Welcome back everyone, I'm Charles Clark. And now we're can continue our discussion
of the hydrogen atom as seen from the perspective of classical mechanics,
picking up where we left off. Now this is going to be the most detailed
mathematical derivation so far, but I emphasize that it's simple vector algebra
and calculus just taking time to relatives.
So I hope it'll be accessible in any of then I have a number of inline quizzes
that will pause during the derivation and give you some simple questions to help
you keep pace. So good luck and let's see how it goes.
Just to remind you where we left off last time.
So we started, we're working on equation of motion for a hydrogen-like system that
is a nucleus with a positive charge, a mass of capital M, and electron of
negative charge, a mass of little m. And then we, we got, we just wrote down
the Newtonian equations. Using the Coulomb interactions we found
that we are able to get a cut an equation that involved only, the relative
separation of the two particles. So that's a big step forward.
here is that equation in a simpler form. So it's a, just a motion of a, of a
particle in the Coulomb field but with a reduced mass, and with a, a relative
co-ordinate. Here's the expression for the reduced
mass. We'll be using this later on, in order to
replicate the. The investigations that led to the
discovery of deuterium which is a heavier isotope of the hydrogen atom.
It's the, the, neutrium nucleus consisted of proton and a neutron.
So it has the same charge as hydrogen but about twice the mass.
And then by denoting this momentum in terms of the, the velo, the mass times
the velocity, the particle, very much the conventional way, we got to these two
first order equations. And I put emphasis on this because this
is really. The general type of evolution equation
that one wants to have. In other words if we have initial values
of the position, sorry, the position and the momentum, we have initial values.
Then we can calculate the first derivative of those, because we have,
basically we just plug in the initial values to these equations.
That first derivative, let's say we have t equals 0, that first derivative allows
us to propagate a little way to, a next high step and so on, so, with just the
initial values of position of the momentum, we can integrate these,
equations of motion and time indefinitely.
Now, whether we can do that in a practical sense with a simple looking
solution or not remains to be determined. The answer is yes, but the hydrogen atom
is quite unusual in this degree and it shows aspects that are not necessarily
encountered in most other dynamic systems.
But more about that later. So now we continue by seeing, we're
going to try to integrate this equation to the best degree by finding so called
constants of the motion. That is, things that stay constant in
time. So, so one of these, one such constant in
the motion which characterizes. They were really all dynamical systems
where the, the the, the, the, the, the potentials that govern our system are
independent of time. Is the total energy.
We put a bunch of particles together and they just have neutral interactions.
The, and they're not influenced by an external environment.
Their energy will be the same. So lets
Let's see how that's done for this current system.
And, very natural procedure is to start from the dynamical variables themselves,
form simple functions of those variables and see to what degree that those
functions are conserved in time. So if we take, if we just think about
products of p and r, here's, here's a, rather obvious candidate to look at.
p squared over 2 mu. This is the expression for a kinetic
energy operator. So does this correspond to a kinetic
energy, operator in this system. And does it lead to anything useful?
So I'm now going to give you a little in line quiz just to see how well you're
tracking that. So I hope that you did well on that
question and the, the message is that this t that we've identified here is a
kinetic energy operator. It's the, the net kinetic energy of the
system in a frame in which the center of mass is not moving.
So that's basically if you get on board a frame that's you when, both these
electron and the nucleus could be moving along at a million miles an hour.
If you have a fast enough car you can catch up with that and then you're in,
then you're in, in a, basically in a system, in a, in a, in a frame in which
you're only seeing the relative motion of the of the coordinate, the separation
seems to, the weighted, mass weighted separation seems to be constant time.
Well, now how does the kinetic energy depend upon time?
Well, let's take it's time derivative. here we have the first derivative of time
in, of the kinetic energy. Well it's a time derivative of p.p,
that's just over two mu, that's just clearly equal to p.
Dot P dot, it's the over mu. And now so for for P, this term P here we
can, we can use, we can use this identity.
To relate it to r dot. And now, let's, maybe I better choose a
different color for the next one. For the p dot, you see we have the
equation of motion up here. So we substitute that in.
So here's, here's, here's what this. P dot, dot, p dot.
I'm sorry for all the dots, but that's you know, our way of talking.
You can pass for a physicist if you pick some of this lingo up.
you see, it's now reduces from this equation, it's easy to reduce to this, so
that r dot, r dot is equal to the scalar r times scalar r dot.
You can use, basically, maybe you need to go here, if you want to follow that in
detail. Some of you will see this self evidently.
But, I mean, you know, I have done this before.
And, it always pays to be careful-,[INAUDIBLE] carefully.
And now, here is the whole point. Is that, you can recognize that this,
this whole term here, is actually the time derivative of something else.
It's basically, let's, you know? It's got this minus ze squared.
Drdt divide by r squared. Is, so, the rdt times 1 over r squared is
equal to minus ddt 1 over r. If you know how to, how to differentiate,
if you know how that ddt of x to the n is equal to n x dot x to the n minus 1, you
can work all this. This is sort of the, kind of the
fundamental identity of functional differentiate, how to differentiate any
power of a quantity. So this is just a, this general rule
gives you the example here. So in other words the time derivative of
the potential energy and, can use another color, time derivative of potential
energy is minus the time derivative of this thing that we've denoted by V of r,
okay. So now I am going to give you another set
of questions to ponder. Okay I hope that you've grasped the
difference and that second question. So now V of r, is the total potential
the, energy of the system. The potential energy doesn't actually
depend upon the reference frame, at least in, in non-relativistic quantum
mechanics. So, let's, let's go on and buy a vowel
and complete the puzzle. I've decided to buy the vowel E, okay?
Because from, from this, from this relationship that the time derivative of
t is minus the time derivative of the potential v we find that t plus v.
The sum of these two, is a constant of motion.
And that is, in fact, from the yo, your previous analyses.
an energy. Well now like, what energy is it, and
that's going to be sorry, just another little brief.
I hope that was straightforward and that you understand that the energy here is
the energy of the system. In the frame that moves with the center
of mass. Once again, the fact that the center of
mass strength could be moving at any speed whatever, we would still have the
same equation of motion to solve, to describe this, this relative in
coordinate. Which is the important thing.
So it's, this is one of these initial condition problems.
You're, we're not really. If we're talking about an atom, we
recognize, it could be moving at any speed, but what we're really interested
in is the apparent constancy of its internal structure.
This is really the important issue. and this is, you know, this is consistent
with the picture of quantum mechanics. Okay.
So now, just to recapitulate what we've done.
we've got ourselves a nice, first order equations of motion.
And we found the conserved energy. Okay.
Now are there other constants in motion this system well what about a something
having to do with who, we, we started with p.p as a, to see whether if.
When you took its time derivative, anything interesting?
Well this would just be r squared. So that's, that's not likely to offer
anything more because, you see, we already have an equation of motion for r.
how about r dot p? Well that's, that's not without interest,
but it doesn't really lead to. lead to much.
Okay, now, how about this, I think many of you will have seen this before.
Here's a, you know, we're sort of going in order.
We really sort of, we did the dot products, r-dot-r, r-dot-p, p-dot-p.
Now, here's the here's the cross product. What about this l, r cross p?
What's the time derivative? Well that's easy.
It's r dot cross p plus r cross p dot. That's a basic rule of differentiation.
And now for this term, we look up here. Okay.
And we see that r dot is just proportional to p.
So that means we get a p cross p, term. That's automatically equal to zero.
We don't even have to worry about the, whether the mu is there or not.
Cross product of a vector with itself, of a vector with itself is zero.
Okay, and now for p dot. Once again, a very nice, you know, we
have a very nice equation. Let's see, I'm going to switch colors
here so it doesn't get too confusing. Apologize to the color blind,
So p dot you see is just proportional to r.
So when we bring that back that's also equal to 0.
So, we have the result that L is a constant of the motion.
Very nice. And you might say it's actually three
constants of the motion because it's a vector quantity.
Now I have a, and I have a question for you about this vector L that we've
discovered. So L is in fact an angular momentum and
I, in the, in the quiz that you just took it, it defines, you know, the, the
location of the coordinate origin which that hanging[UNKNOWN] is defined.
And it's very significant, you see, that that the issue of the momentum and the
position always being in a plane perpendicular to l.
So this is a, this is actually a, it's a property of any, force law.
That, that turns out to be proportional. that is proportional.
Where the force is proportional to the, the vector r.
So, This Coloumb force is a 1 over r squared
law. One over r law 1 over and it's a linear
law in r and so on. All those yield the same result that the
angular minimum is conserved and that means I guess you can, you can now see
with this. r dot l equals 0 always.
And p dot l equals 0 always. So r and p are always perpendicular to l,
and there's, there's only one plane that's perpendicular to l.
So the position the momentum evolve in that plane, so this is, you might say
this is a very important simplification. We started out with two particles they
can each move in 3 dimensions so we started out with a 6 dimensional problem
in some sense and by now we've reduced it.
To, Well, how many dimensions is it?
I mean, the, we have two particles that move in a plane.
So, we, we, we know. We know.
We've, we've really located the orbit. we, we know that, that, that, basically
the, the system has gotta be evolving. Somehow in this plane.
And in the next lecture, we're going to take, and we're going to actually find,
the specific origin orbit, and that is made possible by the discovery of a
hidden symmetry.