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**E**,

**B**or ##F##, what does it mean to say about the "classical electromagnetism" to be U(1) invariant?

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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.

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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.

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

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Because from the point of view of classical physics, there's no physical reason to consider several "species" of electromagnetic potentials.How is Yang-Mills theory only a consequence of quantum theory? [...]

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?[...]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.

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That it does not describe something we observe does not necessarily make it uninteresting. In addition, classical EMBecause from the point of view of classical physics, there's no physical reason to consider several "species" of electromagnetic potentials.

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I don't think the detour into Yang-Mills was helpful. So let's back up a bit.using only the electromagnetic fields and potentials, how do I know that the gauge symmetry is U(1)?

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.:"

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

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

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Do you, perhaps, have a reference for this (book, article)? I'm interested in this calculation, too.[...]

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

Thank you very much for your effort.

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

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

Thank you very much for your effort.

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

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

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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).

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

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Here is my try. Classical electromagnetism (without matter) is invariant under the transformationwhat does it mean to say about the "classical electromagnetism" to be U(1) invariant?

$$A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\lambda.$$

Since ##\lambda## is a

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Do you mean Scheck's "Classical Field Theory"?A very good book is vol. III of Scheck's textbook on theoretical physics

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Mathematically, it makes no difference what so ever, if you replace my 3-form [itex]j[/itex] by “their” [itex]~^*\!J[/itex]. 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 [itex]j^{\mu}[/itex] be a conserved vector current contained aWell, 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,

I still don’t know what you meant by “de Rham bundle”, and how would this change the algebraic definition of the [itex]\mbox{d}[/itex]-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.But this operator (exterior derivative) makes use of the de Rham bundle over ##\mathcal{M}^{(1,3)}##

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

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 [itex]U(1)[/itex] connection 1-form once such connection is properly defined on a bundle (vector or principal).not of the vector bundle you had mentioned in the 2 posts above (#12 and #14 of the thread).

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

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1) I amThis 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 [itex] \mathbb{R}[/itex]. 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 [itex] \mathbb{R}[/itex]. 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.

2) If I write [itex]e^{isG} , \ s \in \mathbb{R}[/itex], then it should be clear to you that [itex]G \in \mbox{T}_{e}\left(U(1)\right) \cong \mathfrak{u}(1)[/itex], i.e., the infinitesimal generator of the group of unitary transformation [itex]U(1)[/itex]. Now, If I can define/find such generator [itex]G[/itex] in the (free) Maxwell theory, then my job is done. Well, if you know that Maxwell theory is a

3) Since [itex]\Lambda (x)[/itex] is a Lie algebra

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

5) Lastly, and more importantly, we now understand that the

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From the other points, only #3) requires an answer from my side:

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

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