# Notation

From here on it is assumed that the tracking takes place in a three-dimensional Cartesian coordinate system. In vector notation, the position and direction of motion are defined by two vectors: $\boldsymbol{r} = x\boldsymbol{\hat i} + y\boldsymbol{\hat j} + z\boldsymbol{\hat k}$


and ${\boldsymbol{\hat{\Omega}}} = u\boldsymbol{\hat i} + v\boldsymbol{\hat j} + w\boldsymbol{\hat k}$


where $\boldsymbol{\hat i}$, $\boldsymbol{\hat j}$ and $\boldsymbol{\hat k}$ are the unit vectors defining the three-dimensional Cartesian coordinate system. Direction vector ${\boldsymbol{\hat{\Omega}}}$ is normalized to unity: ${\boldsymbol{\hat{\Omega}}}\cdot{\boldsymbol{\hat{\Omega}}} = 1$


or $u^2 + v^2 + w^2 = 1$


Coefficients $u$, $v$ and $w$ are the direction cosines, i.e. the cosines of the angle that vector ${\boldsymbol{\hat{\Omega}}}$ forms with the positive x-, y- and z-axis, respectively.

# Surfaces

Monte Carlo transport codes are most typically based on the constructive solid geometry (CSG) type, in which the geometry is composed of homogeneous material cells, defined using combinations of elementary and derived surface types.

The most elemental building block is the surface, described using algebraic equations, typically of the quadratic type. The action that puts an arbitrary position $\boldsymbol{r}$ on one or the other side of a surface is based on a simple test carried out by substituting the coordinates into the surface equation: $S(\boldsymbol{r}) = S(x,y,z) \left\{ \begin{array}{ll} < 0 & \text{if the point is inside the surface} \\ = 0 & \text{if the point on the surface} \\ > 0 & \text{if the point is outside the surface} \\ \end{array} \right .$


This surface test also fixes the concepts of inside and outside for each surface type, which is important when forming the cells from the surface combinations.

The general quadratic surface can be written in parametric form as: $S(x, y, z) = Ax^2 + By^2 + Cz^2 + Dxy + Eyz + Fzx + Gx + Hy + Iz + J$


where $A, B, C, D, E, F, G, H, I$ and $J$ are constants. Common examples of quadratic surfaces obtained from the parametrized quadratic equation include the plane perpendicular to x-axis at $x_0$: $S(x, y, z) = x - x_0$


sphere centered at $(x_0,y_0,z_0)$ with radius $r$: $S(x, y, z) = (x - x_0)^2 + (y - y_0)^2 - (z - z_0)^2 - r^2$


and straight infinite cylinder parallel to z-axis centered at $(x_0, y_0)$ with radius $r$: $S(x, y) = (x - x_0)^2 + (y - y_0)^2 - r^2$


There are also non-quadratic surfaces, such as the torus, for which the surface test cannot be performed by substituting the coordinates into a single equation.

Monte Carlo codes often provide additional derived surface types, which are formed by combinations of elementary surfaces. For example, a cuboid with boundaries $[x_1, x_2]$, $[y_1, y_2]$, $[z_1, z_2]$ consists of six planes perpendicular to the coordinate axes, and surface test works by testing the position of the point with respect to all of them. Other derived surface types convenient for reactor modeling include truncated cylinders, and square and hexagonal prisms.

Surface equations are also used for determining the distance to the nearest material boundary in the direction of motion. The points where the particle path intersects the surface are obtained by solving the surface equation with condition: $S(\boldsymbol{r} + \delta{\boldsymbol{\hat{\Omega}}}) = S(x + \delta u, y + \delta v, z + \delta w) = 0$


i.e. by setting a point located at distance $\delta$ from position $\boldsymbol{r}$ in the direction of motion ${\boldsymbol{\hat{\Omega}}}$ on the surface, and solving for $\delta$. When the equation has multiple solutions, the nearest point corresponds to the smallest positive value of $\delta$. If all solutions are negative or no solution exists, the surface is away from the line-of-sight.

As an example, the distance to a plane perpendicular to x-axis at $x_0$ is obtained from $x + \delta u - x_0 = 0 \Longleftrightarrow \delta = \frac{x_0 - x}{u}$


and the distance to a straight infinite cylinder parallel to z-axis centered at $(x_0, y_0)$ with radius $r$ from: $(x + \delta u - x_0)^2 + (y + \delta v - y_0)^2 - r^2 = 0 \Longleftrightarrow \delta = \frac{-L \pm \sqrt{L^2 - KM}}{K}$


where $K = u^2 + v^2$, $L = u(x - x_0) + v(y - y_0)$ and $M = (x - x_0)^2 + (y - y_0)^2 - r^2$.