- #1

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## Homework Statement

Using index-comma notation only, show:

\begin{equation*}

\underline{\bf{v}} \times \text{curl } \underline{\bf{v}}= \frac{1}{2} \text{ grad}(\underline{\bf{v}} \cdot \underline{\bf{v}}) - (\text{grad } \underline{\bf{v}}) \underline{\bf{v}}

\end{equation*}

## Homework Equations

\begin{align*}

\text{ curl } \underline{\bf{v}} &= \epsilon_{ijk} v_{j,i} \underline{\bf{e}}_k \\

\underline{\bf{v}} \times \underline{\bf{u}} &= \epsilon_{ijk} v_i u_j \underline{\bf{e}}_k

\end{align*}

## The Attempt at a Solution

If I let [itex] \underline{\bf{u}} =\text{ curl } \underline{\bf{v}}[/itex] we get:

\begin{align*}

\underline{\bf{u}} &= \text{ curl } \underline{\bf{v}} =\epsilon_{prq} v_{r,p} \underline{\bf{e}}_q \\

u_j &= \epsilon_{prq} v_{r,p} \underline{\bf{e}}_q \cdot \underline{\bf{e}}_j \\

u_j &= \epsilon_{prq} v_{r,p} \delta_{qj} = \epsilon_{prj} v_{r,p} \\

\underline{\bf{v}} \times \underline{\bf{u}} &= \epsilon_{ijk} v_i u_j \underline{\bf{e}}_k \\

&= \epsilon_{ijk} \epsilon_{prj} v_i v_{r,p} \underline{\bf{e}}_k

\end{align*}

Is this correct so far? Trouble is, I'm not sure what to do next. I'm wondering if trying to simplify [itex]\epsilon_{ijk} \epsilon_{prj}[/itex] would be a fruitful approach. Appreciate hints/tips. Thanks!

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