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I see that it appears as a constant in the relation for the running of the strong coupling constant. What is its significance? Does it have to be established by experiment? Is it somehow a scale for quark confinement? If yes, how? I ask because I saw this in Perkins' Particle Astrophysics

After kT fell below the strong quantum chromodynamics (QCD) scale parameter ∼ 200 MeV, the remaining quarks, antiquarks, and gluons would no longer exist as separate components of a plasma but as quark bound states, forming the lighter hadrons such as pions and nucleons.

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Dear dbrane, $\Lambda_{\rm QCD}$ is the only dimensionful parameter of pure QCD (pure means without extra matter).

It is dimensionful and replaces the dimensionless parameter $g_{\rm QCD}$, the QCD coupling constant. The process in which a dimensionless constant such as $g$ is replaced by a dimensionful one such as $\Lambda$ is called the dimensional transmutation:

The constant $g$ isn't quite constant but it depends on the characteristic energy scale of the processes - essentially logarithmically. Morally speaking, $$ \frac{1}{g^2(E)} = \frac{1}{g^2(E_0)} + K \cdot \ln (E/E_0), $$ at least in the leading approximation. Because $g$ depends on the scale, it is pretty much true that every value of $g$ is realized for some value of the energy scale $E$. Instead of talking about the values of $g$ for many specific values of $E$, one may talk about the value of $E$ where $g$ gets as big as one or so, and this value of $E$ is known as $\Lambda_{\rm QCD}$ although one must be a bit more careful to define it so that it is 150 MeV and not twice as much, for example.

Yes, it is the characteristic scale of confinement and all other typical processes of pure QCD - those that don't depend on the current quark masses etc. In most sentences about the QCD scale, including your quote, the detailed numerical constant is not too important and the sentences are valid as order-of-magnitude estimates. However, given a proper definition, the exact value of $\Lambda_{\rm QCD}$ may be experimentally determined. With this knowledge and given the known Lagrangian of QCD - and the methods to calculate its quantum effects - one may reconstruct the full function $g(E)$.

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Since $g(\Lambda_{QCD})\sim 1$, one can say that for $E\ll\Lambda_{QCD}$ the theory is strongly coupled ($\Rightarrow$ confinement), and for $E\gg\Lambda_{QCD}$ it is weakly coupled ($\Rightarrow$ asymptotic freedom) – Stan May 4 '14 at 15:57

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