How Can You Simplify Problems Involving Variables in Three Dimensions?

[timscully]

1. Variables

Given a generalized basis in three dimensions: e_{1},e_{2},e_{3} and the standard Kronecker delta \delta_{ij}, and using Einstein summation.
With the vector \textbf{x},\textbf{y},\textbf{z} I'm trying to simplify this problem:

2. Problem
\delta_{il} . \delta_{jm} . x_{j}

3. My attempt
\delta_{il}.\delta_{jm} . \textbf{x} . e_{j}<br /> = (e_{i}. e_{l}) . (e_{j} . e_{m}) . \textbf{x} . e_{j}<br /> = (e_{j}. e_{j}) . (e_{l} . e_{m} . e_{j}) . \textbf{x} <br /> = 1 . (e_{l} . e_{m} . e_{j}) . \textbf{x}

Surely this leads to \delta_{il} . \delta_{jm} . x_{j} = 0 as e_{l} , e_{m} , e_{j} are all orthagonal ?

Ultimately I'm trying to prove that
(\delta_{il} . \delta_{jm} - \delta_{jl} . \delta_{im}).x_{j}.y_{l}.z_{m}<br /> = y_{i}.x_{j}.z_{j} - z_{i}.x_{j}.y_{j}

 

[tiny-tim]

Welcome to PF!

I'm trying to simplify this problem:
\delta_{il} . \delta_{jm} . x_{j}
…
Ultimately I'm trying to prove that
(\delta_{il} . \delta_{jm} - \delta_{jl} . \delta_{im}).x_{j}.y_{l}.z_{m}<br /> = y_{i}.x_{j}.z_{j} - z_{i}.x_{j}.y_{j}

Hi timscully! Welcome to PF!

(have a delta: δ )

Forget about the basis vectors!

All δ_ij does is replace i by j (or vice versa) in anything else.

So δ_ijx_j = x_i, δ_ijx_i = x_j.

So just plug 'n' play! ​

 

[timscully]

Thanks for the welcome.

It looks like a great forum.

So, is \delta_{ij} . x_{m} zero, because m is neither i nor j ?

 

[tiny-tim]

not a sum

Thanks for the welcome.

It looks like a great forum.

So, is \delta_{ij} . x_{m} zero, because m is neither i nor j ?

Nooo …

δ_ijx_j is a sum over all values of j, so it only depends on i: δ_ijx_j = x_i.

But δ_ijx_m is not a sum; there is no "contraction"; it still depends on i j and m: δ_ijx_m = x_m if i = j and = 0 if i≠j.

 

[timscully]

Got it. Much appreciated.