Anomaly cancellation triangle

[ChrisVer]

If I have a triangle diagram for $Z \rightarrow W^-W^+$, do all fermions contribute the same for giving the contraint $\sum_i Q_i =0$??
I think that eg top quarks will be absent for such a diagram...

any help?

 

[ChrisVer]

The contributions will not be the same either. Not only do the masses differ, but the couplings to Z as well as the W CKM elements will differ from diagram to diagram.

So how can anomaly cancelation occur if so many things differ?

 

[Hepth]

I'm not sure now to be honest. I thinks in the limit that the masses are smaller than some higher scale?

http://arxiv.org/pdf/hep-ph/0003143.pdf

Page 6 has some info on it. I'll think about it some until someone else comes along.

 

[ofirg]

So how can anomaly cancelation occur if so many things differ?

I think the dangerous part contributing to the anomaly in these triangle diagrams is the divergent part. This will force you to add a non gauge invariant term to the bare Lagrangian if it doesn't cancel out. Since the divergent part comes from the UV, the masses don't come in, only the couplings.

 

[vanhees71]

You must have both quarks per family, and you must not forget the color-degeneracy factor. You'll then see that the anomaly of the local gauge symmetry indeed cancels due to the charge pattern of the matter particles in the Standard Model. Without this, it would be obsolete ("not even wrong")!

 

[ChrisVer]

No my problem is that:

Are those two diagrams contributing the same so that they can cancel each other out?
Because I think that $\sum_i Q_i$ is assuming that they do...