Third (and final hopefully) correction to TP rate
From:
Date: 12/18/97
Time: 9:35:02 AM
Remote Name: 137.138.115.189
Comments
Dear Colleagues,
I hope that this is my last mistake on the rate business. Guy Wilkinson
pointed out that his result differes from my last conclusion 10) of the previous
e-mail. I think that my conclusion on the point 10) is wrong. The mistake I made
was that I mixed up the number of interactions and number of bunch crossing. One
of correct ways to cimpute is: Pb = probability to find a b-bbar pair in one pp
interaction sigma(b-bbar)/sigma(inelastic) P0 = probabilty to have 0 pp
interaction in one bunch crossing exp(-m), where m=sigma(inelastic)*L/f L:
luminosity f: number of bunch crossings in one second P1 = probability to have 1
pp interaction in one bunch crossing m*exp(-m) P2 = m^2 * expe(-m)/2! etc. i.e.
Poisson statistics Therefore Pb1 = probability to have 1 pp interaction in 1
bunch crossing which produces a b-bbar pair,
P1*Pb=[sigma(b-bbar)/sigma(inelastic)]*[sigma(inelastic)*L/f] *exp(-m)
=[sigma(b-bbar)*L/f]*exp(-m) and Nb1 = number of b-bbar pair produced in the
bunch crossings with one pp interaction per second
f*P1*Pb=sigma(B0bbar)*L*exp(-m)
So here is the new list: 1) L = 2*10^32 cm^-2 s^-1 2) sigma(inelastic)= 80
mb 3) N(interaction) = L*sigma(inelastic) = 1.6*10^7 / sec 4) N(real bunch
crossing) = 3*10^7 /sec (74.4% of 40 MHz) 5) N(average number of pp interaction
in one real bunch crossing) = 0.5333 6) Probability to have no pp interaction in
one real bunch crossing = 0.5867 7) Probability to have ONE pp interaction in
one real bunch crossing = 0.3129 8) sigma(b anti-b) = 500 microb 9) N(b anti-b)
= L*sigma(b anti-b) = 10^5 / sec 10) N(b anti-b in one pp interaction bunch
crossing) = 10^5 * 0.5867 = 5.867 * 10^4 / sec (i.e. fraction on b anti-b
produced in one pp interaction bunch = 0.587)
I hope now they are finally all correct. I appriciate discussions with Roger
Forty, Olivier Schneider and Guy Wilkinson. yours Tatsuya
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