Proving the triple curl identity

[anonymous188]

Homework Statement

Prove that $$\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - (\nabla \cdot \nabla)\vec{A}$$ using Einstein notation.

Homework Equations

$$\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - (\nabla \cdot \nabla)\vec{A}$$

$$\varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}$$

The Attempt at a Solution

$$\begin{equation*} \begin{split} \nabla \times (\nabla \times \vec{A})& = \nabla \times (\varepsilon_{ijk}\frac{\partial A_k}{\partial_j})_i\\ & = [\varepsilon_{lmn} \frac{\partial}{\partial_m}(\varepsilon_{njk}\frac{\partial A_k}{\partial_j})_n]_l\\ & = [-\varepsilon_{nml}\varepsilon_{njk} \frac{\partial}{\partial_m}(\frac{\partial A_k}{\partial_j})_n]_l\\\ & =[(\delta_{mk}\delta_{lj} - \delta_{jm}\delta_{lk})\frac{\partial}{\partial_m}(\varepsilon_{ijk}\frac{\partial A_k}{\partial_j})_n]_l\\ & = [\frac{\partial}{\partial_k}(\frac{\partial A_k}{\partial_j})_n - \frac{\partial}{\partial_m}(\frac{\partial A_k}{\partial_m})_n]_l \end{split} \end{equation}$$

One step that I'm not sure about was pulling $\epsilon_{njk}$ out in step 3. Although it's a constant, it seems like it would depend on $n$, 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!

 

[anonymous188]

Hmm.. I made a few typographical errors, but it seems my editing hasn't changed anything. Maybe it takes a while to update.

 

[StatusX]

Your notation $\partial/\partial j$ is confusing. I would suggest either $\partial/\partial x_j$ or $\partial_j$.

Anyway, you're very close. Just try writing out the RHS of the original equation in component form.

 

[anonymous188]

I wonder if I'm allowed to switch the differential operators in the first term of the last line. And then, since the second term is a constant, maybe I can pull it out of the vector. Doing that,

$$\begin{equation*} \begin{split} \nabla \times (\nabla \times \vec{A})& = \nabla \times (\varepsilon_{ijk}\frac{\partial A_k}{\partial_j})_i\\ & = [\varepsilon_{lmn} \frac{\partial}{\partial_m}(\varepsilon_{njk}\frac{\partial A_k}{\partial_j})_n]_l\\ & = [-\varepsilon_{nml}\varepsilon_{njk} \frac{\partial}{\partial_m}(\frac{\partial A_k}{\partial_j})_n]_l\\\ & =[(\delta_{mk}\delta_{lj} - \delta_{jm}\delta_{lk})\frac{\partial}{\partial_m}(\varepsilon_{ijk}\frac{\partial A_k}{\partial_j})_n]_l\\ & = [\frac{\partial}{\partial_k}(\frac{\partial A_k}{\partial_j})_n - \frac{\partial}{\partial_m}(\frac{\partial A_k}{\partial_m})_n]_l\\ & = [\frac{\partial}{\partial_j}(\frac{\partial A_k}{\partial_k})_n]_l - \frac{\partial^2 A_k}{\partial^2_m} \end{split} \end{equation}$$

Now it looks closer to the answer, I think. The only problem is that the first derivative of the first term should have the same index as the the component of the vector, i.e. $n$ should be $j$. And it seems like the second term isn't a constant since it also has that subscript $n$. Hence my confusion.

 

[Dick]

You are using the partial derivative notation a bit funny. I would write your final expression as
$$\partial_j \partial_k A_k-\partial_m \partial_m A_k$$. And that's sort of what you want. I've dropped the funny 'component' subscripts, you don't need them, e.g.
$$(\varepsilon_{ijk} \partial_j A_k)$$ is the ith component of a vector. I can tell that because j and k both occur twice, which makes them 'dummy' indices. Which is what's wrong with your final expression. In one term k is a dummy variable, in the other it's not. Can you figure out how that happened? Keep track of what is 'dummy' in any given step and make sure it stays that way.