# What are some ways to derive $\left( \boldsymbol{E}\cdot \boldsymbol{E} \right) \nabla =\frac{1}{2}\nabla \boldsymbol{E}^2$?

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)

• Is this a maths question? Sep 29, 2023 at 12:25
• 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 Sep 29, 2023 at 12:31
• 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?) Sep 29, 2023 at 13:26

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.
• I am used to the notation $(\nabla \circ \vec A)_{ij} = \partial_i A_j$. Sep 29, 2023 at 12:42
• 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 $A = A_{ij} \vec e_i \circ \vec e_j$. Sep 29, 2023 at 13:00