# QCD, Yang-Mills

1. Sep 28, 2007

### Leonid

Why a magnetic current obviously must be C-
parity odd in the (exact) SU(2) theory? (C^{-1} J_M C=-J_M).
Best regard. Leonid

2. Sep 28, 2007

### OOO

Just a wild guess as a source for further discussion:

1) the magnetic current is a pseudovector since it is the curl of the dual SU(2) field tensor
2) thus it changes sign under PT
3) due to CPT invariance it also changes sign under C

Don't know if this is correct.

3. Oct 20, 2007

### blechman

It also follows from the generalized Maxwell's equations with magnetic monopoles. Gauss's law reads $$\nabla\cdot\vec{E}=\rho$$. Under C, charge changes sign and since derivatives don't care about C, it must be that $$\vec{E}$$ also changes sign. Applying this rule to the generalized Faraday law $$\nabla\times\vec{E}=J_M + \ldots$$, it must be that the magnetic current is also odd if C is a good symmetry.

4. Jun 19, 2008

### ghery

Hi:

I have heard that in QCD, there is something called the beta function, How is that function usefull and does that have to do with the asymptotic freedom of quarks?

5. Jun 19, 2008

### blechman

In any interacting quantum field theory, there is a beta function - this is the function that explains how a coupling constant varies with the energy you measure it.

For example: consider the electric charge of an electron. It turns out that this number depends on the energy you are probing the electron with. The reason in that in quantum field theory, you can make particle-antiparticle pairs and this has the effect of "screening" the charge. As you probe at higher and higher energies (shorter and shorter distances) you are able to cut through more of that "quantum interference" and so the measured charge increases. The beta function is a way of quantifying that idea.

Now on to QCD: it turns out (due to the nature of nonabelian gauge theories) that QCD has the opposite effect - it has ANTI-screening. So as you probe at higher and higher energies (smaller and smaller distances) the measured charge actually DEcreases! This is the phenomenon known as "asymptotic freedom."

As to how it's useful: it is absolutely VITAL! You cannot do QCD computations without it (and get the right answer!).

Hope that helps!