# FeaturedA U(1) invariance of classical electromagnetism

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1. Aug 20, 2017

### dextercioby

This is an interesting question that popped through my mind. Some of us should know what is meant by „gauge transformations”, „gauge invariance/symmetry” and are used to seeing these terms whenever lectures on quantum field theory are read. But the electromagnetic field in vacuum (described in a specially relativistic fashion by the tensor $F_{\mu\nu}$) is a classical one (i.e. it exists also in a non quantum setting), so one has the right to ask. Given E,B or $F$, what does it mean to say about the "classical electromagnetism" to be U(1) invariant?

2. Aug 20, 2017

### Orodruin

Staff Emeritus
You really should be working in terms of the 4-potential and not the field strengths as this is your dynamical field. The gauge invariance you are looking for is the same as that typically introduced in classical electromagnetism. Without a gauge fixing condition your 4-potential will not be uniquely determined (the gauge symmetry being a local U(1) symmetry). This should be described in detail (not only for EM but also for general non-commutative gauge fields) in any textbook covering classical Yang-Mills theory.

3. Aug 20, 2017

### vanhees71

A very good book is vol. III of Scheck's textbook on theoretical physics, where he treats electromagnetism right away from the modern point of view, which is, of course, that it is describing a massless spin-1 field, which is necessarily a gauge field (if you don't want unobserved continuous intrinsic spin-like degrees of freedom).

Classcial electrodynamics is gauge invariant also under the presence of sources (electric charge-current distributions).

As Orodruin said, it's described in terms of the potentials of the field, which is introduced using the homogeneous Maxwell equations
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E} + \frac{1}{c} \partial_t \vec{B}=0.$$
From the first equation you introduce the vector potential
$$\vec{B} = \vec{\nabla} \times \vec{A}.$$
Plugging this into the 2nd equation yields
$$\vec{\nabla} \times \left (\vec{E}+\frac{1}{c} \partial_t \vec{A} \right)=0,$$
which means that the expression in the parantheses is a gradient:
$$\vec{E}=-\vec{\nabla} \Phi - \frac{1}{c} \partial_t \vec{A}.$$
Now, obviously for given $\vec{E}$ and $\vec{B}$ you can add to $\vec{A}$ an arbitrary gradient,
$$\vec{A}'=\vec{A} - \vec{\nabla} \chi,$$
and then in order to get also the same $\vec{E}$ you must have
$$\vec{E}=-\nabla \Phi'-\frac{1}{c} \partial_t \vec{A}'=-\nabla (\Phi'- \frac{1}{c} \partial_t \chi') - \frac{1}{c} \partial_t \vec{A} \; \Rightarrow \; \Phi'=\Phi+ \frac{1}{c} \partial_t \chi.$$
Thus everything must be unchanged under the gauge transformation
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi + \frac{1}{c} \partial_t \chi.$$
The inhomogeneous Maxwell equations are
$$\vec{\nabla} \times \vec{B} - \frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho,$$
or
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})+\frac{1}{c} \left (\vec{\nabla} \Phi +\frac{1}{c} \partial_t \vec{A} \right) = \frac{1}{c} \vec{j}, \quad -\Delta \Phi - \frac{1}{c} \partial_t \vec{\nabla} \cdot \vec{A}=\rho.$$
Using
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A},$$
you can see that it is advantegeous to make use of your gauge freedom by partially fixing the gauge field $\chi$ introducing the Lorenz-gauge condition (note it's Lorenz here, not Lorentz for historical justice!):
$$\frac{1}{c} \partial_t \Phi + \vec{\nabla} \cdot \vec{A}=0,$$
because then the four vector fields decouple into
$$\Box \Phi=\rho, \quad \Box \vec{A}=\frac{1}{c} \vec{j},$$
where the d'Alembert operator is devined as
$$\Box=\frac{1}{c^2} \partial_t^2 - \Delta.$$
Note that for this to be consistent without any reference to matter-field (or classical point-mechanics) equation of motion, you must have the continuity equation for electric charge as an integrability condition,
$$\partial_t \rho + \vec{\nabla} \cdot \vec{j}=0.$$

Another important point of gauge invariance is that the variation of the action for the motion of charged particles,
$$S=\int \mathrm{d} t [-m c^2 \sqrt{1-\dot{\vec{x}}^2/c^2} + q \Phi - q \dot{\vec{x}} \cdot \vec{A}]$$
is gauge invariant.

It's also very easy to see that verything can be made manifestly covariant by making
$$(A^{\mu})=(\Phi,\vec{A}), \quad (j^{\mu})=(c \rho,\vec{j})$$
to four-vectors, leading to the covariant form of Maxwell's equations
$$F_{\mu \nu} = \partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}, \quad \partial_{\mu} F^{\mu \nu}=\frac{1}{c} j^{\nu}.$$
The obvious convenience of the Lorenz gauge comes not the least from the fact that it's a Lorentz-invariant constraint,
$$\partial_{\mu} A^{\mu} =0.$$

The full beauty of gauge invariance comes into view in quantum theory, where you can heuristically argue for the introduction of vector fields by making global intrinsic symmetries like the invariance of the Schrödinger, Klein-Gordon, or Dirac equation under changes of the phase of the fields (building the symmetry group U(1)) local.

4. Aug 20, 2017

### dextercioby

All this is well-known to me, the question remains as: using only the electromagnetic fields and potentials, how do I know that the gauge symmetry is U(1)?

5. Aug 20, 2017

### vanhees71

The electromagnetic four-potential is the affine connection of local U(1) symmetry. I think, this becomes obvious in a natural way only in quantum theory.

6. Aug 20, 2017

### Orodruin

Staff Emeritus
I would say it is also rather evident in the structure of classical Yang-Mills theory as it is appearing in precisely the appropriate manner in the covariant derivative. Specialising that to a U(1) symmertry of course directly gives you classical electromagnetism.

7. Aug 20, 2017

### dextercioby

No, but ignoring the existence of Yang-Mills fields (which is only a consequence of quantum theory), how would you prove that the gauge invariance of the fields is U(1)?

8. Aug 20, 2017

### Orodruin

Staff Emeritus
How is Yang-Mills theory only a consequence of quantum theory?

As always. You would take your Lagrangian and show that it is invariant under gauge transformations. Again, note that the gauge field is the 4-potential, not the electric and magnetic fields.

9. Aug 20, 2017

### dextercioby

Because from the point of view of classical physics, there's no physical reason to consider several "species" of electromagnetic potentials.

I suspected both of you didn't get what I am actually asking here. How does U(1) as a group appear in classical electromagnetism?

10. Aug 20, 2017

### Orodruin

Staff Emeritus
That it does not describe something we observe does not necessarily make it uninteresting. In addition, classical EM is a (Abelian) classical Yang-Mills theory. Calling YM theory uninteresting classically is calling EM uninteresting classically.

11. Aug 20, 2017

Staff Emeritus
I don't think the detour into Yang-Mills was helpful. So let's back up a bit.

The central idea of group theory in physics is that the same equations have the same solutions, and the encountered symmetries are reflected in these solutions. Thus, it makes sense to use these symmetries as labels in categorizing these equations (and their solutions).

Ultimately, a human being needs to be able to identify the symmetries - just as it takes a human being to notice that an equilateral triangle is symmetric about 120 degree rotation, and symmetric about exchange of any two sides. Some human being has to look at the equations of motion and say "the properties of X mean it has the Y symmetry". There is no "operator" that you feed equations of motion and out pops "U(1)".

These effects are not subtle.

Take E&M, and add the light-by-light scattering term E dot B into the classical Lagrangian. All of a sudden ε0 and μ0 stop being numbers and start being tensors (and problems quickly become unsolvable).If I go to an SU(2) theory, I not only have that, but my fields themselves become charged, and charge is no longer expressible as a number - it needs to be a matrix. The equations of motion will be very different than the Maxwell equations, and an expert can recognize them as "aha - that's what SU(2) looks like.:"

12. Aug 21, 2017

### samalkhaiat

Since the (EM) gauge transformations are parameterised by the arbitrary real number $\Lambda$, the underlying group (if any) must be a Lie group, call it $G$. Denote its Lie algebra by $\mathcal{L}(G) \cong T_{e}(G)$. Now, the infinitesimal transformation $\delta_{\Lambda}A_{\mu} = \partial_{\mu}\Lambda$ implies that $$[\delta_{\Lambda} , \delta_{\Omega}] A_{\mu} = 0 .$$ This means that $\mathcal{L}(G)$ is an Abelian Lie algebra. Since $\Lambda : \mathbb{R}^{(1,3)} \to \mathbb{R}$, then $(i \Lambda)$ must map an open subset of space-time into the complex line $i \mathbb{R}$. But, we know that $\mathcal{L}\left(U(1)\right) \cong i \mathbb{R}$. We therefore conclude that $i \Lambda \in \mathcal{L}\left(U(1)\right)$: indeed, topologically $U(1)$ may be identified with circle $S^{1}$, and the tangent space at the identity is obtained by differentiating the curves $t \to e^{i t \theta}$ at $t = 0$ for all real numbers $\theta (x)$, giving as the Lie algebra the complex line $i \mathbb{R}$; thus $\mathcal{L}\left(U(1)\right) \cong i \mathbb{R}$. Now, exponentiation (of the Lie algebra element $i \Lambda$) leads to $$g(x) \equiv e^{i \Lambda (x)} \in U(1) .$$ This allows us to rewrite the EM-gauge transformation as $$A_{\mu} \to A_{\mu} + i g(x) \partial_{\mu} g^{-1}(x),$$ where $g \partial_{\mu} g^{-1} \in \mathcal{L}\left(U(1)\right)$.

Of course a proper (but very lengthy) way to answer your question, is by deriving the Maxwell’s equations from the action of $U(1)$(considered as the set of all sections which are vector bundle isomorphisms) on the fibre bundle over $\mathcal{M}^{(1,3)}$.

Last edited: Aug 24, 2017
13. Aug 21, 2017

### dextercioby

This
Do you, perhaps, have a reference for this (book, article)? I'm interested in this calculation, too.
Thank you very much for your effort.

14. Aug 21, 2017

### samalkhaiat

I believe, many textbooks on differential geometry do explain the $U(1)$-connection and its curvature on $\mathcal{M}^{(1,3)}$. The most elementary textbook which treats the subject is

R.W.R. Darling, “Differential Forms and Connections”, Cam. Uni. Press (1994).

But, your question has a simple one-sentence-answer: If the current is conserved, i.e., $\mbox{d}j = 0$, then the following Maxwell’s action $$S[A] = \int_{\mathbb{R}^{(1,3)}} \left( \frac{1}{2} F \wedge ~^*\!F + j \wedge A \right) ,$$ is invariant under $$A \to A + e^{\Gamma} \mbox{d} e^{-\Gamma} ,$$ where $e^{\Gamma}$ is a function on spacetime with values in the group $U(1)$. Clearly, the above answer is a lot easier than showing that the $U(1)_{p \in \mathcal{M}}$ transformation of the fibre $E_{p}$ does not depend on the local trivialization for the vector bundle at $p \in \mathcal{M}$.

Last edited: Aug 21, 2017
15. Aug 21, 2017

### dextercioby

The fiber bundle formulation of classical field theory has always been a subject that gave me headaches which could not be treated by quite a number of books. For example, the simplest question I still have as „unanswered” is: OK, even if F,j,A „live” in a (vector, principle, associated) bundle, their functional dependence is still on $x^{\mu}$, that is a parameter set for the base space (4D Minkowski spacetime). Therefore, when you write $dj = 0$, is that d the exterior differential in the gauge bundle, or the exterior differential in the de Rham bundle over space-time?

16. Aug 21, 2017

### samalkhaiat

I am not sure, I understand your question. By $\mbox{d}$, one always means the unique map $\mbox{d}: \Lambda^{p} U \to \Lambda^{p+1}U$, ($U$ is open subset in $\mathbb{R}^{n}$), with its usual (coordinate independent) properties. And, by $j$, I meant the current density 3-form $$j = \frac{1}{3!} \epsilon_{\mu \nu \rho \sigma} \ j^{\mu} \ \mbox{d}x^{\nu}\wedge \mbox{d}x^{\rho}\wedge \mbox{d}x^{\sigma},$$ and $$\mbox{d}j = \partial_{\mu}j^{\mu} \ d^{4}x ,$$ where $d^{4}x \equiv \mbox{d}x^{0} \wedge \mbox{d}x^{1} \wedge \mbox{d}x^{2} \wedge \mbox{d}x^{3}$. The above exterior derivative and its coordinate independent properties can also be introduced on differential forms on arbitrary smooth manifold.

17. Aug 22, 2017

### dextercioby

Well, in the literature the matter current is a 1-form (that's why Maxwell's equations are neatly written as dF=0 and $\delta F =j$), you are actually using its Hodge dual as the 3-form,

But this operator (exterior derivative) makes use of the de Rham bundle over $\mathcal{M}^{(1,3)}$, not of the vector bundle you had mentioned in the 2 posts above (#12 and #14 of the thread).

18. Aug 22, 2017

### vanhees71

Another book which treats electromagnetism in the Cartan calculus formulation and carefully examines the mathematical structure underlying good old Maxwell is

F. W. Hehl, Y. N. Obhukov, Foundations of classical electrodynamics, Birkhäuser Boston (2003)

19. Aug 22, 2017

### Demystifier

Here is my try. Classical electromagnetism (without matter) is invariant under the transformation
$$A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\lambda.$$
Since $\lambda$ is a single continuous function, this is a symmetry under a one-parameter continuous transformation. This means that the group of transformation is a one-parameter continuous group. Every one-parameter continuous group is locally isomorphic to U(1). If, in addition, we require that the group must be compact, I think that U(1) is the only possibility even globally.

20. Aug 22, 2017

### Demystifier

Do you mean Scheck's "Classical Field Theory"?