What is divergence and curl of vector $R \times (A \times R)$

Question

What is divergence and curl of vector

$$R \times (A \times R),$$

where $R$ is radius vector and $A$ is constant vector. For divergence I tried to use $b(ac)-c(ab)$ identity, and used identity $\nabla \cdot (fA)$ where $A$ is vector and $f$ is scalar. But I got awrong answer. How do I expand divergence after $b(ac)-c(ab)$? UPD. So i guess i solved divergnce if only $(r\nabla)r=2r$

Answer 1

\begin{align*} \boldsymbol \nabla \boldsymbol \cdot \mathbf F &= \boldsymbol \nabla \boldsymbol \cdot \Big(\mathbf r \boldsymbol \times(\mathbf a \boldsymbol \times \mathbf r)\Big) \\ &= \boldsymbol \nabla \boldsymbol \cdot \Big((\mathbf r \boldsymbol \cdot \mathbf r)\mathbf a\boldsymbol -(\mathbf a \boldsymbol \cdot \mathbf r)\mathbf r\Big) \\ &= (\mathbf r \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \cdot \mathbf a + \boldsymbol \nabla(\mathbf r \boldsymbol \cdot \mathbf r) \boldsymbol \cdot \mathbf a -(\mathbf a \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \cdot \mathbf r - \boldsymbol \nabla(\mathbf a \boldsymbol \cdot \mathbf r)\boldsymbol \cdot \mathbf r \tag{$1$}\\ &= 0 + 2 \mathbf r \boldsymbol \cdot \mathbf a - 3 \mathbf a \boldsymbol \cdot \mathbf r - \mathbf a \boldsymbol \cdot \mathbf r \tag{$2$} \\ &= -2 \mathbf a \boldsymbol \cdot \mathbf r\\\\ \boldsymbol \nabla \boldsymbol \times \mathbf F &= \boldsymbol \nabla \boldsymbol \times\Big(\mathbf r \boldsymbol \times(\mathbf a \boldsymbol \times \mathbf r\Big) \\ &= \boldsymbol \nabla \boldsymbol \times \Big((\mathbf r \boldsymbol \cdot \mathbf r)\mathbf a\boldsymbol -(\mathbf a \boldsymbol \cdot \mathbf r)\mathbf r\Big) \\ &= (\mathbf r \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \times \mathbf a \boldsymbol + \boldsymbol \nabla(\mathbf r \boldsymbol \cdot \mathbf r) \boldsymbol \times \mathbf a \boldsymbol -(\mathbf a \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \times \mathbf r \boldsymbol - \boldsymbol \nabla(\mathbf a \boldsymbol \cdot \mathbf r) \boldsymbol \times \mathbf r \tag{$3$}\\ &= \mathbf 0 \boldsymbol+ 2\mathbf r \boldsymbol \times \mathbf a \boldsymbol-\mathbf 0 \boldsymbol- \mathbf a \boldsymbol \times\mathbf r \tag{$4$}\\ &= -3\mathbf a \boldsymbol \times \mathbf r, \end{align*} where

• $(1)$ follows since $\boldsymbol \nabla \boldsymbol \cdot (\mathbf F + \mathbf G) = \boldsymbol \nabla \boldsymbol \cdot \mathbf F + \boldsymbol \nabla \boldsymbol \cdot \mathbf G$ and $\boldsymbol \nabla \boldsymbol \cdot (\phi \mathbf F) = \phi \boldsymbol \nabla \boldsymbol \cdot \mathbf F + \boldsymbol \nabla \phi \boldsymbol \cdot \mathbf F$;
• $(2)$ follows since $\begin{aligned}[t]&\boldsymbol \nabla \boldsymbol \cdot \mathbf r = \boldsymbol \nabla \boldsymbol \cdot (x,y,z)=3, \\ &\boldsymbol \nabla(\mathbf a \boldsymbol \cdot \mathbf r) = \boldsymbol \nabla(a_1x + a_2y+a_3y)=(a_1,a_2,a_3)=\mathbf a, \text{ and}\\ &\boldsymbol \nabla(\mathbf r \boldsymbol \cdot \mathbf r) = \boldsymbol \nabla\left(x^2+y^2+z^2\right) = (2x,2y,2z)=2\mathbf r;\end{aligned}$
• $(3)$ follows since $\boldsymbol \nabla \boldsymbol \times (\mathbf F + \mathbf G) = \boldsymbol \nabla \boldsymbol \times \mathbf F + \boldsymbol \nabla \boldsymbol \times \mathbf G$ and $\boldsymbol \nabla \boldsymbol \times (\phi \mathbf F) = \phi \boldsymbol \nabla \boldsymbol \times \mathbf F + \boldsymbol \nabla \phi \boldsymbol \times \mathbf F$; and
• $(4)$ follows since $\boldsymbol \nabla \boldsymbol \times \mathbf r = \boldsymbol \nabla \boldsymbol \times(x,y,z)=\mathbf 0$.