Exploring the Charge-less Boson of the Electromagnetic Force

[Rob Hoff]

Why is it that the boson for the electromagnetic force does not have a charge? I apologize if this question is rudimentary, as all of you guys are WAY out of my league. Thanks!

 

[ChrisVer]

I don't know if you can find an answer to this question. If I recall well, we don't know why particles have the charge they do.. however it doesn't appear with a charge in the end.

 

[bhobba]

Why is it that the boson for the electromagnetic force does not have a charge?

Well it is the carrier of the electric charge so it 'sort of' seems reasonable it doesn't have a charge.

That said there is one 'charge' this applies to - gravity. One of the deep things about gravity is gravity gravitates so the graviton itself carries gravitational 'charge' so to speak.

Because of that I would say its because the EM field is itself not a source of the EM field.

Thanks
Bill

 

[Drakkith]

Well it is the carrier of the electric charge so it 'sort of' seems reasonable it doesn't have a charge.

Bhobba, are you aware that the force carrier of the color force, the gluon, carries color charge?

 

[bhobba]

Bhobba, are you aware that the force carrier of the color force, the gluon, carries color charge?

Ahhhh.

Yes - forgot about that - that's true.

Sorry.

Thanks
Bill

 

[tom.stoer]

Why is it that the boson for the electromagnetic force does not have a charge?

We cannot answer this question.

What we can do is to construct a theory to describe electromagnetic fields (for which we know experimentally that they do not carry charge). And we can quantize this theory to describe photons (plus electrons and positrons). So the electromagnetic fields and the photons are uncharged by construction b/c this agrees with our experiments. And the quantization procedure does not introduce a charge for photons.

 

[vanhees71]

...and this indeed answers the OP's question in the sense of physics, where "why questions" are only answered in the sense that we can "derive" something from fundamental laws of nature. One should be aware that a law of nature is an empirically found fact about the world, which is assumed to hold true, and that natural laws may be modified when future observations disprove a hitherto established natural law as not being correct.

That said, let's see what contemporary physics has to say about the question, why photons have no charge. The underlying natural law, on which this answer is based is the Standard Model of elementary particles (SM). The SM tells us that all so far observed matter consists of elementary particles that are described as Dirac fields (fermions) and are called quarks and leptons. Further there are three elementary forces, the strong, the weak and the electromagnetic interaction that are all described by socalled gauge fields, and last but not least there's a scalar field, the Higgs field and the associated Higgs boson that plays a special role. Gravity is not described by the SM, and there is no consistent quantum theory of gravitation yet (this shows, that my caveat is well-deserved, because for sure the SM is not complete, and the lack of a consistent description of gravity is not the only reason to believe that, but that's another story.

The SM is nevertheless a very successful model of the elementary particles and their interactions (except gravity). The SM is based upon a few principles, which all have to do with the idea of symmetries of the mathematical laws describing nature.

First of all any theory of physics should obey the symmetries underlying the mathematical description of space time, and for high-energy particle physics you must use the relativistic description of space time. Leaving out gravity, it is sufficient to use special relativity. Further we know that on the microscopic scale of particles everything follows the laws of quantum theory. Thus, one has to build a quantum theory that obeys the symmetries of special-relativistic space time. As it turns out from this assumption alone follows that we have to deal with a many-body theory, and indeed observation shows that at relativistic collision energies between interacting particles you can create and destroy particles easily.

Now quantum theory with a non-fixed number of particles is most easily described as a socalled quantum field theory. Further, all physical theories must be causal, i.e., nothing can cause an effect which would afford propagating a signal of any kind faster than the speed of light in a vacuum. Already in classical physics, the only way to accommodate actions among bodies was to introduce the field concept (first done by Faraday in the context of electromagnetism), i.e., one assumes that there are only local interactions, i.e., a charged body feels a electric force from a nother charged body not instantaneously as assumed in Newtonian physics but due to the electric field caused by the presence of the other charge at the distant place, and changes of this field due to changes in the charge distribution (including the motion of charges) can only propagator with the speed of light in a vacuum. The force acting on a test charge is thus due to the electric field at the present position of this particle, and in this sense the interaction is local.

This concept has been generalized to the case of quantum theory, and the SM is thus based on the concept of a local microcausal relativistic quantum field theory.

The symmetries of spacetime now tell us even more! E.g., space and time are "translation invariant". That means that all natural laws are valid no matter where I am or when I perform an experiment. One can show that any symmetry like that necessarily implies the validity of a conservation law (and vice versa, i.e., any conservation law implies a certain symmetry of your mathematical equations). These are Emmy Noether's theorems which form the basis of all modern physics.

The good thing with this very abstract ideas is that now you can play around and systematically look for quantum-field theoretical models describing elementary particles! The symmetries of space and time lead to a first classification of all possible kinds of particles. Together with the causality principle it tells us that first of all any particle must be characterized by a mass and a spin. The mass can take any real value, including 0 and the spin takes values 0, 1/2,1,3/2, etc.

Further the causality principle forces us to quantized particles with integer spin as bosons and such with half-integer spin as fermions. This has to do with the strict indistinguishability of particles that do not differ by some intrinsic quantum number (mass, spin, charges; we'll come to the charges in a moment). Fermions obey the Pauli exclusion principle, i.e., fermions of the same kind (say electrons) can not be in the same single-particle quantum state. E.g., there can only be 2 electrons in the single-electron ground state of an atom (it's two, because the electron has spin 1/2 and this implies that there is an electron state with spin-z component +1/2 and one with -1/2, and this makes two states that distinguish the two electrons in the lowest-energy level of an atom). Bosons do not obey the Pauli-exclusion principle but to the contrary like to stay together in a very specific sense.

The starting point for the development of the SM was classical electromagnetism, which in fact is a classical local relativistic field theory. It turns out that you can understand classical electrodynamics, i.e., Maxwell's theory as the interaction between electric charges and currents due to a massless vector field. Now quantizing a massless vector field immediately leads to the conclusion that such a field leads only to a consistent quantum theory if you describe it as what's called a "gauge theory". Quantzing the electromagnetic field from Maxwell's theory implies the existence of photons, which can in a very rough sense understood as particles, but with a lot of caveats.

Of course, we also have to describe the interaction of charged particles due to the presence of the electromagnetic field and the creation or electromagnetic fields due to the presence of electric charges and currents. Now the mathematics tells us that this is only possible without contradictions when there is another kind of symmetry, involving not space and time but the "matter fields" that represent particles. So far we have discovered only spin-1/2 particles (except of the Higgs boson which is a spin-0 particle) and the field discribing them is called a Dirac field. This Dirac field obeys a symmetry which amounts to multiplying it with a complex number of modulus 1 (a "phase factor"). This symmetry implies the conservation of electric charge (a la Noether).

But now, if you want to couple the photon field to the Dirac field you must extend this symmetry by not only allowing to multiply the Dirac field with a constant phase factor but you must demand that the theory must stay even symmetric under multiplication of the Dirac field with a space-time dependend phase factor. The mathematics tells us that this is only possible by introducing precisely a massless spin-1 field! This is perfectly for what we need to describe electrodynamics since the electromagnetic field is exactly such a massless spin-1 field (at least as long as classical Maxwell theory is concerned). Now you can play with this model, which is called QED. The most simple kind is to just have one Dirac field (representing electrons and their anti-particles, called the positron; the phase-factor symmetry dictates that the positron has exactly the same mass and spin as the electron but opposite electric charge) and the massless spin-1 field (the gauge field) guaranteeing the local gauge symmetry described above. This gauge symmetry also dictates the way how we are allowed to couple the electron field with the gauge field, and this leads precisely to a quantum version of Maxwell's theory (with the little but important extension that the spin-1/2 nature of the electrons and positrons implies that they carry not only an electric charge but also a magnetic dipole moment, i.e., they are also little magnets).

Another more technical issue is that relativistic quantum field theories can only be handled in the sense of perturbation theory, i.e., you first describe the particles as non interacting and then include the interactions by formal expansions in powers of the coupling constant (which in the case of QED is the elctric charge of electrons and positrons). Everything looks nice in the lowest orders of perturbation theory and also the predictions of cross sections for scattering processes like electron-electron elastic scattering, scattering of electrons with positrons, annihilation of an electron-positron pair to two photons or the scattering of a photon with an electron (Compton scattering) look not bad in comparison to high-precision measurements. But if you go to higher orders of perturbation theory, all of a sudden certain integrals involved in this calculation become meaningless, because they lead to divergent results. It has turned out, however, that if you restrict the interaction terms between the electron-positron and the gauge field to the lowest order, you get what's called a Dyson-renormalizable theory. The idea is that the charges and masses of the free particles used in perturbation theory to start with (the bare charge and mass) are unobservable but only charges and masses of the interacting particles have physical significance and take the values measured in experiments. In this way you can, at any order of perturbation theory, lump the divergences of the integrals into the unobservable bare quantities, adjusting the observable physical quantities to experiment. A Dyson-renormalizable theory is special in the sense that you only need a finite number of such adjustable constants (for QED the wave-function normalization constants, the mass of the electron and positron and the electric charge (-e for an electron +e for a positron)).

What comes out of this very complicated procedure is the best model of a certain part of nature created ever! QED predicts quantities like the electromagnetic moment of the electron to a precision of 12 (or perhaps even more) digits in agreement with the measured value. The same holds true for the energy levels of the hydrogen atom (including the socalled Lamb shift which is a relativistic QFT effect).

Now we can answer the original question! The local gauge symmetry underlying QED, i.e., the mustliplication of the Dirac field with a space-time dependent phase factor (and the corresponding gauge transformation of the electromagnetic gauge field, i.e., the massless spin-1 field) is Abelian, i.e., performing two such mathematical operations on the fields commute. That's easy to understand: The multiplication of complex numbers commutes, i.e., it doesn't matter in which order you multiply the Dirac field with the two space-time dependent phase factors. This is called a U(1) symmetry (U(1): unitary group of multiplication of phase factors). It turns out that this implies that the photon field does not interact with itself. This is also the deeper reason why the Maxwell equations are linear partial differential equations.

Now there are also the strong interactions. It took the physicists quite a while to establish the correct theory for strong interactions. The search for it lead to the discovery of quarks. Nowadayw we know that there are six sorts of quarks (called up, down, charm, strange, top, and bottom), each of which comes in three varieties which we label as "color" (red, green, blue). Of course it's not color in the literal sense but just a way to label the otherwise indistinguishable quarks. Now you can build a model very similar to electrodynamics, but you use a more complicated gauge symmetry, called a non-Abelian gauge symmetry. The strong force is absolutely symmetric for the three colors, and you can group the spin-1/2 fields describing quarks (it turns out that quarks are also spin-1/2 particles as are the electrons in QED) in color triplets and act with unitary $3 \times 3$ matrices on these triplets. Now you demand that the theory is symmetric under such socalled SU(3) gauge transformations and further demand that this symmetry should be local, i.e., the SU(3) matrices (building a group in the mathematical sense also called SU(3)) may depend on space and time. The locality again implies that you must introduce massless spin-1 fields. This time, however 1 such gauge field is not enough, but the SU(3) group demands 8 massless spin-1 fields. This field is called the gluon field, and to make the theory gauge invariant under the SU(3) gauge transformations the gluons, differently from the case of photons which where introduced to make the Abelian U(1) gauge symmetry local, the gluons must get interacting among themselves, i.e., gluons carry color charge, however a different one than the quarks; a gluon is more like an object carrying a charge and an anticharge, and this makes 8 combinations of colors (not 9 as one might think first, because you can form 9 combinations of three colors and three anticolors, but 1 of those is totally neutral, and doesn't interact; so it's only 8).

This color charge of gluons has profound consequences making the strong interaction very different from the electromagnetic interaction although the mathematics looks pretty similar. One profound difference is that quarks and gluons (or any other color-charged particle) are never observed as free particles as are electrons and photons that are not strongly interacting and thus carrying 0 color charge. We observe them only bound in what we call hadrons. So far we have only found either mesons which are understood as colorless bound states of a quark and an anti-quark or as baryons, which consist of three quarks (red+green+blue makes white, i.e., a colorless object). This socalled confinement is, BTW, not yet really understood from the underlying gauge theory of the strong interactions, which is called Quantum Chromodynamics (QCD) from the greek word "chromos" which means color.

Last but not least there are the weak interactions, which are also a gauge theory of another non-Abelian gauge group, called SU(2). Here, the most puzzling problem for the physicists was that gauge bosons must be kept massless, if you don't want to destroy the gauge symmetry, which is mandatory not to spoil the mathematical consistency of such a model. On the other hand there was a lot of evidence from the study of the weak interaction that they should be describable by a theory which is built much like QED and QCD but with massive gauge bosons. The way out of this dilemma was finally found by Englert, Brout, Higgs, Kibble, Hagen, and Guralnik, who figured out how you can get a gauge symmetry with massive gauge bosons and not destroying the gauge symmetry by giving this mass to the gauge bosons. Higgs was the one who pointed out that the way the physicist accomplished this by introducing addtional spin-0 bosons inevitably leads to at least one additional elementary spin-0 particle, which is now called the Higgs boson after its discoverer. The existence of this very elusive particle is so important that it was the main reason to spend a huge amount of money to look for it in building the Large Hadron Collider at CERN in Geneva. Finally the Higgs boson was experimentally found by two collaborations at the LHC on July 4, 2012. The more the scientists evaluate their data the better the evidence gets that the observed signal really satisfies all the properties predicted by the Salam-Glashow-Weinberg model of the weak interactions (which in some sense also unifies the electromagnetic and weak interactions and is thus also called the electroweak sector of the SM), which includes the Higgs mechanism and thus also the Higgs particle to make the weak gauge bosons (called $W^+$, $W^-$, and $Z^0$) massive without violating the underlying gauge symmetry.

Another great step in the development of the SM was also the proof of the Dyson renormalizability of non-Abelian gauge symmetries by 't Hooft and Veltman in 1971. Only after that proof of renormalizability the SM, including the Higgs mechanism, was really taken seriously by the physicists, and we can say it's one of the greatest achievement of physics since it describes with high precision everything we have observed about elementary particles so far.

As already said above, nevertheless the SM is not complete for various reasons, and thus physicists look for "physics beyond the SM". One reason (despite the lack of a consistent treatment of gravitation within quantum theory) is that the indications for the existence of dark matter become more and more convincing, and there's no clue in the SM, what this dark matter may be made of. There are extensions of the SM, called supersymmetric (SUSY) theories, introducing a symmetry between bosons and fermions, absent in the SM, but so far no hint of SUSY particles are seen at the LHC although they physicists looked for them vigorously. Maybe some hints for such particles (or any other deviation from the predictions of the standard model which can help the theorists to find an even better model) are found when the LHC is started again after its upgrade to higher beam energies.

 

[samalkhaiat]

Why is it that the boson for the electromagnetic force does not have a charge? I apologize if this question is rudimentary, as all of you guys are WAY out of my league. Thanks!

Because the symmetry group [U(1)] which gives rise to the EM-field (i.e., photon field) says it must be masseless, chargeless and spin-1 field. If we allow the “photon” to have “charge”, the symmetry group will no longer be U(1) and the theory will be very different from QED.

 

[ChrisVer]

How does U(1) say it must be chargeless?

 

[Matterwave]

[STRIKE]In my contact with this field, the fact that the gauge group of E&M is U(1), an Abelian group, would only require the force carriers to be mass less.[/STRIKE] That the force carriers are charge less is reflected in the fact that the equations of motion of the field (the Maxwell equations) are linear. A charged force carrier would lead to non-linear field equations. But I don't think you can say which is more basic, that the E.O.M. are linear, or that the force carriers are charge less. These are just two facts which imply each other.

EDIT: Redacted my first comment. I have to think about that point more (as it deals with the spontaneous symmetry breaking) before I would feel comfortable making it with conviction.

 

[atyy]

If I understand this article correctly, it seems to say that the U(1) gauge symmetry can be preserved even if the photon is massive. http://arxiv.org/abs/0809.1003

I'm not sure whether this also allows for nonlinear equations of motion.

 

[The_Duck]

How does U(1) say it must be chargeless?

Charged particles are those that transform nontrivially under a global U(1) rotation. But the photon field is invariant under a global U(1) rotation:

$A_\mu \to A_\mu + \partial_\mu \Lambda$

but $\Lambda$ is a constant function for a global rotation, so $A_\mu \to A_\mu$. So a U(1) gauge boson is uncharged under its associated symmetry.

 

[ChrisVer]

In fact under global U(1) nothing is charged because everything transforms trivially (even the fermionic field equations -dirac equation).

*editted since I read you wrote "global"*

 

[samalkhaiat]

In fact under global U(1) nothing is charged because everything transforms trivially (even the fermionic field equations -dirac equation).

*editted since I read you wrote "global"*

You will never understand QFT if you have poor understanding of Noether theorem. So, I suggest you study Noether theorem and pay attention to the following.

The free photon field Lagrangian is
$$\mathcal{ L }_{ em } = - \frac{ 1 }{ 4 } F_{ \mu \nu } F^{ \mu \nu }$$
This Lagrangian is invariant under the following GLOBAL $U(1)$ transformation
$$A_{ \nu } \rightarrow A_{ \nu } , \ \mbox{ or } \ \ \delta A_{ \nu } = 0 .$$
The Noether current associated with this symmetry is
$$J^{ \mu } = \frac{ \partial \mathcal{ L }_{ em } }{ \partial ( \partial_{ \mu } A_{ \nu } ) } \delta A_{ \nu } = 0 .$$
Thus the free photon field does not carry $U(1)$ charge.

Now consider the free fermion Lagrangian
$$\mathcal{ L }_{ D } = i \bar{ \psi } \gamma^{ \mu } \partial_{ \mu } \psi .$$
This Lagrangian is also invariant under the following GLOBAL $U(1)$ transformations
$$\delta \psi = - i \beta \psi , \ \ \ \delta \bar{ \psi } = i \beta \bar{ \psi } .$$
The Noether current of this GLOBAL $U(1)$ symmetry is then given by
$$J^{ \mu } = \frac{ \partial \mathcal{ L }_{ D } }{ \partial ( \partial_{ \mu } \psi ) } \delta \psi + \frac { \partial \mathcal{ L }_{ D } }{ \partial ( \partial_{ \mu } \bar{ \psi } ) } \delta \bar{ \psi } = \beta \bar{ \psi } \gamma^{ \mu } \psi .$$
The global U(1) fermionic charge is then given by
$$Q = \int d^{ 3 } x \bar{ \psi } ( x ) \gamma^{ 0 } \psi ( x ) .$$
This can be electric charge, lepton number, baryon number etc.

Finally, by gauging the $U(1)$ symmetry of $\mathcal{ L }_{ D }$ we obtain the following QED Lagrangian
$$\mathcal{ L } = i \bar{ \psi } \gamma^{ \mu } \partial_{ \mu } \psi - \bar{ \psi } \gamma^{ \mu } \psi A_{ \mu } - \frac{ 1 }{ 4 } F_{ \mu \nu } F^{ \mu \nu } .$$
Again, this Lagrangian is (obviously) invariant under the GLOBAL $U(1)$ transformations
$$\delta A_{ \nu } = 0 , \ \ \ \delta \psi = - i \beta \psi , \ \ \ \delta \bar{ \psi } = i \beta \bar{ \psi } .$$
And the Noether current in this case is given entirely by the fermionic part of the Lagrangian: $J^{ \mu } = \beta \bar{ \psi } \gamma^{ \mu } \psi$. Therefore, even in the full interacting QED the photon field remain chargless. As a rule, every real vector field has a vanishing $U(1)$ charge.

Sam

 

[samalkhaiat]

[STRIKE]... the fact that the gauge group of E&M is U(1), an Abelian group, would only require the force carriers to be mass less.[/STRIKE]

Local gauge invariance (of any group not only Abelians) rquires the force carriers to be massless vector fields.

That the force carriers are charge less is reflected in the fact that the equations of motion of the field (the Maxwell equations) are linear.

Now this follows from the Abelian nature of the gauge group.

A charged force carrier would lead to non-linear field equations.

Non-lineariry of the EOM and the charged nature of the force carries both follow from the non-abelian nature of the gauge group.

Sam

 

[Matterwave]

Local gauge invariance (of any group not only Abelians) rquires the force carriers to be massless vector fields.

And indeed that is why I redacted my comment. But it is interesting that after spontaneous symmetry breaking the photon turns out to be the mass less one while the W and Z bosons are the massive ones. I have to look closer at the derivations, but have not had time to yet.

 

[atyy]

Local gauge invariance (of any group not only Abelians) rquires the force carriers to be massless vector fields.

And indeed that is why I redacted my comment. But it is interesting that after spontaneous symmetry breaking the photon turns out to be the mass less one while the W and Z bosons are the massive ones. I have to look closer at the derivations, but have not had time to yet.

http://arxiv.org/abs/0809.1003 (Eq 27-29, and Eq 30) seems to argue that the photon can be massive without spoiling gauge invariance.

 

[geoduck]

Local gauge invariance (of any group not only Abelians) rquires the force carriers to be massless vector fields.

Would it be correct to say that local gauge invariance of the quantum action, and not the classical action, requires force carriers to be massless vector fields. It doesn't matter if the classical action has gauge symmetry or not - only that the quantum action has the symmetry?

One can prove using functional integral methods that the quantum action inherits the same symmetries as the classical action. But something must go wrong with spontaneously broken symmetries because the W and Z particles have mass and hence the quantum action is not gauge invariant (an $m^2 A^\mu A_\mu$ term in the quantum action violates gauge invariance). Would it be correct to say that the vacuum does not have SU(2) symmetry, but still has U(1) symmetry, and a semiclassical calculation of the quantum action requires that at tree level, you plug in the ground state solution to the classical action, and this is where the mass of the W and Z come in, but the photon retains its mass because the ground state expectation value of the Higgs, $\nu$, which appears in the tree level calculation of the quantum action is U(1) invariant?

And once you have the tree level quantum action (by plugging in the minima $\nu$ of the Higgs field into the classical action), you use this new action as your starting point for QFT, and calculate your loops diagrams, and everything turns out to be renormalizeable? But for consistency, don't you have to calculate your loops not with Feynman diagrams built from the classical action with the minima inserted, but from the determinant of two field derivatives of the classical action evaluated at the Higgs condensate $\nu$?

 

[vanhees71]

It's in fact not entirely correct. Indeed, in the non-abelian case the gauge bosons must be massless and can become massive only via the Higgs mechanism, because to preserve the local gauge invariance is mandatory to keep the model consistent. With broken gauge invariance unphysical degrees of freedom become part of the physical spectrum and thus the Hilbert-space structure gets destroyed because the inner product becomes no longer positive definite.

The case is different for Abelian gauge theories. Here you can use the Stückelberg mechanism to make the gauge field massive without destroying gauge invariance. Start from the Lagrangian
$$\mathcal{L}=-\frac{1}{4} V_{\mu \nu} V^{\mu \nu} + \frac{m^2}{2} V_{\mu} V^{\mu} + (\partial_{\mu} \phi)(\partial^{\mu} \phi) + m \phi \partial_{\mu} A^{\mu}.$$
This is invariant under the local gauge transformation
$$V_{\mu}'=V_{\mu} + \partial_{\mu} \chi, \quad \phi'=\phi+m \chi.$$
You can add matter fields in the usual way via minimal coupling. One application of this Stückelberg model is the model to describe neutral $\rho$ mesons (and $\omega$ and [/itex]\phi[/itex] mesons if you wish) and charged pions in an extended vector-meson-dominance model:

Kroll, N. M., Lee, T. D., Zumino, B.: Neutral Vector Mesons and the Hadronic Electromagnetic Current, Phys. Rev. 157, 1376, 1967
http://link.aps.org/abstract/PR/v157/i5/p1376

The nice thing is that this theory is renormalizable and gauge invariant. The usual analysis (e.g., with the path-integral formalism a la Faddeev and Popov) shows that the Stückelberg field $\phi$ does not appear as a particle in the physical spectrum but can be absorbed into the gauge field in a kind of unitary gauge, very similar as you can absorb the would-be Goldstone bosons into the gauge fields when using the Higgs mechanism. In any case the additional field gives the third spin component of a massive vector field, while the massless vector field has only two helicity states.

Of course, the model is not manifestly renormalizable in the unitary gauge, as is the case for the unitary gauge in the Higgs model, but you can choose a renormalizable gauge. As in the Higgs case you can choose a kind of 't Hooft $R_{\xi}$ gauge, which eliminates the mixing term between the Stückelberg ghost $\phi$ and the vector field. With this you can prove that the theory is renormalizable for the Green's and vertex functions and that the S-matrix elements are invariant under a change of the gauge constraint and thus also unitary, because the unitary gauge is approach as an appropriate limit of the $R_{\xi}$ gauge.

Further one should say that the Higgs mechanism provides (in both the Abelian as well as in the non-Abelian case) a manifestly gauge invariant and renormalizable theory. Of course, the local gauge symmetry must not be explicitly broken in any way, particularly there must be no anomaly of this symmetry. The standard model is free of an anomaly of its gauge group thanks to the particle content and the corresponding charge pattern of the quarks and leptons (this freedom from anomalies holds for each family separately). The point of the Higgs mechanism indeed is that it provides mass to the gauge bosons. It also provides mass to the quarks and leptons via Yukawa couplings since the quarks and leptons also must not be given masses in the naive way, because the local gauge symmetry of the electroweak model is a chiral one, and you'd destroy this chiral symmetry by writing down naive mass terms. Coupling the (flavor eigenstates of the) quarks and leptons via gauge-invariant couplings to the Higgs doublet leads the gauge symmetry intact and provides the observed (current) masses also to all the "matter particles".

 

[tom.stoer]

Interesting discussion, but I am afraid we lost Rob Hoff

 

[dextercioby]

Why is it that the boson for the electromagnetic force does not have a charge? I apologize if this question is rudimentary, as all of you guys are WAY out of my league. Thanks!

Charge comes from a non-trivial coupling to the e-m field. For the photon to have a charge, it would mean to self-couple. If it self-couples, the theory would be non-linear, hence it wouldn't describe electrodynamics.

 

[Matterwave]

http://arxiv.org/abs/0809.1003 (Eq 27-29, and Eq 30) seems to argue that the photon can be massive without spoiling gauge invariance.

Equation 30 looks to be the standard Higgs mechanism, but applied only for the EM field so it would be a photon Higgs boson I guess? Eqn 27-29 looks like a pretty artificial way to preserve gauge invariance by introducing a scalar field that transforms in exactly the opposite way to the mass QED term to cancel that gauge breaking effect out.

That's about all that I can comment on that paper. Perhaps wait until Sam has a chance to take a look.

Of course, the ultimate answer to this question is experimental verification. If the photon were experimentally found to have some small mass, then we would be looking for a model that gives it mass rather than leaves it mass-less.

 

[Vanadium 50]

The OP is a senior in high school. Do you think he understands these answers?

 

[ChrisVer]

There are mechanisms which can give photons mass (eg into a superconducting material). However the photon is pretty much massless...

 

[samalkhaiat]

The OP is a senior in high school. Do you think he understands these answers?

That is why there is post#8. He can ignore what he cannot understand.

 

[samalkhaiat]

Why is it that people in here think that models with mass generating mechanisms are counter examples to my statement? When I said “Demanding local gauge invariance gives rise to massless gauge fields” I was talking about the following programme:

1) We have a CLASSICAL matter field theory described some Lagrangian $\mathcal{ L }$.

2) This CLASSICAL theory (i.e., its Lagrangian as well as its GROUND state) is invariant under some global COMPACT group of transformations $G$.

2’) The symmetry is NOT broken spontaneously by Higgs or Stueckelberg type mechanisms.

3) We enlarge the COMPACT global symmetry group to a LOCAL one, i.e., to each space-time point $x^{ \mu }$, we associate an independent group element $g(x) \in G$.

4) Then, we demand that our theory is invariant under the enlarged (Local) group $G(x)$, i.e., demanding gauge invariance.

5) This programme NECESSARILY gives rise to a MASSLESS, spin-1 vector field $A_{ \mu }$ which couples in a well-defined way to the original matter field, $J^{ \mu } A_{ \mu }$, with $J^{ \mu }$ being the Noether current associated with the GLOBAL COMPACT group $G$ of the original Lagrangian $\mathcal{ L }$.

The PDF bellow explains the details of this programme. At least read my conclusions there.

So, please do not take my statement out of its intended context.

I also noticed references were made to Stueckelberg model of massive and gauge invariant vector field. Does this contradict my statement regarding the massless nature of the gauge fields? No, there is absolutely no contradiction. To see this let us describe so-called Stueckelberg model. Consider a free massless scalar field theory
$$\mathcal{ L } = \frac{ 1 }{ 2 } \partial_{ \mu } \phi \ \partial^{ \mu } \phi .$$
As we will see below, the massless nature of the scalar field is essential here. Clearly, this Lagrangian is invariant under the following global translation in the field space
$$\phi^{ ' } = \phi + g \ \chi , \ \ \ \ (1)$$
with $g$ is some real number (“coupling constant”) and $\chi$ is the real constant parameter of the translation. Notice that we are already out side the above mentioned 5 points programme, because the field translations (1) form a NON-COMPACT group (see point 2 above). Let us ignore this and continue describing the model.
The conserved Noether current associated with is global symmetry (1) is given by
$$J^{ \mu } = g \ \partial^{ \mu } \phi , \ \ \Rightarrow \ \partial \cdot J = \partial^{ 2 } \phi = 0 .$$
Now, let us enlarge the symmetry (1) to a local translation, i.e., we let the translation parameter to be space-time dependent one, $\chi \rightarrow \chi ( x )$. This implies that our original Lagrangian $\mathcal{ L }$ is no longer invariant under this local translation. This is because of
$$\partial_{ \mu } \phi \rightarrow \partial_{ \mu } \phi + g \ \partial_{ \mu } \chi .$$
However, if we introduce a field $A_{ \mu }$ transforming according to
$$A_{ \mu } \rightarrow A_{ \mu } + \partial_{ \mu } \chi ,$$
then we can form the following invariant combination
$$\left( \partial_{ \mu } \phi - g \ A_{ \mu } \right) \rightarrow \left( \partial_{ \mu } \phi - g \ A_{ \mu } \right) .$$
Thus, the Lagrangian
$$\mathcal{ L }_{ g } = \frac{ 1 }{ 2 } \partial_{ \mu } \phi \ \partial^{ \mu } \phi -g \ A_{ \mu } \ \partial^{ \mu } \phi + \frac{ 1 }{ 2 } g^{ 2 } A_{ \mu } A^{ \mu } - \frac{ 1 }{ 4 } F^{ 2 } , \ \ (2)$$
with $F_{ \mu \nu } = \partial_{ \mu } A_{ \nu } - \partial_{ \nu } A_{ \mu }$, is invariant under the following local gauge transformations
$$\phi^{ ' } = \phi + g \ \chi ( x ) , \ \ \ \ (3)$$
$$A^{ ' }_{ \mu } = A_{ \mu } + \partial_{ \mu } \chi ( x ) . \ \ \ (4)$$
Because of the mass dimensions, $[ A ] = [ \phi ] = 1$, equations (3) and (4) imply that $[ \chi ] = 0$ and $[ g ] = 1$. Thus we can take $g = m$ to be the mass of the field $A_{ \mu }$ and rewrite our gauge invariant Lagrangian (2) as
$$\mathcal{ L } ( \phi , A ) = - \frac{ 1 }{ 4 } F^{ 2 } + \frac{ 1 }{ 2 } m^{ 2 } \ A^{ 2 } - A_{ \mu } \ J^{ \mu } + \frac{ 1 }{ 2 } \partial_{ \mu } \phi \ \partial^{ \mu } \phi . \ \ (5)$$
Clearly, this is a gauge invariant Lagrangian describing massive vector field. But the story does not end here. Indeed, what had happened is the following: The massless scalar field becomes combined with the massless gauge field to form a massive vector field. This is because the symmetry is spontaneously broken, i.e., the invariance of the theory (5) under (3) CAN NOT be UNITARILY implemented in Hilbert space of the states. To prove this, let us assume the opposite. That is, assume that there exists Hermitean operator $Q = Q^{ \dagger } ( \sim a^{ \dagger } a )$, $Q | 0 \rangle = 0$ such that
$$\phi^{ ' } ( x ) = \phi ( x ) + \chi ( x ) = e^{ i Q \chi } \ \phi ( x ) \ e^{ - i Q \chi } .$$
Infinitesimally, we may write
$$\delta \phi \equiv \phi^{ ' } - \phi = \chi = \chi \ [ i Q \ , \ \phi ]$$
Thus, on taking the vacuum expectation value, and using $Q | 0 \rangle = 0$ and $\langle 0 | 0 \rangle = 1$, we find
$$\langle 0 | \delta \phi | 0 \rangle = \chi = 0 .$$
This cannot be true because $\chi$ is an arbitrary function. Thus, the symmetry can not be represented by unitary operator (i.e., Q does not exist) on Hilbert space of the states, i.e., the symmetry is spontaneously broken (check what I said in point 2’ above). This hidden symmetry mechanism combined with the fact that $\phi$ is massless is responsible for the mass term in the Lagrangian (5).

Sam

 

[cosmik debris]

Why is it that the boson for the electromagnetic force does not have a charge? I apologize if this question is rudimentary, as all of you guys are WAY out of my league. Thanks!

Man, there are some complicated answers to this. I think a simple observation would be that if the force carrier (boson) had a charge then the interaction with electrons and protons would be very non-linear and photons would strongly interact with themselves. This behaviour is not observed.