Question about the decay of the W boson to tau lepton

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SUMMARY

The decay of the W boson to tau leptons is often excluded from calculations of processes like pp->W-W+->l- v l+ vb, where l typically represents electrons and muons. This exclusion is primarily due to the instability of tau leptons, which decay before reaching particle detectors, complicating the reconstruction of their decay paths. While W decays to taus are studied, they require different methodologies and are treated separately from decays to electrons and muons. Theoretical calculations for stable taus are nearly equivalent to those for electrons and muons, but practical considerations necessitate a distinct approach for tau decays.

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TL;DR
decays of W boson
Why is generally the decay of the W boson to tau lepton not taken in the calculations?
 
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Calculations of what?
 
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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.
 
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thanks for your response.
 
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.
 
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I'm still a little fuzzy on the question.

If the question is "If we are interested in W \rightarrow e \nu, why do we not consider W \rightarrow \tau \nu. the answer is "because electrons are not taus". If the question is instead, why do we not consider W \rightarrow \tau (\rightarrow e \nu \nu) \nu, the answer is "sometimes we do and sometimes we don't; it depends on the problem we are working."
 
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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.
 

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