Post on 25-Dec-2015
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.
Tool: GEANT4
1
II. Exercise to check basic kinetics: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.
Today’s contents
0,tsin,tcosS aa
Center-of-Mass system
taX,
Y
Z
Momentum
sinsin,cossin,cosp
p mcm
mcm
mcm
e
eDirection of decay-positron
mcm
ZB,0,0B
Magnetic field
Spin-direction
mcm
mcmm
Lab cos1
coscos
,mLabWe measure . Lorentz boost
2
tcoscostsincossinSp
pcos a
mcma
mcm
e
eScm
Scm
Angle between spin-axis and momentum direction of decay-e+ at the center-of-mass system:
2
mp
ppy
maxe
maxee
.y23
1y2yA ,y23yyn
,sincosyA1yn2
1
dyd
dP
2
Monte Carlo
Expected 2D event yield distribution as functions of y and cmS
3
Monte Carlo
4
P=300MeV/c , =3,Tc =7.4nsec, R=333mm,Ta=2/a=2.2sec.
Positron energies28 ~191 MeV
GEANT4B3 T
Condition:
5
8.6MeV positron
50.4MeV positron
GEANT4
102MeV positron
B = 3T
6
II. Check basic kinetic values from GEANT4
7
Probing Spin-dependent Decay Info.
I. To be more simple, I set 100% ! 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!!
ee
8
2
mp
ppy
maxe
maxee
GEANT4
XL
2Z
2YT
pp
ppp
2Z
2Y
2Xe pppp
X axis is always -momentum direction.
9
Monte Carlo vs. GEANT4
y = pcme/pmax
is an angle between spin-axis and momentum direction of decay-e+ at the center-of-mass system.
10
Monte Carlo
GEANT4
11
III. Wiggle plots made by GEANT4
12
“Usual” wiggle plot and “Beam-loss free” wiggle plot
tcosA1t
expNtF a
4 free parametersCovariant matrix is OK.9.5 105 , E> 200 MeV 1.3105e+
13
tcoscosA12
N
tcoscostdydsinyayndyyn2
1tN
tcoscosya12
yn
dydsin
dP
dyd
dP
.tcoscostsincossincos ,here
,cosya1yn4
1
dyd
dP
amcm
amomcm
y
Scm
Scm
y
amomcmS
cmScmcm
amomcma
momcm
Scm
Scm
cm
Scm
tcosAtRtL
tRtLtAsym
,tcosA12
NtR ,tcosA1
2
NtL
a
aa
“Beam loss free” wiggle plot by knowing
momcm
Measure!
An angle between + and e+ momentum direction in the center-of-mass system.
No exponential term! 14
LEFT RIGHT
tcosAtAsym a
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 0.7RIGHT:1 cos 1 0.7
15
A big advantage to measure
.y
Ecos
,cosddyyn
cosdydcosyayn
A
thLabS
cm
cos
Scm
y
y cos
Scm
Scm
LAB
Scm
Scm
.
dyyn
dyyayn
A
y
yCM
Lab-frame
Center-of-mass frame
”Effective Analyzing Power” is smeared by cos cm S
If we can measure cm S event-by-
event, ”Effective Analyzing Power” is NOT smeared by cos cm
S!
Spincm
We have bigger effective Analyzing Power 16
Next things….
I. Study the relationship between measured momentum accuracy/precision and a, Analyzing power <A>.
II. Estimate the required performance of the detector.
Now, I am ready to think about detector performance.
I, also, will play with G4-beamline to think about -beam line. (Need a time to learn it, though.)
17
18
How many positrons we need for EDM ?Value [e cm] statistics comment
Exp. results
( 3.7 3.4 ) 1019
(0.04 1.6 0.17) 1019
( 0.1 0.2 1.07 )1019
11.4106 e+, e
9.4106 e+
975 106 e
CERN (1974~76)E821 (1999, 2000, Trace back detector, Fig .7)*E821 ( 2001, PSD1-5, Tbl. IV)*
Predic-tion
(1.4 1.5 ) 1025 Mass scale of lepton EDMs
> 10 -23 Extended SM model
Our goal
1022 level 1024 level
~1013 @ magic=29.3
~1017@ magic=29.3
~ 31014 @ =3~ 31018@ =3
cme107.4cm2
e
2EDM 14
e
m
Bt
2 t
0
e
t
0e
sume
sume1
0e1
NtexpN~N
,N2AA
1
N2AA
1
EDM sensitivity:
m2
eBA1
“Improved Limit on the Muon Electric Dipole Moment “ 2EAPS/123-QCD
y vs. cos
.y23
1y2yA ,y23yyn
cosyA1yn2
1
cosdyd
dP
2
y
cos
19
20
a
0
0
etotal
a0e
AN
21
NdttNN
,tcost
expNtN
*a
*a
*
20
*2*0
**a
0**0
**00total
,AN
21
AN
21
,NN
,NNN
ミュービーム強度はによらず、一定だとし、(Ntotal=const.)
I checked with Toy Monte Carlo
Relationship between a and