## Saturday, October 8, 2011

The vector relation

$$\text{curl}(\text{curl}\mathbf{A}) = \text{grad}(\text{div}\mathbf{A}) - \text{div}(\text{grad}\mathbf{A})$$

which is the same as

$\nabla \times \left( \nabla \times \mathbf{A} \right) = \nabla \left( \nabla \cdot \mathbf{A} \right) - \nabla^2 \mathbf{A}$

where $$\nabla^2$$ is the Laplacian or Laplace operator and also equals $$\nabla^2 \mathbf{A} = \nabla \cdot \nabla \mathbf{A}$$.

The relation can also be written as

$\nabla^2 \mathbf{A} = \nabla \left( \nabla \cdot \mathbf{A} \right) - \nabla \times \left( \nabla \times \mathbf{A} \right)$

This relation is very useful in areas such as fluid and electrodynamics.

To show that this relation is true we simply expand the operations utilizing the three dimensional coordinate Cartesian system.

First we have,

$$\nabla \times \left( \nabla \times \mathbf{A} \right)$$

so we take

$$\nabla \times \mathbf{A}$$

which equals the determinant

$\nabla \times \mathbf{A} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix}$

Expanding the determinant produces

$\left( \dfrac{\partial A_z}{\partial y} - \dfrac{\partial A_y}{\partial z} \right) \mathbf{i} - \left( \dfrac{\partial A_z}{\partial x} - \dfrac{\partial A_x}{\partial z} \right) \mathbf{j}+ \left( \dfrac{\partial A_y}{\partial x} - \dfrac{\partial A_x}{\partial y} \right) \mathbf{k}$

which can be condensed and thought of as (I apologize for the hard to read formulas!  I am not sure what is causing it to do this.)

$\boldsymbol{\omega} = \boldsymbol{\omega} \left( \omega_x, \omega_y, \omega_z \right)$

where

$\boldsymbol{\omega} = \underbrace{\left( \dfrac{\partial A_z}{\partial y} - \dfrac{\partial A_y}{\partial z} \right)}_{\omega_x} \mathbf{i} - \underbrace{\left( \dfrac{\partial A_z}{\partial x} - \dfrac{\partial A_x}{\partial z} \right)}_{\omega_y} \mathbf{j} + \underbrace{\left( \dfrac{\partial A_y}{\partial x} - \dfrac{\partial A_x}{\partial y} \right)}_{\omega_z} \mathbf{k}$

Then continuing the expansion gives

$\nabla \times \boldsymbol{\omega} = \nabla \times \nabla \times \mathbf{A}$

and the determinant to find is

$= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \omega_x & \omega_y & \omega_z \end{vmatrix}$

which expanded sees

$\left( \dfrac{\partial \omega_z}{\partial y} - \dfrac{\partial \omega_y}{\partial z} \right) \mathbf{i} - \left( \dfrac{\partial \omega_z}{\partial x} - \dfrac{\partial \omega_x}{\partial z} \right) \mathbf{j}+ \left( \dfrac{\partial \omega_y}{\partial x} - \dfrac{\partial \omega_x}{\partial y} \right) \mathbf{k}$

Now subbing in the definitions for $$\omega_x$$,  $$\omega_y$$, and  $$\omega_z$$ for each component

$\dfrac{\partial \omega_z}{\partial y} = \dfrac{\partial}{\partial y} \left( \dfrac{\partial A_y}{\partial x} - \dfrac{\partial A_x}{\partial y} \right) = \dfrac{\partial^2 A_y}{\partial x \partial y} - \dfrac{\partial^2 A_x}{\partial y^2}$

$\dfrac{\partial \omega_y}{\partial z} = \dfrac{\partial}{\partial z} \left( \dfrac{\partial A_z}{\partial x} - \dfrac{\partial A_x}{\partial z} \right) = \dfrac{\partial^2 A_z}{\partial y \partial z} - \dfrac{\partial^2 A_x}{\partial z^2}$

$\dfrac{\partial \omega_z}{\partial x} = \dfrac{\partial}{\partial x} \left( \dfrac{\partial A_y}{\partial x} - \dfrac{\partial A_y}{\partial x} \right) = \dfrac{\partial^2 A_y}{\partial x^2} - \dfrac{\partial^2 A_x}{\partial x \partial y}$

$\dfrac{\partial \omega_x}{\partial z} = \dfrac{\partial}{\partial z} \left( \dfrac{\partial A_z}{\partial y} - \dfrac{\partial A_y}{\partial z} \right) = \dfrac{\partial^2 A_x}{\partial x \partial y} - \dfrac{\partial^2 A_y}{\partial z^2}$

$\dfrac{\partial \omega_y}{\partial x} = -\dfrac{\partial}{\partial x} \left( \dfrac{\partial A_z}{\partial x} + \dfrac{\partial A_x}{\partial z} \right) = \dfrac{\partial^2 A_x}{\partial x \partial z} - \dfrac{\partial^2 A_z}{\partial x^2}$

$\dfrac{\partial \omega_x}{\partial y} = \dfrac{\partial}{\partial y} \left( \dfrac{\partial A_z}{\partial y} - \dfrac{\partial A_y}{\partial z} \right) = \dfrac{\partial^2 A_z}{\partial y^2} - \dfrac{\partial^2 A_y}{\partial y \partial z}$

Now combining

$\left( \dfrac{\partial^2 A_y}{\partial x \partial y} - \dfrac{\partial^2 A_x}{\partial y^2} + \dfrac{\partial^2 A_z}{\partial y \partial z} - \dfrac{\partial^2 A_x}{\partial z^2} \right) \mathbf{i}$

$- \left( \dfrac{\partial^2 A_y}{\partial x^2} - \dfrac{\partial^2 A_x}{\partial x \partial y} + \dfrac{\partial^2 A_x}{\partial x \partial y} - \dfrac{\partial^2 A_y}{\partial z^2} \right) \mathbf{j}$

$\left( \dfrac{\partial^2 A_x}{\partial x \partial z} - \dfrac{\partial^2 A_z}{\partial x^2} + \dfrac{\partial^2 A_y}{\partial y \partial z} - \dfrac{\partial^2 A_z}{\partial y^2} \right) \mathbf{k}$

In progress...to be continued.