The electromagnetic field tensor
The Maxwell's equations are
$$(1)\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}$$
$$(2)\nabla\cdot\mathbf{B}=0$$
$$(3)\nabla \times\mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$$
$$(4)\nabla \times\mathbf{B}=-\mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}+\mu_0\mathbf{J}$$
From (3) we get:
$$\nabla\times\mathbf{E}=-\frac{\partial}{\partial t}(\nabla\times\mathbf{A})[\because \mathbf{B}=\nabla\times\mathbf{A}]\Rightarrow\nabla\times \left(\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}\right )=0$$$$\Rightarrow\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}=-\nabla\phi\Rightarrow\mathbf{E}=
-\left(\frac{\partial\mathbf{A}}{\partial t}+\nabla\phi\right)$$
Now electromagnetic field tensor defined as $$F^{\mu\nu}=\partial ^{\nu}A^{\mu}-\partial ^{\mu}A^{\nu}=\frac{\partial A^{\mu}}{\partial x_{\nu}}-\frac{\partial A^\nu}{\partial x_{\mu}}$$
Electromagnetic Four Potential $$ \space A^{\mu\nu}=\{\frac{\phi}{c},A_x,A_y,A_z\}$$
$$F^{00}=\frac{\partial (\phi c )}{\partial (ct)}-\frac{\partial (\phi c)}{\partial (ct)}=0$$
Similarly all diagonal elements of electromagnetic field tensors are equal to zero.
$$F^{01}=\frac{\partial (\phi/c )}{\partial(-x)}-\frac{\partial A_x}{\partial (ct)}=-\frac{1}{c}\left( \frac{\partial \phi }{\partial x}+\frac{\partial A_x}{\partial t} \right)=\frac{E_x}{c}$$
And so on we get the following matrix of electromagnetic field tensor:
$$F^{\mu\nu}=\begin{bmatrix} 0 & E_x/c & E_y/c & E_z/c\\-E_x/c & 0 & B_z & -B_y\\-E_y/c & -B_z & 0 & B_x\\-E_z/c & B_y & -B_x & 0\end{bmatrix}$$
now $$F_{\mu\nu}=\eta_{\mu\alpha}F^{\alpha\beta}\eta_{\beta\nu}$$ as
$$Minkowski\space metric\space tensor\space\space \eta_{\mu\nu}=\eta^{\mu\nu}=\begin{bmatrix} 1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\0 & 0 & 0 & -1\end{bmatrix}[+\space-\space-\space-]$$ so
$$F_{\mu\nu}=\begin{bmatrix} 0 & -E_x/c & -E_y/c & -E_z/c\\E_x/c & 0 & -B_z & B_y\\E_y/c & B_z & 0 & -B_x\\E_z/c & -B_y & B_x & 0\end{bmatrix}$$
$$(1)\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}$$
$$(2)\nabla\cdot\mathbf{B}=0$$
$$(3)\nabla \times\mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$$
$$(4)\nabla \times\mathbf{B}=-\mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}+\mu_0\mathbf{J}$$
From (3) we get:
$$\nabla\times\mathbf{E}=-\frac{\partial}{\partial t}(\nabla\times\mathbf{A})[\because \mathbf{B}=\nabla\times\mathbf{A}]\Rightarrow\nabla\times \left(\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}\right )=0$$$$\Rightarrow\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}=-\nabla\phi\Rightarrow\mathbf{E}=
-\left(\frac{\partial\mathbf{A}}{\partial t}+\nabla\phi\right)$$
Now electromagnetic field tensor defined as $$F^{\mu\nu}=\partial ^{\nu}A^{\mu}-\partial ^{\mu}A^{\nu}=\frac{\partial A^{\mu}}{\partial x_{\nu}}-\frac{\partial A^\nu}{\partial x_{\mu}}$$
Electromagnetic Four Potential $$ \space A^{\mu\nu}=\{\frac{\phi}{c},A_x,A_y,A_z\}$$
$$F^{00}=\frac{\partial (\phi c )}{\partial (ct)}-\frac{\partial (\phi c)}{\partial (ct)}=0$$
Similarly all diagonal elements of electromagnetic field tensors are equal to zero.
$$F^{01}=\frac{\partial (\phi/c )}{\partial(-x)}-\frac{\partial A_x}{\partial (ct)}=-\frac{1}{c}\left( \frac{\partial \phi }{\partial x}+\frac{\partial A_x}{\partial t} \right)=\frac{E_x}{c}$$
And so on we get the following matrix of electromagnetic field tensor:
$$F^{\mu\nu}=\begin{bmatrix} 0 & E_x/c & E_y/c & E_z/c\\-E_x/c & 0 & B_z & -B_y\\-E_y/c & -B_z & 0 & B_x\\-E_z/c & B_y & -B_x & 0\end{bmatrix}$$
now $$F_{\mu\nu}=\eta_{\mu\alpha}F^{\alpha\beta}\eta_{\beta\nu}$$ as
$$Minkowski\space metric\space tensor\space\space \eta_{\mu\nu}=\eta^{\mu\nu}=\begin{bmatrix} 1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\0 & 0 & 0 & -1\end{bmatrix}[+\space-\space-\space-]$$ so
$$F_{\mu\nu}=\begin{bmatrix} 0 & -E_x/c & -E_y/c & -E_z/c\\E_x/c & 0 & -B_z & B_y\\E_y/c & B_z & 0 & -B_x\\E_z/c & -B_y & B_x & 0\end{bmatrix}$$
