202105241107 Homework 1 (Q2)

Prove that the divergence of a curl is always zero; and the curl of a gradient is always zero.

Solution.

First,

Let $\mathbf{A}=A_x\,\hat{\mathbf{i}} + A_y\,\hat{\mathbf{j}} + A_z\,\hat{\mathbf{k}}$ .

$\begin{aligned} & \quad \nabla \cdot (\nabla\times \mathbf{A}) \\ & = \nabla \cdot \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} \\ & = \nabla \cdot \Bigg( \bigg( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \bigg)\,\hat{\mathbf{i}} - \bigg( \frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z} \bigg)\,\hat{\mathbf{j}} + \bigg( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \bigg)\,\hat{\mathbf{k}} \Bigg) \\ & = \frac{\partial^2 A_z}{\partial x\partial y} - \frac{\partial^2 A_y}{\partial x\partial z}-\frac{\partial^2 A_z}{\partial y\partial x} + \frac{\partial^2A_x}{\partial y\partial z} + \frac{\partial^2A_y}{\partial z\partial x}-\frac{\partial^2A_x}{\partial z\partial y} \\ & = 0 \end{aligned}$

$\therefore$ The divergence of a curl is always zero.

Second,

$\nabla f(x,y,z) = \displaystyle{\bigg( \frac{\partial f(x,y,z)}{\partial x} ,\enspace \frac{\partial f(x,y,z)}{\partial y},\enspace \frac{\partial f(x,y,z)}{\partial z}\bigg)}$
$\mathbf{F}\stackrel{\textrm{def}}{=} \nabla f$

$\begin{aligned} \nabla \times \mathbf{F} & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{vmatrix} \\ & =\bigg( \frac{\partial^2f}{\partial y\partial z} - \frac{\partial^2f}{\partial z\partial y},\enspace \frac{\partial^2f}{\partial z\partial x} - \frac{\partial^2f}{\partial x\partial z},\enspace \frac{\partial^2f}{\partial x\partial y}-\frac{\partial^2f}{\partial y\partial x} \bigg) \\ & = 0 \\ \end{aligned}$

$\therefore$ The curl of a gradient is always zero.

M	T	W	T	F	S	S
					1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29	30
31

Leave a Reply Cancel reply