WEBVTT
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PROFESSOR: Velocity.
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So we assume that we
have an omega of k.
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That's the assumption.
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There are waves in which, if
you give me k, the wavelength,
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I can tell you what is omega.
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And it may be as simple
as omega equal to kc,
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but it may be more complicated.
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In fact, the different waves
have different relations.
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In mechanics, omega would be
proportional to k squared.
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As you've seen, the energy
is proportional to p squared.
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So omega will be
proportional to k squared.
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So in general, you
have an omega of k.
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And the group velocity is the
velocity of a wave packet--
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packet--
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constructed--
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by superposition of waves.
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Of waves.
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All right.
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So let's do this here.
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Let's write a wave packet.
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Psi of x and t is going to
be done by superposing waves.
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And superposition means
integrals, summing over
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waves of different values of k.
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Each wave, I will construct
it in a simple way
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with exponentials.
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ikx minus army of kt.
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And this whole thing, I will
call the phase of the wave.
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Phi of k.
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So that's one wave.
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It could be sines or cosines,
but exponentials are nicer.
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And we'll do with
exponentials, in this case.
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But you superimpose them.
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And each one may be superimposed
with a different amplitude.
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So what does it mean?
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It means that there is a
function, phi of k, here.
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And for different
k's, this function
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may have different values.
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Indeed, the whole assumption
of this construction
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is based on the statement
that phi of k peaks.
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So phi of k--
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as a function of k is
0 almost everywhere,
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except a little bump around some
frequency that we'll call k0.
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Narrow peak.
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That is our wave.
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And depending on how
this phi of k looks,
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then we'll get a different wave.
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We're going to try to identify
how this packet moves in time.
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Now--
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There is a quick way
to see how it moves.
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And there is a way to
prove how it moves.
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So let me do, first, the
quick way to see how it moves.
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It's based on something
called the principle
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of stationary phase.
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I doubt it was said to you
in [? A03 ?] in that way.
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But it's the most
powerful wave to see this.
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And in many ways, the quickest
and nicest way to see.
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Takes a little bit of
mathematical intuition,
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but it's simple.
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And intuition is something that.
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I think, you have.
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If you're integrating--
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a function--
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multiplied-- a function, f of x,
multiplied by maybe sine of x.
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Well, you have f of
x, then sine of x.
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Sine is 1/2 the times positive,
1/2 the times negative.
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If you multiply
these two functions,
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you're going to get the function
that is 1/2 time positive
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and 1/2 the time negative.
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And in fact, the integral
will contribute almost nothing
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if this function
is slowly varying.
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Because if it's slowly varying,
the up peak and the down peak
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hasn't changed
much the function.
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And they will cancel each other.
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So the principle
of stationary phase
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says that if you're integrating
a function times a wave,
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you get almost no contribution,
except in those places where
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the wave suddenly becomes
of long wavelength
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and the phase is stationary.
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Only when the wave doesn't
change much for a while,
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and then it changes again.
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In those regions, the function
will give you some integral.
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So that's the principle
of stationary phase.
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And I'll say it here,
I'll write it here.
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Principal--
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of stationary phase.
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We're going to use that
throughout the course.
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Phase.
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And I'll say the following.
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Since--
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phi anyway only
peaks around k0--
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This is the principle of
stationary phase applied
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to this integral.
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Since-- since only for
k roughly equal to k0.
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The integral--
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has a chance--
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To be non-zero.
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So here is what I'm saying.
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Look.
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The only place where this
integral contributes-- it
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might as well integrate
from k0 minus a little delta
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to k0 plus a little delta.
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Because this whole
thing vanishes outside.
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And if we're going to integrate
here, over this thing,
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it better be that this wave
is not oscillating like crazy.
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Because it's going
to cancel it out.
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So it better stop
oscillating there
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in order to get
that contribution,
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or send in another way.
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Only when the phase stops,
you get a large contribution.
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So on the phase stops varying
fast with respect to k.
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So you need-- need--
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that the phase
becomes stationary--
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with respect to k, which is
the variable of integration--
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at k0.
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So around k0, better be that the
phase doesn't change quickly.
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And the slower it changes,
the better for your integral.
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You may get something.
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So if you want to
figure out where
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you get the most
contribution, you
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get it for k around
k0, of course.
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But only if this thing
it's roughly stationary.
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So being roughly stationary
will give the following result.
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The result is that
the main contribution
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comes when the phase, phi of k,
which is kx minus omega of kt,
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satisfies the condition
that it just doesn't vary.
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You have 0 derivative at k0.
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So the relative
with respect to k
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is x minus d omega of
k dk at k0 t must be 0.
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Stationary phase.
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Function phase has
a stationary point.
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Look what you get.
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It says there that you
only get a contribution
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if this is the case.
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So the value of x, where you
get a big bump in the integral,
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and the time, t, are
related by this relation.
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The hump in this packet will
behave obeying this relation.
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So x is equal to d
omega dk at k0 t.
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And it identifies the packet
as moving with this velocity.
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x equal velocity times times.
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This is the group velocity.
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End of the answer
by stationary phase.
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Very, extremely simple.