Vector field identity derivation using Einstein summation and kronecker delta.

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SUMMARY

The discussion focuses on deriving the vector field identity using the Einstein summation convention and the Kronecker delta. The identity in question is expressed as \vec{\nabla}\bullet(\vec{A}\vec{B})=(\vec{A}\bullet\vec{\nabla})\vec{B}+\vec{B}(\vec{\nabla}\bullet\vec{A}), as stated in Chapter 3, Problem 2 of Schwinger's "Classical Electrodynamics." The solution involves utilizing the dyadic product, represented as \vec{A}\otimes\vec{B} = A_i B_j \, e_i \otimes e_j, to clarify the divergence operation on both vector fields.

PREREQUISITES
  • Understanding of vector calculus, specifically divergence and tensor products.
  • Familiarity with the Einstein summation convention.
  • Knowledge of dyadic products and their applications in vector fields.
  • Basic concepts from classical electrodynamics, particularly as presented in Schwinger's textbook.
NEXT STEPS
  • Study the application of the Einstein summation convention in vector calculus.
  • Learn about dyadic products and their significance in physics.
  • Explore divergence operations in vector fields using examples from classical electrodynamics.
  • Review Chapter 3 of Schwinger's "Classical Electrodynamics" for deeper insights into vector identities.
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, as well as mathematicians and engineers working with vector fields and tensor analysis.

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Homework Statement



Let [tex]\vec{A}(\vec{r})[/tex]and [tex]\vec{B}(\vec{r})[/tex] be vector fields. Show that

Homework Equations



[tex]\vec{\nabla}\bullet(\vec{A}\vec{B})=(\vec{A}\bullet\vec{\nabla})\vec{B}+\vec{B}(\vec{\nabla}\bullet\vec{A})[/tex]
This is EXACTLY how it is written in Ch 3 Problem 2 of Schwinger "Classical Electrodynamics"

The Attempt at a Solution



The only thing I can think of doing is starting from the right hand side and trying to get back to right. This is because I have never seen the left side written like that.
[tex](A_{j}\partial_{j})_{i}B_{i}+B_{i}(\partial_{j}A_{j})_{i}[/tex]
and I don't know where to go from there. I have scoured Google and Wolfram but I was unable to find any help.
 
Last edited:
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I figured it out! You have to use what is called the dyadic.
The unit dyadic is 1=ii+jj+kk
 
Well, the tensor product between A and B is

[tex]\vec{A}\otimes\vec{B} = A_i B_j \, e_i \otimes e_j[/tex]

The divergence is acting on both A and B, so I don't get why the other 2 terms are missing.
 

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