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