Question

Let $\varphi$ be an arbitrary smooth real valued scalar function and $\overrightarrow{V}$ be an arbitrary smooth vector valued function in a three-dimensional space. Which one of the following is an identify?

• A. $curl\left(\varphi \overrightarrow{V}\right)=▽\left(\varphi Div\overrightarrow{V}\right)$
• B. $Div\overrightarrow{V}=0$
• C. $Div\left(Curl\overrightarrow{V}\right)=0$
• D. $Div\left(\varphi \overrightarrow{V}\right)=\varphi Div\overrightarrow{V}$

Answer 1

The correct option is C $Div\left(Curl\overrightarrow{V}\right)=0$

$\overrightarrow{V}=V_1\hat{i}+V_2\hat{j}+V_3\hat{k}$
$Curl\overrightarrow{V}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ V_1& V_2& V_3\end{array}\right|$
$=\hat{i}(\frac{\partial V_3}{\partial y}-\frac{\partial V_2}{\partial z})-\hat{j}(\frac{\partial V_3}{\partial x}-\frac{\partial V_1}{\partial z})+\hat{k}(\frac{\partial V_2}{\partial x}-\frac{\partial V_1}{\partial y})$
$Div\left(Curl\overrightarrow{V}\right)$
$=\frac{\partial }{\partial x}(\frac{\partial V_3}{\partial y}-\frac{\partial V_2}{\partial z})-\frac{\partial }{\partial y}(\frac{\partial V_3}{\partial x}-\frac{\partial V_1}{\partial z})+\frac{\partial }{\partial z}(\frac{\partial V_2}{\partial x}-\frac{\partial V_1}{\partial y})$
$=\frac{{\partial }^2{V}_3}{\partial x\partial y}-\frac{{\partial }^2{V}_2}{\partial x.\partial z}-\frac{{\partial }^2{V}_3}{\partial y.\partial x}+\frac{{\partial }^2{V}_1}{\partial y\partial z}+\frac{{\partial }^2{V}_2}{\partial z.\partial x}-\frac{{\partial }^2{V}_1}{\partial z.\partial y}$
$=0$

Alternative Solution:
$Div\left(Curl\overrightarrow{V}\right)=▽.(▽\times \overrightarrow{V})=0$ (obvious result using vector identities)
$(\because ▽\times \overrightarrow{V}$ is perpendicular to the plane of $▽$ and $\overrightarrow{V})$