In summary, the discussion is about the factor of ##m## in the momentum integral of the internal fermion propagator in a given Feynman diagram. It is argued that the factor of ##m## is necessary to obtain a quantity with the dimension of mass in the diagram. The argument is supported by the fact that ##\gamma_{\mu} k^{\mu}## can be set to 0 in the integral due to the integrand being odd, but this only holds for convergent integrals and a regularization method is needed to make sense of it. The most convenient method is dimensional regularization, which preserves Poincare invariance.
#1
Siupa
29
5
Where does the ##m## in ##(3.2)## come from? It doesn’t seem to enter anywhere in Feynman rules for the given diagram
Anyway, it comes from the internal fermion propagator - you can not just look at the interaction Feynman rules.
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#3
Siupa
29
5
malawi_glenn said:
Source?
Anyway, it comes from the internal fermion propagator - you can not just look at the interaction Feynman rules.
Source is A. Pich, Effective Field Theory, beginning of chapter 3.
I understand that the momentum integral of the internal propagator comes from that loop, but why specifically is there a factor of ##m## in front of it? Or better, why isn‘t it inside the integral in the numerator summed with ##\not\! k##? Is it because the integral with ##k## in the numerator vanishes due to the integrand being odd? Isn’t this argument only valid for convergent integrals though?
in the numerator you have ##\not\! k - m ## in the fermion propagator.
Yes, it will vanish because of odd integrand see https://arxiv.org/abs/2006.16285 eq. 18 page 14
(note the typo, it should read ##(\gamma^5)^2 = 1##)
Look at it this way. How would you get something that has the dimension of mass, in that diagram, unless you had a factor ##m## in front? Can you come up with anything?
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#5
Siupa
29
5
malawi_glenn said:
in the numerator you have ##\not\! k - m ## in the fermion propagator.
Yes, it will vanish because of odd integrand see https://arxiv.org/abs/2006.16285 eq. 18 page 14
(note the typo, it should read ##(\gamma^5)^2 = 1##)
Look at it this way. How would you get something that has the dimension of mass, in that diagram, unless you had a factor ##m## in front? Can you come up with anything?
Seems to be a textbook making the "confusing issue" of divergencies even more confusing. Of course your integral must be regularized first, before you can make any sense of it. The most convenient, but a bit unintuitive, regularization is "dimensional regularization" since it keeps Poincare invariance, and then indeed it's correct to conclude that the part of the integrand ##\gamma_{\mu} k^{\mu}## can be set to 0.
