Examine whether $b: \mathbb R ^3 → \mathbb R^3 $ is a vortex field and determine a vector potential $a$.
$b=\begin{pmatrix} 4x^4y^3z^2-3x^4z^2 \\ -4x^3y^4z^2 \\ 4z^3x^3 \end{pmatrix} $
$a=?$
Hint: Take $a_1$ as $0$.
My idea:
I calculated $curl$ $b$ and because of the fact that $curl$ $b$ $\neq0$ I came to the conclusion that this vector field is "swirly" (vortex field).
Is this correct?
How do I calculate vector potential of this vector field? I found in the book that vector potential is equal the gradient when the vector field is not swirly. But here it is, and how to calculate it in this case?
$\endgroup$ 21 Answer
$\begingroup$According to the definition of the vector potential (given here: ) $a$ is the vector potential of $b$ if
$$b=\nabla \times a.$$
Then
$$a(x,y,z)=\begin{bmatrix} 0\\ x^4z^3\\ x^4y^4z^2 \end{bmatrix}.$$
Indeed,
$$b=\begin{bmatrix} \frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\\ \frac{\partial}{\partial z}\\ \end{bmatrix} \times\begin{bmatrix} 0\\ x^4z^3\\ x^4y^4z^2 \end{bmatrix}= \begin{vmatrix} i&j&k\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ 0&x^4z^3&x^4y^4z^2 \end{vmatrix}= $$
$$=\begin{bmatrix} 4x^4y^3z^2-3x^4z^2\\ -4x^3y^4z^2\\ 4x^3z^3 \end{bmatrix}.$$
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