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Like my topic, how does the fine structure constant vary with energy? I looked it up and found: $$\alpha_{eff}(q^2) = \frac{\alpha}{1-\frac{\alpha}{3\pi}\ln(\frac{-q^2}{Am_e^2})},$$ where $\alpha^{-1} \approx 137$, $A=\exp(5/3)$, and $m_e$ is the electron mass. Consider two beams of particles with the same mass with the momenta: $$p_1=(E,\vec{p}), p_2 = (E,-\vec{p}),$$ what would "$-q^2$" be in this case?

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First, the standard formula would be

$$\alpha(q^2) = \frac{\alpha(\mu_0^2)}{1-\frac{2\alpha(\mu_0^2)}{3\pi}\ln\frac{q^2}{\mu_0^2}}$$

where the reference energy scale $\mu_0$ can indeed be chosen to be the electron mass, as in your formula, except for that $\exp\frac{5}{3}$. I have never seen that: where did you get your formula?

Then, I could write several paragraphs of theory to explain to you that in all rigour $q^2$ is arbitrary, and so is the reference point $\mu_0$, but I am not sure how useful that would be as if you had the theoretical baggage to understand what is called renormalisation, you would not be asking this question.

So I'll cut the chase and tell you that a pragmatic value would be to choose $q$ as the centre of mass energy of your two particles.

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    $\begingroup$ I believe OP uses the Peskin and Schroeder equation 12.89, page 424. $\endgroup$ – SRS Jul 10 '17 at 15:23

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