Difference between revisions of "Sadiq IPAC 2013"

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= convert to latex =
 
= convert to latex =
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A version which assumes small angles is given in Eq 7.35 of the same reference as the triple differential cross section:
  
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:<math>\frac{d \sigma}{d \epsilon_1 d \theta_1 d \theta_2} = 8 \left ( \frac{\pi a}{\sinh (\pi a)} \right )^2 \frac{a^2}{2 \pi} \frac{e^2}{\hbar c} \left ( \frac{\hbar}{m_e c }\right )^2 \frac{\epsilon_1 \epsilon_2}{k^3} \theta_1 \theta_2 </math>
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The Cross section formula is given in Formula 3Cs, pg 928 of reference [http://www.physics.isu.edu/~tforest/Classes/NucSim/Day8/BremXsectFormula_Rev.Mod.Phys_vol31_pg920_1959.pdf H.W. Koch & J.W Motz, Rev. Mod. Phys., vol 31 (1959) pg 920] as
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: <math>\times \left \{ \frac{V^2(x)}{q^4} \left [ k^2 (u^2 + v^2) \xi \eta - 2 \epsilon_1 \epsilon_2 (u^2 \xi^2 + v^2 \eta^2 ) + 2 (\epsilon_1^2 + \epsilon_2^2)uv \xi \eta cos(\phi) \right ]   \right . </math>
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:<math>\left . + a^2W^2(x) \xi^2 \eta^2 \left [ k^2(1 - (u^2+v^2)\xi \eta - 2 \epsilon_1 \epsilon_2 (u^2 \xi^2 + v^2 \eta^2) -2 (\epsilon_1^2 + \epsilon_2^2) u v \xi \eta \cos(\phi)\right ]\right \}</math>
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;Note: Bethe & Heitler first calculated this radiation in 1934 which is why you will sometimes hear Bremsstrahlung radiation refererd to as Bethe-Heitler.
 
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:<math>d \sigma = 4 Z^2r_e^2 \alpha \frac{d \nu}{\nu} \left \{ \left (1 + \left( \frac{E}{E_0} \right )^2 \right ) \left [ \frac{\phi_1(\gamma)}{4} - \frac{1}{3} \ln Z -f(Z)\right ] - \frac{2E}{3E_0} \left [ \frac{\phi_2(\gamma)}{4} - \frac{1}{3} \ln Z -f(Z)\right ] \right \} </math>
 
  
 
where
 
where
  
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: <math>E_0</math> = initial total energy of the electron
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:<math>k =</math> photon momentum/energy
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:<math>E</math> = final total energy of the electron
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:<math>\theta_1</math> = scattering angle of <math>e^+</math>
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: <math>\nu = \frac{E_0-E}{h}</math> = energy of the emitted photon
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:<math>\theta_2</math> = scattering angle of <math>e^-</math>
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:<math>Z</math> = Atomic number = number of protons in target material
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: <math>\phi = \phi_1 - \phi_2 = \phi</math> angle between the <math>e^+</math> and <math>e^-</math> pair
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: <math>\gamma = \frac{100 m_ec^2 h \nu}{E_0 E Z^{1/3}}</math> = charge screening parameter
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:<math>\epsilon_1 = \sqrt{p_1^2 + m_e^2}</math> = Energy of the positron
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Coulomb correction to using the Born approximation (approximation assumes the incident particle is a plane wave interacting with a static E-field the correction accounts for changes iin the plane wave due to the presence of the field)  Charge screening and the coulomb correction are different effects that have been shown to be additive/independent. [[File:Haug_2008.pdf]]
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:<math>\epsilon_2 = \sqrt{p_2^2 + m_e^2}</math> = Energy of the electron
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:<math>f(Z) = (Z \alpha)^2 \sum_1^{\infty} \frac{1}{ n [ n^2 + (Z \alpha)^2]}</math>  
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:<math>u = \epsilon_1 \theta_1</math>
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: <math>\sim (Z \alpha)^2 \left \{ \frac{1}{1+(Z \alpha)^2} +0.20206 - 0.0369(Z \alpha)^2 + 0.0083 (Z \alpha)^4 - 0.002 (Z \alpha)^6\right \}</math>
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:<math>v=\epsilon_2 \theta_2</math>
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:<math>\alpha = \frac{1}{137}</math>
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:<math>\xi = \frac{1}{1+u^2}</math>
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: <math>\phi_1</math> and <math>\phi_2</math> = screening functions that depend on Z
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:<math>\eta= \frac{1}{1+v^2}</math>
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:<math>q^2 = u^2 + v^2 + 2 u v \cos(\phi)</math>
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if <math>Z \ge 5</math>
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: <math>x= 1-q^2 \xi \eta</math>
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:<math>V(x) = 1 + \frac{a^2}{(1!)^2} +  \frac{a^2 (1+a^2) x^2}{(2!)^2} + \frac{a^2 (1+a^2)(2^2+a^2)x^4 x^2}{(3!)^2} + \cdots</math>
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:<math>\phi_1(\gamma) = 20.863 - 2 \ln[1+(0.55 \gamma)^2] - 4[1-0.6e^{-0.98} - 0.4e^{-3 \gamma/2}]</math>
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:<math>W(x) = \frac{1}{a^2} \frac{d V(x)}{d x}</math>
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:<math>\phi_2(\gamma) = \phi_1(\gamma) - \frac{2}{3}(1+6.5 \gamma + 6 \gamma^2)</math>
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: <math>a = \frac{Ze^2}{\hbar c}</math>
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For Z<5 see [http://www.physics.isu.edu/~tforest/Classes/NucSim/Day8/Tsai_ScreeningFunctions_Rev.Mod.Phys._vol46_pg815_1974.pdf Tsai, Rev.Mod. Phys., vol 46 (1974) pg 815]
 
−
 
 
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:if <math>3 \ge Z < 5</math> use Equation 3.46 and 3.47
 
  
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:if <math> Z < 2</math> use Equation 3.25 and 3.26
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;Note: The above equations for the differential cross section are using "natural" units where <math>c  \equiv 1</math>

Latest revision as of 03:51, 18 May 2013

Title: Linac Based Positron Production

Positron production using S-band High Repetition Rate Linac (HRRL) was performed at Idaho State University's Idaho Accelerator Center (IAC). Positrons were produced by impinging electrons to a tungsten foil. Bremsstrahlung photons generated in the tungsten foil pair produces electron and positrons. In this paper, we describe the production, transportation and detection of positrons when electron beam energy is 15 MeV.


convert to latex

A version which assumes small angles is given in Eq 7.35 of the same reference as the triple differential cross section:

[math]\frac{d \sigma}{d \epsilon_1 d \theta_1 d \theta_2} = 8 \left ( \frac{\pi a}{\sinh (\pi a)} \right )^2 \frac{a^2}{2 \pi} \frac{e^2}{\hbar c} \left ( \frac{\hbar}{m_e c }\right )^2 \frac{\epsilon_1 \epsilon_2}{k^3} \theta_1 \theta_2 [/math]
[math]\times \left \{ \frac{V^2(x)}{q^4} \left [ k^2 (u^2 + v^2) \xi \eta - 2 \epsilon_1 \epsilon_2 (u^2 \xi^2 + v^2 \eta^2 ) + 2 (\epsilon_1^2 + \epsilon_2^2)uv \xi \eta cos(\phi) \right ] \right . [/math]
[math]\left . + a^2W^2(x) \xi^2 \eta^2 \left [ k^2(1 - (u^2+v^2)\xi \eta - 2 \epsilon_1 \epsilon_2 (u^2 \xi^2 + v^2 \eta^2) -2 (\epsilon_1^2 + \epsilon_2^2) u v \xi \eta \cos(\phi)\right ]\right \}[/math]

where

[math]k =[/math] photon momentum/energy
[math]\theta_1[/math] = scattering angle of [math]e^+[/math]
[math]\theta_2[/math] = scattering angle of [math]e^-[/math]
[math]\phi = \phi_1 - \phi_2 = \phi[/math] angle between the [math]e^+[/math] and [math]e^-[/math] pair
[math]\epsilon_1 = \sqrt{p_1^2 + m_e^2}[/math] = Energy of the positron
[math]\epsilon_2 = \sqrt{p_2^2 + m_e^2}[/math] = Energy of the electron
[math]u = \epsilon_1 \theta_1[/math]
[math]v=\epsilon_2 \theta_2[/math]
[math]\xi = \frac{1}{1+u^2}[/math]
[math]\eta= \frac{1}{1+v^2}[/math]
[math]q^2 = u^2 + v^2 + 2 u v \cos(\phi)[/math]
[math]x= 1-q^2 \xi \eta[/math]
[math]V(x) = 1 + \frac{a^2}{(1!)^2} + \frac{a^2 (1+a^2) x^2}{(2!)^2} + \frac{a^2 (1+a^2)(2^2+a^2)x^4 x^2}{(3!)^2} + \cdots[/math]
[math]W(x) = \frac{1}{a^2} \frac{d V(x)}{d x}[/math]
[math]a = \frac{Ze^2}{\hbar c}[/math]
Note
The above equations for the differential cross section are using "natural" units where [math]c \equiv 1[/math]
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