McXtrace - An X-ray ray-trace simulation package

A.1  Transformation of random numbers

In order to perform the Monte Carlo choices, one needs to be able to pick a random number from a given distribution. However, most random number generators only give uniform distributions over a certain interval. We thus need to be able to transform between probability distributions, and we here give a short explanation on how to do this.

Assume that we pick a random number, \(x\), from a distribution \(\phi (x)\). We are now interested in the shape of the distribution, \(\Psi (y)\), of the transformed \(y=f(x)\), assuming \(f(x)\) is monotonous. All random numbers lying in the interval \([x; x+dx]\) are transformed to lie within the interval \([y; y+f'(x)dx]\), so the resulting distribution must be \(\Psi (y) = \phi (x) / f'(x)\).

If the random number generator selects numbers uniformly in the interval \([0; 1]\), we have \(\phi (x) = 1\) (inside the interval; zero outside), and we reach \begin {equation} \Psi (y) = \frac {1}{f'(x)} = \frac {d}{dy} f^{-1}(y) . \end {equation} By indefinite integration we reach \begin {equation} \label {e:randtrans} \int \Psi (y) dy = f^{-1}(y) = x , \end {equation} which is the essential formula for random number transformation, since we in general know \(\Psi (y)\) and like to determine the relation \(y=f(x)\). Let us illustrate with a few examples of transformations relevant for the McXtrace components.

The circle For finding a random point within the circle of radius \(R\), one would like to choose the polar angle, \(\phi \), from a uniform distribution in \([0; 2\pi ]\), giving \(\Psi _\phi = 1/(2\pi )\). and the radius from the (normalised) distribution \(\Psi _r=2r/R^2\).

For the radial part, eq. (A.2) becomes \(y/(2 \pi ) = x\), whence \(\phi \) is found simply by multiplying a random number (\(x\)) with \(2\pi \).

For the radial part, the left side of eq. (A.2), gives \(\int \Psi (r) dr = \int 2 r/R^2 dr = r^2/R^2\), which from (A.2) should equal \(x\). Hence we reach the wanted transformation \(r = R\sqrt {x}\).

The sphere For finding a random point on the surface of the unit sphere, we need to determine the two angles, \((\theta , \phi )\).

\(\Psi _\phi \) is chosen from a uniform distribution in \([0; 2\pi ]\), giving \(\phi = 2\pi x\) as for the circle.

The probability distribution of \(\theta \) should be \(\Psi _\theta =\sin (\theta )\) (for \(\theta \in [0; \pi ]\)), whence by eq. (A.2) \(\theta =\cos ^{-1}(x)\).

Exponential decay

In a simple 2-state model a molecule may be said to relax into its ground state at a time which is exponetially distributed after the initial excitation at \(t=0\). We thus want to pick a relaxation time from the normalised distribution \(\Psi (t) = \exp (-t/\tau ) / \tau \). Use of Eq. (A.2) gives \(x = 1 - \exp (-t/\tau )\). For convenience we now use the random variable \(x_1 = 1-x\) (with the same distributions as \(x\)), giving the simple expression \(t = - \tau \ln (x_1)\).

Normal distributions The important normal distribution can not be reached as a simple transformation of a uniform distribution. In stead, we rely on a specific algorithm for selecting random numbers with this distribution.