# Vector calculus - Are these expressions equivalent?

In a book I have come across the following expression

$E(\nabla \cdot E)=E\cdot(\nabla E)$, where $E=\sum_i E_i \vec{e}_i$.

Unfortunately I could not prove this, when I calculate both expression using the sum notation I end up with:

$E(\nabla \cdot E)=\sum_{i,j} E_i\partial_jE_j\vec{e}_i$

$E\cdot(\nabla E)=\sum_{i,j} E_j\partial_jE_i\vec{e}_i$

Both expressions are obviously not equivalent, have I missed something?

Kind regards, Max

• If $E = \sum E_i \vec{e_i}$, then I'm not sure what they mean by $\nabla E$. I usually see the $\nabla$ operator applied to real-valued functions $f$, and it produces a vector $\nabla f$. Sep 9, 2015 at 8:37
• I think what they want to express by that is $\nabla E=\nabla \otimes E$. (Tensor product) Sep 9, 2015 at 8:56
• I have seen $\nabla \times E$ for the curl, but never $\nabla \otimes E$, although I am familiar with the tensor product. Could you clarify the meaning of this? Sep 9, 2015 at 9:00
• This would be the tensor product of two vectors, i.e. $\nabla \otimes E = \sum_{i,j} \partial_i E_j \vec{e}_i \otimes \vec{e}_j$. Sep 9, 2015 at 9:07

If for example $E = v|v|^{-3}$ (a gravitational style field) you will have $\nabla\cdot E = 0$, but $\nabla E$ is not perpendicular to $E$.
At fx $(1,0)$ we have $E=(1,0)$, $\partial_x E = (-1, 0)$, and $\partial_y E = 0$. I see no interpretation where $E\cdot\nabla E$ would be zero.
• I have not found any conditions that would contradict your answer. Even if I had assumed a vector field (as in electrostatics; EQS) for which $\nabla \times E = 0$ applied, the expression would still be incorrect. Sep 9, 2015 at 14:43