measuring polarization of high energy gamma rays. · 2009. 1. 20. · gamma rays....
TRANSCRIPT
Measuring polarization of high energy
gamma rays.
Polarized photon processes in Geant4.
Gerardo O. Depaola
Universidad Nacional de Córdoba
Member of the collaboration.
Polarized Gamma Astrophysics
Emission Mechanisms (Synchrotron Radiation, Bremsstrahlung, Compton Scattering, Photon Splitting)
Astronomical Sites (Solar Flares , Galactic BH candidates,
Pulsars, AGN, GRB,etc.)COMPTEL
Coburn and Boggs (2003)
The role of Simulations Development and Optimization of instrument concepts
Detailed characterization of instrument response for analysis
Possibility of comparisons between different configurations
Verification of Scientific Objectives and Potential risks
Data processing test-bed and Validation of Physics results
The technique to measure polarization depend
to the energy: Four Main Process
Photoelectric Effect (~ keV to ~50 kev).
Compton Effect (~50 keV to ~10 Mev).
in the nuclear field.
Pair Production:
(>~50 MeV) in the electron field.
Methodology
• Study the cross section
• Convert the cross section into a PDF (Probabilistic Distribution Function).
• Develop an algorithm.
• Write the algorithm into a Computational language (C++) and insert it into the code (GEANT4).
• Test -> Made a simple simulation to reproduce de distribution and/or made simulation of experimental set up and compare results.
Photoelectric Effect
Introduce polar angle distribution.
The polar angle is sampled from the K-shell cross
section derived by Sauter.
Introduce Polarization. The cross section was obtained using the Stokes parameters.
Introduce azimuthal angle distribution.
Polar angle distribution
The polar angle is sampled from the K-shell cross section
derived by Sauter (1955) using K-shell hydrogenic electron
wave function
( )
( )( )( )
θβ−−γ−γγ+
θβ−
θ
γ
β
α=
Ω
σ
cos1212
11
cos1
sin
k
Zr
d
d4
235
2
e
4
Where: k is the photon energy in mc2 units,
E is the electron energy
( )2mc/E1+=γ ( )2
2
mc2E
mc2EE
+
+=β
Azimuthal angle distribution
The “Stokes” matrix for the Photoelectric effects is:
( )
−
−+
θβ−
θα=
0000
B000
A000
00DD1
cos1
sin
k
ErZT
3
2
2
2
o
54
So the cross section for
polarized incident radiation is:
[ ]
φ
φ
φσ
2cos21
0
2sin
2cos
1
0001
D
Td
+≈
−
−=
( )
−
θβ−= 1
cos1Ek
2
k
1D
Example of this
in Geant4
Comparison with experiment
Comparison with experimental data published in Nature (vol. 411,
2001) by E. Costa et. al. : “An efficient photoelectric X-ray
polarimeter for the study of black holes and neutron stars” and
related publications.
The experiment
Gas: 80% Ne, 20% dimethylether at 1
atm.
GEM hole geometry: 40 µm diameter, 60 µm pitch.
128 anodes with a pitch of 200 µm
Drift/absortion gap: 6 mm.
Experimental results
Measure of a single
photoelectron track.
The dimension of the hexagons
is proportional to the energy
deposited
Auger Electron
Photoelectron
Azimuthal distribution of
the charge barycentre.
BA2
B
CC
CC
minmax
minmax
+=
+
−=µ
In this case, µ = 0.44
Modulation factor for 100% linearly polarized radiation:
Azimuthal distribution of barycentre
-1.5 -1.0 -0.5 0.0 0.5 1.0 1.50
100
200
300
400
500
600
700
800
Angular distribution for randomize linearly polarized x-ray
φ [rad]
-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5
200
400
600
800
Angular distribution for 100% linearly polarized x-ray
along the x axis.
Model: a + b*cos2(φ+c)
a = 249 ± 4
b = 547 ± 6
c = 0.0014 ± 0.0056
φ [rad]
Unpolarized Polarized
Modulation factor µ = 0.41Impact point: x = 0, y = 0.Charge Barycentre:
∑=T
iib
E
Exx ∑=
T
iib
E
Eyy
=> the azimuthal angle ϕ = atan(yb/xb)
COMPTON SCATTERING
The Klein Nishina cross section:
Θ+−
ν
ν+
ν
ν
ν
ν=
Ω
σ 2
0
0
2
0
22
0 cos42h
h
h
h
h
hr4
1
d
d
Where,
hν0 : energy of incident photon.hν : energy of the scattered photon.Θ : angle between the two polarization vectors
Angles in the Compton Effect
θ Polar angle
φ Azimuthal angle
ε Polarization vector
y
x
O z
ξ
θα
φhν
hν0ε A
C
ε’⊥
ε’||
CO
ε
Ahν
ε’ ξβ Θ
ξ
x
Polar Angle distribution
Summing over the 2 components:
Sampling method (G4LowEnergyPolarizedCompton)
Integrate over φ
Sample θ
Energy from the Compton θ-E relation Sampling φ from P(φ) = a (b – c cos2 φ ) distribution
−+=
Ωφθ
ν
ν
ν
ν
ν
νσ 22
0
0
2
0
22
0 cossin22
1
h
h
h
h
h
hr
d
d
Scattered Photon Polarization Geometrical relations for obtaining the Scattered Photon Polarization.
β is obtained from cos Θ = cos βN and Θ is sampled from the Klein Nishina cross section.
=−=
⇒=
φθξ
φθξ
22 cossin1sin
cossincos
β
φθθ−φφθ−=ε coskcoscossin
N
1jcossinsin
N
1iN 2'
||
( ) βφθ−θ=ε⊥ sinksinsinjcosN
1'
ε’⊥
ε’||
CO
ε
Ahν
ε’ ξβ Θ
ξ
x
MEGA (courtesy of A. Zoglauer)Marx Planck Instituto of Munich
Polarization 2 MeV
-0,3
-0,2
-0,1
0
0,1
0,2
0,3
0 45 90 135 180
Azimuthal scatter angle [°]
rel. counts
0
5
10
15
20
25
30
35
0 20 40 60 80 100Compton scatter angle [deg]
Modulation[%]
Dependence of the degree of modulation
on the Compton scatter angle
Azimuthal scatter angle distribution
(geometry and efficiency corrected)
The MEGA prototype, which consists
of l/12 of the volume of a full
telescope, has been calibrated at the
High Intensity Gamma Source of the
Free Electron Laser facility at Duke
University
Reconstruction of incoming direction
from scattering angle θ and the the scatter photon direction.
Ellipses obtain for 20 100 keV photons
-15 -10 -5 0 5 10 15-15
-10
-5
0
5
10
15
Coordenada Y [cm]
Coordenada X [cm]
Intersection point between the
ellipses.
θ
The reconstruction method
Pair Production
The existing methods of analyzing can be
divided into three groups
Uses the photoproduction of particles on amorphous targets and investigate the correlations between the photon polarizations and the kinematics of the final particles.
Uses the coherent effects of the interaction of the photon with single crystal.These lead to a dependence of the photon absorption on the angle between the photon polarization vector and the crystallographic axes.
Uses the correlations between the polarization of the particles in the initial and final state. The exceptional complexity of such experiment means that can not be realized at the present times.
In the first and second groups, the differential cross section
has an azimuthal asymmetry in the form:
[ ]ϕσϕσσϕ
σπ 2cos12cos2 )()()( Λ−=−= PP
d
d tlt
Where:
σ(t): is the total cross section on unpolarized photon.
σ(l): is the part due to polarization.
P: is the degree of linear polarization.
Λ= σ(l)/ σ(t): is the asymmetry.
ϕ: is the azimuthal angle.
For high energy photon, the production of e+e- pairs on nuclei and electrons is promising:
1) γ +A → A + e++ e-
2) γ + e- → e- + e++ e-
The cross sections of the processes 1) and 2) are proportional, respectively, to Z2 and Z and increase with photon energy. At very large energy, this growth is logarithmic.
The process 1) is difficult to applicate for photon energies > 1 GeV due to the small opening of the created pair.
The process 2) is very convenient for measuring the linear polarization of photon beams in a very wide range of photon energies. A significant proportion of the recoil electron have a kinetic energy sufficient for their detection.
Gamma conversion in the nuclear field
Modification polar angle
distribution.
Different approach
Sample directly the trigonometric values
Introduce Polarization.
Calculation of the cross section (E>100 MeV)
Modification azimuthal angle distribution
Important for Pair production telescopes
e-
e+
hν
Pair production telescopes
Polar angle sampling
Sauter-Gluckstern-Hull formula
( )2cospEp2
sin
d
dP
±±±±
±
± θ−
θ=
θ
From this distribution the sampled angles are:
±±
±±±
±±
±+−
+−=
+−
−=
Erndp
prndE
Erndp
rndrnd
)12(
)12(cos;
)12(
)1(2sin θθ
This formulae produce the same distribution as in G4GammaConvertion
( ) ±±−− θ=+= Eu;eudeuC)u(f au2au
The advantage of the first formula is to sample directly sinθ and cosθ
Azimuthal distribution and PolarizationBethe-Heitler Cross Section:Unscreened point nucleus
( )( ) ( )
( )
]m)1)(cosE(
)1(cosE)coscoscossinsin1)(E(E[2q
cos2sin
sin
E
)E(
sin
sin
E
E
cos1
sin
cos1
sin
cos1
)cos(sin
cos1
cossinq
cos1
)cos(sinE
cos1
cossinE4
q
EEddEd
mr
2
Z2d
2
2
2
22
2
43
2
0
2
2
+−θ−ωω+
+−θω+θθ−φθθ−−ω−=
φ+
θ
θ−ω+
θ
θ
−ω×
×θ−
θ
θ−
θω−
θ−
φ+ψθ−
θ−
ψθ−
−
θ−
φ+ψθ−ω+
θ−
ψθ−ωΩΩ
ωπ
α−=σ
−
+−+−+
+
−
−
+
+
+
−
−
+
+
−
−
+
+
−
−−+
Validity: First Born approximation, no screening, negligible nuclear recoil
Angles in pair productionz
y
xεεεε
k
p-p+
Ψ
φ
θ-θ+
The polarization dependence
is represented trough the
angle Ψ.
The angle ϕ is the angle between the projection of the
particle moments in the x-y
plane.
For unpolarized beam => Ψ is isotropic, not ϕ which is related to the azimuthal
distribution of the pair.
Integration over the energy and polar angles => azimuthal
distribution of a pair created by 100 MeV photon.
0.00 0.52
1.05 1.57
2.09 2.62
3.14 3.00 3.02
3.04 3.06
3.08 3.10
3.12 3.14 4
6
8
10
12
14
16
1 d σ
α (Zr 0 ) 2 d φ d ψ
φ [rad] ψ [rad]
This surface can be
parameterized and
its parameters can
be put in function of
the photon energy.
Integrating over Ψone obtain the φ distribution .
Azimuthal distribution
0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 3 .5 0 .7
0 .8
0 .9
1 .0
1 .1
1 .2
1 .3
1 .4
1000 M eV
R
0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 3 .5 0 .7
0 .8
0 .9
1 .0
1 .1
1 .2
1 .3
1 .4
500 M eV
0 .0 0 .5 1 .0 1 .5 2 .0 2 .5 3 .0 3 .5 0 .7
0 .8
0 .9
1 .0
1 .1
1 .2
1 .3
1 .4
100 M eV
∆ ε [rad ]
Ratio of nº of pairs contained in plane parallel to the
polarization vector with nº of pairs perpendicular plotted
against experimental azimuthal resolution
Asymmetric ratio for pair production
High Energy Gamma-ray Polarimeter
Polarization capabilities
evaluated with G4
Pixelized gas MicroWell detectors
for a 3D recon of the pair
“A Concept for a High Energy
Gamma - Ray Polarimeter” P. F.
Bloser (NASA), S. D. Hunter (NASA),
G. O. Depaola (UNC), F. Longo
(INFN).
SPIE 5165, “X- ray and Gamma –
ray Instruments for Astronomy XIII”.
(2003) 322.
Pair production by linearly polarized γγγγ-rays on electrons: Angular distribution for the electron recoil.
Feynman diagrams of Triple Production
The first study of this process were done by:
A. Borsellino and V. Votruba.
A. Borsellino, Nuovo Cimento 4, 112 (1947).
A. Borsellino, Rev. Univ. Tuc. 6, 37 (1947)
Calculate the pair production cross section if the field produce by a particle if mass M and charge Ze; he used only the a and b diagrams.
V. Votruba, Bull. Int. Akad. Tcheque Sci. 49, 19 (1948).
Calculate the all eight diagrams and take the limits for low andhigh energy.
( )−+
−+
−+ −−−+= ppppkEEE
pdpdpdM
rd 1
4
1
33
1
32
2
2
0
4δ
κπ
ασ
22
2
1dcba AAAAM −+−=
Mork’s calculations (Phys. Rev. 1967) of the contribution to the
total cross section of the considered process from different
groups of diagrams separately are very instructive:
XB Borsellino diagrams
Xγ γ-e- diagrams
The other are the interference terms.
∆i =σi / σB => σ = σB(1-∆)
From the figure => for
ω≥16m
σ is described by B diagrams with an accuracy better than 1.2%.
EBEBEBB XXXXXXM γγγγ +++++=2
Momentum and angles of final state particles
z
x
y
p1
p2
k
θ1
θ2
θ21
ϕ1
ϕ21
ϕ2
z
y
x
k
p-
p+
θ-θ+
ϕr
pr
ϕ+
ϕ-
θr
=≤≤
ωθθ
m4arccos0 max
The polar angle θ with which is emitted any of the electrons (whatever will be electron or positron), is
limited by the energy ω of photon according to:
Differential cross sections for triple production
by linearly polarized photons
From this θ, p relation one can see that for θ =0, p ∞, this is consequence of the approximation. The kinematics of the process impose limits to θ maximum.
−−
−−
−=
2cotln
cos
sin12cos
2cotln
cos
cos511
cos
sin
3
22
1
1
1
2
1
1
1
2
1
3
1
2
11
0
θ
θ
θϕ
θ
θ
θ
θ
θα
θϕ
σπ
P
r
dd
d
mqEmp +== 2
1
2
1
sin
cos2
θ
θ
In the high energy gamma rays limits, it is possible to find the following angular
distribution expression for the recoil electron, in which the terms of order m/ωcan be ignored.
p
θ
θmax
pmax(θ)
pmin(θ)
p+(∆,θ)
p-(∆,θ)
θ
∆min = 2m
p(θmax)
∆ = constant arbitrary value
D = 0
π/2( )m
mmp
−=
ω
ωθω
2,
*
max1
From these figure one can
see that for p > p* the
recoil electro are emitted at
angles less than
Where p* is:
( )max1 ωθ
In this figures we represent schematically the recoil electron momentum as
function of ∆2 and θ1.
( ) ( )
( ) 1
222
2
2/1
2
1
2
1lim
sin4
:where
cos,,
θ
θθω
mSSmD
D
mSmp
−+=
−=∆
( )
++
−= −
1
1
1
11
max11 cos,p
mEm
p
mEp
ωωθ
p
1000
100
10
1
0,1
0,01
0,001
0,2 0,4 0,6 0,8 1,0 1,2 1,4 θ
plim(θ)
10 m
50 m
100 m
ω = 1000 m
In this figure we shown the region of allowed values of the linear
momentum (tridimensional) against the polar angle for several
values of the energy of the incoming photon ω.
θ
θ2lim
sin
cos2mp =
Probability density of linear (three dimensional) momentum values, for several polar angle values
of the emitted particle, for ω = 100 m. The dotted line, corresponding to θ = 0.6 for ω = 1000
m, shows the increasing ω typical behavior: probability concentrates strongly near the boundary
of the allowed region (∆ ≅ 2 m), slightly shifting toward the higher p border.
1 2 3 4 5 6 7 p
θ = 0,5
0,6
0,7
0,80,9
1,01,1
1,21,3
d2σdp dθ
1
α r02
10
1
0.1
1000 m
)cos(2Pdσdσdσ (l)(t) ϕ−=
)(
)(
t
l
d
d
σ
σ=Λ
∫=max
00
)(2)( )(
p
p
ii
p dpdpd
dF
θ
σθ
( )( )
−−= 2cotln
cos
cos511
cos
sin
3
2 2
3
2
0)( θθ
θ
θ
θαg
rF t
a
( )( )
−= 2cotlncos
1cos
sin
3
2 2
3
2
0)( θθ
θ
θ
θαg
senrF l
a
Angular Distribution
We define the asymmetry as:
We also definewhere the upper index ‘i’ that can be
‘t’ or ‘l’
p0 is some threshold momentum value for the detection of recoil electrons
Asymptotic functions
calculated by Boldyshev et.al.
They arrived to this results
calculating the dσ2/dpdϕassuming the recoil electron
acquires the maximum angle
compatible with its momentum
p0 = 1
p0 = 0,5
p0 = 0,2
ω = 100 m
0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 1,1 θ
7
6
5
4
3
2
1
p0 = 1
p0 = 0,5
p0 = 0,2 ω = 1000 m
0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 1,1 θ
7
6
5
4
3
2
1
p0 = 1
p0 = 0,5
p0 = 0,2
ω = 100 m
1,0
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 1,1 θ
ω = 1000 m
1,0
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 1,1 θ
Integral of the differential cross section (in units of ⅔αr02) over the p values from the
three p0 threshold values indicated, plotted as function of the polar angle of recoil electron.
Superimposed are shown the asymptotic functions The upper figures belong to the non
polarized radiation case. The polarized radiation case is shown below. For ω = 1000 m all the graphics appear superimposed to the asymptotic function.
)()(0
θi
pF
)()(0
θt
pF
)()(0
θl
pF
Asymmetry as a function of the polar angle of the electron recoil
for the asymptotic values of cross section
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.60.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
θ [rad]
Λ
)(
)(
t
a
l
aa
F
F=Λ
From this figure one can
see that the maximum
value of Λa is around 0.14 for θ maximum.
MONTECARLO SIMULATION OF THE
ASYMPTOTICS EXPRESIONS
We must generate random values of θ and ϕ distributed with
probability proportional to the following function f(θ,ϕ), for θrestricted inside of its allowed interval value, not too close to its
extremes (either 0, or θmax):
where:
As we will see, for θ < 90º, F1 is several times greater than FP, and since both are positive, it follows that f is positive for any possible
value of P (0 ≤ P ≤ 1).
( ) ( ) ( ) ( )( )θϕθθ
θϕθ PFFf 2cosP
cos
sin, 13
−=
( ) ( )( )2cotlncos
cos511
2
1 θθ
θθ
−−=F
( ) ( )( )2cotlncos
sin1
2
θθ
θθ −=PF
Algorithm for non polarized radiation.
We must generate random values of θ between 0 and θmax = arccos( ), distributed with probability proportional to the following function f1(θ):
ωm4
( ) ( )( ) ( )θθ
θθ
θ
θ
θ
θθ 13
2
31cos
sin2cotln
cos
cos511
cos
sinFf ×=
−−=
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.60
5
10
15
20
f1(θ)F
1(θ)
θ [rad]
Graphics of f1(θ) and F1(θ). The infinity of F1 at extreme θ = 0 is very weak (logarithmic), and thus,
multiplying by sinθ, shows that f1 →0. The f1 divergence at π/2 is not important because the interval of
allowed values of θ ends before this point.
To simulate the f1 function, we
decomposed in two factors: the first,
senθ/cos3θ, easy to integrate, and the other, F1(θ), which may constitute a reject function, on despite of its θ = 0 divergence
( ) 2if,cos35
331cos
3
14 22
1 πθθθθ →
++→ KF
Expanding F1 for great values of θ, we see it is proportional to cos2θ:
Thus F1 divided by cos2(θ) will be
a better reject function, because it
tends to constant value 14/3 =
4,6666... for large θs, whereas for small θs, cos(θ)→ 1.
It seems adequate to choose θ0near 5º, and, after some
manipulation looking for round
numbers we obtain:
( )( )
00.14º47.4cos
º47.42
1 ≅F
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.60.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.60.0
0.2
0.4
0.6
0.8
1.0
4.47º
r(θ)
1/3
θ [rad]
( )( )
≤θ
≥θ
θ−
θ+
θ
θ−−
θ=
θ
θ
=θ
4.47ºfor;1
4.47ºfor;cos1
cos1ln
cos2
cos511
cos14
1
cos14
1 2
22
1F
r
Finally we define a reject function :
Now we have a probability distribution function (PDF) for θ, p(θ) = C f1(θ), expressed as a product of another PDF, π(θ), by the reject function:
( ) ( ) ( ) ( )θθπθθ rCCfp '
1 ≅=
( )θcos
θsin14Cθπ π=Where:
the cumulative
probability ( )ω4mln
θ)ln(cos2
m4
ωln7
θ)ln(cos14θd)θπ(P
θ
0π =
−=′′= ∫
Finally we sample a random number ξ1 (between 0 and 1), and obtain the corresponding θ value:
ω=θ
ξ
2
1
m4arccos
Another random number ξ2 is sampled for the reject process: the θ value is accepted if ξ2 ≤ r(θ), and reject in the contrary. For θ ≤ 4,47º all values are accepted.
Algorithm for polarized radiation.
The azimuthal dependence of the differential cross section is given by:
( ) ( ) ( )( )θF)cos(2PθFθcos
θsinθ,f P13
ϕϕ −=
( )( )2θcotgln
cosθ
θsin1)(θF
2
P −=
The expansion of FP for θ near π/2 shows that it is proportional to
cos2(θ), and FP/cos2(θ) tends to a non
null value, 2/3. This value is exactly
7 times the value of F1/cos2(θ).
This suggests applying the
combination method, rearranging the
whole function as follows:
( ) ( ) ( ) ( )( )
−=
θF
θFP)cos(21
θcos
θFθtgθ,f
1
P
2
1 ϕϕ
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.60.0
0.2
0.4
0.6
0.8
1.0
1.2
π/2
2/3
FP(θ)
θ [rad]
FP(θ)/cos(θ)
2
The conditional probability of ϕ given θ is p(ϕ / θ) :
( ) ( )( ) ( )
( ) ( ) ( )( )
−==
θF
θFP2cos1θF
θcos
sinθC
θcos
θFtgθCπ2
1
θq
θ,pθ/p
1
P13
2
1
ϕϕ
ϕ
θ
θϕ−
π=
)(F
)(FP)2cos(1
2
1
1
P
Where: ( ) ( )θcos
θFθtgCπ2d),p(θq
2
12π
0== ∫ ϕϕθ
ϕ
2π
p(ϕ / θ)Reject function for
the ϕ values.
1.-We begin sampling a random number ξ1 and obtain θ from
2.- Then we sample a second random number ξ2 and accept the values of θ if ξ2 ≤ r(θ),
( ) 4.47ºθfor;θcos1
θcos1ln
θcos2
θ5cos11
θcos14
1θr
2
2≥
−
+−−=
For θ ≤ 4,47º all values are accepted .
3.- Now we sample ϕ. We sample a third random number ξ3 (which is defined as ϕ/2π) and evaluate the reject function
( ) ( ) ( )( )
−=
θF
θFPξπ4cos1
2π
1ξr
1
P33θ
( )
−−
−
−=
2
θcotglnθcos51cosθ
2
θcotglnθsincosθ
P)4cos(12π
1
2
2
3ξπ
4.- Finally, with a fourth random number ξ4 , we accept the values of ϕ = 2πξ3 if ξ4 ≤ rθ(ξ3).
The sample technique
ω=θ
ξ
2
1
m4arccos
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.60
1000
2000
3000
4000
5000
6000
7000
8000
9000
Arbitrary Units
θ [rad]
50 m
100 m
1000 m
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.60.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0.18
Arbitrary Units
θ [rad]
50 m
100 m
1000 m
Example of polar angle distribution for the electron recoil for three
differences energies generated with the method (Distribution
obtained with 100000 histories). On the right, the same as left but
renormalized to show that the dependence with the energy is only in
the θmax.
0 1 2 3 4 5 60
250
500
750
1000
1250
Arbitrary Units
ϕ [rad]
Azimuthal angle distribution for the electron recoil. Continuous
line, data fit with expression p(ϕ) =a(1-b(cos(2ϕ-c)) => dσintegrated over all θ.
( )
137.0θdσ
θdσb
θ)cos(2dσPdσ)(d
max
max
max
θ
0
(t)
θ
0
(l)
t
θ
0
(l)(t)
==⇒
ϕ−=ϕσ
∫
∫
∫
d
d
d
b=0.138+/-0.001 for ω=1000.
The theoretical value is:
This result also shows that one does
not lost much asymmetry taking into
account all θ, since the maximum asymmetry is for θ = θmax and for that θ, Λ = 0.143.
Comparison of the cross section with experimental distribution with
respect to θ1.1
)( θσ dd t
Experimental data with:
0.4≤ q ≤ 1 MeVc,
q ≥0.4 MeV
20 ≤ ω ≤ 600 MeV
45 ≤ ω ≤ 75 MeV.
The dot-dash curve is the
asymptotic cross section.
The solid curve is the cross
section for ω = 1400m and q0 = 0.8 m.
Simulation example of electron recoil in 1.0 m of 85%Xe 25%CO2
Same as before but includind the pair created. Blue lines are positron, red lines are electron and green lines are photons.
Simulation results
-0,5 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5-0,5
0,0
0,5
1,0
1,5
2,0
2,5
3,0
3,5
4,0
θ
100 MeV
Simulated distribution
Theoric distribution
0,0 0,5 1,0 1,5 2,0 2,5 3,0
0
100
200
300
400
500
ϕ
Comparison between the theoretical polar angle distribution and simulate results.
( ) ( )θ θ θ= +
arctan tan tanX Y
2 2
Reconstructed azimuthal distribution; superimpose a cos2 ϕ.
X
Y
θ
θϕ
tan
tanarctan=
The End