# If the running coupling constant $\alpha(\mu)$ of QED becomes of order one at high $\mu$, why not changing $\mu$?

In the (modified) MS renormalization scheme, after dimensional regularization, we introduce some parameter $$\mu$$ with power of mass to keep the dimensionality of integrals under control. The parameters in the lagrangian end up being functions of $$\mu$$, and from the beta functions we can compute how they change when we change the "scale" $$\mu$$.

Now that's the point: we say that perturbation theory "breaks down" at high energy in QED because, if we identify $$\mu^2 = s$$, we end up with an effective fine structure constant which is of order one, hence of course we can no longer make use of perturbation theory.

If that's the case, why do we choose such a value of $$\mu$$? If it's a fake parameter that we put in our theory to "fix" an issue we have with dimensional regularization, why don't we pick a different value of $$\mu$$ which is very different from $$s$$, rendering $$\alpha(\mu)$$ very small?

Forget about the RG flow for a moment. Pick a value of $$\mu$$ so that $$\alpha(\mu)$$ is small. Now, fixing that small $$\alpha$$, take a look at a scattering process at high enough energy $$M$$ so that $$\alpha(M)$$ would be large (but still use your small $$\alpha$$). You will find that the 1-loop correction to your scattering process is large compared to the tree level diagram. This must happen because calculating 1-loop corrections is exactly how we derived the beta function after all. So perturbation theory is still breaking down even if you keep a small $$\alpha(\mu)$$.

• So if we choose $\mu$ such that $\alpha(\mu)$ is small, the actual contributions from loop diagrams (coefficients multiplying $\alpha$, $\alpha^2$, ...) become large? – user35319 Jan 6 at 20:58
• @user35319, Exactly, you typically get something like a factor $\log (M/\mu)$ that ends up making perturbation theory still break down – octonion Jan 7 at 19:15

It's the $$\alpha(Q)$$ at the high momentum/energy scale $$Q$$ (not the renormalization scale $$\mu$$!) of the physical process that invalidates the perturbation theory.

"Running" is actually in terms of the tangible scale of the scattering/physical process, which is $$Q$$ rather than some obscure "renormalization scale $$\mu$$" (text book QFT view) or "cutoff scale $$\Lambda$$" (Wilsonian view). See “Running with momentum p” v.s. “running with renormalization scale μ” for more details.

In QFT, there are 5 different mass scales (See here for more details), namely,

• m: the mass of the particle in concern
• $$\Lambda$$: the UV cutoff scale of the regularization scheme
• Q: the energy scale of the incoming/outgoing particles involved in a scattering process.
• $$\mu$$: the renormalization scale
• M:the mass scale where beyond standard model physics effect comes into the picture.

If you manage to appreciate the differences and relationships between these 5 mass scales, you would have a clearer picture of QFT.