Recent content by bb13

I think I'm starting to understand. So the ϒ is generated when the e+e- annihilate into quark and antiquark via electromagnetic interaction, but the B mesons are generated via the weak interaction on ϒ, so they are not taken into account for the rule of R=10/3, thus increasing that value?

The weak interaction leads to the two B mesons, right? But I still don't see why the R value doesn't work. Until know we've seen that the e+e- decays into ϒ(4s) which at the same time decays into B mesons thus increasing the cross section for the decay into hadrons right? I don't understand...

Looking in the PDG website I can see that the decat into B Bbar is the most favourable decay. Is that enough to say why the hadronic cross section is so large? Or should I also look the decays of B Bbar? According to PDG the most favourble decay for B is a leptonic decay, which I don't see how...

Could you give some insight about what decay modes could increase the value of σ(e+e-→hadrons) and how would that increase the cross section please?

The exact center of mass energy is √s=10.539 GeV σ(e+e-→μ+μ-)=782 pb σ(e+e-→hadrons)=3917 pb

Using Monte Carlo simulation and finding the efficiency for the annihilation. Selecting only the events that passa a certain cut, I used σ=Nselected/(Lε) where L is the luminosity and epsilon the efficiency found with Monte Carlo. I think I'm suposed to get this value and explain it with the ϒ...

In the calculation of R=σ(e+e-→hadrons)/σ(e+e-→μ+μ-) from BaBar experimental data at a center of mass energy of √s≈10 GeV i obtain R=5. Theoretically I should get a value of R=10/3. I know it has something to do with the resonances of ϒ mesons shown in the plot attached, but I don't know how to...