Coupling constants run with the energy scale $\mu$. But what is exactly this energy scale. My question is, if I have a physical process, how do I compute $\mu$?


2 Answers 2


There are at least two answers possible to give, but both, in the end, amount to the same thing: There is no "right" way to fix the energy scale of a process, but that doesn't matter, except that your perturbation theory will probably break if you choose the scale badly.

The old answer: The renormalization scale is arbitrarily defined to fix some parameters of the theory to measured values and get rid of infinities, e.g. a $\phi^4$ scalar coupling $\lambda$ is sometimes fixed by looking at the $\phi^4$ interaction at the channel $s^2 = 4\mu^2, t^2 = 0, u^2 = 0$, but you could as well look at the channel at $s^2=t^2=u^2=\mu^2$. We need to just take any kind of condition to fix our counterterms and get finite answers for our amplitudes.

There is, generically, no real meaning to the renormalization scale, but if you define it at diagram channels like $s^2=t^2=u^2=\mu^2$, then it will reflect the "typical" energy scale of the process. By doing renormalized perturbation theory, the counterterms in the Lagrangian will be fixed by the renormalization conditions, and yield finite answers for amplitudes regardless of how the scale is chosen. However, if you are at processes which have vastly different channels from the scale, the perturbative corrections start to grow large incredibly fast, so that you gain nothing by having amplitudes that are "theoretically" independent of the scale, and this motivates seeking running couplings - by being able to vary the renormalization scale without doing order-by-order perturbation explicitly on the diagram, we save a lot of effort.

The modern, Wilsonian answer: A quantum field theory is viewed intrinsically as an effective theory possessing an inherent momentum cutoff $\Lambda_0$. This isn't a "scale" anymore at which we do some shenanigans because we want to get rid of infinities, it is a property of the theory, telling you how high the excited momenta go. The Fourier modes of the quantum fields are literally cut-off above this scale, and we view the path integral measure (which is "defined" by a limit process on a lattice) as only containing the measures for the momenta below $\Lambda_0$.

"Renormalizing" now means that we can integrate out even more Fourier modes, going to a new cut-off $\Lambda<\Lambda_0$, making them couplings in the Wilsonian effective action rather than dynamical objects of the theory. Then $\Lambda$ is clearly the scale above which we do not expect many modes of the fields to be excited, it is indeed the order of the momenta occuring in the process.

Perhaps a bit disappointing to your question, this way of looking at the cut-off is inherently approximative, and there is no exact way to get the "energy scale of a process" - but again, it doesn't matter, since all the renormalization does is shift the way our perturbation theory works by running the couplings, and we are again in principle free to choose any kind of scale in the process to be our renormalisation scale - but if you take one that deviates much from the intuitive meaning of "order of momenta in the process" too much, you will again have no fun trying to get the counterterms perturbatively.


You go to the center of mass frame to find that $\sum_i \vec{p}=\vec{0}$, and the total momentum four vector is thus $$P_{\text{tot}}^{\mu}=\left(\frac{1}{c}\sum_{i}E_i^{\text{COM}}, \vec{0}\right)$$ then we define the energy scale covariantly as $\mu=\sqrt{-s}$ where $s$ is the mandelstam variable $s\equiv P_{\text{tot}}^{\mu}P_{\text{tot}\mu}$ in units of $c=1$, so that in COM frame $$ \mu = \sum_i E_i^{\text{COM}} = \sum_i \frac{m_{oi} c^2}{\sqrt{1 - v_i^{\text{COM }2}/c^2}} $$

  • $\begingroup$ Rather than focusing on $s$ in particular it might be better to take the more general approach of looking at the mass of the exchange line of the leading diagram be it $s$, $t$, or $u$. $\endgroup$ Apr 27, 2015 at 23:20

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