Here are the decays each B meson goes through to get to a D meson:
Number of B Mesons: 1 billion |
B => B0 Bbar0: 500 million |
B => B+ B- =500 million |
B+- => K+- psi/g =1% |
B0 => K0 psi/g= 1% |
psi/g => D2(2460) D |
psi/g => D2(2460) D* |
psi/g => D1(2420) D |
psi/g => D1(2420) D* |
So, as you can see, we start off with one billion B mesons. 500 million mesons decay to B neutral Bbar neutral mesons, and 500 million decay to B+ and B- mesons. The charged B mesons then decay to charged kaons and psi/g particles, and the neutral B mesons decay to neutral kaons and psi/g particles. The psi/g particles then decay to our four different types of particles: the D2(2460), D1(2420), D, and D* mesons. (The number in parentheses just tells the mass of the meson.)
The BaBar detector isn't 100% efficient, and its efficiency varies depending on the particle. So, while the efficiency for the charged kaon was a solid 95%, the efficiency for a neutral kaon was only 40%. We had to take all of this into account when making our predictions. Here are all of our efficiencies:
B+- => K+- psi/g =1% | efficiency for K+-: 95% |
B0 => K0 psi/g= 1% | efficiency for K0: 40% |
However, just because the detector will be able to detect a certain decay doesn't mean that that decay will actually happen in the first place. We also have to take the branching ratios of the decays into account. As I explained in an earlier blog post, the branching ratio is the probability that a certain decay will actually take place. Here are the branching ratios we came up with:
psi/g => D2(2460) D | BR: 25% |
psi/g => D2(2460) D* | BR: 20% |
psi/g => D1(2420) D | BR: 25% |
psi/g => D1(2420) D* | BR: 20% |
Once I had all the assumptions, I could finally get started on the actual calculations! This was probably the easiest part of the whole process; all I had to do was multiply the right percentages together to get the final number of D mesons produced. Here are the calculations:
Calculations: | ||
(5.0 x 10^8 charged B mesons)(1%)= 5.0 x 10^6 K+- psi/g | (5 x 10^8)(1% BR)(40% efficiency)= 2.0 x 10^6 K0 psi/g | |
(.95)(5.0 x 10^6)= 4.75 x 10^6 K+- psi/g | ||
psi/g => D2(2460) D | ||
(.25)(.01)(.05)(4.75 X 10^6)= 593.75 | (.25)(.01)(.05)(2.0 x 10^6)= 250 | |
psi/g => D2(2460) D* | ||
(.2)(.01)(.03)(4.75 x 10^6)= 285 | (.2)(.01)(.03)(2.0 x 10^6)= 120 | |
psi/g => D1(2420)D | ||
(.25)(.05)(.01)(4.75 x 10^6) = 593.75 | (.25)((.05)(.01)(2.0 x 10^6)= 250 | |
psi/g => D1(2420)D* | ||
(.2)(.03)(.01)(4.75 x 10^6)= 285 | (.03)(.01)(4.0 x 10^5)= 120 |
Our numbers were actually a lot better than we were expecting. From one billion B mesons, about 2500 D mesons will be produced (and actually detected). This is a good enough number to continue on with our work.
Of course, this all depends on if the assumptions we made were correct. To get a second opinion, Dr. Bellis is going to send the assumptions and calculations we made to the BaBar researchers at Stanford. If they agree with what we have come up with, then they will send us the rest of the BaBar data, which is when the real fun will begin. That's when all my Python and data analysis skills will come in handy. Fingers crossed!