Let xij be component i of the deformed version of basis vector j. When there is no rotation or strain then xij = δij. Pure rotations produce no strain and the deformed versions of the basis vectors remain orthogonal. xij carries all of the strain information but also carries the irrelevant rotation information. Since the zone must remember its unstrained shape but not its unstrained orientation, we choose a coordinate system where the zone edge coordinates form a lower triangular matrix. This is the Gram Schmidt process. Here is how to efficiently extract the strain information.
The matrix xij may be expressed as sikukj, where s is the symmetric strain, and u is an orthogonal rotation.
(I use Kronecker deltas and the summation convention in place of sigma signs which don't work well in my word processor.)
xij = sikukj
( = α from the previous page)
α+α = δjkxmjxnk = δjk smquqjsnpupk = (δjkuqjupk)smqsnp = δpqsmqsnp
(δjkuqjupk) = δpq follows directly from the definition of orthogonality.
xij is available and we compute δpqsmqsnp which is symmetric, from it.
Smn = (δpqsmqsnp–δmn)/2 and take it as our approximation to the strain εnp = snp – δnp.
((x2–1)/2 ≅ x–1 when x ≅ 1. Everything commutes.)
The strain tensor εij causes stress σij = Cijklεkl. Cijkl is the elasticity tensor which is characteristic for the material and may vary from one point to another in an inhomogeneous material. Cijkl = Cjikl = Cijlk = Cklij. For isotropic material εij = μεij+ λδijδklεkl where μ is the shear modulus which quantifies resistance to shape change, and λ is the bulk modulus which quantifies resistance to changes in volume (bulk). δklεkl is the volume difference and λδij denotes a stress that pushes equally in all directions, i.e. ordinary pressure. For isotropic material the first equation also holds where Cijkl = μδijδkl + λδikδjl. For non viscous fluids μ = 0.