Welcome back. In this lesson, we're going to develop the equations to describe heterogeneous nucleation, much like we did in the development of the equations for homogenous nucleation. Remember in the case of homogeneous nucleation, we said that there were in fact two competing terms. One term was a volume term, and the other was a surface area term. When we describe the process of heterogeneous nucleation, we're going to see that those type of terms reappear, but with a bit more complexity. I've written the equation for the heterogeneous nucleation process in the screen above. And what it tells me is I have a volume term, which is related to how much undercooling I have, and the volume of the particular structure that has nucleated. So, in this case, we're talking about the volume of the spherical cap. Now when we look at the next term, which is the term in red, that term has to do with the surface area of the spherical cap, where we have the interface between the solid phase and the liquid phase. Then we have a term that's associated with the solid and the mold and its area times the solid mold interfacial energy. And the last term we have, again, is the solid mold and the interfacial energy between the mold and the liquid. And notice, what we are able to do is to relate the solid and mold to both the appearance and the disappearance of the interface where the spherical cap is present. So now what we've done is to go down and rewrite those expressions. And we've done that as a result of the diagram that lies to the right. We have three surface energy terms that are important. One of the surface energy term is between the mold and the liquid. The other is between the solid and the liquid. And the third is between the solid and the mold. So what we're going to do is to sum all of the horizontal forces associated with those interfacial energy terms. When we do that, what we'll find is that we can express the interfacial energy between the mold and the liquid and the interfacial energy of the solid in the mold, and come up with and expression that tells us that the interfacial energy associated with the solid and the liquid times the cosine of theta is equal to that difference in the two interfacial energies. And when we include this into our equations above, what we have is the surface area of the spherical cap, and the volume of the spherical cap. And remember that the geometry of the spherical cap depends on the angle theta. So it's going to tell us something about the surface area between the solid and the mold, and the surface area associated with the solid and the liquid. So using that geometry, now we have these two equations and those are the equations that we need to pay particular attention to, to help us develop an understanding of the spherical cap approximation to determine heterogeneous nucleation. So I've written the equations for heterogeneous nucleation in terms of two terms, the volume term and the associated geometric term. So those are the result of inserting those equations that we had on the previous, visual. And now what I've done is to multiply and divide the equation by 4. The reason that I've done that is, if you look at the first term, that first term is four-thirds pi r cubed delta G sub v + 4 pi r squared times the interfacial energy between the solid and the liquid. That term should look familiar to you. That term is, in effect, the homogeneous nucleation equation that we described in the earlier lessons. So that's our delta G of homogeneous nucleation. The next term is what we're calling our wedding term, which is f of theta. And so it's going to be, as we change the angle theta we're going to change the values of f of theta. And now we need to examine how those variables change, that is with f of theta. So now what we've done is we're going to write the barrier to the heterogeneous nucleation process in terms of first, homogeneous nucleation, and secondly, times the associated wedding of that particular interface. And what we're going to see is that that function, f of theta, lies between two values. It goes between 0 and 1 as you change the wedding angle from 0 to 180 degrees. So what this tells us is, if we look at the bottom of the equation, what that's telling [COUGH] us is that in one case, when the value of f of theta is equal to 1, it means we have no wedding associated with that particular interface. And consequently, the only type of nucleation that we can have in the absence of something that is wedding is homogenous nucleation. Alternatively, if we look at the lower limit, when that value is equal to 0, what that tells me is I have effectively complete wedding, and the barrier to the nucleation process, is essentially equal to 0. And one of the things that becomes important is the fact that the radius is not going to be changed or altered that we calculated for homogeneous nucleation, it's going to be the same r star. And the reason, of course, for that is, even though we have introduced the substrate, the behavior associated with the interface separating the liquid and the solid has not changed. But what has changed is the barrier to the nucleation process. And the barrier looks exactly the same as the homogeneous barrier that we described previously, but this time it's modified by the presence of that wedding angle f of theta. So by beginning the process of heterogeneous nucleation, we're able to use homogeneous nucleation as a starting point and develop how we can modify that particular behavior with respect to the effectiveness of the wedding of the material as the substrate f of theta. Thank you.