
#1
Apr3008, 04:50 PM

P: 2

Why was is it needed to include the Z boson along with the W's... is the theory nonrenoramalizable without it?




#2
Apr3008, 06:39 PM

Sci Advisor
HW Helper
P: 2,886





#3
Apr3008, 07:34 PM

PF Gold
P: 2,885





#4
Apr3008, 07:38 PM

Sci Advisor
HW Helper
P: 2,886

Z boson 



#5
Dec1008, 05:10 PM

P: 3

Isn't that so? 



#6
Dec1008, 07:44 PM

P: 527

I'm not sure about unitarity or renormalizability but one way to think about it was if the theory was SU(2) the theory has no mutually commuting generators so it's can describe theory of electromagnetism.




#7
Dec1108, 03:53 PM

P: 986

You can describe lowenergy limit of weak interaction without intermediate bosons, that's called Fermi theory, but it is not renormalizable. A theory with spontaneously broken SU(2) x U(1) symmetry group nicely describes everything, and SU(2) x U(1) just happens to have 4 generators, which become a photon and three new gauge bosons.
I don't think that unitarity enters in any way. 



#8
Dec1108, 07:49 PM

P: 3

For example: [tex]\sigma(e \nu \rightarrow e \nu) \propto {G_F}^2 s[/tex] Since cross sections express the likelihood of interaction between particles, what happens is that at sufficient high energies the probability of some process happening is greater than 1. In the Fermi theory this energies are around [tex]\sqrt{s}[/tex]=300 GeV. That's why I thought that there was a problem with unitarity. After looking into it now I'd say that the theory had both problems, it wasn't renormalizable and it violated unitarity. 


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