Lorentz transformation of electromagnetic 4-potential I'm looking for the exact correspondence between
Lorentz transfer four vector 

and the four vector of scalar and vector potential $A^\mu = (\phi(t,\vec{x}),\vec{A}(t,\vec{x}))^{T}$.
Does $ct=A(t), x=\phi(t),y=\phi(x), z=\phi(x)$?
 A: Covariant notation is a simple way to say how something transforms under Lorentz. An object with an index, e.g. $z^\mu=(z^0,\vec z$), transforms under Lorentz as,
$$
{z^{\prime}}^{\nu} = {\Lambda^\nu}_\mu z^\mu
$$
where $\Lambda^\mu_\nu$ is the matrix you have written down in your question. $z^\mu$ is called a four-vector. This transformation property is true regardless of whther $z^\mu$ represents position $z^\mu=x^\mu=(ct,\vec x)$, momentum $z^\mu=p^\mu=(E/c,\vec p)$, the four-vector potential $z^\mu=A^\mu=(\phi/c,\vec A)$, or anything else that is a four-vector.
The correct identification for similar components of a four-vector between position and four potential would be,
$$ 
ct \leftrightarrow \phi/c\\
x \leftrightarrow A_x\\
y \leftrightarrow A_y\\
z \leftrightarrow A_z\\
$$
i.e. $ct \leftrightarrow \phi/c$ and $\vec x \leftrightarrow \vec A$.
I advise you read a few chapters on covariant notation in special relativity. You should be able to figure out how $A^\mu$ transforms because you know how $x^\mu$ transforms and you know that all four-vectors transform in the same way.
