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I am attempting to follow a derivation from a physics paper relating to covariant electromagnetism. It is given that,

$$ F^{\mu \nu} = u^{\mu}E^{\nu} - E^{\mu} u^{\nu} + \epsilon^{\mu \nu \alpha \beta} u_{\alpha} B_{\beta}$$

Also,

$$ B^{\mu} = - \frac{1}{\omega} \epsilon^{\mu \nu \alpha \beta} k_{\nu} u_{\alpha} E_{\beta}$$

Ultimately, I want to show that

$$ F^{\mu \nu} = \frac{1}{\omega} (k^{\mu} E^{\nu} - E^{\mu}k^{\nu})$$

Inserting the second equation into the first gives,

$$ F^{\mu \nu} = u^{\mu}E^{\nu} - E^{\mu} u^{\nu} + \frac{1}{\omega}\epsilon^{\mu \nu \alpha \beta} \epsilon_{\mu \nu \alpha \beta} k^{\nu} E^{\mu}$$

since $u_{\alpha}u^{\alpha} = -1$. But I am confused as to how to progress form here, especially in relation to the 4 dimensional levi-cevita tensors.

Thanks in advance

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    The paper you link to presents the equation above for $F^{\mu \nu}$ but also gives and equation for $* F^{\mu \nu}$. You need to insert $B^{\mu}$ into the one above (equation $(3)$ in the paper) *and* in the dual equation I mention here (Equation $(4)$ in the paper). Then a little moving around like terms will give you your answer. – 2017-02-09

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You must be careful to distinguish free indices from dummy indices. In your second equation you have

$$ B^{\mu} = - \frac{1}{\omega} \epsilon^{\mu \nu \alpha \beta} k_{\nu} u_{\alpha} E_{\beta}$$

and here the indices $\nu, \alpha $ and $\beta $ are already "in use", and they should not be mixed with $\nu, \alpha $ and $\beta $ from the first equation. This is avoided by using different indices:

$$ B^{\mu} = - \frac{1}{\omega} \epsilon^{\mu \sigma \rho \phi} k_{\sigma} u_{\rho} E_{\phi}$$

Now you can substitute to get

$$ F^{\mu \nu} = u^{\mu}E^{\nu} - E^{\mu} u^{\nu} + \frac{1}{\omega}\epsilon^{\mu \nu \alpha \beta} \epsilon_{\mu \sigma \rho \phi} k^{\sigma} u^{\rho} E^{\phi}$$

The tensors share one index and their product can be simplified to

$$\epsilon^{\mu \nu \alpha \beta} \epsilon_{\mu \sigma \rho \phi} = -\begin{vmatrix} \delta^{\nu}_{\sigma} & \delta^{\nu}_{\rho} & \delta^{\nu}_{\phi} \\ \delta^{\alpha}_{\sigma} & \delta^{\alpha}_{\rho} & \delta^{\alpha}_{\phi} \\ \delta^{\beta}_{\sigma} & \delta^{\beta}_{\rho} & \delta^{\beta}_{\phi} \end{vmatrix} $$

Expand this determinant and see what you get.

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    Thanks that is helpful! To clarify: the expression for the tensor product at the end of you answer, does that apply only if the indices run over 3 values, e.g. 1,2,3. What about if the indices run 0,1,2,3? – 2017-02-13
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    @user1887919 I figured as much. When you use Greek letters, it is understood that you are summing over {0,1,2,3}, as opposed to Roman letters when you sum over {1,2,3}. – 2017-02-13