"Running with momentum $p$" v.s. "running with renormalization scale $\mu$" 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 $p$ or $p^2$ 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 $\mu$.
 A: No, you can't simply identify the renormalization scale $\mu$ with the momentum $p$. 
To recap, many renormalization schemes depend on a parameter $\mu$ with the dimensions of energy/momentum. The quantity $\mu$ need not have any physical interpretation. However, it turns out that if the typical momentum scale of a process is $O(\mu)$, then higher-order contributions (loop diagrams) will be smaller. 
Hence the seemingly useless and confusing parameter $\mu$ is actually one of the greatest advantages of continuum RG over Wilsonian RG. By choosing $\mu$, we can make the calculation of a physical observable much more efficient. For instance, the coupling $e^2(\mu)$ describes the generic strength of all interactions involving particles with momentum $O(\mu)$. (For more detail, see this question.) That's why continuum RG is also called "resummation". It moves around the terms within a series to put most of the contribution in the leading terms.
You can't just say $\mu$ is "the momentum" because even the simplest processes have multiple momentum scales. For example, consider your typical $2 \to 2$ QED scattering, where particles with momenta $p_{1i}, p_{2i}$ scatter to momenta $p_{1f}, p_{2f}$. Which of these four momenta is supposed to be $\mu$? Actually, none of them! It's usually taken to be the momentum of the exchanged photon, i.e. $p_{1f} - p_{1i}$ for $t$-channel scattering.
Picking $\mu$ is a seriously nontrivial issue. Hundreds of papers have been written on the topic of "scale setting in QCD", which is the question of how to pick $\mu$ for QCD processes. This is extremely important for getting accurate results and completely opaque. I was told once that for any given $\mu$ you should treat the results you get for $\mu' \in [\mu/2, 2 \mu]$ as "theoretical uncertainty".
Can you use renormalization schemes without $\mu$? Absolutely, just use Wilsonian RG (for an overview, see here). It is indeed conceptually clearer, but it's never used for precision calculations in particle physics for exactly the reasons above.
