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

(1)

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).