Deriving Maxwell's Equations from Electromagnetic Tensor 
Given 
$ F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} $
It is obvious that the diagonals are zero, as 
$ F_{ii} =\partial_{i}A_{i} - \partial_{i}A_{i} = 0 $
And, setting $0$ as time and $1,2,3$ as $x,y,z$ respectively, then 
$F_{01} = F_{0x} =\frac{\partial}{\partial x^{0}}A_{1}-\frac{\partial}{\partial x^{1}}A_{0} = E_{x}
$
continuing with $F_{03}$ we obtain $E_{x} ...E_{z}$ in the first row of $F$
$F_{12} = F_{xy} = \partial_{x}A_{y}-\partial_{y}A_{x} = B_{z}$
also continuing with $F_{23} = F_{yz} $ and $F_{31} = F_{zx}$ we obtain $B_{x}$ and $ >B_{y}$
Thus we get $\triangledown \times \vec{A} = \vec{B}$
So getting the signs straight, we finally have this. 
$F_{\mu\nu} = \begin{pmatrix}
0 &  E_{x}&E_{y}  &E_{z} \\ 
-E_{x} & 0 & B_{z} &-B_{y} \\ 
 -E_{y}& -B_{z} & 0 &B_{x} \\ 
-E_{z} & B_{y} & -B_{x} & 0
\end{pmatrix}$
I understand there should be "over c" for $E$ components. 

Two Questions:


*

*Why does 


$F_{01} = F_{0x} =\frac{\partial}{\partial x^{0}}A_{1}-\frac{\partial}{\partial x^{1}}A_{0}$ result in $ E_{x}$?
$A$ is a vector potential, and I've learned that $\vec{E}$ can be represented by a $-\triangledown \phi$ where $\phi$ is a scalar potential. 


*I don't understand what I am supposed to do to with this matrix to get the two Maxwell's equations below. 


$\triangledown \cdot \vec{E} = \rho$
and 
$\triangledown \times \vec{B} = \vec{J} + \frac{\partial\vec{E}}{\partial t}$
Apparently, this can be solved by 
$\sum_{\mu}^{3}\partial_{\mu}F^{\mu\nu} = J^{\nu}$
where,
$\nu = 0, \triangledown \cdot \vec{E} = \rho$
and 
$\nu = i, \triangledown \times \vec{B} = \vec{J} + \frac{\partial\vec{E}}{\partial t}$
But where did $\sum_{\mu}^{3}\partial_{\mu}F^{\mu\nu} = J^{\nu}$ come from?
 A: The most general form of Maxwell's equations are (setting $\mu_0 = \varepsilon_0 = 1$)
\begin{align}
\vec{\nabla} \cdot \vec{B} &= 0 \\
\vec{\nabla} \times \vec{E} &= - \frac{ \partial \vec{B} }{ \partial t} \\
\vec{\nabla} \cdot \vec{E} &= \rho \\
\vec{\nabla} \times \vec{B} &=  \vec{J} +  \frac{ \partial \vec{E} }{ \partial t}
\end{align}
The first equation implies
$$
\boxed{ \vec{B} = \vec{\nabla} \times \vec{A} } 
$$
Plugging this into the second equation, we find
$$
\vec{\nabla} \times \left( \vec{E} + \frac{ \partial \vec{A} }{ \partial t} \right) = 0 
$$
This equation then solves to
$$
\boxed{ \vec{E} = - \vec{\nabla} \phi - \frac{ \partial \vec{A} }{ \partial t}  } 
$$
Plugging the boxed equations into the last two Maxwell's equations, we get
$$
\nabla^2 \phi + \frac{ \partial  }{ \partial t} (\vec{\nabla} \cdot \vec{A} ) = - \rho ~~~~~~~~ ...... (1)
$$
and
$$
\frac{ \partial^2 \vec{A} }{ \partial t^2} + \vec{\nabla} \times ( \vec{\nabla} \times \vec{A} ) + \frac{\partial }{\partial t} (\vec{\nabla}\phi) =  \vec{J}   ~~~~~~~~ ...... (2)
$$
Note that we have a total of 4 equations. In the covariant formalism, the define the 4-vectors
$$
A^\mu : = ( \phi , \vec{A}),~~~ J^\mu : = (\rho, \vec{J})
$$
All you have to do is show that the equation
$$
\partial_\mu F^{\mu\nu} = J^\nu
$$
are in fact identical to (1) and (2). [The Minkowski sign convention is here assumed to be $(+,-,-,-)$.] For instance, if I choose $\nu = 0$ in the equation above, I find
$$
J^0 = \partial_\mu F^{\mu0}  = \partial_0 F^{00} + \partial_i F^{i0} = \partial_i ( \partial^i A^0 - \partial^0 A^i )
$$
Using the definitions above, we find
$$
\rho = -\nabla^2 \phi - \frac{ \partial  }{ \partial t} (\vec{\nabla} \cdot \vec{A} )
$$
which is precisely (1).
I will leave it to you to show that if I choose $\nu = i = 1,2,3$, then I reproduce the 3 equations (2).
Thus, the equation $\partial_\mu F^{\mu\nu} = J^\nu$ "comes from" the Maxwell equations themselves. They are simply a convenient rewriting of the Maxwell equations.
A: Maxwell equations are not "derived" from this tensor.
1) This is a definition of $E_x$ via $\partial A_x/\partial t -(\nabla\phi)_x$.
The other equations are derived from the corresponding action or postulated, if you like. In the former case you can use the equations for $A_{\mu}$ if you have them from the action.
