Prove that $\operatorname{curl}({\vec a\over r})=\frac{[\vec a \times\vec r]}{r^3}$

Question

Prove that $\operatorname{curl}({\vec a\over r})=\frac{[\vec a \times\vec r]}{r^3}$ where $r$ is the radius vector and $a$ is a constant vector. I break this problem down into $\nabla \times ({1\over r}\vec a)=\vec a(\nabla\times{1\over r})+(\nabla\vec a)\times{1\over r}$, then the second member with $\nabla\vec a$ should be equal to zero, but I am not sure how to continue with the first member.

Answer 1

Some care is required here: if $\;r\;$ is the radius vector, then what you want is $\;\text{curl}\left(\frac a{|r|}\right)\;$, and also $\;|r|^3\;$ in the right side's denominator.

Directly by definition. If $\;a=(a_1, a_2, a_3)\;$ , then

$$\frac a{|r|}=\left(\frac{a_1}{\sqrt{x^2+y^2+z^2}},\,\frac{a_2}{\sqrt{x^2+y^2+z^2}},\,\frac{a_3}{\sqrt{x^2+y^2+z^2}}\right)\;,\;\;\text{so:}$$

$$\text{curl}\left(\frac a{|r|}\right)=\begin{vmatrix}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\frac{a_1}{\sqrt{x^2+y^2+z^2}}&\frac{a_2}{\sqrt{x^2+y^2+z^2}}&\frac{a_3}{\sqrt{x^2+y^2+z^2}}\end{vmatrix}=$$$${}$$

$$\left(\frac{-a_3y+a_2z}{(x^2+y^2+z^2)^{3/2}}\,,\,\,\frac{-a_1z+a_3x}{(x^2+y^2+z^2)^{3/2}}\,,\,\,\frac{-a_2x+a_1y}{(x^2+y^2+z^2)^{3/2}}\right)$$

And now

$$\frac{(a\times r)}{|r|^3}=\frac1{|r|^3}\begin{vmatrix}i&j&k\\a_1&a_2&a_3\\x&y&z\end{vmatrix}=\frac1{|r|^3}\left(a_2z-a_3y\,,\,\,a_3x-a_1z\,,\,\,a_1y-a_2x\right)$$

Check both expressions above are identical.

Answer 2

Let $\boldsymbol u(\boldsymbol x)=\boldsymbol a/|x|$. The curl is $$ (\operatorname{curl} u)_i=\epsilon_{ijk}\nabla^ju^k$$ Taking $a$ to be a constant vector, $$\nabla_ju^k=\nabla_j\left(\frac{a^k}{|x|}\right)=\nabla_j\left(\frac{a^k}{\sqrt{x^lx_l}}\right)=a^k\left(\frac{-1}{x^lx_l}\right)\nabla_j\left(\sqrt{x^lx_l}\right)=\frac{-a^k}{x^lx_l}\frac{x_j}{\sqrt{x^lx_l}} \\ =\frac{-a^kx_j}{(x^lx_l)^{3/2}}=\frac{-a^kx_j}{|x|^3}$$

So $$(\operatorname{curl}u)_i(\boldsymbol x)=\epsilon_{ijk}~\frac{-a^kx^j}{|x|^3}=\frac{-1}{|x|^3}~\epsilon_{ijk}x^ja^k$$ But the cross product is defined as $$(p\times q)_i:=\epsilon_{ijk}p^jq^k$$ And thus $$(\operatorname{curl}u)_i(\boldsymbol x)=\frac{-1}{|x|^3}~(x\times a)_i$$ Flipping the cross product (anti-symmetry) and rewriting in vector form, $$\nabla\times\boldsymbol u(\boldsymbol x)=\frac{1}{|x|^3}\boldsymbol a\times\boldsymbol x$$