IVB theory

1. Apr 22, 2008

indigojoker

Would anyone know of any textbook that has explicit calculations of cross section using intermediate vector boson theory? I've looked in Perkins and Halzen+Martin but I do not see any in those texts.

2. Apr 22, 2008

humanino

I just looked into my Halzen & Martin, and I find there the muon decay rate calculated explicitely (just one example).

If Halzen & Martin's introductory textbook does not go deep enough, you probably need any textbook entitled "introduction to quantum field theory". Or "Dynamics of the Standard Model" (Donoghue, Golowich & Holstein @ Cambridge University Press) might do if you want phenomenology. If you are enclined towards theory, try Pokorsky's "Gauge Field Theories" @ same editor.

3. Apr 22, 2008

indigojoker

In H+M, then give the amplitude using ivb theory (12.14) then they makes the approximation that $$q^2<<M_W^2$$ allowing the use of $$\frac{G}{\sqrt{2}}=\frac{g^2}{8M^2_W}$$ which then they eventually conclude with 12.35

Are there examples where they do not make this assumption?

4. Apr 22, 2008

humanino

Later on when they talk about interferences in electron-positron annihilation for instance. But as you reach the end of this book, you probably need a thicker one

Therefore, congrats to you, because Halzen & Martin is a nice summary of basic particle physics.

5. Apr 23, 2008

indigojoker

Correct me if I'm wrong, but calculating the cross section using IVB theory is very similar to using V-A theory. For example:

Say the cross section for some process is: $$\sigma = \frac{G^2 s}{\pi}$$ (H+M 12.60)

Then using IVB theory, we take out the $$\frac{G^2}{4}$$ constant when calculating the amplitude and replace it with $$\left(\frac{g^2}{M_W^2+q^2}\right)^2$$ so when all is said and done, we are left with the cross section as: $$\sigma=\left(\frac{g^2}{M_W^2+q^2}\right)^2 \frac{4s}{\pi}$$

I think this makes sense because the first cross section allows $$\sigma$$ to go to infinity as s becomes large while using the IMV theory, the cross section is corrected at large s by the q^2 on the denominator, thus giving a finite total cross section.

6. Apr 25, 2008

nrqed

Hi indigojoker,

It's definitely true that the IVB model improves greatly the high energy behavior.
It turns out, though, that it is nonrenormalizable if I recall correctly. I am not sure about this but I think that essentially, the difference between the IVB and the Standard Model weak interaction is that in the IVB model there is no relation between between the couplings to the charged and neutral carriers. In the weak interaction, there is of course a definite relation between them. Another difference (I think but am not sure) is that I think the original IVB model had only vector coupling ($$\gamma^\mu$$ ).

The book by Aitchison and Hey discusses a little bit the IVB model (Gauge theories in Particle Physics)