Utility of gauge four-potential $A_{\mu}$ (as opposed to electric and magnetic fields ${\bf E}$ and ${\bf B}$) in E&M?

Question

The action for an electromagnetic field with source charges is given by

$$S= \int \left\{ \frac{1}{4\mu_0}F^{\mu\nu}F_{\mu\nu} - J^\mu A_\mu \right\}dx$$

By setting $dS=0$ and taking the Lorenz gauge, the resulting field equation is

$$-\Box A_\mu = \mu_0 J_\mu$$

I can see that there are a lot of conceptual advantages to thinking about E&M in terms of electromagnetic four-potentials. However, is this approach actually useful for working out problems? That is to say, is there any computational advantage to using this equation, or is it usually simpler to solve the standard form of Maxwell's equations?

Answer 1

If your question is asking whether the four-potential is more useful in classical electromagnetism from a purely computational standpoint, the answer would be no. It's not to say that it isn't useful, it's just that it only groups together two equations in the Lorentz gauge that are already useful themselves. The Lorentz gauge,

$$ \Box\phi = -\frac{\rho}{\epsilon_0} $$

$$ \Box \textbf{A} = - \mu_0 \textbf{J} $$

is a very useful gauge due to reasons that Lubos Motl pointed out. However, saying that the four-potential is any more useful would be the same as saying $\textbf{F} = m\textbf{a}$ is more useful than its components. In situations where you're solving for the fields, given the distribution of charges and currents, it's equivalent to the two equations above. For most purposes, it's only a notational convenience.

Answer 2

It isn't classical field theory, but there are a few features of using of 4-potential in QFT.

The first one is that 4-potential as 4-vector can't be used for describing massless photons. It is because the fact that it must describe massless particles leads to its transformations not as 4-vector under the Lorentz group. Specifically, $$ A^{\mu} \to \Lambda^{\mu}_{\ \nu}A^{\nu} + c\Omega_{\mu}, $$ which is the consequence of transformation of the polarization vector: $$ \varepsilon_{\mu}(p) \to \Lambda_{\mu}^{\ \nu}\varepsilon_{\nu}(p) + c p_{\mu}. $$ If all EM scattering amplitudes $M = \varepsilon_{\mu}(p)M^{\mu}$ hadn't satisfied relation $p_{\mu}M^{\mu} = 0$, we wouldn't have used 4-potential as field which describe EM field, because all of processes won't be lorentz-invariant. But it has.

The second one is that we must use 4-potential for describing the interaction with EM field due to experimental fact. If we had use field tensor $F_{\mu \nu}$ for building interaction with matter, we would get the matrix elements which become smaller with energy-momentum growth faster in compare with case of using of 4-potential. Correspondingly, in the coordinate dependence we wouldn't get the inverse square law.

Сompletely similar words can be said about the other massless fields of integer spins (helicities). For example, the correct field (I'm talk about the quantum version of linearized GR) for describing of gravitational interactions is the Weyl tensor. But we use metric (which is also transformed not as 4-tensor under the Lorentz group), because we also have interactions which is described by the inverse square law.