G4Beamline PbBi

Development of a Positron source using a PbBi converter and a Solenoid

Conclusions

1. A 0.3 (0.6) Tesla Solenoid with a diameter to allow a 9.74 (3.94) cm diameter pipe would collect a positron per thousand incident electrons on a 2mm thick PbBi target with 0.125 mm thick SS windows.
2. A 4 Tesla Solenoid will remove beam pipe heating from scattered electrons downstream of the target when using a 3.94 cm diameter beam pipe.

Reports

Niowave_Report_11-30-2015

Task List

1.) Create a positron (10,000 positrons) and electron event file containing t,x,y,z,Px,Py,Pz for positrons exiting the solenoid and an incident Gaussian beam 1cm in diameter and with a sigma of 1cm.

2.) Determine the back ground when using a 3.48 diameter beam pipe and Solenoid field of 0.2 for a NaI detector placed at

3.) Experiment, install dipole and solenoid in the tunnel.

Beam Pipe Heating

A 10 MeV electron beam with a radius of 0.5 cm was incident on a 2 mm thick PbBi target. The target is positioned at Z = -902 mm.

Element	dimension
Inner beam pipe radius	1.74 cm
Inner beam pipe thickness	0.165 cm
water jacket thickness	0.457 cm
outer beam pipe radius	2.362 cm
outer beam pipe thickness	0.165 cm
Solenoid inner radius	2.527 cm
Solenoid outer radius	4.406 cm

Max power deposited in beam pipe from uniform beam heating

If you assume a 1mA beam then the beam power incident on the target is

Beam Power = E(MeV) [math] \cdot [/math]I ([math]\mu[/math] A) = 10 MeV [math]\times[/math] 1000 mA = 10 kW

If the beam does not interact with the target and all the beam power is distributed uniformly along a 100 cm long beam pipe with a diameter of 3.48 cm then the power deposited per area would be

[math]10000 \mbox{W} \times \frac{1}{100 \mbox{cm}}\times \frac{1}{ \pi \times 3.48 \mbox {cm} } = 2.3 \frac{\mbox{W}}{\mbox{cm}^2}[/math]

A simulation predicts that about 8 out of 20 electrons will interact with the target and intercept a 34.8 mm diameter beam pipe surrounding the target.

[math]\mbox{P}_{\mbox{max}} = \left ( \frac{2}{5} \right ) 2.3 \frac{\mbox{W}}{\mbox{cm}^2}\lt 1 \frac{\mbox{W}}{\mbox{cm}^2}[/math]

BUT the beam does not intersect the pipe uniformly and instead can have a hot spot.

Heating along the Z-axis

GEANT4 predicts that scattered electrons, photons, and positrons (mostly scattered electrons) deposit

Energy deposited (MeV) in a 1 m long 3.48 cm diameter beam pipe surrounding a 2 mm target located at Z=-900 mm.

According to the above figure, GEANT4 predicts a total of [math]3.08\times 10^7[/math] MeV (the integral adds up the energy in each 1cm bin) of energy will be deposited in a 1m long beam pipe surrounding a 2 mm thick PbBi target located at Z=-902 mm when 20 million electrons impinge the target. The peak energy deposition is 0.3 MeV/e[math]^-[/math]

If this energy were uniformly distributed along the 5 mm thick beam pipe having a diameter of 3.48 cm then I would see

[math]3.08 \times 10^{10} \mbox{keV} \times \frac{1}{100 \mbox{cm}}\times \frac{1}{ \pi \times 3.48 \mbox {cm} } \times \frac{1}{2 \times 10^7 \mbox{e}^-}=3.5\frac{\mbox{keV}}{\mbox{cm}^2 \;\;\mbox{e}^-}[/math]

if you assume a 1 mA beam of electrons then this becomes

[math]\left ( \frac{3.5 \; \mbox{W} }{ \mbox{cm}^2 } \right) [/math]

I converted the above histogram to deposited power by 1000 mA, divide by the number of incident electrons, divide by the circumference of the beam pipe, convert the number of electrons to Coulombs, and use a unit conversion from MeV to W-s per MeV.

[math]\left(Counts \frac{\mbox{MeV}}{\mbox{cm}}\right) \times \left( \frac{1}{2 \times 10^{7} \mbox{e}^-} \right ) \times \left( \frac{1}{ \pi \times 3.48 \mbox{cm}} \right ) \times \left( \frac{1. \times 10^{-3}\mbox{ C}}{\mbox{s} }\right )\times \left( \frac{\mbox{e}^- }{1.6 \times 10^{-19}\mbox{ C}}\right ) \left( \frac{1.6 \times 10^{-13}\mbox{W} \cdot \mbox{ s}}{\mbox{MeV} }\right ) \times [/math]

If you use the above factors to weight the histogram, then the figure below shows that GEANT4 predicts a power deposition density of [math]= 4 \frac{W}{cm^2}[/math], 1 cm downstream of the target. Back scattered electrons appear to create the hottest spot of [math]= 15 \frac{W}{cm^2}[/math] about 1cm upstream of the target.

Power Deposition Zoomed in and 902 mm offset applied	Power deposition over the 1 m long beam pipe

BeamPipeHeating_4mmthick_PbBi_PositronTarget

Unit conversion

The energy deposited by photon, electrons, and positrons is predicted by GEANT4 and recorded in energy units of keV per incident electron on the PbBi target. To convert this deposited energy to a power you need to assume a beam current. Assuming 1 beam current of 1 mA, the conversion is given easily as

[math]\left( \frac{\mbox{keV}}{\mbox{cm}^2 \mbox{e}^-}\right) \times \left( \frac{ \mbox{e}^-}{1.6 \times 10^{-19}\mbox{C}} \right ) \times \left( \frac{1 \times 10^{-3} \mbox{C}}{\mbox{s}} \right ) \times \left( \frac{1.6 \times 10^{-16}\mbox{W} \cdot \mbox{ s}}{\mbox{keV} }\right )[/math]

[math]\left( \frac{\mbox{keV}}{\mbox{cm}^2 \mbox{e}^-}\right) = \left( \frac{\mbox{W} }{\mbox{cm}^2 } \right )[/math]

Results Table

Beam Pipe Diameter (mm)	Hot Spot ([math]MeV/e^-[/math])	Hot Spot ([math]keV/cm^2/e^-[/math])
34.8	0.35
47.5	0.24
60.2	0.20
72.9	0.16
97.4	0.12

Converter target properties

Definition of Lead Bismuth

1cm diameter target 2 mm thick PbBi

0.5 Tesla solenoid

Desire to know

Emmittance (mrad * mm)

dispersion (Delta P/P) (mradian/1000th mm/1000th)

of electrons after the PbBi target.

pole face rotation in vertical plane.

G4BeamLine and MCNPX

Target thickness optimization

Energy Deposition in Target system (Heat)

MCNPX simulations of energy deposition into different cells are below. There is a slight overestimate (they add up to about 120%). Positrons contribute less than 1% of electrons' contribution. No magnetic filed is assumed.

Solenoid

Uniform ideal Solenoid

Beam Pipe Heating with Solenoid

The energy deposited by electrons scattered into a 3.48 diameter stainless steel beam pipe (1.65 mm thick) from a PbBi target as a function of a uniform Solenoidal magnetic field.

B-field (Tesla)	Hot Spot ([math]MeV/e^-[/math])
0.0	0.35
0.3	0.35
1.0	0.35
1.5	0.22
2.0	0.10
4.0	0.002

To convert this deposited energy per incident electron on the target to a heat load in the pipe you need to divide by the area of the pipe.

A histogram is filled with 1 cm bins along the Z axis. The surface area becomes [math]1 cm \times 2 \pi 3.48/2 = 10.933 cm^2[/math]. The beam pipe diameter assumed is 3.48 cm.

When filling the histogram binned 1 cm in Z, you should weight it by the amount of depositred energy divided by the circumference of the pipe and divided by the number of incident electrons on the target (5 million). The energy units are converted to keV by multiplying the numberator by 100 or in this case dividing by 5000 instead of 5 million.

TH1F *T00N=new TH1F("T00N","T00N",100,-1000.5,-0.5)

Electrons->Draw("evt.EoutPosZ>>T00N","evt.DepE/10.088/5000")

To convert From Mev/ e- to kW/cm^2 assuming a current of 1mA (10^-3 C/s) you

[math]\left( \frac{\mbox{MeV}}{\mbox{cm}^2 \mbox{e}^-}\right) \times \left( \frac{ \mbox{e}^-}{1.6 \times 10^{-19}\mbox{C}} \right ) \times \left( \frac{1 \times 10^{-3} \mbox{C}}{\mbox{s}} \right ) \times \left( \frac{1.6 \times 10^{-13}\mbox{W} \cdot \mbox{ s}}{\mbox{MeV} }\right )[/math]

[math]\left( \frac{\mbox{keV}}{\mbox{cm}^2 \mbox{e}^-}\right) = \left( \frac{\mbox{W} }{\mbox{cm}^2 } \right )[/math]

Energy deposited (MeV) along a 1 m long beam pipe of stainless steel 1.65 mm thick.

With SS windows

Positrons->Draw("sqrt(evt.BeamPosPosX*evt.BeamPosPosX+evt.BeamPosPosY*evt.BeamPosPosY)","evt.BeamPosMomZ>0 && evt.BeamPosPosZ>-500 && sqrt(evt.BeamPosPosX*evt.BeamPosPosX+evt.BeamPosPosY*evt.BeamPosPosY)<97.4/2");

Positron Collection rates with 60 cm long Solenoid

Sample Positron Production Events

When the solenoid is 1.5 Tesla, a 10 MeV electron produces a 6.5 MeV photon that pair produces a 4.4 MeV positron and a 1 MeV electron	Same Event but this time the solenoid is 0.3 Tesla and the positron hits the beam pipe, annihilates and makes two 511 keV photons

Sample Brem event producing no positrons

When the solenoid is set to 1.5 Tesla, a 10 MeV electron produces three photons less than 1 MeV in the target, two of them compton scatter in the beam pipe	The same event but this time the electron produces only 1 photon than ionizes in the target

With SS windows

Positrons->Draw("sqrt(evt.BeamPosPosX*evt.BeamPosPosX+evt.BeamPosPosY*evt.BeamPosPosY)","evt.BeamPosMomZ>0 && evt.BeamPosPosZ>-500 && sqrt(evt.BeamPosPosX*evt.BeamPosPosX+evt.BeamPosPosY*evt.BeamPosPosY)<97.4/2");

B-field (Tesla)		34.8 mm diameter pipe	47.5	60.2	72.9	97.4
0.0	0.35	1,2,4,4,5	2,3,4,4,6	4,4,6,7,9	6,8,9,10,11	16,14,15,16,17
0.1		225,236,250,246,249=241[math] \pm[/math] 10	282,282,293,294,306=291[math] \pm[/math] 10	373,366,370,364,373=369[math] \pm[/math] 4	451,437,440,438,451=443[math] \pm[/math] 7	602,584,563,558,570=575[math] \pm[/math] 18
0.3	0.35	626,619,596,619,611 =614[math] \pm[/math] 11	720,726,706,730,717=720[math] \pm[/math] 9	871,864,840,841,834 =850[math] \pm[/math] 16	987,968,939,943,952 =958[math] \pm[/math] 20	1118,1106,1069,1067,1080=1088[math] \pm[/math] 23
0.6		929,935,949,969,961=949[math] \pm[/math] 17	1022,1031,1046,1059,1052 =1042[math] \pm[/math] 15	1120,1130,1152,1154,1146 =1140[math] \pm[/math] 15	1168,1184,1210,1221,1206 =1198[math] \pm[/math] 21	1212, 1218,1240,1254,1242=1233[math] \pm[/math] 18
1.0	0.35	1117,1085,1083,1061,1085=1086[math] \pm[/math] 20	1188,1154,1140,1111,1134=1145[math] \pm[/math] 28	1225,1190,1178,1149,1172 =1183[math] \pm[/math] 28	1243.1208,1195,1164,1184=1199[math] \pm[/math] 30	1252,1219,1206,1178,1200=1211[math] \pm[/math] 27
1.5	0.22
2.0	0.10	1198,1210,1215,1223,1176=1204[math] \pm[/math] 18	1216,1227,1235,1241,1196 =1223[math] \pm[/math] 18	1237,1243,1252,1257,1214=1241[math] \pm[/math] 17	1249,1252,1262,1266,1225 =1251[math] \pm[/math] 16	1257,1262,1270,1276,1234=1260[math] \pm[/math] 16
4.0	0.002

Positron Rates -vs- Solenoid Field for 2mm thick PbBi target and several Beam pipe diameters

Positron & Electron event files

Event files were generated assuming an ideal solenoid having an inner radius of 2.527 cm surrounding a beam pipe with a radius of 1.74 cm. Electrons impinge a 2mm thick PbBi liquid target that has a surface area of 2.54 cm x 2.54 cm. Stainless steel windows 0.25 mm thick sandwhich the PbBi target at locations Z= -90.325 and Z= -89.875 cm. The target is located at Z =-90.1 cm and the beam begins 20 cm upstream at Z = -110.1 cm. The incident electron beam is a 0.5 cm radius cylinder.

Positrons exiting the Solenoid

The solenoid design has changed such that the max field is 0.20 Tesla (0.22) and its length is 150 mm.

In other words I should generate position and momentum files for positrons and electrons at the Z location 15 cm downstream from the middle of the LBE target and within a 3.48 cm diameter beam pipe.

/vis/viewer/zoom 2

/gps/pos/centre 0.0 0.0 -150.

/vis/viewer/panTo -90.1 0 cm

/vis/viewer/reset

The incident electron beam is a 0.5 cm radius cylinder. The LBE target is 2mm and located at Z = -106 mm. The Z location of positrons exiting the beam pipe at the end of the 15 cm long solenoid is 84.275 mm. The positrons are 21.725 mm from the middle of the LBE target to allow for windows and pipe connections.

A comma delimited text file with the above events in the format of

t(ns),x(cm),y,z,Px,Py,Pz(GeV)

in units of cm for distance and MeV for momentum is located at

for positrons

http://www2.cose.isu.edu/~foretony/Positrons_2mm10MeVCyl.dat

and

http://www2.cose.isu.edu/~foretony/Electrons_2mm10MeVCyl.dat

Solenoid Map

Inner Radiusu=

Outer Radius =

Length =

Current=

Magnetic Field Map in cylindrical coordinates (Z & R) from Niowave

Beam Line Design

PbBi_BeamLine_Elements

goals for JLab

Positrons#Simulations