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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?

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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.

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  • $\begingroup$ Thanks for your reply. For the Klein-Gordon equation, however, $j_\mu$ has dimensions of $ecp_\mu/V$, where $V$ is volume, so how does that work? $\endgroup$ – dgwp Feb 19 '18 at 15:35
  • $\begingroup$ In which Klein-Gordon equation? $\endgroup$ – ndrearu Feb 19 '18 at 15:42
  • $\begingroup$ For the Klein-Gordon equation\begin{equation*} (p_\mu p^\mu-m^2c^2)\phi=0 \end{equation*}it is common to define\begin{equation*} j^\mu=2p^\mu c|N|^2 \end{equation*}where $N=1/\sqrt{V}$. $\endgroup$ – dgwp Feb 19 '18 at 16:12
  • $\begingroup$ In the KG equation you wrote there's no electromagnetic interaction and the last definition of $j^{\mu}$ is not the same current as before. $\endgroup$ – ndrearu Feb 19 '18 at 17:27
  • $\begingroup$ Yes, that is the free K-G equation, but if we add an EM interaction via minimal substitution and define the current as in the last definition, what happens with the Maxwell equation? $\endgroup$ – dgwp Feb 19 '18 at 18:17
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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 $$[A] = \mathrm{J/(Am)}.$$

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