1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $\bar{MS}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.