- So for the remainder of the talk, we're going to discuss a theorem that describes the conductivity of Archie rocks. And the theorem is that the bulk conductivity of Archie rocks comprises three factors, or is a product of three factors. The factors are the conductivity of the brine in the rock, that's what does the conducting, the amount of brine in the rock, and that's either equal to the porosity or equal to the fractional brine law, you can see SW. And the distribution, the geometrical distribution of the brine in the rock, which we give a name as a parameter, it's E sub-zero, or, in the case of hydrocarbon-bearing rock, E sub-T. So now we'll prove the theorem. Factor one, the brine conductivity. You'll recall from your high school geometry that there are certain things that can't be proved in geometry like for example, the shortest distance between two points is a straight line. Euclid doesn't prove that, that's taken as self-evident. We take, in this theory, the axiom that the bulk conductivity of the Archie rock is proportionate to the conductivity of the brine. What does that mean? Well it means if you double the conductivity of the brine then you double the conductivity of the rock and so on, that's asserted as self-evident, we don't try to prove it. The fractional volume of the brine is the second factor, and we begin with the Lima. The Lima is a small theorem that's proved on the way to a larger theorem, and the Lima is in Archie rocks, the conductivity is proportional to the fractional volume of the brine, that is the porosity to the first power. The point of confusion here is that pour geometry and porosity are coupled in a rock. You can't change the porosity of the rock without simultaneously changing the pour geometry, if digenesis is the mechanism of change, so to think clearly about what's happening, you have to decouple the pour geometry from the pour space, from the amount of porosity. This is difficult to do in a physical experiment, but we can use thought experiments to do it. So here's two demonstrations of thought experiments that prove this point, and I want you to pay close attention. This is more than just a cartoon, although it looks like a cartoon. Here's a representation of a core plug with a porosity that's all collected into a single, sinuous channel. So it has some bulk conductivity, it's proportional to the brine conductivity, and it's also proportional to the conductivity in the single channel. Now if I hold the pour geometry constant, so what do I mean by that? It means that I take that pour, and I just duplicate it, so the geometry hasn't changed, but the amount of porosity is now changed by a factor of two. So the bulk conductivity of this new core plug is twice the conductivity of the original core plug. It will not matter if the channel has complications in it, and it will not matter if there are more than one or two channels, there could be any number of channels, as long as the pour geometry doesn't change, then the bulk conductivity of this core plug is going to be proportional to the conductivity of the brine, but also the porosity to the first power. Now this is a little cartoony demonstration, so one that's a little less cartoony, because I want you to think of two quarter cylinders, one has zero percent porosity and the other has 30 percent porosity, and now imagine that you duplicate these cylinders in every respect, including the pour geometry, four times and you put the cylinders together, in these five ways that vary from zero percent porosity to 30 percent porosity. So I think you can see that, as the porosity is varying linearly, as the high-porosity quarter cylinder is increased, the conductivity is varying linearly as well. And so for constant pour geometry, the conductivity scales directly with porosity, so that constitutes our proof of the second proposition in the theorem. And by the way, part of the object here is to engender discussion. If you were going to find something wrong with this theorem, probably it's in this step. The third factor is brine geometrical distribution. Consider a core plug with conductivity given by sigma naught. It has an intermediate conductivity, well, intermediate between what? If you imagine taking all of the porosity out of this core plug and putting it into a cylinder, taking the rock-forming minerals and making the walls of a tube, then you can imagine this situation on the left, where you have a minimum conductivity because we interrupt the connection of the cylinder from one side to the other, so it has no conductivity. On the other hand, if we make a connection of the cylinder, then it will have a maximum conductivity, and we know what that conductivity will be, it'll be the product of the brine conductivity and the porosity of the rock. Also note that the porosity is the fractional area of the end of the cylinder, with respect to the entire area of the cylinder. So we have that the observed conductivity, sigma naught, falls between a minimum and a maximum, and so now we make the proposition that the bulk conductivity of the rock is proportional to these three factors, and we make it an equality, the brine conductivity, the amount of brine, and the geometrical distribution of the brine. And we also know, from the arguments that we've just made, that the values that the geometrical factor can take are between zero and one. And what is the geometrical factor? Well, if you give me a core plug, I can measure all three of the quantities in the equation that I can solve for the geometrical factors, so for a core plug, I know how to find the geometrical factor, for an ensemble of core plugs, I can find the geometrical factor for each of the core plugs.