Question

If A and Bare vector fields, prove the following:

nabla(A* B)=(B* nabla)A+(A* nabla)B+B*( nabla* A)+A*( nabla* B) .

Answer 1

Consider $\nabla(\vec{A} \cdot \vec{B})=\sum \vec{i} \frac{\partial}{\partial x}(\vec{A} \cdot \vec{B})=\sum \vec{i}\left(\frac{\partial \vec{A}}{\partial x} \cdot \vec{B}+\vec{A} \cdot \frac{\partial \vec{B}}{\partial x}\right)$

$=\sum \vec{i}\left(\frac{\partial \vec{A}}{\partial x} \cdot \vec{B}\right)+\sum_{i}\left(\vec{A}, \frac{\partial \vec{B}}{\partial x}\right) \ldots \ldots(i)$

Now $\vec{B} \times\left(\vec{i} \times \frac{\partial \vec{A}}{\partial x}\right)=\left(\vec{B} \cdot \frac{\partial \vec{A}}{\partial x}\right) \vec{i}-(\vec{B}, \vec{i}) \frac{\partial \vec{A}}{\partial x}$

$\Rightarrow\left(\vec{B} \cdot \frac{\partial \vec{A}}{\partial x}\right) \vec{i}=\vec{B} \times\left(\vec{i} \times \frac{\partial \vec{A}}{\partial x}\right)+(\vec{B} \cdot \vec{i}) \frac{\partial \vec{A}}{\partial x}$

$\therefore \sum\left(\vec{B} \cdot \frac{\partial \vec{A}}{\partial x}\right) \vec{i}=\sum \vec{B} \times\left(\vec{i} \times \frac{\partial \vec{A}}{\partial x}\right)+\sum(\vec{B} \cdot \vec{i}) \frac{\partial \vec{A}}{\partial x}$

$=\vec{B} \times \sum\left(\vec{i} \times \frac{\partial \vec{A}}{\partial x}\right)+\left(\vec{B} \cdot \sum \vec{i} \frac{\partial}{\partial x}\right) \vec{A}$

$=\vec{B} \times(\nabla \times \vec{A})+(\vec{B} \cdot \nabla) \vec{A}$

$\therefore \sum \vec{i}\left(\frac{\partial \vec{A}}{\partial x} \vec{B}\right)=\vec{B} \times(\nabla \times \vec{A})+(\vec{B} \cdot \nabla) \vec{A}$

Similarly, by interchanging the roles of $\vec{A}$ and $\vec{B}$ , we can prove

 $\sum \vec{i}\left(\vec{A} \cdot \frac{\partial \vec{B}}{\partial x}\right)=\vec{A} \times(\nabla \times \vec{B})+(\vec{A} \cdot \nabla) \vec{B}$  

Substituting (ii) and (iii) in (i), we get

 $\nabla(\vec{A} \cdot \vec{B})=\vec{B} \times(\nabla \times \vec{A})+(\vec{B} \cdot \nabla) \vec{A}+\vec{A} \times(\nabla \times \vec{B})+(\vec{A} \cdot \nabla) \vec{B}$