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For asymptotic freedom of QCD, people give the argument that alpha_strong decreases at high energy (thus there is freedom at low distance).

But if we make the reasoning on the QCD potential, for which experimental measurements exist, we do see that at low distance (r), the potential is decreasing more and more. Thus \vec{F}=-\vec{grad} V(r), we deduce that the force is more and more important at low distance, thus the particles are 'not free' : they undergo a more and more important force, which is in contradiction with the reasoning on alpha_strong.

Where is the mistake in my reasoning ? Thank you

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  • $\begingroup$ The mistake seems to be the same one we discussed a couple of weeks ago: At asymptotically high energies the system is ultra-relativistic, and the non-relativistic effective potential $V(r)$ is not a good parameterization of it. Asymptotic freedom is a statement about the QCD coupling $\alpha_s$ and the non-relativistic effective potential is not relevant. $\endgroup$
    – David Schaich
    Commented Aug 18, 2019 at 8:24
  • $\begingroup$ Ah... ok. Sorry, I had not completely understood this point. Now it is clear thanks to you. ok, just to be sure, an additional point (which is different to my question) : is the relationship Lagranian L=T - V wrong for relatistic particles ? $\endgroup$
    – Mathieu Krisztian
    Commented Aug 18, 2019 at 11:16
  • $\begingroup$ Yes. Here is a sketchy way to think about it: In relativistic settings the lagrangian (or better the action) is a functional that must be invariant under Lorentz transformations (i.e., rotations and boosts). Boosting into or out of a massive particle's rest frame will change its kinetic energy $T$ and hence $V=T-L$. Both $T$ and $V$ are therefore ill-defined unless you discard relativity and pick a particular inertial frame. (For example, in lattice QCD the static potential is defined by inserting infinitely massive external probes and working in their rest frame.) $\endgroup$
    – David Schaich
    Commented Aug 18, 2019 at 17:02
  • $\begingroup$ All right. Thank you very much David $\endgroup$
    – Mathieu Krisztian
    Commented Aug 18, 2019 at 19:42

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