Dual Faraday Tensor - Levi Civita Problem

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SUMMARY

The discussion focuses on calculating the Dual Faraday Tensor using the Levi-Civita symbol, specifically the equation *F^{ab} = \frac{1}{2}E^{abcd}F_{cd}*. Participants clarify the correct application of the Levi-Civita symbol, noting the values of E^(1234) and its implications for the 4x4 matrix representation of the Faraday tensor. The conversation highlights the importance of correctly identifying permutations and their associated signs, leading to a resolution of the initial confusion regarding the number of terms in the matrix.

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  • Understanding of tensor notation and operations
  • Familiarity with the Faraday tensor in electromagnetism
  • Knowledge of the Levi-Civita symbol and its properties
  • Basic concepts of permutation parity (odd/even)
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MidnightR
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Ok I'm trying to calculate the Dual Faraday Tensor, but I'm having trouble with the notation I think...

*F^{ab} = \frac{1}{2}E^{abcd}F_{cd} where E is the levi-civita symbol.

I'm correct in that

E^(1234) =

1 for (1234,3412,2341,4123)
-1 for (4321,3214,2143,1432)
0 otherwise

But this only gives me 8 terms in my 4x4 matrix? Which is incorrect?

F_cd is the Faraday tensor, which is also a 4x4 matrix.
 
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MidnightR said:
I'm correct in that

E^(1234) =

1 for (1234,3412,2341,4123)
-1 for (4321,3214,2143,1432)
0 otherwise

But this only gives me 8 terms in my 4x4 matrix? Which is incorrect?

You are both missing permutations like 1324 and you have the wrong sign on others. For example, 2341 and 4123 are both odd, while 2143 and 4321 are even.
 
You're right, thanks. It works now.
 
So 1324 is odd and 3214 is even. Correct?
 
No, because \epsilon^{1324} = - \epsilon^{3124} = - \left( - \epsilon^{3214}\right), so that when the first is odd (and it is), the last is odd as well.
 
Thank's.
 

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