example of vector potential

If the solenoidal vector $ \vec{U} = \vec{U}(x,\,y,\,z)$ is a homogeneous function of degree $ \lambda$ ($ \neq -2$ ), then it has the vector potential

$\displaystyle \vec{A} = \frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r},$ (1)

where $ \vec{r} = x\vec{i}\!+\!y\vec{j}\!+\!z\vec{k}$ is the position vector.

Proof. Using the entry nabla acting on products, we first may write

$\displaystyle \nabla\times(\frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r}) =
\...
...}\cdot\nabla)\vec{r}-(\nabla\cdot\vec{U})\vec{r}
+(\nabla\cdot\vec{r})\vec{U}].$

In the brackets the first product is, according to Euler's theorem on homogeneous functions, equal to $ \lambda\vec{U}$ . The second product can be written as $ U_x\frac{\partial\vec{r}}{\partial x}+
U_y\frac{\partial\vec{r}}{\partial y}+U_z\frac{\partial\vec{r}}{\partial z}$ , which is $ U_x\vec{i}+U_y\vec{j}+U_z\vec{k}$ , i.e. $ \vec{U}$ . The third product is, due to the sodenoidalness, equal to $ 0\vec{r} = \vec{0}$ . The last product equals to $ 3\vec{U}$ (see the first formula for position vector). Thus we get the result

$\displaystyle \nabla\times(\frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r}) =
\frac{1}{\lambda\!+\!2}[\lambda\vec{U}-\vec{U}-\vec{0}+3\vec{U}] = \vec{U}.$

This means that $ \vec{U}$ has the vector potential (1).



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