Help with Levi-Civita manipulation

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 1K views
user1139
Messages
71
Reaction score
8
TL;DR
Confused as to how I can obtain a divergence term by manipulating using Levi-Civita
How do I write the following expression

$$\epsilon_{mnk} J_{1n} \partial_i\left[\frac{x_m J_{2i}}{|\vec{x}-\vec{x}'|}\right]$$

back into vectorial form?

Einstein summation convention was used here.

Context: The above expression was derived from the derivation of torque on a general current distribution. It is part of an expression obtained by considering
$$\left[\left(\vec{x}\times\vec{J}_1(\vec{x}')\right)\left(\vec{J}_2(\vec{x})\cdot\vec\nabla\frac{1}{|\vec x-\vec x'|}\right)\right]_k$$

My source of confusion is that I am suppose to obtain a divergence term from the first expression but there is the $x_m$ term in the square brackets. As such, I am unsure of how to proceed.
 
Physics news on Phys.org
Apply the product rule, and use Kronecker delta to express partial of ##x_m##:
[tex]\partial_i \left[ x_m J_{2i} \frac{1}{|\vec{x}-\vec{x}'|}\right] = \delta_{m i} J_{2i}\frac{1}{|\vec{x}-\vec{x}'|} + x_m J_{2i}\partial_i \left[ \frac{1}{|\vec{x}-\vec{x}'|}\right][/tex]
where ##\partial_i x_j = \delta_{ij}## and ##\delta_{ij} = 1## if ##i=j## otherwise ##=0##.

You should then be able to resolve the result in vector form.
 
  • Like
Likes   Reactions: vanhees71 and BvU