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Indicial notation - Levi-Cevita and Tensor

wmccunes

New member
Jan 30, 2013
8
Use indicial notation to show that:
$$
\mathcal{A}_{mi}\varepsilon_{mjk} + \mathcal{A}_{mj}\varepsilon_{imk} + \mathcal{A}_{mk}\varepsilon_{ijm} = \mathcal{A}_{mm}\varepsilon_{ijk}
$$
I'm probably missing an easier way, but my approach is to rearrange and expand on the terms:
$$
\mathcal{A}_{mi}\varepsilon_{mjk} + \mathcal{A}_{mj}\varepsilon_{mki} + \mathcal{A}_{mk}\varepsilon_{mij} = \mathcal{A}_{mm}\varepsilon_{ijk}
$$
Expanding the first term
$$
\mathcal{A}_{mi}\varepsilon_{mjk} = \varepsilon_{1jk}\mathcal{A}_{1i} + \varepsilon_{2jk}\mathcal{A}_{2i} + \varepsilon_{3jk}\mathcal{A}_{3i} =\\

\varepsilon_{123}\mathcal{A}_{11} + \varepsilon_{132}\mathcal{A}_{11} + \varepsilon_{231}\mathcal{A}_{22} + \varepsilon_{213}\mathcal{A}_{22} + \varepsilon_{312}\mathcal{A}_{33} + \varepsilon_{321}\mathcal{A}_{33} = \\

\mathcal{A}_{11} - \mathcal{A}_{11} + \mathcal{A}_{22} - \mathcal{A}_{22} + \mathcal{A}_{33} - \mathcal{A}_{33} = 0
$$
If this were correct I believe the pattern would hold for the other two terms, and the equation would equal zero...
 

wmccunes

New member
Jan 30, 2013
8
there is an easier way, of course, using indicial.
$$
\mathcal{A}_{mi}\varepsilon_{mjk} + \mathcal{A}_{mj}\varepsilon_{imk} + \mathcal{A}_{mj}\varepsilon_{ikm} = \mathcal{A}_{mk}\varepsilon_{ijk}\\
$$
multiplying all by $\varepsilon_{ijk}$ leads to kroniker delta rules, whereupon the expression can be quickly simplified...