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PROFESSOR: I want to demonstrate
two ways in which you

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can see phase shift.

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So basically, the reason
we use phase shift

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is that these are the things
that you can calculate.

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Calculating phase
shifts is possible.

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So how do you do that?

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We'll be a two-step procedure.

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We'll only finish
that next time.

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But let's get started.

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So suppose you have
your wave again.

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And for a fixed l, a
given partial wave,

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this is the full solution
for this scattering problem.

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It is an Al Jl of kr
plus bl at Nl of kr.

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y l0.

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That's your solution.

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The signal that
you got scattering,

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and you have
something right here,

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is the existence of this
term, because when there's

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no potential and no
scattering, the solution

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is valid all the
way to r equals 0,

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and therefore, this
has no singularity.

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But this term is saying
that this solution

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doesn't extend all
the way to r equals 0,

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because this diverges.

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So something un-trivial
is happening.

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So bl is the signal
that they're scattering.

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Now, expand for large R. So this
is proportional to Al sine kr

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minus l pi over 2 minus bl
cosine kr minus l pi over 2

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y l0 1 over a kr.

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It's the same.

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I'm going to drop all
constants very fast.

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Now, here is my claim.

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You were thinking
of phase shifts.

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Well, the phase shift is
nothing else than bl over Al

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is minus the tangent
of the phase shift.

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This is a claim.

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Or you could say this is another
definition of a phase shift,

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and I'm going to argue that it's
the same, actually, than what

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we did.

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To do that, I have to
just expand a little more.

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So what I'm going to
do is divide by a,

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so take the a out, akr.

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Now you have sine kr minus l
pi over 2 minus bl over Al.

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Bl over Al is tangent
delta, so plus tan delta l

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cosine kr minus
l pi over 2 y l0.

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And that is proportional to
a over kr tangent delta l

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is sine over cosine.

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So let's put the 1
over cosine delta l

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here, sine kr minus l
pi over 2 cosine delta l

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plus cosine kr minus l pi over
2 sine delta l y l0 of theta.

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So that ratio tangent, I
put a sine over cosine,

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I have it here.

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But this, your favorite
trigonometric identity,

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is equal to a over kr
1 over cosine delta l

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sine, a single sine of kr
minus l pi over 2 plus delta l.

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So if this is the phase
shift, the solution

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looks like this far away, a
sine of kr minus l pi over 2,

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plus a delta l.

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That's one way of
identifying the phase shift.

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But I want to show that's the
same phase shift we had before,

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but that's clear already.

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Up to constants, this is--

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I'm sorry, y l0 of theta.

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And up to constants, this is e
to the ik r minus l pi over 2

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plus i delta minus
e to the i minus ik

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r minus l pi over
2 minus i delta.

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And now I drop everything else.

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And now I multiply or take
out this phase, take it out.

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I'm just working up to
proportionality, which

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is all you care at this moment.

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And this is e to the i
kr minus l pi over 2.

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If I take that out, this
becomes plus 2i delta l.

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And this becomes minus e to the
minus i kr minus l pi over 2.

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And those are the waves we had
before, somewhere here, here.

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Here they are.

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You see them?

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Still, they were here.

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This wave and that wave with
delta here has showed up.

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So this delta that
I've defined here

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is the same phase shift
we introduced before.

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So now you have three ways
of recognizing a phase shift.

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A phase shift can be recognized
in the partial wave expansion.

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A phase shift can be
recognized by looking

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at the scattering wave
far away and seeing

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that it takes this
form, and you say,

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oh, here is the phase shift.

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And the phase shift
can be recognized

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by looking at the solution
in terms of spherical Bessel

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functions, and it's the
ratio of these coefficients.

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Those are the three definitions
of the phase shift, three ways

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of seeing your phase shift.

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Instead of elaborating
more on this,

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let's do one example
to convince you

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that this is solvable
and doable in fact.

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So the example is a
hard sphere example.

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This is the object that
you're scattering off.

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The potential is equal to
infinity for r less than a

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and 0 for r greater than a.

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This is the origin and radius
a sphere, the waves come in.

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You want the cross-section.

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OK.

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It might look like
this is hard, OK.

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How are we ever
going to solve this?

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In fact, will be very easy.

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We have everything
ready to solve.

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So let's remember what we have.

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Well, there's going to
be a radial solution,

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Rl Remember Rl is Ul over
R. And that takes the form

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Al Jl of kr plus Bl Nl of kr.

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That is the general
solution for a radial thing,

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and therefore your general
solution for your wave function

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or of theta is what
we were writing here,

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except that the
superposition of them.

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So I said here is
how you recognize

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what is the phase shift
for a given partial wave,

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but we're going to have
all the partial wave.

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So this is going to
be the sum over l

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of these things, Al
of Jl of kr plus Bl

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Nl of kr times Pl of theta.

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I could write y l0, but
that's up to a constant Pl.

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So that's your general solution.

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And in fact, I didn't
even have to write this,

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because you have there on
that blackboard, that's

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the general solution for the
full wave away from this sphere

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that's not valid
for R less than 0.

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But this is your full solution
for a given partial wave

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for all the waves it's there.

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OK, so that equation
solves a problem.

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pl of cosine theta, yes.

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It's really better
written like that.

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Yes, that's more rigorous.

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Yes.

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OK, so what do we do now?

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Somehow you have to use it.

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You have a sphere.

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We haven't used a sphere yet.

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So what does the
sphere tell you?

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It tells you that
the wave function

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must vanish on the sphere,
because it's infinitely hard.

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It's like an infinite wall.

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So this wave function
should vanish psi

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at a theta, which
is equal to sum over

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l Al Jl of ka plus Bl of Nl
of ka times Pl of cosine theta

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should be 0.

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OK.

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One equation for
infinitely many unknowns,

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but the Pl's are a complete set.

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If you expand the function of
theta in terms of Pl's, you

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can determine every coefficient.

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They're linearly
independent, the Pl's.

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So if you can think
of this, this is an a.

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This is just a number.

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So this is a sum of numbers
times Pl's must be 0.

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All the numbers must
be 0, because the Pl's

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are independent.

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Therefore, here we have that
a Al Jl of ka plus Bl Nl of ka

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must be 0 for all l.

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And therefore, tangent delta
l, which is minus Bl over Al

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has been determined.

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The tangent delta l is Bl
over Al, and that's equal to--

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it's Jl of ka over Nl of ka.

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Done.

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All phase shifts computed.

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The Bessel functions are known.

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You look them up.

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You calculate them
to any accuracy.

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But you have here
all the phase shifts.

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Therefore, you have the
cross section of this sphere.

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You have the differential
cross section.

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You have anything you want.

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It has all been determined.

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We can do a little
bit of algebra

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if you want to calculate
what the sine squared

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delta of something.

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Sine squared can be expressed
in terms of tan squared.

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It's tan squared delta l over
1 plus tan squared delta l.

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So it's our Jl squared over
Jl squared plus Nl squared.

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Tan l, you substitute
this, and you get that.

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Therefore, the cross
section can be calculated,

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and the differential cross
section can be calculated.

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It's pretty interesting.

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Let's do it here.

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The cross section, remember
sigma is 4 pi over k,

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sum over l equals 0 to
infinity 2l plus 1 times

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this sine squared.

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So Jl of ka squared over
Jl squared of ka plus Nl

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squared of ka.

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So that's your
full cross section.

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If k is very small, ka much
less than 1, that's small k.

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That's small energy, long
wavelength approximation.

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This formula is interesting for
high energy, for low energy,

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for intermediate energies.

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The angular dependence
is interesting.

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It's a lots of things
you could look at.

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00:15:48,580 --> 00:15:52,490
Let's look at low energy.

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You need the expansions of this
quantities for small things,

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and they're easy to find.

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Basically, this remain finite.

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This becomes infinite.

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It dominates it.

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00:16:03,940 --> 00:16:09,110
It's not difficult.
I'll write what 1 gets.

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It's 4 pi over k
squared, sum over l

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00:16:13,040 --> 00:16:18,620
equals 0 to infinity,
1 over 2 l plus 1 2

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00:16:18,620 --> 00:16:22,460
to the l l factorial
2l factorial.

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So a mess of factorials.

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I'm sorry.

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00:16:25,940 --> 00:16:29,890
ka to the 4l plus 2.

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00:16:29,890 --> 00:16:33,290
And here is to the
question of convergence

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if ka is much smaller
than 1, in this case--

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this was the approximation--
that will sure converge.

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00:16:41,200 --> 00:16:45,860
The powers go up and up.

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00:16:45,860 --> 00:16:50,480
So for l equals 0,
it is interesting,

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00:16:50,480 --> 00:16:52,440
is the dominant one.

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00:16:52,440 --> 00:17:01,280
So l equals 0 only sigma turns
out to be 4 pi over k squared.

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00:17:01,280 --> 00:17:07,599
And all these factors just
give you a 1 for l equals 0.

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The ka squared, a
squared, ka squared,

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00:17:15,119 --> 00:17:20,990
which is 4 pi a squared,
which is a pretty cute answer.

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00:17:20,990 --> 00:17:25,380
This is the low
frequency cross section.

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00:17:25,380 --> 00:17:28,910
If it's a long wave
length approximation,

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the cross section of the object
is not the apparent cross

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00:17:34,100 --> 00:17:38,660
section, which is the diameter,
but it's actually the full area

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00:17:38,660 --> 00:17:41,060
of the sphere 4 pi a squared.

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00:17:41,060 --> 00:17:44,240
People try to
imagine why this is.

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00:17:44,240 --> 00:17:47,270
It's almost like the
wavelength is so big

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00:17:47,270 --> 00:17:51,120
that the wave wraps around the
sphere and gets stuck there.

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00:17:51,120 --> 00:17:56,420
So the sphere captures
proportional to the area

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00:17:56,420 --> 00:17:57,600
of the whole thing.

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00:17:57,600 --> 00:18:01,960
So we'll see more
of that next time.