# Fundamental theories are gauge theories

1. Jan 6, 2014

### metroplex021

Hi folks -- I was reading some (non-technical) work by Frank Wilczek, in which he stated that any fundamental theory -- that is, well behaved in the E →∞ limit -- must be a local gauge theory. Does anyone know of the reasons for why this is thought to be the case?

Even sketchy remarks appreciated! Thanks!

2. Jan 6, 2014

### RGevo

I believe this is a comment about the re normalization. How parameters of the theory change as the scale (energy) changes.

Though im not sure what is meant by a fundamental theory. For example I believe one can have a field theory with global symmetries describing condensed matter systems.

And the SM is known to be effective. So is there then no known correct fundamental theorem?

3. Jan 10, 2014

### tom.stoer

How? Why?

4. Jan 10, 2014

### DarMM

The reason for this is that essentially matter is described by Fermions/Spinors (spin 1/2), roughly speaking and the interactions between matter are mediated by Spin-0 or Spin-1 particles.

In four dimensions, spin-0 particles are trivial on their own (don't interact) and so the theory needs spin-1 particles. However if these spin-1 particles are fundamentally massive (that is, if their mass is not generated by their interactions with another field), then the theory will contain irremovable infinities.

Only if the spin-1 particles are massless will there be no infinities. So we need massless spin-1 particles.

Finally massless spin-1 particles are not well suited to being described by fields (due to them have too few degress of freedom) however. Whenever we try the field has artificial unphysical extra degrees of freedom. The fact that these degrees of freedom can be altered without affecting the physics is known gauge symmetry.

So basically in four dimensions an interacting theory must be a gauge theory, because only the interaction of massless spin-1 particles is infinity/divergence free.

5. Jan 10, 2014

### RGevo

That was a good explanation DarMM, thanks.

Regarding the SM being effective, the evidence is neutrino masses and dark matter. Clearly an extension is required, even if we do not care about naturalness. There is a nice talk on this by Strumia, http://workshops.ift.uam-csic.es/WMH126/strumia.pdf. Where towards the end he discusses running the SM to Mplanck and above.

6. Jan 12, 2014

### tom.stoer

I still think that this is a (too) bold claim; yes, there are indications that the SM cannot be fundamental, but I would not dare to claim that we know this for sure.

7. Jan 12, 2014

### RGevo

Okay,

then, -> "And the *minimal* SM is known to be effective" ;)

8. Jan 12, 2014

### tom.stoer

I still don't agree, but perhaps this is due to my understanding of "effective theory".

For me an effective theory is a theory formulated in terms of non-fundamental d.o.f. which are sufficient to explain physics within a certain energy range, but which are to be replaced by new d.o.f. beyond a certain energy scale. Examples are solid stage physics (with phonons and other effective d.o.f.), chiral effective theories (with pions and nucleons), etc.

What about asymptotic safety approach to quantum gravity, Connes non-commutative geometry? We know for sure that we need an UV completion of SM + gravity, but it can very well be that there are no new fundamental d.o.f. from which the known particles of the SM do emerge (as bound states or whatever) but simply a new mechanism for UV completion whereas the d.o.f. known from the SM remain "fundamental". I would not call such a theory "effective".

All what I am saying is that currently we really do not know how such an UV completion will look like.

9. Jan 12, 2014

### strangerep

It has long puzzled me that such unwanted degrees of freedom of the photon field (arising from the so-called "continuous spin" operators in the Poincare unirrep) should automatically hook up with local phase changes of the fermion field. The photon "continuous spin" operators are essentially just left over from the rotation group as we contract the invariant mass to zero. But a massive fermion retains physically non-trivial 3-rotation degrees of freedom. So it seems they're not the same degrees of freedom that are being matched up when we minimally-couple fermion and photon to obtain a gauge-invariant interaction. Something else seems to be going on, but I don't understand what.

But I doubt that nature "cares" whether we can/can't solve our full interacting theory by perturbation around a free theory.

10. Jan 13, 2014

### DarMM

The thing is however, that the extra degrees of freedom are not just unwanted, but unphysical. The photon is simply not an object suited to being described by fields, since it is a rep of the little group $ISO(2)$, but fields can only naturally carry reps of the little group $SO(3)$. Only constrained fields can described the photon. So, when one uses an unconstrained field like $A_{\mu}$, then as you know, we get two extra nonphysical polarisations that cause unitarity to break down. This extra polarisations can then also interact with the electron, either through scattering, but more importantly, they form part of its Coloumb field.

In the field formalism, these extra polarisations manifest as gauge degrees of freedom, so the electron also picks up a guage variance. Since the photons form the electrons Coloumb field or electric charge, which is transformed by global phase changes, these non-physical photons induce a gauged phase.

It's nothing to do with solving the theory. Theories with divergences do not exist. So perhaps a clearer way to say it is that there doesn't exist a field theory in four dimensions that doesn't contain a massless spin-1 particle and hence has gauge symmetry in the field formalism.

11. Jan 13, 2014

### strangerep

There seems to be an inconsistency in what you're saying: the extra polarisations are unphysical, but... the Coulomb field is physical.

Oh,.... perhaps you're alluding to the schemes for dressing the electron field with a particular coherent photon state which causes (some of) the gauge degrees of freedom to disappear, while also giving the bare electron a Coulomb field, and banishing the QED IR divergences?

12. Jan 14, 2014

### DarMM

Actually, this might sound odd, but the Coloumb fields aren't physical in the standard approach to QED.

I'm talking about if we do QED with $A_{\mu}$ an unphysical object, containing two extra, unitarity breaking degrees of freedom. If we do this, instead of using a constrained field, or even better loop like operators that actually represent the photon, then these unphysical photons will contribute to the Coloumb field of the electron in the theory.

In other words, the theory contains electron states with unphysical Coloumb fields, unitarity breaking Coloumb fields. This Coloumb field is unphysical.

Only on a restricted subspace $\mathcal{H}_{phys}$ of the space of states of the theory will the Coloumb field not contain these extra polarisation, however since the electron field is an operator over the whole space it picks up the extra gauge phase regardless.

Last edited: Jan 14, 2014
13. Jan 14, 2014

### tom.stoer

It is perfectly derived in http://www.sciencedirect.com/science/article/pii/S0003491684710591

14. Jan 14, 2014

### RGevo

Hi sorry I come back into this a bit late.

What I meant was that, we must introduce at least new neutrino states. Thus, extending the SM beyond its current form without right handed guys.