Question

prove that div ( curl v) -0​

Answer 1

Proof:

Let, $\mathsf{\vec{v}=x\hat{i}+y\hat{j}+z\hat{k}}$

Now, div (curl $\vec{v}$)

$\mathsf{=\nabla. (\nabla\times \vec{v})}$

$=\nabla.\left|\begin{array}{ccc}\mathsf{\hat{i}}&\mathsf{\hat{j}}&\mathsf{\hat{k}}\\ \mathsf{\frac{\partial}{\partial x}}&\mathsf{\frac{\partial}{\partial y}}&\mathsf{\frac{\partial}{\partial z}}\\\mathsf{x}&\mathsf{y}&\mathsf{z}\end{array}\right|$

$\mathsf{=\nabla.[(\frac{\partial z}{\partial y}-\frac{\partial y}{\partial z})\hat{i}-(\frac{\partial z}{\partial x}-\frac{\partial x}{\partial z})\hat{j}+(\frac{\partial y}{\partial x}-\frac{\partial x}{\partial y})\hat{k}]}$

$\mathsf{=(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}).[(\frac{\partial z}{\partial y}-\frac{\partial y}{\partial z})\hat{i}-(\frac{\partial z}{\partial x}-\frac{\partial x}{\partial z})\hat{j}+(\frac{\partial y}{\partial x}-\frac{\partial x}{\partial y})\hat{k}]}$

$\mathsf{=\frac{\partial}{\partial x}(\frac{\partial z}{\partial y}-\frac{\partial y}{\partial z})-\frac{\partial}{\partial y}(\frac{\partial z}{\partial x}-\frac{\partial x}{\partial z})+\frac{\partial}{\partial z}(\frac{\partial y}{\partial x}-\frac{\partial x}{\partial y})}$

$\mathsf{=\frac{{\partial}^{2}z}{\partial x\:\partial y}-\frac{{\partial}^{2}y}{\partial x\:\partial z}-\frac{{\partial}^{2}z}{\partial y\:\partial x}+\frac{{\partial}^{2}x}{\partial y\:\partial z}+\frac{{\partial}^{2}y}{\partial z\:\partial x}-\frac{{\partial}^{2}x}{\partial z\:\partial y}}$

= 0 ,

where, $\mathsf{\partial x\:\partial y=\partial y\:\partial x}$ ,

$\mathsf{\partial y\:\partial z=\partial z\:\partial y}$

and $\mathsf{\partial z \:\partial x=\partial x \:\partial z}$

Hence, proved.

Rules:

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

• $\mathsf{\hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1}$

• $\mathsf{\hat{i}.\hat{j}=\hat{j}.\hat{k}=\hat{k}.\hat{i}=0}$

• $\mathsf{\vec{a}.\vec{b}=\vec{b}.\vec{a}}$