Hello everyone, welcome back. So now we're going to go and solve the problem for the classical motion of the hydrogen atom. Now some of you may have done this before, but we're going to do it in a way that leads to some of the material that's essential for the development of quantum mechanics, particularly understanding the angle momentum operator and some other symmetry operators that are very important in quantum theory and give us the actual insight into how to solve equations in general. So to return where we left off last time, we have these first-order equations of motion. Now let's try to make some progress in solving them by isolating invariant quantities, that is, constants of the motions, things that don't change in time. I mean, you know what some of these are, for example, the total energy. So now, since we're given equations of motion for these two vector quantities, it makes sense to try to find constants of the motion by looking at various other products. So here is the kinetic energy of the system, something of which you're familiar with. Write it as p squared/2 mu, it's the same as one half mu r dot squared. And now, just as a check to see how well you're following things, here's a little online quiz on the actual meaning of this T, the kinetic energy. So I hope it's clear that the kinetic energy is expressed here is in the frame in which the center of mass is at rest. Because it just deals, this p has to do with the motion of the relative coordinate. So the electron and the nucleus together could be moving as a million miles an hour with respect to your frame of reference, but it's in the frame of the center mass that the expressions become most economical. And then here is a, this is actually something that happens quite generally in any system of particles interacting by pairwise forces. You can see that the time derivative, the kinetic energy, just following the Newtonian force laws, is minus the timed derivative of this quantity minus the timed derivative of the potential energy. So, the potential energy here, when added to the kinetic energy, gives a constant of motion, meaning that when you start the system with an arbitrary configuration of the two particles, that is the positions and the velocities are what they are at some initial time. Then at any later time, you then compute the total energy in terms of the sum of the kinetic and potential energies. And at any later time, that sum adds up to exactly the same thing. Now I'll have a quick check to see about your understanding of this point. So I hope it was clear that from the construction, the potential energy depending upon the relative coordinate, it actually doesn't matter, at least in non-relativistic mechanics, whether that's in the frame at rest or in the frame moving with the center of mass, it's the same. Okay, so the result of that is we found one constant of motion, meaning this helps reduce the complexity of the system since that we know that however the particles may evolve from their initial configuration, it's always the case that their energy has the same value. So they can't move all over configuration space. You might say that this defines a sort of a surface in the multi-dimensional space on which the particles are constrained to move. Not to belabor the point, but why don't we have another quick check to see how well you're following things? So I hope the results of that last quiz were pretty transparent. I mean, once again, the energy is the energy in the frame that's fixed on the center of mass. Okay, so we had a process of starting with a product of variables and finding a sort of a conservation law that related its time derivatives to another function of position. So why not proceed systematically on those lines and see if there are other products that have a fairly simple tiny dilution? So to put it another way, this, the kinetic energy is proportional to p dot p. So then there are two other scalar products of these two vectors, r dot r and r dot p. And I think you will find that there's nothing simple about the time evolution of either of those. So how do we proceed in some systematic fashion? Well there are three possible scalar products of the vectors r and p, and then there's one vector product. Because r cross r as 0 as p cross p. Now here I am assuming, Basic knowledge, Of vector algebra. So I mean, I hope you do understand. If you don't understand what the vector cross product and dot product are, this would be a good time to pause and take a refresher. But the basic idea is that things like the dot product is something that's invariant under rotation of all the coordinates in the system, that's a so called scalar on the rotation. The vector product is sort of the next degree of complexity. It acts like a vector under rotation. So, a vector, when you rotate it, assumes a different orientation. In general, a rotation changes a vector into another vector. The dot product is proportional of the cosine of the angle between two vectors. And if you rotate the system, that clearly stays the same. So here is this, Quantity, the cross product of the position momentum. What is its equation of motion? Well, it's straightforward to compute the time derivative of that, L dot by definition is r dot cross p + r cross p dot. Okay, so now, what about that? Well r dot, Is just proportional to p, so that means that this is equal to 0. Now, what about the second term? Well, p dot is proportional to the vector r, not as a complicated coefficient, but in terms of the cross product, this becomes proportional r dot r, so also equal to 0. So here's the second constant of motion, the angular momentum r cross p. So, let me say two things about this. And that is this result that the angular momentum is a constant of motion is generic for systems in which the force is the gradient of a radial function. There's no dependence of the force on the relative angle between the two particles. And the second is, this puts a very strong constraint on the possible motions that can occur from given initial conditions, and we'll review that in the next online quiz, which is going to be the last for this part. So I hope it was clear to you in thinking about that, that for example, here's the defining, here's the definition of angular momentum. And so this means that r dot L is equal to 0. Let's see, how do we see that? Well, we know that r cross p is perpendicular to r and p, very simple result of vector algebra. So that means that L, which is, we've determined is fixed in space for all time. So let's say at any given time, we have a value of the position and the value of the momentum. From those, we determine the value of the angular momentum just by taking the cross product, very simple. And then forever after, the value of the angular momentum stays the same, so that r is always perpendicular to L. Well that means it lies, I mean, if you draw, let's orient L, so it's sort of standing up straight. That means that r is moving in the plane that must always be in the two dimensional plane that's perpendicular to L and that the same thing is true with p. So in other words, the conservation of angular momentum takes us from a problem, which is sort of three dimensional in its formulation and reduces it to motion in two dimensions. That's a huge reduction in complexity. So we now know that all we have to do is to enumerate the allowed values of angular momentum and then to solve the equations of motion for the particular values of the angle of momentum. And that is going to pass through into quantum mechanics in a very important way.