Does the property "equal charges repel" still work for non abelian theories?

[arivero]

I recall that there was an argument from Born expansion showing that exchange of odd spin between equal sign charges generates a repulsive potential, and if the charges are different or the spin is even the potential is attractive.

I wonder, how does it work for non abelian gauge theory charges? Is it still the same?

 

[jambaugh]

Consider that you can always find a one-parameter abelian subgroup to the non-abelian gauge group. As we see with broken iso-spin the U(1) subgroup of the SU(2) weak isospin group manifests as a component of electric charge with the other contribution being hyper-charge. This seems to indicate to me that, if abelian gauge groups (with spin-1 boson gauge mediators) necessarily repel like charges, then the same must also happen (with the same qualifications) in the non-abelian gauge case.

Qualifier: I have not investigated this deeply and so take my argument with a few grains of salt.

 

[malawi_glenn]

Non abelian vector bosons changes the "charge" of the particles in the interaction (consider QCD/SU(3) as an example).

Furthermore, non-abelian gauge bosons have self interactions, and you can have confinement.

 

[ohwilleke]

I recall that there was an argument from Born expansion showing that exchange of odd spin between equal sign charges generates a repulsive potential, and if the charges are different or the spin is even the potential is attractive.

I wonder, how does it work for non abelian gauge theory charges? Is it still the same?

My intuition is that this is not true at that level of generality.

This flows from three concerns discussed below the "UPDATE" section.

UPDATE: It occurs to me that by "exchange of odd spin" you mean gauge theories with a carrier boson that has an integer spin 1, 3, 5, . . . I hadn't caught that in the initial read of the quoted language. So, disregard the following three paragraphs.

As a counter-example, suppose that you devised a non-abelian gauge theory with a massless spin-2 carrier boson in which the carrier boson has interactions via that force with other carrier bosons. This is essentially the notional basis of canonical attempts of quantize gravity. In such a theory, the force described by the gauge theory and carried via the carrier boson could be attractive at all times between particles with equal sign and identical charges (e.g. between two particles with identical mass-energy in the quantum gravity context).

Of course, if your hypothetical theory were true, that would be a powerful "no go" theory for lots of kinds of quantum gravity theories. The fact that I've never seen such a "no go" theory articulated in that case also argues against its validity.

At a minimum, I think you need a non-abelian gauge theory with a spin-1 (mod 2) carrier boson for this to be the case (the graviton needs to be spin-2 rather than spin-1 since it is a force that is always attractive.

END UPDATE

1. I'm not convinced that even in that case (i.e. an "odd, integer spin carrier boson"), that it is impossible to imagine a theory in which, for example, identical charges are attractive and opposite charges are repulsive.

@Vanadium 50 states below that "The spin of the exchanged particle determines whether like-like interactions are attractive or repulsive" and if correct, then this basically overcomes this concern.

2. I am also not convinced that there could not be a case in which "equal sign" is ill defined because there are multiple kinds of charges that never present in isolation (as is the case for the eight color combinations of gluons) - although perhaps it is better defined than "opposite sign." So, perhaps this concern isn't insurmountable either.

3. I'm also not entirely convinced that it is impossible to have a gauge theory in which charged of the same sign sometimes repel each other and sometimes attract each other.

For example, suppose that you had a gauge theory identical to QED except that photons has a positive or negative electric charge, and that were X-rays or more ultraviolet equal signs were attractive and opposite signs repelled each other, while for photons with lower frequencies the usual attraction/repulsion rules of electromagnetism applied. I don't see why a theoretically possible non-abelian gauge theory couldn't have those properties.

I'm not sure if this is overcome by the definition of "gauge theory" which restricts itself to a narrower class of possible force carrying bosons than my expansive imagination of possibilities in theory space could allow. If one accepts that the carrier boson's spin determines attraction/repulsion then indeed, this shouldn't be possible.

It could be that some or all of three of these propositions can indeed be addressed, but I the reasoning to support these points is not self-evident.

 

[malawi_glenn]

I guess this could be a fun hobby project for me. $q_1q_1 \to q_1 q_2$ scattering and work out the potential. Or just a toy model ala "scalar qcd" with scalar particle fields transforming under some representation of SU(2) $\frac{1}{2} (D_\mu \phi_i)(D^\mu \phi^i)$

Will become problematic once I go beyond tree-level though :)

 

[Vanadium 50]

Boy there is a lot of brush here. Let me attempt to sweep some away:

(1) In non-U(1) theories, "opposite charge" is not well defined. However, "same charge" is. I presume the OP knows this, and that's why the question is formulated the way it is.
(2) Scattering at LO looks exactly the same for repulsive and attractive potentials. You need to look at higher orders.
(3) The spin of the exchanged particle determines whether like-like interactions are attractive or repulsive.

I will argue that this question has in general no answer for particles who change identities - and therefore charges - because of this interaction. If you have A + B → C + D did A turn into C or turn into D? Furthermore, it conflates kinematics and dynamics: if A and B are bound (i.e. the potential is attractive) and C and D are much lighter so have enough energy to escape the well, they can "repel", even if the potential between C and D is identical between A and B. The only way you will get an answer is if it comes from the definition.

I hope this clears things up and we can go on.

We have an example of a non-Abelian theory where we know the answer: the strong nuclear force. Not QCD - the remnant force that holds nuclei together. Furthermore, since the neutron and proton have (more or less) the same mass, we are in a special case where mass doesn't matter. And here we know the answer because we can measure it: at large distances, the force is attractive (carried by pions), and at small distances the force is repulsive (carried by rhos and omegas).

One last step. This tells us the answer for QCD for physical states. Two quarks attract (or repel) because the hadrons they are in attract (or repel).

 

[ohwilleke]

We have an example of a non-Abelian theory where we know the answer: the strong nuclear force. Not QCD - the remnant force that holds nuclei together. Furthermore, since the neutron and proton have (more or less) the same mass, we are in a special case where this doesn't matter. And here we know the answer because we can measure it: at large distances, the force is attractive (carried by pions), and at small distances the force is repulsive (carried by rhos and omegas).

Interesting example, although I don't know that it disproves the hypothesis.

I'd note that the pion carried force, the rho carried force, and the omega carried force probably count as separate gauge theories in the context of the OP rather than the strong nuclear force counting as a single gauge theory. Each carrier is really its own gauge theory in the sense described by the OP.

Also, the attractive pion as a spin-0 meson is probably not a gauge theory with an "exchange of odd spin" (which seems to mean a force carrying boson with spin 1, 3, 5, . . . ). while the rho meson and the omega meson are each spin-1 and hence involve an 'exchange of odd spin" and are repulsive.

 

[malawi_glenn]

I was thinking about providing the nucleon force example myself.

It is an effective non-abelian gauge field theory (chiral perturbation theory).

But, I was not sure how to account for same vs. opposite charges there.
Like is nn force mediated by rho and omega repulsive and np force mediated by rho and omega attractive? That is what the question was, same charge spin 1: repulsive, opposite charge spin 1: attractive.
Not so sure about that.

 

[Vanadium 50]

But, I was not sure how to account for same vs. opposite charges there.

"Oposite" can be confusing. "Same" is not so bad:" - the neutron is the same as the neutron.

 

[malawi_glenn]

"Oposite" can be confusing.

Perhaps I should have written "different" because that was the phrasing in the OP

 

[Vanadium 50]

"Different" is false. Go to the U(1) case: Q = +2 is different from Q = +1 but they both repel. :"Opposite" is, as I argued above, ill-defined at best. "Same" we can discuss.

 

[malawi_glenn]

"Different" is false. Go to the U(1) case: Q = +2 is different from Q = +1 but they both repel. :"Opposite" is, as I argued above, ill-defined at best. "Same" we can discuss.

Do you have any input regarding my "toy model" with SU(2) charged scalars?

 

[Vanadium 50]

Not really. I think we can agree that we have one example of a non-Abelian theory we can test experiemntally and its a mess. To me that answers the question -= "you get a mess".

 

[arivero]

"Different" is false. Go to the U(1) case: Q = +2 is different from Q = +1 but they both repel. :"Opposite" is, as I argued above, ill-defined at best. "Same" we can discuss.

Hmm yep I failed precision here. But let's go with sameness.

It is also funny that the argument is frequently used to explain gravity (spin 2 is even, so attractive for similar charges, says the lore) but the proof is about quantum mechanics.

 

[Vanadium 50]

If you say "gravity is spin-2" you are talking about quantum mechanics.

But it is fairrly simple to understand why spin-2 is the simplest possibility. Spin-0 would couple to a Lorentz scalar, so you would have no Pound-Rebka effect or gravitational lensing. Spin-1 would mean planets and stars repel each other rather than attract. So spin-2 is the simplest.

(We know more than that, but let's not drift the thread)

 

[vanhees71]

No, there are classical relativistic spin-2 fields. If you make a classical theory of classical point particles with a classical spin-2 field you get attraction between particles with equal-signed charges.

 

[arivero]

About the argument with the nuclear force, it is true that it is a mess, but I had always thought that it was because of this spin argument. Mesons with odd spin causing repulsive force, mesons with even spin causing attractive force.

Now I wonder if it even happens with U(1) if you consider exchanges of muonium, positronium etc. Of course very short lived, so the mess start at distances even smaller than the nuclear force radius.

 

[malawi_glenn]

Now I wonder if it even happens with U(1) if you consider exchanges of muonium, positronium etc. Of course very short lived, so the mess start at distances even smaller than the nuclear force radius.

You mean some kind of effective field theory of QED?

 

[arivero]

You mean some kind of effective field theory of QED?

A scenery could be a pair of protons. Besides their exchange of mesons, they could also exchange muonium states. Hmm, except the issue of size of course... but what about a stronger coupled U(1) theory?

 

[Vanadium 50]

Well, this is getting into a more mathematical physics which holds no real interest to me. The question is answered to my satisfaction: at zeroth order, it is "its the same" and at the next order it's "its a mess". Can we come up with a toy theory that is just messy enough to make the answer different? Maybe - and that will interest a class of physicists. Just not me.

 

[malawi_glenn]

I guess you are not a potential reader and buyer of my QFT book then :(

 

[Vanadium 50]

I guess you are not a potential reader and buyer of my QFT book then :(

Probably not.

I'm an experimenter. I am better with QFT than most of my colleagues, so it's not holding me back. But it is unlikely to make me a better experimenter compared to other things I might spend my time on.

 

[ohwilleke]

I guess you are not a potential reader and buyer of my QFT book then :(

I might bite on that one (if not too expensive).

 

[malawi_glenn]

I might bite on that one (if not too expensive).

Might take a while, hope you are still around the next decade!

 

[ohwilleke]

hope you are still around the next decade!

Me too!

 

[arivero]

(3) The spin of the exchanged particle determines whether like-like interactions are attractive or repulsive.

...
We have an example of a non-Abelian theory where we know the answer: the strong nuclear force. Not QCD - the remnant force that holds nuclei together. Furthermore, since the neutron and proton have (more or less) the same mass, we are in a special case where mass doesn't matter. And here we know the answer because we can measure it: at large distances, the force is attractive (carried by pions), and at small distances the force is repulsive (carried by rhos and omegas).

Pions being spin zero and rhos and omegas spin 1?

 

[Demystifier]

I guess you are not a potential reader and buyer of my QFT book then :(

What book?

 

[malawi_glenn]

What book?

Should have written "future" book ;)