So I am wondering about one thing. The charged propagators in weak theory are W+- bosons. The mathematical expression for them, while drawing the Feynman diagrams is:(adsbygoogle = window.adsbygoogle || []).push({});

[itex]-i\frac{g_{\mu\nu}-\frac{q_\mu q_\nu}{m_W^2}}{q^2-m_w^2}[/itex].

The problems that are usually given to me are simple and involve cases where either [itex]q^2>>m_w^2[/itex] or [itex]q^2<<m_w^2[/itex], so I can simplify the propagator and carry on with the calculation.

But what happens if the impulse transfer q is the same as the mass of the W boson?

Griffiths only says: "However, when a process involves energies that are comparable to [itex]M_wc^2[/itex] we must, of course, revert to the exact expression."

How does that help if they are the same? I'll still have a divergent expression!

Is this the point of renormalization? I just put, by hand, some small parameter down there and everything is fine? But that's kinda like cheating :\

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# The propagator divergence in weak theory

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