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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.

6
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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.

32
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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?

35
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It means that there is a
function, phi of k, here.

36
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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

39
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is based on the statement
that phi of k peaks.

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So phi of k--

41
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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.

48
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Now--

49
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There is a quick way
to see how it moves.

50
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And there is a way to
prove how it moves.

51
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So let me do, first, the
quick way to see how it moves.

52
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It's based on something
called the principle

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of stationary phase.

54
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I doubt it was said to you
in [? A03 ?] in that way.

55
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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.

57
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Takes a little bit of
mathematical intuition,

58
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but it's simple.

59
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And intuition is something that.

60
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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

72
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hasn't changed
much the function.

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And they will cancel each other.

74
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So the principle
of stationary phase

75
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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.

81
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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.

83
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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.

86
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We're going to use that
throughout the course.

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Phase.

88
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And I'll say the following.

89
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Since--

90
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phi anyway only
peaks around k0--

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This is the principle of
stationary phase applied

92
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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.

99
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The only place where this
integral contributes-- it

100
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might as well integrate
from k0 minus a little delta

101
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to k0 plus a little delta.

102
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Because this whole
thing vanishes outside.

103
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And if we're going to integrate
here, over this thing,

104
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it better be that this wave
is not oscillating like crazy.

105
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Because it's going
to cancel it out.

106
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So it better stop
oscillating there

107
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in order to get
that contribution,

108
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or send in another way.

109
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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.

111
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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--

114
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at k0.

115
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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.

121
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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.

132
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It says there that you
only get a contribution

133
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if this is the case.

134
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So the value of x, where you
get a big bump in the integral,

135
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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.

137
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So x is equal to d
omega dk at k0 t.

138
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And it identifies the packet
as moving with this velocity.

139
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x equal velocity times times.

140
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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.