EGSnrc
Collimated source normalization
EGSnrc normalizes collimated sources by fluence
When using the collimated source, EGSnrc normalizes the quantity of interest \(X\) by the fluence \(\Phi\) across the target shape and outputs \(X\!/\Phi\).
target shape \(S\)
source shape
\Omega
\,d\,
N
A
(sets the fluence scale)
A Monte Carlo result is typically tallied over the \(N\) independent emissions, as in:
\frac{X}{\Phi} = \frac{\Omega d^2}{N} \cdot \sum_{i=1}^N X_i
Consider a source emitting \(N\) particles towards a target shape \(S\) of total surface \(A\). The result \(X\) normalized per fluence \(\Phi\) on
\(\)the target shape is:
\frac{X}{\Phi} = \frac{X}{N/\Omega d^2}
d^2 \equiv {\int_A r^2 \text{d}\Omega}/{\int_A \text{d}\Omega}
The goal is to model particles as if emitted isotropically
target shape \(S\)
source shape
\Omega
\,d\,
N
\(d\) is defined
(set the fluence scale)
\frac{X}{\Phi} = \frac{\Omega d^2}{N} \cdot \sum_{i=1}^N X_i
A
Problem 1: it is difficult to calculate the solid angle
target shape \(S\)
source shape
\Omega
\,d\,
N
\(d\) is defined
(set the fluence scale)
\frac{X}{\Phi} = \frac{{\color{red}\Omega} d^2}{N} \cdot \sum_{i=1}^N X_i
Problem 1:
\(\)It is in general difficult to determine analytically
\(\)the solid angle \(\Omega\) of an arbitrary shape, which involves an integral over the target shape \(S\).
A
\Omega \equiv {\int_S r^2 \sin\!\theta \text{d}\theta \text{d}\phi}
Problem 2: sampling points isotropically on the target
source shape
\Omega
N
\delta a
\frac{X}{\Phi} = \frac{\Omega d^2}{N} \cdot \sum_{i=1}^N X_i
Problem 1:
\(\)It is in general difficult to determine analytically
\(\)the solid angle \(\Omega\) of an arbitrary shape, which involves an integral over the target shape \(S\).
Problem 2:
It is in general difficult to pick points \(r_i\)
on surface \(S\) according to a distribution \(\rho(\bm r_i)\) that \(\)is isotropic with respect to the source, i.e.,
{\color{red} \rho(\bm r_i) } \, \delta a \equiv \dfrac{\delta \Omega (\bm r_i)}{\Omega}
\bm r_i
\delta \Omega
A
target shape \(S\)
\Omega \equiv {\int_S r^2 \sin\!\theta \text{d}\theta \text{d}\phi}
source shape
\Omega
N
\delta a
\(d\) is defined
(set the fluence scale)
\frac{X}{\Phi} = \frac{\Omega d^2}{N} \cdot \sum_{i=1}^N X_i
\bm r_i
\delta \Omega
g (\bm r_i) \, {\color{red} \Rho (\bm r_i) }\, \delta a = \dfrac{\delta \Omega (\bm r_i)}{\Omega}
{\color{red} \rho(\bm r_i) } \, \delta a \equiv \dfrac{\delta \Omega (\bm r_i)}{\Omega}
\(\)Pick the points uniformly on the surface according to \( \Rho(\bm r_i) \equiv 1/A\), and apply weight \(g(\bm r_i)\) to compensate:
target shape \(S\)
A
Solve problem 2: pick points on target, uniformly !
Solve problem 2: pick points on target, uniformly !
target shape \(S\)
source shape
\Omega
N
\delta a
\(d\) is defined
(set the fluence scale)
\frac{X}{\Phi} = \frac{\Omega d^2}{N} \cdot \sum_{i=1}^N X_i
\bm r_i
\delta \Omega
g (\bm r_i) \, {\color{red} \Rho (\bm r_i) }\, \delta a = \dfrac{\delta \Omega (\bm r_i)}{\Omega}
\(\)Pick the points uniformly on the surface according to \( \Rho(\bm r_i) \equiv 1/A\), and apply weight \(g(\bm r_i)\) to compensate:
g(\bm r_i) = \dfrac {A}{\delta a} \cdot \dfrac{\delta \Omega (\bm r_i)}{\Omega}
A
{\color{red} \rho(\bm r_i) } \, \delta a \equiv \dfrac{\delta \Omega (\bm r_i)}{\Omega}
g(\bm r_i)
\left( \frac{A}{\delta a} \frac{\delta \Omega}{\Omega} \right)
This also solves problem 1: the solid angle is eliminated!
target shape \(S\)
source shape
\Omega
N
\delta a
\(d\) is defined
(set the fluence scale)
\frac{X}{\Phi} = \frac{{\color{red}\Omega} d^2}{N} \cdot \sum_{i=1}^N X_i
\bm r_i
\delta \Omega
g (\bm r_i) \, {\color{red} \Rho (\bm r_i) }\, \delta a = \dfrac{\delta \Omega (\bm r_i)}{\Omega}
\(\)Pick the points uniformly on the surface according to \( \Rho(\bm r_i) \equiv 1/A\), and apply weight \(g(\bm r_i)\) to compensate:
g(\bm r_i) = \dfrac {A}{\delta a} \cdot \dfrac{\delta \Omega (\bm r_i)}{\Omega}
A
{\color{red} \rho(\bm r_i) } \, \delta a \equiv \dfrac{\delta \Omega (\bm r_i)}{\Omega}
\left( \frac{A}{\delta a} \frac{\delta \Omega}{\color{red} \Omega} \right)
target shape \(S\)
source shape
\Omega
N
\delta a
\frac{X}{\Phi} = \,\,\, \frac{d^2}{N} \cdot \sum_{i=1}^N X_i
\bm r_i
\delta \Omega
A
\left( A \frac{\delta \Omega}{\delta a} \right)
This also solves problem 1: the solid angle is eliminated!
The differential solid angle to area ratio is geometrical
target shape \(S\)
source shape
\Omega
N
\delta a
\frac{X}{\Phi} = \,\,\, \frac{d^2}{N} \cdot \sum_{i=1}^N X_i
\bm r_i
\delta \Omega
A
\left( A {\color{red} \frac{\delta \Omega}{\delta a} }\right)
\hat \bm u_i
\hat \bm n_i
r_i^2 \delta \Omega = \delta a_\perp = \delta a | \hat \bm u_i \cdot \hat \bm n_i |
Geometrically:
Therefore:
{\color{red}\dfrac {\delta \Omega}{\delta a}} = \dfrac {| \hat \bm u_i \cdot \hat \bm n_i |}{r_i^2}
Sample uniformly, and apply weight to compensate
target shape \(S\)
source shape
\Omega
N
\delta a
\bm r_i
\delta \Omega
A
\hat \bm u_i
\hat \bm n_i
\frac{X}{\Phi} = \,\,\, \frac{d^2}{N} \cdot \sum_{i=1}^N X_i
\color{red} \ \left(\! A \,\frac {| \hat \bm u_i \ \cdot \ \hat \bm n_i |}{r_i^2} \right)
w_i \equiv w(\bm r_i) \equiv A \,\frac {| \hat \bm u_i \ \cdot \ \hat \bm n_i |}{r_i^2}
The collimated source samples points\(\)
uniformly on the target shape surface\(\) and applies the statistical weight \(w_i\)
to each source particle.
This is valid in general, even when source points are sampled in space within an extended source shape.
Define
EGSnrc
Recovering normalization by N
Divide by 4\(\pi d^2\) to recover normalization by particle
Given a collimated isotropic source simulation result, it is possible to recover
per particle normalization by dividing by 4\(\pi d^{\,2}\), where \(d\) is the distance specified in the collimated source definition.
\dfrac{X}{N_{4\pi}} =
what we want
Divide by 4\(\pi d^2\) to recover normalization by particle
\dfrac{X}{N_{4\pi}} = {\color{red}\left( \dfrac{X}{\Phi_{\Omega}} \right) } \left( \dfrac{\Phi_\Omega}{N_{4\pi}}\right)
= {\color{red}\left( \dfrac{X}{\Phi_{\Omega}} \right) } \left( \dfrac{N_\Omega}{\Omega d^2}\dfrac{1}{N_{4\pi}}\right)
= {\color{red}\left( \dfrac{X}{\Phi_{\Omega}} \right) } \left( \dfrac{1}{4\pi d^2}\right)
what we want
what EGSnrc outputs
“fix” it !
for an isotropic source:
\frac{N_\Omega}{N_{4\pi}} = \frac{\Omega}{4\pi}
Given a collimated isotropic source simulation result, it is possible to recover
per particle normalization by dividing by 4\(\pi d^{\,2}\), where \(d\) is the distance specified in the collimated source definition.
Solution: apply this factor to the result or, equivalently, specify \(d=1/\sqrt{4\pi}\).
EGSnrc
Calculating a solid angle
The EGSrc collimated source can calculate the solid angle
target shape \(S\)
origin
\Omega
N
\delta a
\bm r_i
\delta \Omega
\hat \bm u_i
\hat \bm n_i
\Omega = \int_S \text{d}\Omega = \int_S \frac{\delta\Omega}{\delta a} \text{d}a
Recall that \(w = A \dfrac{\delta\Omega}{\delta a}\)
A
The EGSrc collimated source can calculate the solid angle
\Omega = \int_S \text{d}\Omega = \int_S \frac{w(\bm r_i)}{A} \text{d}a
Recall that \(w = A \dfrac{\delta\Omega}{\delta a}\)
This integral is simply the average value of the weight over the surface, which is straightforward numerically.
target shape \(S\)
origin
\Omega
N
\delta a
\bm r_i
\delta \Omega
\hat \bm u_i
\hat \bm n_i
\Omega \approx \frac{1}{N} \sum_{i=1}^N w_i = \langle w \rangle
The solid angle is the average weight:
This is only valid if rays intersect the target once (target is isomorphic to a sphere centered on the origin).
A
EGSnrc
Recovering the probability density
The probability density normalized on the target surface
Given an isotropic source collimated towards a target shape, what is the probability density \(\rho_{\,\Omega}(\bm u_i)\) of particles, normalized over the target solid angle?
But \(w = A \dfrac{\delta\Omega}{\delta a}\) and \(\Omega = \langle w \rangle\), hence
Consider the fundamental relation that defines an isotropic source:
\rho_{\,\Omega}(\bm u_i) \, \delta a \equiv \dfrac{\delta \Omega (\bm r_i)}{\Omega}
target shape \(S\)
\Omega
\delta a
\bm r_i
\delta \Omega
\rho_{\,\Omega}(\bm u_i) = \dfrac{w_i}{\langle w \rangle A}
origin
N
This is only valid if rays intersect the target once (target is isomorphic to a sphere centered on the origin).
A
\bm u_i
The probability density normalized in \(4\pi\)
Given an isotropic source collimated towards a target shape, what is the probability density \(\rho_{4\pi}(\bm u_i)\) of particles, normalized isotropically?
But \(w = A \dfrac{\delta\Omega}{\delta a}\), hence the density:
Consider the fundamental relation that defines an isotropic source:
\rho_{4\pi}(\bm u_i) \, \delta a \equiv \dfrac{\delta \Omega (\bm r_i)}{4\pi}
target shape \(S\)
\Omega
\delta a
\bm r_i
\delta \Omega
\rho_{4\pi}(\bm u_i) = \dfrac{w_i}{4\pi A}
origin
N
This is only valid if rays intersect the target once (target is isomorphic to a sphere centered on the origin).
A
\bm u_i
EGSnrc collimated source normalization
By ftessier
EGSnrc collimated source normalization
Explains how EGSnrc normalized collimated sources by unit fluence on the target shape. Also explains how to renormalize the result per source particle.
