The renormalized charge/coupling in QFT is usually phrased as renormalization scale $\mu$ dependent $\alpha(\mu)$ in the renormalization group setting. But can we take the more elucidating angle of "momentum $p$ dependent" $\alpha(p^2)$? The renormalization scale $\mu$, as it is taught in most QFT text books (often introduced un-intuitively as the scale parameter in dimensional regularization), is baffling to new learners rather than clarifying.
Let's shed some light on the renormalization scale $\mu$ with a simple example of $$ x(t) = ln(t/t_0) + x_0. $$ (in the physics context, translated to $$ \alpha(p) = ln(p/\mu) + \alpha_0 $$ with $\alpha$ being the coupling constant, $p$ being momentum , $\mu$ being renormalization scale, respectively)
The variable $x$ is the solution to a first-order differential equation ($\beta$-function) of $$ \beta (x) = dx(t)/dln(t) = 1, $$ with the initial condition $$ x(t)|_{t = t_0} = x_0. $$
The "running with renormalization scale $\mu$" approach is tantamount to regarding $x(t, t_0, x_0)$ as the solution to an alternative differential equation (differentiating against the initial condition point $t_0$, which is $\mu$ in physics context) $$ \beta '(x) = dx(t_0)/dln(t_0) = -1, $$ with the initial condition $$ x(t_0)|_{t_0 = t} = x_0. $$ Is this wicked and naughty way of looking at the original differential equation really helpful (or just add to the confusion)?
Let's take a look at another example of self-energy $\Sigma(\not{p})$ in the fermion propagator $$ G = \frac{i}{\not{p}-m_0 - \Sigma(\not{p})+i\epsilon} $$ where self-energy $\Sigma(\not{p})$ can be generally expressed as $$ \Sigma(\not{p}) = a(p^2) + b(p^2)\not{p}. $$ To simplify our discussion, let's assume that (which means there is no wave function renormalization) $$ b(p^2) = 0. $$ If we further expand self energy as $$ \Sigma(p^2) = a(p^2) = m_0' + c_1p^2 + c_2p^4 + ... $$ we will find out that $m_0'$ is divergent, while $c_1$ and $c_2$ are finite. The whole (mathematically shady) mass renormalization business is hinging on the assumption that $$ m_r = m_0 + m_0' $$ is finite (or equivalently, $m_0 = m_r - m_0'$, regarding $m_0'$ as mass counter term), so that the fermion propagator $$ G = \frac{i}{\not{p}-m_0 - \Sigma(p^2)+i\epsilon} $$ $$ = \frac{i}{\not{p}- (m_r + c_1p^2 + c_2p^4 + ...) + i\epsilon} $$ is finite and well defined.
Note that while $m_0$ and $m_0'$ are divergent, finite $m_r$ (it's not the physical pole mass $m_p$, unless $c_1= c_2 = 0$) can be determined by experiment.
On the other hand, the finite coefficients $c_1$ and $c_2$ can be calculated ($d\Sigma(p^2)/dp^2$ and $d^2\Sigma(p^2)/(dp^2)^2$ are finite, is that cool! It has to do renormalizability/local counter terms of renormalizable QFT), so that we know how self-energy $\Sigma(p^2)$ (or more precisely, the finite and well defined $m_0 + \Sigma(p^2) = m_r + c_1p^2 + c_2p^4 + ...$) runs with momentum/energy $p^2$.
The whole discussion above about running of $\Sigma(p^2)$ does NOT depend on the renormalization scale $\mu$ at all!
Update:
"Can you use renormalization schemes without $\mu$"? Surely one can, without resorting to any kind of RG (be it Wilsonian/Polchinskian/Wetterichian RG or perturbative QFT RG). Just resume the geometric series (that is how Landau pole was found by Landau!) of Feynman diagrams a la, 1/N (t'Hooft), rainbow/ladder approximation, etc. There are tons of alternative ways of achieving this so called RG enhancement without invoking RG accompanied by the illusive $\mu$.
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