- #1

anonymous188

- 17

- 0

## Homework Statement

Prove that [tex]\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - (\nabla \cdot \nabla)\vec{A}[/tex] using Einstein notation.

## Homework Equations

[tex]\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - (\nabla \cdot \nabla)\vec{A}[/tex]

[tex] \varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}[/tex]

## The Attempt at a Solution

[tex] \begin{equation*}

\begin{split}

\nabla \times (\nabla \times \vec{A})& = \nabla \times (\varepsilon_{ijk}\frac{\partial A_k}{\partial_j})_i\\

& = [\varepsilon_{lmn} \frac{\partial}{\partial_m}(\varepsilon_{njk}\frac{\partial A_k}{\partial_j})_n]_l\\

& = [-\varepsilon_{nml}\varepsilon_{njk} \frac{\partial}{\partial_m}(\frac{\partial A_k}{\partial_j})_n]_l\\\

& =[(\delta_{mk}\delta_{lj} - \delta_{jm}\delta_{lk})\frac{\partial}{\partial_m}(\varepsilon_{ijk}\frac{\partial A_k}{\partial_j})_n]_l\\

& = [\frac{\partial}{\partial_k}(\frac{\partial A_k}{\partial_j})_n - \frac{\partial}{\partial_m}(\frac{\partial A_k}{\partial_m})_n]_l

\end{split}

\end{equation}

[/tex]

One step that I'm not sure about was pulling [itex]\epsilon_{njk}[/itex] out in step 3. Although it's a constant, it seems like it would depend on [itex]n[/itex], so I don't know if I'm allowed to do that. If everything is right up to the last step, then I'm confused about what to do next. What I have in the last line definitely doesn't look close to the answer.

Thanks in advance for any help with this problem. If you could just provide hints as to the next step or errors in my logic rather than giving away the solution, that would be great. I usually don't like asking for help, but I'm in a difficult situation at the moment.

Thanks!

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