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I'm asked to prove the following identity, using index notation:

$(\nabla\times A)\times A=A \cdot\nabla A - \nabla(A \cdot A)$

However, when I work it out, I find that the actual solution should be:

$(\nabla\times A)\times A=A \cdot\nabla A - \frac{1}{2}\nabla(A \cdot A)$

Am I missing something, or is the book wrong?

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Here, A is a vector, of course. –  Grave Oct 9 '12 at 5:28
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3 Answers

Expanding everything out componentwise, I get

$(\nabla\times A)\times A$ $$ \begin{align} &{\lower{1pt}{\Large(}}(\partial_1,\partial_2,\partial_3)\times(A_1,A_2,A_3){\lower{1pt}{\Large)}}\times(A_1,A_2,A_3)\\ &=(\partial_2A_3-\partial_3A_2,\partial_3A_1-\partial_1A_3,\partial_1A_2-\partial_2A_1)\times(A_1,A_2,A_3)\\ &=(A_3\partial_3A_1+A_2\partial_2A_1-A_3\partial_1A_3-A_2\partial_1A_2,\\ &\hphantom{=(}A_1\partial_1A_2+A_3\partial_3A_2-A_1\partial_2A_1-A_3\partial_2A_3,\\ &\hphantom{=(}A_2\partial_2A_3+A_1\partial_1A_3-A_2\partial_3A_2-A_1\partial_3A_1) \end{align} $$ $(A\cdot\nabla)A$ $$ \begin{align} &(A_1\partial_1+A_2\partial_2+A_3\partial_3)(A_1,A_2,A_3)\\ &=(A_1\partial_1A_1+A_2\partial_2A_1+A_3\partial_3A_1,\\ &\hphantom{=(}A_1\partial_1A_2+A_2\partial_2A_2+A_3\partial_3A_2,\\ &\hphantom{=(}A_1\partial_1A_3+A_2\partial_2A_3+A_3\partial_3A_3) \end{align} $$ $\frac12\nabla(A\cdot A)$ $$ \begin{align} &\tfrac12(\partial_1,\partial_2,\partial_3)(A_1A_1+A_2A_2+A_3A_3)\\ &=(A_1\partial_1 A_1+A_2\partial_1 A_2+A_3\partial_1 A_3,\\ &\hphantom{=(}A_1\partial_2 A_1+A_2\partial_2 A_2+A_3\partial_2 A_3,\\ &\hphantom{=(}A_1\partial_3 A_1+A_2\partial_3 A_2+A_3\partial_3 A_3) \end{align} $$ And from these, I get $$ (\nabla\times A)\times A=(A\cdot\nabla)A-\tfrac12\nabla(A\cdot A) $$

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I think you forgot to change the $3$s to $2$s in the middle column of the result for $\frac12\nabla(A\cdot A)$. –  joriki Oct 9 '12 at 10:48
    
@joriki: fixed, thanks. –  robjohn Oct 9 '12 at 10:50
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A similar proof in spirit to Shuhao Cao's, using geometric calculus.

In GC it's easier not to deal with the cross product or curl, but with the more fundamental entities: the generalized dot and wedge products, and the corresponding interior and exterior derivatives.

The basic relation needed to relate curl/cross product to these other operations is $a \times b = -i (a \wedge b)$. The quantity $i$ is called the unit pseudoscalar; it might also be called a volume form, and using it finds for you the Hodge dual of a given object. Here, that duality is captured in the natural multiplication operation called the geometric product. I won't worry over too many of these details; just know that $i$ commutes with all objects in 3d, and $i^2 = -1$.

Hence we see that

$$\begin{align*}(\nabla \times A) \times A &= -i [(\nabla \times A) \wedge A] \\ &= -i [(-i \{\nabla \wedge A\}) \wedge A] \\&= i^2 (\nabla \wedge A) \cdot A\end{align*}$$

where in the last step we've used the duality of the dot and wedge: $(iX) \wedge Y = i (X \cdot Y)$.

To analyze $(\nabla \wedge A) \cdot A$, we can use an analogue of the BAC-CAB rule.

$$(\nabla \wedge A) \cdot A = \dot \nabla (\dot A \cdot A) - (A \cdot \nabla) A$$

But by symmetry, $\dot \nabla (\dot A \cdot A) = \frac{1}{2} \nabla (A \cdot A)$. This basically completes the proof; in fact, nothing really exotic has been done except changing notation and using the BAC-CAB rule! Certainly this was a surprise to me, or else I never would've bothered to change notation in the first place.

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Nice answer, +1. –  Shuhao Cao Jun 1 '13 at 20:03
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There is a nice identity for smooth vector fields: $\newcommand{\b}{\boldsymbol}$

$$ \nabla (\b{u}\cdot \b{v}) = (\b{u}\cdot \nabla) \b{v} + (\b{u}\cdot \nabla) \b{v} - (\nabla \times\b{u})\times\b{v} - (\nabla \times\b{u})\times\b{v} .\tag{1} $$

For a component wise proof like robjohn gave, you can check Balanis's book. A more physical and geometrical proof would be using Feynman subscript notation: $$ \nabla(\b{u} \cdot \b{v})= \nabla_{\b{u}}(\b{u} \cdot \b{v}) + \nabla_{\b{v}} (\b{u} \cdot \b{v}) , $$ where the notation $\nabla_{\b{u}}$ means the gradient is applied only on $\b{u}$ and we keep $\b{v}$ fixed.

Geometrically, $\nabla_{\b{u}}(\b{u} \cdot \b{v}) $ can be decomposed into the normal derivative and the tangential derivative along $\b{v}$: $$ \nabla_{\b{u}}(\b{u} \cdot \b{v}) = (\b{v}\cdot\nabla)\b{u}- (\nabla \times \b{u})\times \b{v} , $$ hence (1) holds.

Now in your case $\b{u} = \b{v} = \b{A}$: $$ \nabla (\b{A}\cdot \b{A}) = (\b{A}\cdot \nabla) \b{A} + (\b{A}\cdot \nabla) \b{A} - (\nabla \times\b{A})\times\b{A} - (\nabla \times\b{A})\times\b{A}, $$ thus $$ (\nabla \times\b{A})\times\b{A} = (\b{A}\cdot \nabla) \b{A} - \frac{1}{2}\nabla (\b{A}\cdot \b{A}) . $$

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There's also a similar notation using overdots: i.e. Feynman's $\nabla_u (u \cdot v) = \dot \nabla (\dot u \cdot v)$. –  Muphrid Jun 1 '13 at 19:39
    
@Muphrid Thanks, good to know. –  Shuhao Cao Jun 1 '13 at 19:41
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