Exercise to treat spin-dependent decays I. Goal: – – Study the relationship between momentum pe accuracy/precision and a, Analyzing power <A>. Estimate the required performance of the detector. II. Exercise to check basic kinetics: Today’s contents 1. Energy and momentum conservation, 2. 2D event yield distribution as functions of y and cmS y = pcme/pmax cmS is an angle between spin-axis and momentum direction of decay-e+ at the center-of-mass system. ( see next page) III. Check wiggle plots: “usual” wiggle plot, “Beam-loss free” wiggle plot. 1 Center-of-Mass system Magnetic field Z Y Scm a t m cm Spin-direction B 0,0, BZ S cos a t, sin a t,0 Direction of decay-positron pe m m m cos cm , sin cm cos , sin cm sin pe X, Momentum Angle between spin-axis and momentum direction of decay-e+ at the center-of-mass system: cos Scm pe m m S sin cm cos sin a t cos cm cos a t pe m We measure Lab , . Lorentz boost cos m Lab m cos cm m 1 cos cm 2 1 dP n y 1 Ay cos sin , dyd 2 2y 1 2 . n y y 3 2 y , Ay 3 2y y p e p e max p e max m 2 3 4 Condition: 3T B P=300MeV/c , =3, Tc =7.4nsec, R=333mm, Ta=2/a=2.2sec. Positron energies 28 ~191 MeV 5 8.6MeV positron B=3T 50.4MeV positron 102MeV positron 6 II. Check basic kinetic values from GEANT4 7 Probing Spin-dependent Decay Info. I. To be more simple, I set 100% e e ! II. Probe “decay process” information in the lab frame directly. (I use “UserSteppingAction”.) Spin vector, momentum of at previous step of decay process. Momentum and energies of daughters. III. Check momentum/energy conservation. Within few eV at =1, within few keV at =3. why? IV. Apply Lorentz transformation to get values in the center-of-mass system. V. Cook values as I want!! 8 p T p 2Y p 2Z pL pX y p e p e max p e max m 2 p e p 2X p 2Y p 2Z X axis is always momentum direction. 9 y = pcme/pmax is an angle between spin-axis and momentum direction of decay-e+ at the center-of-mass system. 10 11 III. Wiggle plots made by GEANT4 “Usual” wiggle plot and “Beam-loss free” wiggle plot 12 t Ft N exp 9.5 105 , E> 200 MeV 1.3105e+ 1 A cosa t 4 free parameters 13 Covariant matrix is OK. mom “Beam loss free” wiggle plot by knowing cm An angle between + and e+ momentum direction in the center-of-mass system. dP 1 n y 1 a y cos Scm , dyd cm 4 mom mom here , cos Scm sin cm cos sin a t cos cm cosa t . dP dP n y mom 1 a y cos cm cosa t S S dyd cm sin cm dydcm 2 1 S S mom Nt n y dy n y a y sin cm dydcm t cos cm cosa t 2 y y Scm N m Measure! 1 A cos cm cosa t 2 Lt N 2 1 A cosa t , R t N 2 Lt R t Asym t A cosa t Lt R t 1 A cosa t , No exponential 14 term! LEFT RIGHT Asym t A cosa t No worry about -beam loss! But, need to handle left-right detector asymmetry. 9.5 105 , 1.9 105 e+ y> 0.6 , LEFT: 1 cos 0.7 RIGHT:1 cos 1 0.7 15 A big advantage to measure Lab-frame A LAB S S n y a y cos dyd cos cm cm y cos Scm n y dy d cos cos th E Lab . y If we can measure cm S event-byevent, ”Effective Analyzing Power” is NOT smeared by cos cm S! n y a y dy y n y dy y , ”Effective Analyzing Power” is smeared by cos cm S Center-of-mass frame A CM S cm cos Scm y S cm Spin cm . We have bigger effective 16 Analyzing Power Next things…. Now, I am ready to think about detector performance. I. II. Study the relationship between measured momentum accuracy/precision and a, Analyzing power <A>. Estimate the required performance of the detector. I, also, will play with G4-beamline to think about beam line. (Need a time to learn it, though.) 17 How many positrons we need for EDM ? statistics Value [e cm] comment Exp. ( 3.7 3.4 ) 1019 11.4106 e+, e CERN (1974~76) results (0.04 1.6 0.17) 1019 9.4106 e+ E821 (1999, 2000, Trace back detector, Fig .7)* ( 0.1 0.2 1.07 )1019 975 106 e E821(2001, PSD1-5, Tbl. IV)* Predic (1.4 1.5 ) 1025 -tion > 10 -23 Our goal Mass scale of lepton EDMs Extended SM model ~1013 @ magic=29.3 ~1017@ magic=29.3 1022 level 1024 level ~ 31014 @ =3 ~ 31018@ =3 e 14 EDM 4.7 10 e cm 2 2m c EDM sensitivity: 2t m Bt e 1 A1A 2 N 0e 1 A1A 2 N sum e 0 0 N sum ~ N exp t N e e e t “Improved Limit on the Muon Electric Dipole Moment “ 2EAPS/123-QCD , eB A1 2m 18 y vs. cos cos dP 1 n y 1 Ay cos dyd cos 2 2y 1 2 n y y 3 2y , A y . 3 2y y 19 t N e t N 0 exp cosa t , Relationship between a and N total N e t dt N 0 0 a 1 2 N 0 A 2 ミュービーム強度はによらず、一定だとし、(Ntotal=const.) N total N 0 N*0 *, N*0 N0 , * 1 2 1 2 *a * N*0 A 2 * N 0 A 2 *a a * * , I checked with Toy Monte Carlo