The invers of fixed Levi-Civita symbol's element

In summary: If so, can you also solve for ##\delta^\rho_\beta##?Yes, \epsilon_{\mu\nu\rho\sigma} \; \rho^{\alpha\beta\delta\lambda} ~=~3! \delta^\lambda_\sigmaWhere \mu,\nu,,... are the diffeomorphism indices, once one has fixed this indices, how can I find the tensor \rho?You can use the same approach to find ##\delta^\rho_\beta##.
  • #1
merhalag
2
0

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.
 
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  • #2
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:
$$ \epsilon_{abcd} \; \rho^{abce} ~=~ \delta^e_d$$?
 
  • #3
strangerep said:
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. :)
 
  • #4
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...

merhalag said:
[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?
 
  • #5


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,
 

Related to The invers of fixed Levi-Civita symbol's element

1. What is the inverse of the fixed Levi-Civita symbol's element?

The inverse of the fixed Levi-Civita symbol's element is defined as the element that when multiplied by the original element results in the identity element of the group.

2. Why is the inverse of the fixed Levi-Civita symbol's element important?

The inverse of the fixed Levi-Civita symbol's element is important because it allows for the solution of equations involving the Levi-Civita symbol and its inverse, which are commonly used in physics and mathematics.

3. How is the inverse of the fixed Levi-Civita symbol's element calculated?

The inverse of the fixed Levi-Civita symbol's element can be calculated by using the properties of the symbol and its inverse, such as its anti-symmetry and the fact that it takes on values of 0, 1, or -1.

4. Can the inverse of the fixed Levi-Civita symbol's element be a scalar?

No, the inverse of the fixed Levi-Civita symbol's element cannot be a scalar. It is a special type of element known as a tensor, which has both magnitude and direction.

5. Are there any applications of the inverse of the fixed Levi-Civita symbol's element?

Yes, the inverse of the fixed Levi-Civita symbol's element has many applications in physics and mathematics, such as in the calculation of determinants, cross products, and the solution of differential equations.

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