# 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

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

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 (http://planetmath.org/PositionVector) for position vector).  Thus we get the result

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

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

Title example of vector potential ExampleOfVectorPotential 2013-03-22 15:42:56 2013-03-22 15:42:56 pahio (2872) pahio (2872) 7 pahio (2872) Example msc 26B12