Reflection seismic example

Assume for the sake of argument that all three Thomsen parameters, $\gamma$, $\epsilon$, and $\delta$, have been determined for a given reservoir and that the reservoir exhibits VTI characteristics. If the reservoir does not exhibit VTI symmetry, then I might need to consider HTI (horizontal transverse isotropy) or some still more complicated type of anisotropy. But, for VTI, we need to know something about the variety of behaviors that are possible in the presence of fractures. Equations (4) and (5) show the results expected if the fractures are vertical and randomly oriented. But there are obviously other possibilities as well, and to have a better chance of making a valid interpretation of the observed behavior, we need to know more about the range of possibilities for the Thomsen parameters. I will not attempt to be exhaustive here, but just present one other result that can clearly be distinguished by such data.

Consider the case of horizontal fractures. Then, the axis of symmetry is vertical, and so the reservoir would exhibit VTI symmetry again, just as in the case of vertical fractures randomly oriented. But the resulting expressions for the Thomsen parameters in terms of the Sayers and Kachanov (1991) parameters are quite different.

I find  

$$\gamma = \frac{c_{66}-c_{44}}{2c_{44}} = \eta_2\rho G,$$ (19)

and  

$$\epsilon = \frac{c_{11} - c_{33}}{2c_{33}} = [(1+\nu)\eta_1 ... ..._2]\rho\frac{E}{(1-\nu^2)} \simeq \eta_2\rho\frac{2G}{1-\nu}.$$ (20)

The background shear modulus is G, and corresponding Poisson ratio is $\nu$. Again, I find that $\delta = \epsilon$ to the lowest order in the crack density parameter. Also, I have neglected the term in $\eta_1$ in the final expression as this is on the order $1\%$ of the term retained. So I find that the magnitude of the coefficients in this case differs by a factor of 2 from those of randomly oriented vertical fractures as in Eqs. (4) and (5). But more importantly, the sign of these expressions is opposite that of this other case.

Equations (19) and (20) could have been found in a very simple way from the results of previous sections by using the following argument: We know that an isotropic distribution of fractures is represented in the Sayers and Kachanov (1991) formalism by the correction matrix (1) and, furthermore, that (1) is also the weighted sum of (2) and (3) -- specifically, $2/3 \times$ Eq. (3) plus $1/3 \times$ Eq. (2). So the similarly weighted averages of the individual Thomsen parameters for these two cases must add up to zero (since Thomsen's parameters vanish for isotropic media). In fact, this is exactly what we have found to be true. We could have used this fact to provide a quicker (and much more elegant) derivation of (19) and (20) than the method I actually used. Of course, the utility of this type of argument is limited to the linear contributions of crack density to the Thomsen parameters that I have been considering here.

In both examples, the Thomsen parameter measurements may be used to estimate the magnitude of the $\eta_2\rho$ product assuming the background shear modulus G and the background Poisson ratio $\nu$are known, or can be estimated. But, horizontally fractured systems can also be easily distinguished from vertically fractured systems, since the sign of the constants is opposite in these two scenarios.

It would be helpful for interpretation purposes to enumerate other related scenarios that could be distinguished from these two by using the anisotropy parameter data. I will leave such problems, especially those involving azimuthal dependence (and therefore not VTI), to future work. There is no fundamental problem with computing the relations between the Thomsen parameters and the Sayers and Kachanov parameters for arbitrarily complicated choices of fractured reservoir scenarios. All can be studied, but making good choices about which of the necessarily limited number of scenarios time permits us to consider are also the most fruitful ones will be one of the key steps in the process.