Can anyone give a physical reason that $U(1)$ is the gauge group for classical electromagnetism?

I am familiar with the principal bundle formalism for Yang-Mills theory and see that since the Lie algebra of $U(1)$ is $\mathbb{R}$, the Yang-Mills equations reduce to Maxwell's equations. However, I am looking for a plausible physical reason that $U(1)$ is related to classical electromagnetism.

This seems to be the same question: Classical electrodynamics as an $\mathrm{U}(1)$ gauge theory , but does not have a satisfactory answer.

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    $\begingroup$ it depends on what you mean by "physical". To me, rigorous group theoretical arguments are as physical as it comes... (and, ultimately, the fact that the model works very well is the best argument I can think of) $\endgroup$ Commented May 27, 2016 at 10:58
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    $\begingroup$ Sorry, @auxsvr - but gauge transformations cannot map electric fields to magnetic ones or vice versa. You must have confused gauge transformations with (S-) dualities. Otherwise, the answer to the OP is that $U(1)$ is the only connected Abelian simple Lie group. Electromagnetism has to be Abelian because the light doesn't interact with itself (linearity, which implies the absence of higher-order interactions from F mu nu F mu nu), and it is a long-range force which bans breaking. But in particle physics, U(1) does arise as a leftover of a larger group. $\endgroup$ Commented May 27, 2016 at 14:07
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    $\begingroup$ Sorry but $F+i*F$ (just like $F$ itself) is gauge-invariant, so the gauge transformation doesn't change a damn thing about it. $\endgroup$ Commented May 27, 2016 at 14:41
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    $\begingroup$ I'm not sure what you are looking for. The four-potential transforms as $A\mapsto A+\mathrm{d}\chi$ under a gauge transformation, so the Lie algebra of your gauge group is $\mathfrak{u}(1)=\mathbb{R}$, and so the gauge group can be either $\mathrm{U}(1)$ or $\mathbb{R}$. Since classical electromagnetism doesn't have any other fields transforming in a representation of the gauge group, but just the current $j^\mu$ with $\mathrm{d}{\star}j = 0$ that transforms trivially, you can't decide, classically, which one to choose. We take $\mathrm{U}(1)$ because this is what get in the quantum theory. $\endgroup$
    – ACuriousMind
    Commented May 27, 2016 at 15:04
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    $\begingroup$ Thanks for these comments, guys. A $U(1)$ transformation turning an electric field into a magnetic field is the kind of answer I would like, but doesn't seem to be correct. Must be Abelian because light doesn't interact with itself is also fairly satisfying. @auxsvr, in general $F$ transforms as $F \mapsto gFg^{-1}$ for $g \in G$ (the gauge group) which reduces to $F \mapsto F$ for $G$ Abelian. $\endgroup$
    – Bendy
    Commented May 27, 2016 at 15:12

2 Answers 2


When Maxwell formulated his equations he did so using quaternions, which BTW is a more elegant formalism, and Heaviside formulated them as we normally read them. Our standard vector forms of the Maxwell equations are more convenient for electrical engineering. These equations linearly add the electric and magnetic fields. This linear property is a signature of the abelian nature of the Lie group $U(1)$ for electrodynamics.

Let me argue this in somewhat more modern language with quantum mechanics. Suppose we have a quantum field (or wave) that transforms as $\psi(\vec r) \rightarrow e^{i\theta(\vec r)}\psi(\vec r)$. Now act on this with the differential operator $\hat p = -i\hbar\nabla$ and we find $$ \hat p\psi(\vec r) = -i\hbar\nabla\psi(\vec r) = -i\hbar\left(\nabla\psi(\vec r) + i\psi(\vec r)\nabla\theta\right) $$ We then see this does not transform in a homogeneous fashion, so we change the operator in to a covariant one by $\nabla \rightarrow \nabla - ie\vec A$ where $\vec A$ is the vector potential that subtracts out the $\nabla\theta$. We can now perform quantum mechanical calculations that include the electromagnetic field in a consistent manner.

We now consider the covariant differential one-form ${\bf D} = {\bf d} - ie{\bf A}$ such that ${\bf d} = dx\cdot\nabla$ and ${\bf A} = {\vec A}\cdot dx$. We can now look as the action of ${\bf D}\wedge {\bf D}$ on a unit or constant test function $$ {\bf D}\wedge {\bf D}\odot\mathbb I = {\bf d}\wedge{\bf d} \odot\mathbb I - e^2{\bf A}\wedge{\bf A} \odot\mathbb I - ie{\bf d}\wedge{\bf A} \odot\mathbb I - ie{\bf A}\wedge{\bf d} \odot\mathbb I , $$ where the wedge product of a p-form with itself is zero and elementary manipulations gives $$ {\bf D}\wedge {\bf D}\odot\mathbb I = ie\left({\bf d}\wedge{\bf A}\right) \odot\mathbb I . $$ Breaking out the differential form on the vector potential gives the magnetic field by $B = -\nabla\times A$ that the vector potential one-form wedged with itself is zero is a signature of the abelian or $U(1)$ symmetry of the electromagnetic field. For other gauge fields there is a color index, where there are different sorts of charges, and so ${\bf A}\wedge{\bf A}$ is nonzero and this is a signature of the nonabelian nature of other gauge fields, in particular the weak and strong nuclear forces.

This is done in a three dimensional or nonrelativistic manner, and of course this must be generalized to relativistic QM or the Dirac equation for the quantum dynamics of a fermion. So this is a bit of an elementary introduction. The main upshot though is the reason electromagnetism is abelian or $U(1)$ is there is only one electric charge $e$, with positive and negative values, while other gauge fields have an array of color-charges.

  • $\begingroup$ Thanks for this. Are you basically saying that, if the gauge group were nonabelian, there would be terms corresponding to different charges in the Lagrangian? $\endgroup$
    – Bendy
    Commented May 27, 2016 at 15:36
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    $\begingroup$ In effect that is the case. If there were several charges there would be different vector potentials, or what in more advanced language are called gauge connections. These would form 2-form products and physically it would means the photon would carry these charges. In QCD this happens and is why the field is confining. The gluons attract each other and instead of forming a dipole field similar to the B field of a magnet they tend for form a sort of flux tube that is self binding or attracting. $\endgroup$ Commented May 27, 2016 at 15:45
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    $\begingroup$ Yes, but this question is asking about classical E&M, not quantum (QED). $\endgroup$ Commented Jun 23, 2020 at 1:24

A speculative classical reason is that the existence of classical magnetic monopoles requires a $U(1)$ gauge group. It is well-known that generic electromagnetic fields in non-trivial spacetimes only admit an electromagnetic potential $A$ that is defined locally, meaning that you have one expression for $A$ on one patch of spacetime and another in another. On the overlaps between these patches, the expressions must be related to one another by a gauge transformation. In the case of a magnetic monopole, the gauge transformation required is not single valued but rather takes values on $U(1)$. This is explained on the first few pages of professor Tong's notes https://www.damtp.cam.ac.uk/user/tong/gaugetheory/gt.pdf.

It would be interesting to see other classical configurations of the electromagnetic field where this happens and without monopoles. I feel like there should be some out there (maybe in condensed matter physics books). Let me know if someone finds them!


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