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I'm trying to prove the following identity using Einstein's Summation Convention:

$$\vec{\nabla}\times(\vec{\nabla}\times\vec{F})=\vec{\nabla}(\vec{\nabla}\cdot\vec{F})-\vec{\nabla}^2\vec{F}$$

Where, in Cartesian coordinates: $\vec{\nabla}^2\vec{F}=(\vec{\nabla}^2F_x,\vec{\nabla}^2F_y,\vec{\nabla}^2F_z)$.


My Approach: To make calculations easier, I chose to use the following notation:

$$\partial_{x_i}\equiv\frac{\partial}{\partial x_i},\quad (x_1,x_2,x_3)\equiv(x,y,z)$$

$\delta$ is Kronecker's Delta, $\varepsilon$ is the Levi-Civita tensor. I will denote vectors by capital letters with an arrow, and scalars with lowercase letters. Thus, using Einstein's notation:

$$\vec{G}=\vec{\nabla}\times\vec{F}=\varepsilon_{ijk}\partial_{x_j}F_k\vec{e}_i\implies G_c=\varepsilon_{cjk}\partial_{x_j}F_k\\ (LHS)_a=(\vec{\nabla}\times\vec{G})_a=\varepsilon_{abc}\partial_{x_b}G_c=\varepsilon_{abc}\varepsilon_{cjk}\partial_{x_b}\partial_{x_j}F_k$$

Since $\varepsilon_{abc}=\varepsilon_{cab}$ and $\varepsilon_{abc}\varepsilon_{cab}\equiv\delta_{aj}\delta_{bk}-\delta_{ak}\delta_{bj}$, we conclude that:

$$(LHS)_a=\delta_{aj}\delta_{bk}\partial_{x_b}\partial_{x_j}F_k-\delta_{ak}\delta_{bj}\partial_{x_b}\partial_{x_j}F_k$$

As for the RHS:

$$g=\vec{\nabla}\cdot\vec{F}=\partial_{x_j}F_j\\\vec{H}=\vec{\nabla}g=\partial_{x_a}g\vec{e}_a=\partial_{x_a}\partial_{x_j}F_j\vec{e}_a\implies H_a=\partial_{x_a}\partial_{x_j}F_j\\\vec{P}=\vec{\nabla}F_a=\partial_{x_b}F_a\vec{e}_b\implies P_b=\partial_{x_b}F_a\\\vec{R}=\vec{\nabla}^2\vec{F}\implies R_a=\vec{\nabla}^2F_a=\vec{\nabla}\cdot(\vec{\nabla}F_a)=\vec{\nabla}\cdot\vec{P}=\partial_{x_b}P_b=\partial_{x_b}^2F_a$$

Since $RHS=\vec{H}-\vec{R}$:

$$(RHS)_a=H_a-R_a=\partial_{x_a}\partial_{x_j}F_j-\partial_{x_b}^2F_a$$

In conclusion, I need to show that:

$$\delta_{aj}\delta_{bk}\partial_{x_b}\partial_{x_j}F_k-\delta_{ak}\delta_{bj}\partial_{x_b}\partial_{x_j}F_k=\partial_{x_a}\partial_{x_j}F_j-\partial_{x_b}^2F_a$$

Where $j,k,b\in\left\{1,2,3\right\}$ are summation variables, and $a$ represents the index of each side (meaning a is not a summation variable). The sides of the equation seem similar, but maybe I was wrong somewhere. Nonetheless, I couldn't simplify, unfortunately, the LHS in order to show it is equal to the RHS.

Thank You!

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  • $\begingroup$ Get rid of the delta functions (i.e. perform the sum over $j$ in $\delta_{aj}\ldots$ and so on) and you have your results. $\endgroup$
    – Winther
    Mar 4, 2020 at 21:00
  • $\begingroup$ @Winther Oh I got you. Thank you! $\endgroup$
    – Amit Zach
    Mar 4, 2020 at 21:06
  • $\begingroup$ In the first term you will end up with a $\partial_a\partial_k F_k$ which is identical to the first term on the RHS. You have done all the hard parts, what is left summing kroneker delta's is the simplest thing. $\endgroup$
    – Winther
    Mar 4, 2020 at 21:06

1 Answer 1

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Well since this took ages to type I feel bad to delete this so I would just complete the proof, maybe someone would need this someday.

Well, let's simplify the LHS. $\delta_{aj}\delta_{bk}$ is nonzero only when $a=j$ and $b=k$. This would eliminate the second (negative) term, and then the first (positive) term would be:

$$a=j,b=k\implies\partial_{x_a}\partial_{x_j}F_j$$

Now, $\delta_{ak}\delta_{bj}$ is nonzero only when $a=k$ and $b=j$. This would eliminate the first (positive) term, and then the second (negative) term would be:

$$a=k,b=j\implies-\partial_{x_b}^2F_a$$

Summing them up we get:

$$(LHS)_a=\partial_{x_a}\partial_{x_j}F_j-\partial_{x_b}^2F_a$$

And this completes the proof.

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