The propagator divergence in weak theory

[dingo_d]

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:

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

The problems that are usually given to me are simple and involve cases where either q^2>>m_w^2 or q^2<<m_w^2, 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 M_wc^2 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 :\

 

[ofirg]

By divergence are you referring to the zero in the denominator when q^{2}\approxm_{w}^{2}? in those cases, the finite width of the W boson plays a role and the denominator of the propogator is q^{2}-m_{w}^{2}+im_{w} \Gamma_{w}, where \Gamma_{w} determines the width of the resonance.

 

[ofirg]

\Gamma_{w} is the decay width of the W boson. It is related to the lifetime of the W boson,i.e., \Gamma_{w}\tau_{w}=1

 

[dingo_d]

Hmmm, so there is some real physics behind that. I thought that they just made the renormalization so that they would just avoid divergences, as a mathematical trick...

Thanks for the clarification :)

 

[ofirg]

In some cases, the propogator really does diverge. It happens when a stable particle (and thus \Gamma=0) emit a very soft massless particle (E~0) or a colinear massless particle (if the particle is also massless). For example, an electron emiting a very soft photon or a colinear photon (if the electron mass in neglected). Then, it is needed to regularize the propogator, but this divergences always cancel when calculating observable quantities.

But all this is not needed in the W case.