Question

1. Verify the identity: ∇ × (A × B) = (B · ∇)A − (A · ∇)B + A(∇ · B) − B(∇ · A) where A and B are vector functions.

Answer 1

$\mathrm{\nabla=\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}}$

$\mathrm{A=A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k}}$

$\mathrm{B=B_{x}\hat{i}+B_{y}\hat{j}+B_{z}\hat{k}}$

first, we must know what is $\nabla\times(\vec{\mathrm{A}}\times\vec{\mathrm{B}})$

$\vec{\mathrm{A}}\times\vec{\mathrm{B}}=\vec{\mathrm{C}}=\mathrm{C_{x}\hat{i}+C_{y}\hat{j}+C_{z}\hat{k}}$

$\mathrm{(A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k})\times(B_{x}\hat{i}+B_{y}\hat{j}+B_{z}\hat{k})=(A_{y}B_{z}-A_{z}B_{y})\hat{i}+(A_{z}B_{x}-A_{x}B_{z})\hat{j}+(A_{x}B_{y}-A_{y}B_{x})\hat{k}}$

so,

$\mathrm{C_{x}=A_{y}B_{z}-A_{z}B_{y}}$ ;   $\mathrm{C_{y}=A_{z}B_{x}-A_{x}B_{z}}$ and   $\mathrm{C_{z}=A_{x}B_{y}-A_{y}B_{x}}$

and

$\nabla\times\vec{\mathrm{C}}=\vec{\mathrm{D}}$

$\mathrm{\left(\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}\right)\times(C_{x}\hat{i}+C_{y}\hat{j}+C_{z}\hat{k})=\left(\frac{\partial C_{z}}{\partial y}-\frac{\partial C_{y}}{\partial z}\right)\hat{i}+\left(\frac{\partial C_{x}}{\partial z}-\frac{\partial C_{z}}{\partial x}\right)\hat{j}+\left(\frac{\partial C_{y}}{\partial x}-\frac{\partial C_{x}}{\partial y}\right)\hat{k}}$where,

$\mathrm{\frac{\partial C_{z}}{\partial y}=\frac{\partial}{\partial y}\left(A_{x}B_{y}-A_{y}B_{x}\right)=A_{x}\frac{\partial B_{y}}{\partial y}-\frac{\partial A_{y}}{\partial y}B_{x}}$$\mathrm{\frac{\partial C_{y}}{\partial z}=\frac{\partial}{\partial z}\left(A_{z}B_{x}-A_{x}B_{z}\right)=\frac{\partial A_{z}}{\partial z}B_{x}-A_{x}\frac{\partial B_{z}}{\partial z}}$

$\mathrm{\frac{\partial C_{x}}{\partial z}=\frac{\partial}{\partial z}\left(A_{y}B_{z}-A_{z}B_{y}\right)=A_{y}\frac{\partial B_{z}}{\partial z}-\frac{\partial A_{z}}{\partial z}B_{y}}$

$\mathrm{\frac{\partial C_{z}}{\partial x}=\frac{\partial}{\partial x}\left(A_{x}B_{y}-A_{y}B_{x}\right)=\frac{\partial A_{x}}{\partial x}B_{y}-A_{y}\frac{\partial B_{x}}{\partial x}}$

$\mathrm{\frac{\partial C_{y}}{\partial x}=\frac{\partial}{\partial x}\left(A_{z}B_{x}-A_{x}B_{z}\right)=A_{z}\frac{\partial B_{x}}{\partial x}-\frac{\partial A_{x}}{\partial x}B_{z}}$

$\mathrm{\frac{\partial C_{x}}{\partial y}=\frac{\partial}{\partial y}\left(A_{y}B_{z}-A_{z}B_{y}\right)=\frac{\partial A_{y}}{\partial y}B_{z}-A_{z}\frac{\partial B_{y}}{\partial y}}$

therefore,

$\nabla\times(\vec{\mathrm{A}}\times\vec{\mathrm{B}})=\vec{\mathrm{D}}$

$\vec{\mathrm{D}}\mathrm{=\left[A_{x}\left(\frac{\partial B_{y}}{\partial y}+\frac{\partial B_{Z}}{\partial Z}\right)-B_{x}\left(\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{Z}}{\partial Z}\right)\right]\hat{i}+\left[A_{y}\left(\frac{\partial B_{x}}{\partial x}+\frac{\partial B_{Z}}{\partial Z}\right)-B_{y}\left(\frac{\partial A_{x}}{\partial x}+\frac{\partial A_{Z}}{\partial Z}\right)\right]\hat{j}+\left[A_{z}\left(\frac{\partial B_{x}}{\partial x}+\frac{\partial B_{y}}{\partial y}\right)-B_{z}\left(\frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}\right)\right]\hat{k}}$

now, we see what is

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

where,

$(\vec{\mathrm{B}}\cdot\nabla)\vec{\mathrm{A}}=\mathrm{\left[(B_{x}\hat{i}+B_{y}\hat{j}+B_{z}\hat{k})\cdot\left(\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}\right)\right](A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k})=B_{x}\frac{\partial A_{x}}{\partial x}\hat{i}+B_{y}\frac{\partial A_{y}}{\partial y}\hat{j}+B_{z}\frac{\partial A_{z}}{\partial z}\hat{k}}$

$(\vec{\mathrm{A}}\cdot\nabla)\vec{\mathrm{B}}=\mathrm{\left[(A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k})\cdot\left(\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}\right)\right](B_{x}\hat{i}+B_{y}\hat{j}+B_{z}\hat{k})=A_{x}\frac{\partial B_{x}}{\partial x}\hat{i}+A_{y}\frac{\partial B_{y}}{\partial y}\hat{j}+A_{z}\frac{\partial B_{z}}{\partial z}\hat{k}}$

$\small \vec{\mathrm{A}}(\nabla\cdot\vec{\mathrm{B}})=\mathrm{(A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k})\left[\left(\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}\right )\cdot(B_{x}\hat{i}+B_{y}\hat{j}+B_{z}\hat{k})\right]}=\mathrm{A_{x}\left(\frac{\partial B_{x}}{\partial x}+\frac{\partial B_{y}}{\partial y}+\frac{\partial B_{z}}{\partial z}\right)\hat{i}+A_{y}\left(\frac{\partial B_{x}}{\partial x}+\frac{\partial B_{y}}{\partial y}+\frac{\partial B_{z}}{\partial z}\right)\hat{j}+A_{z}\left(\frac{\partial B_{x}}{\partial x}+\frac{\partial B_{y}}{\partial y}+\frac{\partial B_{z}}{\partial z}\right)\hat{k}}$

and

$\small \vec{\mathrm{B}}(\nabla\cdot\vec{\mathrm{A}})=\mathrm{(B_{x}\hat{i}+B_{y}\hat{j}+B_{z}\hat{k})\left[\left(\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}\right )\cdot(A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k})\right]}=\mathrm{B_{x}\left(\frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{z}}{\partial z}\right)\hat{i}+B_{y}\left(\frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{z}}{\partial z}\right)\hat{j}+B_{z}\left(\frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{z}}{\partial z}\right)\hat{k}}$

when these expressions are putting into the equation, the result is $\small \vec{\mathrm{D}}$ . This verify the identity