THE SCHOENBERG-MUIR CALCULUS

We now apply these steps to the the layer-averaging techniques of Backus (1962), Helbig and Schoenberg (1987) and Schoenberg and Muir (1989). These methods consider the effect of a static stress on a welded stack of infinite elastic layers. Although derived for the static case, the results are also applicable to the case of propagating waves so long as the wavelength is much longer than the scale of the layering.

Step 1: Hooke's law (equation (5)) generalizes to  

$$\left[ \matrix { \sigma_1 \cr \sigma_2 \cr \sigma_3 \cr \sig... ...ilon_3 \cr \epsilon_4 \cr \epsilon_5 \cr \epsilon_6 } \right] ,$$ (13)

where

$$\left[ \matrix { \sigma_1 \cr \sigma_2 \cr \sigma_3 \cr \sig... ...ilon_{23} \cr 2 \epsilon_{31} \cr 2 \epsilon_{12} } \right] \ .$$

Note that equation (13) is not exactly the three-dimensional equivalent of equation (5). In equation (5) the ``strain'' $\Delta x$is measured in units of displacement, whereas in equation (13) the strain is a dimensionless proportionality of the form $\Delta \mbox{\rm length} / \mbox{\rm length}$.Similarly, in equation (5) the left hand side is in units of force, whereas in equation (13) the left hand side is in units of force per unit area. We will have to take these length normalizations into account in step 2c.

Step 2a: The layer parameters are the stiffness matrix $\bf C$, the layer thickness h, and the layer density $\rho$.

Step 2b: Instead of ``springs in series'' we have a stack of infinite flat layers with welded interfaces. (We will assume the layering is normal to the z axis.) Instead of a constant scalar spring tension we have that the stress normal to the layering must be the same in all layers. Thus equation (6) becomes  

$$\sigma_{3_{\,i}} \ \equiv\ \sigma_3 \ ,\ \ \ \ \sigma_{4_{\,... ...iv\ \sigma_4 \ ,\ \ \ \ \sigma_{5_{\,i}} \ \equiv\ \sigma_5 \ .$$ (14)

Because the layers are infinite and welded together and so cannot accommodate any differential horizontal expansion, we also find that the strain tangential to the layering must be the same in every layer:  

$$\epsilon_{1_{\,i}} \ \equiv\ \epsilon_1 \ ,\ \ \ \ \epsilon_... ...psilon_2 \ ,\ \ \ \ \epsilon_{6_{\,i}} \ \equiv\ \epsilon_6 \ .$$ (15)

Step 2c: The first two additive parameters are obvious ones. The thickness h of each layer adds, as does the areal density $h \, \rho$; thus the first two layer-group elements are trivially h and $h \, \rho$.

To obtain vertical displacements, which sum through the stack, we must multiply the vertical components of strain by the layer thickness h. Thus equation (7) becomes  

$$h_{\mbox{\rm\scriptsize total}} \, \epsilon_{3_{\,\mbox{\rm\... ...ox{\rm\scriptsize total}}} = \sum h_i \; \epsilon_{5_{\,i}} \ ,$$ (16)

where h_i is layer thickness and $h_{\mbox{\rm\scriptsize total}} = \sum h_i$is the total thickness of the stack of layers.

There is one more additive equation to find. Think of squeezing our layercake horizontally in a vise; all layers are forced to undergo the same horizontal distortion. The various layers will exert differing forces on the vise blades; to find the total traction, sum over the horizontal forces exerted by each of the layers. Stress is force per area, so force is area times stress. The (infinite) horizontal dimension tangential to the vice blades is the same for all layers and so can be factored out, leaving only the vertical thickness h multiplied by the horizontal components of stress. We thus have  

$$h_{\mbox{\rm\scriptsize total}} \, \sigma_{1_{\,\mbox{\rm\sc... ...mbox{\rm\scriptsize total}}} = \sum h_i \; \sigma_{6_{\,i}} \ .$$ (17)

Step 3: We can now rearrange equation (13) to segregate the additive and constant elastic terms, obtaining  

$$\left[ \matrix { \sigma_1 \cr \sigma_2 \cr \sigma_6 \cr \cr ... ..._6 \cr \cr \epsilon_3 \cr \epsilon_4 \cr \epsilon_5 } \right] .$$ (18)

Written explicitly as a block matrix equation in terms of the additive and constant parameters,

$$\sigma_{\mbox{\rm\scriptsize add}} = h \left[ \matrix {\sig... ...t[ \matrix {\epsilon_1 \cr \epsilon_2 \cr \epsilon_6 } \right],$$ (19)

equation (18) becomes  

$$\left[ \matrix { {1 \over h} \sigma_{\mbox{\rm\scriptsize ad... ...cr {1 \over h} \epsilon_{\mbox{\rm\scriptsize add}} } \right] .$$ (20)

After a little matrix algebra we finally obtain Hooke's law in the desired form, the elastic equivalent of equation (11):  

$$\left[ \matrix { \sigma_{\mbox{\rm\scriptsize add}} \cr \cr ... ...st}} \cr \cr \epsilon_{\mbox{\rm\scriptsize const}} } \right] .$$ (21)

Equation (21) is really nothing more than another form of Hooke's law, one better suited to the geometry of layered media.

Step 4: The somewhat complicated matrix coefficient on the constant vector parameter in equation (21),  

$$h \; \left[ \matrix { {\Bigl({\mbox{\bf C}}_{TN} {\mbox{\bf ... ...mbox{\bf C}}_{TN} {\mbox{\bf C}}_{NN}^{-1}\Bigr)}^T } \right] ,$$ (22)

defines the Schoenberg-Muir layer-group representation of the elastic stiffness matrix $\bf C$.Note this 6 by 6 block matrix has 21 independent elements, the same number as $\bf C$;the complete layer group also includes two more independent parameters, h and $h \, \rho$, for a total of 23 group elements.

Does the Schoenberg-Muir system form an Abelian group? Yes. Association and commutativity are obviously satisfied. The inverse can be formed simply by changing the sign of h, and the identity element is the layer of zero thickness. The group is closed in the layer-group domain, although it is possible to create representations in the layer-group domain by layer subtraction that correspond to physically disallowed elastic media (Schoenberg and Muir, 1989). The group concept of subtraction coupled with the Schoenberg-Muir calculus does provide a useful theoretical framework for decomposing observed elastic properties into geophysically meaningful components; for example observed orthorhombic anisotropy can be decomposed into a transversely isotropic background with fractures (Hood, 1991) and (Hood and Schoenberg, 1992).