(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) |
(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:
(2.41) |
(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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