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For each of the two reference books the constant equations are as follows: $$ \boldsymbol{E}\times \left( \nabla \times \boldsymbol{E} \right) =-\left( \boldsymbol{E}\cdot \nabla \right) \boldsymbol{E}+\frac{1}{2}\nabla \boldsymbol{E}^2 $$

$$ \boldsymbol{E}\times \left( \nabla \times \boldsymbol{E} \right) =-\left( \boldsymbol{E}\cdot \nabla \right) \boldsymbol{E}+\left( \boldsymbol{E}\cdot \boldsymbol{E} \right) \nabla $$ We can reason about the terms that correspond to the same in both equations. $$ \left( \boldsymbol{E}\cdot \boldsymbol{E} \right) \nabla =\frac{1}{2}\nabla \boldsymbol{E}^2 $$ What are some ways to reason about the above equation to arrive at it? (I would like to have access to a variety of reasoning options)

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  • $\begingroup$ Is this a maths question? $\endgroup$ Sep 29, 2023 at 12:25
  • $\begingroup$ I think it's a physics problem that I came across in an electrodynamics book.@MariusLadegårdMeyer, If there's a mistake I'll post it to the math community $\endgroup$
    – Vancheers
    Sep 29, 2023 at 12:31
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    $\begingroup$ could you add the reference? It looks like you're mixing vectors and operators (like $|\mathbf{E}|^2 \nabla$. This is an operator, since you need to ask what is nabla acting on?) $\endgroup$
    – basics
    Sep 29, 2023 at 13:26

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Your second equation with its $({\bf E}\cdot {\bf E})\nabla$ isnot really correct as the nabla has to act on one of the ${\bf E}$'s. The correct form is best written as $({\bf E} \cdot \nabla {\bf E})$ with the dot understood being a product between the two ${\bf E}$'s and not a scalar product with the $\nabla$. There is no standard vector notation way of saying this, but I've put the parentheses where I have to distinguish it from $({\bf E}\cdot \nabla) {\bf E}$. In components, however, the $j$-th component of the term is $$ ({\bf E} \cdot \nabla {\bf E})_j \stackrel{\rm def}{=} E_i \partial_j E_i = \frac 12 \partial_j (E_i E_i), $$
where a sum on the "$i$" index is understood.

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  • $\begingroup$ I am used to the notation $(\nabla \circ \vec A)_{ij} = \partial_i A_j$. $\endgroup$ Sep 29, 2023 at 12:42
  • $\begingroup$ @Sebastian Reese Where did you find that notation? It's not one I know. $\endgroup$
    – mike stone
    Sep 29, 2023 at 12:44
  • $\begingroup$ It was commonly used in the physics lecture at my university (in Germany; I think I've also seen it in some German books but I can't find an example right now). It was used in general as the "dyadic product", as in $A = A_{ij} \vec e_i \circ \vec e_j$. $\endgroup$ Sep 29, 2023 at 13:00

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