A Question about the decay of the W boson to tau lepton

Click For Summary
The decay of the W boson to tau leptons is generally excluded from calculations, such as the cross-section for processes involving W bosons, because tau particles decay quickly before reaching detectors, unlike electrons and muons. This rapid decay complicates the reconstruction of tau decays, requiring different methodologies and leading to less clean signals due to missing energy from neutrinos. While W decays to taus are studied, they are often treated separately from decays to electrons and muons. In specific contexts, such as when considering stable tau approximations, calculations can be similar but involve additional complexities due to the instability of the tau. Overall, the focus remains on more stable decay channels for clearer experimental results.
ribella
Messages
6
Reaction score
0
TL;DR
decays of W boson
Why is generally the decay of the W boson to tau lepton not taken in the calculations?
 
Physics news on Phys.org
Calculations of what?
 
  • Like
Likes vanhees71, Vanadium 50 and topsquark
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.
 
  • Like
Likes artis, vanhees71, topsquark and 1 other person
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.
 
  • Like
Likes artis, vanhees71 and topsquark
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."
 
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
23
Views
4K
Replies
30
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
Replies
4
Views
4K