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.