Deriving the Electromagnetic Tensor The electromagnetic tensor is given as:
\begin{equation}
F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu
\end{equation}
How do you derive this? And how come there is a partial derivative in front of $A_\mu$? Do you multiply the derivatives or what? 
 A: I will assume that you know that we can write the magnetic and electric field as:\begin{equation}
\mathbf{B}= \mathbf{\nabla} \times \mathbf{A}
\end{equation}
\begin{equation}
\mathbf{E}=-\mathbf{\nabla}\Phi-\frac{\partial \mathbf{A}}{\partial t}
\end{equation}
where $c=1$ and I am using Heaviside-Lorentz conventions (i.e. "lazy physics" conventions). The important properties the fields have is that they are invariant under a $\mathrm{U(1)}$ gauge transformation of the form:
\begin{equation}
\mathbf{A}'=\mathbf{A}-\mathbf{\nabla} \alpha = \mathbf{A} + i e^{-i \alpha} \mathbf{\nabla} e^{i \alpha} 
\end{equation}
\begin{equation}
\Phi'=\Phi+\frac{\partial \alpha}{\partial t} = \Phi - i e^{-i \alpha} \frac{\partial}{\partial t} e^{i \alpha}
\end{equation}
We now define the define the four-vector potential, $A^\mu$, and the four-vector electric current, $j^\mu$, as follows (and I am using Minkowski signature $(+,-,-,-)$):
\begin{equation}
\begin{array}{cc}
\{A^\mu\} = \begin{pmatrix} \Phi \\ \mathbf{A}  \end{pmatrix} \; ,& \{j^\mu\} =\begin{pmatrix} \rho_e  \\ \mathbf{J}_e \end{pmatrix} 
\end{array}
\end{equation}
and we define (i.e. we do not derive) the antisymmetric electromagnetic field strength tensor as:
\begin{equation}
F^{\mu \nu} = \partial^\mu A^\nu - \partial^\nu A^\mu
\end{equation}
Now, we can evaluate:
\begin{equation}
\begin{aligned}
F^{0 i } & = \partial^0 A^i - \partial^i A^0 \\&
= \partial_0 A^i + \partial_i A^0 \\&
= - E^i
\end{aligned}
\end{equation}
and:
\begin{equation}
\begin{aligned}
F^{ij} & = \partial^i A^j - \partial^j A^i \\&
= (\delta^i{}_l \delta^j{}_m - \delta^j{}_l \delta^i{}_m) \partial^l A^m \\&
= \varepsilon_{klm}\varepsilon^{kij}\partial^l A^m \\&
= -\varepsilon_{kij} [\mathbf{\nabla} \times \mathbf{A}]^k \\&
= -\varepsilon_{ijk} B^k
\end{aligned}
\end{equation}
Therefore, we can write:
\begin{equation}
\left\{ F^{\mu \nu} \right\} = \begin{pmatrix} 
0 & -E^1 & -E^2 & -E^3 \\
E^1 & 0 & -B^3 & B^2 \\
E^2 & B^3 & 0 & -B^1 \\
E^3 & -B^2 & B^1 & 0
 \end{pmatrix}
\end{equation}
Note that the observable electromagnetic field strength tensor is invariant under a $\mathrm{U(1)}$ gauge transformation:
\begin{equation}
A'^\mu=A^\mu + \partial^\mu \alpha
\end{equation}
as is required.
Furthermore, we can define the dual of the electromagnetic field tensor, ${}^*F^{\mu \nu}$, as:
\begin{equation}
{}^*F_{\mu \nu} = \frac{1}{2} \varepsilon_{\mu \nu \sigma \tau} F^{\sigma \tau}
\end{equation}
such that the Maxwell equations can be written in a more compact and manifestly Lorentz invariant manner:
\begin{equation}
\begin{array}{cc}
\partial_\mu F^{\mu \nu } = j^\nu \; , & \partial_\mu {}^* F^{\mu \nu} =0 
\end{array}
\end{equation}
To see that these last two equations indeed satisfy the Maxwell equations is a bit of a tedious (but worthwhile) exercise that I will leave up to you.
