Analytical solutions to intensity calculation in scintillation detectors and their application to the scintillation flash coordinate reconstruction

Abstract

Detectors with continuous scintillator and matrix of photomultipliers attached are widely used in modern particle physics and various medical tomographs. There are two common ways to calculate the coordinates of the scintillation flash via photomultipliers energy outcome: Anger’s method and Monte Carlo simulations. In this article, we derive an analytical solution for the computation of the energy outcome from several photomultipliers built into the bottom face of the scintillation camera with a continuous scintillator shaped as a rectangular parallelepiped. The lateral faces of the scintillation camera are considered specular reflectors with reflective indexes $\sim 1$ or 0. The analytical solution is used as a reference point for the scintillation flash coordinate reconstruction. The achieved results are similar to those the Monte Carlo simulation provides, but require less computational time.

Appendix: Integrals

In this appendix, we derive Eqs. (5) and (18). To derive Eq. (5), the integral over y should be computed first in Eq. (4). To do that, the indefinite integral should be considered:

$$\begin{aligned} \int \frac{\textrm{d}v}{(v^2+k)^{3/2}}&=\int \frac{\textrm{d}v}{k} \left( \frac{1}{\sqrt{v^2+k}}-\frac{v^2}{(v^2+k)^{3/2}}\right) ,\nonumber \\ \int \frac{v^2}{(v^2+k)^{3/2}}&=-\frac{v}{\sqrt{v^2+k}}+\int \frac{\textrm{d}v}{\sqrt{v^2+k}},\nonumber \\ \int \frac{\textrm{d}v}{(v^2+k)^{3/2}}&=\frac{v}{k\sqrt{v^2+k}}+C. \end{aligned}$$

Thus,

$$\begin{gathered} \int\limits_{{y^{{(L)}} }}^{{y^{{(L)}} + b}} {\frac{{{\text{d}}y}}{{((x - x_{0} )^{2} + (y - y_{0} )^{2} + z_{0}^{2} ))^{{3/2}} }}} \hfill \\ = \frac{{y^{{(L)}} - y_{0} + b}}{{((x - x_{0} )^{2} + z_{0}^{2} )\sqrt {(x - x_{0} )^{2} + (y^{{(L)}} - y_{0} + b)^{2} + z_{0}^{2} } }} \hfill \\ - \frac{{y^{{(L)}} - y_{0} }}{{((x - x_{0} )^{2} + z_{0}^{2} )\sqrt {(x - x_{0} )^{2} + (y^{{(L)}} - y_{0} )^{2} + z_{0}^{2} } }}. \hfill \\ \end{gathered}$$

(28)

Now the integral over x can be computed. The corresponding indefinite integral is

$$\begin{aligned}\int \frac{\textrm{d}u}{(u^2+c)\sqrt{u^2+p}}= &\begin{bmatrix} u=\sqrt{p}\tan {q}\\ \textrm{d}u=\sqrt{p}\frac{\textrm{d}q}{\cos ^2 q} \end{bmatrix}\nonumber \\ = & \int \frac{\textrm{d}q}{(p\tan ^2 q+c)\cos {q}}= \int \frac{\textrm{d}q\cot }{(p+c \cot ^2 q)\sin {q}}\nonumber \\ = & \begin{bmatrix} t=\frac{1}{\sin q}\\ \textrm{d}t=-\frac{\cos q}{\sin ^2 q}\textrm{d}q=-t \cot q \textrm{d}q \end{bmatrix}=-\int \frac{\textrm{d}t}{p+c(t^2-1)}=\nonumber \\ &-\frac{1}{p-c}\int \frac{\textrm{d}t}{1+\left( \sqrt{\frac{c}{p-c}}t\right) ^2}=\nonumber \\ & - \frac{1}{\sqrt{c(p-c)}}\tan ^{-1}\left( \sqrt{\frac{c}{p-c}}t\right) +C=\nonumber \\ & -\frac{1}{\sqrt{c(p-c)}}\tan ^{-1}\left( \sqrt{\frac{c}{p-c}}\frac{1}{\sin q}\right) +C=\nonumber \\ & -\frac{1}{\sqrt{c(p-c)}}\tan ^{-1}\left( \sqrt{\frac{c}{p-c}}\sqrt{\frac{u^2+p}{u^2}}\right) +C\nonumber \\ = & \frac{1}{\sqrt{c(p-c)}}\tan ^{-1}\left( \sqrt{\frac{p-c}{c}}\sqrt{\frac{u^2}{u^2+p}}\right) +{\tilde{C}}. \end{aligned}$$

(29)

Here, $u=x-x_0$, $p=(y^{(L)}-y_0)^2+z_0^2$, and $c=z_0^2$. The last equality follows from $\tan ^{-1}x+\tan ^{-1}\frac{1}{x}=\frac{\pi }{2}$. Thus,

$$\begin{aligned}{} \int \limits _{x^{(L)}-x_0}^{x^{(L)}-x_0+a}\frac{\textrm{d}u}{(u^2+c)\sqrt{u^2+p}}= \frac{\tan ^{-1}\left( \frac{u\sqrt{p-c}}{\sqrt{c}\sqrt{p+u^2}}\right) }{\sqrt{c}\sqrt{p-c}}\Bigg |_{x^{(L)}-x_0}^{x^{(L)}-x_0+a}. \end{aligned}$$

(30)

Finally, the integral from Eq. (4) equals

$$\begin{gathered} I_{A} = \Bigg|\frac{{{\text{sign}}(y^{{(L)}} - y_{0} + b)}}{{4\pi }}I \hfill \\ \quad \times \Bigg[\tan ^{{ - 1}} \left( {\frac{{(x^{{(L)}} - x_{0} + a)|y^{{(L)}} - y_{0} + b|}}{{z_{0} \sqrt {(x^{{(L)}} - x_{0} + a)^{2} + (y^{{(L)}} - y_{0} + b)^{2} + z_{0}^{2} } }}} \right) \hfill \\ - \tan ^{{ - 1}} \left( {\frac{{(x^{{(L)}} - x_{0} )|y^{{(L)}} - y_{0} + b|}}{{z_{0} \sqrt {(x^{{(L)}} - x_{0} )^{2} + (y^{{(L)}} - y_{0} + b)^{2} + z_{0}^{2} } }}} \right)\Bigg] \hfill \\ \quad - \frac{{{\text{sign}}(y^{{(L)}} - y_{0} )}}{{4\pi }}I \hfill \\ \quad \times \Bigg[\tan ^{{ - 1}} \left( {\frac{{(x^{{(L)}} - x_{0} + a)|y^{{(L)}} - y_{0} |}}{{z_{0} \sqrt {(x^{{(L)}} - x_{0} + a)^{2} + (y^{{(L)}} - y_{0} )^{2} + z_{0}^{2} } }}} \right) \hfill \\ - \tan ^{{ - 1}} \left( {\frac{{(x^{{(L)}} - x_{0} )|y^{{(L)}} - y_{0} |}}{{z_{0} \sqrt {(x^{{(L)}} - x_{0} )^{2} + (y^{{(L)}} - y_{0} )^{2} + z_{0}^{2} } }}} \right)\Bigg]\Bigg|. \hfill \\ \end{gathered}$$

(31)

Equation (18) can be derived in the same manner. First, the integral over y should be computed in Eq. (17)

$$\begin{gathered} \iint_{B} {\frac{{{\text{d}}x{\text{d}}y}}{{((x - x_{0} )^{2} + (y - y_{0} )^{2} + z_{0}^{2} )^{{3/2}} }}} \hfill \\ \quad = \int_{{x_{1} }}^{{x_{2} }} {\int_{{y_{0} + L}}^{{y_{0} + \sqrt {R_{{cr}}^{2} - (x - x_{0} )^{2} } }} {\frac{{{\text{d}}x{\text{d}}y}}{{((x - x_{0} )^{2} + (y - y_{0} )^{2} + z_{0}^{2} )^{{3/2}} }}} } \hfill \\ \quad = \left[ {\begin{array}{*{20}c} {\tilde{x} = x - x_{0} } \\ {\tilde{y} = y - y_{0} } \\ \end{array} } \right] = \int_{{x_{1} - x_{0} }}^{{x_{2} - x_{0} }} {\int_{L}^{{\sqrt {R_{{cr}}^{2} - \tilde{x}^{2} } }} {\frac{{{\text{d}}\tilde{x}{\text{d}}\tilde{y}}}{{(\tilde{x}^{2} + \tilde{y}^{2} + z_{0}^{2} )^{{3/2}} }}} } \hfill \\ \quad = \int_{{x_{1} - x_{0} }}^{{x_{2} - x_{0} }} {\text{d}} \tilde{x}\frac{{\sqrt {R_{{cr}}^{2} - \tilde{x}^{2} } }}{{(\tilde{x}^{2} + z_{0}^{2} )\sqrt {R_{{cr}}^{2} + z_{0}^{2} } }} - \int_{{x_{1} - x_{0} }}^{{x_{2} - x_{0} }} {\text{d}} \tilde{x}\frac{L}{{(\tilde{x}^{2} + z_{0}^{2} )\sqrt {\tilde{x}^{2} + L^{2} + z_{0}^{2} } }}. \hfill \\ \end{gathered}$$

(32)

The second integral is already computed above

$$\begin{gathered} L\int\limits_{{x_{1} - x_{0} }}^{{x_{2} - x_{0} }} {\frac{{{\text{d}}u}}{{(u^{2} + c)\sqrt {u^{2} + p} }}} = L\frac{{\tan ^{{ - 1}} \left( {\frac{{u\sqrt {p - c} }}{{\sqrt c \sqrt {p + u^{2} } }}} \right)}}{{\sqrt c \sqrt {p - c} }}\Bigg|_{{x_{1} - x_{0} }}^{{x_{2} - x_{0} }} = \frac{{\tan ^{{ - 1}} \left( {\frac{{uL}}{{z_{0} \sqrt {L^{2} + z_{0}^{2} + u^{2} } }}} \right)}}{{z_{0} }}\Bigg|_{{x_{1} - x_{0} }}^{{x_{2} - x_{0} }} . \end{gathered}$$

(33)

The first integral can be computed as follows:

$$\begin{gathered} \int\limits_{{x_{1} - x_{0} }}^{{x_{2} - x_{0} }} {\text{d}} u\frac{{\sqrt {R_{{cr}}^{2} - u^{2} } }}{{(u^{2} + z_{0}^{2} )\sqrt {R_{{cr}}^{2} + z_{0}^{2} } }} \hfill \\ \quad = \frac{1}{{\sqrt {R_{{cr}}^{2} + z_{0}^{2} } }}\left.\left(\frac{{\sqrt {R_{{cr}}^{2} + z_{0}^{2} } }}{{z_{0} }}{\textrm{tan}}^{{{{ - 1}}}} \left( {\frac{{u\sqrt {R_{{cr}}^{2} + z_{0}^{2} } }}{{z_{0} \sqrt {R_{{cr}}^{2} - u^{2} } }}} \right) - {\textrm{tan}}^{{{{ - 1}}}} \left( {\frac{u}{{\sqrt {R_{{cr}}^{2} - u^{2} } }}} \right)\right)\right|_{{x_{1} - x_{0} }}^{{x_{2} - x_{0} }} . \hfill \\ \end{gathered}$$

(34)

Substituting Eqs. (34) and (33) in Eq. (32), we get Eq. (18) for the function $F(u,R_{cr},z_0,L)$.