Today we are going to talk about wave guides and about
resonance cavities. Last Thursday I discussed the
boundary conditions of electromagnetic waves on the
surface of an idea conductor in vacuum.
And today we will see some of the amazing consequences.
And I will return to something that I have discussed with you
before but never fully explained, and that is the setup
whereby we have met a metal plate.
This is my coordinate system, x, y z.
This is the z direction. And then I have here another
plate, two parallel plates, and this will be the x
direction. I call this x equals zero.
And this x, zero, a.
And this then is the y direction.
That is the setup. You see it there.
I am going to try to send electromagnetic radiation
through this gap. And what holds always,
so what now current holds always, that k is always kx x
roof plus ky y roof plus kz times z roof.
K is the direction then of propagation.
Also lambda equals 2 pi divided by that k.
And the magnitude of that k is the square root of kx2 plus ky2
plus kz2. And omega equals kc if this is
in vacuum. I am going to send through this
gap linearly polarized radiation in the y direction.
That is a choice that I make. I do not have to do that,
but that is a choice that I make.
And think about the y direction and the z direction,
very, very large, infinitely large.
Or, in practice, many, many, many times the
wavelength of the radiation. To meet the boundary
conditions, the electric field, which is only in the y
direction, must becomes zero here and must become zero there
because the electric field cannot be in this plane.
Remember E of t, which was the tangential
component of the surface of the conductor, must be zero.
E of y must vanish here and it must vanish there.
And so for this case whereby we only have radiation linearly
polarized in the y direction, so E of y must become zero for
x equals a and for x equals zero.
Since there is no dependence of the E field in the y direction,
k of y will be zero. And you will see that come back
many times in this problem. Now I would like to take a look
at this geometry from above looking down on the xz plane.
Here is the z direction. And this is the x direction.
And the y direction is coming straight out of the blackboard
and pointing in your direction. I am looking down on that plane
from above. Let the k vector be this.
This is a vector. This k vector has a component
in the x direction, which is k of x.
And it has a component in the z direction, which is k of z.
But it does not have a component in the y direction.
K of y is zero.
Think of them as water waves, as far as I am concerned.
This is the direction of propagation in the gap.
The water waves would be perpendicular to the direction
of propagation which are the planes with constant E fields.
I am going to put on in here. This is 90 degrees.
And I am going to put another one in here.
And so this is now propagating with speed c in this direction.
And the separation between these two crests is lambda.
That is my definition of lambda.
Think of this surface, which is a surface
perpendicular to the blackboard, as having a constant phase of
the wave. It means then,
for instance, that the E vector is pointing
in your direction and has reached a maximum.
It would also have reached a maximum here then.
We go through one complete cycle, 2pi radiance from here to
here. That is the definition of
wavelength. Now, in one period that this
wave moves from here to here, in that same amount of time
this wave here moves from here to there.
And so I will give this here a name, as I have done before,
and I call that L of z. And so it is immediately
obvious, by the definition of phase velocity,
that the phase velocity in the z direction must be Lz divided
by lambda times c. Because it goes from here to
here with the velocity z. But here it goes much further
in the same amount of time, so now that must be times c
knot. So it is larger than c.
And this, by the way, and that follows home the
geometry, is also k divided by k of z times c.
And that is larger than c. The phase velocity is larger
than c, and I have discussed that with you at length before.
Now, look at what this wave is doing.
These waves hit the wall and they then reflect off the wall.
I will try to make you see that.
They reflect off the wall, they come back this way and
then they reflect off the wall again.
And so these waves which zigzag slowly work their way through
the gap in the z direction. And you can immediately see,
therefore, that the speed with which they work their way
through the gap in the z direction must be less than c.
Because when they go from this point A to this point D,
in that same amount of time in the z direction they only go
from B to D. And so the group velocity in
the z direction is then BD divided by AD times c.
And that follows from this geometry.
That equals k of z divided by k times c, and that is less than
c. And it better be less than c.
Because you can just see how difficult it is for this wave
that reflects to work its way through in the z direction.
We discussed earlier, and I explained that using the
analogy with water waves hitting the shore, that a phase velocity
larger than c has really no meaning.
It is a geometrical effect of the consequences that the water
slams on the wall everywhere at the same moment in time.
However, the group velocity -- What is that?
What did I do wrong? Excuse me?
Thank you very much. This is larger than zero.
Boy, it is not my day. Is this what you want?
Thank you. All right.
Notice that in this specific case, the product of group
velocity and phase velocity happens to be c squared.
It is a special case. It is not always the case.
But what is important that you recognize, and this picture is
so nice, you can really digest that picture and relate to it,
that you immediately recognize that the phase velocity,
which is the speed of this distance, is larger than c and
that the group velocity is smaller than c.
Now we can do the math, and I can write down the wave
equation dell square E. There is only a component in
the y direction because that is the radiation I have chosen.
I get D2Ey/Dx2 plus D2Ey/Dy2 plus D2Ey/Dz2 equals epsilon
zero mu zero times D2Ey/Dt2. And this term does not exist
because there is no dependence of the E vector in the y
direction. The solution for the wave that
goes from here to here immediately presents itself,
because we have a lot of experience now,
as a standing wave in the x direction and is going to be a
traveling wave in the z direction.
11:44
And so you see here, it is staring you in the face
because I knew, of course, the answer that this
is a traveling wave and that this is a standing wave.
And if I use that solution, which has to be correct,
and I put it in that equation, which I can,
it will meet the three-dimensional wave equation,
you substitute that in here, you will find something that,
of course, you also know already, namely that omega
squared, which is k2 times c2. But k now only has a kx and has
a kz, and so you get this kx2 plus kz2 times c2.
That is what you will find if you substitute that in the wave
equation. And you will find that c2
equals one over epsilon zero mu zero.
Now the boundary conditions, which we haven't met yet,
require that the vector, which is only in the y
direction, and once more to remind you, it is linearly
polarized in the y direction, now must vanish when x equals
zero and when x equals a. And so that now means that kx
must now be n times pi divided by a.
And n then can be one or two or three or four.
And, if we look in a different direction, I make another
cross-section, this is y and this now is x,
so this is zero and this is a. And the z direction is now
coming straight out of the blackboard.
Then for n equals one kx is going to be pi over a,
and that is this wave. That is the sinusoid.
And so at a particular moment in time the E vectors would be
like this. And you see they vanish here
and they vanish here. There can be no tangential
component in this plane here, and there can be no tangential
component in that plane here. And so this is then n equals
one. And, of course,
it oscillates like this. It is a standard wave.
For n equals two there is another possibility,
kx would be 2pi divided by a. And I can put that in a
different color. That would be like this.
And so now the E vector has right here a nodal plane.
And now it goes like this. And so those are possibilities
which are only restrictions in the x direction.
We can write down now the k vector.
The k vector now is kx times x, but kx is n pi divided by a in
the x direction. And then I have a kz in the z
direction. And there is no ky.
Ky is zero. And omega, which is kc,
is therefore c times the square root of n pi over a squared plus
kz squared. And this equation is one that I
had on the blackboard earlier before we understood why it was
that way when I discussed with you dispersion.
This is called the dispersion relation.
This, by the way, under the square root is simply
k. And so now you can ask yourself
what is now the phase velocity in the z direction.
We already know what it is. It must be that.
We already reasoned that purely from geometry,
but it must come out now, too.
Well, the phase velocity in the z direction is omega divided by
kz. 17:29
Now you can also calculate what the group velocity is.
The group velocity in the z direction is d omega dkz.
And, since you have this term here, you should be able to do
that. That is an 18.01 problem.
If you take d omega dkz, you will find now that that
becomes, and I will leave you with that, that becomes kz
divided by k times c. And that is always less than c.
This is what we also knew already.
I had already predicted that based on geometry arguments.
That comes out now if you use this dispersion relation.
And this dispersion relation then has the absurdity that if
kz goes to zero that the phase velocity goes to infinity but
that the group velocity goes to zero.
And so that now I want to address further.
And the best way that we do that is to make what we call an
omega kz diagram. I am going to plot for you this
curve. It is extremely useful to
always think of this as an omega kz diagram.
Here is kz and here is omega. And this straight line,
which would be a non-dispersive medium, would be omega is kz
times c. But that is not what we have.
What we have is for n equals one there is a particular
frequency omega c, we call that the cutoff
frequency, below which no radiation can go through the
gap. And that cutoff frequency is
when n equals one and when kz becomes zero.
And so that is c times pi over a.
And this curve then, for n equals one,
goes like this. This is n equals one.
If I radiate electromagnetic radiation linearly polarized in
this direction and I know the frequency.
Let us say this is the frequency that I radiate,
that is a given, then radiation will go through
beautifully. And this will then be the value
of kz. The value of kx is
nonnegotiable. That value for kx must be the
value that I had earlier. Where is it?
That must be pi divided by a. Let me write that down again.
For n equals one k of x must be pi divided by a.
But it is kz that adjusts itself so that it meets the
boundary conditions. Kx does not adjust itself.
Kx has no choice. Kx must meet the boundary
conditions so that the E vector vanishes at x equals zero and x
equals a. And so any solution must lie on
this curve. And so it is kz that pays the
price and that adjusts itself. Now, keep in mind that all
radiation with frequency above this value here can go through
this gap. There is often a
misunderstanding among students. They think that there are
resonance frequencies. There are no resonance
frequencies. Any frequency above that value
can go through the gap. You can immediately see here
that the phase velocity is larger than c because the phase
velocity is omega divided by kz. And so when I draw this line
then this slope here, omega divided by kz,
is larger than this slope. And that corresponded to a
speed c. And so you see,
since this slope is steeper, the phase velocity is larger
than c. And you see that when you reach
the situation that kz becomes zero that then you have this.
22:30
I now want to write down for you something that may help you
later if you want to understand your notes.
I will put this back up again. I am going to do the following
experience now with you in my head.
I start with a certain frequency omega,
and I am going to lower that frequency.
I do it in my head. It is a Duncan experiment.
I am going to write down step by step what is going to happen.
This point will slowly go down. And so we will come down this
line here. And finally we reach here.
I probably can write it here, it is nicer for you because
then you have it all together. I start with omega,
which is larger than omega c. That is how I start.
It is a given. And now I am going to lower
omega. The next thing I am going to do
is to lower omega. When you lower omega kx cannot
change. It must remain pi over a
because it must meet the boundary conditions.
Therefore, kz gets smaller. You are going down this line
and the kz gets smaller. Omega is kz.
Omega is always k times z. Clearly, if omega comes down k
must come down. And that means kz comes down in
this case because kx is not going to give up.
And so k gets smaller. And if k gets smaller then,
of course, lambda increases because lambda is 2pi over k.
Until disaster strikes omega becomes omega c.
And when you reach that point kz has become zero and kx is
still pi over a. And so now k itself is kx.
And for lambda now is 2pi divided by that value kx.
And that means 2pi divided by kx equals 2a.
It means if I have a given frequency, which we do have here
that I cannot change, I have 10 gigahertz,
and 10 gigahertz has a wavelength of three centimeters
in vacuum. It means the moment that I make
a any smaller than 1.5 centimeters there is no
propagation anymore in that direction, because now I can no
longer meet the boundary conditions for kx.
And I will demonstrate that, as I did before.
This is not the first time you will see that.
Before I do that, I want to discuss with you the
meaning now of n equals two. What does that mean now?
In other words, we had here the n equals one.
And we also have n equals two. And so I am going to make a new
plot because it becomes otherwise too cluttered.
I will leave this on. Make sure you get that
sequence. That is very important,
and I will put it up later again.
I am going to make a new plot. Otherwise, it becomes too
confusing, too many lines. kz.
Omega. Here is your line known
dispersion. Omega is c times k of c.
And here now is your n equals one, omega c equals pi divided
by a times c. And here, exactly twice as
high, is n equals two. Whereby, another omega c is 2pi
times c divided by a. And so I have one line here and
I have another line here. This is now the plot of this
equation for n equals one, and independently for n equals
two. This is n equals one and this
is n equals two. Imagine now that this is our
omega. That is a given.
That is the one that we happen to have.
This frequency can obviously go through the gap.
But now it has a choice out of two possibilities.
It can either go through this way with this value for kz,
which corresponds to n equals two.
That is the mote in green there.
Or, it can go through in this mote with this value for kz for
n equals one. Or, it can do both.
It can do any linear combination between these two,
because each one is a solution to Maxwell's equations.
Each one obeys the boundary condition and each one meets the
wave equation. And so there are now two ways
that it can go through. If it decides to choose this
route then kx is a given, is pi over a.
If it decides to go this way then kx is 2pi over a.
The only difference between these two is that kx is
different. K is not different because I
have the same omega. It is the same omega,
it has the same k, the same wavelength,
but this one has a different k of x, therefore,
a different k of z. And you see that.
And this one has a different k of x, therefore,
a different value for k of z. I cannot be more specific.
Before I want to demonstrate to you, which is a repeat,
that if I have here electromagnetic radiation,
three centimeter waves polarized in this direction and
I am going to send it to this receiver, which can receive it,
before I am going to demonstrate that when I make the
gap smaller than 1.5 centimeters that it will all of a sudden
disappear, the radiation. Before I do that,
I want to ask you a question which is important so can test
whether you understand this concept.
My radiation is linearly polarized in this direction.
Out of that came all this misery.
Is there a way that I can send electromagnetic radiation
through that gap when the gap spacing for my transmitter is
smaller than 1.5 centimeters? I know I cannot do it when it
is polarized in this direction. It will immediate disappear.
You will hear that. Is there a way that I can still
get radiation through when the opening of the gap is smaller
than 1.5 centimeters? I realize that it may take you
a little bit of time to think of it.
And I helped you by saying all this misery is the consequence
of the fact that I chose the direction of linear polarization
in the y direction.
Exactly. I will do that,
believe me. If I rotate my transmitter by
90 degrees and I send in radiation whereby the E vector
is linearly polarized in the x direction there is no misery at
all because all boundary conditions will always be met by
Mother Nature. Because all that Mother Nature
has to do is to make sure that the normal component at the
conductor is sigma S divided epsilon zero.
All it has to do is rearrange the charges on the surface,
but nothing has ever to go to zero.
The moment that I make the radiation polarized in this
direction, there is no dispersion.
The phase velocity is c. The group velocity is c,
and this is the line that holds now.
It is only the component in the y direction that gives you this
misery. That is very interesting,
isn't it? Because what that means is that
if the radiation that you send in had a component both in the y
and in the x direction, I can do that.
I can send radiation through that is linearly polarized like
this. Then something unbelievable is
going to happen. The component in the x
direction will go straight through with the speed of light.
And the component in the y direction has to suffer like
this. And so they completely
decouple. Your x direction is
nondispersive, group velocity is c,
phase velocity is c, and it is the component in the
y direction that decouples. And that can go so slowly that
it can even have a group velocity of zero.
When I show you that I can change the size of that gap,
what that means in terms of this line is the following.
33:05 Because, when I make a smaller,
it is this point that omega c goes up.
And so I make omega c go up until it hits my 10 to the 10
hertz. In our thinking here,
we lowered omega. And I showed you what happens.
Now, to do the experiment, I lower a, and I move this line
up until I hit that point. I think we are ready for that
now. You will be able to hear it.
I would like someone to witness it.
Could you come and witness it? Because I want you to see that
the separation between the gap is now two centimeters.
You stand a little bit on the side here and no in the beam.
Look here. What is your name?
Chrystally, wonderful name. My name is Walter,
by the way. [LAUGHTER]
This is 2.0 centimeters. You see that?
This is a transmitter. It is polarized like this.
Here is the receiver which receives the polarized radiation
like this. We modulate it with 550 hertz
so that you can hear it. You cannot hear 10 to the 10
hertz. That is not the way nature
designed you. Nor can I hear it.
This is 550 hertz. The gap is 2.0 centimeters.
I am going to make the gap smaller with my hand.
And you watch it when it gets to 1.5 centimeters.
It is gone. All of a sudden it disappears.
That is the consequence of the fact that this point hits here
now. But now it is 1.5 centimeters
apart. I can now rotate this 90
degrees and rotate this 90 degrees so that now we have the
E vector in this direction. There are no longer any
restrictions. And the radiation goes through
because now I have this red stuff.
And now I can make the gap smaller.
And what do you think when I make the gap smaller and smaller
and smaller? Well, very slowly you will hear
nothing. Because, if the gap has no
cross-sectional opening anymore, no radiation can go through.
Don't expect miracles. If I now make the gap smaller
and smaller and smaller, nature has no problem sending
radiation through, but nature is stuck through the
cross-sectional opening. That is, of course,
you make this opening narrower and narrower and narrow then
less and less and less radiation will go through.
And so I will now make the gap smaller.
She is my witness. I will put this on top.
It is now about 1.5 centimeters.
I will make it smaller. It is now about 1.0 centimeter.
Still going well. About 0.7 centimeters.
About 0.5 centimeter. About 3.0 millimeters.
And obviously we slowly lose intensity.
Thank you very much for assisting me.
My pleasure, actually.
And so you see how bizarre it is that the behavior is so
fundamentally different for these two different modes.
And then the idea of the decoupling, if you oscillate in
a direction that has an x and a y component that the radiation
decouples. I think this is a nice natural
moment to have our break. And I was just wondering why
the attendance was so high. But then I was reminded of the
fact that it is Tuesday. I am very flattered that you
all came. And we will hand them out.
And then we will not start yet until the whistle blows.
I will raise all the blackboards so that you can see
everything. Can you help handing this out?
Can you also help? Could you also hand this out
and make sure everyone has one?
Now I am going to change the situation very dramatically.
And now I am going to give you a closed box conductor.
There is a conductor. Our coordinate system is the
same, x, y, z. And I simply call this a,
I call this b and I call this c.
That is the length. And that is the coordinate
system. And now comes the question,
if I now have electromagnetic radiation in this box,
what is going to happen? Now there is no way that it can
get in or out if it is in there. Clearly, you now have a
situation that you have a cavity with very discrete resonances.
It is totally different from what we did before.
You only get standing waves with very discrete resonance
frequencies. Well, I will write down for
you, because I want to ram it down your throat,
but I could have left it here, k equals kx times x plus ky
times y plus kz times z. K equals the square root kx2
plus ky2 plus kz2. Omega equals k times c.
Lambda is 2pi divided by k. None of this is negotiable.
Now, in this box, I can have an E vector which
has an x, y and z component. I have an E vector which is
something in the x direction plus something in the y
direction plus something in the z direction.
And our task is now to find the resonance frequency for this
cavity. Let us only look for now at the
component of the E vector in the x direction.
Everything that follows now is only true for the x direction.
E of x, which is the component of the E field indirect
direction, must be zero. E of x is this component in the
x direction. It must be zero at y equals
zero and at y equals b. If it were not zero there would
be a tangential electric field in that top plate.
It must be zero when y equals zero and also when y equals b.
But it must also be zero when z is zero and when z is c.
Because, if the E vector is like this, it must be zero not
only on the bottom plate and on the top plate,
but it must also be zero in the front plate which is this plate
and in the back plate because it cannot have any E vector in that
surface either. The E of x alone must meet now
four special boundary conditions.
41:35 And, to remind you that it is
in this direction, I put the x there.
What you have now are two standing waves,
one standing wave in the y direction and one standing wave
in the z direction. And, to meet the boundary
conditions which I wrote down here, it is immediately obvious
that k of y must now be m pi divided by b.
And k of z must be n pi divided by c.
And then m equals, in this case it is sine,
so it is 1, 2, 3, etc.
And n equals 1, 2, 3, etc.
And, as long as I meet this condition, the E of x component
is quite happy. It means that k now only has a
component in the z and in the y direction.
K of x is zero for this component.
And so we get k of y times y plus k of z times z.
The wave equation dell squared E, if I apply the wave equation
now only to the x component -- 43:32
And this term is not there for that x component.
But the other terms are there. And if I substitute my
solution, which I know deep in my belly has to be correct,
if I substitute that in this wave equation then what you
find, of course, is no surprise that omega
squared equals k2 times c2. And so that is going to be ky2
plus kz2 times c2. But my ky and my kz are
quantized because they are now only allowed for certain values.
And so I can write then now that omega squared is then c
squared time m pi over b squared plus n pi over c squared.
B is for the y direction and c is for the z direction.
What is the problem? Only one person,
please.
The z direction is c, the x direction is a and the y
direction is b.
B is in the y direction. We have a boundary condition in
the y direction and a boundary condition in the z direction.
Why don't you come here and just make the change because I
missed something. [LAUGHTER]
Oh, I leave it c. I cannot change that.
Because, if I change that, then it becomes a terrible
chain reaction. That is c.
Oh, I am very sorry that I called that c.
Thank you very much. Can I get my chalk back,
by the way? [LAUGHTER] We agree there was
no mistake on the blackboard, but there is confusion about c.
My apologies. If I had called it d that it is
another problem. Dx, dy and then you have
another d. Indeed I have followed this
convention. This was a c.
This is a b. You see the same c here?
That is that same c. Z equals c.
It has nothing to do with the speed of light.
It is one of those things that happen.
M stands for Mary and n stands for Nancy.
It doesn't stand for normal this time.
It stands for Nancy. My apologies.
Can we go on? I have your permission?
Yeah. OK.
Thank you for trying because, indeed, it could have been a
mistake. But it wasn't meant that way.
What you see there now is that you have a whole family of
frequencies which meet all the conditions for the x direction.
And that whole family then you can give Mary and Nancy here as
subscript. There is an omega 1,
1 there is an omega 1, 2, an omega 2,
1, an omega 2, 2 and so on.
I tried to see what actually this system was doing.
It was awfully difficult, but I think I can make an
attempt. Suppose we look at this system
in this direction. We are looking at the yz plane.
Here is the z direction which has this length c.
Sorry, I will repeat it once more to avoid confusion.
This is b so this is the y direction, and this is the z
direction. And so x is coming straight out
of the blackboard. We are looking now from the
side like this. That means that the component
of the E vector in the x direction must be zero
everywhere. It must be zero here because
this is a plane. And there cannot be any
tangential component in that plane, but this is also a plane
and there cannot be any tangential component in that
plane. And this is also a plane,
and there cannot be any tangential component here.
Everywhere here must be the E vector and the x direction must
be zero. And so what does it mean,
the 1, 1 mode? It means that the whole thing,
think of it as a membrane, that the whole membrane comes
to you, the E vector is like this, it goes away from you,
E vector is like this, it goes towards you and goes
away from you. That is what the 1,
1 mode is. Think of it as a membrane.
But, keep in mind that, of course, it extends in the x
direction. And it is everywhere the same
in the x direction over the whole length a.
I cannot change that and I cannot make you see that.
For instance, the 2, 1 mode would then be,
if I make m two and n one then, in the y direction,
I would have here a complete nodal surface which runs all the
way through the box, entirely through the box.
No x component of the E field. And this one would then be an E
field in your direction and this would be an E field in this
direction and it would go like this.
And then, if you have the 2, 2 mode, so x is in this
direction, then you would get this.
You would get one whole nodal surface here,
no E component, no x component of the E vector,
no x component of the E vector throughout the entire box,
throughout complete x. And then this would come to
you, this would come to you, this would go away from you and
there would be this standing oscillating wave.
Now, independently, I can now look at the y
component of the E field. Of course, I only looked at the
x component. So let's now look in the y
direction and let's now ask what is the problem with E of y?
Well, E of y also has to become zero when it,
in a plane, becomes the tangential component.
And, therefore, E of y must be zero when x
equals zero and when x equals a and when z equals zero and when
z equals c. If E of y in this direction
must vanish for x equals zero, it must vanish for x equals a
and it must also vanish in this front plate z equals zero and in
the back plate z equals c. Now, by parallel,
you can immediately come up with a relationship.
I will call it now l as in lion for x.
And then, for z, I will keep the n as in Nancy
so that we don't get confused. So that is now c.
I will do it in terms of squares.
And then I will get l pi over a squared plus n pi over c
squared. And that is it.
Now we have another infinite family of values from l as in
lion, n as in Nancy, which gives me the resonance
condition in the y direction. And so if I had linearly
polarized radiation, I could have in the x direction
linearly polarized radiation which would then have to obey
this. I could have linearly polarized
radiation in the y direction which would have to obey this
sequence of frequencies. And I could do one in the z
direction which, of course, you can make up for
yourself. And any linear combination of
those would be fine. But, of course,
it is also possible that I have an E vector which has a
component x, y and z. And that simultaneously all
boundary conditions are met. I don't just radiate linearly
polarized radiation only in the x direction and only in the z
direction, but now I say ah-ha. If now I make omega squared l,
m, n, if that now is c squared times l pi over a squared plus m
pi over b squared plus n pi over c squared then I can have an E
vector, which is in some random direction.
Well, maybe not too random because I have to meet this
condition. And now I have a whole family
of an infinite number of resonance frequencies,
which are not only E vectors in xy or in z direction.
Did I make a slip? Thank you very much.
I have a c squared downstairs, right?
Isn't that one over epsilon zero mu zero?
Thank you. I appreciate it.
Really, I do. Because it is so nasty,
these slips. We are happy now?
Yes, I am. All right.
Now I am going to change gears because I wanted to find a way
to demonstrate this to you. I am going to change gears and
I am going to make this a box. And I am going to ask myself
what are the resonance frequencies for sound?
Sound is a longitudinal wave, so sound has no linear
polarization. And so now I can immediately
write down the function of the over pressure,
p, over and above one atmosphere or below one
atmosphere, because now I know that at all surfaces I must have
pressure antinodes. The particles cannot move.
They are stuck against the wall.
The pressure can build up, pressure can get low,
pressure can build up, but the particles at the wall
themselves cannot move through the wall.
And so now we have pressure antinodes at the walls.
And so it is easy now for me, without any further thinking,
to write down the general solution for the pressure in the
box as a function of the dimensions of that box.
And so it is p, which is either a little bit
above one atmosphere or a little bit below one atmosphere,
it is some amplitude p zero. And now we get the cosine kx
times x times the cosine ky times y times the cosine of kz
times z. And then the whole thing times
cosine omega t, which is now the frequency of
my sound source. And so now, in complete analogy
with what I wrote on the blackboard there,
I must have pressure antinodes at x equals zero and at x equals
a, at y equals zero and y equals b, at k equals zero and k equals
c. And so now I have that my
boundary conditions require that kx is l pi over a,
that ky equals m pi over b and that kz is n pi over c.
Then I am happy. Then at all surfaces I have met
the boundary conditions that the pressures are a maximum.
And so what always holds is that omega squared is always k
squared time v squared. But this v is now the speed of
sound. And the wavelength lambda is
2pi divided by k. I am going to rewrite that
equation on the blackboard in terms of frequency in hertz
because it is going to play a key role in the demonstration
that is coming. It is exactly this same result
that you see here. But I am going to write it now
as f as a function of l, m, n, which is omega l,
m, n divided by 2pi, which is therefore v,
which is the speed of sound, divided by 2pi.
And then I get here the square root of l pi over a squared plus
m pi over b squared plus n pi over c squared.
Notice that you lose your pi. This pi here eats up this pi,
eats up this pi, eats up that pi,
and this is now in hertz. And so now you have an equation
that if you make a box and you know the dimensions of that box
that you know a is the dimension in the x direction,
b in the y direction and c in the z direction.
You can now predict at what sound frequencies this box will
resonate. Markos Hankin,
who is sitting there, and I have spent the last two
weeks of our lives, he more time than I did,
to build such a box. And the box is here.
It is a marvel. He even gave it colors.
He loved it so much, he could hardly sleep at night
he told me. And there is the box.
And here is a loud speaker on the side.
On this side he has a whole in there of one inch.
And so the sound comes in there, and inside the box is a
microphone. And this microphone will tell
us whether we hit resonance, because then we will see a very
high signal from this microphone.
You have no idea how much fun we have had this with.
And I will show you here. You have plenty of time to copy
it because this is going to take at least 15 minutes,
this demonstration. You see here the dimensions of
the box. They were meant to be 30 by 40
by 50 centimeters, but the accuracy of our
measurements became so stunningly high that Markos
said, well, maybe it is not 40, maybe it is 39.9 because there
was one particular frequency that was a little bit off.
And he was right. We measured it,
and it was a little bit less than 40 centimeters.
You see the sizes there and you see the uncertainties.
The accuracy of our measurement was so enormous that we began to
worry about what the speed of sound was.
And so that means we have to know the room temperature.
Markos and I agreed that we would just accept for now that
it is probably very close to 344 meters per second,
which is the speed of sound at room temperature of 20 degrees
centigrade. But, keep in mind,
it is linearly proportional with the square root of the
temperature in degrees Kelvin. That means one degree
centigrade difference will make a difference.
Two degrees centigrade will be a noticeable difference.
We would be surprised, though, if it is any more
different than 1%, but it could be different by
1%. Here you see the motes that
Markos computerized and predicted with increasing
frequency. Notice that when n equals one,
that is when c, which is the largest dimension
in the z direction, when that is the largest value
that gives you the lowest frequency.
That should be immediately obvious.
Because if you make l zero and m zero and n is one,
if this one is 50 centimeters, which is half a meter,
you will get the largest value possible.
Oh, we are here now. If c is 0.5 meters,
and you multiply that by two, then you get one meter.
This is zero and this is zero. And so it is 344 divided by
one, which is 344. And so that is why this one is
344, because one dimension happens to be half a meter.
And then from that moment on it becomes way more difficult to
see what is happening in the box.
The next frequency that we predict is 0,
1, 0. And we went all the way up to
812. In fact, Markos went even past
4,000, but they become so enormously close together that
it is no longer fun. Here they are quite well
separated. We are going to make you see
the result and we are going to make you hear the result.
Hearing is easy because we turn on the sound as we receive it
from the microphone inside the box.
And I am going to turn that one now.
Now you, in principle, should be able to hear the
frequency of the microphone inside the box.
We are also going to show you the results.
One at the bottom is the speaker.
That is the driving frequency of the speaker.
The frequency, as I have said it now,
you can also see here. It is very important that you
see that. That frequency is now,
I think, 325 hertz. You will see it come up.
Way off resonance. The one on top is the one that
the microphone records. And, if I speak,
you see nonsense at the top because the microphone inside
can hear me. It is important that we are
relatively quiet. And I am going to increase the
frequency and search for the first resonance.
And you can see how accurate it becomes because I will go over
and under the resonance, and you can see there the
response of the microphone.
That is it. Did you notice that I went over
and I went under it? Look at what incredible signal.
I shouldn't be talking. 344.
I am going to find the next one.
434. Less than 0.5% off.
Next one is 552.
548. Less than 1% off. Next one.
575. Amazing what Mother Nature is
doing all the time making sure that there are pressure
antinodes of all the volts. And only then do you see the
resonance. The next one.
670. The next one.
You may get bored, but Markos and I don't.
We love this.
691. Unbelievable.
If you say physics works, that is the understatement of
the day.
721. Next one.
792. Only about 0.5% off.
And now I will go only to the next one.
812.
What Markos also did, he scanned very slowly from 300
hertz to 850 extremely slowly so you don't see the wave structure
anymore, and he recorded only the amplitude recorded by the
microphone. That means the largest value
then of the sinusoid there, the peak value.
And he made a plot of them. And you see that plot here,
which will blow your mind. Horizontal frequencies.
This is the experiment you have just seen.
It was presented in a different way.
Here you see, as we scan with the frequency
very slowly over it, which we did but now you see it
all in one picture, when you hit the 344 hertz that
is the amplitude that is recorded inside by the
microphone due to the resonance. And you see the 434.
These are the values that we measured this morning.
547, OK, it was one hertz off. 574, one hertz difference.
696. You see them all there.
And you also get a feeling for the height.
And we tried to understand the difference in height,
but we are not acoustic engineers.
We contacted some people at MIT who are acoustic engineers,
and they did not have a clue either.
We are not alone. It is very, very difficult.
Because now you have to understand where your microphone
is located in the box. This is already so amazing.
I mean absolutely so amazing. Then one afternoon when we knew
this was going to work and when both of us were very proud
Markos called me. He said, Walter,
you have to come down. I am going to show you
something that you won't believe.
And I said, well, you have got to tell me first.
He said we can show the student transient behavior in a way that
no one has ever demonstrated in class.
What did Markos do? He said when we turn on all of
a sudden the sound then it is like starting to drive all of a
sudden a coupled oscillator which has resonance frequencies.
What happens when all of a sudden you start driving a
coupled oscillator? Well, then you get steady-state
solution plus a transient solution.
The system is going to oscillate in its normal modes.
And then, ultimately the transients will die out because
the driver, of course, is the only one that survives
because that is the frequency that you impose on the system.
And so what Markos then did, as a function of time,
chose a frequency. We will show shortly what he
has chosen, but it is somewhere around 580 hertz.
That is the driving frequency. And here he shows the amplitude
of the microphone during a 1.5 second pulse,
and he shows only the amplitude when it is at its highest and
when it is at its lowest value of the sinusoid.
Imagine that at this moment in time this is the zero value,
this is the highest value and this is the lowest value.
This is a huge sinusoid, highest and lowest value.
But a little later in time because of the transient
phenomenon, which is going to interfere with the driver,
often you get even beat frequencies.
A little later it may be that amplitude of the microphone is
lower here and, therefore, lower here,
too. And then ultimately,
given enough time, this one and this one will no
longer change because this now the stead-state solution.
Now we have the amplitude of the steady-state solution.
But in the beginning you will see the interference of the
transient with the steady-state solution.
And so we were wondering what the transient frequencies would
be that would start up all of a sudden when we start with this
pulse. And we had no clue.
We had no prediction. We had acoustic experts who
said, well, just try and see what happens.
And we are going to show you what happens.
And we have now all reasons to believe that the moment you turn
on the 580 hertz that this one, which is only 7 hertz away,
apparently becomes very potent, not so clear why,
and that one is going to set up a transient oscillation.
But that transient oscillation, which ultimately has to die
out, is going to mix now with our 780, and it will cause a
beat phenomenon. The two beat against each
other, and so you will see this do this.
Of course, it goes in time. You will see it do this,
this and then it will die out. You will be able to see,
for the first time, a transient phenomenon.
73:03 [APPLAUSE]
Markos, I don't think we need that any more.
Should I turn that off? You do your thing.
Here is the frequency that he is going to use as the driver.
You are going to do roughly near 580 or 581 or whatever.
It is very important, again, that none of you talk
and that I don't talk, because talking,
all the nonsense you see now is due to my talking.
Can you just freeze it? Now, from here to here is 1.5
seconds. He has 581 as a driver.
And here you see a transient phenomenon.
From here to there we measured it.
It is exactly what you would expect from the beat.
In other words, if he has 581 and we measure
here 575, which is this resonance, then the difference
is six hertz. And we measured that from here
to here is indeed one-sixth of a second.
One-sixth of second means six hertz.
The oscillation that you see right at the beginning when we
start the driver, that oscillation is six hertz.
That means it must be the beat frequency between the 581 driver
and a normal mote frequency at which the system is going to
fight back, so to speak. That must be the 575 hertz.
And so the 575 hertz and the 581 give you a beat frequency of
six hertz, but the 575, of course, cannot survive.
It dies out. And the 581 is the one that
survives. Now there is something else,
which is an extra bonus. And that is something you
haven't missed here. At the end, we turn all of a
sudden the driver off. The moment that the driver goes
off, the system can only oscillate in a superposition of
its normal modes. Now, which normal modes I do
not know. And the experts may not even
know. But we can look at the moment
that we turn it off. And what do we see there?
We see a signal that is about 27 hertz.
We have measured that it is very close to 27 hertz.
I, therefore, conclude that it must be the
beat frequency between the 575 normal mode and the 548 hertz
normal mode. That difference is 27 hertz.
And you see that beat phenomenon obviously dying out.
It is a transient. Tell your parents that you have
seen this. Tell your friends that you have
seen this. Have a good vacation.
I am leaving this afternoon for Amsterdam. 964 1:17:04,107 I will be back Sunday morning.