Actually I know definition of both div and curl, but I do not know how to approach this example:
$$\vec{A} = \frac{\vec{r}}{|\vec{r}-\vec{a}|^4}$$
where $r$ is a motion vector so it looks like:
$$\vec{A} = \frac{(x,y,z)}{|(x-a_x, y-a_y, z-a_z)|^4}$$
Well, I guess module as a length could reduce the power to $2^{nd}$ but it does not make the life easier. Hints/ suggestions?
On
You could, of course, just apply the definition and compute. Otherwise the following identities will help you:
$$\nabla \cdot \left( \frac{\mathbf{v}}{f} \right) = \frac{f \nabla \cdot \mathbf{v} - (\nabla f) \cdot \mathbf{v}}{f^2} $$
$$\nabla \times\left( \frac{\mathbf{v}}{f} \right) = \frac{f \nabla \times\mathbf{v} - (\nabla f) \times\mathbf{v}}{f^2} $$
For simplicity I will denote a vector of $\mathbb{R}^3$ by $\vec{x}=(x_1,x_2,x_3)$, and set $\vec{A}=(A_1,A_2,A_3)$. Then $$ A_i=\left[\sum_{k=1}^3(x_k-a_k)^2\right]^{-2}x_i. $$ Since \begin{eqnarray} \dfrac{\partial A_i}{\partial x_j}&=&-4\left[\sum_{k=1}^3(x_k-a_k)^2\right]^{-3}x_i(x_j-a_j)+\left[\sum_{k=1}^3(x_k-a_k)^2\right]^{-2}\delta_{ij}\\ &=&-4|\vec{x}-\vec{a}|^{-6}x_i(x_j-a_j)+|\vec{x}-\vec{a}|^{-4}\delta_{ij}\\ &=&\dfrac{-4x_i(x_j-a_j)}{|\vec{x}-\vec{a}|^6}+\dfrac{\delta_{ij}}{|\vec{x}-\vec{a}|^4} \end{eqnarray} we have \begin{eqnarray} \text{div}A&=&\sum_{i=1}^3\dfrac{\partial A_i}{\partial x_i}=\dfrac{3}{|\vec{x}-\vec{a}|^4}-\dfrac{4}{|\vec{x}-\vec{a}|^6}\sum_{i=1}^3x_i(x_i-a_i)\\ &=&\dfrac{3|\vec{x}-\vec{a}|^2-4\vec{x}\cdot(\vec{x}-\vec{a})}{|\vec{x}-\vec{a}|^6}=-\dfrac{(\vec{x}-\vec{a})\cdot(\vec{x}+3\vec{a})}{|\vec{x}-\vec{a}|^6}\\ \text{curl}A&=&\left(\dfrac{\partial A_2}{\partial x_3}-\dfrac{\partial A_3}{\partial x_2},\dfrac{\partial A_3}{\partial x_1}-\dfrac{\partial A_1}{\partial x_3},\dfrac{\partial A_1}{\partial x_2}-\dfrac{\partial A_2}{\partial x_1}\right)\\ &=&\dfrac{-4}{|\vec{x}-\vec{a}|^6}\left(x_2(x_3-a_3)-x_3(x_2-a_2),x_3(x_1-a_1)-x_1(x_3-a_3),x_1(x_2-a_2)-x_2(x_1-a_1)\right)\\ &=&\dfrac{-4}{|\vec{x}-\vec{a}|^6}\left(a_2x_3-a_3x_2,a_3x_1-a_1x_3,a_1x_2-a_2x_1\right)=-\dfrac{4\vec{a}\times\vec{x}}{|\vec{x}-\vec{a}|^6} \end{eqnarray}