PlanetPhysics/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 $$\begin{matrix} \vec{A} = \frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r}, \end{matrix}$$ 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 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).