# Einstein Summation Convention, Levi-Civita, and Kronecker delta

1. Sep 7, 2009

### tony873004

1. The problem statement, all variables and given/known data
Evaluate the following sums, implied according to the Einstein Summation Convention.
$$\begin{array}{l} \delta _{ii} = \\ \varepsilon _{12j} \delta _{j3} = \\ \varepsilon _{12k} \delta _{1k} = \\ \varepsilon _{1jj} = \\ \end{array}$$

3. 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!

2. Sep 8, 2009

### gabbagabbahey

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

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

Same problem with this one.

Good.

3. Sep 8, 2009

### tony873004

Thank you very much!

4. Sep 8, 2009

### latentcorpse

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

5. Sep 8, 2009

### tony873004

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

6. Sep 8, 2009

### 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!