1. ## Vector Calculus Identity

Show that $\displaystyle \nabla \times (\Phi \vec{A})=\Phi \nabla \times \vec{A}-\vec{A}\times \nabla \Phi$, where $\displaystyle \Phi$ represents a scalar field.

2. Originally Posted by davesface
Show that $\displaystyle \nabla \times (\Phi \vec{A})=\Phi \nabla \times \vec{A}-\vec{A}\times \nabla \Phi$, where $\displaystyle \Phi$ represents a scalar field.
Let $\displaystyle \vec{A}=f\vec{i}+g\vec{j}+h\vec{k}$

Note I will just the subscript notation for partials e.g $\displaystyle \displaystyle \frac{\partial f}{\partial x}=f_x$

$\displaystyle \displaystyle \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ \phi f & \phi g & \phi h \end{vmatrix} = \left( \phi_y h+\phi h_y - \phi_z g -\phi g_z \right)\vec{i}-\left( \phi_x h +\phi h_x-\phi_z f -\phi f_z\right)\vec{j}+\left( \phi_x g +\phi g_x -\phi_y f -\phi f_y\right)\vec{k}$

Now we need to break this up into what we want, so lets collect all of the $\displaystyle \phi$ with no derivatives

$\displaystyle \phi \left[ (h_y-g_z)\vec{i}-(h_x-f_z)\vec{j}+(g_x-f_y)\vec{k}\right-\left[(g\phi_z-h\phi_y)\vec{i}-(f\phi_z-h\phi_x)\vec{j}+(f\phi_y-g\phi_x)\vec{k} \right]$

$\displaystyle \phi \nabla \times \vec{A}-\vec{A}\times \nabla \phi$