Sadiq IPAC 2013

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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.


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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:

[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]