Einstein Summation Convention, Levi-Civita, and Kronecker delta

[tony873004]

Homework Statement

Evaluate the following sums, implied according to the Einstein Summation Convention.
$$\begin{array}{l} \delta _{ii} = \\ \varepsilon _{12j} \delta _{j3} = \\ \varepsilon _{12k} \delta _{1k} = \\ \varepsilon _{1jj} = \\ \end{array}$$

The Attempt at a Solution

$$\begin{array}{l} \delta _{ii} = \delta _{11} + \delta _{12} + \delta _{13} + \delta _{21} + \delta _{22} + \delta _{23} + \delta _{31} + \delta _{32} + \delta _{33} = 1 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 1 =3 \\ \varepsilon _{12j} \delta _{j3} = \left( {\varepsilon _{121} + \varepsilon _{122} + \varepsilon _{123} } \right)\left( {\delta _{13} + \delta _{23} + \delta _{33} } \right) = \left( {0 + 0 + 1} \right)\left( {0 + 0 + 1} \right) = \left( 1 \right)\left( 1 \right) = 1 \\ \varepsilon _{12k} \delta _{1k} = \left( {\varepsilon _{121} + \varepsilon _{122} + \varepsilon _{123} } \right)\left( {\delta _{11} + \delta _{12} + \delta _{13} } \right) = \left( {0 + 0 + 1} \right)\left( {1 + 0 + 0} \right) = \left( 1 \right)\left( 1 \right) = 1 \\ \varepsilon _{1jj} = \left( {\varepsilon _{111} + \varepsilon _{122} + \varepsilon _{133} } \right) = 0 + 0 + 0 = 0\\ \end{array}$$

Am I doing these right? Thanks!

 

[gabbagabbahey]

$$\delta _{ii} = \delta _{11} + \delta _{12} + \delta _{13} + \delta _{21} + \delta _{22} + \delta _{23} + \delta _{31} + \delta _{32} + \delta _{33} = 1 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 1 =3$$

Your final result is correct, but $\delta_{ii}=\delta_{11}+\delta_{22}+\delta_{33}$

$$\varepsilon _{12j} \delta _{j3} = \left( {\varepsilon _{121} + \varepsilon _{122} + \varepsilon _{123} } \right)\left( {\delta _{13} + \delta _{23} + \delta _{33} } \right) = \left( {0 + 0 + 1} \right)\left( {0 + 0 + 1} \right) = \left( 1 \right)\left( 1 \right) = 1$$

Not quite, $\varepsilon _{12j} \delta _{j3} =\varepsilon _{121} \delta _{13}+\varepsilon _{122} \delta _{23}+\varepsilon _{123} \delta _{33}$

$$\varepsilon _{12k} \delta _{1k} = \left( {\varepsilon _{121} + \varepsilon _{122} + \varepsilon _{123} } \right)\left( {\delta _{11} + \delta _{12} + \delta _{13} } \right) = \left( {0 + 0 + 1} \right)\left( {1 + 0 + 0} \right) = \left( 1 \right)\left( 1 \right) = 1$$

Same problem with this one.

\varepsilon _{1jj} = \left( {\varepsilon _{111} + \varepsilon _{122} + \varepsilon _{133} } \right) = 0 + 0 + 0 = 0[/tex]

Good.

 

[tony873004]

Thank you very much!

 

[latentcorpse]

$\epsilon_{12j} \delta_{j1}=\epsilon_{121}=0$ by defn.

 

[tony873004]

$\epsilon_{12j} \delta_{j1}=\epsilon_{121}=0$ by defn.

I'm not sure if I get this then. Am I supposed to be multiplying the epsilon and delta results together? Or is the answer gabbagabbahey gave for #2 the final answer. By multiplying, I get 1 for your example. Here's my latest attempt at the original questions:
Thanks, gabbagabbahey and latentcorpse!
$$\begin{array}{l} \delta _{ii} = \delta _{11} + \delta _{22} + \delta _{33} = 1 + 1 + 1 = 3 \\ \varepsilon _{12j} \delta _{j3} = \varepsilon _{121} \delta _{13} + \varepsilon _{122} \delta _{23} + \varepsilon _{123} \delta _{33} = 0\left( 0 \right) + 0\left( 0 \right) + 1\left( 1 \right) = 1 \\ \varepsilon _{12k} \delta _{1k} = \varepsilon _{121} \delta _{11} + \varepsilon _{122} \delta _{12} + \varepsilon _{123} \delta _{13} = 0\left( 1 \right) + 0\left( 0 \right) + 1\left( 0 \right) = 0 \\ \varepsilon _{1jj} = \left( {\varepsilon _{111} + \varepsilon _{122} + \varepsilon _{133} } \right) = 0 + 0 + 0 = 0 \\ \end{array}$$

 

[gabbagabbahey]

For some reason (perhaps just as an example?), latentcorpse calculated $\varepsilon_{12j} \delta_{j1}$ instead of $\varepsilon_{12j} \delta_{j3}$

And your latest attempt looks good to me!