Electromagnetism and Energy-Momentum Tensor [closed]

I'll begin with the question:

A plane electromagnetic wave propagating in the z-direction has fields $E = E_0 \hat{x}cos[\omega(t-z)], B = E_0 \hat{y}cos[\omega(t-z)]$. Find all components of the energy momentum tensor for this wave.

My attempt at a solution:

The energy momentum tensor is given by:

$$T^{\mu\nu} = -\frac{1}{4\pi}(F^{\mu\alpha}{F^{\nu}}_{\alpha} - \frac{1}{4}\eta^{\mu\nu}{F^{\alpha\beta}}F_{\alpha\beta})$$

I know $F^{\mu\alpha}$. It is $$F^{\mu\alpha} = \left[\begin{array}{cccc} 0 & -E_x & 0 & 0 \\ E_x & 0 & 0 & B_y \\ 0 & 0 & 0 & 0 \\ 0 & -B_y & 0 & 0 \end{array}\right]$$

Now in order to get ${F^{\nu}}_{\alpha}$ I need to apply the minkowski metric, right? That is ${F^{\nu}}_{\alpha} = \eta_{\alpha\mu}F^{\mu\nu}$ Where I am using the (+---) form of the metric. In which case this would leave ${F^{\nu}}_{\alpha} = F^{\mu\alpha}$, since the diagonal in the above matrix is composed of all 0s and therefore the metric has no effect. Is this correct? I still have another question about going from ${F^{\alpha\beta}}$ to $F_{\alpha\beta}$, but I would like to make sure I am on the right track first.

closed as off-topic by ACuriousMind♦, Kyle Kanos, Sebastian Riese, user10851, Daniel GriscomMar 2 '16 at 17:29

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• Since the metric is Minkowski, this is not GR. – Ryan Unger Feb 26 '16 at 4:18
• Note that $F^{\mu\nu}\ne F^{\nu\mu}$. Also, the metric does have an effect, even though the diagonal of $F^{\mu\nu}$ is all zeroes. To see that, do the matrix multiplication. – Red Act Feb 26 '16 at 5:35