Question

Problem 1.26 Prove that the divergence of a curl is always zero. Check it for function Va in Prob. 1.15

Answer 1

Let $\bold{A}=A_x\hat{x}+A_y\hat{y}+A_z\hat{z}$ be a vector function.

Curl of $\bold{A}$ is $\nabla\times\bold{A}=\begin{vmatrix} \hat{x} & \hat{y} & \hat{z}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ A_x & A_y & A_z \end{vmatrix}$

Divergence of $\bold{A}$ is $\nabla\cdot\bold{A}=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}$

$\nabla\times\bold{A}=\begin{vmatrix} \hat{x} & \hat{y} & \hat{z}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ A_x & A_y & A_z \end{vmatrix}=\hat{x}\left[ \frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right ]+\hat{y}\left[ \frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right ]+\hat{z}\left[ \frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right ]$

Divergence of curl of $\bold{A}$ is $\nabla\cdot(\nabla\times\bold{A})=\frac{\partial}{\partial x}\left[ \frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right ]+\frac{\partial}{\partial y}\left[ \frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right ]+\frac{\partial}{\partial z}\left[ \frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right ]$

$\nabla\cdot(\nabla\times\bold{A})=\left[ \frac{\partial^2 A_z}{\partial x\partial y}-\frac{\partial^2 A_y}{\partial x\partial z}\right ]+\left[ \frac{\partial^2 A_x}{\partial y\partial z}-\frac{\partial^2 A_z}{\partial y\partial x}\right ]+\left[ \frac{\partial^2 A_y}{\partial z\partial x}-\frac{\partial^2 A_x}{\partial z\partial y}\right ]$

Since order of partial differentiation can be changed,

$\nabla\cdot(\nabla\times\bold{A})=\left[ \frac{\partial^2 A_z}{\partial x\partial y}-\frac{\partial^2 A_y}{\partial z\partial x}\right ]+\left[ \frac{\partial^2 A_x}{\partial y\partial z}-\frac{\partial^2 A_z}{\partial x\partial y}\right ]+\left[ \frac{\partial^2 A_y}{\partial z\partial x}-\frac{\partial^2 A_x}{\partial y\partial z}\right ]$

$\nabla\cdot(\nabla\times\bold{A})=\left[ \frac{\partial^2 A_z}{\partial x\partial y}-\frac{\partial^2 A_z}{\partial x\partial y}\right ]+\left[ \frac{\partial^2 A_x}{\partial y\partial z}-\frac{\partial^2 A_x}{\partial y\partial z}\right ]+\left[ \frac{\partial^2 A_y}{\partial z\partial x}-\frac{\partial^2 A_y}{\partial z\partial x}\right ]=0$

Hence divergence of a curl is always zero.

$\bold{v}=x^2\hat{x}+3xz^2\hat{y}-2xz\hat{z}$

Curl of $\bold{v}=x^2\hat{x}+3xz^2\hat{y}-2xz\hat{z}$ is

$\nabla\times\bold{v}=\begin{vmatrix} \hat{x} & \hat{y} & \hat{z}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ v_x & v_y & v_z \end{vmatrix}=\hat{x}\left[ \frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z}\right ]+\hat{y}\left[ \frac{\partial v_x}{\partial z}-\frac{\partial v_z}{\partial x}\right ]+\hat{z}\left[ \frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}\right ]$

$\nabla\times\bold{v}=\hat{x}\left[ \frac{\partial (-2xz)}{\partial y}-\frac{\partial (3xz^2)}{\partial z}\right ]+\hat{y}\left[ \frac{\partial (x^2)}{\partial z}-\frac{\partial (-2xz)}{\partial x}\right ]+\hat{z}\left[ \frac{\partial (3xz^2)}{\partial x}-\frac{\partial (x^2)}{\partial y}\right ]$

$\nabla\times\bold{v}=\hat{x}\left[ 0-6xz\right ]+\hat{y}\left[ 0+2z\right ]+\hat{z}\left[ 3z^2-0\right ]=-6xz\hat{x}+2z\hat{y}+3z^2\hat{z}$

Divergence of curl of $\bold{v}=x^2\hat{x}+3xz^2\hat{y}-2xz\hat{z}$ is

$\nabla\cdot(\nabla\times\bold{v})=\frac{\partial (-6xz)}{\partial x}+\frac{\partial (2z)}{\partial y}+\frac{\partial (3z^2)}{\partial z}=-6z+0+6z=0$