# Asymptotic Freedom in QCD

1. Mar 29, 2008

### DMESONS

Hi,

As you knew "Asymp. freedom in QCD" did explain th strong interactions...
I want anyone to simplify this theory and show that:

why g factor above 1?

and
Is Feynman rules still work for strong interactions, I mean as it successfully interpret QED interactions?

2. Mar 29, 2008

### BenTheMan

I can answer the second question quickly, the first may take a while :)

Feynman diagrams are just a pneumonic for writing down a perturbation series. So, for example, you know that the Taylor series of something like

$$\frac{1}{1-x} = 1+ x +x^2 + \ldots$$

works for x < 1. If you try to plug larger values of x in, you end up with nonsense, like negative fraction = infinity (unless you're doing string theory!).

You can think of each Feynman diagram like a term in the above series. So, roughly,

physics process = $$g^2$$ easy Feynman diagram + $$g^4$$ slightly harder Feynman diagram + $$g^6$$ very hard Feynman diagram + ...

The important thing is that each term in the series is smaller than the preceding term, if g < 1. But if g >= 1, then the Feynman diagram method for doing calculations fails, and you have to invent elaborate methods to do calculations.

3. Mar 29, 2008

### kdv

I think you meant "mnemonic"

Just to add to what Ben said. Because of asymptotic freedom and infrared slavery, the Feynman diagram approach to QCD works only at high energies (when the energies exchanged are much larger than the QCD scale) because at low energy the coupling constant is large and the whole perturbation approach in terms of Feynman diagrams with more and more axchanges of gluons breaks down. Then one needs a nonperturbative approach (lattice QCD) or a phenomenological approach (MIT bag model, Yukawa potential, etc)

Last edited: Mar 29, 2008
4. Mar 29, 2008

### George Jones

Staff Emeritus
But this isn't enough to guarantee that the series doesn't sum to nonsense. For example, many (most?) of the renormalized perturbation expansions in QED that agree with experiment so well to a few orders are actually divergent series.

I guess this just means that perturbative QED isn't an ultimate theory.

5. Mar 29, 2008

### BenTheMan

Hi George---

All perturbation series are eventually divergent. The way to see this is that, even though you get another one or two powers of g at every order (which is less than 1), you get another n! Feynmann diagrams to compute. I have never proved this to myself, but I take it on good authority that the perturbation series is actually divergent at order 1/g. So this means QED amplitudes diverge if you compute too many loops (i.e. 137). This also means that if g is close to 1, the answers start to diverge very quickly.

The other way to see that QED can't be fundamental is the presence of a Landau pole---the coupling constant in QED becomes stronger at high energies (opposite behavior to QCD), and becomes infinite eventually. The place where this happens is the Landau pole, and I think it's at something like 10^38 GeV. Again, this is ok because QED is subsumed by the electroweak physics at 100 GeV.

6. Mar 29, 2008

### BenTheMan

Damned google. I knew it wasn't `neumonic', so I typed it into google.

7. Mar 31, 2008

### lbrits

First question:

Why is g large? Why is g in QED small? When you start computing these diagrams, you end up with integrals that look like they are infinite. But our theory never claims to be a complete theory (valid at arbitrarily short distances), so we hope that we can cut off our integrals at some energy scale $$\Lambda$$. If the theory is to be predictive (renormalizable), we might be able to do calculations without knowing what the high energy behaviour is. This isn't so bizarre as it sounds. We can do fluid dynamics because our equations are insensitive to the properties of atoms, save for one or two parameters like viscosity. In the same way, our QCD is insensitive to short-distances, except that we need to specify one or two parameters, like g. Since we don't have a more complete theory, we have to go out and make some measurements in the lab. In the same way, if you didn't know anything about atoms, you could only determine the density of a fluid from experimental measurement. If you knew about atoms, you could, in principle, calculate it.

Ok, now, the fun part. Renormalizable theories like QCD behave well when you scale $$\Lambda$$. What I mean by that is, if you change $$\Lambda$$, you end up with the same theory, but different values of g. This is reminiscent of fluids also. If you change the size of atoms, you still end up with fluid dynamics, but your viscosities and densities change. Therefore, since we don't actually know what value of $$\Lambda$$ to use, we pick one at random, and go measure g in the lab. Any prediction you make shouldn't depend on which $$\Lambda$$ you picked. This means that there is a relationship between g and $$\Lambda$$ such that the theory is invariant under changes in $$\Lambda$$ (they are accompanied by changes in g), but predictions like scattering cross-sections are unchanged.

Now, g is only a parameter in your theory, like viscosity. But you could measure the strength of an interaction between two particles in a lab, and call the strength $$g_{lab}$$. This isn't the same as g, (because, in principle, it is the result of infinitely many Feynman diagrams instead of just one), and is called the renormalized coupling constant. However, it turns out that because $$g$$ and $$\Lambda$$ obey a certain differential equation that reflects the insensitivity of the theory to changes in $$\Lambda$$, the renormalized coupling obeys a similar differential equation that contains the energy scale of the *experiment*. This equation tells you how strong $$g_{lab}$$ gets if you crank up the particle energies.

Furthermore, it turns out that $$g_{lab}$$ is small when energies are large, and large when energies are small. Quite the opposite of QED, and is very sensitive to the number of colours and flavours of quarks.

For a (much) better explanation, I suggest you take a look at the field theory textbook by Zee.

Second question:

Like what has been said, Feynman diagrams are a way of enumerating terms in an expansion of something more complicated. The problem with g being large is not so much that the diagrams don't converge, (since they don't converge anyway), but that diagrams at any order contribute in essentially non-trivial ways, whereas, for example, in QED, high order diagrams contribute very little by themselves. In QED, it's the fact that you have infinitely many orders to sum that makes it diverge. But as long as you only want an approximate answer, it's ok.

It is often possible to re-sum these series so that they converge. This is when the series is so-called Borel resummable. I belive this isn't the case when the theory has instantons, in which case you have to take care of those separately.

To get a sensible answer out of QCD, then you would need to sum all the diagrams. Or, at least, staggeringly many. This can be done by using tools in string theory, for example. Another possibility is taking the large-N limit, where the number of colours of gluons is taken to be large. Diagrams that can be drawn on a plane without any intersections are the ones that contribute most, and diagrams with more complicated topologies are suppressed by a factor of $$(g^2 N)^{-1}$$. In two dimensions, this can be done exactly, as was done by 't Hooft I believe.

Last edited: Mar 31, 2008
8. Mar 31, 2008

### lbrits

Well, depending on the theory, it is possible that the series is (Borel) resummable. So even though term by term the series diverges, we can recover the original expression that the series approximates, once we know all the terms. I think this process fails when you have instantons.

But in principle, it doesn't mean that perturbative QED is bad. It just means that you should be careful when interpreting the answer. For example, if I mistakenly gave you the series $$1 + x^2 + x^3 + ...$$ and told you that $$x = 5$$, you could dig deeper and figure out/guess that what I *meant* to write was $$1/(1-x)$$.

In the same way, if I gave you an infinite series representing the QED partition function, you could use some mathematical trickery to infer the original expression. But anyway I guess this is a technical discussion rather than a practical one. I believe this process only works for highly symmetrical (e.g. conformal) theories.

As was pointed out by BenTheMan, there are other, more pressing reasons to worry about QED (not just of the perturbative kind). But I don't know if the Landau pole is real or if is just an artifact of finite order perturbation theory.

9. Apr 6, 2008

### DMESONS

Hi all,

I am very grateful for your concern and help.

Thanks a lot