In 7.9 of Griffith's Particle Physics textbook, he defines the renormalized coupling constant as $$g_R=g_e\sqrt{1-\frac{g_e^2}{12\pi^2}ln\frac{M^2}{m^2}}$$ where $g_e$ is the nonrenormalized coupling constant and $m$ is either the mass of the muon or the mass of the electron (not sure which) because he's talking about electron-muon scattering. He then rewrites the ampplitude for the scattering process using this coupling constant instead. However, he defined $M$ as a cutoff that would be in the limit of infinity at the end of the calculation. Taking the limit as $M$ goes to infinity of $g_e$ would, as far as I can tell, yields $\pm i\infty$. It grows very slowly, but it doesn't seem like this method successfully "sweeps infinities under the rug." So my question is essentially: why is the coupling constant finite (and, I would assume, real) even though its definition makes it appear otherwise?
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$\begingroup$ when discussing electron-muon scattering, one often neglects the mass of the electron, because $m_\mu\sim200m_e$. Therefore, the mass $m$ is probably the muon mass. $\endgroup$ – AccidentalFourierTransform Feb 12 '17 at 19:57
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Because on the right-hand-side, $g_e$ depends on $M$, and you are supposed to take $M\to\infty$ while keeping $g_R$ fixed: $$ g_R\equiv\lim_{M\to\infty}g_e(M)\sqrt{1-\frac{g_e^2(M)}{12\pi^2}ln\frac{M^2}{m^2}} $$
In this sense, you can rest assured that $g_R$ is finite and real, and independent of $M$, because it is defined to be finite, real, and independent of $M$.
answered Feb 12 '17 at 19:56
AccidentalFourierTransformAccidentalFourierTransform
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