The one-loop contribution of the vacuum polarisation of the photon after using dimreg is given by

$$\Pi_2^{\mu\nu}= e^2 J(q) \left(\eta^{\mu\nu} - \frac{q^\mu q^\nu}{q^2}\right),$$

with the metric tensor $\eta$, coupling constant $e$ and momentum $q$. Also

$$J(q) = -\frac{q^2}{2\pi^2} \int_0^1 dx x(1-x)\left(\frac{2}{\epsilon}-\ln\Delta-\gamma+\ln(4\pi) + O(\epsilon)\right)$$

with $\epsilon=d-4$ and $\Delta=m^2-x(1-x)q^2$.

The renormalised electric charge can be expressed as

$$e^2 = e_0^2\left(1-\frac{1}{2}J^{\prime\prime}(0) + O(e_0^4)\right)$$

Dimreg then yields

$$e^2 = e^2_0\left(1-\frac{e_0^2}{6\pi^2\epsilon}+O(e_0^4)\right)$$

Where did the $\gamma$ and $\ln 4\pi$ go?


It's implicitly performing the modified minimal subtraction $\bar{MS}$ renormalization scheme, only in a sloppy way.

One should retain the finite terms at the intermediate stage: $$ e^2 = e^2_0\left(1-\frac{e_0^2}{12\pi^2}[\frac{2}{\epsilon} - \ln(m^2) - \gamma+\ln(4\pi)]+O(e_0^4)\right) $$

The usual minimal subtraction $MS$ renormalization scheme will only subtract the divergent term (the $O(\frac{1}{\epsilon})$ part), while the modified minimal subtraction $\bar{MS}$ renormalization scheme will subtract the finite terms as well ($\ln(m^2)$, $\gamma$, and $\ln(4\pi)$). The difference between $MS$ and $\bar{MS}$ all boils down to whether the counterterm should include only a divergent term ($MS$) or both divergent and finite terms ($\bar{MS}$).

The OP's formula is implicitly performing $\bar{MS}$ at different stages, causing unnecessary confusion. At any rate, the final physical results should be the same, being it $MS$, $\bar{MS}$, or sloppy $\bar{MS}$.


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