Minimal substitution, four-potential and units When we make the minimal substitution
\begin{equation*}
p^\mu\rightarrow p^\mu+\frac{e}{c}A^\mu
\end{equation*}
the four-potential $A^\mu$ must be proportional to $1/e$ in order to ensure the whole term has units of momentum. However, in the Maxwell equation
\begin{equation*}
\partial^\mu\partial_\mu A^\mu=\frac{j^\mu}{c}
\end{equation*}
it seems that $A^\mu$ must be proportional to $e$, in order to account for the factor of $e$ in $j^\mu$. Can anyone tell me what is going on here? What am I missing?
 A: You're just missing that there is something proportional to $e$ also in the four-potential $A_\mu$ since $A_\mu=(\phi,\vec{A})$ where $\phi$ and $\vec{A}$ are the scalar and vector potentials.
The dimensions are OK. We have that


*

*$cp_\mu$ and $eA_\mu$ are energies;

*$j_\mu/c$ is a charge density, 


then, if you multiply for $e$ at both the sides of the last equation, we have
$$
\partial_\mu\partial^\mu eA^\nu = \frac{ej^\nu}{c}
$$
and from the left member we see that both sides must be energies divided by square meters. We check that this is true for the r.h.s.: it is a square charge density and, keeping in mind that $e^2/r$ is an energy, than it is an energy divided by square meters.
A: This is a result of people using different systems of units in different parts of the physics course. Rewriting in SI,
$$p^\mu \mapsto p^\mu + e A^\mu,$$
$$\partial^\mu \partial_\mu A^\nu = \mu_0 j^\nu,$$
we can see that the vacuum permeability $\mu_0$ appears. One could say this has the potential to cancel two units of $e$ per, for example, Ampère's force law, where $\mu_0 I_1 I_2$ appear on one side and only mechanical quantities on the other. The discrepancy vanishes.
Or, better, you can easily find the exact units of all the quantities present here. Some are easy to remember or rederive from basic definitions,
$$\begin{aligned}\
[p] &= \mathrm{Js/m}, \\
[e] &= \mathrm{As}, \\
[j] &= \mathrm{A/m^2}, \\
[\mu_0] &= \mathrm{Vs/(Am)} = \mathrm{J/(A^2m)}, \\
\end{aligned}$$
and $\partial^\mu \partial_\mu$, being a second derivative by coordinates, applies $1/\mathrm{m}^2$. From either equation one can easily derive
$$[A] = \mathrm{J/(Am)}.$$
