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I Levi-Civita symbol in Minkowski Space

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  1. Apr 14, 2017 #1
    I set eyes on the next formulas:
    \begin{align}
    E_{\alpha \beta \gamma \delta} E_{\rho \sigma \mu \nu} &\equiv g_{\alpha \zeta} g_{\beta \eta} g_{\gamma \theta} g_{\delta \iota} \delta^{\zeta \eta \theta \iota}_{\rho \sigma \mu \nu} \\
    E^{\alpha \beta \gamma \delta} E^{\rho \sigma \mu \nu} &\equiv g^{\alpha \zeta} g^{\beta \eta} g^{\gamma \theta} g^{\delta \iota} \delta^{\rho \sigma \mu \nu}_{\zeta \eta \theta \iota} \\
    E^{\alpha \beta \gamma \delta} E_{\rho \beta \gamma \delta} &\equiv -6 \delta^{\alpha}_{\rho} \\
    E^{\alpha \beta \gamma \delta} E_{\rho \sigma \gamma \delta} &\equiv -2 \delta^{\alpha \beta}_{\rho \sigma} \\
    E^{\alpha \beta \gamma \delta} E_{\rho \sigma \theta \delta} &\equiv -\delta^{\alpha \beta \gamma}_{\rho \sigma \theta} \,.
    \end{align}
    in https://en.wikipedia.org/wiki/Levi-Civita_symbol#Levi-Civita_tensors I want to know for what signature (+---) or (-+++) it is given. Is there simple way to check signature?
    In my opinion there are need plus rather than minus at least in last three formulas:
    \begin{align}
    E^{\alpha \beta \gamma \delta} E_{\rho \beta \gamma \delta} &\equiv 6 \delta^{\alpha}_{\rho} \\
    E^{\alpha \beta \gamma \delta} E_{\rho \sigma \gamma \delta} &\equiv 2 \delta^{\alpha \beta}_{\rho \sigma} \\
    E^{\alpha \beta \gamma \delta} E_{\rho \sigma \theta \delta} &\equiv \delta^{\alpha \beta \gamma}_{\rho \sigma \theta} \,.
    \end{align}
     
  2. jcsd
  3. Apr 14, 2017 #2

    andrewkirk

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    In Minkowski Space, under the standard basis, changing the signature of the metric simply changes the sign of the metric tensor components, between (-1, 1, 1, 1) and (1, -1, -1, -1). Since formulas (1) and (2) above involve the multiplication of four metric components, changing the sign of all of them will make no difference (as ##(-1)^4=1##).

    So I think the formulas are correct regardless of which signature is chosen.
     
  4. Apr 15, 2017 #3
    What about (3)-(5)? I think that it is incorrectly.
     
  5. Apr 15, 2017 #4

    vanhees71

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    The - signs come from the fact that, ##\det \eta_{\mu \nu}=-1## (no matter which convention is used). There's endless trouble with the Levi-Civita symbol because of this damn sign, but you can't help it. Causality structure dicatates that Minkowski space's fundamental form must have signature (1,3) or (3,1). The trouble with the Levi-Civita symbol comes from the fact that the absolute sign depends on whether you define the fully contravariant or covariant components as the sign of the permutation of the indices from lexical order. In most HEP books one uses
    $$\epsilon^{\mu \nu \rho \sigma}=\delta^{\mu \nu \rho \sigma}_{0123}$$
    and
    $$\epsilon_{\mu \nu \rho \sigma}=g_{\mu \alpha} g_{\nu \beta} g_{\rho \gamma} g_{\sigma \delta} \epsilon^{\alpha \beta \gamma \delta}=-\delta_{\mu \nu \rho \sigma}^{0123}.$$
    Again, the sign doesn't depend on whether you use west or east-coast metric.
     
  6. Apr 15, 2017 #5
    Thank you for answer, but what about (3)-(5)?. My question appeared from a fact that if I use (5) than I come to a negative square of an amplitude. And if I use (8) than square of an amplitude is positive one.
     
  7. Apr 15, 2017 #6

    vanhees71

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    The minus sign must be there because of the definition of the Levi-Civita tensor components (I wrote ##\epsilon## instead of ##E##). For an amplitude an overall sign doesn't matter. It's just a phase factor.

    Where this sign matters is in relative contributions to amplitudes. Famously there was a sign error in some evaluation of the anomalous magnetic moment of the muon, and different groups of theoreticians had a hard time to find out who is right. At the end it turned out that someone had used another sign convention for the Levi-Civita tensor than it was implemented in the computer algebra system FORM, and that's where the confusion came from. As I said, this sign conventions are a nuissance and can drive you nuts, looking for the correct signs, but it's necessary, and one must just check again and again whether ones sign conventions are used consistently, particularly when working with different textbooks or papers from different authors.
     
  8. Apr 15, 2017 #7
    Yes, for an amplitude an overall sign doesn't matter but my an amplitude is proportional Levi-Civita tensor and its square proportional Levi-Civita tensor on Levi-Civita tensor. Or for square of the amplitude does overall sign not matter too?
     
  9. Apr 15, 2017 #8

    vanhees71

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    It's the modulus squared. So it's always positive semidefinite!
     
  10. Apr 15, 2017 #9
    My square of the amplitude is $$ \epsilon_{0}^{\sigma\lambda\rho} \epsilon_{0}^{\mu\nu\alpha}g_{\rho\alpha}p_{\sigma}q_{\lambda}p_{\mu}q_{\nu}$$ and its sign depends of defenation (5) or (8). Don't I understand something?
     
  11. Apr 15, 2017 #10

    vanhees71

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    Isn't this the amplitude? Anyway, which amplitude is this in which QFT model?
     
  12. Apr 15, 2017 #11
    No, my amplitude is $$ \epsilon_{0}^{\sigma\lambda\rho} q_{\sigma} p_{\lambda} e_{\nu}$$ where $$e_{\rho} e_{\alpha}=g_{\rho,\alpha}$$ and my model is effective Wess-Zumino-Witten action.
     
  13. Apr 15, 2017 #12

    vanhees71

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    But then you take the modulus squared, and all ambiguity of signs are gone.
     
  14. Apr 15, 2017 #13
    But is module from $$ \epsilon_{0}^{\sigma\lambda\rho} q_{\sigma} p_{\lambda} e_{\nu}$$ just $$ \epsilon_{0}^{\sigma\lambda\rho} q_{\sigma} p_{\lambda} e_{\nu}$$?
     
  15. Apr 20, 2017 #14
    Do you understand what I'm saying?
     
  16. Apr 21, 2017 #15

    vanhees71

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    Obviously not. The usual way is

    (a) define a QFT model in terms of a Lagrangian
    (b) derive the Feynman rules from it
    (c) define the process you want to evaluate the S-matrix element to and use the Feynman rules from (b)

    Try to explain your problem in this way to me. I'm pretty sure that after that you'll have answered your question yourself ;-).
     
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