Prove there exists no vector that $\nabla\times A=\frac{\hat r}{r^2}$

Question

I need to prove that there exists no vector $A$ such that $\nabla\times A=\frac{\hat r}{r^2}$ ($r$ is the position vector, and $\hat r$ the unit position vector.).

Answer 1

Firstly observe that the necessary condition is satisfied:

$$\bigtriangledown \cdot \left( \bigtriangledown \times A \right)=\bigtriangledown \cdot \frac{\hat r}{r^2}=0$$

infact:

$$\frac{\hat r}{r^2}=\left(\frac{x}{(x^2+y^2+z^2)^\frac32},\frac{y}{(x^2+y^2+z^2)^\frac32},\frac{z}{(x^2+y^2+z^2)^\frac32} \right)$$

thus

$$\bigtriangledown \cdot \frac{\hat r}{r^2}=0$$

infact

$$\frac{\partial{\frac{x}{(x^2+y^2+z^2)^\frac32}}}{\partial x}=\frac{-2x^2+y^2+z^2}{(x^2+y^2+z^2)^\frac52}$$

http://www.wolframalpha.com/input/?i=partial+derivative+x%2F(x%5E2%2By%5E2%2Bz%5E2)%5E(3%2F2)

Even if the necessary condition is satisfied, if you consider a sphere centered in the origin, the flux of $\frac{\hat r}{r^2}$ is not equal to zero, thus in general it is not possible to find $A$ such that:

$$\nabla\times A=\frac{\hat r}{r^2}$$

Answer 2

Hint: What is the divergence of $$ \frac{\hat{r}}{r\cdot r} $$ and what is the divergence of the curl of a vector field?