Question: How can one go about showing an empirically-derived parameter, which was originally developed as an isotropic parameter (i.e., a scalar), to be a tensor quantity based on other relationships that relate it to another known 2nd-rank tensor parameter?
Quick Background: The parameter in question is called the $\beta$-factor. The parameter is used in the following equation,
$$\frac{dp}{dx}=\frac{\mu}{k}v+\beta \rho v^2$$
which is an equation that models fluid flow through a porous medium. This equation assumes the flow to be "one-dimensional" or to be isotropic in permeability $(k)$ and $\beta$, thus both parameters are typically taken as a scalar quantity when using the equation above.
However, for "three-dimensional" flow through anisotropic porous media, permeability $(k)$ is known to be a 2nd-rank tensor, realized using a 3 by 3 matrix being both symmetric and positive definite. Because both $k$ and $\beta$ are parameters that characterize a porous medium's ability to transmit fluids, I believe $\beta$ too should be a 2nd-rank tensor quantity, yet I do not have a mathematical theory nor do I have any experimental data to support my hunch. What I would like to do is use "one-dimensional" relationships developed for $k$ and $\beta$ and see if I can develop a tensorial relationship for $\beta$ without breaking the rules of matrix algebra (of which I am not well versed).
What I have tried: To make a connection between the permeability tensor and the $\beta$-factor, I made use of the one-dimensional relationship known as the Kozeny-Carman equation (*nomenclature of variables used and their tensor rank are listed at end of post), written in the following form: $$\tag{1} k=\frac{\phi}{c \tau^2 S_p^2}; \ \ \tau=L_e/L$$
and the experimentally derived one-dimensional relationship between permeability and $\beta$ found by Liu et al., which is given as,
$$\tag{2} \beta=\frac{\tau}{k \phi}$$
Rearranging Eq(1) to solve for tortuosity ($\tau$), $$\tag{3} \tau=\frac{\phi^{1/2}}{c^{1/2}k^{1/2}S_p}$$ and then substituting this equation into Eq(2), $$\tag{4} \beta=\frac{1}{c^{1/2}\phi^{1/2}k^{1/2}kS_p}$$ Define a scalar constant $C=1/c^{1/2}\phi^{1/2}S_p$ for simplification ($c, \phi$, and $S_p$ are all scalar quantities), $$\tag{5} \beta=\frac{C}{k^{1/2}k}$$ To write Eq(5) in tensorial form, denote the inverse of the permeability tensor as: $\bar k^{-1}=\bar K$ and denote the square root of the inverse of the permeability tensor as: $\bar K^{1/2}=\bar D$ where $\bar D \cdot \bar D = \bar K$.
And now write Eq(5), taking $C=1$, as either: $$\tag{6A} \bar \beta= \bar D \cdot \bar K$$ or $$\tag{6B} \bar \beta = \bar K \cdot \bar D$$ since multiplication of matrices is not commutative.
To apply numerical values and see how the above derivation might work, I assigned the 3 by 3 matrix of the permeability tensor with the elements: $$\tag{a} \bar k=\begin{bmatrix}200 & 50 & 40\\50 & 150 & 30\\40 & 30 & 100\end{bmatrix}$$
As a check to see if the permeability matrix is symmetrical and positive definite, the eigenvalues of the permeability matrix are (rounding): 82.71, 119.58, and 247.71, which are all positive eigenvalues, thus I think I can conclude the permeability tensor is symmetric and positive definite.
The inverse of the permeability matrix (Eqn(a)) is, $$\tag{b} \bar K=\begin{bmatrix}0.0058 & -0.0016 & -0.0018\\-0.0016 & 0.0075 & -0.0016\\-0.0018 & -0.0016 & 0.0112\end{bmatrix}$$
And the square root of the inverse permeability matrix is, $$\tag{c} \bar D=\begin{bmatrix}0.0744 & -0.0103 & -0.0108\\-0.0103 & 0.0856 & -0.0092\\-0.0108 & -0.0092 & 0.1050\end{bmatrix}$$
Performing the calculation of Eq(6A), I obtain: $$\tag{d} \bar \beta=\begin{bmatrix}0.0005 & -0.0002 & -0.0002\\-0.0002 & 0.0007 & -0.0002\\-0.0002 & -0.0002 & 0.0012\end{bmatrix}$$
which I note is symmetrical. The eigenvalues of Eqn(d) are: 0.0003, 0.0008, 0.0013, thus if what I have done up to this point is correct, I believe I can conclude that $\bar \beta$ is symmetric and positive definite 2nd-rank tensor.
Further, performing the calculation of Eq(6B), I obtain: $$\tag{e} \bar \beta=\begin{bmatrix}0.0005 & -0.0002 & -0.0002\\-0.0002 & 0.0007 & -0.0002\\-0.0002 & -0.0002 & 0.0012\end{bmatrix}$$
which is equivalent to Eq(d). Thus, for this instance the multiplication of the matrices IS commutative, i.e., $\bar D \cdot \bar K = \bar K \cdot \bar D$.
Request: Perhaps my question isn't clearly defined as stated at the beginning. I'm hoping to get some feedback with regards to the above (e.g., whether it is wrong or right), and perhaps some insight on any sort of identities/axioms/laws for matrix algebra that I am trying to apply here that may (or may not) support the derivation.
$k, \bar k$ = permeability, permeability tensor (2nd-rank, symmetrical, positive definite); $\phi$ = porosity (scalar, 0-rank); $c$ = empirical numerical factor (scalar, 0-rank); $\tau$ = tortuosity (symmetrical, 2nd-rank); $L_e, L$ = total length fluid travels through the tortuous pathways in the porous medium, external/bulk length of the porous medium (scalar, 0-rank); $S_p$ = wetted surface area of the pores per unit pore volume of the porous medium