example of vector potential
ExampleOfVectorPotential
2009-04-18 06:49:24
2009-04-18 06:49:24
Definition
parent
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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
\begin{align}
\vec{A} = \frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r},
\end{align}
where\, $\vec{r} = x\vec{i}\!+\!y\vec{j}\!+\!z\vec{k}$\, is the position vector.
{\em 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 \PMlinkname{first formula}{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).