I'm trying to rewrite $\left ( \vec{\nabla} \times \vec{A} \right ) \cdot \left ( \vec{\nabla} \times \vec{A} \right )$ in some other way. I tried using Levi-Civita symbol and Kronecker delta, but I'm stuck. Here is what I did: $$\left ( \vec{\nabla} \times \vec{A} \right ) \cdot \left ( \vec{\nabla} \times \vec{A} \right ) = \left ( \vec{\nabla} \times \vec{A} \right )_i \left (\vec{\nabla} \times \vec{A} \right )_i = \epsilon_{ijk} \frac{\partial A_k}{\partial x_j} \epsilon_{imn} \frac{\partial A_n}{\partial x_m} = \left ( \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km} \right ) \frac{\partial A_k}{\partial x_j} \frac{\partial A_n}{\partial x_m}$$ $$= \left ( \frac{\partial A_k}{\partial x_m} \right )^{2} - \frac{\partial A_m}{\partial x_j} \frac{\partial A_j}{\partial x_m} = \left ( \frac{\partial A_k}{\partial x_m} \right )^{2} - \frac{\partial A_j}{\partial x_j} \frac{\partial A_m}{\partial x_m} $$ And I'm stuck with both these terms. (I'm sorry for no rigour switching order of partials, but I couldn't come up with anything else). Where I messed up?
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3$\begingroup$ $\frac{\partial A_j}{\partial x_j} \frac{\partial A_m}{\partial x_m} \neq \nabla^{2} \vec{A}$, because the square of the derivative is not the same as the second derivative. Why not start from $( \vec{\nabla} \times \vec{A} ) \cdot ( \vec{\nabla} \times \vec{A} ) = \|\vec{\nabla} \times \vec{A}\|^2$ ? $\endgroup$– AbezhikoCommented Dec 31, 2022 at 21:51
2 Answers
You've got up to $$(\nabla\times A)\cdot(\nabla\times A)=\left(\frac{\partial A_j}{\partial x_m}\right)\left(\frac{\partial A_j}{\partial x_m}\right)-\left(\frac{\partial A_j}{\partial x_m}\right)\left(\frac{\partial A_m}{\partial x_j}\right).$$ Note that this is a double sum over $j$ and $m$ (this is why I replaced your $k$ by a $j$, to make the Einstein summation notation correct).
The Jacobian $J$ has components $J_{ij}=\frac{\partial A_i}{\partial x_j}$. So we may write $$(\nabla\times A)\cdot(\nabla\times A)=J_{jm}J_{jm}-J_{jm}J_{mj}.$$ Now $J_{jm}J_{jm}$ is the sum of the diagonal elements of the matrix $JJ^\top$, i.e. $\operatorname{Tr}(JJ^\top)$ and likewise $J_{jm}J_{mj}=\operatorname{Tr}(J^2)$.
So what we have is $\operatorname{Tr}(JJ^\top)-\operatorname{Tr}(J^2)$.
Your work so far is correct. The answer given here (by user15546) gives one path to condensing this expression. Let $J$ denote the Jacobian, so that $$ J_{ij} = \frac{\partial A_i}{\partial x_j}. $$ Then we can write $$ \sum_{k,m=1}^3\left ( \frac{\partial A_k}{\partial x_m} \right )^{2} - \sum_{j,m = 1}^3\frac{\partial A_j}{\partial x_j} \frac{\partial A_m}{\partial x_m} = \operatorname{trace}(JJ^T) - \operatorname{trace}(J^2). $$ We could alternatively derive a similar formula directly, avoiding the index juggling. To start, note the non-zero entries of $J-J^T$ are the components of the curl $\nabla \times A$, with each component appearing twice. Thus, $$ (\nabla \times A) \cdot (\nabla \times A) = \frac 12 \|J - J^T\|_F^2 = \frac 12 \operatorname{tr}[(J - J^T)(J^T - J)]. $$ From there, we can expand $$ \frac 12 \operatorname{tr}[(J - J^T)(J^T - J)] = \\ \frac 12 \left(\operatorname{tr}[JJ^T] - \operatorname{tr}[J^2] - \operatorname{tr}[J^TJ^T] + \operatorname{tr}[J^TJ]\right) = \\ \frac 12 \left(2\operatorname{tr}[JJ^T] - 2\operatorname{tr}[J^2]\right) = \\ \operatorname{tr}[JJ^T] - \operatorname{tr}[J^2]. $$
