Reduced Quadrature SetΒΆ
In 1D or 2D problems, the resulting angular flux will not depend on the symmetric dimensions. In a 1D slab geometry, we will have
In a 2D slab geometry we similarly have
For the transport operator, using a full 3D quadrature set is therefore significantly more expensive than necessary: the same \(\Omega_x\) will result in the same resulting \(T^{-1}\) application. Therefore, we can use a reduced quadrature set where we only have one quadrature point per unique \(\Omega_x\) (and \(\Omega_y\) in 2D).
This reduced quadrature set does have to provide a method to generate the \(w_k Y_{lm}(\Omega_k)\) terms required for the moment calculation.
The normal quadrature will have the following moment calculation:
Let \(\{\Omega_k\}\) be the full, 3D quadrature set. We define a set of orbits \(\mathcal{K}\) which groups the quadrature points by their unique \(\Omega_x\) (and \(\Omega_y\) in 2D) values. Let \(\kappa ` index an orbit, with :math:`f\) indexing a orbit member. Let :math`tilde{Omega}_{kappa}` be a single representative from the orbit. The sum can be written as
However, by the definition of an orbit, we have that \(\psi(\Omega_{\kappa f}) = \psi(\tilde{\Omega}_{\kappa})\) for all \(f\). Therefore, we can write the sum as
Thus, if we modify the quadrature set to use \(\tilde{\Omega}_{\kappa}\) as the quadrature points and \(\tilde{A}_{lm, \kappa}\) as the moment calculation coefficients, we will get the same result as the full quadrature set.
The reverse expansion can be the same; that maps from the moments to the angular flux and is not affected by the reduction of the number of angles.