- #1
S. Moger
- 53
- 2
I'm new to working with tensors, and feel a bit uneasy about the nomenclature. I picture words like antisymmetry in terms of average random matrices where no symmetry can be found at all. However, if I understand it correctly, antisymmetry is a type of symmetry, but where signs are inverted. So with this nomenclature Reagan, in that famous joke, is right to label anticommunists as a type of communists. But, anyway that's just linguistic mishmash.
1. Homework Statement
[itex]A_{ij} = k \epsilon_{ijk} a_k[/itex]
k is a constant. [itex]\vec{a}[/itex] is a vector.
What must k be to fulfill [itex]A_{ij} A_{ij} = |\vec{a}|^2[/itex] ?
The dual use of k also confuses me somewhat but from what I understand the only constant is the non-index k.
Is it correct to assume that [itex]\vec{a}[/itex] is taken to exist in three dimensions, from the use of the 3-index levi cevita?
[itex]A_{ij} A_{ij} = k \epsilon_{ijk} a_k \cdot k \epsilon_{ijk} a_k = k^2 \epsilon_{ijk} \epsilon_{ijk} a_k a_k = 6 k^2 a_k a_k[/itex]
So if k is [itex]\pm \frac{1}{\sqrt{6}}[/itex] it appears to fulfill the requirements, but this isn't the correct answer.
1. Homework Statement
[itex]A_{ij} = k \epsilon_{ijk} a_k[/itex]
k is a constant. [itex]\vec{a}[/itex] is a vector.
What must k be to fulfill [itex]A_{ij} A_{ij} = |\vec{a}|^2[/itex] ?
The Attempt at a Solution
The dual use of k also confuses me somewhat but from what I understand the only constant is the non-index k.
Is it correct to assume that [itex]\vec{a}[/itex] is taken to exist in three dimensions, from the use of the 3-index levi cevita?
[itex]A_{ij} A_{ij} = k \epsilon_{ijk} a_k \cdot k \epsilon_{ijk} a_k = k^2 \epsilon_{ijk} \epsilon_{ijk} a_k a_k = 6 k^2 a_k a_k[/itex]
So if k is [itex]\pm \frac{1}{\sqrt{6}}[/itex] it appears to fulfill the requirements, but this isn't the correct answer.