D1 - four-vectors

in the Lorenz gauge:

◻ \vec{A} = μ_{0} \vec{j}

◻ ϕ = \frac{ρ}{ε_{0}} ◻ (\frac{ϕ}{c}) = \frac{ρ c}{ε_{0} c^{2}} = μ_{0} ρ c

note that $\vec{j}$ and $ρ$ are frame dependent
creating new 4-vectors for electromagnetism
4-current density:

j^{α} = (ρ c, \vec{j})

4-potential:

A^{α} = (\frac{ϕ}{c}, \vec{A})

∴ ◻ A^{α} = μ_{0} j^{α}

not ready to prove that $A^{α}$ behaves as a 4-vector, but can do for $j^{α}$
considering 'proper charge density', $ρ_{0}$ , in the rest frame of particles, where $\vec{j} = 0$
in a frame moving at $+ \vec{v}$ , the volume is contracted by $γ$ due to length contraction

ρ_{0} = \frac{Q}{V_{0}} \to ρ^{'} = \frac{Q}{V^{'}} = \frac{γ Q}{V_{0}} = γ ρ_{0}

\vec{j} = 0 \to {\vec{j}}^{'} = - γ ρ_{0} \vec{v}

at the reference frame, the charges move at $- \vec{v}$
since $u^{α} = γ (c, \vec{v}) :$

j^{α} = ρ_{0} u^{α}

$j^{α}$ is a 4-vector, as $ρ$ is an invariant scalar, and $u^{α}$ is a known 4-vector