Why is QCD non-perturbative at low energies?

1. Mar 21, 2005

Norman

Why is QCD non-perturbative at low energies?

In QCD isn't the expansion parameter the fine structure constant?
$$\alpha_s \approx \frac{1}{137}$$
Does this somehow depend on energy, which wouldn't make sense to me. I cannot seem to find a good answer anywhere, so I thought I would ask here.
Thanks.

2. Mar 21, 2005

juvenal

I think you have QED and QCD mixed up. Do you have access to a basic particle physics textbook like Halzen and Martin?

3. Mar 21, 2005

Norman

of course, of course, it was a foolish statement...

but what is used in QCD as the expansion parameter? If must (by implication) be dependent on the reciprocal energy somehow. I will check a particle physics book tomorrow, I guess.
Thanks Juvenal.

4. Mar 21, 2005

juvenal

QCD has a running coupling constant highly dependent on the energy scale.
There's something called Lambda_QCD which is around 200 MeV. At energies much larger than Lambda_QCD, the coupling constant is under 1, and one can do perturbative expansions. Unfortunately, hadronization, meaning free quarks becoming hadrons and which cannot be ignored in high energy physics, falls in the non-perturbative regime and there isn't a very good understanding of it. But there are QCD factorization theorems which allow one to separate out the long and short distance effects, so that one can still test QCD in the short-distance, and fit for the long-distance physics via experiment.

I would recommend the Halzen and Martin (or Griffiths) particle physics textbook as a place to start.

5. Mar 21, 2005

Norman

Ok, so I have consulted some online texts and I have began to look into the issue of the running coupling constant. It seems, at first glance anyways, that the issue of the strong coupling constant has something to do with the gluon being able to couple to itself. Fluctuations in the gluon propagator over small time periods are somehow absorbed into the coupling constant as an energy dependent term. Is this correct? Is there a simple way to explain this or is it just as complicated as it seems to me (very that is)?

6. Mar 21, 2005

Norman

Are you sure Lambda_QCD is so small? Perturbative QCD cannot (I thought) be used below 1 GeV. The whole reason I came upon this is because I am looking into explainations of meson production through inelastic nucleon-nucleon scattering in the energy range of 1-10 GeV. And in the literature it explains that below 10 GeV perturbative QCD is not used due to the nonperturbative nature of QCD at those energy scales. But above about 1 GeV, the methods used in the low-energy regions, namely the One Boson Exchange Models or the effective field theory Chiral Perturbation Theory, do not make the correct predictions of cross sections.

References for above:
http://arxiv.org/abs/nucl-th/0311002
and
R. Machleidt and I. Slaus, "The nucleon-nucleon interaction" J. Phys. G: Nucl. Part. Phys. 27 (2001)

7. Mar 21, 2005

juvenal

From Halzen and Martin - alpha_S is on the order of 0.1 for momentum transfers on the order of 5-6 GeV, so you do need to be at relatively high energies if you don't want to include too many terms in your expansion.

And yes, 200 MeV is about right.

You do know that the Nobel Prize in Physics was awarded this year for QCD, right?

I'm not sure what answer you want for the theory, but let me quote Zee's book p 369: "We cannot follow the renormalization group flow all the way down to the low momentum scale characteristic of the quarks bound inside hadrons since the coupling g becomes ever stronger and our perturbative calculation of beta(g) is no longer adequate. Nevertheless, it is plausible although never proven that g goes to infinity and that the gluons keep the quarks and themselves in permanent confinement."

8. Mar 22, 2005

marlon

QCD is non-perturbative at low energies because the strong interaction gets stronger and stronger when energy lowers. Perturbation theory needs a small coupling constant because you actually write down 'the formula's' in terms of a series with rising powers in the coupling constant. This is just like an expansion.

If the coupling constant (which expresses the strength of an interaction and examples are charge and mass) get's bigger and bigger, you can prove that your expansions won't converge but diverge, they will 'explode' and become infinite. So no perturbationtheory, if the coupling constant is too big and thus low energy QCD is non-perturbative.

This years Nobel Prize was awarded to three scientists that discovered this principle of asymptotic freedom (ie the interaction get's stronger as energy lowers). In a posh language, you can say that the beta function is negative...

and read the 'on the origin of confined species'

marlon

9. Mar 22, 2005

Norman

marlon and juvenal, thanks for the help.
I have skimmed through Griffiths and Bailin & Love and some papers and have found a large difference in the value of Lambda_QCD, in Bailin & Love they quote it at .065 GeV and Griffiths says it is between 100 MeV and 500 MeV while Machleidt & Slaus say it is approximately equal to 1 GeV. Is this large range of values simply do to some choice made in renormalization (which I use without completely understanding the process- I will be learning it in the next month or so, please bare with me)?

I do know that the Nobel prize was given for QCD and Asymptotic freedom this year and like marlon says that it is related to the beta function used in the coupling constant. I understand, at least superficially, that confinement is true- since we never witness free quarks, at least not yet. I also realize that the current search for the elusive Quark-Gluon Plasma at RHIC is right at the center of this whole argument.

10. Mar 22, 2005

Haelfix

The flip side to asymptotic freedom is infrared slavery. This is why the study of nuclear forces was so difficult for thirty years, and why the theory is still patchwork at low energies.

11. Mar 22, 2005

PhysicsFan

Another potentially useful book is The Quantum Quark by Andrew Watson.