Chapter MA, Problem 4.3MA

To determine

Show the vector identity $\nabla \times \nabla \times A=\nabla \nabla \cdot A-{\nabla }^{2}A$.

Explanation of Solution

Description:

Consider the del operator $(\nabla )$.

$\nabla =\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)$

Consider the vector A.

$A=A_x{a}_x+A_y{a}_y+A_z{a}_z$

Find $\nabla \times \nabla \times A$.

$\begin{array}{c}\nabla \times A=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ A_x& A_y& A_z\end{array}\right|\\ =\left[\left(\frac{\partial }{\partial y}{A}_z-\frac{\partial }{\partial z}{A}_y\right)a_x-\left(\frac{\partial }{\partial x}{A}_z-\frac{\partial }{\partial z}{A}_x\right)a_y+\left(\frac{\partial }{\partial x}{A}_y-\frac{\partial }{\partial y}{A}_x\right)a_z\right]\end{array}$

$\begin{array}{c}\nabla \times \nabla \times A=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \left(\frac{\partial }{\partial y}{A}_z-\frac{\partial }{\partial z}{A}_y\right)& -\left(\frac{\partial }{\partial x}{A}_z-\frac{\partial }{\partial z}{A}_x\right)& \left(\frac{\partial }{\partial x}{A}_y-\frac{\partial }{\partial y}{A}_x\right)\end{array}\right|\\ =\left\{\begin{array}{l}\left[\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}{A}_y-\frac{\partial }{\partial y}{A}_x\right)+\frac{\partial }{\partial z}\left(\frac{\partial }{\partial x}{A}_z-\frac{\partial }{\partial z}{A}_x\right)\right]a_x-\\ \left[\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}{A}_y-\frac{\partial }{\partial y}{A}_x\right)-\frac{\partial }{\partial z}\left(\frac{\partial }{\partial y}{A}_z-\frac{\partial }{\partial z}{A}_y\right)\right]a_y+\\ \left[-\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}{A}_z-\frac{\partial }{\partial z}{A}_x\right)-\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}{A}_z-\frac{\partial }{\partial z}{A}_y\right)\right]a_z\end{array}\right\}\end{array}$

$\nabla \times \nabla \times A=\left\{\begin{array}{l}\left[-\frac{{\partial }^2{A}_x}{\partial y^2}+\frac{{\partial }^2{A}_y}{\partial x\partial y}+\frac{{\partial }^2{A}_z}{\partial x\partial z}-\frac{{\partial }^2{A}_x}{\partial z^2}\right]a_x+\\ \left[-\frac{{\partial }^2{A}_y}{\partial x^2}+\frac{{\partial }^2{A}_x}{\partial x\partial y}+\frac{{\partial }^2{A}_z}{\partial y\partial z}-\frac{{\partial }^2{A}_y}{\partial z^2}\right]a_y+\\ \left[-\frac{{\partial }^2{A}_z}{\partial x^2}+\frac{{\partial }^2{A}_x}{\partial x\partial z}+\frac{{\partial }^2{A}_y}{\partial y\partial z}-\frac{{\partial }^2{A}_z}{\partial y^2}\right]a_z\end{array}\right\}$        (1)

Find $\nabla \nabla \cdot A$:

$\begin{array}{c}\nabla \nabla \cdot A=\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(A_x{a}_x+A_y{a}_y+A_z{a}_z\right)\\ =\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\left(\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)\\ =\left[\frac{{\partial }^2{A}_x}{\partial x^2}+\frac{{\partial }^2{A}_y}{\partial x\partial y}+\frac{{\partial }^2{A}_z}{\partial x\partial z}\right]a_x+\left[\frac{{\partial }^2{A}_x}{\partial x\partial y}+\frac{{\partial }^2{A}_y}{\partial y^2}+\frac{{\partial }^2{A}_z}{\partial y\partial z}\right]a_y+\left[\frac{{\partial }^2{A}_z}{\partial x\partial z}+\frac{{\partial }^2{A}_y}{\partial z\partial y}+\frac{{\partial }^2{A}_z}{\partial z^2}\right]a_z\end{array}$

Find ${\nabla }^{2}A$:

$\begin{array}{c}{\nabla }^{2}A={\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)}^2\left(A_x{a}_x+A_y{a}_y+A_z{a}_z\right)\\ =\left(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}\right)\left(A_x{a}_x+A_y{a}_y+A_z{a}_z\right)\\ =\left[\frac{{\partial }^2{A}_x}{\partial x^2}+\frac{{\partial }^2{A}_x}{\partial y^2}+\frac{{\partial }^2{A}_x}{\partial z^2}\right]a_x+\left[\frac{{\partial }^2{A}_y}{\partial x^2}+\frac{{\partial }^2{A}_y}{\partial y^2}+\frac{{\partial }^2{A}_y}{\partial z^2}\right]a_y+\left[\frac{{\partial }^2{A}_z}{\partial x^2}+\frac{{\partial }^2{A}_z}{\partial y^2}+\frac{{\partial }^2{A}_z}{\partial z^2}\right]a_z\end{array}$

Find $\nabla \nabla \cdot A-{\nabla }^{2}A$:

$\nabla \nabla \cdot A-{\nabla }^{2}A=\left\{\begin{array}{l}\left[\frac{{\partial }^2{A}_x}{\partial x^2}+\frac{{\partial }^2{A}_y}{\partial x\partial y}+\frac{{\partial }^2{A}_z}{\partial x\partial z}\right]a_x+\left[\frac{{\partial }^2{A}_x}{\partial x\partial y}+\frac{{\partial }^2{A}_y}{\partial y^2}+\frac{{\partial }^2{A}_z}{\partial y\partial z}\right]a_y\\ +\left[\frac{{\partial }^2{A}_z}{\partial x\partial z}+\frac{{\partial }^2{A}_y}{\partial z\partial y}+\frac{{\partial }^2{A}_z}{\partial z^2}\right]a_z-\left[\frac{{\partial }^2{A}_x}{\partial x^2}+\frac{{\partial }^2{A}_x}{\partial y^2}+\frac{{\partial }^2{A}_x}{\partial z^2}\right]a_x\\ -\left[\frac{{\partial }^2{A}_y}{\partial x^2}+\frac{{\partial }^2{A}_y}{\partial y^2}+\frac{{\partial }^2{A}_y}{\partial z^2}\right]a_y-\left[\frac{{\partial }^2{A}_z}{\partial x^2}+\frac{{\partial }^2{A}_z}{\partial y^2}+\frac{{\partial }^2{A}_z}{\partial z^2}\right]a_z\end{array}\right\}$

$\nabla \nabla \cdot A-{\nabla }^{2}A=\left\{\begin{array}{l}\left[-\frac{{\partial }^2{A}_x}{\partial y^2}-\frac{{\partial }^2{A}_x}{\partial z^2}+\frac{{\partial }^2{A}_y}{\partial x\partial y}+\frac{{\partial }^2{A}_z}{\partial x\partial z}\right]a_x+\\ \left[-\frac{{\partial }^2{A}_y}{\partial x^2}-\frac{{\partial }^2{A}_y}{\partial z^2}+\frac{{\partial }^2{A}_x}{\partial x\partial y}+\frac{{\partial }^2{A}_z}{\partial y\partial z}\right]a_y+\\ \left[-\frac{{\partial }^2{A}_z}{\partial x^2}-\frac{{\partial }^2{A}_z}{\partial y^2}+\frac{{\partial }^2{A}_z}{\partial x\partial z}+\frac{{\partial }^2{A}_y}{\partial z\partial y}\right]a_z\end{array}\right\}$        (2)

Substitute Equation (2) in Equation (1).

$\nabla \times \nabla \times A=\nabla \nabla \cdot A-{\nabla }^{2}A$

Conclusion:

Thus, the vector identity $\nabla \times \nabla \times A=\nabla \nabla \cdot A-{\nabla }^{2}A$ is shown.