3D Vectors

1. Nov 23, 2008

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
$$\delta_{il}.\delta_{jm} . \textbf{x} . e_{j} = (e_{i}. e_{l}) . (e_{j} . e_{m}) . \textbf{x} . e_{j} = (e_{j}. e_{j}) . (e_{l} . e_{m} . e_{j}) . \textbf{x} = 1 . (e_{l} . e_{m} . e_{j}) . \textbf{x}$$

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
$$(\delta_{il} . \delta_{jm} - \delta_{jl} . \delta_{im}).x_{j}.y_{l}.z_{m} = y_{i}.x_{j}.z_{j} - z_{i}.x_{j}.y_{j}$$

2. Nov 24, 2008

tiny-tim

Welcome to PF!

Hi timscully! Welcome to PF!

(have a delta: δ )

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

So δijxj = xi, δijxi = xj.

So just plug 'n' play!

3. Nov 24, 2008

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 ?

4. Nov 25, 2008

tiny-tim

not a sum

Nooo

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

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

5. Nov 25, 2008

timscully

Got it. Much appreciated.