U(1) invariance of classical electromagnetism

[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?

 

[Orodruin]

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.

 

[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.

 

[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)?

 

[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.

 

[Orodruin]

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.

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.

 

[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)?

 

[Orodruin]

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

how would you prove that the gauge invariance of the fields is U(1)?

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.

 

[dextercioby]

How is Yang-Mills theory only a consequence of quantum theory? [...]

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

[...]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.

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?

 

[Orodruin]

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

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.

 

[Vanadium 50]

using only the electromagnetic fields and potentials, how do I know that the gauge symmetry is U(1)?

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.:"

 

[samalkhaiat]

How does U(1) as a group appear in classical electromagnetism?

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 $\mathcal{M}^{(1,3)}$.

 

[dextercioby]

This

[...]
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 $\mathcal{M}^{(1,3)}$.

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

 

[samalkhaiat]

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

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 $p \in \mathcal{M}$.

 

[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?

 

[samalkhaiat]

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?

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.

 

[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).

 

[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)

 

[Demystifier]

what does it mean to say about the "classical electromagnetism" to be U(1) invariant?

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.

 

[Demystifier]

A very good book is vol. III of Scheck's textbook on theoretical physics

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

 

[vanhees71]

https://www.amazon.com/dp/3662555778/?tag=pfamazon01-20

 

[samalkhaiat]

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,

Mathematically, it makes no difference what so ever, if you replace my 3-form $j$ by “their” $~^*\!J$. However, it is economical and more natural to associate a 3-form with the vector current and not 1-form. And here is my reasoning for it: Let $j^{\mu}$ be a conserved vector current contained a 4-dimensional region $D$ of spacetime with smooth 3-dimensional boundary $\partial D$. I want to express $j^{\mu}, \ \partial_{\mu}j^{\mu}$ and the total charge $Q$ in the language of forms. Well, the problem at hand provides me with the following dada: I have the invariant volume element on $D$ $$d^{4}x = \epsilon_{0123} \ \mbox{d}x^{0} \wedge \mbox{d}x^{1} \wedge \mbox{d}x^{2} \wedge \mbox{d}x^{3} ,$$ the hyper-surface differential element on $\partial D$ $$d^{3}\sigma_{\mu} = \frac{1}{3!} \epsilon_{\mu\nu\rho \tau} \ \mbox{d}x^{\nu} \wedge \mbox{d}x^{\rho} \wedge \mbox{d}x^{\tau} ,$$ and of course the Stokes’ theorem for my problem $$\int_{\partial D} \ \left( \cdot \right) = \int_{D} \ \mbox{d} ( \cdot ) .$$ Now, the dimensionality of $D$ and $\partial D$ forces us to insert a 3-form $j_{(3)}$ in Stokes’ formula, and the only 3-form available in my problem is $$j = j^{\mu} (x) \ d^{3}\sigma_{\mu} (x) . \ \ \ \ \ \ \ \ (1)$$ Now everything fits in place, the total charge is given by the familiar integral in field theory $$Q = \int_{\partial D} \ j = \int_{\partial D} \ j^{\mu} (x) \ d^{3}\sigma_{\mu} (x) ,$$ and $$\mbox{d}j = \partial_{\eta}j^{\mu} \ \mbox{d}x^{\eta} \wedge d^{3}\sigma_{\mu} = \partial_{\mu}j^{\mu} \ d^{4}x .$$ Thus, current conservation translates to the statement $\mbox{d}j = 0$, which in turns leads to constancy of the scalar charge $Q$. In many of the literatures, you find $~^*\!j$ instead of $j$ on the left hand side of (1), because they usually use Maxwell’s theory as example, but the problem stated above is independent of any field theory and the *-operation is neither needed nor included in the data.

But this operator (exterior derivative) makes use of the de Rham bundle over $\mathcal{M}^{(1,3)}$

I still don’t know what you meant by “de Rham bundle”, and how would this change the algebraic definition of the $\mbox{d}$-operator? You should know me by now. English is not my first language, mathematics is. So, please write down some equations so that I understand what you mean. I don’t blame you for that, because I know the subject is sick notations and terms wise. In fact, the number of different notations and terms is of order of the number of literature on the subject, and this is why I don’t like to post stuff about it.

So, did you mean the inhomogeneous Hodge-de Rham operator $\hat{\mbox{d}} \equiv \mbox{d} + \delta ,$ or the (abelian) de Rham cohomology groups $$H^{p}(\mathcal{M}) = \frac{ \{ \mbox{closed p-cocycles} \}}{\{ \mbox{exact p-coboundaries} \}} .$$ In both cases the operator $\mbox{d}$ sees only the variables of $\mathcal{M}$. The groups $H^{p} (\mathcal{M})$ determine the topological invariants of the space. So, they are interesting when $\mathcal{M}$ is topologically non-trivial. But $\mathbb{R}^{n}$ is topologically trivial, i.e., it can be covered by a single coordinate patch. Thus, $H^{p \neq 0}(\mathbb{R}^{n}) = 0$ which is nothing but the Poincare’s lemma: On contractible manifold any closed form is exact.

not of the vector bundle you had mentioned in the 2 posts above (#12 and #14 of the thread).

Please note that I did not need to introduce any concept from differential geometry into my posts in this thread. I used forms to write the Maxwell’s Lagrangian simply because it saves me from writing indices. So, differential forms played no part in my answer to your original question. The two sentences that I made about fibre bundle in posts #12 & #14, were just to tell you how people “derive” the source-free Maxwell equation as compatibility condition for the gauge transformation of the $U(1)$ connection 1-form once such connection is properly defined on a bundle (vector or principal).

 

[dextercioby]

[...] Since $\Lambda : \mathbb{R}^{(1,3)} \to \mathbb{R}$, then $(i \Lambda)$ must map an open subset of space-time into the complex line $i \mathbb{R}$ [...]

This is something that just came to me: what is "i" doing there? Just because it exists, one obtains U(1) by exponentiation, but actually, without the "i", it is simply $\mathbb{R}$. I know that the "i" is requested by passage to quantum mechanics or QFT, but, from a purely mathematical standpoint, and also from the point of view of classical physics, the gauge symmetry should be simply $\mathbb{R}$. Actually, mathematicians do not put the "i" in the exponential which links an element in the vecinity of the identity of a Lie group to the element of the Lie algebra, only physicists do, as per the Stone's theorem which is fundamental in the implementation of symmetry groups in quantum physics.

 

[samalkhaiat]

This is something that just came to me: what is "i" doing there? Just because it exists, one obtains U(1) by exponentiation, but actually, without the "i", it is simply $\mathbb{R}$. I know that the "i" is requested by passage to quantum mechanics or QFT, but, from a purely mathematical standpoint, and also from the point of view of classical physics, the gauge symmetry should be simply $\mathbb{R}$. Actually, mathematicians do not put the "i" in the exponential which links an element in the vecinity of the identity of a Lie group to the element of the Lie algebra, only physicists do, as per the Stone's theorem which is fundamental in the implementation of symmetry groups in quantum physics.

1) I am free to write $\delta A_{\mu} = -i \partial_{\mu}(i\Lambda)$.

2) If I write $e^{isG} , \ s \in \mathbb{R}$, then it should be clear to you that $G \in \mbox{T}_{e}\left(U(1)\right) \cong \mathfrak{u}(1)$, i.e., the infinitesimal generator of the group of unitary transformation $U(1)$. Now, If I can define/find such generator $G$ in the (free) Maxwell theory, then my job is done. Well, if you know that Maxwell theory is a constraint system, then you should also know that $G$ exists and it is called Gauss’ law, the generator of gauge transformation. Here is how one can find it. Construct the following conserved current $$J_{\Lambda}^{\mu} = - F^{\mu\nu}\partial_{\nu} \Lambda \ ,$$ and consider its integrated charge $$G_{\Lambda} = \int_{\mathbb{R}^{3}} d^{3}x \ J_{\Lambda}^{0} = \int_{\mathbb{R}^{3}} d^{3}x \ \pi^{\nu}(x) \partial_{\nu}\Lambda (x) \ ,$$ where $$\pi^{\nu} (x) \equiv \frac{\delta}{\delta (\partial_{0}A_{\mu})} \left( - \frac{1}{4} \int_{\mathbb{R}^{3}} d^{3}x \ F^{2} \right) = - F^{0 \nu}(x) ,$$ and $\pi^{0}(x) \approx 0$ is the primary constraint. Now, we first observe that $G_{\Lambda}$ generates an Abelian Lie algebra (via Poisson bracket) $$\{ G_{\Lambda} , G_{\Omega} \} = 0 \ .$$ The second, since the current $J_{\Lambda}^{\mu}$ is conserved, its charge $G_{\Lambda}$ must generate infinitesimal symmetry transformations on the coordinate pair $(A_{\mu} , \pi^{\mu})$. Indeed, it is an easy exercise to show that $G_{\Lambda}$ does generate the correct gauge transformations: $$\{ G_{\Lambda} , \pi^{\mu}(x) \} = 0 = \delta \pi^{\mu} \ ,$$ $$\{ G_{\Lambda} , A_{\mu}(x) \} = \frac{\delta G_{\Lambda}}{\delta \pi^{\mu}(x)} = \partial_{\mu}\Lambda = \delta A_{\mu}\ .$$ The last equation, can be rewritten as $$\delta A_{\mu} = -i \frac{d}{ds} \{ e^{isG_{\Lambda}} , A_{\mu}(x) \} |_{s = 0} , \ \ e^{isG_{\Lambda}} \in U(1) \ .$$

3) Since $\Lambda (x)$ is a Lie algebra element taking values in $\mathbb{R}^{(1,3)}$, your “$\mathbb{R}$” must be the 1-dimensional Lie algebra $\mathbb{R}$. Now, you are stuck because, as a Lie algebra, $\mathbb{R}$ has many locally isomorphic Lie groups. Indeed, it is the Lie algebra of the following four (among many more) Lie groups: $U(1) = \{ z \in \mathbb{C} : \ |z| = 1 \}$, the positive reals $( \mathbb{R}^{+} , \times )$, the multiplicative group of $\mathbb{R}$ with two connected components, also known as, the non-zero reals $(\mathbb{R}^{\times} , \times )$ and the additive group $( \mathbb{R} , +)$. So, tell me, which one of those four (locally isomorphic) groups is “your” gauge group and why?

4) In the presence of a point source, if you take the gauge group to be $\mathbb{R}$, you end up violating the Dirac charge quantization $e \ g = 2 \pi n$. Indeed, in this case one can compute the de Rham cohomology groups to be $$H^{(p \ = \ 0 , 2)}( \mathbb{R}^{4} - \mathbb{R}_{\tau}) = \mathbb{R} \ ,$$ where $\mathbb{R}_{\tau}$ is the world line of the point source. The Dirac relation $e \ g = 2 \pi n$ is satisfied if and only if $U(1)$ is the gauge group.

5) Lastly, and more importantly, we now understand that the full Maxwell theory emerges naturally from gauging the global $U(1)$ symmetry of an arbitrary matter field action, i.e., by sticking an independent $U(1)$ group to each space-time point (associating a $U(1)$ fibre to each point $x^{\mu}$).

 

[dextercioby]

Thank you for your comments. Indeed, point 4) specifically addresses what I stated above, that U(1) is required by a natural passage to quantum mechanics.
From the other points, only #3) requires an answer from my side:

[...]So, tell me, which one of those four (locally isomorphic) groups is “your” gauge group and why?[...]

It is undecided, if only classical electromagnetism (i.e. $\mathcal{L} = -1/4 F^2$) is taken into account.

 

[bhobba]

what does it mean to say about the "classical electromagnetism" to be U(1) invariant?

Very nice thread - loved it.

Personally though I prefer the reverse - getting EM from local U(1) invariance of QM rather than asking what does it mean to be U(1) invariant.

It's even found in popular books eg Victor Stenger - The Comprehensible Cosmos - page 251 under Electromagnetism:.
https://www.amazon.com/dp/1591024242/?tag=pfamazon01-20

He covers the other stuff too (symmetry breaking and all that), again at the populist level using just a little bit of math and calculus.

This is an advanced thread, but I mention it for those not advanced but would like a gentle introduction - you just need a bit of calculus.

Thanks
Bill

 

[vanhees71]

I also think that QT clarifies many troubles of classical physics, not only phenomenologically but also from a mathematical point of view. While gauge invariance is pretty straight forward also in classical electrodynamics, the true trouble is the idea of "point charges", which just doesn't fit into the classical field-theoretical framework. The mechanics should be rather treated in the sense of continuum mechanics, which is the natural setup for classical field theory.

Another example is statistical mechanics. It starts already with the definition of the relation between phase-space volume and "number of microstates", for which you obviously need a "natural" measure for phase-space volume, which is immediately delivered by QT in terms of Planck's constant $h=2 \pi \hbar$.

In many respects QT is easier than classical physics, but of course from a somewhat advanced point of view ;-)).

 

[Ben Niehoff]

Classically, it is true that there is no reason for the gauge group to be $U(1)$ rather than $\mathbb{R}$ (which is simply the universal cover of $U(1)$).

For non-Abelian groups, the kinetic term $F^2$ has its gauge indices contracted via the group's Killing form, and then questions of compactness translate into questions of whether the kinetic term has the correct sign (to prevent runaway solutions, or, quantum-mechincally, in order to have a minimum energy).

For Abelian groups, however, the Killing form is flat. So to determine whether such groups are compact or noncompact requires further input (in this case, the quantization of angular momentum).

Note that classical Yang-Mills theory (with non-Abelian gauge group) exists just fine. It just doesn't describe reality very well.

 

[thierrykauf]

$$\bar{\psi}\gamma^\mu \partial_\mu \psi$$ becomes $$\bar{\psi}e^{-i\phi(x)}\gamma^\mu[\partial_\mu + e\frac{\partial\phi(x)}{\partial x}]e^{i\phi(x)}\psi$$ under an internal rotation with one angle, hence the (1) of U(1). When the electron is placed in an external electromagnetic field, the coupling shows up as $$j^\mu A_\mu$$ that is $$\bar{\psi}\gamma^\mu ieA_\mu\psi$$. This has the same form as the U(1) transformed Dirac derivative term. So if you started with an electron already coupled to an E&M field, the Lagrangian would automatically be U(1) invariant. The word gauge is a misnomer. Weyl introduced it because he had conformal invariance in mind. But the name stuck. Even when it means invariance by rotation. Here gauge refers to invariance of the second kind, that is when the phase is space dependent. There are more thorough answers on this thread.

 

[cartan]

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)?

No this is incorrect. Yang-Mills field theory is classical, everything you write down is a classical field and describes classical field theory. It's not until you introduce the path integral that you obtain quantum field theory. Furthermore, classical electromagnetism is actually indistinguishable as a U(1) Yang-Mills theory or a $\mathbb{R}$ Yang-Mills theory. It's not hard to see why this should be true, first they have isomorphic Lie algebras, secondly they are both abelian, and one dimensional, so the Bianchi identity and the Yang-Mills equation simplify in the same way. Finally, the gauge transformations end up working out the same way for a potential $A$, as one can easily check in any local gauge.