D.1 Generic vector identities

The rules of engagement are as follows:

The first identity to be derived involves the “vectorial triple product:”

\begin{displaymath}
\nabla\times\nabla\times\vec v
= \nabla(\nabla\cdot\vec v) - \nabla^2 \vec v %
\end{displaymath} (D.1)

To do so, first note that the $i$-​th component of $\nabla$ $\times$ $\vec{v}$ is given by

\begin{displaymath}
v_{{\overline{\overline{\imath}}},{\overline{\imath}}} - v_{{\overline{\imath}},{\overline{\overline{\imath}}}}
\end{displaymath}

Repeating the rule, the $i$-​th component of $\nabla$ $\times$ $\nabla$ $\times$ $\vec{v}$ is

\begin{displaymath}
(v_{{\overline{\imath}},i} - v_{i,{\overline{\imath}}})_{{...
...line{\overline{\imath}}},i})_{{\overline{\overline{\imath}}}}
\end{displaymath}

That writes out to

\begin{displaymath}
v_{i,ii} + v_{{\overline{\imath}},{\overline{\imath}}i} + ...
...\overline{\overline{\imath}}}{\overline{\overline{\imath}}}}
\end{displaymath}

since the first and fourth terms cancel each other. The first three terms can be recognized as the $i$-​th component of $\nabla(\nabla\cdot\vec{v})$ and the last three as the $i$-​th component of $-\nabla^2\vec{v}_i$.

A second identity to be derived involves the “scalar triple product:”

\begin{displaymath}
(\vec a \times \vec b)\cdot \vec c
= \vec a \cdot (\vec b\times \vec c) %
\end{displaymath} (D.2)

This is easiest derived from simply writing it out. The left hand side is

\begin{displaymath}
a_yb_zc_x - a_zb_yc_x + a_zb_xc_y - a_xb_zc_y + a_xb_yc_z - a_yb_xc_z
\end{displaymath}

while the right hand side is

\begin{displaymath}
a_xb_yc_z - a_xb_zc_y + a_yb_zc_x - a_yb_xc_z + a_zb_xc_y - a_zb_yc_x
\end{displaymath}

Inspection shows it to be the same terms in a different order. Note that since no order changes occur, the three vectors may be noncommuting operators.