# How many fundamental fields / constraints are in Maxwell's Equations?

I've seen electromagnetics formulated in the potentials ($\bf A$, $\varphi$) (and their magnetic counterparts) in the Lorenz gauge for a long time. Justifications for using these include that they are less singular than the fields ($\bf E$, $\bf H$) and reduce the redundancy of Maxwell's equations. All the fields derive from two vector and two scalar potentials (total of 8 potentials). In older engineering works (1950s-1969), I've found they don't use the ($\bf A$, $\varphi$) potentials at all, and instead use a set of two vector potentials called the Hertz vectors by the authors, usually noted by ${\bf\Pi}$, from which all the fields derive. They seem to have an additional level of regularity to them because the ($\bf A$, $\varphi$) potentials derive from them. There are also a total of only 6 scalar quantities between the two, from which all the fields derive.

Modern authors seem to prefer the dyadic Green's functions. I haven't studied these in any depth, but it appears to me that in this formulation, all the fields derive from 5 scalar quantities? Not sure about this one so feel free to correct me. But it got me thinking about my questions:

Is 3D classical EM, how many fundamental constraints are there in Maxwell's equations? What is the minimum number of quantities that can be defined, from which all the fields can be said to derive? I suspect the differential forms / tensor formulation will be necessary to get to the bottom of this, but if there is an interpretation in terms of 3D vector calculus, that'd be great.

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Off the top of my head you can set $\varphi=0$ as a gauge-fixing condition and derive everything from the vector potential so it's at most 3 quantities. – alexarvanitakis Feb 28 '13 at 22:40
For the counting of physical d.o.f. in the standard formulation of EM with one $4$-potential, see also this Phys.SE post. However, OP seems to be asking about non-standard EM-dual models with two $4$-potentials. A question about EM with two $4$-potentials was also asked here in the context of magnetic monopoles. – Qmechanic Feb 28 '13 at 23:38
Qmechanic, I had read your post in what you've linked. Is the short answer "3" for standard EM without magnetic charge and current, and "6" for models that include them? – rajb245 Mar 1 '13 at 0:00
@alexarvanitakis: Can you cite that? I think it's called the Weyl gauge, but I can't find any material to demonstrate that everything can be derived from the vector potential. – jvriesem May 9 '15 at 22:40
@jvriesem it seems to me that Maxwell's equations imply a conservation type relationship between $\varphi$ and the vector potential (something relating $\nabla\cdot\mathbf{A}$ and $\frac{\partial\varphi}{\partial t}$). So I think the scalar potential can always be calculated from knowledge of the vector potential. – rajb245 May 11 '15 at 13:51

An electromagnetic wave with a well-defined frequency and direction, i.e. $\vec k$, only has two possible truly physical i.e. transverse polarizations, i.e. the linearly polarized waves in the $x$ and $y$ direction (or the two circularly polarized ones). That implies that a truly physical counting of polarizations gives you 2, more generally $D-2$ in $D$ spacetime dimensions.

Starting from the $A_\mu$ potential fields, one component is unphysical because it's pure gauge, $A_\mu\sim\partial_\mu\lambda$, and one of them is forbidden due to Gauss' constraint $\rm div\,\vec D=0$ etc. that already constrains the allowed initial state of the electromagnetic field. Both of these killed polarizations are ultimately linked to the $U(1)$ gauge symmetry.

If one is allowed to count off-shell and unphysical fields, there may be many more components than two. But it's always possible to deduce that there are two physical polarizations at the end. For example, when we view $\vec B,\vec E$ as basic fields, there are six components, a lot. But these fields only enter Maxwell's equations through first derivatives, and not second as expected for "normal" bosonic fields, so these fields are simultaneously the canonical momenta for themselves. This brings us to three polarizations but one of them is killed by the constraints, the Maxwell's equations that don't contain time derivatives.

The Hertz vector is just the most famous "non-standard" example how to write the electromagnetic field as a combination of derivatives of some other fields. One must understand that the room for mathematical redefinitions etc. is unlimited and it is a matter of pure maths. All these descriptions may describe the same physics. At the end, the only "truly invariant because measurable" number of "fields" that all these approaches must agree about is the number of linearly independent physical polarizations of a wave/photon with a given $\vec k$.

If you can analyze any mathematical formulation of electromagnetism or another field theory and derive that there are $D-2$ physical polarizations (this usually boils down to the difference of the number of a priori fields minus the number of independent constraints and the number of parameters defining identifications i.e. gauge symmetries – but the independence is sometimes hard to see and requires you to make many steps of the counting), then you have proved everything that is "really forced to be true". Various formalisms may offer you other ways to count the number of off-shell fields (with different answers) and they may be useful (because they satisfy certain conditions or enter some laws) but to discuss them, one has to know what the laws where they enter actually are.

A truly physical approach is only one that counts the physical polarizations. The gauge symmetry is just a redundancy, a mathematical trick to get the right theory with 2 physical polarizations out of a greater number of fields with certain extra constraints or identifications. The precise number of constraints or identifications may depend on the chosen mathematical formalism and it is not a physically meaningful question – it is a question of a subjectively preferred mathematical formalism because the physics is equivalent for all of them.

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When you say a wave with a well defined frequency (say $\omega$) and $\vec k$, you are referring to one harmonic plane wave in a Fourier integral over $\omega$, and $\vec k$, correct? I understand that such a field fundamentally has two polarizations. So is it possible to formulate the physical fields, using mathematical tricks, as the derivatives of just two scalar functions of space and time? – rajb245 Mar 1 '13 at 17:24
Yes, it's of course possible to encode all physical degrees of freedom of the electromagnetic field to two functions of space and time. Just choose a convention for the polarizations for each $\omega,\vec k$ and calculate the Fourier combinations. The problem is that the Hamiltonian for these two "scalar" (they're not really scalar!) fields of the spacetime will be nonlocal. The introduction of at least some auxiliary components is needed for the Lagrangian to be nicely local and manifestly Lorentz-invariant: gauge invariance's reason of existence is to make Lorentz inv. manifest. – Luboš Motl Mar 1 '13 at 18:59
The entire basis of this answer is based on a wave. What about static cases in which there are no waves and no polarizations to speak of? – jvriesem May 9 '15 at 22:41
What do you exactly want to know about them? "What about" isn't a well-defined question. I said that the auxiliary components are needed for the Lorentz-invariant dynamic of the spin-one field. They are needed for the most general situations. If you pick some less clever, special, e.g. static, situations only, you may become unable to prove what I proved (or sketched a proof) above. But that doesn't mean that my proof has a problem. – Luboš Motl May 10 '15 at 5:43