Summary of what's gone on before. In the previous set of notes from the 6th, (there were no notes on the 7th), it was pointed out that the 'convenient' comment on page 11 of the notes was to cryptic. Today's entire half hour was spent figuring out the following derivation that sprang from the convenient comment. We want to derive:
$\epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm}  \delta_{im}\delta_{jl}$
Here's what to do. First remember sum notation gives you
$\epsilon_{ijk}\epsilon_{lmk} = \epsilon_{ij1}\epsilon_{lm1} + \epsilon_{ij2}\epsilon_{lm2} +\epsilon_{ij3}\epsilon_{lm3}$
Here's the first use of the big trick for the day. Because of the properties of the LeviCivita symbol, $\epsilon_{ijk}$, on the indices 2 and 3 will make the first term nonzero, while only the pairs 1,3 and 1,2 will make the other two terms nonzero. Once any of these combinations is chosen however, the other two terms will vanish. Given that, let's get to work on the first term, ignoring the other two that evaluate to zero.
$\epsilon_{ij1}\epsilon_{lm1}$
For this expression, the only index choices that will evaluate as nonzero are shown in the table below:
i

j

l

m

Result

2

3

3

2

1

2

3

2

3

+1

3

2

2

3

1

3

2

3

2

+1

The dudes:
No comments:
Post a Comment