The invers of fixed Levi-Civita symbol's element

[merhalag]

Homework Statement

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]

Homework Equations

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

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 spatial 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.

 

[strangerep]

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:
$$ \epsilon_{abcd} \; \rho^{abce} ~=~ \delta^e_d$$?

 

[merhalag]

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:
$$ \epsilon_{abcd} \; \rho^{abce} ~=~ \delta^e_d$$?

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. :)

 

[strangerep]

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...

[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]...

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?

 

[paranormal]

Dear student,

The inverse of the Levi-Civita symbol element \epsilon^{tabc} can be calculated using the following steps:

1. Rewrite the Levi-Civita symbol in terms of the Kronecker delta symbol \delta^{\mu\nu} as \epsilon^{tabc} = \frac{1}{4!}\epsilon_{\mu\nu\rho\sigma}\delta^{\mu}_{t}\delta^{\nu}_{a}\delta^{\rho}_{b}\delta^{\sigma}_{c}.

2. Use the equation you mentioned, \epsilon_{\mu\nu\rho\sigma}\epsilon^{\mu\nu\alpha\beta}=\delta^{\rho}_{\alpha}\delta^{\sigma}_{\beta}-\delta^{\rho}_{\beta}\delta^{\sigma}_{\alpha}, to simplify the expression.

3. Substitute the values of the indices in the equation to get \epsilon_{taef}\epsilon^{tabc}=\delta^{b}_{e}\delta^{c}_{f}-\delta^{b}_{f}\delta^{c}_{e}.

4. Rearrange the terms to get \epsilon^{tabc}=\frac{1}{4!}\epsilon_{taef}\epsilon^{tabc}=\frac{1}{4!}(\delta^{b}_{e}\delta^{c}_{f}-\delta^{b}_{f}\delta^{c}_{e}).

5. Finally, solve for \epsilon^{tabc} to get \epsilon^{tabc}=\frac{1}{4!}(\delta^{b}_{e}\delta^{c}_{f}-\delta^{b}_{f}\delta^{c}_{e})^{-1}.

I hope this helps to clarify the process for calculating the inverse of the Levi-Civita symbol element. Please let me know if you have any further questions or concerns. Keep up the good work with your studies!

Best regards,