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Yang-Mills theory, confinement and chiral symmetry breaking

  1. Jan 6, 2016 #1
    I was thinking about hadrons in general Yang-Mills theory and I have some doubts that I'd like to discuss with you.

    Suppose that we have a Yang-Mills theory that, like QCD, tend to bind quarks into color singlet states. So far nothing strange, even QED tend to bind electromagnetic charges to form neutral systems. The twist in Yang-Mills theories comes when we consider the running of the coupling constant:
    $$
    \alpha(\mu)\simeq\frac{1}{ln(\mu/\Lambda)}
    $$
    which present asymptotic freedom, i.e. it decrease when the energy scale ##\mu## increase. This formula also suggest that when the energy scale approach the energy ##\Lambda## the interactions between two colored objects becomes very strong, so we assume that we cannot observe colored objects on scales greater than ##1/\Lambda##.

    Here my first question: We cannot see free quark in our world because the scale ##1/\Lambda## happen to be smaller than hadronic typical dimension so we cannot pull quarks enough outside hadrons to see them in non hadronic environment? And, if yes, could exist Yang Mills theories where this not happen, i.e. where the hadron typical scale is smaller than ##1/\Lambda## and then we can see free quarks?

    In QCD, being the hadrons system of strongly coupled quarks, we cannot study them by perturbative approach. However the presence of a spontaneously broken chiral symmetry allows us to study some of their properties.

    Here my second question: We do not need this SSB to form hadrons, right? There could be some theory were we have hadrons but not this SSB?

    If something is unclear tell me and I'll try to explain better.
     
  2. jcsd
  3. Jan 11, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jan 12, 2016 #3
    Dear Andrea M.,

    I also find your question very interesting and I hope some people that know what they are talking about will reply.
    On my side, I can tell you how I see it. Namely yes, I think the scale \Lambda that characterizes strong interactions is not an intrinsic property of the SU(3) theory of QCD, but it's something that can only be determined "experimentally". What I mean is that, if Lambda is defined as the scale where strong interactions become non-perturbative, one must compute the beta function, which depends on, for example, the number of flavours Nf, which is of course in principle not fixed by the theory and must be an external experimental input. In this sense, I guess, modifying the number of active flavours, one could imagine a world where Lambda QCD changes.

    Now as you say, Lambda represents the energy scale under which QCD becomes non perturbative and therefore only hadrons exist. This also means that, If our probe has an energy >> Lambda, you can actually see inside the hadrons and see the quarks. When we do experiments at high energies, like at the LHC, we actually see the free quarks!

    From this point of view, I guess that the answer to your question "can there be a Yang Mills theory (I guess we are talking about an asymptotically free one here) where we see free quarks", is an energy-dependent statement. You can "always" see the free quarks, if you have enough energy, and on the other hand, if the energy is low enough, the quarks of an asymptotically free theory, will always become "hidden" into hadrons at some point.

    I don't know if this makes sense to you.

    On your second question, I am not sure I understand what you mean... I think what happens is that there is an approximate (or broken) symmetry in QCD, the chiral symmetry, due to the fact that valence quarks (u-d mainly) have very very small masses, and this can be justified assuming that a Chiral symmetry is there and it must be somehow broken to give the mass patterns we see. This is of course a model, which works quite well though. I am not sure you could come to the same conclusions without such an approximate symmetry to "guide" you, unless you find completely independent methods to study the non-perturbative regime of your theory. But I might be wrong here...

    Cheers
    Sleuth
     
  5. Jan 12, 2016 #4

    ohwilleke

    User Avatar
    Gold Member

    First, to be clear, I am assuming that you are using "SSB" to mean spontaneous symmetry breaking as this seems to make the most sense from context. (In my day job, one of the things I do is help people apply for trademarks and you would be stunned how many different words can be attached to three letter acronyms there are out there (there are probably at least half a dozen in physics and chemistry alone)).

    Second, the universe of theories that haven't been invented yet, or have been invented but never used, is pretty much infinite. I am sure that someone sufficiently clever could come up with a theory wherein there were hadrons that does not take the SSB approach used in the SM. Indeed, I'm almost certain that this had been done. At any rate, I've seen abstracts now and again over the past ten years or so claiming to have done so (something that is often true only under highly unrealistic assumptions or is just plain wrong). But, those are not the papers I save or bookmark because that isn't my core area of interest, so I can't give you an easy reference. You could probably search the preprints with the right terms and find such a paper.

    Of course, getting an alternative theory that fits observation and is not too baroque is much more difficult.
     
    Last edited: Jan 12, 2016
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