Running of the fine structure constant: meaning of $\alpha(Q^2=0)$?

Question

In Measurement of the Running of the Fine-Structure Constant, the L3 collaborations writes

At zero momentum transfer, the QED fine structure constant $\alpha(0)$ is very accurately known from the measurement of the anomalous magnetic moment of the electron and from solid-state physics measurements: \begin{equation*} \alpha^{-1}(0)=137.03599976(50)~~. \end{equation*} In QED, vacuum polarization corrections to processes involving the exchange of virtual photonsresult in a $Q^2$ dependence, or running, of the effective fine-structure constant, $\alpha(Q^2)$.

However, elsewhere I find that the accepted value of the fine structure constant $\alpha\approx1/137$ is really $\alpha(511\text{keV})$, not $\alpha(0)$. Since L3 is measuring the momentum transfer $Q$ in GeV, are they simply ignoring the 511keV? If not, what is the difference between the true $\alpha(0)$ and the $\alpha(511\text{keV})$? Although I understand the difference is small, which $Q^2$ is measured to satisfy the usual value $$\alpha(Q^2)= \frac{e^2}{4 \pi \varepsilon_0 \hbar c}~~?$$ Furthermore, I would like some clarification on the meaning of momentum transfer "in the timelike region" or "in the spacelike region." From my reading, I understand that these are the cases of annihilation or scattering in an $e^-e^+$ scattering experiment where the photon exchanged moves through a timelike or spacelike region (pictured below). However, I would like some clarification on what they mean by "momentum transfer in one region or the other."

Answer 1

As I discuss also in this answer of mine, there is no unique definition of "the energy scale of a process", i.e. the $Q^2$, and hence there is also not a single function $\alpha(Q^2)$ - there are a lot of different $\alpha(Q^2)$, depending on

1. how you define the $Q^2$ associated with a certain diagram with certain values for its Mandelstam variables

2. what renormalization scheme you are using

so without taking care to compare these conventions between the sources that you are using, there isn't really any meaning to asking whether $\alpha(0)$ or $\alpha(Q_0^2)$ for some $Q_0^2\neq 0$ should be the fine-structure constant.

The "momentum transfer" terminology for $Q^2$ refers to a popular way of assigning energy scales to tree-level diagrams according to the momentum the virtual particle "should have had" if we impose energy-momentum conversation: In terms of Mandelstam variables (see also this answer by JamalS), for an s-channel diagram, this is $Q^2 = s$ and $s$ is positive, for a t-channel diagram, this is $Q^2 = t$, and $t$ is negative. If you now think about $Q$ as an actual 4-momentum, $Q^2>0$ means it's timelike and $Q^2<0$ means it's spacelike (or the other way around, check your metric sign convention!), but in any case this leads to calling s-channel contributions "timelike" and t-channel contributions "spacelike".