Vector Potential&o=10616

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 negative gradient is a given vector field.

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

mathbf{v} = nabla times mathbf{A}.

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.

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

Theorem

Let

mathbf{v} : 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{A} (mathbf{x}) = frac{1}{4 pi} nabla times int_{mathbb R^3} frac{ mathbf{v} (mathbf{y})}{left|mathbf{x} -mathbf{y} right|} , dmathbf{y}.
Then, A is a vector potential for v, that is,
nabla times mathbf{A} =mathbf{v}.

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{A} + nabla m
where m 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

References

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

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