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Z boson decay partial width

by Catria
Tags: boson, decay, partial, width
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Catria
#1
Nov10-13, 05:03 PM
P: 52
Hello everyone,

I have read about the theoretical values of the Z boson decay partial width and how well they agreed with experiment. However there is something I do not quite understand: since these theoretical calculations were performed with the hypothesis that the masses of the decay products were negligible with respect to the mass of the Z boson, what changes would have to be effected if one did not neglect the mass of the bottom quark (~4-5 GeV) when computing the partial width of the Z boson into a bottom and anti-bottom quark pair?
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mfb
#2
Nov10-13, 05:19 PM
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If the theoretical calculations would not take the mass into account, they would predict the same values for all quarks. They do not, and the predicted difference has been confirmed by experiment.
Catria
#3
Nov10-13, 05:22 PM
P: 52
Then, what... form would these corrections take? Would those corrections for the bottom quark have the effect of increasing or decreasing the partial width?

mfb
#4
Nov10-13, 05:25 PM
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Z boson decay partial width

Based on the PDG values, it looks like mass is increasing the partial decay width. Hadronization makes it tricky to disentangle the light quarks, but bb gets more than 1/5 of the hadronic decay width.
Vanadium 50
#5
Nov10-13, 06:13 PM
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The theoretical effect will be something around 1%. The observational difference is about 1% with an uncertainty of about half a percent.
mfb
#6
Nov11-13, 12:43 PM
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How did you get those numbers? Γ(Z → bb¯ )/Γ(Z → hadrons ) is given as 0.21629 ± 0.00066 where bold highlights matching digits. The uncertainty is tiny compared to the difference to 0.2.
Vanadium 50
#7
Nov11-13, 01:06 PM
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The phase space factor is about a percent. This can be accurately calculated, since its a ratio.

The absolute branching fraction to hadrons has loop effects (color connection) of the order alpha_s/16 pi, or again a good fraction of a percent. So one would need to compare Γ(Z → bb)/[Γ(Z → bb) + Γ(Z → dd) + Γ(Z → ss)] which what I did and is good to half a percent. Your suggestion to lump all the hadrons together is interesting, but it would require taking the weak mixing angle measurement from some other experiment, so one could constrain Γ(Z → uu)/[Γ(Z → dd). I'd have to look more closely to see if you win or not with this.


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