 So, the 3D particle in a box model has served us well as a model for ideal gases. It's been able to make pretty good predictions about things like the pressure and the internal energy and the enthalpy and the heat capacity for various gases. But now we've seen that there's some chinks in the armor. It does a great job of making predictions for monatomic ideal gases, but significantly less good job at predicting the heat capacity for diatomic ideal gases. So, let's see if we can understand the reasons why that would be true, and it boils down to the fact that the gases, as we've modeled them with the 3D particle in a box model, only have kinetic energy, so remember when we modeled the gases, we modeled their energy as consisting only of the kinetic energy, and for a monatomic gas molecule or atom of gas, that's reasonable. The atom bouncing around this box has no energy other than the kinetic energy and maybe a little bit of potential energy between the atoms themselves, but kinetic energy is the main form of energy in this gas, if that gas is something like argon or neon or helium. If that gas is an N2 or an O2 or some diatomic molecule, then if I draw that molecule as a diatomic molecule, it not only has kinetic energy of translation as it bounces around the box, as we've described with the 3D particle in a box, but it also has additional types of energy like rotational energy, as that molecule tumbles around in the box, or might have vibrational energy as the bond length vibrates. So there's a few different types of energy that we haven't allowed that molecule to exhibit to us in the Schrodinger equation, so that's what we're going to have to prepare to do a better job of predicting the properties of diatomic molecules. So let's start by making sure we understand how to describe the mathematics of what a diatomic molecule would look like. So let me draw a very large diatomic molecule. So here's a diatomic molecule with two atoms. Let's say atom one has mass M1 and atom two has a mass M2, so it doesn't need to be a homonuclear diatomic molecule like N2 or O2, it could be NO, or it could be HCl, it could be a diatomic molecule with two atoms of different masses. So atom number one, atom number two have masses that are different from one another. Atom number two is at some position R2 in space, and atom number one is at some position R1 in space, and the vector that describes the vector from atom one to atom two, that's R2 minus R1, that's the vector describing the bond of the molecule. There's another important quantity we can describe which is the center of mass of the molecule, the position of the center, not atom one or not atom two, but the center of the molecule. Homonuclear diatomic, the center is in the middle. The center of mass, though, if I weighted a little heavier towards whichever atom is heavier, in particular if I say M1 over M1 plus M2 times the position of atom one, and M2 over the sum of those two masses times R2, that's the definition of the center of mass. That'll give me a position in space that's closer to the heavier atom. If we say, let's go ahead and make sure the center of mass of the molecule is at the origin, so here's the origin, there's, let's see, let's draw the axes in a different color. There's the X and Y and Z axes. I've put the center of mass of the molecule, if the center of mass of the molecule is right there, I've made sure I've put the center of the mass of the molecule at the origin. The reason I've done that is because now as this molecule rotates, it'll tumble around the center of the molecule, which I've placed at the origin, so by not allowing it to move away from the origin, I've ignored any possibility that the molecule is moving around in space translationally, moving in X, Y, and Z, so the center of the mass is staying right where it is. I want to just consider the rotation of the molecule, primarily, and perhaps the vibration of the molecule, neither of which changes its center of mass away from the origin. So now the next piece of information we need is, we need to be able to talk about the geometry of the molecule and how this orientation is described in space. If I give you R1 and R2, that tells you where the atoms are, but it might be easier to specify the orientation of the molecule using this polar angle theta by how much is the molecule bent down from the Z axis and the azimuthal angle, which is if I project this vector into the X, Y plane by how much is it swept away from the X axis in a clockwise direction. So if I give you the bond length and theta and phi, that also tells you the geometry of the molecule just as well as telling you the coordinates R1 and R2. So, with that information, to start getting set up to solve Schrodinger's equation, I can say the energy of this molecule, if I want to talk about the kinetic energy of this molecule, kinetic energy we're familiar with, that's one half MV squared, because this molecule has two atoms in it. This atom has some kinetic energy, this atom has some kinetic energy. The total kinetic energy is the kinetic energy of atom one, combined with the kinetic energy of atom two. So these equations all describe the geometry of the molecule as well as its kinetic energy. But now there's a simplification that we can make, and I'm going to skip the algebra for this step, but it's not too difficult to do. If I combine these equations together, in particular, since I know there's some connection between R1 and R2, that bond length tells me the distance between atoms one and two. If I use that information in the center of mass, which I've restricted to be at the origin, I can rewrite this kinetic energy equation instead of being the sum of these two terms. If I rewrite v2 in terms of v1, and rewrite them both in terms of the bond length itself. So just like v is the rate of change of the position R1, v1 is the rate of change of R1, v2 is the velocity of atom two or the rate of change of R2, this v that I'm writing here, that's the rate of change of the bond length R. So that velocity vector, if I square it, I can rewrite the kinetic energy not as two terms, but as a single term, but where the constant in front of this v squared is something new. This letter mu, that quantity, is called the reduced mass. It's not the mass of atom one, it's not the mass of atom two, it's this particular combination of the two atoms, multiply them in the numerator, add them together in the denominator. That quantity is called a reduced mass, and that quantity, one half reduced mass times bond velocity squared, gives me the kinetic energy of the molecule. So again, I'm claiming that this equation, if I plug in reduced mass and I rearrange it a little bit, is exactly the same as this equation. You can double check me on the algebra, or if that doesn't sound like fun, you can take my word for it that this equation is true, and we can use this expression. There's an important reason to use this expression for the kinetic energy, which is that now in order to describe the molecule, let me redraw the x, y, and z axes, a molecule with this geometry and this kinetic energy and these bond lengths and locations can also be described with this kinetic energy and this mass, and what that means is if I position a molecule a distance r away from the origin, this bond length r away from the origin, with the mass equal to the reduced mass, with the same geometry bent down by this angle theta away from the z axis and swept clockwise by the single phi away from the x axis, then the energy and the properties of this molecule, not molecule, but a particle at a distance r with mass mu has the same kinetic energy, it also has similar other properties like angular momentum, it has exactly the same properties as this system, which has two atoms and two coordinates. The reason we prefer to write it this way is because to describe this system to you, I only have to give you three coordinates. I can tell you r and theta and phi, and I've described this system. So three degrees of freedom, three numbers will suffice to describe this system. In order to describe this system, I have to give you six numbers, x, y, and z for atom two and the same x, y, and z, three different coordinates for atom one. So I would have to give you six coordinates to specify this system, only three to specify this system. When we get around to solving Schrodinger's equation, it'll be much easier to solve Schrodinger's equation for a problem with only three variables than it will be for a problem with six variables. So this extra work we've done to rewrite the geometry in these reduced mass coordinates is going to save us a lot of work when we solve Schrodinger's equation, which is what we'll tackle next.