A Intuition behind asymptotic freedom/slavery

  • A
  • Thread starter Thread starter paralleltransport
  • Start date Start date
  • Tags Tags
    Intuition
paralleltransport
Messages
128
Reaction score
97
TL;DR Summary
I'd like an intuitive way to see the coupling growth vs. scale
It is well known that ϕ3ϕ3 is d= 6 is asymptotically free, while ϕ4ϕ4 in d=4 is asymptotically slave (or "trivial" or marginally irrelevant, or has a QED style pole). The standard way is to compute the 1 loop correction to the 4 point (or 3 point) vertex respectively, renormalize (based on some physical thought measurement or lab scale), and then deduce the growth/reduction of the coupling strength vs. momenta. However is there an intuitive way to visualize this? I feel like to get the sign of the running coupling, most of the integration over feynman parameters just give some constant multiple and get substracted out based on the renormalization scheme anyways so there should be an easier way to get whether the coupling grows weaker or stronger at high momenta without the difficult integrals.

Source https://www.physicsforums.com/forums/high-energy-nuclear-particle-physics.65/post-thread
 
Physics news on Phys.org
Having worked out the beta function in gory detail, I conclude there's no intuitive explanation for asymptotic freedom/slavery.

- For $\phi^3$ in 6d, it involves a non-trivial 1-loop cancellation between the renormalization effect of the kinetic term vs. the coupling term: the anomalous dimension partially cancel the quantum fluctuation causing increased coupling, so it is not obvious which way it goes from naive intuition.

- For $\phi^4$ theory in 4d, the fact the anomalous dimension at 1-loop vanish could be "seen" by the fact 1-loop correction to the propagator does not depend on momentum. Then one "sees" that the 1 loop vertex correction cause marginal irrelevance, but that's highly non-trivial (anomalous dimension gets corrected at higher loops...)
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top