When the speed of propagation of a wave and a medium is
independent of the frequency, thereby independent of the
wavelength, we call that a non-dispersive medium.
When the speed of propagation depends on frequency,
we call that a dispersive medium.
And that is on our plate today. I want to revisit the case that
we have N beats, each mass M,
and they are connected with little strings,
little L. And total length,
say, is capital L. Perhaps you remember,
you even had a problem whereby we had five beats.
It was also demonstrated by me in my lectures.
This end was fixed. And this end was fixed.
So this was a very special case whereby N was five.
And we derived the normal mode frequencies.
The general result is that the normal mode frequencies in terms
of N, N being one, two, three, etc.,
equals two omega zero times the sine of N pi divided by 2N plus
one. I don't want you to remember
that. I certainly don't remember it
when we removed this. I certainly don't want to
remember it, but this is something that today I will
need, 2N plus one.
Omega zero was the square root of T divided by ML.
So, this little m, let me make it a capital M.
Each one of those beats has Mass capital M.
You can see that the maximum possible frequency is two omega
zero because that's when the sine of this function equals
one. The boundary conditions,
that it is fixed here and that it is fixed there demands that
there are certain values of K of N which are allowed which is N
pi divided by L if the length here is L.
Now, omega N, as you see here,
does not increase linearly with little m because it increases
with the sine of N. But, the speed of propagation,
V, is of course the ratio of omega divided by K.
So, that is omega N divided by K of N.
And, since K of N goes up linearly with N,
little N, but omega N does not go up linearly with little N,
it is clear that the speed of propagation is lower for higher
frequencies because when you go to higher values of little n,
K goes linearly up, but omega N goes slower.
And so, you see here an example of dispersion,
namely that the speed of propagation is lower for higher
frequencies than for lower frequencies.
And therefore, it also depends on wavelength.
A very nice way of seeing that is to make a plot.
And, you will see many of these plots today of omega versus K.
Let me make the plot here. So, this vertical is going to
be omega. And this is K.
This is omega.
So this is K1. This is K2, K3,
K4, K5. Let me move the K.
And then, here is the end of the string for this value here.
It's K5, and this value here is K2.
It was [those?] normal mode solutions.
And so, the maximum value that omega can have is then two omega
zero. And if I connect,
if I draw this curve now, it looks like this.
It's a sine curve. So, for K1, this is omega one.
For K2, this is omega two, and so on.
So, you can immediately see that the speed of propagation in
the fifth mode is lower than the speed of propagation at the
lower modes because this point here, which is K5 connected with
omega five, I can connect that with the origin.
And then, this angle is a measure for the speed of
propagation because the speed of propagation is omega divided by
K. So, omega divided by K,
the slope is a measure for the propagation speed.
And you see that if I go to this point here,
and I draw this line that the slope is larger.
And so, the speed is larger. So, you see here in a graphical
way, which is always very nice, to use omega K diagrams.
What I already argued with you here, that the higher the
frequency, the lower the speed of propagation.
And, that is what we call dispersion.
Now, dispersion is very common in physics.
As you will see today, water waves under certain
conditions are dispersive. Electromagnetic waves,
which we will begin to think about today, and next lecture we
will go further into that, radio waves and lights and
infrared are electromagnetic waves.
They can be dispersive and certain media,
not in vacuum, but in media.
Waves on a string, as I will discuss very shortly,
can also be dispersive, are in fact dispersive.
The consequences of dispersion are very nonintuitive.
And this is what I want to discuss with you first.
I will start with two waves which have a different
wavelength, and I will give them purposely a different speed of
propagation. Think of this as being on a
string because that's always very easy to see because that's
a transverse motion. So, I have Y1,
which is a certain amplitude. And I'm going to make a
traveling wave. So, we get the sine of K1 X
minus omega 1T, and so the speed of propagation
is omega one divided by K1. And then, I have a second one,
which for simplicity I give the same amplitude the sine of K2
times X minus omega 2T. And this is V1.
And V2 is omega T2 divided by K2.
And, the two are not the same. If they are the same,
then it's a non-dispersive medium.
But, they are not the same. And so, now I want to add them.
So, I want to know what Y is, which is the sum of the two.
So, Y is Y1 plus Y2. So, that becomes 2A times the
sine of half the sum times the cosine of half the difference.
And so, that gives me the sine of K1 plus K2 divided by two
times X plus this term. So, that becomes minus omega
one plus omega two divided by two times T.
So, that is the sine of half the sum.
I should put brackets here so the minus sine holds for here
and for there. And, I have to multiply by the
cosine of half the difference. So, I get K1 minus K2 divided
by two times X. And, now I get minus omega one
minus omega two divided by two times T.
So, look closely what I have done here.
I've added two sine curves. I get twice the amplitude,
the sine of half the sum times the cosine of half the
difference. Now, let us take,
as an example whereby K1 is approximately K2.
And, omega one is approximately omega two.
So, we take the frequencies close together.
You would agree with me, then, that K1 plus K2 over two
is then effectively K to the sum of the two divided by two.
You would also agree we've made it omega one plus omega two
divided by two is omega, sort of the mean value between
the two. So, I can rewrite this now that
Y is approximately 2A times the sine.
And, now I will just write KX minus omega T,
K being now the mean value between the two,
omega being the mean value between the two times the cosine
of delta K divided by two times X minus delta omega divided by
two times T. And now, look at what we have
here. This all by itself is a
traveling wave. The sine alone is a traveling
wave. And, that sine alone has a
velocity. This wave is a traveling wave,
which has a velocity which I'm going to call the phase velocity
with a P. That's our definition of phase
velocity, which is omega divided by K.
So, when you see that wave moving, it will move with that
speed. But the cosine term has a very
different velocity. It has a velocity delta omega
divided by delta K. And, that is the same as omega
divided by K if the waves travel with the same speed.
So, that V, which we call the group velocity,
is delta omega divided by delta K.
And, if the two are the same, then it's a non-dispersive
medium. But, if the two are different,
then you're going to see that this wave moves with a different
speed than that wave. The consequences are even more
bizarre. The wavelength of the sine
curve is, of course, given by 2 pi divided by K.
That's the definition of wavelength.
And, if you think of the wavelength, the repetition
pattern of the cosine function, that is 2 pi divided by this K.
So, that is 4 pi divided by delta K.
And, if delta K is very small, then this wavelength,
the repetition of the cosine is way longer than the wavelength
of the sine curve. If I made a curve of the sine
curve, you would see this. And, that moves with the face
velocity, say, in the positive direction.
If I make a plot of the cosine, that would be like this.
And, that would move with the group velocity.
And, if the two are the same, then this overall pattern,
which is the product of the two, which I will put in
shortly, moves with the same speed.
And if the two are the same, then this overall pattern,
which is the product of the two, which I will put in
shortly, moves with the same speed.
If I multiply this with this, and I will try to do that here,
you will get this.
So, what you are looking here at is exactly this function.
The sines are in here, and the cosine is there,
some kind of a beat phenomenon, of course.
And, it is this envelop that moves with the group velocity as
I already have here. It is the individual sines that
move with the phase velocity. And, if the two are the same,
it's a non-dispersive medium. If, however,
the two are not the same, and that's what I wanted,
I wanted the two to be different, then you will see
that the group velocity is different from the phase
velocity. Group velocity can be larger
than the phase velocity, and so then this red envelope
moves faster than the individual sines.
It can also be smaller than the phase velocity.
I will demonstrate that to you. When you look at omega K
diagrams, in general you get the full story.
I'll make you an omega T diagram, omega K diagram of a
nondispersive medium, which is very boring.
So, this is omega versus K. And, this is a nondispersive
medium, straight line. It's clear that V
[UNINTELLIGIBLE], which is omega divided by K is
this indicated by this angle by the slope, and is the same.
Independent of omega, it's everywhere the same.
That's a nondispersive medium. That is the definition of a
nondispersive medium that always, regardless of frequency,
have the same phase velocity. And so, you will see since the
group velocity is the omega DK, which is the tension of this
line, you'll see that the group velocity is the same,
of course, because if you take the tension of this line,
no matter where you take it that is the same slope.
So this is also the group velocity.
And so, this is a nondispersive medium.
When we dealt with strings, so far, we always made the
assumption that the speed of propagation was independent of
frequency. And so we always treated string
as if they were nondispersive. And, we derive that from the
wave equation. I can show you another K
relationship now. For instance,
the omega K relationship could go like this.
That would be a dispersive medium now.
You should be able to tell me now whether the speed is higher
or lower at high frequencies. Well, the speed of propagation
is clearly lower at high frequency because it's dictated
by this angle. And this angle is smaller here
than it is there. But the group velocity is also
dependent, now, on frequency.
For instance, right at this frequency here,
so right at this frequency, the group velocity is given by
the slope, the omega DK. And that group velocity,
which is the omega DK, has a different angle,
the slope, than this slope. And so, you see that for all
these points here, the group velocity is different
from the face velocity. And, the exact values of the
phase velocity, and the exact values of the
group velocity follows from the shape of that curve.
So, you have to know that curve.
And that curve is called, in general, the dispersion
relation. So, any relation between omega
and K gives the whole show away. Suppose you had an omega K
relationship that goes like this.
And, you will see today that that's possible.
So, now, you have enough knowledge to tell me that at
high frequency, the phase velocity is higher
than at low frequency because the slope is higher.
You also can see that the group velocity at high frequency,
which is the tangent, is higher than the phase
velocity. Pick any point here,
and you'll see that the tangent has a higher slope than the line
straight through here. So, the group velocity at any
point is larger than the phase velocity.
It is even possible in physics, but it is a rare case,
that you have this situation. And I will demonstrate it you,
but it's a computer simulation. So, it's always easy to do that
on a computer. It is even possible that there
are parts of the omega K diagram which go like this.
They may go up here, but they go down again.
If that's the case, that is remarkable what you
have. This slope would indicate the
phase velocity. But the slope would now
indicate the group velocity. And that group velocity now is
in the opposite direction of the phase velocity.
So, the sine wave goes like this.
And the group goes back. Even that is possible.
That's a rare case, but it does happen.
So what you have here is that the wavelength,
remember wavelength increases in this direction because K and
lambda are inversely proportional.
So, the wavelength increases now with increasing frequency.
If my frequency increases, the wavelength increases,
which is something absurd. We always think that if the
frequency increases, that the wavelength will get
shorter. Well, there are media whereby
that is not the case. That is what dispersion is all
about. I can create a pattern like
this with two way this review, which I will do.
And I do that in a graphical sense, that I have two
transparencies. One transparency has black bars
on it. Many, as you'll see shortly,
and they have a separation, D.
And, I have many of them, maybe 100.
I didn't count them. And I have another
transparency, which has bars which are 5%
thicker. And also, the opening between
them is 5% larger. So here, we have 1.05 D.
And, I line these up. So, this is the very first one
on my sheet. This is number one,
and this is number one. So, they line up.
This one, the opening, the separation is 5% larger.
So that means when I reach here on my sheet number 20,
then I reach here at number 19. But they are,
again, on top of each other. They are in phase so to speak.
So, my number 20 coincides with number 19 here because this one
is 5% larger spacing. The wavelength is 5% larger.
That means halfway in between, the black lines from this sheet
will exactly eclipse the open spots of this one.
And so, these black ones here will all [occurlt?],
to use the astronomical phrase, occurlt the openings.
And so this whole central portion will [do?] black.
Black will be on top of light, and light will be on top of
black. And black will top light.
Here, however, it's back to what was here.
So, you are beginning to see this kind of pattern.
Think of this, for now, as being dark.
And think of this, now, as the light.
This kind of pattern is what you're going to see if I put the
two transparencies on top of each other as I will do very
shortly. Now comes the wonderful thing.
Not only will you see such a pattern, but imagine now that I
move one of these sheets, say the upper sheet,
over a distance half D. So I move it only over this
distance. Remember, here the black ones
were occurlting the open ones. But, if I shift that by half D,
it will be reversed. The black ones will no longer
occult the open ones. And so, this whole black area
will instantaneously turn to light again.
And, that means that this overall pattern moves 20 times
faster than the motion of one of these sheets.
If I move this sheet over a full distance,
D, this entire pattern that you see here, which I think of as
being the group velocity, moves 20 times faster.
And that so I'm going to show you.
And, the reason why I get 20 times, of course,
because I have chosen the wavelength is to be 5% apart.
And I am going to give them a different phase velocity.
The first thing that I want to do is give them the same phase
velocity. So here, you see those two
sheets on top of each other. I've each marked them,
one with a red mark and one with a blue mark.
I think a blue one is the one that has 5% larger spacings.
And you see exactly what I predicted.
You see those dark bends. So these are the areas here
where, these are the areas, the dark areas,
where the black bend of one occurlts the openings of the
other. And then, in between,
you see light here and light here.
And you see light there. If I can manage to move them
both with the same speed, so I take them firmly in my
hand. And, I'm going to move them
both with the same speed. Then, notice that the bars,
the individual bars move with exactly the same speed as the
whole pattern. So that means the group
velocity and phase velocity are the same.
I'll try that once more. There we go.
I have it firmly in my hands now.
So, you see the group velocity and the phase velocity are the
same. But now, I'm going to move one
relative to the other. Watch closely.
Now I'm going to move one relative to the other.
And you can tell by preparing the red spot with the blue spot.
Compare the red spot with the blue spot, and you can see the
relative motion. I'm moving them now.
And, look how fast the group velocity is compared to the
motion between the two. You can actually hardly see
that the red one is moving relative to the blue one.
You can hardly see that. But look how fast the group
velocity is. You see that?
20 times faster, you see how the red one
separates from the blue one? It's a very low-speed.
That is the phase velocity. It's the difference between the
phase velocity. That's really what I should
have said. It's the difference between the
phase velocity. That's really what I should
have said. It's the difference between the
phase velocity. And so, then the group velocity
goes 20 times faster. So now, I want to return to our
continuous strings. When we dealt with continuous
strings, we derived the wave equation.
And I still remember how we derived it.
We took into account that the tension is responsible for the
restoring force. So, we made the [wire?] like
this, and then we had tension. And then, we worked out the
math. And then, we found that in that
case, D2Y DX squared was one over V squared times D2Y DT
squared.
And we found by substituting the solution into the wave
equation that the speed of propagation was the square root
of T divided by mu, T being the tension.
Mu was the mass per unit length.
In this speed is no K. There is no omega.
So, it's a nondispersive medium.
It says that, you tell me what omega is,
and this is going to be the speed.
You tell me what omega is. That's going to be the speed.
It's independent of frequency. It's a nondispersive medium.
And the reason for that is because we only took into
account that the tension in the string was responsible for the
restoring force. So that gave us,
then, that omega was VK that came out of this wave equation.
And so, we found that omega squared is V squared times K
squared. That is the result of the
calculation when the tension is exclusively responsible for the
restoring force. And you get a nondispersive
medium. However, there is something
that we did not take into account.
And that is the stiffness of the wire.
Imagine for now that the tension is zero.
So you take a piano string or you take the string from a
violin. And assume there is no tension
at all. And you take it in your hands
and you bend it. There's no tension.
You just bend it. It wants to straighten out.
That's the result of stiffness. And it is due to the stiffness
now that you get an extra restoring force which we have
ignored. We didn't take that into
account. And, the restoring force,
due to stiffness turns out to be proportional to K to the
power four. It's inversely proportional to
the wave lengths to the power of four.
It's an approximation. In other words,
our wave equation that we had is no longer valid because this
wave equation only took into account the tension.
If now you go through the whole procedure again,
which is slightly more complicated, you'll find now
that you get a different relationship between omega and
K. And now, you see that the
medium becomes dispersive. Now you get that omega squared
equals V squared times K squared plus alpha times K to the power
of four. And this is the result of that
stiffness. Beckafee and Barrett call this
alpha A squared. That's fine,
of course. That's just a matter of
definition. A is a positive number.
And, what this tells you is that the higher the frequency,
the higher the speed of propagation.
I can make for you an omega K diagram.
So, let's have here omega. And let's have here K.
So, what we have before when we only took the tension into
account, we had this. Omega is linearly [with?] K,
and that slope is the phase velocity.
It's also the group velocity. All wavelengths have the same
speed. But now, because of this term,
it's going to curve up because Alpha is positive.
And now you're going to get this.
And now you see in front of your own eyes that the higher
the frequency, when you're here,
the phase velocity here is higher than at low-frequency
here. This slope is lower than that
slope. And so, now we have a
dispersive medium. And also, the group velocity
right here is even higher than the phase velocity because this
slope here which is the tangent is higher than the connection to
zero. So, a piano string is
dispersive. And, that has major
consequences. Now, the values for alpha,
as you can imagine, depend, of course,
entirely on your string. If you have a very thick
string, then it is very stiff. It's very hard to bend.
So, alpha will be high. Also, if Young's modulus of the
wire is very high, then it is extremely difficult
to deform it. So, alpha will also be high.
So, there is no such thing as one value for alpha for all your
piano strings. They will all be different for
all the strings. But, I do want to do a
calculation to give you an order of magnitude idea what effect
this would have on a piano. So, I choose a value for alpha
which is not entirely absurd, although way higher values of
alpha are possible. And then, I want to calculate
with you quantitatively what difference that would make for a
particular string in the piano. And so, the case that I have
chosen, I take alpha is ten to the minus two.
I take the tension, which is very common for piano
strings, which is 250 N. I take the length of a piano
spring which is one meter, just for simplicity.
And I give it a mass per unit length of 10 g per meter.
So, that would be ten to the minus two in our SI units,
ten to the -2 kg per meter. And, I want to explore with you
the 10th harmonic of this string.
Well, T divided by mu, by the way, this can be
written, I will write it down again in, this is also T divided
by mu times K squared plus alpha K to the fourth.
This V, remember, was T divided by mu.
So, T divided by mu, that's easy enough.
That is 2.5 times ten to the fourth.
I have to know what K is. Well, lambda ten,
lambda one is two meters, right, fixed at both ends.
If it's 1 m, the length, then lambda one is
2 m. So, lambda ten is ten times
smaller. So, that is two divided by ten.
So, K ten, which is, of course, by definition,
2 pi divided by lambda ten, is then ten pi.
And so, now I have all the ingredients to compare this
omega squared, this term with that term to see
by how much the frequency changes over the case where we
had no dispersion. So this first term,
now, and you can check that, of course, for yourself.
So I will write down omega squared, again,
which is that term, is T over mu.
And then, I have to multiply it by K squared.
And then, I find 2.5 times ten to the 7th.
And then, I can calculate alpha K to the fourth,
and I have all the ingredients on the blackboard.
And, you'll find that this term is ten to the fourth.
If I calculate roughly what omega is, I'm not interested in
the exact value of omega. You'll see shortly why.
You will find that omega is about 5,000 rad per second in
this case, which translates into a frequency of about 800 Hz.
The value itself is not so important.
But what is important now, that as a result of this extra
number, the frequency will go up by 0.02%.
And this, of course, you can check immediately for
yourself with your calculator. And, 0.02% increase means,
in this case, 1/6 of the hertz.
So at the 10th harmonic if you expected exactly 800 Hz,
this is of course not exact, but I take around off number,
then it would be 1/6 of the hertz higher because of the
dispersion relation. But, if you make alpha ten to
the minus one, which is by no means
impossible, then it would go up by 1.6 Hz.
And so the bottom line, now, is that for a real piano
string, omega N is no longer N times omega one.
And, that is the reason. That is why pianos go sharp,
as musicians say. That means at the higher
harmonics, the frequency is a little higher than linearly
proportional with N. And that's what's called,
makes the piano go sharp. Professor Wyslouch developed
several years ago, when he was lecturing 8.03,
a toy model, which he uses on the computer.
And the toy model has the following dispersion relation.
It has no direct connection with the string,
no direct connection with any physical thing.
But it is simply used for the purpose of demonstration.
And that is that omega squared equals V squared times K squared
times one plus alpha K squared. And, if you're not careful,
you may think that it is the same as this.
But it's not. It's very different.
The difference is that this term here has a V squared in it.
And this term here doesn't have this V squared.
So, totally different dispersion relationship.
In this program you are going to see, [Bolech?] Wyslouch shows
you six waves. All six have the same
amplitude. But, they differ.
The shortest wavelengths and the longest wavelengths differ
by 12%. So, two neighboring wavelengths
are two and a half percent apart.
And, the velocities of the individual waves,
the phase velocity, is then dictated by this
dispersion relation. That is omega divided by K.
So, these velocities are not the same if you put in alpha
nonzero, of course. If you make alpha zero,
then they are the same.
First, we have to understand what you're going to see when
the waves are not moving. If I take six waves with
different wavelengths, there I showed you only two
waves with different wavelengths.
And you saw these beautiful beats.
If you do that with six waves and you lined them up,
say, at the center of your screen, that the amplitudes all
line up with each other, you will see six times the
amplitude at the center of the screen.
But when you go away from the center, you do not see anything
like this. What you are going to see,
the computer will show you. It's not so intuitive.
But you'll see something like this.
And then, it bulges a little and then bulges again.
And that's simply a matter of geometry.
That has nothing to do with dispersion.
We haven't even touched dispersion.
We simply take six waves and we draw six waves all with
different wavelengths, to a half percent apart between
adjacent wavelengths. We put them on top of each
other like I did there, and you see a pattern.
And that pattern you'll see. And, to make sure that you
remember that pattern for the next ten minutes,
I'm first going to show you what happens when alpha is zero.
That means all the waves will move with exactly the same
speed. And, the envelope,
crazy as the envelope will be, is moving with exactly that
same speed because the envelope is the group velocity.
And then, when you have really remembered, when you really
remember that envelope, then I'm going to change alpha.
If I make alpha positive, it will turn up.
That's the general idea [of your plot?] goes like this.
If you make alpha negative, which is not the case for a
string, it's just in this toy model, I can put alpha minus,
right? If you put alpha minus,
then you're going to get this. And, if you go to very high
values of K, it will even turn over because if alpha becomes
[high enough negative?], ultimately it will turn over.
And, if I can do that, if I can choose a value for K
very high with a negative alpha, this is the situation that was
required, remember, for the phase velocity to be in
the opposite direction as the group velocity.
So, I can also show you that, which is really cute.
So you see, the envelope move in a different direction than
the individual waves. So, Marcos, did we decide on
five? So, the first thing that I want
to run with you is, then, alpha equals zero.
You are going to see the six waves.
There it comes. Now, look very closely.
The red one is the superposition of six waves.
The blue ones are the waves. The blue ones all travel with
the same speed. Take my word for it.
I put alpha zero. And so, this red curve,
which is the superposition, which has nothing to do with
dispersion, it has to do with geometry, that red curve moves
with exactly the same velocity as the blue ones.
The group velocity is the same as the phase velocity.
And this crazy shape of the red one, [UNINTELLIGIBLE] geometry
to the overlay of the six waves, that shape is not changing.
So, hold onto it. Make sure you remember that
five minutes from now. That shape is not changing.
And, the reason why it's not changing is because the waves
move with the same speed, all of them.
So, that shape cannot change. All right.
So now I'm going to make alpha 1,000.
And I make it positive. And, I also want you to see it
a little longer in time because I want you to see that now that
the phase velocities are different for the individual
waves. I want you to see that the
overall envelope is not going to change, which is no surprise,
of course. That is now the result of
dispersion. So, this overall envelope will
not change. Because now you change the
geometry of how they line up with each other.
So, what you are going to see here now is an example whereby
the six waves all have different velocities.
The group velocity is larger than the phase velocity because
I made alpha positive, and the shape of the envelope
is slowly going to change. You cannot see,
of course, with your naked eye. Neither can I,
that the six waves don't move with the same speed.
It's impossible to see that. Neither can you see that the
group velocity is different from the phase velocity.
You just don't have the resolution in her eyes.
But notice, now, that the overall [red?] shape
already has changed somewhat. It's no longer as crispy.
It's no longer as sharp. That's why I'm giving it a
little more time than the first one so that you can actually
begin to see that the shape, the red shape,
is changing in time. And that is the result of
dispersion. Here, look.
Look when this red one is coming out.
It's way more drawn out now. That is the result of
dispersion.
Alpha, by the way, is only 1,000 in this toy
model. Hey, look at that.
You see that? Very, very different.
And now, I want you, oh, let's look at the omega K
plane. This program also shows you the
omega K plane. Now, this is not omega zero
here, so be very careful when you look at this.
You see only a very short part of the omega K plane.
Something here, you see, a very high value for
K. You see a short section there.
So you see omega vertically plotted, and you see K
horizontally plotted. This end, the fact that it
looks like a straight line there doesn't mean that it's
nondispersive because it looks that so many very small
sections. So this looks like a straight
line. But the whole thing,
of course, is on a curve. And that's why it's
nondispersive. And so, now I'm going to show
you a value for alpha which is negative.
And I'm going to give it the same amount of time.
So, we start with the same original shape that you are used
to, now, which is that crispy shape.
And now if you look very closely at the red one,
look very closely. I will keep pointing at the top
of the red one. You see it's moving in the
opposite direction. You see that?
It's moving back.
So the group velocity is now moving in the opposite direction
as the phase velocity. And the shape is changing.
Look. Look how fast the shape is
changing, much faster even than in the first case.
So, we are now in our omega K diagram.
We are now here. So, we have a linear portion
there, what it looks like linear here.
But it is, of course, part of this curve.
Look how dispersive this is. So, the change of the envelope
is the result of dispersion, and the fact that the group
velocity is opposite the direction of the phase velocity.
Boy, this whole envelope is falling apart,
right? It's totally falling apart.
And, that's only due to the fact that these six waves no
longer have the same phase velocity.
And so, we are going to look at the omega K plane when this is
coming to a halt. I expect it's very closely,
now, hey, boy, yeah, there we go.
So, here's your omega K plane. Again, it's a little section
here. You can see it's slightly
curved. But, even if it looks like a
straight line, it's because of this curve.
So, that's where [UNINTELLIGIBLE] very high value
for K that he picked. And it was the negative value
for alpha, then, that gives you that result.
So, as we just saw, now, the fact that there is
dispersion changes the shape if you have the superposition of
many different waves. Now, suppose you plucked a
string, or as we did during the last lecture,
you have a square wave on top of a string.
And you do a Fourier analysis of this square wave.
And you say to the Fourier analysis, OK,
show me what you're going to do if we look in time,
that each one of those Poirier components is going to oscillate
with its own frequency. And the cause that was
nondispersive, we always use that omega N was
N times omega one. And therefore,
if we waited one full period of the first harmonic,
the shape, whatever it was, was back to what it was before.
It never changed because after one full oscillation of Nancy
one, Nancy two has made two oscillations.
Nancy three had made three oscillations.
So, you're exactly back where you were.
That is no longer the case when you have dispersion because now
one wave takes a little longer or a little shorter than the
other wave. So, they no longer meet up with
each other. Another way of putting it is
that the traveling waves, of course, standing waves are
the result of traveling waves, the traveling waves have
different velocities because of the dispersion.
And that, now, will make any sharp features in
a pulse disappear because the sharp features,
if you have a rectangular pulse, the sharp features
require a high frequency to make that steep.
And there will be a large difference there between the
phase velocity of the high frequencies and the low
frequencies. And so, you're going to see
your sharp pulse fall apart. And that's what I'm going to
demonstrate to you. This toy model will be used to
do the Fourier analysis of a square wave.
I will first show you the square way of the way I showed
it last Tuesday when there was no dispersion.
You will see the evolution in time, and I waited half the
period of the fundamental when the square wave came out upside
down. I will show you that to remind
you of what that was like. And then I will add,
all right. I think I have to find that
program moving triangle. OK, this is the one.
This is it, moving square at the center.
So, are we ready for this? So, this is what I showed you
last Tuesday. The square is half the length
of the string. We Fourier analyze it.
All these Fourier components are going to oscillate in their
own way, which is like saying we have two pulses moving in the
opposite direction. And they are reflecting,
and they come back. I have 100 terms here this
time. Last time I only had 50 terms.
I have 100. And only the odd ones
contribute, remember? So, we effectively have 50
terms: Mary one, Mary three, Mary five,
all the way to Mary 99. And so, look.
This is one half period of the fundamental.
And your shape is back, well, upside down.
But that's because I only ran it half a period.
The shape is back from where it was.
And the reason is, it is a nondispersive medium.
And so, the reason is that the Fourier components oscillate in
this fashion. But now, I'm going to use [one
Bolex?] toy model. And, I am going to do in the
these are the Fourier components, which are not so
important for you right now. And so, now I'm going to add
dispersion. And the dispersion that I'm
adding is alpha is 0.01, one hundredth,
and we do the same pulse. We get exactly the same Fourier
analysis. The blue curves are identical,
except they oscillate no longer according to omega N equals N
omega one. But, they now oscillate
according to our new toy relation.
And now look what's going to happen.
So, the higher frequencies, now, the traveling speed of the
higher frequencies is different from the traveling speed of the
lower frequencies. And if now you wait one half.
Of the fundamental, you are going to see when the
thing is upside down, it really doesn't look like a
square anymore. You can sort of still recognize
it a little. The Fourier components,
the blue components, have not changed.
They are exactly the same as they were.
And this is now what you have. So, that is the consequence of
dispersion. So, I think this is a great
moment for a break. We will reconvene in five
minutes. [SOUND OFF/THEN ON] So I
summarize for you here, which is really something
you'll have to remember. And there is really not much
more to it than that, that the phase velocity is
omega divided by K, and that the group velocity is
D omega DK. And so, there is no such thing
as one phase velocity for a nondispersive medium.
There is no such thing as one group velocity.
It depends on the curve omega versus K.
So, you have to know what we call the dispersion relation in
order to say what the phase velocity is for what frequency.
And, dispersion is extremely common and physics.
If we take deep water waves, and I would imagine that what
is meant by that is that the water is deeper than the length
of your wavelength. That is probably the
operational definition. It's very dispersive,
the dispersion relation, which I will not derive.
I didn't derive it either for the string.
Omega squared equals GK. G is just the gravitational
acceleration plus S divided by rho times K to the power three.
And, S is the surface tension, which is for water about 0.72 N
per meter. I don't know whether you've
dealt with surface tension in 8.01.
But if not, you just can accept this S.
And, the density of water, rho, is about 1,000 kg per
cubic meter. That, of course,
is a little different for seawater, but that's the nice
number sort of to use. Now, this S term is only
important for a very short wavelength.
If you take the case that lambda is much,
much larger than one centimeter, then I can show you
that the S term is unimportant. So, let's take lambda equals
one meter. If lambda is one meter,
and K is 2 pi divided by lambda, so you can calculate
what GK is, it's about 62 in SI units because G is 9.8.
And, you can also calculate now S divided by rho times K to the
third. Remember, K is 2 pi divided by
lambda. And this, now,
is 0.02. So, it's insignificant.
It's very small. And so, for this case,
you might as well, for first order approximation
forget that term. And so, you can write down now
that omega divided by K, which is the phase velocity,
in this case, comes out to be 1.25 m per
second. And what you also find,
and I will show you that very shortly, but the group velocity
is half the phase velocity. And so it's half this value.
And if you want to know what frequency, the water is hopping
up and down. The frequency,
of course, is the phase velocity divided by lambda.
So, that is about 1.25 Hz. So, it's a little faster than
one oscillation per second.
If we have the case that lambda is much larger than 1 cm,
I will continue here. So, lambda is much,
much larger than 1 cm. I can use the dispersion
relation, and I can calculate now what the phase velocity and
the group velocity is in these general terms.
So, I know now that omega, to a high degree of
approximation, is the square root of GK
because I forget the S term. And so, omega divided by K,
which is the phase velocity, is there for the square root of
G divided by K. Do we agree?
And, notice that this is, therefore, proportional to the
square root of the wavelength because one over K is
proportional with lambda. So, what it means is that the
larger the wavelength, the higher the phase velocity.
If lambda is larger, K is smaller,
the higher the phase velocity. What now is the group velocity?
Well, the group velocity is the omega DK.
So, you take this one and you take the derivative of the
square root of G times K. So, you first get your one
half. And then, you get your square
root of G, but now we have to divide by the square root of K.
So, you get times the square root of G over K.
But, the square root of G over K was the phase velocity.
So, it's half the phase velocity.
And, that's what I already anticipated here.
So, for water waves, which are much larger than 1
cm, you see that the higher the velocity, the larger the
wavelength. And you also see that the group
velocity is half the phase velocity.
So, let's now take a situation that lambda is,
say, much smaller than 1 cm. It doesn't even have to be all
that much smaller, but just smaller than 1 cm.
Then, it's the second term that completely dominates.
So, now you get that omega squared, to a very good
approximation is S divided by rho times K to the third.
So, now, the phase velocity, which is omega divided by K is
now the square root of S divided by rho times K,
which now is proportional to the square root of one over
lambda. So, now you have to reverse.
So now you see that the larger the wavelength,
the smaller the phase velocity. None of this,
of course, is intuitive. You have no feeling for that.
I have no feeling for that. What I'm asking you in problem
set number six is to prove that in this case,
the group velocity is 1 1/2 times the phase velocity,
not at all obvious, but that's what will come out.
Here, the group velocity was half the phase velocity.
And, the phase velocity was larger for larger wavelengths.
Here, it is reversed. And the group velocity is 1 1/2
times the phase velocity. Shallow water waves,
which is also part of your problem set, are nondispersive,
not so intuitive. So, if you have very shallow
water waves, which means, then, that the amount of water
that you have, the height of the water is
substantially lower than the wavelengths.
Then, you get a nondispersive relation, which is not at all
obvious either of course. Sound in this room is
nondispersive. Thank goodness.
Imagine for a minute that the high-frequency would travel with
substantially higher speed than my low frequencies.
Then only the people in front here knew what I was saying.
And the one in back had no clue because the higher frequencies
might arrive much earlier than the lower frequencies.
And you could not make sense of what I'm saying.
Imagine you go to a concert. If sound were very dispersive,
then you would have to pay a very high price for the tickets
in the front row. And, the tickets in the
balcony, they would give them away for free because you would
have no clue what you are listening at anyhow because the
different frequencies of the violins and of the bass would
reach you at different times. So, we are fortunate that
sounds in this lecture hall is nondispersive.
The next lecture, we will enter the domain of
electromagnetic waves. In other words,
you're going to love it. You're going to see Maxwell's
equations. And I know you love Maxwell's
equations. You remember 8.02,
the good old days. And we will derive using
Maxwell's equations that there is such a thing as
electromagnetic waves, and that they move in a vacuum
with a speed which we call [C?] in physics, which is 300,000 km
per second. That can actually be derived.
So, in free space, that means in vacuum,
electromagnetic waves are nondispersive.
Omega equals C times K. That's a nondispersive
relationship. So, the frequency of
electromagnetic waves is C, the speed of light,
divided by lambda. And so we will encounter with
8.03 radio waves. And we will actually produce
them. And let's say we take a
frequency of about 1 MHz, ten to the sixth hertz,
that would give you a wavelength of about 300 m.
You can use this equation. If we go to radar,
which we have right here in front of us, the frequency is
about 10 GHz, ten to the tenth hertz,
and so you have a wavelength of about 3 cm.
And if you go to infrared radiation, you have a wavelength
of tens of microns. And if you go to visible light,
you get frequencies up to five times ten to the 14th hertz.
And, you get a wavelength of about half a micron.
And when you go to ultraviolet and X-rays and gamma rays,
the frequencies go higher and higher.
And the wavelengths get shorter and shorter.
Nondispersive [in vacuum?], they all travel with exactly
the same speed. Light in glass,
or in water, for that matter,
is not nondispersive. It's very dispersive.
As we will derive in 8.03, the speed of electromagnetic
radiation is C divided by the square root of the dielectric
constant, kappa E, and the relative permeability,
kappa N, which you may or may not recognize from 8.02 days.
And in vacuum, this is one,
and this is one. But that's not the case for
water, and that's not the case for glass.
And in fact, this kappa E is strongly
frequency dependent. So if I take water as an
example for its kappa M is very close to one because water is
not ferromagnetic material, and all materials that are not
ferromagnetic have a kappa M or relative permeability very close
to 1.0000. So, forget kappa M now.
But I'll show you the enormous frequency dependency of kappa E
of the dielectric constant. It's huge for water.
If you look in the frequency range, F is about zero to ten to
the tenth hertz. So, that is our 10th gigahertz.
That is radar. Kappa E is about 78.
So, that means the speed of radar in water is one ninth,
roughly, of C. So, it's only 10%,
roughly. It's ten times slower.
If we go to visible light, so this is the rate,
and this is, of course, an even lower
frequency that goes to the radio.
If you look at visible light, you may say that light by
definition has to be visible. Well, we physicists make a
distinction between light in general, which is all
electromagnetic radiation, and a certain part of the
spectrum that we can see. And that, we call the visible
part. So that's visible light,
which has a frequency of very roughly five times ten to the
14th hertz. It's huge.
And, kappa E there is roughly 1.77.
Look at the enormous difference.
At that very high frequency, they make the electric dipole
of water cannot follow the changing electric fields.
And therefore, it just sits there and it has
almost no impact. And that's why the number is so
low. So, that means that in water,
visible light has a speed which is about 30% lower than C.
That is dispersion. And, later in the course,
we will see how a prison made of water or made of glass makes
it possible to decompose white lights into colors.
That is only possible cause of dispersion.
It is only possible because the speed of light,
for blue light, is different from the speed of
light for red light.
We will also explore in 8.03 the transports of
electromagnetic waves through metal pipes.
We call them waveguides. Here, I have a metal pipe.
The opening here is about 1 1/2 centimeters across.
Light has no difficulties getting through.
I can look at you and I can see you.
And I can see you. You may be able to see me.
I don't know. I have no problems.
Lights has no difficulty getting through here.
But, radar, with a wavelength of 3 cm would have great
difficulties getting through here.
I will derive in the future what happens with radar when you
send it through metal plates. You will see a dramatic example
of dispersion. And this is the result of the
boundary conditions of the electromagnetic waves with the
pipe. But I want to tell you the
results because it is such a remarkable example of
dispersion. And I want to demonstrate it to
you because it's very relevant in this lecture.
Because it all comes down to dispersion.
We have there two aluminum metal plates that you see there.
Here is one, and here is the other.
Let's call this the direction of Z.
And, let's say they are separation A apart.
So, that's the separation between them.
We have a 10 GHz transmitter, which is a radar transmitter
here. And here we have a receiver,
which can receive the 10 GHz. And so, we're going to send it
from here to the other end. The wavelengths,
lambda, is 3 cm. So, this is ten to the tenth
hertz. So, lambda is 3 cm.
When A is less than 1.5 cm, radar cannot go through
anymore. It's larger than 1.5 cm.
It can, and I will demonstrate that.
[UNINTELLIGIBLE] behind that is, and that's really what I
want to get across is to dispersion relation which I will
derive, but not now. But I will derive it a few
weeks from now when we reach that point.
But I want to show you what that dispersion relationship
will be if we have done our homework: omega N,
whereby N equals one, two, three, four,
five equals C times the square root of N pi divided by A
squared plus K of Z squared. Let's first look at this term
even though you don't see directly where it comes from.
You can connect to it already. It has to do with boundary
conditions. Remember, when we had a string
that is fixed at both ends, we got terms like this.
Remember, we had sound in a box.
We had N pi divided by L in the X direction, and pi divided by L
in the Y direction. So, you recognize here that it
has to do with the boundary condition of the electromagnetic
radiation without exactly understanding why it is the way
it is. And so, the phase velocity,
V phase, in the Z direction, is omega divided by KZ.
And, the group velocity in the Z direction is D omega DKZ.
I will concentrate now only on the case that N equals one.
So I make this a one because that gives me the lowest
frequency possible in this dispersion relation.
And I'm now going to graph that thing, omega versus K.
This is nondispersion. [SOUND OFF/THEN ON]
I can write here omega equals KZ times C.
That is the case when K is huge, very short wavelength,
way shorter than the dimension of the [pi?].
Then, that system acts like a nondispersive media.
You're somewhere here. But, if you go to the low
frequencies, this is the curve. And what you see here,
there is a value which we call the cutoff frequency omega.
It's normally called omega C. Below it, no radiation can go
through the pipe. This is not a solution anymore
to my wave equation. You are stuck to this line for
the mode N equals one.
So, what is required: that the frequency for this
radiation, well, first let me give you what
omega minimum is. Omega minimum is when you make
in the extreme case KZ zero, and N equals one.
So, you get C pi divided by A. That is the minimum value for
omega that can propagate through these two plates.
And so, a required frequency is that F is larger than C divided
by 2A. That means A has to be larger
than 1.5 cm. And, that is so nonintuitive.
And this is what I want to demonstrate to you.
So we have there those two plates.
And the plates are now something like 2 cm apart.
You can check this after the lecture.
We modulate the 10 GHz with a nasty triangular sound of 550
Hz. And the reason why that is
nasty: because you now understand that anything that is
modulated triangular requires high harmonics because think in
terms of Fourier space, and higher harmonics,
not so nice in your ears. So, if this were a nice
sinusoid, you would hear a beautiful 550 Hz tone.
But because we modulated with a triangle, it's a nasty tone.
But the only reason why we wanted you to hear something is
that when this receiver receives the radar signal,
we want to hear this modulated signal.
Unfortunately, you and I cannot hear 10 GHz.
Our ears are not designed to follow a frequency of
10,000,000,000 Hz. It cuts off at 20 kHz,
as you remember. So, we are going to show you
this transmitter first. The transmitter is the signal
upstairs. So, it is 10 to your hertz,
but it is a triangular modulation which is 550 Hz.
Here's the receiver. When I turn on the receiver,
you'll hear that signal coming out here.
[SOUND PLAYS] And at the bottom, you see this signal
received by the receiver. And the fact that it is a
little larger than the upstairs is simply a matter of how we
adjust our amplifiers. Nicole, can you be my witness
and look here so that you can tell the class what I'm doing?
I'm going to take these two plates now.
Please come up here and just watch here.
And tell them that what I'm doing is completely honest.
You see the opening here now of these two plates?
And these plates are about 1.8 cm apart, almost two.
And now, I'm going to make it a little less than 1 1/2 cm.
Watch there and listen. [SOUND PLAYS]
It's gone. It's completely gone,
and you tell the class that the opening is still sizable,
right? The opening is just a little
less than 1 1/2 cm. And it instantaneously goes
away. [SOUND PLAYS] There it is.
Why don't you squeeze it, Nicole?
And it's gone. [SOUND PLAYS] Thank you,
Nicole. You are an expert.
Now, there's one more thing that is important.
And that has to do with your sleepless nights because,
you know, I'm very concerned about you getting enough sleep.
Look at that dispersion relation.
Look at this curve. This line indicates the speed
of light, C. That line indicates that when
omega divided by K is C, that's the speed of light.
So, what now is the phase velocity, for instance,
at this frequency so that it has this value for K?
Well, it's this. This flow is larger than this
one. It's higher than the speed of
light. Not only is it higher than the
speed of light, the phase velocity,
that by the time I reach this point here, the phase velocity
goes to infinity because look at the slope.
It's 90°. So, we now have a case that the
phase velocity is larger than C. No.
The phase velocity can be much, much, much, much,
much larger than C approaching infinity.
And what now is the group velocity when we reach this
point of the cutoff frequency? That is the tangent.
And so, this is the slope of the group velocity.
So, the group velocity goes to zero in that extreme case.
I sure as hell hope that you can sleep tonight. 1132 1:18:54,186 See you next Tuesday.