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

  • $\begingroup$ 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 $\nabla f$. $\endgroup$
    – Eric Auld
    Sep 9, 2015 at 8:37
  • $\begingroup$ I think what they want to express by that is $\nabla E=\nabla \otimes E$. (Tensor product) $\endgroup$
    – massmog
    Sep 9, 2015 at 8:56
  • $\begingroup$ 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? $\endgroup$
    – Eric Auld
    Sep 9, 2015 at 9:00
  • $\begingroup$ This would be the tensor product of two vectors, i.e. $\nabla \otimes E = \sum_{i,j} \partial_i E_j \vec{e}_i \otimes \vec{e}_j$. $\endgroup$
    – massmog
    Sep 9, 2015 at 9:07

1 Answer 1


Maybe it's because they are not equal, why do you expect them to be equal with your interpretation?

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.

But if it's in a book it must be right, right? Anyway the book is either wrong, or they have definitions that doesn't coincide with the one you're trying to use. Check the definitions in the book.

  • $\begingroup$ Thank you very much for your answer! I will check the results in my book again and I will correct my question if any conditions were not met. $\endgroup$
    – massmog
    Sep 9, 2015 at 9:10
  • $\begingroup$ 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. $\endgroup$
    – massmog
    Sep 9, 2015 at 14:43

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