9.4.4 General form of the frictional stress tensor

In Cartesian coordinates, the stress tensor given in equation (9.54) can be written as the sum of two tensors, each defined on orthogonal subspaces

 

$$\tau^{ij} \tau^{ij}_{tran} + \tau^{ij}_{vert},$$ = (9.100)

where

 

$$\tau^{ij}_{tran} \rho \, A \, ( g^{ik} \, g^{jl} + g^{il} \, g^{jk} - g^{ij} \, g^{kl} ) \, e_{kl}$$ =

$$\rho \, A \, ( 2 \, e^{ij} - g^{ij} \, e^{k}_{k} )$$ = (9.101)

is the transverse stress tensor, defined over the transverse coordinates i,j,k,l = 1,2 and set to zero if one of the indices is 3.

$$\tau^{ij}_{vert} 2 \, \rho \, \kappa \, e^{i3} \, g^{j3}$$ = (9.102)

is the vertical stress tensor, where i=1,2 and $\tau_{vert}^{ji} = \tau_{vert}^{ij}$. In these expressions, g^ij are the components to the inverse metric tensor, which is the Kronecker delta in Cartesian coordinates. The transformation to curvilinear coordinates of interest here maintains orthogonality of the coordinates and transverse isotropy about the third direction. Such a coordinate transformation maintains the form of the stress tensor given here, where the metric tensor is now generally nontrivial, and the strain tensor is computed using covariant derivatives as described in the next section. It is notable that such a form for the stress tensor could have been ``guessed'' given the three constraints: (A) symmetry $\tau^{mn} = \tau^{nm}$, (B) separately trace-free in the two lateral directions and in the vertical direction; $\tau^{1}_{\; 1}+\tau^{2}_{\; 2}=0=\tau^{3}_{\; 3}$, and (C) laterally isotropic. The previous analysis of the viscosity tensor, although more tedious than starting from these three assumptions, exposed more of the underlying properties of the stress tensor.