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}}})_{{\...
...rline{\overline{\imath}}},i})_{{\overline{\overline{\imath}}}}
\end{displaymath}

That writes out to

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