1. Dec 5, 2015

### BiGyElLoWhAt

So I've been trying to figure this out, and I've come to 2 possible conclusions, either QCD is broken, or it behaves completely different, with different characteristics than I am used to (such as possibly not obeying superposition?)

If blue is attracted to both red and green, then shouldn't a blue quark see a "color neutral" object as having 2 attractive forces and 1 repulsive? Also, as I understand it, quark confinement is a consequence of the coupling constant being a function of r, as opposed to 1/r, like the other forces. So if you superimpose the fields from each quark, a blue quark, I would think would see a large attractive force, as there is 2*(growth factor associated with distance) as strong of a force acting on the blue quark attracting it as there is repelling it (from the other blue quark in the white object).

Thanks for the indulgences.

2. Dec 5, 2015

Staff Emeritus
You've gotten off to a very bad start. Your "QCD is broken" option means that you are so incomparably brilliant that you can tell, even though you can't do a single QCD calculation, that it is wrong, and that the thousands of scientists who have learned how to do these calculations are fools unable to tell that they are completely wrong. It's arrogant and deeply disrespectful. It was also unnecessary - you could have asked this question in a straightforward way, without going out of your way to be offensive.

Ask yourself what would happen if the same-color repulsive force was twice as big as the other-color attractive force.

3. Dec 5, 2015

### Staff: Mentor

Well, it's pretty clear that QCD is not broken, and it is equally clear that QCD is a valid quantum field theory... So we might consider a third possibility, namely that you're understanding of it is in some way incomplete. Where have you studied/learned about it?

4. Dec 5, 2015

### BiGyElLoWhAt

Well, I didn't mean it like that, I was assuming I was wrong, I suppose I should have said that. However, I do know that QCD is generally accepted to be incomplete.

What would be the reasoning for the repulsive force to be twice as strong as the attractive force? That was one possibility that I ran through, but I can't see any reason that the strong force should behave in such a way, unless there's some sort of dependence on other quantum states, like spin, or something.

Again, I'm pretty sure I've gathered that you work in particle physics, and judging by your post, it seems that it might be the case. Not trying to offend you, or anyone else, so sorry about that. Thanks for the response, anyways.

5. Dec 5, 2015

### BiGyElLoWhAt

The internet and an undergrad particle physics class.

6. Dec 5, 2015

### BiGyElLoWhAt

So needless to say, I do NOT have an in depth understanding of what makes this force behave differently than the other forces.

7. Dec 5, 2015

### Orodruin

Staff Emeritus
Without realising it, you have hit the nail on the head here. Chromodynamics is a non-Abelian gauge theory and even classically that would imply that the equations of motion are non-linear and therefore do not satisfy the superposition principle. On a quantum level, it makes little sense to talk about forces at all - and as QCD is confining, it really is only viable to think of it in quantum terms.

8. Dec 5, 2015

### BiGyElLoWhAt

Ok, but I don't see how that necessarily answers the question. That was kind of just a random thing that I threw out there. Are you saying that one "strong force" is dependant on another "strong force"? Because that seems to explain it, but I don't see under what circumstances that would be something that happened. Also, I've read the definition and wiki page for albian and non abelian a few times but it never seems to stick. I'll check into it again.

9. Dec 5, 2015

### dextercioby

Can you just remove the word "force" from your statements about quantum theories? This word makes sense (in physics) only when Newton's laws are applicable, which is not the case of a theory whose name starts with "quantum". One more comment: to have misconceptions about a theory requires (in my mind) a minimum knowledge of that theory. I'm afraid this this doesn't apply to you.

10. Dec 5, 2015

### BiGyElLoWhAt

Ok, then let me rephrase. How does the binding mechanism behave? How does the creation (or whatever word you want to use) of the strong field occur, and how does the coupling to said field behave?

11. Dec 5, 2015

### Orodruin

Staff Emeritus
The point is that, being a non-Abelian theory, the strong field also acts a source of the strong field. Because of this, the superposition principle does not apply.

12. Dec 7, 2015

### BiGyElLoWhAt

Is this a consequence of gluons possessing color charge? And if that's the case, would it not be possible to sum those with the quarks' generated fields? Assuming we had some way to know approximately where the gluons were at, (I don't think we've ever observed one, at least directly)

13. Dec 7, 2015

### Orodruin

Staff Emeritus
Well, rather the other way around. The quarks carry colour charge because the field theory is non-Abelian. When you quantise it, the gauge field excitations will carry a colour charge.

Of course! ... But doing so will lead to new contributions to the field which will also act as a field source, which you can again take care of by adding their contribtion and so on. This is the entire basis of perturbation theory. The problem here is that QCD is strongly coupled and therefore perturbation theory breaks down (at low energies).

I suggest not looking at gluons as small billiard balls, this will only lead to misconceptions. I completely avoided the quantised theory above because the main features you are looking for are present already in the classical theory.

14. Dec 7, 2015

### BiGyElLoWhAt

I'm assuming you mean gluons vs. quarks. This, I suppose, is more of a chicken or the egg question, but is the reason we choose a non-abelian theory a consequence of our conceptual representation of the gluon, or was there something else that sparked the use of a non-abelian theory? If it's the latter, then what was the reasoning?

We can only observe the field where there are excitations, no? So, I can understand this coming out in a calculated field, but, as you said, this doesn't work for small distances (energies). Wouldn't this imply that there is a limited (energy? gluon? not sure what word to use here) emission?

I suggest not looking at gluons as small billiard balls, this will only lead to misconceptions. I completely avoided the quantised theory above because the main features you are looking for are present already in the classical theory.[/QUOTE] I understand. I'm not very well versed, and sometimes use words interchangeably, that probably shouldn't be.

15. Dec 7, 2015

### Orodruin

Staff Emeritus
I meant to write gluons in that paragraph. The mind is slower than the fingers.

We use a non-Abelian theory because it describes what we observe.

This depends on what you mean by excitations. If you put an equal sign between excitation and particle, this is not the case. Classical electromagnetic fields are an example. I do not understand your point about energy emission.

16. Dec 7, 2015

### BiGyElLoWhAt

I'm attempting to put an equal sign between excitations and force carrying particles, not all particles, if that makes a difference.

The point I was trying to make is this: it seems that gluons could only act as field sources to an extent. If we keep summing all these contributions, to some theoretical field, sure, it could increase forever, as where we calculate the field, we are essentially (as I see it, but feel free to correct me) putting a gluon there, which will then in turn contribute to the field, and etc. etc. That doesn't correspond to reality, at least at low energies, so this makes me think that there are really a finite number of gluons, and in turn a finite amount of energy (obviously) in the field, as infinities aren't something that we see. Maybe that's not really a point, I'm just trying to understand what's going on as best I can.

17. Dec 7, 2015

### Orodruin

Staff Emeritus
Your thinking here is way to classical. The contributions add and add and add and ... well you get the idea. These are virtual gluons and you have to add over all possible states. You do this in QED as well (although not with pure gluon diagrams) but the higher order corrections are much more suppressed due to the smaller coupling constant (also, there are fewer diagrams due to QED being Abelian). You simply cannot think of these as particle excitations or free particles.

18. Dec 7, 2015

### BiGyElLoWhAt

Hmmm... Ok, then. Where would be a good place to start looking? I have been on again off again working through QFT for the gifted amateur, but that's probably the only thing that I have that might be close. With most of the things I find online, they are either too layman, or too over my head. Virtual particles were touched on in class, but we spent probably close to a day or 2 on them, and then we were over it when we couldn't rectify them with conservation of momentum. We don't have a strong theory department where I am.