PlanetPhysics/Example of Vector Potential

If the [[../SolenoidalVectorField/|solenoidal]] [[../Vectors/|vector]] \,

$\overrightarrow{U}=\overrightarrow{U}(x,y,z)$

\, is a homogeneous [[../Bijective/|function]] of degree

$\lambda$

(

$\ne -2$

),\, then it has the [[../SolenoidalVectorField/|vector potential]]

$\begin{array}{c}\overrightarrow{A}=\frac{1}{\lambda +2}\overrightarrow{U}\times \overrightarrow{r},\end{array}$

where\, $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$\, is the [[../PositionVector/|position vector]].

Proof. \, Using the entry nabla acting on products, we first may write $\mathrm{\nabla }\times (\frac{1}{\lambda +2}\overrightarrow{U}\times \overrightarrow{r})=\frac{1}{\lambda +2}[(\overrightarrow{r}\cdot \mathrm{\nabla })\overrightarrow{U}-(\overrightarrow{U}\cdot \mathrm{\nabla })\overrightarrow{r}-(\mathrm{\nabla }\cdot \overrightarrow{U})\overrightarrow{r}+(\mathrm{\nabla }\cdot \overrightarrow{r})\overrightarrow{U}].$ In the brackets the first product is, according to Euler's [[../Formula/|theorem]] on homogeneous functions, equal to $\lambda \overrightarrow{U}$.\, The second product can be written as\, </math>U_x\frac{\partial\vec{r}}{\partial x}+ U_y\frac{\partial\vec{r}}{\partial y}+U_z\frac{\partial\vec{r}}{\partial z}$,whichis$U_x\vec{i}+U_y\vec{j}+U_z\vec{k}$,i.e.$\vec{U}$.Thethirdproductis,duetothesodenoidalness,equalto$0\vec{r} = \vec{0}$.Thelastproductequalsto$3\vec{U}$(seethe[[../PositionVector/|firstformula]]forpositionvector).Thuswegettheresult<math>\mathrm{\nabla }\times (\frac{1}{\lambda +2}\overrightarrow{U}\times \overrightarrow{r})=\frac{1}{\lambda +2}[\lambda \overrightarrow{U}-\overrightarrow{U}-\overrightarrow{0}+3\overrightarrow{U}]=\overrightarrow{U}.$ This means that $\overrightarrow{U}$ has the vector potential (1).