In summary: However, in summary, the decay of the W boson to tau lepton is not always taken into account in calculations because it requires a different approach and the resulting signal is not as clean as with electrons and muons. Depending on the specific problem being studied, W decays to taus may or may not be considered, but in general, they are not as commonly included as the other two leptonic decays. This is due to the fact that taus decay before reaching the particle detector, making it more difficult to study their decays. Additionally, the use of hadronic calorimeters and the presence of missing energy from neutrinos adds further complications. However, for stable taus, the calculations are equivalent up to minor corrections.
#1
ribella
6
0
TL;DR Summary
decays of W boson
Why is generally the decay of the W boson to tau lepton not taken in the calculations?
for example, I want to calculate the cross-section of the process pp->W-W+->l- v l+ vb (Here, l is generally taken as electron and muon). Therefore, why is generally the decay of the W boson to tau lepton not taken in the calculations?
Tau decays before reaching the particle detector while electrons and muons cross it without decay (in almost all cases, for the muon). It's easy to study decays to electrons and muons together but reconstructing decays to taus needs a completely different approach. We do study W decays to taus, but separately from the other two leptonic decays.
In tau decays, we have to use the hadronic calorimeters of the detector, and we have missing energy from the neutrino. This is, as mentioned, not as clean signal as two niceley charged tracks.
If the question is "If we are interested in [itex]W \rightarrow e \nu[/itex], why do we not consider [itex]W \rightarrow \tau \nu[/itex]. the answer is "because electrons are not taus". If the question is instead, why do we not consider [itex]W \rightarrow \tau (\rightarrow e \nu \nu) \nu[/itex], the answer is "sometimes we do and sometimes we don't; it depends on the problem we are working."
#8
RGevo
138
31
Assuming a stable tau, the calculations are equivalent up to corrections of order (mtau/mw)^2, and therefore almost numerically equivalent.
As the others alluded to, since this particle is not stable, the experimental (and theoretical) reality is more complicated.