ΛQCDΛQCD from the beta function of QCD coupling constant

In summary: I am not sure you can say that the theory is completely undefined below that scale.In summary, 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.## By using separation of variables, we can solve the beta function equation and obtain the relation ##\displaystyle{\frac{1}{\alpha_{3}} = \frac{23}{6\pi
  • #1
spaghetti3451
1,344
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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}.}##

--------------------------------------------------------------------------------------------------------------------------------------

What is the physical significance of ##\Lambda_{\text{QCD}}## as obtained by solving for the the strong coupling ##g_3##?
 
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  • #2
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
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
spaghetti3451 said:
Does this mean that the theory of QCD is undefined below the QCD scale?

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:

What is ΛQCD?

ΛQCD is the energy scale at which the strong nuclear force, described by the theory of Quantum Chromodynamics (QCD), becomes strong enough to bind quarks together to form hadrons.

What is the beta function of the QCD coupling constant?

The beta function of the QCD coupling constant is a mathematical expression that describes how the strength of the strong nuclear force changes as the energy scale is varied. It is an important tool for studying the behavior of QCD at different energy scales.

How is ΛQCD related to the beta function of the QCD coupling constant?

ΛQCD is related to the beta function of the QCD coupling constant through its value at a particular energy scale. The value of the coupling constant at this energy scale is used to determine the value of ΛQCD.

Why is ΛQCD an important parameter in QCD?

ΛQCD is an important parameter in QCD because it helps us understand the behavior of the strong nuclear force at different energy scales. It allows us to make predictions and calculations about the interactions between quarks and gluons, and is essential for understanding the structure of protons and neutrons.

How is ΛQCD determined from the beta function of the QCD coupling constant?

ΛQCD is determined from the beta function of the QCD coupling constant by solving for the value of the coupling constant at the energy scale where the beta function vanishes. This value is then used to calculate ΛQCD using theoretical and experimental methods.

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