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I read some older posts about this question, but I don't know if I'm getting it. I'm working with a Lagrangian involving some Levi Civita symbols, and when I calculate a term containing $\epsilon^{ijk}$ I obtain the contrary sign using $\epsilon_{ijk}$. I always apply the normal rules: $\epsilon_ {ijk}=\epsilon^{ijk}=1$; $\epsilon_ {jik}=\epsilon^{jik}=-1$ etc. I believed that there is no difference between covariant and contravariant Levi-Civita symbol. What do you know about this?

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Comment to the question (v2): Apart from the issue of various overall sign conventions found in the literature, note that:

  1. On one hand, there is the Levi-Civita symbol with upper (lower) indices, whose entries are only $0$s and $\pm 1$s; it is a contravariant (covariant) pseudotensor density, respectively.

  2. On the other hand, there is the Levi-Civita tensor with upper (lower) indices, whose definition differs from the Levi-Civita symbol by a factor of $\sqrt{|\det(g_{\mu\nu})|}$; it is a contravariant (covariant) pseudotensor, respectively.

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    $\begingroup$ I think the most common trend in the literature I see is for the lower-indexed symbol defined with signs given by the parity of the index permutation in the normal way, while the upper-indexed symbol differs from the lower-indexed one by $\mathrm{sgn}(\det g)$. But the OP seems to be in Euclidean $\mathbb{R}^3$ anyway. $\endgroup$
    – user10851
    Commented Jul 5, 2014 at 12:22
  • $\begingroup$ @Chris White: Possibly. Sean Carroll's GR notes agree with your sign convention, cf. Trimok's comment above. $\endgroup$
    – Qmechanic
    Commented Jul 5, 2014 at 12:29
  • $\begingroup$ Hi there, without going into the exact reasons is $\epsilon_ {ijk}=\epsilon^{ijk}=1$ or $\epsilon_ {ijk}=\epsilon^{ijk}=-1$? I'm only asking as I need to perform calculations with the contravariant and covariant forms so I need to know which one is correct, thanks! $\endgroup$
    – Electra
    Commented Oct 5, 2023 at 1:57
  • $\begingroup$ Hi @Electra. Thanks for the feedback. Be aware that different authors might have different conventions. $\endgroup$
    – Qmechanic
    Commented Oct 5, 2023 at 6:13

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