(2.35) |

For the first term we note that by the assumption of isotropy of the symmetric part we have

By using partial integration on the second term we obtain

(2.37) |

With symmetric isotropy we get

(2.38) |

In summary we get the following set of equations: For and :

(2.39) |

The derivation does not need the full isotropy condition for its validity. Instead, we only need two reduction conditions:

A third reduction condition

(2.41) |

is not needed if the sixth moment is redefined as

(2.42) |

So by using the isotropy condition we can eliminate the moment . For the validity of Equation 2.36 the isotropy assumption is a sufficient, but not a necessary condition. A simple counter example is a function of the form . However, this form does not fulfill cylindrical symmetry. Examples which fulfill cylindrical symmetry can be constructed too, but one has to consider multivariate polynomials up to sixth order for a proof.

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R. Kosik: Numerical Challenges on the Road to NanoTCAD