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Vector potential of a three-dimensional vector field




vectorPotential(j, [x1, x2, x3], <Test>)


vectorPotential(j, x) returns the vector potential of the vector field with respect to . This is a vector field with .

The vector potential of a vector function j exists if and only if the divergence of j is zero. It is uniquely determined.

If the vector potential of j does not exist, then vectorPotential returns FALSE.

If j is a vector then the component ring of j must be a field (i.e., a domain of category Cat::Field) for which definite integration can be performed.

If j is given as a list of three arithmetical expressions, then vectorPotential returns a vector of the domain Dom::Matrix().


Example 1

We check if the vector function has a vector potential:

delete x, y, z:
  [x^2*y, -1/2*y^2*x, -x*y*z], [x, y, z], Test

The answer is yes, so let us compute the vector potential of :


We check the result:

curl(%, [x, y, z])

Example 2

The vector function does not have a vector potential:

vectorPotential([x^2, 2*y, z], [x, y, z])



A list of three arithmetical expressions, or a 3-dimensional vector (i.e., a 3×1 or 1 ×3 matrix of a domain of category Cat::Matrix)

x1, x2, x3

(indexed) identifiers



Check whether the vector field j has a vector potential and return TRUE or FALSE, respectively.

Return Values

Vector with three components, i.e., an 3 ×1 or 1×n matrix of a domain of category Cat::Matrix, or a boolean value.

See Also

MuPAD Functions

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