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PROFESSOR: Let me discuss, for
a second, effective potential.

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And-- actually, let me ask you
a question to get you started.

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We're going to look at high n.

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N-- so n's, say, 100.

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And then l will go
from 0 up to 99.

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Very high l's.

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What's going to be
different from them?

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Well here is one
possibility, and that's

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a possibility that happens.

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The orbits are going to go
from being circular to being

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elliptical as you change l.

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And I ask you--

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which l would correspond
to circular orbit,

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and which l would correspond
to the most elliptical orbit?

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So let's take a poll/
so most circular orbit--

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is it l equals 0 or l equals 99?

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Who votes for l equals 0
for the most circular orbit?

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I have about half [INAUDIBLE],
a little bit more than half.

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OK.

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Who votes for l equals 99?

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[INAUDIBLE]

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Which is the most [INAUDIBLE]?

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OK, this is funny,
because we all

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associate most
circular with l equals

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0, spherically symmetric.

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But it's wrong.

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This is the most circular.

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AUDIENCE: What if l equals 100?

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PROFESSOR: You cannot
reach l equal 100.

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Actually-- and this
is most elliptical.

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and you can have the
intuition for that.

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Let's avoid the calculation
and be intuitive.

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And the intuition it is kind
of interesting, because we

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said at effective potential.

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And indeed, I want to discuss--

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that's what I want
to understand.

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Effective potential.

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So maybe I'll draw this
line a little lower.

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And what is the
effective potential?

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We've written it
already many times.

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It's h squared l times l
plus 1 over 2mr squared

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minus e squared over r.

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So the minus e squared
over r is here.

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Maybe I should draw it in a
way that gives me more room.

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Like that.

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OK.

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And then there is the energy
level that I'm looking at.

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It's some high energy level.

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So I'll put it near
here, not too near

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that I don't see
it, but near here.

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Here is-- my en is here.

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OK.

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So the answer is
actually already there.

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Not too clearly, but here it is.

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What does a particle
do in this potential?

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It goes with lots of kinetic
energy, very fast here,

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it goes from here to here.

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Here to here.

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But what does that mean?

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That the radius is changing
from r to some other value of r.

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From r equals 0 to
some other value of r.

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So what looks like an orbit?

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If you have a planet
going around the sun,

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and at some point it
seems to reach the sun,

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and at some point it goes far--

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here is the sun--

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then the orbit
must be like this.

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That's likely to be the orbit.

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On the other hand, as
soon s is equal to 1,

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you have the part
of the contributions

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that affect the potential.

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You have a 1 over
r square like this.

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And then the total
potential, which

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is the original
term plus that, will

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have something that looks
like this, and go like that.

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And then this orbit is going to
go from some small value of r,

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but different from
0, to another one.

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So those are all
elliptical orbits,

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because the radius
is not constant.

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And then this orbit,
it goes like that.

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Now look at it--

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if you have some
elliptical orbit-- so then

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after that time, it just
becomes a little more

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like this, in which the minimum
radius for, say l equal 1

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is here, and the maximum
radius has been reduced.

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l equal 2, l equal 3, they
all produce different ones.

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My graph is not perfect.

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Different ones.

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Different ones.

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And finally, one
that just touches it.

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And that one, the
radius is fixed.

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The particle is
moving like that.

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And that is your
spherical orbit.

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And if you increase l a
little more, no more solution.

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So it's the top
l that produces--

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this is already almost circular,
and this is perfectly circular.

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So then it goes like this, and
then eventually goes like this,

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and finally at the
end it goes like that.

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And it's circular
at the last case.

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For the most l, you finally
get your circular orbit.

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And also the intuition is clear.

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If you have a
circular orbit, you

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have lots of angular momentum--

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r cross p is very good.

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When two vectors are
orthogonal, the cross-product

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of angular momentum is large.

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But look here.

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Here is r, for the
shortest orbit,

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and p is in that direction.

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R cross p almost like that.

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This is a very low
angular momentum orbit.

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So l equals 0 is the
most elliptical orbit.

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L equal very high is
the most circular orbit.

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There is one thing--

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a calculation-- maybe I'll
use 10 minutes next time--

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is to show that there's
two turning points here--

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an r plus and an r minus.

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So there is an r plus for
one solution, and an r minus,

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and it goes like that.

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So the r pluses
and the r minuses

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corresponds to an ellipse that--

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here is the center.

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That's where the proton sits.

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And you go r plus out is the
maximum, and r minus is here.

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So the two turning points
that define the maximum r

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and minimum r tell
you about the lips

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on these r plus plus r minus.

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And when you find those critical
points and do the calculation,

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you in fact can show
that r plus plus r minus

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is indeed equal
to n squared a 0.

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I think that number
comes out exactly there.

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R plus plus r minus over 2.

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The average is exactly that.

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So the roots, the two roots,
are such that their average

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is always the same.

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So this is not--

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I'm not sure I get it
right with my graph,

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but it should be
that if you take

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the average of that
intersection points,

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they're always at
the same point.

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And so that's how it
happens under an analog in--

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so you get the intuition here.

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Here are your orbits.

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And these are
equal energy orbits

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of different eccentricity
as you go in there.

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And even Kepler knew that,
because Kepler figured out

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that the time that it takes
the planet to go around

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depends only on the sum of these
two things, the major diameter.

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So all these
solutions over here--

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if they were planets--

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would take the same time
to go around the sun.

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So they all, in
quantum mechanics,

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have the same energy, and
they're therefore degenerate.

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So the mystery of the
quadratic potential

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that makes all of these
periods to be the same

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is the mystery that is making
the energies the same in all

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of these orbits.

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So we'll finalize
this story next time

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and do some other fun things.