# Prove $\vec{\nabla}\times(\vec{\nabla}\times\vec{F})=\vec{\nabla}(\vec{\nabla}\cdot\vec{F})-\vec{\nabla}^2\vec{F}$ using Einstein's notation

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!

• 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. Mar 4, 2020 at 21:00
• @Winther Oh I got you. Thank you! Mar 4, 2020 at 21:06
• 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. Mar 4, 2020 at 21:06

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.