# A R value in electron positron annihilation

Tags:
1. Dec 8, 2017

### bb13

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 dont know how to explain it. Help please.

2. Dec 8, 2017

### Staff: Mentor

How did you get 5?

3. Dec 8, 2017

### bb13

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 meson, but I don't know how and I didn't find any explanation in any book. Thank you.

4. Dec 8, 2017

### Staff: Mentor

Then something is wrong with your Monte Carlo or the efficiency treatment.
What are the cross sections you got?

The ϒ mesons don't matter if you are at exactly 10 GeV or generally not inside one of the resonances.

5. Dec 8, 2017

### bb13

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

6. Dec 8, 2017

### Staff: Mentor

That is inside the ϒ(4s) peak. Sure, that changes the values. You probably get something between the R-value outside the resonances and the ϒ(4s) decay modes.

7. Dec 8, 2017

### bb13

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?

8. Dec 8, 2017

### Staff: Mentor

Check the decay modes and see which interaction is leading to them. It is not the electromagnetic interaction.

9. Dec 9, 2017

### bb13

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 would increase the hadronic cross section, it should in fact lower R as the cross section into muons would increase, wouldn't it?

10. Dec 9, 2017

### Staff: Mentor

The R value considers the direct collision process only. B mesons are hadrons, their decay modes later don’t matter.

Which interaction leads to the decay to two B mesons? Can you see why the derivation of the R value doesn’t work there?

11. Dec 9, 2017

### bb13

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 why the derivation for R doesn't work under this procedure.

12. Dec 10, 2017

### Staff: Mentor

Why would you expect the weak interaction to be relevant?
Even if it would: Why would you expect it to follow a rule that is based on electric charges?

e+e- doesn’t “decay” to ϒ(4s).

The R-value of 10/3 comes from the assumption that the e+e- pair interacts via a virtual photon. Only in this case are the electric charges relevant. An ϒ is not a photon.

13. Dec 10, 2017

### 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?

14. Dec 10, 2017

### Staff: Mentor

There is no weak interaction involved.

The $\Upsilon$ is a resonance, it has a cross section way above the process you consider to get R=10/3 for other energies.
It decays via the strong interaction with nearly 100% probability, and that decay produces hadrons.