# Proof of $\nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)} - k^2 \mathbf{A} = \mathbf{0}$

I have this exercise and I don't know how to do it.

These are the instructions:

Show that if the vector A satisfy the equation:

## 1.$$\nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)} - k^2 \mathbf{A} = \mathbf{0}$$

It will also satisfy:

and

## 3. $$\nabla \cdot \mathbf{A} = 0$$

Notice that $$k$$ is a constant.

Hint: apply $$\nabla \cdot$$ to the first equation.

So I tried to apply the $$\nabla \cdot$$ to the first equation and I had this:

$$\nabla\cdot \mathbf{(} \nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)}- k^2 \mathbf{A} \mathbf{)}$$

$$\delta_{ij}$$ $$\partial_i \mathbf{(} \nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)}- k^2 \mathbf{A} \mathbf{)}_j$$

$$\partial_i \epsilon_{ijk} \partial_j (\nabla \times \mathbf{A})_k - \partial_ik^2 \mathbf{A}_i$$

$$\epsilon_{ijk} \epsilon_{klm}\partial_i \partial_j \partial_l \mathbf{A}_m - \partial_ik^2 \mathbf{A}_i$$

$$(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})(\partial_i \partial_j \partial_l \mathbf{A}_m) - \partial_ik^2 \mathbf{A}_i$$

$$(\delta_{il}\delta_{jm} \partial_i \partial_j \partial_l \mathbf{A}_m-\delta_{im}\delta_{jl}\partial_i \partial_j \partial_l \mathbf{A}_m) - \partial_ik^2 \mathbf{A}_i$$

$$(\partial_i \partial_j \partial_i \mathbf{A}_j - \partial_i \partial_j \partial_j \mathbf{A}_i) - \partial_ik^2 \mathbf{A}_i$$

So from here I just don't know how to change the index notation to vector notation, also, I can't see how this satisfies the equations 2 and 3. So please let me know if I made a mistake in any of the previous steps and help me telling me what I should do next.

Thank you very much.

• the divergence of a curl is always 0; this is a very ueful fact you should always keep in mind (easily proven in index notation, similar, but simpler to what you've done). Anyway, it seems that from what you've written, the bracketed term is $0$ (use equality of mixed partials, resolve the summation convention/rename indices if it helps). The final term is $-k^2 \nabla \cdot \mathbf{A}$. – peek-a-boo May 1 '20 at 6:51

## 1 Answer

For the first one, it holds for all vector $$\mathbf{X}$$ that $$\nabla\cdot (\nabla \times \mathbf{X})=0$$ Applying this to your equation, we get $$0=\nabla\cdot 0=\nabla\cdot[\nabla \times(\nabla \times \mathbf{A})]-k^2\nabla \cdot\mathbf{A}=0-k^2\nabla \cdot\mathbf{A},$$ so if $$k\neq 0$$ then you have that $$\nabla\cdot \mathbf{A}=0$$.

Now, another well-known identity is that $$\nabla \times (\nabla\times \mathbf{X})=\nabla(\nabla\cdot \mathbf{X})-\nabla^2\mathbf{X}.$$ For our case, and provided that $$\nabla\cdot \mathbf{A}=0$$, then $$0=\nabla \times (\nabla\times \mathbf{A})-k^2\mathbf{A}=\nabla(\nabla\cdot \mathbf{A})-\nabla^2\mathbf{A}-k^2\mathbf{A}=0-\nabla^2\mathbf{A}-k^2\mathbf{A}$$ and you get that $$\nabla^2\mathbf{A}+k^2\mathbf{A}=0.$$

• Thanks you very much, omg, I just didn't know what to do. Thanks :3 – Camilo Vargas May 2 '20 at 4:34