Vector potential

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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 negative gradient is a given vector field.

Formally, given a vector field v, a vector potential is a vector field A such that

[ \mathbf = \nabla \times \mathbf. ]

If a vector field v admits a vector potential A, then from the equality

[\nabla \cdot (\nabla \times \mathbf) = 0]

(divergence of the curl is zero) one obtains

[\nabla \cdot \mathbf = \nabla \cdot (\nabla \times \mathbf) = 0,]

which implies that v must be a solenoidal vector field.

An interesting question is then if any solenoidal vector field admits a vector potential. The answer is affirmative, if the vector potential satisfies certain conditions.

Contents

1 Theorem
2 Nonuniqueness
3 See also
4 References

Theorem

Let

[\mathbf : \mathbb R^3 \to \mathbb R^3]

be solenoidal vector field which is twice continuously differentiable. Assume that v(x) decreases sufficiently fast as ||x||→∞. Define

[ \mathbf (\mathbf) = \frac \nabla \times \int_ \frac (\mathbf)} -\mathbf \right\|} \, d\mathbf. ]

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

[\nabla \times \mathbf =\mathbf. ]

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.

Nonuniqueness

The vector potential admitted by a solenoidal field is not unique. If A is a vector potential for v, then so is

[ \mathbf + \nabla m ]

where m is any continuously differentiable scalar function. This follows from the fact that the curl of the gradient is zero.

References

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

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Vector potential

Theorem

Nonuniqueness

See also

References