Derivation of the Electromagnetic Stress-Energy Tensor in Flat Space-time I am working on deriving the electromagnetic stress energy tensor using the electromagnetic tensor in the $(-, +, +, +)$ sign convention. However, I have hit a snag and cannot figure out where I have gone wrong.
$$ F^{\mu \alpha}= 
    \begin{bmatrix}
    0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\
    -\frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\
    -\frac{E_{y}}{c} & -B_{z} & 0 &  B_{x} \\
    -\frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\
    \end{bmatrix}
$$
$$
F^{\mu}_{\alpha} = 
    \begin{bmatrix}
    0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\
    \frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\
    \frac{E_{y}}{c} & -B_{z} & 0 &  B_{x} \\
    \frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\
    \end{bmatrix}
$$
$$ T^{\mu\nu} = \frac{1}{\mu_0}(F^{\mu \alpha}F^{v}_{\alpha} - \frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha \beta})$$
Doing matrix multiplication of the matrices $F^{\mu \alpha}$ and $F^{\nu}_{\alpha}$ from above gives
$$
F^{\mu \alpha}F^{\nu}_{\alpha} =
 \begin{bmatrix}
 (\frac{E}{c})^{2} & -B_{z}\frac{E_{y}}{c} + B_{y}\frac{E_{z}}{c}  & \frac{E_{x}}{c}B_{z} - \frac{E_{z}}{c}B_{x} & -\frac{E_{x}}{c}B_{y} + \frac{E_{y}}{c}B_{x} \\
 B_{z}\frac{E_{y}}{c} - B_{y}\frac{E_{z}}{c} & -B_{z}^{2} - B_{y}^{2} - (\frac{E_{x}}{c})^{2} & -\frac{E_{x}}{c}\frac{E_{y}}{c} + B_{y}B_{x} & \frac{E_{x}}{c}\frac{E_{z}}{c} + B_{z}B_{x} \\
 -B_{z}\frac{E_{x}}{c} + B_{x}\frac{E_{z}}{c} & -\frac{E_{y}}{c}\frac{E_{x}}{c} + B_{x}B_{y} & -(\frac{E_{y}}{c})^{2}-B_{z}^{2}-B_{x}^{2} & -\frac{E_{y}}{c}\frac{E_{z}}{c} + B_{z}B_{y} \\
 B_{y}\frac{E_{x}}{c} - B_{x}\frac{E_{y}}{c} & -\frac{E_{z}}{c}\frac{E_{x}}{c} + B_{x}B_{z} & -\frac{E_{z}}{c}\frac{E_{y}}{c} + B_{y}B_{z} & -(\frac{E_{z}}{c})^{2}-B_{y}^{2}-B_{x}^{2} \\
 \end{bmatrix}
$$
Subtracting the $\frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}= \frac{1}{4}\eta^{\mu\nu}[2(B^{2} - (\frac{E}{c})^{2})]$ and multiplying by $\frac{1}{\mu_{0}}$ gives
$$ T^{\mu\nu}=\frac{1}{\mu_{0}}  
   \begin{bmatrix}
 (\frac{E}{c})^{2} + \frac{1}{2}(B^{2} - (\frac{E}{c})^{2}) & -B_{z}\frac{E_{y}}{c} + B_{y}\frac{E_{z}}{c}  & \frac{E_{x}}{c}B_{z} - \frac{E_{z}}{c}B_{x} & -\frac{E_{x}}{c}B_{y} + \frac{E_{y}}{c}B_{x} \\
 B_{z}\frac{E_{y}}{c} - B_{y}\frac{E_{z}}{c} & -B_{z}^{2} - B_{y}^{2} - (\frac{E_{x}}{c})^{2} - \frac{1}{2}(B^{2} - (\frac{E}{c})^{2}) & -\frac{E_{x}}{c}\frac{E_{y}}{c} + B_{y}B_{x} & \frac{E_{x}}{c}\frac{E_{z}}{c} + B_{z}B_{x} \\
 -B_{z}\frac{E_{x}}{c} + B_{x}\frac{E_{z}}{c} & -\frac{E_{y}}{c}\frac{E_{x}}{c} + B_{x}B_{y} & -(\frac{E_{y}}{c})^{2}-B_{z}^{2}-B_{x}^{2} - \frac{1}{2}(B^{2} - (\frac{E}{c})^{2}) & -\frac{E_{y}}{c}\frac{E_{z}}{c} + B_{z}B_{y} \\
 B_{y}\frac{E_{x}}{c} - B_{x}\frac{E_{y}}{c} & -\frac{E_{z}}{c}\frac{E_{x}}{c} + B_{x}B_{z} & -\frac{E_{z}}{c}\frac{E_{y}}{c} + B_{y}B_{z} & -(\frac{E_{z}}{c})^{2}-B_{y}^{2}-B_{x}^{2} - \frac{1}{2}(B^{2} - (\frac{E}{c})^{2}) \\
 \end{bmatrix}
$$
However, the textbook definition of the electromagnetic stress energy tensor is:
$$ T^{\mu\nu} =
    \begin{bmatrix}
    \frac{1}{2}(\epsilon_{0} |E|^{2} + \frac{1}{\mu_{0}}|B|^{2}) & \frac{S_{x}}{c} & \frac{S_{y}}{c} & \frac{S_{z}}{c} \\
    \frac{S_{x}}{c} & -\sigma_{xx} & -\sigma_{xy}  & -\sigma_{xz} \\
    \frac{S_{y}}{c} & -\sigma_{yx} & -\sigma_{yy}  & -\sigma_{yz} \\
\frac{S_{z}}{c} & -\sigma_{zx} & -\sigma_{zy} & -\sigma_{zz} \\
    \end{bmatrix}
$$
with $\vec{S} = \frac{1}{\mu_{0}}(\vec{E} \times \vec{B})$ and $\sigma_{ij} = \epsilon_{0} E_{i}E_{j} + \frac{1}{\mu_{0}}B_{i}B_{j} - \frac{1}{2}(\epsilon_{0} E^{2} + \frac{1}{\mu_{0}}B^{2})\delta_{ij} $
So, I know my $T^{01} = T^{10}$, $T^{02} = T^{20}$, and $T^{03} = T^{30}$ but they do not. They are of opposite signs. What did I do incorrectly?

Ok...I have done a little more work and I think I have it.
If 
$$ F^{\mu \alpha}= 
    \begin{bmatrix}
    0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\
    -\frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\
    -\frac{E_{y}}{c} & -B_{z} & 0 &  B_{x} \\
    -\frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\
    \end{bmatrix}
$$
and
$$ T^{mv} = \frac{1}{\mu_0}(F^{\mu \alpha}F^{\nu}{}_{\alpha} - \frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha \beta})$$
then letting c = 1
$$
F^{\mu \alpha}F^{v}{}_{\alpha} = \eta_{\beta\alpha}F^{\mu\alpha}F^{\nu\beta}
$$
If we let $\mu = 0,1,2,3$, $\nu = 0, 1, 2, 3$, and $\alpha=\beta$ summing over repeated indexes gives
for $\mu = 0$ and $\nu = 0$
$$
\eta_{\beta \alpha}F^{0\alpha}F^{0\beta} = \eta_{00}F^{00}F^{00}+\eta_{11}F^{01}F^{01}+ \eta_{22}F^{02}F^{02} + \eta_{33}F^{03}F^{03} = E_{x}^{2} + E_{y}^{2} + E_{z}^{2}
$$
for $\mu = 1$ and $\nu = 0$
$$
\eta_{\beta \alpha}F^{1\alpha}F^{0\beta} = \eta_{00}F^{10}F^{00}+\eta_{11}F^{11}F^{01}+ \eta_{22}F^{12}F^{02} + \eta_{33}F^{13}F^{03} = B_{z}E_{y} - B_{y}E_{z}
$$
and so on...
Is this correct? When looking at the indexes for the metric will $\alpha=\beta$ always be true when working in general relativity?
