boldsymbol{x}-\boldsymbol{p})\times ( \boldsymbol{v}( s \boldsymbol{x} + (1-s) \boldsymbol{p} ))\ ds }[/math]

Nonuniqueness

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