I Computing F with Nabla Identity

[rakso]

TL;DR Summary

Nabla identity

Hi!

The topic is electrodynamic but it's a question about Nabla identity. Given $$ F = (p \cdot \nabla)E $$

How does one compute F? Is this correct?

$$ F = \sum_{i} p_i \partial_{i} E_{i} e_{i} $$

 

[kuruman]

Not quite. You missed the "cross" terms.
$$\vec F=\sum_i p_i \partial_i \left(\sum_j E_j \hat e_j\right).$$

 

[rakso]

Ah, I see.

So for example, if we're in $R^3$, $\vec{F}$ would then be

$\vec{F} = (p_1 \partial_1 + p_2 \partial_2 + p_3 \partial_3) \vec{E} = (p_1 \partial_1 E_1 + p_2 \partial_2 E_1 + p_3 \partial_3 E_1)\hat{e}_1 + (p_1 \partial_1 E_2 + p_2 \partial_2 E_2 + p_3 \partial_3 E_2)\hat{e}_2 + (p_1 \partial_1 E_3 + p_2 \partial_2 E_3 + p_3 \partial_3 E_3)\hat{e}_3$

Assuming that the basis vectors are constant?

 

[vanhees71]

Sure, if you have a Cartesian basis, they are independent of position and thus in this case you simply have
$$f_i=p_j \partial_j E_i.$$