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The invers of fixed Levi-Civita symbol's element

  1. Jun 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi every one, I am really confused on how to calculate the inverse of this Levi Civita symbole's element [itex]\epsilon^{tabc}[/itex], I tried to used this equation [itex]\epsilon^{\mu\nu\rho\sigma}\epsilon_{\mu\nu\alpha\beta}=\delta^{\rho}_{\alpha}\delta^{\sigma}_{\beta}-\delta^{\rho}_{\beta}\delta^{\sigma}_{\alpha}[/itex]

    2. Relevant equations

    In fact I am searching [itex]\left(\epsilon^{tabc}\right)^{-1}[/itex] where the indices should be lower.


    3. The attempt at a solution

    Based on the equation above, is it correct if I write: [itex]\epsilon_{taef}\epsilon^{tabc}=\delta^{b}_{e}\delta^{c}_{f}-\delta^{b}_{f}\delta^{c}_{e}[/itex]

    I recall that [itex]\mu\nu\rho\sigma...[/itex] are the diffeomorfism indices on curved space-time, "t" is the temporal indice and where "a,b,c..." are the spacial ones.

    I apologize for my bad english once again. I have tried to do my best to make the problem statement as clear as possible.
     
  2. jcsd
  3. Jun 20, 2014 #2

    strangerep

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    Is this part of a problem that was given to you?

    If so, you should post the full problem so helpers can see the context.

    If not, then what do you mean by the "inverse" of a 4th-rank tensor. Are you searching for a tensor (which I'll denote as ##\rho##) which satisfies:
    $$ \epsilon_{abcd} \; \rho^{abce} ~=~ \delta^e_d$$?
     
  4. Jun 21, 2014 #3
    Yes, [itex]\epsilon_{\mu\nu\rho\sigma} \; \rho^{\alpha\beta\delta\lambda} ~=~3! \delta^\lambda_\sigma[/itex]

    Where [itex]\mu,\nu,,...[/itex] are the diffeomorphism indices, once one has fixed this indices, how can I find the tensor [itex]\rho[/itex]?

    For exemple:

    [itex]T^{a}=P_{a'}\epsilon^{tabc}\epsilon^{ta'}\:_{bc}[/itex], how can I find the expression of [itex]P_{a'}[/itex] alone in fonction of [itex]T_{a}[/itex] and probably the inverse of [itex]\epsilon^{tabc}[/itex]....

    May I made the context clear? I apologize once again for my bad english. :)
     
  5. Jun 21, 2014 #4

    strangerep

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    You didn't answer my first question, so I'll ask one more time: did someone else give you this problem to solve, or did you make it up yourself?

    (I think you're still not telling me the entire question. If so, my answers might not be helpful.)

    Anyway...

    If I understand your opening post, "t" is a fixed index, so... can't your equation be written just in terms of the 3-index ##\epsilon## symbol, i.e.,

    $$T^{a}=P_{a'}\epsilon^{0abc}\epsilon^{0a'}\:_{bc} ~~?$$
    Using the earlier ##\epsilon## formula, replacing ##\sigma## and ##\beta## by 0, we get
    $$\epsilon^{\mu\nu\rho 0}\epsilon_{\mu\nu\alpha 0}=\delta^{\rho}_{ \alpha}\delta^{0}_{0}-\delta^{\rho}_{0}\delta^{0}_{\alpha} ~.$$Now, we need consider only ##\rho\ne 0## and ##\alpha\ne 0##, otherwise the equation degenerates to ##0=0##.
    Therefore, the last term above is 0, leaving ##\delta^\rho_\alpha##.

    Can you apply this to your ##T^a## equation?
     
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