Electromagnetic four-potential

Remarks

Lowering the index gives $A_\mu = (\phi / c, -A_1, -A_2, -A_3)$, and the electromagnetic tensor is $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu.$$ For instance $F_{0i} = \frac{1}{c} \big( -\partial_t A_i - \partial_i \phi \big) = E_i / c$. In static situations $\phi$ is the electric potential of electrostatics.

Example: a charge $q$ at rest at the origin has the four-potential $A^\mu = \big( \frac{q}{4 \pi \varepsilon_0 c \abs{x}}, 0, 0, 0 \big)$, which gives the Coulomb field and $B = 0$.

Non-example: for the uniform magnetic field $B = (0, 0, B_0)$ with $B_0 \neq 0$, no four-potential with spatial part $A = 0$ exists, since then $\nabla \times A = 0 \neq B$; one can take $A = \frac{B_0}{2} (-y, x, 0)$ instead.

For any $C^2$ function $\chi$, the four-potential $A^\mu - \partial^\mu \chi$, that is $\phi - \partial_t \chi$ and $A + \nabla \chi$, gives the same fields.