# Monoclinic anisotropy

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Monoclinic anisotropy has a single plane of symmetry, without rotational symmetry about the normal to that plane. In the sedimentary crust, the plane is usually the bedding plane, and there are two or more non-orthogonal sets of fractures oriented perpendicular to this plane. Such fractures are caused by the orthogonality of the stress tensor, in places where the tectonic history is not simple, but the maximum principal stress has always been vertical. This situation is common in shale resource areas.

Monoclinic joints in the Utica Shale, Ohio

In seismics, the corresponding elastic stiffness matrix (symmetric) has 12 independent components:

${\displaystyle \{c_{\alpha \beta }\}={\begin{pmatrix}c_{11}&c_{12}&c_{13}&0&0&c_{16}\\c_{12}&c_{22}&c_{23}&0&0&c_{26}\\c_{13}&c_{23}&c_{33}&0&0&c_{36}\\0&0&0&c_{44}&0&0\\0&0&0&0&c_{55}&0\\c_{16}&c_{26}&c_{36}&0&0&c_{66}\\\end{pmatrix}}}$

The indices in this matrix refer to components of the principal coordinate system. Normally, this is not known in a field context, and (unlike the case of orthorhombic anisotropy) it is not easy to determine from the data.

Assuming that the one symmetry plane is horizontal, and that the survey coordinate system has one axis vertical, it is common that the horizontal axes of the survey coordinate system are not aligned with those of the principal coordinate system. In this case, the stiffness matrix (referred to the survey coordinate system) has the same form as above, but with ${\displaystyle c_{45}=c_{54}}$ non-zero, so that effectively there are 13 elements to determine. Dealing with such systems is beyond the limit of feasibility in 2020; the same is true for all lower symmetries.[1]