<s>[INST] The Electromagetic Tensor and Minkowski Metric Sign Convention I am trying to figure out how to switch between Minkowski metric tensor sign conventions of (+, -, -, -) to (-, +, +, +) for the electromagnetic tensor $F^{\alpha \beta}$. 
For the convention of (+, -, -, -) I know the contravariant and covarient forms of the electromagnetic tensor are:
$$ F^{\alpha \beta} =
    \begin{bmatrix}
    0 & -\frac{E_{x}}{c} & -\frac{E_{y}}{c} & -\frac{E_{z}}{c} \\
    \frac{E_{x}}{c} & 0 & -B_{z} & B_{y} \\
    \frac{E_{y}}{c} & B_{z} & 0 &  -B_{x} \\
    \frac{E_{z}}{c} & -B_{y} & B_{x} & 0 \\
    \end{bmatrix}
$$
and
$$ F_{\alpha \beta} = \eta_{\alpha \mu} F^{\mu v} \eta_{v \beta} = 
    \begin{bmatrix}
    0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\
    -\frac{E_{x}}{c} & 0 & -B_{z} & B_{y} \\
    -\frac{E_{y}}{c} & B_{z} & 0 &  -B_{x} \\
    -\frac{E_{z}}{c} & -B_{y} & B_{x} & 0 \\
    \end{bmatrix}.
$$
Now for the convention of (-, +, +, +) are the contravariant and covariant forms of the electromagnetic tensor just switched from above along with signs?:
$$ F^{\alpha \beta}= 
    \begin{bmatrix}
    0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\
    -\frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\
    -\frac{E_{y}}{c} & -B_{z} & 0 &  B_{x} \\
    -\frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\
    \end{bmatrix}
$$
and
$$ F_{\alpha \beta} = \eta_{\alpha \mu} F^{\mu v} \eta_{v \beta} =
    \begin{bmatrix}
    0 & -\frac{E_{x}}{c} & -\frac{E_{y}}{c} & -\frac{E_{z}}{c} \\
    \frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\
    \frac{E_{y}}{c} & -B_{z} & 0 &  B_{x} \\
    \frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\
    \end{bmatrix}~?
$$
Basically, I am trying to figure out how to switch between the two sign conventions.
 [/INST] I use this way:
\begin{equation}\tag{1}
F_{ab} = \partial_a \, A_b - \partial_b \, A_a,
\end{equation}
where 
\begin{equation}\tag{2}
A^a = (\phi, \, A_x, \, A_y, \, A_z), \qquad\qquad A_a = (\phi, - A_x, - A_y, - A_z).
\end{equation}
Then, we have:
\begin{align}
E_i &= \Big( -\, \vec{\nabla} \, \phi - \frac{\partial \vec{A}}{\partial t} \Big)_i, \tag{3} \\[12pt]
B_i &= (\vec{\nabla} \times \vec{A})_i. \tag{4}
\end{align}
(1) and sign convention (2) implies
\begin{equation}\tag{5}
F_{0 i} = \partial_0 \, A_i - \partial_i \, A_0 \equiv E_i.
\end{equation}
Also: $F^{0 i} = -\, E_i$.
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