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Is the Levi-Civita symbol a tensor?

R. A. Sharipov afirm (In "Quick Introduction to Tensor Analysis", page 30) that "...the Levi-Civita symbol is NOT a tensor..."

$\epsilon_{jkq}=\epsilon^{jkq}=\left\{\begin{array}{rl} 0, & \mbox{if among $j$, $k$, $q$ there are at least two equal numbers} \\ 1, & \mbox{if $(j,k,q)$ is even permutation of numbers $(1,2,3)$} \\ -1, & \mbox{if $(j,k,q)$ is odd permutation of numbers} \end{array}\right.$

What does that phrase mean?

Thanks!

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  • $\begingroup$ If you define a rank-2 tensor $A$ as an object that transforms as $A_{ij}' = a_{ik} a_{jl} A_{kl}$ then a pseudotensor transforms as $A_{ij}' = J a_{ik} a_{jl} A_{kl}$ where $J = \det A$. The permutation tensor is a rank-3 pseudotensor, see "Classical Mechanics" by Goldstein for the general form. $\endgroup$ – Biswajit Banerjee Apr 28 '14 at 4:02
  • $\begingroup$ But I need to understand the words of Sharipov in the context of his article Sharipov . ¿Does he want to say: "Despite its form or notation will not be a tensor? $\endgroup$ – Esteban May 5 '14 at 1:31
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I had this same question and I found this: http://www.physics.ucc.ie/apeer/PY4112/Tensors.pdf Page 11, it explains that the Levi-Civita Symbol is a Tensor Density and transforms it using to a tensor the same way that Sharipov did it (and explains the steps).

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    $\begingroup$ Welcome to Math.SE, please avoid external links in your answer (if one day the link doesn't work, your answer becomes useless). $\endgroup$ – user37238 Jul 1 '14 at 13:32

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