(Physics) Is every field theory a gauge theory?

First I preface my question because a field in physics is defined quite differently from the way it it is defined in mathematics. Parenthetically, I have never been able to bring these 2 definitions into register, but that is by-the-by.

So I guess the obvious first answer to my question is: "well it all depends what you mean by a gauge theory".

Agreed. Try this from an amateur: A physical theory will be a gauge theory if it is invariant under arbitrary local symmetry transformations. This seems to imply our theory is a field theory, but I am happy to be corrected here.

Now I am hampered somewhat by the fact the only field theories that I have even nodding acquaintance with are electromagnetism and GR. Certainly, or so I believe, my ad hoc definition applies equally to both.

Now following Drs Yang and Mills, we have a rather richer structure for EM than we do for GR. Specifically, that Yang-Mills gauge theories are modelled on principal bundles, whereas GR, at least in its usual formulation is not - though I can convince myself (with great deal of woolly thinking) that it might be.

Moreover, whereas the curvature in a Yang-Mills theory is a function (via the covariant derivative) of the connection, which is the field of Yang-Mills potentials. in GR the curvature has almost nothing (?) to do with the connection, which is not a field of any description rather curvature is a function of the metric.

So, let's say, for the sake of discussion (or not), that GR is a gauge theory, but not a Yang-Mills gauge theory.

Anyhoo, I asked a question.....why am I rambling on like this?

 

Log in or Sign up to hide all adverts.

Every quantum field, is described more or less by gauge transformations. This is what i have come to understand, so my answer would be a firm ''yes''.

 

Log in or Sign up to hide all adverts.

I'm a bit confused about what you're asking here, and I have a long history of confusing YOU with my responses.

Undeterred by history, let me say that I can certainly think of a field theory that doesn't have any local or global symmetries. For example, suppose I take a real field \(\phi\) and add an interaction that looks like \(\phi^3\). This action is not invariant under any discrete or continuous transformations that I can think of, but is a perfectly well-defined (albeit sick) field theory.

 

Log in or Sign up to hide all adverts.

\(\phi\mapsto-\phi\)?

 

Nevertheless, fundamentally-speaking, the fields we deal with are conformal to gauge symmetries.

 

Not a symmetry as the kinetic term will be quadratic in the field, ie \(\partial_{\mu}\phi \partial^{\mu}\phi + \phi^{3}\) doesn't have parity symmetry.

Ben's example is a particular case of Yukawa couplings, where fields couple to one another (and possibly themselves) but it cannot be formulated in terms of a gauge field connection. For instance, there is no connection A such that under \(\partial_{\mu} \to D_{\mu} = \partial_{\mu} + A_{\mu}\) we have \(\partial_{\mu}\phi \partial^{\mu}\phi + \phi^{3} \to D_{\mu}\phi D^{\mu}\phi\). This is in contrast to the gauge field coupling in QED between a spinor field \(\psi\) and the EM field A, ie \(\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi + e\bar{\psi}\gamma^{\mu}A_{\mu}\psi\) becomes \(\bar{\psi}\gamma^{\mu}D_{\mu}\psi\) under \(D_{\mu} = \partial_{\mu} + e A_{\mu}\).


The \(\phi^{3}\) example involves a single field and thus if you could construct it as a gauged system the connection would have to depend on \(\phi\) as you have no other field to use. Furthermore, since the connection must be vector valued, since it related to \(\partial_{\mu}\), its not clear how you'd construct such a thing from a scalar field alone. The only fundamental scalar field in the SM is the Higgs and it couples via Yukawa terms, not gauge terms and the same is true for the effective theory scalar fields (which were the original motivation for Yukawa's work). A vector field like the gluon, photon or weak bosons have a natural contribution to a connection and that is indeed how they are used in the SM. The next possibility for a bosonic connection would be spin 2 particles of which the graviton is the only one taken vaguely seriously in must cases. It obviously can't be gauged like \(\partial_{\mu} \to \partial_{\mu} + g_{\mu\nu}\), that's inconsistent tensor indices but it can be done in a consistent way as seen in the usual Christofell symbols and the L-C connection in GR.

Gauge symmetries are important in order to obtain things like charge conservation or protect the mass of certain particles (like the photon) from renormalisation. QH, I think that's why you might be under the impression all field theories are gauge theories, those field theories you can construct which aren't gauge theories are either physically unviable or mathematically 'sick'.

 

Ah, I assumed that interaction term was in the dynamic equation rather than in the Lagrangian.

Cool. Good to know.

 

Did you actually mean a photon in this example?

'' like charge conservation or protect the mass of certain particles (like the photon) ''

 

Oh dear, I seem, like someone else here, to have a habit of asking questions whose answers I don't fully understand. But if it seems I am under the impression "quoted" above, I confused you as much as I did Ben.

As above. I think I meant something like this:

GIVEN a set of physical phenomena that are best described by one or more interacting fields (in the sense that physicists define fields) and certain rules for these interactions, is it ever the case that this description - if one may elevate it to the status of a theory - can be globally invariant without being locally invariant?

Both Ben and Alpha gave examples where a "theory of fields", if this term has any meaning, are neither locally nor globally invariant. And they both, in their own ways admitted these were pathological or non-physical, which I think was to miss my point.

Or did I miss theirs? (On re-reading - always a good plan - I think I may have assumed Alpha to be asserting the opposite of what he actually did. Not sure.....)

 

Sure, but not if you're a string theorist

Please Register or Log in to view the hidden image!

Particle phenomenologists invoke global symmetries all the time---baryon number conservation, for example, or R parity if you're building SUSY models.

A simple example is the complex scalar field:

\(\mathcal{L} = \partial_{\mu} \phi \partial^{\mu} \phi* - m^2 \phi*\phi\)

You can even turn on a phi^4 term if you like. It turns out that this theory has a symmetry if \(\phi \rightarrow e^{i\alpha}\phi\), which you can check. But things go to hell if you make the \(\alpha \rightarrow \alpha(x)\).

If you want a local theory, you have to add a photon. Physicists actually call this ``gauging'' the symmetry. Then, as AN mentioned, you get beautiful conclusions, like conservation of charge and the fact that the photon MUST be massless.

The classic example of a global symmetry is the chiral symmetry of the quark sector in the SM. So, for example, if you turn off quark masses, there is a symmetry of the quark kinetic terms. But this symmetry is broken (slightly) by small quark masses.

 

Let me say, the purpose of adding the photon is to ensure that the symmetry \(\phi \rightarrow e^{\alpha(x)}\phi\) is obeyed.

 

May I ask, since Bohn's experiment proved finite measure capabilities, how one decides mathematically the photon has zero mass?

Your equation implies continuity.

Can you help me understand this?

I have other questions.

 

Mathematically is easy: electric charge is conserved, therefore the photon is massless. This is a direct consequence of quantum mechanics.

Measuring a zero photon mass is impossible, however, because that's how measurements work.

If you have any questions about this, you should start a new thread. If you continue to post off-topic, you will be banned.

 

First let me apologize to Alpha for not responding directly to him. Mostly his posts go way over my head: sometimes I think he is flattering me by assuming I know more than I actually do, sometimes I think I must be really thick. No doubt all here have made up their own minds on the latter....

Yes easily done. Essentially it has to do with the fact that \(\partial_{\mu}e^x = e^x\) and that \(e^0 =1\)

This I cannot quite see. Someone please walk me through it.

Sorry to be so dense

Uh? Say again? This I simply cannot process. Someone please explain in gentle terms. Sorry again

PS I am sitting in a French campsite in rural France connecting via my 3 dongle. This works magic in Vista (I hate MS) but not at all in Ubuntu. Any tips out there?

 

Well, here \(\alpha\) is a constant, so it is even more trivial than that.

Here a IS a function of x. We compute (assuming a is real)

\(\phi \rightarrow e^{i\alpha(x)} \phi, \,\,\, \phi* \rightarrow e^{-i\alpha(x)} \phi*\)

Then, substituting into the lagrangian above, clearly the mass term is invariant. But the kinetic term requires some work. To do this, let's suppose that a is some tiny thing. Then we can write

\(e^{i\alpha(x)}\approx 1 + i\alpha(x) + O(\alpha^2)\)

Then the kinetic term looks like

\(\partial e^{i\alpha(x)}\phi \partial e^{-i\alpha(x)} \phi* = \partial \left[(1+i\alpha(x))\phi\right] \partial \left[(1-i\alpha(x)) \phi*\right]\)

Now we can just multiply things out, taking derivatives. We have

\(= \partial_{\mu} \phi \partial^{\mu} \phi* + i \phi*\partial_{\mu}\phi\partial^{\mu}\alpha(x)- i \phi \partial_{\mu} \alpha(x)\partial^{\mu}\phi* + O(\alpha^2)\)

What's happening here is that the derivative isn't transforming properly. This probably should ring some geometry bells for you---we need a derivative which is invariant under the gauge transformations (covariant derivative).

What we find is that the derivative needs to have some extra piece that absorbs the inconsistent transformation of the flat space derivative. The extra piece of the covariant derivative (which, if memory serves, is the connection) is exactly the photon.

The thing that should just knock your socks off is that we just invent a particle, based on our intuition, that has a specific set of properties to leave the lagrangian invariant under the local U(1) symmetry, and we actually find it in experiments.

This sounds...obscene.

 

Ben I thank you for your patience.

OK, I hadn't thought of that simplification; makes life little easier.

Which is, if I make no mistake, just

\((\partial_{\mu}\phi+\partial_{\mu}i\alpha(x)\phi)(\partial^{\mu} \phi^*- \partial^{\mu} i \alpha(x)\phi^*)\)

Ya well, after several elementary calculus errors, I get something similar to this. So while not identical to your result, close enough to convince me that your general conclusion is right: \(\phi \to e^{i\alpha (x)}\phi\) doesn't transform nicely

Well yes, if my memory serves, the covariant derivative comes from the connection, something like....errr...

\(D\phi = \partial A + i \phi A\) where \(A\) is our connection. Or something, but yeah, they are intimately related.

But lacking any sort of physics "intuition" whatever, I cannot see for the life (or death) of me how you "equate" the connection \(A\) (a mathematical abstraction) with the photon (a physical entity, of sorts).

I guess there is a philosophical discussion to be had here, but I, for one, would not be interested - philosophy is for nerds in my book (in spite of being the son and grandson of professional philos!).

Is there a physics discussion to be had, addressing the same question?

 

Well, one thing to note (that might clue you in) is that the ``connection'' A has the same index structure as \(\partial\). This is a space-time index, so A must be a vector.

Secondly, A has no mass term, so it must be massless, if it is a real thing.

You can also study the interactions. The covariant derivative is

\(D\phi = \partial \phi + i e A \phi\)

If you were dealing with fermions,

\(D\psi = \partial \psi + i e A \psi\),

where e is the fundamental electric charge. Then you can write out the action (blah blah) and you see an interaction term

\(e \bar{\psi} \gamma^{\mu} A_{\mu} \psi\)

which describes two fermions interacting with A with strength e, which is exactly what we expect from the photon. (The gamma is there because it is in the kinetic term. It is the Dirac matrix.)

Finally, you can compute

\(\left[D_{\mu},D_{\nu}\right] = i e F_{\mu\nu}\)

where F is the Maxwell tensor.

After all of these calculations, you feel confident in saying that, if electric charge exists and is conserved, then my theory predicts the existence of a particle with all of the properties of the photon.

If you were to repeat the calculation, assuming that \(\phi\) transformed under some internal symmetry, then you'd find that the connection has to have some ``knowledge'' of this symmetry as well. Moreover, you can work out that A has to transform as an adjoint.

So all of these things should give you a clue that A should be a photon, or (in the more general case) some other gauge boson of some other force.

But the fact that you can equate this connection with a physical particle is indeed amazing, and the more you study these things, the deeper a respect you gain for people like Yang and Mills who were thinking about these things in the 1950's.

 

It means that you can investigate symmetries in physics using Lie algebras rather than Lie groups, as you make the connection Lie algebra valued, \(\alpha(x) = A^{a}(x)T_{a}\) for generators T.

As Ben outlines you have a Lie algebra generator valued connection. This leads naturally to the Maxwell tensor via \([D_{\mu},D_{\nu}] = F_{\mu\nu}\) where \(F_{\mu\nu} = F_{\mu\nu}^{a}T_{a} = \partial_{\nu}A_{\mu} - \partial_{\mu}A_{\nu} + [A_{\mu},A_{\nu}] = \Big(\partial_{\nu}A_{\mu}^{a} - \partial_{\mu}A_{\nu}^{a} \Big)T_{a}+ [A^{b}_{\mu},A_{\nu}^{c}]T_{b}T_{c} = \mathcal{F}_{\mu\nu}^{a}T_{a} + A^{b}_{\mu}A^{c}_{\nu}[T_{b},T_{c}] = \mathcal{F}_{\mu\nu}^{a}T_{a} + f^{a}_{bc}A^{b}_{\mu}A^{c}_{\nu} = F_{\mu\nu}^{a}T_{a} = F_{\mu\nu}\).

For u(1) you get f=0 and that's Maxwell's stuff. Otherwise you get Yang-Mills. When you make the matrix component structure of F manifest you get somethign of the form \(F_{\mu\nu}^{a}T_{a} = F_{\mu\nu}^{a}(T_{a})^{i}_{j} \equiv (F_{\mu\nu})^{i}_{j}\). Compare this to the Riemann tensor which has \([D_{a},D_{b}]\xi^{c} = R^{c}_{abd}\xi^{d}\) and can be viewed as \((R_{ab})^{c}_{d}\). Thus both of them are viewable in terms of commutators of covariant derivatives and as matrix valued 2 form coefficients. More formally (if memory serves) you have \(R^{c}_{abd} \in \Lambda^{2}(T*M) \otimes End(TM)\) and similarly for the gauge construction.

I'll explain more tomorrow when I have time and if anyone actually gives a hoot.

 

Well QH gives a hoot

But give me time to digest Ben's and your posts. They cover topics with which I should be familiar, but will require some heavy-duty thinking.

Put another way, next serve is mine

 

In field theories kinetic terms are usually completely symmetric, so if we make (infinitesimal) transformations which don't change potential (like changing phase), situation remains the same - they are gauge invariant.
So higher dimensional filed theories usually have some gauge invariance, while one dimensional usually don't have such continuous family of symmetries, but can have discrete one - like ( phi -> -phi ) on this nice animation here: en.wikipedia.org/wiki/Topological_defect