Can Maxwell's equations be generalized to all fields?

Question

For having studied both classical and quantum optics, I regard Maxwell's equations as the grand "cheat sheet" from which (almost) all optical/photonic phenomena can be derived. Yet, I also know that wave-particle duality extends to all other fields and particles in the standard model. I'm therefore left with a nagging sense that Maxwell's equations should---up to differing units---be universal (cf. e.g., Gauss' law for gravitation).

I expect that some behaviors (say, the Aharonov-Bohm effect) won't be observable since some particles in the standard model are charged while others aren't, or that monopoles may exist while others don't, etc. That said, don't we have any evidence that the overall template of Maxwell's equations is universal?

Answer 1

Maxwell's equations describe a massless vector (spin-1) U1-gauge field (the photon-field). Other particles have different properties (spin, mass, coupling to other fields), and have different equations of motion.

Answer 2

Maxwell's equations can be written as a massless wave equation and this is a special case of Einstein's energy-momentum relation in wave form. The general case is the Klein-Gordon equation, which is satisfied by any free non-interacting field.

Answer 3

This is because of special relativity. The fact that the speed of light is finite can be packaged nicely im saying that we live in a space where the notion of proper distance makes sense:

\begin{equation} ds^2=-c^2dt^2+ dx^2+dy^2+dz^2 \end{equation} In particular, the only things we can talk about are things which are Lorentz/Poincare invariant/covariant. In other words, the symmetries of Minkowski space fix the form of Maxwell’s equations and their analog for other matter. This is summarized by the so-called representation theory of the Poincare group. These are described by quantum numbers like spin and momentum. Maxwell theory is a spin-1 massless representation for example, and the Maxwell equations are the simplest equations of motion consistent with special relativity. The Dirac equation comes from the spin-1/2 representations, and the Klein-Gordon equation comes from the spin-0 representation. If you loosen up your assumptions and impose the equivalence principle instead, you’ll end up getting Einstein gravity. So your suspicion has a natural explanation, the possibilities consistent with the symmetries we see in nature are quite constrained.

Answer 4

To add a bit to the story:

Any conserved four vector implies the mathematical existence of fields satisfying Maxwell's equations [1]. For this reason, I like to think of Maxwell's equations as an expression of charge conservation.

One example arises in acoustics where pressure and velocity obey a continuity equation $\partial_t p + \vec \nabla \cdot \vec v = 0$, which implies the existence of an electric like field $\vec x$ that acts like a linear displacement field while the magnetic like field $\vec b$ acts something like a rotational displacement field, which together satisfy Maxwell's equations:

$p = -\vec \nabla \cdot \vec x$

$\vec v = \partial_t \vec x + \vec \nabla \times \vec b$

with constraint equations

$\vec \nabla \times \vec x = \partial_t \vec b$

$\vec \nabla \cdot \vec b = 0$

and with pressure and velocity playing roles analogous to charge density and current respectively.

The primary difference is that in acoustics these displacement fields are potentials carrying gauge symmetries (i.e. transformations under which $p$ and $\vec v$ are invariant), so the observable quantities and their force laws are different, even though Maxwell's equations make an appearance [2].

[1] Maxwell's equations are universal for locally conserved quantities
[2] Acoustic versus electromagnetic field theory