Vector Potential

Unless otherwise specified, the operators curl and vector potential apply to scalar or vector functions of a set of exactly three standard basis variables. The default is x, y, z, but you can use other sets of basis variables by choosing Set Basis Variables from the Vector Calculus submenu and changing the default variables.

Start with ∇×(xy, yz, zx) =  $\left[\vphantom{
\begin{array}{c}
-y \\
-z \\
-x
\end{array}
}\right.$$\begin{array}{c}
-y \\
-z \\
-x
\end{array}$$\left.\vphantom{
\begin{array}{c}
-y \\
-z \\
-x
\end{array}
}\right]$ to get the following.


$\blacktriangleright$ Vector Calculus + Vector Potential

 $\left[\vphantom{
\begin{array}{c}
-y \\
-z \\
-x
\end{array}
}\right.$$\begin{array}{c}
-y \\
-z \\
-x
\end{array}$$\left.\vphantom{
\begin{array}{c}
-y \\
-z \\
-x
\end{array}
}\right]$, Vector potential is $\left[\vphantom{
\begin{array}{c}
-%
\frac{1}{2}z^{2}+xy \\
yz \\
0
\end{array}
}\right.$$\begin{array}{c}
-%
\frac{1}{2}z^{2}+xy \\
yz \\
0
\end{array}$$\left.\vphantom{
\begin{array}{c}
-%
\frac{1}{2}z^{2}+xy \\
yz \\
0
\end{array}
}\right]$


Notice that we did not get the original vector field when we asked for a vector potential of its curl. That is because the vector potential is determined only up to a field whose curl is zero. You can verify that this is the case. First, calculate the difference of the two vectors.


$\blacktriangleright$ Evaluate

$\left[\vphantom{
\begin{array}{c}
xy \\
yz \\
zx
\end{array}
}\right.$$\begin{array}{c}
xy \\
yz \\
zx
\end{array}$$\left.\vphantom{
\begin{array}{c}
xy \\
yz \\
zx
\end{array}
}\right]$ - $\left[\vphantom{
\begin{array}{c}
-\frac{1}{2}z^{2}+xy \\
yz \\
0
\end{array}
}\right.$$\begin{array}{c}
-\frac{1}{2}z^{2}+xy \\
yz \\
0
\end{array}$$\left.\vphantom{
\begin{array}{c}
-\frac{1}{2}z^{2}+xy \\
yz \\
0
\end{array}
}\right]$ =  $\left[\vphantom{
\begin{array}{c}
\frac{1}{2}z^{2} \\
0 \\
xz
\end{array}
}\right.$$\begin{array}{c}
\frac{1}{2}z^{2} \\
0 \\
xz
\end{array}$$\left.\vphantom{
\begin{array}{c}
\frac{1}{2}z^{2} \\
0 \\
xz
\end{array}
}\right]$


Then compute the curl of the difference.


$\blacktriangleright$ Evaluate

∇×$\left[\vphantom{
\begin{array}{c}
\frac{1}{2}z^{2} \\
0 \\
xz
\end{array}
}\right.$$\begin{array}{c}
\frac{1}{2}z^{2} \\
0 \\
xz
\end{array}$$\left.\vphantom{
\begin{array}{c}
\frac{1}{2}z^{2} \\
0 \\
xz
\end{array}
}\right]$ =  $\left[\vphantom{
\begin{array}{c}
0 \\
0 \\
0
\end{array}
}\right.$$\begin{array}{c}
0 \\
0 \\
0
\end{array}$$\left.\vphantom{
\begin{array}{c}
0 \\
0 \\
0
\end{array}
}\right]$


Try the same experiment after changing the basis variables to u, v, w with Vector Calculus + Set Basis Variables.


$\blacktriangleright$ Vector Calculus + Vector Potential

$\left[\vphantom{
\begin{array}{c}
-v \\
-w \\
-u
\end{array}
}\right.$$\begin{array}{c}
-v \\
-w \\
-u
\end{array}$$\left.\vphantom{
\begin{array}{c}
-v \\
-w \\
-u
\end{array}
}\right]$, Vector potential is $\left[\vphantom{
\begin{array}{c}
-\frac{1}{2}w^{2}+uv \\
vw \\
0
\end{array}
}\right.$$\begin{array}{c}
-\frac{1}{2}w^{2}+uv \\
vw \\
0
\end{array}$$\left.\vphantom{
\begin{array}{c}
-\frac{1}{2}w^{2}+uv \\
vw \\
0
\end{array}
}\right]$


A vector field can be written either as the triple (v, w, u) or as a column matrix. Try choosing Vector Potential for (v, w, u).


$\blacktriangleright$ Vector Calculus + Vector Potential

(v, w, u), Vector potential is $\left(\vphantom{ \frac{1}{2}w^{2}-uv,-vw,0}\right.$${\frac{{1}}{{2}}}$w2 - uv, - vw, 0$\left.\vphantom{ \frac{1}{2}w^{2}-uv,-vw,0}\right)$