4-Vector Potential Notation How am I supposed to interpret this notation:
$$F^{uv} = \partial^uA^v-\partial^vA^u$$
I know that $\partial^u = (\frac{1}{c}\frac{\partial}{\partial t},-
\vec\nabla)$
So for example for the potential $$A=\left(\begin{matrix}
0 & 0 & 0& E_z\\ 0 & 0 & B_y & 0\\ 0 & -B_x & 0 & 0\\ E_z & 0&0&0\end{matrix}\right)$$
So to compute $F^{23} = \partial^u\left(\begin{matrix}0\\B_y\\0\\0\end{matrix}\right) -\partial^v  \left(\begin{matrix}0&0&B_y&0\end{matrix}\right) = 0 - - B = B$.
Is this the correct way to do it? I'm just getting confused by the notation. 
 A: The indices come with ordering $(0,1,2,3)$ so that $\partial^0=\frac{1}{c}\frac{\partial}{\partial t}$, $\partial^1=-\frac{\partial}{\partial x}$ etc. 
$A^\mu$ is a 4-vector with components $(A^0,A^1,A^2,A^3)$, 
not a matrix as your notation suggest.  Thus, in your specific example,
$$
F^{23}=\partial^{2}A^3-\partial^3 A^2= 
-\frac{\partial}{\partial y}A_z+\frac{\partial}{\partial z}A_y\, .
$$
A: The 4-potential $A_\mu$ is a four-vector, not a matrix. Set the speed of light $c=1$ and it is defined as
$$
A^\mu = (\phi,\vec{A})\,,
$$
in which $\phi$ is the electric potential and $\vec{A}$ vector potential. The magnetic field is given by 
$$
\vec{B} = \nabla \times \vec{A}\,,
$$
and the electric field is given by
$$
\vec{E} = -\frac{\partial \vec{A}}{\partial t}-\nabla \phi\,.
$$
Thus the electromagnetic tensor (as you write in the question description) is 
$$
F^{\mu\nu} = \partial^\mu A^\nu-\partial^\nu A^\mu = \left(\begin{array}{cccc}
0&-E_1&-E_2&-E_3\\
E_1&0&-B_3&B_2\\
E_2&B_3&0&-B_1\\
E_3&-B_2&B_1&0
\end{array}\right)\,.
$$
This tensor is gauge invariant. Under the transformation $A_\mu\rightarrow A_\mu+\partial_\mu\chi$, $F^{\mu\nu}$ will not change.
The Lagrangian of electromagnetic field is 
$$
\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + A_\mu J^{\mu}
$$
and the corresponding Euler-Lagrange equation is Maxwell equation
$$
\frac{\partial \mathcal{L}}{\partial A_\nu}-\partial_\mu\frac{\partial \mathcal{L}}{\partial \partial_\mu A_\nu}=0~~\Rightarrow ~~\partial_\mu F^{\mu\nu} = J^\nu\,.
$$
Actually, the 4-vector potential $A_\mu$ can be viewed as the connection on a fiber bundle (something like the Christoffel symbol in general relativity), and thus the tensor $F_{\mu\nu}$ is the curvature. In quantum field theory we have the covariant derivative $D_\mu=\partial_\mu + ieA_\mu$, it is similar to the covariant derivative on curved spacetime $\nabla_\mu=\partial_\mu+\Gamma_{\mu\nu}^\lambda$.
