Is the photon a gauge boson for neutrinos?

[MichPod]

TL;DR Summary

Photon is a gauge boson for an electron. What about neutrino?

If I put this in technically correct terms, to enable a local symmetry related to electron phase change, we need to introduce a spin 1 field which is identified as electromagnetic field.

Yet there are other spin 1/2 particles. Say, what about neutrino? As it does not couple with electromagnetic field, which boson field is expected to provide its local symmetry related to the phase change specifically?

The same question may probably be raised about any other particles. Like if we demand a phase change symmetry of any particle (with any spin, both bosons and fermions), are there "pair" particles to guarantee it, and as it is not the case, how is this resolved?

Disclaimer: I have no knowledge of particle physics, so this post is not to advertise my "ideas", but rather to learn.

 

[Vanadium 50]

"Gauge boson for neutrinos" makes no sense. A photon is a gauge boson. It does not couple to neutrinos, but it is still a gauge boson.

 

[PeterDonis]

Photon is a gauge boson for an electron.

No, it isn't. It's the gauge boson for the electromagnetic interaction. Electrons respond to this interaction, because they have electric charge, but neutrinos don't, because they are electrically neutral.

Neutrinos and electrons both respond to the weak interaction, which has three gauge bosons that are different from the photon: the W+, W-, and Z bosons.

 

[PeroK]

Disclaimer: I have no knowledge of particle physics ...

On the contrary, your OP reveals a non-trivial knowldege of particle physics!

 

[MichPod]

No, it isn't. It's the gauge boson for the electromagnetic interaction. Electrons respond to this interaction, because they have electric charge, but neutrinos don't, because they are electrically neutral.

Neutrinos and electrons both respond to the weak interaction, which has three gauge bosons that are different from the photon: the W+, W-, and Z bosons.

I am following https://www.rainerhauser.ch/public/scripts/StandardModel1.pdf, page 27 and 28.
They transform the electron field in the Dirac equation by multiplying the wave function by some arbitrary magnitude-one complex value different in each point in the space-time, i.e. by arbitrary changing the phase of the wave function, then to compensate for this (to promise local symmetry in respect to phase change), they introduce a spin 1 massless field which couples with electron and which they identify as electromagnetic field.

So practically my question was what would be the result of the same procedure if applied to other particles, not electron (don't they have the same sort of local symmetry?). Like what field would we need to introduce to promise the same kind of local symmetry for a neutrino? Would it be meaningful if we try to promise a local symmetry of the same kind for a photon or, alternatively to some other massive boson like Z boson and which field would we need to introduce as a result of that?

 

[PeterDonis]

So practically my question was what would be the result of the same procedure if applied to other particles, not electron

The standard model already does apply this procedure to other particles; it applies (in the Abelian version that is described on the pages you reference) to any particle (or more precisely any fermion--there are also electrically charged bosons in the standard model, whose photon interaction term looks a bit different) that has nonzero electric charge. Note that the word "electron" does not appear anywhere on the pages you referenced, and there is nothing there that says they are only applying the procedure described to electrons.

 

[PeterDonis]

what field would we need to introduce to promise the same kind of local symmetry for a neutrino?

As I've already said, the neutrino is electrically neutral so it doesn't interact with the photon. It only responds to the weak interaction. That means it does not have the U(1) (electromagnetic) gauge symmetry that the electron has.

The weak interaction gauge bosons are an example of the non-Abelian version of the procedure described in the paper you reference (the gauge group is SU(2)). That procedure by itself doesn't explain why those gauge bosons have masses; they get their masses as a result of electroweak symmetry breaking, which is something separate from what is described in the paper you reference.

 

[MichPod]

"it applies to ... any fermion ... that has nonzero electric charge"
Ok. So per charged fermions we get a photon gauge field which promises a local symmetry in relation to the wave function phase change, right?
So would can we expect if the same procedure would be applied to bosons or to fermions without electrical charge? Would we get some new fields to promise this symmetry for the mentioned particles?
Like can we identify such a gauge field with Z boson for neutrino? (Let's leave aside the point that they have nonzero mass as a result of symmetry breaking.)
And anyway, what would be a way to promise this kind of symmetry for bosons themselves?

 

[LittleSchwinger]

Ok. So per charged fermions we get a photon gauge field which promises a local symmetry in relation to the wave function phase change, right?

That's a local change in the phase of the fermion field, not a wave function.

 

[PeterDonis]

can we identify such a gauge field with Z boson for neutrino?

Not the Z boson by itself; the 3 weak gauge bosons. All three of them are gauge bosons for the weak interaction. There are three because the gauge group for the weak interaction is SU(2).

 

[Vanadium 50]

Here;'s the thing. You ask an A-level question based on an A-level text and want a B-level answer. That's not going to work ("Please prove the Fundamental Theorem of Calculus without using subtraction or any math more advanced than subtraction".)

The electron was an example. It works for all charged particles.

If you follow exactly the same procedure for neutrinos, you get 0 = 0. True, but unhelpful.

Bringing the Z boson into it is unlikely to be helpful, as it brings in more complications.

 

[MichPod]

Here;'s the thing. You ask an A-level question based on an A-level text and want a B-level answer. That's not going to work ("Please prove the Fundamental Theorem of Calculus without using subtraction or any math more advanced than subtraction".)

The electron was an example. It works for all charged particles.

If you follow exactly the same procedure for neutrinos, you get 0 = 0. True, but unhelpful.

Bringing the Z boson into it is unlikely to be helpful, as it brings in more complications.

Well, well, to the best of my memory I marked my question as "A", no idea how it got "B" (may be it was graded so by the Moderators?). ;-)
I nowhere told I want an answer below the level of the text I was referring to.

 

[PeterDonis]

to the best of my memory I marked my question as "A", no idea how it got "B" (may be it was graded so by the Moderators?

Yes, the thread level was changed because your posts indicated that you did not have the appropriate background knowledge for an "A" level discussion.

I nowhere told I want an answer below the level of the text I was referring to.

It's not just a question of what level of answer you want, but of what level of background knowledge you have.

 

[MichPod]

Well, whatever. If I got to that page of the course I referred to, I can expect an answer on the same level at least. Or I could expect an answer in the same terms and level on which I asked the question, providing my question was put correctly.

 

[PeterDonis]

If I got to that page of the course I referred to, I can expect an answer on the same level at least.

If you are referring to your post #5, you have gotten answers to that post on the same level as your question.

 

[MichPod]

If you are referring to your post #5, you have gotten answers to that post on the same level as your question.

I probably missed some important part in your #7 answer. Do you mean that this symmetry U(1) related to the phase change I was speaking about simply does not apply to neutrino?

Taking photons (as an example of a boson without an electrical charge), do I understand it correctly that a phase change symmetry U(1) does not apply to them either?

 

[PeterDonis]

Do you mean that this symmetry U(1) related to the phase change I was speaking about simply does not apply to neutrino?

The symmetry U(1) is related to the electromagnetic interaction. The neutrino is electrically neutral. So what do you think the answer is?

Taking photons (as an example of a boson without an electrical charge), do I understand it correctly that a phase change symmetry U(1) does not apply to them either?

Now you are getting yourself very mixed up.

The photon is the gauge boson for the U(1) gauge symmetry of the electromagnetic interaction. The fact that the photon is uncharged is a consequence of the fact that the field equations for this interaction, Maxwell's Equations, are linear, meaning that there is no coupling due to the electromagnetic interaction between the gauge bosons themselves.

 

[MichPod]

I was speaking of U(1) as of the local symmetry when the field is multiplied by some phase factor (a complex value with magnitude one) which may be different in each space-time point. Nowhere did I mention that this is necessarily a symmetry of the electromagnetic interaction. Let's may be call it even not U(1), but as a "local symmetry of the field phase change", then we'll arrive to my terms and to the issue I consider I was discussing.
Disclaimer: yes, this symmetry is U(1) in math terms, but if you identify U(1) strictly with EM interaction by definition, then I'd prefer not to mention that it is U(1) and will just call it "a symmetry under a phase change".

 

[PeterDonis]

I was speaking of U(1) as of the local symmetry when the field is multiplied by some phase factor (a complex value with magnitude one) which may be different in each space-time point. Nowhere did I mention that this is necessarily a symmetry of the electromagnetic interaction.

I know you didn't mention it. I did, because it is the symmetry of the electromagnetic interaction, and that is why it is relevant to this discussion.

Let's may be call it even not U(1), but as a "local symmetry of the field phase change"

No, let's not ignore the actual correct physics involved.

if you identify U(1) strictly with EM interaction by definition

I have said no such thing. What I have said is that, in the actual physics that you are asking about, U(1) is the gauge symmetry of the electromagnetic interaction, and that is why it is relevant to this discussion.

You are giving plenty of illustrations of why the level of this thread was downgraded to "B". Your whole approach to this is very superficial and you don't appear to have a good grasp of the actual physics involved.

 

[vanhees71]

I am following https://www.rainerhauser.ch/public/scripts/StandardModel1.pdf, page 27 and 28.
They transform the electron field in the Dirac equation by multiplying the wave function by some arbitrary magnitude-one complex value different in each point in the space-time, i.e. by arbitrary changing the phase of the wave function, then to compensate for this (to promise local symmetry in respect to phase change), they introduce a spin 1 massless field which couples with electron and which they identify as electromagnetic field.

So practically my question was what would be the result of the same procedure if applied to other particles, not electron (don't they have the same sort of local symmetry?). Like what field would we need to introduce to promise the same kind of local symmetry for a neutrino? Would it be meaningful if we try to promise a local symmetry of the same kind for a photon or, alternatively to some other massive boson like Z boson and which field would we need to introduce as a result of that?

QED can be constructed by "gauging" the mentioned global U(1) symmetry, leading to the introdoction of a "gauge connection", i.e., the electromagnetic potential. In principle can do this for any fields of charged particles, e.g., for pions (although there a more sophisticated version, called "vector meson dominance model(s)" where beyond pions also the light vector mesons are involved leads to much better phenomenological results).

As already mentioned above, since neutrinos are electrically neutral you cannot use this approach of gauging the U(1) symmetry. Here you have to read about the extension of the gauge principle to non-Abelian gauge symmetries. Here the "flavor symmetry" is gauged. The symmetry group is a (chiral) $\text{SU}(2) \otimes \text{U}(1)$ symmetry (weak isospin and weak hyper-charged) which is "Higgsed" to the electromagnetic $\text{U}(1)$ symmetry, i.e., of the four gauge bosons three get massive by absorbing the would-be Goldstone field-degrees of freedom (and that's why there's in fact no spontaneous symmetry breaking of a local gauge symmetry although it's still called this way after more than 60 years after discovery of the Anderson-Higgs-Kibble-et-al mechanism). So you have three massive gauge bosons realized in a "hidden local gauge symmetry", which are the $W^{\pm}$ and $Z^0$ vector bosons of the weak interaction and a massless photon. In the physical spectrum you have in addition to the lepons, quarks, and gauge bosons (at least) one massive scalar boson, the Higgs boson(s).

Also note that this electroweak Glashow-Salam-Weinberg model (aka quantum-flavor dynamics, QFD) is only conistent taking into account leptons (consisting of a charged lepton and a neutrino) and the quarks (an "up" and "down" version, each with 3 color-degrees of freedom) in each generation. We have 3 generations in the Standard Model, i.e., the electron, electron neutrino, up- and down-quarks; muons muon neutrinos, charm- and strange-quarks; and the tauon, tau neutrino, and top- and botton-quarks.

A good textbook at "B-level" is

O. Nachtmann, Elementary Particle Physics - Concepts and
Phenomenology, Springer-Verlag, Berlin, Heidelberg, New
York, London, Paris, Tokyo (1990).

It's a bit outdated, because naturally it doesn't contain a discussion of neutrino masses and mixing, but it's an approach which introduces the Standard Model "as simple as possible but not simpler".

 

[Vanadium 50]

So would can we expect if the same procedure would be applied to... fermions without electrical charge

It is surely in your text.

The author introduces the EM interaction, and the Lagrangian picks up a term $-iq\gamma^\mu$. If q=0, this term is zero, and the derivation stops there.