9.3.4 Transverse isotropy

The presence of gravity provides a symmetry breaking from three-dimensional isotropy down to transverse isotropy about the local vertical direction $\hat{z}$. It is important that the stress tensor also respect this symmetry, which in turn provides constraints on the form of the viscosity tensor.

In order to understand what transverse, or axial, isotropy imposes on the viscosity tensor, it is necessary to recall that a fourth order tensor transforms under a change of coordinates in the following manner

$$C^{\overline{a} \, \overline{b} \, \overline{c} \, \overline{d} } \Lambda^{\overline{a}}_{\; a} \, \Lambda^{\overline{b}}_{\; b} \, \Lambda^{\overline{c}}_{\; c} \, \Lambda^{\overline{d}}_{\; d} \, C^{abcd}.$$ = (9.29)

Transverse isotropy means two things. First, the physical system remains invariant under arbitrary rotations about the $\hat{e}_{3}$ direction, where $\hat{e}_{3} = \hat{z}$ is the vertical direction. Second, the physical system remains invariant under the transformation $z \rightarrow -z$, and $x \rightarrow y, y \rightarrow x$, which is a transformation between two right handed coordinate systems, with the vertical pointing up and down, respectively. The transformation matrix for the rotational symmetry takes the form of a rotation about the vertical

 

$$\Lambda^{\overline{a}}_{\; a} \left( \begin{array}{ccc} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{array}\right),$$ = (9.30)

and the transformation matrix between right handed systems takes the form

 

$$\Lambda^{\overline{a}}_{\; a} \left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right).$$ = (9.31)

Imposing the constraint that

$$C^{\overline{a} \, \overline{b} \, \overline{c} \, \overline{d} } \equiv$$ C^abcd

$$\Lambda^{\overline{a}}_{\; a} \, \Lambda^{\overline{b}}_{\; b} \, \Lambda^{\overline{c}}_{\; c} \, \Lambda^{\overline{d}}_{\; d} \, C^{abcd},$$ = (9.32)

where $\Lambda^{\overline{a}}_{\; a}$ is one of the given transformation matrices, provides for relations between the 21 remaining elements of C^abcd. To determine the relations between the elements of C^abcd requires no more than careful enumeration of the possibilities. For example, with a rotation angle of $\pi/2$ about $\hat{z}$, rotational symmetry implies

$$C^{\overline{1} \, \overline{2} \, \overline{2} \, \overline{2} } \equiv C^{1222} = -C^{2111}.$$ (9.33)

However, the transformation between the two right handed coordinate systems implies

C¹²²² = C²¹¹¹. (9.34)

These two results are satisfied only if

C¹²²² = C²¹¹¹ = 0. (9.35)

For a rotation of $\pi/4$, isotropy implies

$$C^{1111} = (C^{1111} + C^{1122} + 2 \, C^{1212})/2,$$ (9.36)

or

C¹²¹² = (C¹¹¹¹ - C¹¹²²)/2. (9.37)

Continuing in this manner implies that the only nonzero elements of C^ijmn are C^ijij, where $i \ne j$, and C^iijj. The relation between these nonzero elements can be written

$$C^{iijj} \equiv c^{ij} \left( \begin{array}{ccc} c^{11} & c^{12} & c^{13} \\ c^{12} & c^{11} & c^{13} \\ c^{13} & c^{13} & c^{33} \end{array}\right),$$ = (9.38)

and

C¹²¹² = (c¹¹ - c¹²)/2 (9.39)

C²³²³ = C¹³¹³ = c⁴⁴/2, (9.40)

which brings to five the total number of independent degrees of freedom.