Computing F with Nabla Identity

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Discussion Overview

The discussion revolves around the computation of the vector field F using the Nabla identity in the context of electrodynamics. Participants explore the mathematical formulation and implications of the identity, particularly focusing on the correct application of derivatives and the treatment of basis vectors.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that F can be computed as $$ F = \sum_{i} p_i \partial_{i} E_{i} e_{i} $$ and questions its correctness.
  • Another participant challenges this by stating that the "cross" terms were missed and provides a revised expression for F as $$\vec F=\sum_i p_i \partial_i \left(\sum_j E_j \hat e_j\right).$$
  • A subsequent reply clarifies the expression for F in three-dimensional space, suggesting that if the basis vectors are constant, the formulation simplifies to $$f_i=p_j \partial_j E_i.$$

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial formulation of F, as there are differing views on the inclusion of cross terms and the treatment of basis vectors. The discussion remains unresolved regarding the correct computation of F.

Contextual Notes

Assumptions regarding the independence of basis vectors in Cartesian coordinates are noted, but the implications of these assumptions on the computation of F are not fully explored.

rakso
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TL;DR
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?

$$ F = \sum_{i} p_i \partial_{i} E_{i} e_{i} $$
 
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Not quite. You missed the "cross" terms.
$$\vec F=\sum_i p_i \partial_i \left(\sum_j E_j \hat e_j\right).$$
 
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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?
 
Sure, if you have a Cartesian basis, they are independent of position and thus in this case you simply have
$$f_i=p_j \partial_j E_i.$$
 
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