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

 A: 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

*

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


*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".
