The topics for the exam are on the Web.
You click on schedules and you can see the topics as we cover
them at the various lectures. There is no way that I can
cover all of these during one exam, and there is no way that I
can cover them during this exam review.
What I do not cover today can, and probably will be on the
exam. And not all that I cover today
will be on the exam. I have to make a choice,
and I make my choice. I want to start with a
traveling wave in a string. Let the string have tension T,
mass per unit length mu, and let the traveling wave have
an amplitude A. Here is the traveling wave.
And this amplitude is then a. And the speed of propagation is
v. And this v is the square root
of T divided by mu. That follows from the wave
equation, but I am not going to address that issue now.
I want to deal with energy. I first will write down the
equation for the traveling wave. If this is the x direction and
this is the y direction then y, as a function of x,
t, would then be the amplitude a that you have there.
And then you can write this down in many ways.
I will write it down as the sine of k times x minus vt.
If you want to write down for kv omega, that is fine,
too. I have no problems with that.
The idea now, what is the energy in one
wavelength, say? There are two parts to it.
There is kinetic energy and there is potential energy.
The kinetic energy is not because the wave is moving in
this direction. There is no material moving in
this direction. The material is just moving up
and down. That is all it is doing.
The kinetic energy is due to the motion in the y direction.
There is potential energy due to the fact that I have to give
this straight wire, this straight string a shape.
And, in order to give it that shape, I have to do work.
There is a tension. And I have to stretch the
string. That is why I have to do work
to give it that shape. It comes in two parts,
and I will only do the kinetic energy part and I will leave you
with the potential energy part. I slice out here a section dx.
And so the mass in that section dm is mu times dx.
And so the little bit of kinetic energy,
k stands for kinetic energy, in this small part is one-half
times the mass. 03:33
And for vy, I can write in dy/dt squared.
And so the question now is what is dy/dt?
Well, that is easy because we have the function there.
Dy/dt equals, first I get A,
then out pops a k, then out pops the minus v,
and then the sine becomes a cosine.
Cosine k times x minus vt. But I have to square that,
so I have a square here and I get a square here.
dk can now be written as one-half times mu.
And I will leave the dx all the way at the end.
And so I will get A2, I will get k2,
I will get v2, and I will get the cosine
squared of k times x minus vt. And then, at the very end,
I get my dx. And I now want to know how much
kinetic energy there is in one wavelength, because the total
amount of energy, if this is infinity long is,
of course, infinity high. I am interested in the amount
of kinetic energy for one wavelength.
And so that k then is the integral of this whole thing
from zero to lambda. That now is no longer dk,
but that now is k, provided that you realize that
it is k, kinetic energy per unit wavelength.
I only do it for one wavelength.
I get one-half. I am going to get here a mu.
I get an A2. I am going to write down for k
2pi divided by lambda. That becomes 4pi squared
divided by lambda squared. For v squared,
I am going to write down t divided by mu.
Remember the velocity was the square root of t divided by mu.
And now I have to do the integral of this cosine squared
dx between zero and lambda. And you will take my word for
it that this is lambda divided by two.
And so I have to multiply. The integral is the only one
that I have to do. These are constant.
Lambda divided by two. And, if you look now,
you lose a mu. Your four goes and you lose
even one lambda. And so you get A squared times
pi squared times t divided by lambda.
That is the kinetic energy per wavelength.
I will leave you with the potential energy.
It is a similar derivation, but now you have to deal with
work that you have to do. And what comes out is perhaps a
bit of a surprise that the kinetic energy is exactly the
same as the potential energy. Not at all obvious,
but that is the way it works out.
That means that the total energy per wavelength,
now I write down E total, again, per wavelength,
WL stands for wavelength. Not for Walter Lewin but for
wavelength. The total energy per wavelength
is now twice that, so it is 2A squared times pi
squared T divided by lambda. Suppose I am generating this
wave. I am standing someone and I am
wiggling this. Then I have to generate this
energy for every period that I go through, because in one
period I generate one wavelength.
If I divide this by the time for one oscillation then I have
the average power over one oscillation.
Now, the period of an oscillation we normally call T.
But I don't want to have another T because we already
have a T. 08:29
And this is in watts, of course.
This is joules per second. If now we turn to standing
waves, the situation is very different.
Here I have a standing wave. There are nodes here.
And at those nodes the string does not move.
And let's assume that the maximum displacement is A.
Again, the tension is T, mu is the same,
but when it comes to a halt the maximum displacement in the
center is then A. And the question now is how
much energy is there in the standing wave?
This was a traveling wave. This is a standing wave.
10:00 This comes up and has a
velocity in that direction. That is then the maximum
kinetic energy that you can have.
E total is U max, is K max.
And so we already know what U is.
All we have to say now is that that is the same as a squared
times pi squared times t divided by lambda.
And so you see that a traveling wave, with the same amplitude as
a standing wave, has twice as much energy per
wavelength. And, as far as power is
concerned, well, you have two waves going
through each other. And so you have a reflection
from the other side. You really don't have to do
anything anymore to drive the system.
You just kind of let it sit. If there is no energy
dissipation the standing wave will just support itself.
You do not have this power that you have to continuously put in
because, in the case of a traveling wave,
you are continuously generating a wave that moves away from you.
That is not the case with a standing wave.
It is generated at one point in time, it reflects and maintains
itself. Now let's turn to
electromagnetic traveling waves. I take plane wave solutions in
its most general form. E as a function of x,
y, z and t can be written as an amplitude, but this is in three
directions so it has an x component, a y component and a z
component. And then I can write down here,
say, for instance, the cosine of omega t minus k
dot r. This is a dot product.
And the meaning of k, k is called propagation vector.
12:33 And omega equals k times v.
And, if it is in vacuum, then v is the same as c,
of course. But v then is c divided by n if
you have a dielectric. The index of refraction n for
dielectric is then square root of kappa E times kappa m.
13:33 That is the position vector.
If you wanted to know what the associated magnetic field is,
associated with this electric field, then your best bet always
is that the curl of E is minus dB/dt.
If you know that whole function, x, y and z,
you can take the curl of E. Sometimes it is time-consuming.
Sometimes it is fast. It depends on the wave.
And then you have to do an integral in time to get the B
vector. And what comes out of this is
actually something that I do remember.
There is very little in physics that I remember.
This is one of the things that I remember so that I don't have
to apply every time Maxwell's equations when I solve these
problems. And what I remember is what I
wrote down earlier point by point on the blackboard.
For traveling waves E is perpendicular to B.
E and B are both perpendicular to k.
Therefore, E cross B, let me give that unit vectors,
is k roof. 15:33
In phase in space and in phase in time.
And what that means is the following.
The location where E reaches a maximum is also the location
where B will reach a maximum. And they reach it at the same
moment in time. If you are somewhere and you
find that the E vector reaches a maximum then you know that the B
vector at that moment will also reach its maximum.
And, if you are somewhere in space where E is zero,
you can be sure that B is also zero.
That is the case for traveling waves.
The pointing vector S is E cross B divided by mu zero.
And if there were magnetic material you would also have a
kappa m here, but I will leave that out.
You can write down, for the magnitude of B,
E divided by v. This pointing vector,
of course, is time variable because E itself varies with
frequency omega. B varies with frequency omega.
The pointing vector is, of course, time dependent.
And so the magnitude of the pointing vector you can also
write down then as the E squared value divided by mu zero and
divided by v. Because the B can then be
replaced E divided by v. Since they are at 90 degrees
relative to each other, I can ignore the cross because
the sine of the angle is then one.
If E is a sinusoidal function, for instance,
E zero times cosine omega t, then it is clear that you get
here the square, E zero squared times the square
or cosine omega t. If you time average that,
I mean this s is variable in time, but if you just want to
know what the time average value is then the time average value
of the cosine squared equals one-half.
And so you can also write down then that S, I will write it
down here, time-averaged would then be E zero squared.
This is now the amplitude divided by 2 mu zero times v.
And the two comes from the fact that the average value of cosine
square omega t equals one-half. And this is then in watts per
square meter. That is the traveling wave.
Now I want to go to standing waves.
The standing electromagnetic wave, just like the one on the
string, has a separation of space and time.
Let us suppose we take a specific example.
We have here a coordinate system.
I will call this x, call this y and call this z.
Suppose we have linearly polarized radiation.
I will make a little bit difficult than normal.
I will linearly polarize it in the yz plane.
In other words, the E vector would be like this
then and oscillating back and forth in this plane
perpendicular to k. The standing wave E,
as a function of x, y, z and t, can then be written
as this vector. This here would be E zero y at
its maximum. This would be E zero z at its
maximum. That is the one that comes
first, so you get E zero y in the y direction plus E zero z in
the z direction. That is the direction of that E
vector. And now you get your spatial
part in the x direction for which you can write down either
the sine of kx times x or, if you prefer the cosine,
I have no problem with that. And then we have here cosine
omega t. There is no dependence of E in
y and z. They are plane waves.
If you take any plane perpendicular to x infinite in
size, independent of y and z, the E vector will be the same.
And so, therefore, k of y is zero and k of z is
also zero. There is only a k of x.
And so k then equals k of x. And so lambda is 2pi divided by
this value then. And omega divided by this k
value -- I write an x,
but you can leave the x out. It would then be v,
or it could be c, of course.
If you are interested in knowing what the B field is,
you know that the curl of E is minus dB/dt.
The fact that we have here the spatial part separated by the
time part means that there are locations for x which never
change. Whereby the E field is always
zero. That is typical for a standing
wave. And, in this case,
they will be nodal planes. The whole plane perpendicular
to the x direction, I will draw one here and will
draw another one here. The E field in that whole plane
at all moments in time will be zero.
That is the case when this sine or whatever the cosine you have
chosen happens to be zero. And so these are nodal
surfaces. And, of course,
the antinodal surfaces fall right in between.
It is not so easy to draw now for you what you will actually
be seeing in terms of this sinusoidal wiggle that goes like
this. That is not so easy to do
because the wave will be in the plane that is coming out like
this. And it is difficult to show you
that in a three-dimensional way. I will make an attempt,
nevertheless. It will be in this plane that
is tilting forward that you then have nodes and antinodes and
they oscillate like this. Any plane perpendicular to x at
any moment in time will see exactly the same thing.
Now, it is interesting to compare our list with the
traveling wave. Again, with the standing wave,
E is perpendicular to B. It is not too useful to talk
about direction of propagation because they are really two
waves going through each other, so the whole idea of k vector
is a little bit bizarre. However, the magnitude of B,
the maximum value will be E divided by v.
The maximum possible value of B is E divided by v.
But now comes the big surprise. E and B are 90 degrees out of
phase in space and time. If you have here the nodal
planes, for A those are the antinodal planes for B.
And, where you have the antinodal planes for E,
you would have the nodal planes for B.
They are 90 degrees out of space and out of time.
No surprise, of course, that the average
value of the pointing vector is then going to be zero.
And that, of course, can also be seen when you think
that there is a traveling wave in this direction and a
traveling wave in this direction.
There is sort of energy flow in this direction and energy flow
in this direction. And the time average will
obviously end up to be zero. But, if you simply take that E
cross B and you time average it, you will see that immediately.
This is the consequence of the 90 degrees out of phase and 90
degrees in space and time. Now I would like to pursue
this. I could clean the blackboard.
Let me go to the other side. I am going on now to
accelerated charges. 25:32
If this charge is a positive charge and there is a sudden
acceleration, a little later in time I will
see an electric field. There is a traveling wave going
in my direction. It is not a plane wave,
but there is a traveling wave going in my direction,
in the direction r. And so that E vector then is in
this direction. I will return to this.
It is in the opposite direction as a perpendicular.
This vector, which is the component of a
perpendicular to r, is called a perpendicular.
If q is positive then this E vector is in the opposite
direction of a perpendicular. The E vector that you
experience at time t is then minus q times a perpendicular at
time t prime. Because the acceleration took
place earlier than when the signal arrives where I am
because it takes time to travel. In other words,
t prime equals t minus r/c. This is the time that it takes
for the signal to reach me. And so t prime is earlier than
t. And then we have this divided
by 4pi epsilon zero r. And I believe there is here a c
squared. Yes, there is.
And this is in volts per meter. Now a can easily be some
amplitude times the sine of omega t.
In other words, I can oscillate this one up and
down harmonically. Clearly, when I do that,
I will get an electric field that will also oscillate here
harmonically. You really do not get plane
wave solutions because, in any direction that you look,
you can change the r vector. And you have here the answer
you see, the E vector. There is really no such thing
as really a plane wave solution. If you are far away you can
probably approximate it by a plane wave.
The associated B field has all the ingredients that we are
familiar with perpendicular to E.
And the B field must also be perpendicular to the direction
of propagation which, in this case,
is r. And E cross B will also be in
this direction. And, if you take all that into
account, then you can write the B vector as the unit vector in
the r direction times E divided by c.
29:06 A perpendicular is a times the
sine of theta. It is this component.
And so you see that the magnitude of the E vector that
you will see when you look at that charge will depend on where
you are in space. If you happen to be at theta 90
degrees you will see the maximum E possible.
If you happen to be in this direction where theta is zero
you will not see any electric field.
And so there is a very strong dependence of the magnitude of
the E vector in the direction. And the same is true for the B
vector. The B and the E vector are
always married to each other, so they get the same sine
theta. The pointing vector is now
going to be proportional with sine squared theta because you
have an E and you have a B. I would like to summarize for
you, in the same way that I did that here, what I would like you
to remember. I will raise this later again
because I want to work at a height so that you can see what
I am doing. A summary which is very good to
remember is that E is in the plane through r and a.
In the plane does not mean into the plane.
It is in the plane. Look here.
The plane through a and r is the blackboard.
And, therefore, the E vector is in the
blackboard in this case. But that is key that E is in
the plane through r and a. Not into the plane.
And E is perpendicular to r. Notice I have that.
And E amplitude is proportional to the amplitude of E
perpendicular. And, therefore,
it is proportional to the sine of theta.
E itself is proportional to one over r.
Notice that. Not one over r squared but one
over r, which is obvious. It has to do with the
conservation of energy. If you make your sphere around
it then the energy that flows through the sphere must be the
same, no matter how larger your sphere is.
And since that energy comes from the pointing vector it has
an E cross B, E is inversely proportional
with r and B is inversely proportional with r so that
energy is conserved. The pointing vector will then
be inversely proportional to r squared.
If you are interested in the total power, so you are not
interested really in the fact that the pointing vector is a
strong function of theta, namely sine square theta.
But if you integrate the pointing vector over one sphere
that you choose and you take the local pointing vectors and you
multiply that by the local area, which becomes an integral,
then you can actually come up with a power.
This is now in joules per second.
In other words, you tell me what q is,
you tell me what a is. You are creating
electromagnetic radiation. I will tell you how many joules
per second work you have to do. And that is called Lamor
result, the Lamor equation. And that becomes q squared
times a squared divided by 6pi epsilon zero c to the power
three. Notice there is no r anymore
because clearly it is independent of where you are in
space. That is how many joules per
second you have to generate. It is obvious that it is
proportional to q squared because E is proportional with
q, so B is also proportional with q, so the pointing vector
will be proportional to q squared.
It is also obvious that there is an a squared because E is
proportional to a perpendicular. 34:00
This is the energy that you have to generate per second in
order to create electromagnetic radiation.
It is not the kinetic energy that you have to put in the mass
of the charge, the one-half mv squared.
That is a whole different story.
This is the price you pay for creating electromagnetic
radiation. 35:00
I will expand on that in a way that is perhaps easier for you
digest. Maybe I went over too fast when
I covered that in class. I demonstrated it.
You remember I did it with the smoke signal and I did it with
the sunset? So we have unpolarized
radiation. Let's first agree what is
unpolarized radiation. There is a beam of light coming
straight at you and it is unpolarized.
The first plane wave, I think of it still in terms of
a very classic 19th century idea, plane wave solutions.
The first plane wave is linearly polarized like that.
The second one is linearly polarized like that.
Then there is one like this, then there is one like this,
one like this and one like
this. It is a zoo of everything.
That is unpolarized radiation. I pick one of those right here.
This one. And here happens to be fine
dust particles which have electrons.
And these electrons are shaken up and down by that electric
field. What I am drawing now here is
the motion of that electron. The electron is going to be
accelerated in a harmonic fashion.
And where are you? Well, you are here.
37:00 When the radiation comes like
this then the blackboard is 90 degrees.
When you are you here it is 90 degrees.
This is 90 degrees. That is also 90 degrees.
This is not 90 degrees, but that all is 90 degrees.
Radiation comes in like this and I am looking there.
But this is in the plane through r, n,
a, so it is in the blackboard, it is perpendicular to r.
That is nice. So it is linearly polarized.
Do we agree? Linearly polarized.
Now there is another one that comes in.
This one comes in. There it is.
This one comes in. It starts to shake the electron
in this direction. I am still where I was before.
I am here, the same location. This angle now is theta.
The second plane wave comes in like this.
What do I see here? E vector is in the plane
through r, n, a.
That is the blackboard. E is perpendicular to r.
I see this. Do I see the same strengths of
the E vector? No, it is much less because a
perpendicular is much less because theta is much less.
But I will still see E vector in this direction following my
recipe. And then there is one
electromagnetic wave that comes in which happens to be in this
direction. Well, in that case,
I will see nothing. That is tough luck.
But any other direction you will always see electric vector
perpendicular to your line of sight.
And that is only true for 90 degrees.
In other words, if I made a circle here and I
have unpolarized light coming in here straight at you.
When you are here and you look down, you will see the E vector
like this. When you are here and you are
looking down, you will see an E vector like
this. If you look here and you look
in this direction then you will see the E vector like this.
And this is only true if you are at 90 degree angles.
I can easily make you see that if you change the angle that it
is not 100% linearly polarized. And the best way for me to show
that is here are these electrons which are being shaken like
this. One comes in and shakes like
this. And let's look now at forward
scattering. Not 90 degrees,
this is 90 degrees, forward scattering.
Here is the electron going up and down.
You, sitting there in the audience.
If this one comes in, you will see electric vector
like this generated by this charge which goes like this.
Now, the next one comes in which goes like this.
You will see an E vector like this.
The next one will come in like this.
Forward scattering, you will see like this.
You see, if unpolarized light comes in, in forward direction,
it remains 100% unpolarized. In any other direction,
it will be partially polarized. But, in the 90% direction,
it is 100% linearly polarized. That is the reason why the sky,
90 degrees away from the sun, is 100% polarized.
And I check that almost every other day to make sure that
physics still works. It is great fun.
You take your linear polarizers, I always carry a
handful with me, and look at the sky 90 degrees
from the sun. Indeed, the sky is practically
100% polarized. It is quite amazing.
I have a movie that is not the greatest movie.
I tried to show this to you earlier.
It is not the greatest. It is trying to make you see
that when you oscillate charges back and forth that something
happens in the E field that you get in the E field kinks.
And students have asked me more than once, why do you only gets
kinks and, therefore, electromagnetic radiation if
you accelerate them? Why don't you get kinks if you
simply move them with a constant velocity?
And I think the best answer that I can give is the
following. Think of the electric field
lines as spokes, rigid spokes that attached to
the charge. They are rigid but are fragile.
Like spaghetti, but they are rigid.
And so they move with constant velocity.
All this spaghetti moves with the charge.
They all have the same velocity.
There is no stress anywhere on the spaghetti because the whole
thing has the same velocity. Now, all of a sudden,
I take the charge and accelerate it.
Now the spaghetti feels the kink.
The spaghetti feels all of sudden like it wants to break
because it is very fragile. And that break caused the kink
in the field line. That is the best way that I can
convince myself why a constant velocity does not cause
electromagnetic radiation but is really the acceleration.
Some students thought when I used the word spaghetti that I
meant cooked spaghetti. No, I did not mean cooked
spaghetti. I meant uncooked spaghetti that
is very brittle and easily breaks just like this.
And, when you acceleration all of a sudden the charges,
these field lines can break like uncooked spaghetti.
And this movie is making an attempt.
It is not the best thing but we will make an effort.
Markos, if you can do the honors there then I will do the
honors here. And you may not get much out of
it, but at least it is an attempt.
Where we going to give it TV or are we going to make it
completely dark? TV.
If you wait a second then I think the part that I want to
show, OK, oscillating charges.
It oscillates but the speed is constant.
It is instantaneous fast acceleration and then the speed
remains constant. And you see these shells here
which are then representative for the electromagnetic
radiation. Not harmonic yet.
We are going to shake them harmonically shortly.
And so these lines that you see are the field lines,
and they are just here in these shells.
They are broken. It is broken spaghetti.
Also, notice it is like a spherical wave.
It has nothing to do with plane wave solutions anymore.
You really get a spherical wave going out.
It is not so clear from here that the strength of the wave is
very different for the different directions of theta.
That is not so clear. And so here you have a simple
harmonic motion. This is about one hertz
electromagnetic radiation. Have you ever heard of one
hertz? We deal normally with megahertz
and gigahertz. This is very slow motion,
but it is simple harmonic. And you will begin to see that,
indeed, there are electromagnetic,
these are E field lines. That are being distorted I the
direction perpendicular to the direction of your line of sight.
Is it a great attempt? No, but it is an attempt.
I think this is a good moment for a break, perhaps a little
bit early. And we will reconvene in five
minutes so you can warm up.
I will now tell you what I will not talk about,
which is a lot. I will not discuss Fourier
analysis. That does not mean that it will
not be on the test. Make sure you are familiar and
comfortable with the examples that I have in the problem sets.
I will not cover today Doppler shift.
Make sure you feel comfortable with Doppler shift.
It is not a very difficult subject but it has very far
reaching consequences as we discuss cosmology and black
holes. I will not discuss Fresnel
equations, even though they are very crucial.
They were at the center of my lecture earlier this week.
I will not discuss Snell's law. I will not discuss today the
Brewster angle, but don't be surprised,
don't be shocked if there was a problem related to the Brewster
angle. I will not discuss critical
angles, total internal reflection, because it may be on
your test. And I will not discuss today
radiation pressure. That does not mean that I will
not be on your test. It is simply not possible to
cover all of this in the available amount of time.
Now, of course, I have not left out purposely
things that will not be on the test.
A lot of stuff that I have covered today will be on the
test, of course, but there will also be stuff
that I cover today that will not be on the test.
I want to discuss now one of the bizarre dispersion relations
that evolved from the boundary conditions of electromagnetic
wave on ideal conductors. 47:30
And I want to propagate through these plates,
which are infinitely large in the y direction,
infinitely large in the z direction, at least very much
larger than the wavelengths. I want to propagate
electromagnetic radiation which is linearly polarized only in
the y direction. That is what I want to do,
and that is what I demonstrated also.
Well, this is beginning to be boring now, k is kx times x plus
ky y roof plus kz z roof. And this, by the way,
is zero because there is no dependence of the E field in the
y direction, so ky is zero. K is the square root of kx
squared plus kz squared. Omega is k times v.
Let's assume this is in vacuum. Omega is k times c.
The boundary conditions demand that at x equals zero and at x
equals a this component must vanish because you cannot have
an electric field in the surface of an ideal conductor.
That was one of the boundary conditions that we derived.
In other words, E of y must become zero for x
equals zero and x equals a. And, as a result of that,
you are going to get a standing wave in the x direction and you
are going to get a traveling wave in the z direction.
And the boundary conditions demand now, in order to meet
this, that the k of x is going to be m pi divided by a.
If then you have your wave then you will see,
substituting for kx, this value, it will always give
you then, if you have the sine of kx times x,
always a zero here and always a zero there.
And so, omega will then be c times k.
It is c times the square root of m pi over a squared plus k of
z squared. And this is what we call the
dispersion relationship. This equation is responsible
for a bizarre behavior. And the bizarre behavior then
can best be shown in a diagram which we call the omega kz
diagram. I will raise this later again
because I want to work over my head so that you can see what I
am doing. This is kz and this is omega.
And I am going to plot for you this relationship.
This line would be omega equals kz times c.
That would be nondispersive. However, this is different
because of this. And so now you have here a
frequency that is the lowest frequency possible for which
radiation can actually go between the plates.
This then is the case for Mary equals one.
And omega is then c times pi divided by a.
And so here, for omega, it is the cutoff
frequency. Pi times c divided by a.
And, if I plot then that curve, you get something like this.
No frequency below that value can propagate through the gate,
so to speak, through the opening because it
cannot meet then the necessary condition that the E field
becomes zero here and zero there which is nonnegotiable.
It has to become zero here and it has to become zero there
because there is an ideal conductor.
K of x must obey this boundary condition.
Let us assume that, at a particular moment in time,
we have a frequency going through these two planes,
the frequency of this linearly polarized radiation in the y
direction. And that this is the value for
omega. That means that the associated
value for kz is then this. Kz must adjust itself so that
kx can remain what it has to be. Everywhere on this line kx is
the same value. And the kx value everywhere is
pi over a. And so, kz is being slaved to
become the value that meets this dispersion relation.
And that is this line. Kz settles there.
The phase velocity in the z direction, v phase in the z
direction is omega divided by kz.
Well, look what that means. Omega divided by kz.
I can draw a line here. And so this angle is an
indication for the ratio omega divided by kz which is larger,
this angle is higher than this one.
And this was c, remember?
And so, you see that the value is always larger than c.
54:30 And, therefore,
the group velocity is smaller than c.
And so, this is smaller than c. If you lower omega and
gradually reach your cutoff frequency below which you can no
longer propagate any radiation between the two plates then you
get a situation which becomes even more bizarre.
That when you reach that point nothing will go in the z
direction anymore. Nothing will come out.
I demonstrated that. The group velocity will,
therefore, have to become zero. Well, you can see that because
the tangent on this slope here becomes horizontal.
The group velocity indeed right here is zero,
but the phase velocity is infinitely high.
Think about it. Because kz now becomes zero.
And so you get an infinitely high phase velocity.
And I spent quite some time during my lectures to explain to
you why that has no physical meaning.
I don't want to go over that now.
There is no such thing as resonance frequencies.
Often you people think that these are resonances.
No, it is not a resonance at all.
It is a mode. It means that if you have
radiation at this frequency that in the x direction kx will
adjust itself such that you get a standing wave in the x
direction. In this case,
you get something like this, a nice little sinusoid,
zero here and zero here. That is where m equals one.
And kz will then be whatever it has to be.
And so when you change omega kz will adjust all the time.
If you go to very high frequencies there is here
another cut off frequency which is twice this number because
that is when Mary equals two. And when kz becomes zero.
That becomes twice as high. Then, of course,
there are two different ways that you can propagate radiation
in the z direction. Because now you have here the
line. And if now your omega is high
enough then you get an intersection with this line and
you get an intersection with that line.
That allows you for two different values of kx and two
different values of kz. It is not a resonance
frequency. Nothing is resonating.
It is a mode but it is not a resonance mode.
It is not a normal mode in that sense.
Now, the most interesting thing for me is, and I stress that
when we discuss this and I even demonstrated that.
We did this with radar, remember, with the ten
gigahertz transmitter, the three centimeter waves.
The most interesting thing is that if you reach omega equals
omega c -- And you can do that if we had a
ten gigahertz transmitter and made a smaller and smaller and
smaller. And, when a became less than
lambda over two, that means omega became less
than omega c, then no radiation would
propagate through anymore. And so what we did to
demonstrate was we had lambda three centimeters.
We started out with a gap of about two centimeters and
radiation went through. And, the moment we hit the 1.5
centimeters, it stopped abruptly.
The interesting thing now is that if your radiation is only
linearly polarized in the x direction there is no such
problem. There is no such thing as this
crazy dispersion relationship. Because an E vector being
perpendicular to this plate and being perpendicular to this
plate is no problem. If the E vector oscillates with
frequency omega, always perpendicular to those
two plates, the only thing that happens is nature like crazy
adjusts the local row S's, the local number of Coulombs
per centimeter, so that you always meet the
normal boundary condition. But nothing ever has to become
zero. And so, therefore,
there is no boundary condition that gives rise to such a crazy
dispersion relationship. If you had radiation linearly
polarized in this direction then it would follow this line.
Nondispersive. Phase velocity is c.
Group velocity is c. Now comes the interesting
point. Suppose you managed to get
radiation that is unpolarized and this you try to send through
there, and you make a smaller than lambda over two,
that means that the vertical component of the E vector,
of each one of those waves, cannot get through.
Only the horizontal component can get through,
but without any difficulty. Speed of light.
That means you have created linearly polarized light out of
unpolarized radiation. I would not say light in this
case. It is radar.
If you could manage to get unpolarized radar going into
this direction and you make a smaller than lambda over two,
all components in this direction are killed.
Only this component can go through.
And so, you have created at the other end linearly polarized
radiation. And I realized after I gave
that lecture how cute that is actually, that this is a way,
in principle, that you can create linearly
polarized radiation simply by squeezing it through a very
narrow opening, a very narrow tunnel,
so to speak. The mother of all
demonstrations was the sound box.
61:04 The box is closed on all sides.
That means if we think of the pressure there must be pressure
antinodes at all surfaces. The air particles cannot go
beyond the wall. They can push on the wall,
they can suck on the wall, they can push on the wall and
suck on the wall. That means antinodes in
pressure. And they have to be antinodes
on all walls. I can write down the pressure
p. That is the pressure over and
above one atmosphere, as some amplitude p zero times
the cosine of kx times x times the cosine of ky times y times
the cosine of kz times z times cosine omega t.
And the reason why I already picked cosines is because I know
I want the antinodes when x equals zero and when x equals a,
I want the antinodes when y equals zero and y equals b,
and the same for the c direction.
And so, the boundary condition now demands that kx can now only
have unique values, discrete values,
which is l pi divided by a. Ky can only be Mary pi divided
by b and kz can only be Nancy pi divided by c.
If not then the boundary conditions are not yet,
and then I don't have antinodes at all the surfaces.
And l, m and n can be zero or one or two or three,
including zero. Make one of those zero.
There is no sine here. If you have a sine there and
you make it zero then everything becomes zero.
But, if you make a cosine zero, then it is one.
Zero is allowed. Except, they cannot all three
be zero. And you will see very shortly
why. Omega, which is always k times
v, v is the speed of sound, so omega equals v then times
the square root of l pi over a squared plus m pi over b squared
plus n pi over c squared. Omega has then unique values.
Now we are dealing with resonance frequencies,
completely different from there.
There were no resonances there. These are the unique discrete
resonances, l, m and n.
And now you can see why you cannot make them all three zero.
Well, of course, then you have omega equal zero.
And that is not a very interesting thing.
Now comes the question what is the lowest possible frequency
that is the lowest resonance frequency?
Well, that depends on which dimension a, b or c is the
largest. And so, if a,
for instance, where ten centimeters and if b
is 20 centimeters and if c is 50 centimeters, which happens to be
0.5 meters then clearly the lowest frequency is l zero,
m zero, n equals one. 1, 1, 0.
For the 0, 0, 1, I then get a frequency f
which is omega divided by 2pi. And that means I lose all these
pi's here, by the way. And so, I get v divided by 2c.
That is my c. Because, look,
if I make Nancy one, my pi is gone,
so I only have a one over c squared.
This is zero. I get one over c,
and the two comes from this. The frequency,
that is the lowest possible frequency is v divided by 2c,
c being now half a meter. And so, if v is 344 meters per
second then the lowest frequency is 344 hertz.
And I proudly demonstrated that you.
Our prediction was accurate to better than one hertz.
We have exactly 344 as the lowest possible frequency.
And then, of course, you can look for higher order
frequencies. And that depends then on the
dimensions of a and b, whether the next one is 0,
0, 2 or whether the next one is 0, 1, 1 or whether it is 1,
0, 1. And we ranked them all.
I showed you eight of those frequencies.
And we had a handout, at least on the Web,
I showed you these wonderful resonance curves that we
generated during our lecture. I think it was Lecture 16.
You can still download it from the website.
Now comes an interesting question.
What would happen if I knock out this panel in front and the
panel in back? I knock out both panels in the
z direction. I make it open in the z
direction. Open at both sides.
What happens now? Well, the pressure now can
never be an antinode at z equals zero and at z equals c.
On the contrary, it is connected with the
universe. No pressure differential can
ever build up. The pressure has to become a
node now for z equals zero and z equals c.
Well, that we can easily do. We just change the cosine into
a sine and nothing else changes. Because, look,
if now I make kz n pi over c then you will see for any value
for n that you choose this will indeed become a pressure node.
You will get zero. And so this now is the right
solution then when the panels in the z direction have been
removed. The only thing that you cannot
allow, you cannot allow n to be zero, because if now you make
zero then no mode could exist. Because, if n is zero,
then this would always be zero. Now you have a situation that,
yes, l can be zero, m can be zero,
but n cannot be zero. What now is the lowest possible
frequency? It is the same,
0, 0, 1. And so, the lowest possible
frequency, when you make it a tunnel, is again 344 hertz.
And that was question number nine.
I asked you for this one which was, of course,
a give-away. If you attended that lecture,
you could not possibly have forgotten that.
I was almost crying when I showed you this demonstration,
so you could not have forgotten that I was crying.
And then I just wanted to test your insight that you realize
the antinodes become nodes but that nothing changes.
At least not in terms of resonance frequency.
Something changes in the box, of course, because now you have
four walls where you have pressure antinodes and you have
non-walls where you have pressure nodes but the resonance
frequency is the same. When Markos and I were working
on this, we got this crazy idea to ask the question what would
be the resonance frequencies of a sphere?
We have a sphere. And, of course,
it is not a perfect sphere. It is more like this.
And it has a diameter, I call it capital D.
And the diameter is about 28.5 centimeters.
Uncertainty is only a few millimeters.
And so we got this crazy idea to put a speaker there,
just like we did with the box, and to have a microphone
inside. And then to see if we increase
the frequency of the speaker whether we could predict and
actually confirm the resonance frequencies of the motion of the
air inside this box. And so I said to Markos,
I know the answer. It is easy.
It is a trivial problem. If this were a perfect sphere
then clearly the wall here must become a pressure antinode.
A complete sphere must become a pressure antinode because there
must be spherical symmetry. A sphere is a sphere,
right? There must be spherical
symmetry. You cannot have an antinode
here and a node here, because nature doesn't know the
difference between upstairs, downstairs, left and right.
So it must be a complete antinode sphere.
But this center must then also be an antinode because,
if the air flows away and pushes onto the wall and then
sucks onto the wall, it must always come back there.
The center itself must also become a pressure antinode.
Knowing that, I said to Markos,
well, that means lambda must be 28.5 centimeters.
Because from node to node to node is one whole wavelength.
And so I made the prediction that the frequency f,
which, of course, is the speed of sound divided
by lambda, would be 340 divided by 0.285.
And that is something like 1190 hertz.
And I even made the prediction what the second harmonic would
be. You must, again,
add an antinodal sphere. Another antinodal sphere where
the pressure reaches maximum, minimum, maximum,
minimum because of the spherical symmetry.
And so that means now the diameter is twice the
wavelength. The wavelength is now half that
value because from here to here to here, one wavelength from
here to here to here is the same wavelength.
And so I predicted that the second harmonic would be twice
this high. And so we set it up.
And full of expectation, I said to Markos,
why don't we just start the system somewhere at 800 hertz
and then we will find these resonances.
And Markos, a little bit more cautious than I am,
said, well, let's start a little lower.
I said, well, why waste our time?
He said, well, let's start low.
Let's start at 45 hertz. And that is what he did.
You will see here the frequency of the speaker,
which is mounted there. And you will then see there,
if that works. Oh, boy, Markos,
it does work. You will see at the bottom the
driver, which is the speaker, and you will see at the top the
response of the microphone which is inside.
And so we started to slowly increase the frequency,
which I thought was going to be a complete waste of time.
45 hertz. 50 hertz.
58 hertz. Oh, I forgot to turn on the
microphone. Thank you, Markos.
There would be no resonance at all.
[LAUGHTER] Let's go back to the 45 then, 48, 51,
52, 53, 56, 58, 60.
And there is, at 61 hertz,
an unbelievable resonance. I couldn't believe my eyes.
I said this is nonsense. This cannot be.
It is impossible. How can you have 60 hertz
resonance frequency? What is wrong with my
reasoning? Well, we were staring at each
other in disbelieve. We went on.
I said to him, you know, there are 60 hertz in
the lights in the building. That is what you pick up.
Some crazy pickup. It is not a sound resonance.
Whenever you see 60 hertz in the United States it is pickup,
if you see 50 hertz in Europe you know it is some pickup,
so I was willing to forget about it.
In any case, we went on and were trying to
find the other frequencies. I will quickly now try to go to
a much higher value. I go in core steps and see
where the first one is. I was hoping to see the 1190.
And now it is 600, 700.
I will now go a little slower. Do you see the frequency at the
lower signal increasing much higher than it was before?
And here you are actually beginning to hear it.
You are getting close to the resonance.
And so the 61 made me sick in my stomach.
And I was not all too happy with 787 hertz.
And so, I insisted to at least start looking for the 1190.
And so, we started looking for the 1190.
And I was beginning to be happy because I think it is coming.
I think it is coming up there, I think it is coming up there,
and it fell a little short but not embarrassingly short.
1164.
I went home and could not sleep that night.
I could not sleep. 60 hertz is absurd.
Keep in mind that anything that is this size cannot have such a
low frequency. I did not sleep the first
night. I did not sleep the second
night. And then the third night I woke
up. And then I remembered when I
was your age, when I was a student in college
that one day I had a little flask like this and I was
blowing air past it. I said to myself what?
Something ten centimeters in size, you expect resonance
frequencies of the order of the speed of sound divided by the
length that is three kilohertz. This is nowhere near three
kilohertz. All of a sudden that image came
back to me, and I remember what I did.
I said there has to be an explanation for that.
I went to the library, looked in books and couldn't
find the solution. I went to the library and found
an article. And that article told me what
the resonance was, this one, only one resonance,
no harmonics provided that you know all dimensions.
You have to know d, you have to know l,
which is our case l is 99.5 centimeters with an uncertainty
of about three millimeters, and you have to know this d,
which in our case is 5.5 centimeters with an uncertainty
of maybe one millimeter. I did my homework.
I turned the crank. I went to the Hayden Library
the next day and found an article.
Maybe not the same that I found 50 years ago,
but I found the article with the same solution.
I plugged in the numbers. What did I find?
62 hertz. And I was very happy.
I had recovered what I had lost 50 years ago.
Now, you are at about the same age that I was then.
And I want to present this to you as a challenge,
something that goes a little bit beyond the standard
textbook. If you present me before
December 7th with a proper solution for the 61 hertz,
I will generously reward you with extra credit for 8.03.
I am not too worried about the 787.
The reason for that is it is not a perfect sphere.
There is no doubt in my mind, if that had been a perfect
sphere, if this thing had not been there, you would not have
seen the 61, which is true. You would not have seen the
787. But the opening makes it very
difficult to calculate. In fact, it is not even so bad
that my 1164 does show up where my prediction was 1190.
But I am not even sure whether that is really the one that I
predicted. In any case,
it is the 61 that is the bizarre one.
And it has no resonances. There is only one answer.
There is only one resonance frequency, no harmonics.
I will be here over the weekend to help you prepare for your
exam if you need me. I will make half-hour
appointments. You will have to indicate what
you want to discuss with me, and I will then agree on a time
with you. I wish you luck,
and I will see you then on Tuesday.