Gradient identity involving cross product

I found this particular identity in an extract of my physics course : $$\nabla \times\nabla \times\vec{F}=-\nabla^2\vec{F}+\nabla(\nabla \vec{F})$$

However, nothing is said about $\vec{F}$, nor any physical interpretation is associated with this particular equation.

How can one prove this ? And what does this identity actually mean ?

• I would suggest trying indical notation – zed111 Mar 15 '15 at 16:05
• As written, this identity isn't valid (note that both terms on the RHS should be second derivatives). See Wikipedia's page on the vector Laplacian for the correct statement and other details, including some applications in physics. – Semiclassical Mar 15 '15 at 16:14
• en.wikipedia.org/wiki/Del#Second_derivatives . The form is indeed invalid, as you pointed out . The correct form can be found on that page, but I still see no way to prove or to interpret this. – stami93 Mar 15 '15 at 16:24

It's pretty straightforward if you use index notation.

A Brief Explanation of Index Notation

The rules for index notation are:

1. summation is always implied when the same index occurs exactly twice in a single term. Such indices are called summing indices or dummy indices. Example: $\vec v \cdot \vec w=\sum_{i=1}^3v_i w_i = v_iw_i$. Here $i$ is the only summing index.
2. if there is an index which is unpaired, then it is not summed over and must appear in every term on BOTH sides of the equation. Such indices are called free indices. Example: $x_a = R_{ab} v_b = \sum_{b=1}^3 R_{ab} v_b$. Here $a$ is the free index and $b$ is the summing index. Notice here that $x_a$ is the $a$th component of the vector $\vec x$ and $R_{ab}$ is a scalar indexed by $2$ indices -- where, in general, $R_{1,2} \ne R_{2,1}$ (this is how you denote the entries of a matrix in index notation).
3. if your sum includes more than $2$ of the same index then you CANNOT use this implied summation convention. Example: $A_{ab}B_{cb}x_bv_c$ is not a valid expression using the implied summation convention because there are $3$ $b$ indices in this single term and thus the summation is ambiguous. If this summand occurs, you'd have to explicitly use the $\sum$ symbol.

A couple of symbols also need to be explained:

1. The permutation symbol: $\epsilon_{pqr} = \begin{cases} 1, & pqr \in \{123, 231, 312\} \\ -1, & pqr \in \{321, 213, 132\} \\ 0, & \text{otherwise}\end{cases}$
2. The Kronecker delta: $\delta_{pq} = \begin{cases} 1, & p=q \\ 0, & \text{otherwise}\end{cases}$
3. The partial derivative wrt the $p$th argument of the function $f(x_1, x_2, \dots)$ will be denoted $\partial_p f$. For example, we'll denote $\dfrac{\partial f(x,y,z)}{\partial y}$ as $\partial_2 f$.

So now, let's start:

$$\begin{array}\ [\nabla \times (\nabla \times \vec F)]_p & \text{Just consider the pth component} \\ = \epsilon_{pqr}\partial_q(\nabla \times \vec F)_r & \text{[a\times b]_p := \epsilon_{pqr}a_qb_r -- this can be used as} \\ & \text{the definition of the cross product} \\ = \epsilon_{pqr}\partial_q\epsilon_{rst}\partial_sF_t & \text{Applying the same definition} \\ = \epsilon_{pqr}\epsilon_{rst}\partial_q\partial_sF_t & \text{\epsilon_{rst} is just a scalar and so it commutes} \\ & \text{with derivative operator} \\ = \epsilon_{rpq}\epsilon_{rst}\partial_q\partial_sF_t & \text{\epsilon_{pqr}=\epsilon_{qrp}=\epsilon_{rpq}} \\ = (\delta_{ps}\delta_{qt} - \delta_{pt}\delta_{qs})(\partial_q\partial_sF_t) & \text{Apply the identity (which you can prove} \\ & \text{if you want) \epsilon_{rpq}\epsilon_{rst} = \delta_{ps}\delta_{qt} - \delta_{pt}\delta_{qs}} \\ = \partial_t\partial_pF_t -\partial_s\partial_sF_p & \text{Apply the identity \delta_{qp}a_p = \delta_{pq}a_p=a_q} \\ = \partial_p\partial_tF_t - \partial_s\partial_sF_p & \text{Clairaut's theorem} \\ = [\nabla (\nabla \cdot \vec F)]_p -[(\nabla \cdot \nabla)\vec F]_p & \text{We're almost done now} \\ = [\nabla(\nabla\cdot \vec F)-\nabla^2\vec F]_p\ \ \ \ \ \ \square\end{array}$$

Once you get used to index notation, you'll see that there wasn't really any thinking required for this -- it was all pretty mechanical.

BTW, this process also shows you that $\vec F$ must have continuous second partials or else this identity doesn't hold (because we had to make use of Clairaut's theorem).

• Looks terrible in local coordinates. (Just kidding.) – Frieder Mar 15 '15 at 17:35
• I got no question regarding the validity of your thinking, but I've been doing some reading on indexing for the last $30$ minutes and I still can't put all the pieces in your computation together. I didn't even know about this up until now, so I can't actually grasp your proof yet – stami93 Mar 15 '15 at 17:53
• @stami93 I've attempted to explain enough so that you can see what I did without any previous experience with index notation. See if you can understand it now. – user137731 Mar 15 '15 at 18:35
• If you'd like to see this notation used in the context of coordinate transformations, you can see my answer to this question. And for some idea why the implied summation only occurs over pairs of indices, you can see my answer to this question. – user137731 Mar 15 '15 at 18:44
• @stami93 You might also want to take a look at some other ways of proving this identity or there's one more here -- I'm particularly partial to any proof using geometric calculus (but I figured index notation would be easier for you). – user137731 Mar 16 '15 at 0:11

First of all I woul like to point out two typo: the right formula is $$\nabla \times\nabla \times\vec{F}=-\nabla^2\vec{F}+\nabla(\nabla \cdot \vec{F})$$ Now, the curl of $F=(F_1,F_2,F_3)$ can be computed by using the standard definition, i.e. $$\triangledown \times F = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} \right) \times \left(F_1,F_2,F_3 \right) = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ F_1 & F_2 & F_3 \end{vmatrix}$$ while $$\triangledown \cdot F = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} \right) \cdot \left(F_1,F_2,F_3 \right) = \frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}$$ When you apply the curl to F you get a new, vector. Then apply once more the definition of curl to compute $\triangledown \times \triangledown \times F$ and use the expression of the divergence of F to simplify the computation.

I want to give a hint:

One may easy proof:

$$a \times (b \times c) = \left( {a \cdot c} \right){\rm{b}} - \left( {a \cdot b} \right)c$$

for any $a,b,c \in {R^3}$, and it does depent on this order to get a linear combination.

In the sense of using differential operators, one need this order: $$\begin{array}{c} a \times (b \times c) = {\rm{b}}\left( {a \cdot c} \right) - \left( {a \cdot b} \right)c\\ \nabla \times (\nabla \times \overrightarrow F ) = \nabla \left( {\nabla \cdot \overrightarrow F } \right) - \left( {\nabla \cdot \nabla } \right)\overrightarrow F \\ = \nabla \left( {\nabla \cdot \overrightarrow F } \right) - {\nabla ^2}\overrightarrow F \end{array}$$

• You have to be careful with this, though. $a,b,c$ are vectors in your identity, but two of them are vector operators in OP's. This type of "well if it works for vectors.." argument doesn't always hold for vector operators because real-valued vectors commute under the dot product but $\nabla \cdot \vec F \ne \vec F \cdot \nabla$. – user137731 Mar 15 '15 at 17:05
• @Bye_World: For sure, your absolutely right. Just wanted to show analogy about such an identity. Bracket's are needed. – Frieder Mar 15 '15 at 17:09