# A ΛQCDΛQCD from the beta function of QCD coupling constant

1. Mar 15, 2017

### spaghetti3451

The beta function for the strong coupling $g_3$ is given by

$\displaystyle{\mu \frac{\partial g_{3}}{\partial\mu}(\mu) = - \frac{23}{3} \frac{g_{3}^{3}(\mu)}{16\pi^{2}},}$

with

$\alpha_{3}(\mu = M_{Z}) = 0.118.$

We can use separation of variables to solve the beta function equation:

$\displaystyle{\int \frac{dg_{3}}{g_{3}^{3}} = - \frac{23}{48\pi^{2}} \int \frac{d\mu}{\mu}}$

$\displaystyle{\frac{1}{g_{3}^{2}} = \frac{23}{24\pi^{2}}\ln\left(\frac{\mu}{\Lambda_{\text{QCD}}}\right)}$

$\displaystyle{\frac{1}{\alpha_{3}} = \frac{23}{6\pi}\ln\left(\frac{\mu}{\Lambda_{\text{QCD}}}\right).}$

Using the physical condition, we then find that

$\displaystyle{\Lambda_{\text{QCD}} = \left(91\ \text{GeV}\right) e^{-6\pi/2.714}.}$

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What is the physical significance of $\Lambda_{\text{QCD}}$ as obtained by solving for the the strong coupling $g_3$?

2. Mar 15, 2017

### ChrisVer

I am not sure I understand your question, but LambdaQCD introduces a natural (lower) cut-off for your theory, as shown by the equation of 1/α3 (or g32)... ? you already used that when you integrated your energy scale.

3. Mar 15, 2017

### spaghetti3451

Does this mean that the theory of QCD is undefined below the QCD scale?

I thought that QCD does not have a Landau pole.

4. Mar 16, 2017

### ChrisVer

I think it's known though that QCD is non-perturbative at low energies? And your LambdaQCD you found (~90MeV) is the energy scale where α3 = 1 (which already whispers that it's not a good idea to expand into powers of α3).
That's why there is Lattice QCD.
Of course there can be more, but I am not very confident to write, for example I think LambdaQCD appears only in some renormalization schemes...

Last edited: Mar 16, 2017