I'm going to talk today about energy in electromagnetic waves.
There must be energy in there, because we know that electric
fields contain energy, and magnetic fields contain
energy. And you may remember that the
electric field energy density is one-half epsilon zero E squared,
this is now in joules per cubic meter, and the magnetic energy
field density, which also has come up
earlier in the course, is one over two mu zero times B
squared. Again, joules per cubic meter.
Now, we -- when we deal with traveling waves,
in vacuum, at any moment in time, the magnitude of B is E
divided by C. So this is also one divided by
two mu zero, I can replace this by E squared
divided by C squared. But C squared is one over
epsilon zero my zero. So this is also one-half
epsilon zero E squared. And when you see this,
this is an absolutely amazing result, because what this tells
you is that the energy density in the magnetic field of a
traveling wave is exactly the same as the energy density in
the electric field of an electromagnetic wave.
That's really an amazing thing, the symmetry is absolutely
beautiful. So the total energy density is
the sum of the two, so I double this one,
so that it's epsilon zero E squared joules per cubic meter,
and of course, I can also write for that,
epsilon zero, just one E, and then the other,
I write B times C. So I write for E,
B times C. So this is, again,
in joules per cubic meter. Yes, I'm happy with that,
that's fine. Now, I want to ask the
question, if electromagnetic waves come by me,
how much energy passes through one square meter?
It's like an energy flux. And so I have here one square
meter, and this one square meter is
perpendicular to the direction that the electromagnetic wave is
going, and I want to know how much energy flows through there.
One square meter every second. In one second,
light travels a distance C, which is a horrendous distance,
three hundred thousand kilometers.
This side of this box is what light travels in one second.
And this is one square meter, and I'm going to calculate how
energy now goes through this one square meter in one second.
Of course, I could have chosen this box a billion times
smaller, would have gotten the same answer of course,
but for convenience, I choose this to be C,
and this to be one square meter.
So the volume of that box is C cubic meters.
And all the energy of that box is going to come out here in one
second, because I know that electromagnetic waves move with
the speed of light, which is C.
And so therefore, the energy that comes out here
per square meter per second is that U total there that I have,
which is the amount of energy for every cubic meter,
but I have so many cubic meters.
And so I can use this result, now, here, and I can substitute
that here, so I get epsilon zero times E times B times C squared.
I have to multiply it by this C here.
And this, of course, is also E times B divided
by mu zero, because C squared, in vacuum, is one over epsilon
zero, mu zero. And this, now,
is joules per square meter, per second.
Because in one second, all that energy comes out,
and I have already chosen one square meter area.
Let me see whether I'm happy with that, yes,
I'm happy with that. We call this the pointing
vector. And we write it,
in general as a vector. We write it,
S with a vector, and it's called the pointing
vector with a Y, and we write that as E cross B
divided by mu zero. E B divided by mu zero.
You don't need the cross, really, because E and
B are always perpendicular to each other in a traveling wave.
The advantage of this notation is that S, which is the energy
flux, goes in a certain direction, and the velocity of
the weave is always in the direction of E cross B,
so it also tells you, then, in which direction the
radiation is flowing, whereas here,
you -- you lack that information.
And so this, then, to remind you,
is in these units, which is watts per square
meter. How many joules per second?
One square meter through a plane perpendicular to the
direction of propagation. Now, E and B are changing with
a frequency omega, cosines omega T or sines omega
T. And so S is changing with the
cosine square of omega T. If you are somewhere in space,
and this electromagnetic wave comes by, there are moments that
S is zero, namely, when E and B happen to be zero.
And there are moments that it is at maximum,
when E and B happen to be at maximum.
And so when we deal with electromagnetic radiation,
it's more meaningful to discuss the time-average value.
And the time-average value of the pointing vector is,
first of all, the time-average of the cosine
square omega T, or the sine square omega T,
whatever the case may be. And the average value of cosine
squared is one-half.
And then I can write now, for E, E zero,
and for B, I can write B zero, which is now the maximum value
possible, divided by mu zero. And if you want to write it
differently, if you want to write it only in terms of E
zero, then you can write down one-half E zero squared divided
by mu zero times C. So this is the more practical
equation, that gives you a number which
is time-averaged over the oscillations.
Let me make sure that there is nowhere, a slip of the pen --
no, there isn't -- I hate to have slips of the pen,
because you can't edit them out later on your videotape.
Slips of the tongue, you can edit it.
The slips of the pen, you can't.
OK, this looks good. So now, we have an average
value for the pointing vector, and so we can calculate,
now, how much energy flows through
one square meter per second. And I can give you an example,
suppose I have a plane electromagnetic wave,
and E zero is one hundred volts per meter.
That's just what it is, how we get it,
that's a different story. And this could be radio
emission, this could infrared, this could be light,
I d- I don't specify the frequency.
I don't have to. Frequency doesn't come up
there. And so the value for -- average
value for the pointing vector, I can pick this equation.
E zero is hundred. I know mu zero,
I know C, so I can calculate what it is.
It's one hundred squared divided by two,
divided by mu zero, divided by C.
And when I do that, I find that this is thirteen
watts per square meter.
So imagine that you would stand in this electromagnetic wave
coming to you, we take all our clothes off and
we let it hit us, and suppose it absorbed us.
Suppose it is radiation that absorbs us -- some radiation may
go through you. Gamma rays may go straight
through you -- they are electromagnetic radiation.
But certainly, light will not go through you.
And radio waves, some of them will not go
through you. So you observe them with your
body. Would you notice that?
I doubt it. You probably have a surface
area that's close to one square meter.
Thirteen watts, thirteen joules per second,
not very noticeable. You radiate,
yourself, hundred watts, hundred joules per second,
so I don't think you will notice that.
But imagine, now, that we increase the value
of E zero, and we make it thousand volts per meter.
Now, this goes up by a factor of a
hundred, because if E goes up by a factor of ten,
automatically, B goes up by a factor of ten --
remember, in electromagnetic waves, they're always coupled.
So the pointing vector, which is the product of the
two, goes up by a factor of a hundred, so now,
you're talking about one point three kilowatts per square
meter. And if you absorb that on your
body, believe me, uh, it may fry you.
Certainly if you do that out on the beach,
you get a -- a deep suntan. And if you do it long enough,
then you can hurt yourself very badly.
So now comes the question, does a light bulb emit plane
waves? Well, not really.
Plane waves have no beginning and they have no end,
they exist at all time and all space.
Look at my plane wave solutions from last lecture.
You can substitute in there any value for X, Y,
and Z, and any moment in time, the year five thousand B.C.,
you get an answer. Doesn't specify when,
doesn't specify where. And of course that's not very
realistic. In the real world,
there is a beginning and there is an end to the electromagnetic
radiation, and therefore, they have also a finite length.
Remember the quarter-nanosecond pulses that we sent to the moon
that we discussed last time. They were only seven
centimeters long. That's not very much like a
plane wave. So I want to discuss with you a
little further what these waves look like, and I want to take a
closer look at how electromagnetic waves are
produced by charges that we begin to shake,
that we are accelerating. Key in the whole process is
that you accelerate a charge. If a charge is just moving at a
constant velocity, it will not produce
electromagnetic radiation. But I want to give you some
feeling, at least some classical physical feeling on how these
electromagnetic waves are produced.
It is a picture that has its own limitations,
but it's still useful. It's not a quantum mechanical
treatment, but it is something that you and I can see and
therefore, perhaps, appreciate.
Suppose I have here a charge which is not moving.
Just sitting there. These are field lines,
I just draw only a few field lines.
And if it's a positive charge, the arrows are outwards,
if it's a negative charge, the arrows are inwards.
And I'm going to accelerate this for a time delta T.
Let's accelerate it in this direction.
And then we bring it, again, to a halt state.
I redraw this. This point is the same as this
point here. I accelerate it,
and delta T seconds later, it happens to be here.
So this is T zero, and this is T equals delta T.
And now I'm going to draw a circle --
it should actually be a sphere, three-dimensionally -- about
this point. And here is that sphere.
And this sphere has radius C delta T.
This field line, which was the field line that
goes with this charge, is here, and this field line is
here, and this one is here,
I only draw three. This one, this one,
and this one. And the message that I
accelerated this charge could not possibly have reached this
location in space, because that message can only
travel with the speed of light. So the electric field here is
exactly the same as it was when it was still here.
So when the object is here, the electric field must still
be like this here, and it must be like this here,
and it must be like this there, because the message hasn't
reached that point. But now, look at this charge,
which is now here at time D delta T.
Now, the electric field is like so, is like so,
and is like so. So this field line must somehow
meet up with this one, it's one and the same field
line. And what does that mean?
That somewhere there, there must be a kink in the
electric field. And there must be a kink
here. Notice there is no kink here,
which is interesting. And so it is the collection of
these kinks that propagate outwards, with the speed of
light, and they produce an electromagnetic disturbance,
a change. If you were out in space here,
and if, for instance, I were to oscillate this charge
back and forth, you would see these kinks go by
all the time, these breaks in the
electric field, and you would experience that
as an electromagnetic wave, if there is a changing electric
field, according to Maxwell's equations, there has to be also
a changing magnetic field. But the interesting thing is,
even though this is an extremely simple picture,
notice that in this direction, if you were here,
you would not see any kinks. So there's no electromagnetic
radiation going in this direction,
nor is there any going in this direction.
And the maximum is going in this direction,
and something in between is going in that direction.
It's not much of a plane wave, for that matter.
I mean, if anything, it's more like a spherical
wave. But it's a very special
spherical wave, not the same strength in all
directions. And so even though this is a
rather classical picture, it helps me,
at least, to see how these changing electric fields -- and
therefore, associated B fields -- are
formed by charges that we accelerate.
And I have a two-minute movie that shows that,
also, in a slightly more detailed way that I was able to
do. And so let's look at that
movie, Marcos, you ready for that?
OK, then you can start that. So this is a computer-generated
movie whereby we accelerate charges.
This is a constant speed, now, and we're going to stop
it. Stopping means there is an
acceleration, right?
You may call it a deceleration, but stopping is an
acceleration. It's going to be stopped.
And you see now, here, these kinks?
And they propagate out with the speed of light,
then, if this was electromagnetic radiation.
You're going to see this several times,
so you get another chance. Your charge is now stopped,
you see the magnetic -- the electric field lines coming out,
and now it's being accelerated. And you see these kinks here?
Moving out with the speed of light.
You're going to see more. There it is,
it's going to be accelerated, and doing the acceleration,
only during the acceleration do you see the formation of the
kinks, when it goes with constant speed,
no longer, it's only during -- now it stops.
Stops means an acceleration. There you see this wave front
moving out. Let's look a little bit more as
you can see the oscillating effect.
So you already have seen this, acceleration,
here is the wave, it stops, that means
deceleration, and here you see the wave
front. And when it's sitting still,
or when it's going with constant speed,
then there is no electromagnetic wave produced.
Now we're going to see some oscillating charges,
which is more realistic when you have an antenna,
and you have current going up and down with frequency omega,
you obviously have stops and starts, it oscillates back and
forth, and that's what you see now.
And look at these beautiful -- there's a wave going out,
here's a wave going out, only during the acceleration.
There's one going out, and there's one going out,
so you accelerated backwards, forwards, backwards, forwards.
I think that's fine, thank you, Marcos.
So the classic picture, even though it has lots of
limitations, it's not a quantum mechanical
treatment, is still very useful.
For instance, if we think of a -- an antenna,
just a straight wire with a current, going up and down,
with high frequency, it could be seventy megaHertz,
could be gigaHertz, and you produce electromagnetic
radiation, then you are accelerating charges up and down
by having current going like this, then from this classical
picture, we know that no radiation will go out in this
direction. But we also know that in the
direction perpendicular to the acceleration,
that is in this direction -- remember, here,
the acceleration was like so -- and in the direction of the
acceleration, no electromagnetic waves go
out. So nothing goes out here,
nothing goes out there. But in the plane perpendicular
to A, which, in this case, was the blackboard,
is a whole plane like this, and in this case,
of the antenna, is the horizontal plane,
here we have a maximum radiation going out everywhere
in this plane.
And in between somewhere, it's not zero and it is not the
maximum value. So now I would like to return
to the idea of the pointing vector.
Because we talked only about the pointing vector,
we derive that in terms of plane wave solutions.
And we now know that plane wave solutions are really not too
realistic. Suppose we have the sun here.
And the sun is a really powerful light bulb,
three times ten to the six watts radiates,
largely invisible light, and in infrared radiation,
which is electromagnetic radiation.
And we are a distance of hundred fifty million
kilometers, here somewhere on Earth, and therefore,
through every square meter in the direction perpendicular to
the line of sight to the sun, through every square meter,
there is about one kilowatt flowing, one kilowatt per square
meter. And you can calculate that at
home to verify that, that's easy calculation,
you know that it is three times ten to the six joules per
second, and if the distance is hundred and fifty million
kilometers, you can calculate how much of that energy flows
through one square meter. And that's very close to one
kilowatt. And we will call that,
nevertheless -- even though it's not really a plane wave
solution in that sense -- we will call that,
nevertheless, the pointing vector.
And I really don't care whether you
think of this as being a plane electromagnetic wave,
or whether you think of it as more like spherical waves,
if you see there, or even if you want to think of
them as individual photons, and I will come back to the
photons. As far as I'm concerned,
it's all of the above. But the result is that there is
an energy flux, every square meter,
aimed in the direction of the sun, one thousand joules per
second will flow through that. If you want to
hang on to the plane wave solution, and if you want to put
any value at the plane wave solution, then you have to draw
the consequence that you can calculate, then,
the E zero of that plane wave. And I really don't know how
meaningful that is, but then you would then get the
equation that one thousand, that is, the average value of
the pointing factor, that's a given,
is then E zero squared divided by two mu zero C.
We derived that there. And so you can calculate,
now, what E zero is, and you'll find that E zero is
approximately eight hundred and seventy volts per meter.
To be frank with you, I don't quite know what to do
with that number. It is some equivalent electric
field. But it's really not an electric
field, of course, that you could measure,
because it's not really a plane wave in the sense that we
derived our pointing vector for the plane wave solutions.
So the bizarre thing is that in physics, you can think of
light, for instance, as a plane wave,
you can think of it as spherical waves,
you can think of it as photons, and you can choose which ever
fits you best to explain a certain phenomenon.
And it is only in quantum mechanics where the wave and the
photon picture merge. Let's talk about photons now.
That is one aspect of the electromagnetic waves,
then. Photons are really individual
packages. Wave trains.
You can think of them as bullets.
Well-defined in time and space. [Wssshhht], there goes a
photon. That photon has a certain
energy which it carries with it. That means it has a certain
momentum. And the momentum of a photon,
which comes from one of -- the great work by Albert Einstein --
the momentum of a photon P is the energy of one photon divided
by C. You just have to take my word
for that. If you ever take special
relativity you will understand this a little better.
But give it now -- take it now at face value.
And now, if I shoot many photons at a target,
and these photons, if they are absorbed by that
target, then the target will experience a force.
Because if the photons carry momentum, and if the momentum
-- photon, photon, photon, photon,
photon -- I feel a force. You remember from eight oh one,
a force is D P D T, is the transfer of momentum.
If I throw rotten tomatoes at you, then when you get these
tomatoes on your face, they hit you like this and then
they go [pffft] like this, so they come in a certain
direction, and they lose all their momentum in that
direction. That gives a force on you,
in this direction. Because that
momentum in this direction is destroyed, but momentum is
conserved and so you get that momentum.
If, for instance, I threw one kilogram tomatoes
at you each second -- so one after another -- on average
time, average one kilogram per second, and if each tomato had a
speed of five meters per second -- and I cal that the X
direction -- and here is your face --
they hit your face, and then they go [plllt] down.
So all the momentum in the X direction is destroyed,
then you will experience a force in that X direction --
it's in the X direction -- which is one times five,
which is five newtons. It's a time-average force.
If, for some reason, the pot- the tomatoes would
bounce back with a perfect elastic collision,
like tennis balls would do, then the momentum transfer is
twice, comes in with momentum, and comes back.
So now the momentum transfer is twice this number,
it's not just destroyed, no, it's reversed.
So in that case, the force on your case would be
double that, would be ten newtons.
And so we can now carry this further, to our electromagnetic
radiation, and we can return to our pointing vector.
Here, I have one square meter. And radiation is coming in.
And I know exactly how much radiation is coming in,
for every square meter, there is this value S.
So S, which is an energy, per square meter,
per second. Remember, that's -- did I
mention, of the pointing vector? If I divide this by C,
and I divide this by C, then look what I have here.
I have energy divided by C. But the energy,
in electromagnetic radiation, divided by C,
according to Einstein, is momentum.
And I don't care whether this is hundred billions of
individual photons, that's OK with me.
That is a momentum. And so I have now a momentum
per unit time that makes it a force.
But a force per square meter makes it a pressure.
And so this is pressure. And we call that radiation
pressure. It means that if you are
exposed to bombardment of electromagnetic radiation,
and you absorb it, that you feel a pressure.
The radiation comes to you, and you're being pushed
backwards. And this is the pressure that
you will experience. It makes a difference whether
the radiation is absorbed,
or whether you are capable of reflecting that radiation,
just like with the tomatoes. And so if I'm trying to be as
general as I can -- let me, first of all,
specify the mean value of S, that's the only thing in
electromagnetic radiation was mattered.
It's the mean value. So the mean value then,
if I divide that by C, and I multiply that by an --
a vector alpha, U will come up shortly and tell
you what alpha is -- that is now the radiation pressure.
An alpha, if alpha is one, then I have full absorption.
All the radiation is absorbed. If there is no absorption,
if it's completely transparent -- and there is some radiation
that goes straight through you -- then alpha is zero.
But if there is hundred percent reflection --
which can happen, you can reflect radio waves
from metals with almost hundred percent reflection -- then alpha
is two. So that you get twice the
pressure. If you take the sun,
whereby the mean value for S is one thousand,
you can calculate now, if you absorb that radiation,
over one square meter, what the pressure would be that
you would experience. So your body,
one square meter, you expose it to the sun,
in the direction of the sun. That force is insignificantly
small. The pressure would only be
three times ten to the minus six newtons per square meter,
because you have to multiply that number thousand -- you have
to -- not multiply, divide it by C.
There's not much left over. And if you hold your hand up in
the direction of the sun, your hand would only have an
area maybe of one-hundredth of a square meter,
that force is unnoticeable. So in our daily lives,
we don't experience radiation pressure very much.
Regardless of whether you think of electromagnetic radiation as
plane waves, or you want to think of them as spherical
waves, or you want to think of them as photons -- I really
don't care -- but what matters as far as radiation pressure is
concerned, is how much energy flow through one square meter,
and which fraction of that energy is absorbed,
or maybe, some of it even reflected.
So the radiation pressure is not
that important in our daily lives, but is important in
astronomy. And in fact,
you can see it. Some -- some of you can see it.
A comet has two tails, and one of those tails is the
result of radiation pressure. A comet is made of carbon
dioxide, and it has also dust, it's about the size of
Manhattan. And when it gets close to the
sun, the radiation pressure of the sun, which is this radiation
pressure that we just talked about,
pushes onto the dust particles, forming the tail,
and this tail is white light, white-yellowish light,
which is reflected sunlight off these dust particles.
And then there is a second tail which is bluish,
which is hard to see with your naked eye, and that one is the
result of the solar winds. We discussed the solar wind
earlier, causes the aurora in the upper atmosphere of the
Earth. The solar winds are protons and
electrons which are emitted by the sun in a rather erratic way,
sometimes a lot, sometimes not so much,
and they move with a speed of two hundred fifty miles per
second or so, and they ionize the C O 2.
So they excite the molecules when the [unintelligible]
excitation of the molecules, they emit blue light.
So you get two tails, and we'll show you these two
tails. The tails can be a hundred
million kilometers in size, they can be huge.
As some of you may remember, or should remember,
in nineteen ninety seven, we had a fabulous example of a
comet, which was Hale-Bop.
I watched it every night for months on end.
This is a picture that you're going to see of Hale-Bop which
is a time-exposure, it's not what you can see with
your naked eye. And, in fact,
frankly speaking, I have never seen the blue
tail, but the white tail, which is the one due to
radiation pressure, I have seen many,
many times. So let's take a look at
Hale-Bop, and there you see it. And you see clearly the two
tails. You see the blue tail here,
which is the result of the interaction with the solar wind,
and then here you see the dust which is the result -- the tail
is the result of radiation pressure.
The sunlight is absorbed, fully absorbed by these dust
particles, and is pushed away. I just wonder,
who has seen Hale-Bop, three years ago?
Yes? Just hold your hands up,
I'm just curious, it was a
spectacular sight, and it was easy to see for
months on end, you could see it.
You don't see very bright comets all that often.
So without being very qualitative in terms of the
electric and the magnetic field strengths, it is easy to
determine the direction of the oscillating electric field,
if we know in what direction the charges were being
oscillated. And that has to do with the
polarization of the radiation, and so I would like to send the
remaining time today on that. And let's stick to the center
board. So we have a oscillating
charge, being accelerated, and I will choose a coordinate
system like so, and let's suppose we oscillate
the charge in this direction.
Acceleration [wssshhht] all the time, and the frequency is
omega. And we are somewhere in space,
here, and we receive electromagnetic radiation,
we are at point P -- like to make you see it
three-dimensionally. And at this angle,
B theta, and this is the position vector R,
from the origin of the oscillating charge to where you
are, you are at point P. And when you are at P,
I'm going to give you some very simple rules which will always
allow you to determine the direction of the electric field.
The electric field is always perpendicular to the direction
position vector, the direction of propagation.
So you're looking in this direction to the charge,
the electric field that you experience, if the charge is
doing this, is always perpendicular to R.
And A and R and E are always in one plane.
And notice that's the way I drew it in the blackboard.
E and A and R are in one plane. So it's extremely simple.
So if this is oscillating with angular frequency omega,
then this electric field will also be oscillating with that
angular frequency omega. If you double the charge,
the electric field will double. If
you double the acceleration, the electric field will also
double. That's reasonably intuitive.
So the electric field strength is proportional with the charge
that you oscillate, with the acceleration.
And then, there comes this effect that we discussed there,
that nothing goes out in the direction of the acceleration,
no energy goes out in this direction, so no electric field
is produced in this direction, and the maximum is produced in
this direction, and somewhere in between here,
and that is then reflected by the sine of theta.
If theta is zero, nothing goes out.
If theta is ninety degrees, you get a maximum.
That's the whole plane, perpendicular to this
oscillating charge, this whole plane,
maximum electric field. And then -- and that may not be
so obvious to you now -- it's also inversely
proportional to the distance R. If you double R,
the electric field strength goes down by a factor of two.
And the pointing vector, which is the product of E and B
-- but E is always proportional to B -- so the pointing vector,
then, is proportional to Q squared, proportional to A
squared, proportional to the sine square of theta,
and inversely proportional with R square.
And that is obvious,
that it has to be inversely proportional with R squared.
Because if you go -- if you have a sphere,
and radiation goes out, and you're twice as far away,
then you know that the area of the sphere is four times larger,
and so the amount of energy per square meter must be four times
lower. So the pointing vector must
fall off as one over R squared, that's the conservation of
energy. And if you accept the fact that
the pointing vector falls off as one
over R squared, then the E vector must fall off
as one over R, because the pointing vector is
the product of E and B. E and B must both fall off as
one over R. I want to show you a picture
that may help, to see you -- how the radiation
-- how the electric field is oriented relative to a charge
that we are accelerating.
In the middle here, we accelerate a charge,
up and down. Frequency omega.
Whatever radiation you want to make, fine with me.
You can make omega as large as you want to.
For one thing, notice, these -- these waves
here, these snakes, represent electromagnetic
radiation. And nothing is going out in
this direction. And we understand why.
The maximum is going out in a plane perpendicular to this
direction. That is this,
this, this and this. And notice that the E-vector
that you receive when you're here is perpendicular to R --
this is R -- and it is perpendicular.
And notice what I said, that A and R and E are in one
plane. A, which is the acceleration,
R and E are in one plane. A, R, and E,
are in one plane. A, R, E, are in one plane.
So you can always determine the
direction of the oscillating electric field.
And if you go at an angle theta, which is neither zero --
which it is here -- nor ninety degrees -- which it is here,
here, and here -- but if you have something in between,
then notice that the E vector here is drawn a little smaller.
And that is that sine theta here.
And consequently, the pointing vector in that
direction will be smaller than in
this direction, and in this direction,
the pointing vector will be zero.
We call this radiation -- whether you're here,
or there, or there, or there -- we call that
linearly polarized radiation, for the simple reason that the
electric field is oscillating in one direction.
It's linear. And I'm going to produce for
you linearly polarized radiation, and I have two
demonstrations for that.
So I'll get the lights back on again, and let's discuss these
demonstrations. I have, here,
a transmitter which transmits at ten gigaHertz.
That is, uh, a wavelength of three
centimeters. You wouldn't call that radio,
you would call that radar, but that's just a matter of
names. And we have also a receiver.
We have a transmitter and we have a receiver.
The transmitter is here, and the receiver is here.
If I give you a three-dimensional picture --
this is my coordinate system, this is coming straight out of
the blackboard to you, that's what I meant by this --
then the transmitter is aimed like so, just like this.
Current is going to be oscillating like this.
And the receiver -- think of this as my radio,
which is receiving -- is here. And the antenna of the radio is
also in this direction. And so, the electric field
which you see here, which comes from these
oscillating charges -- which oscillate, by the way,
at a horrendous frequency, ten billion times per second --
their electric field is perpendicular to R,
this is R, and the electric fields R and A are in one plane.
Ah, so that means that the electric
field that arrives here is oscillating like so.
And therefore, this antenna,
this receiver, is very happy.
The radiation comes in exactly in the right way,
and so it will receive it. We have modulated this signal
with an audio signal, amplitude modulation,
as we discussed that earlier, that you also use with radios.
And we modulated with approximately one kiloHertz
audio signal. And we'll make you listen to
that audio signal. And you can tell,
then, that this receiver is, indeed, receiving the
three-centimeter radar from this transmitter.
That's what we will first do. So I'll just turn on the
transmitter [tone], and here you hear [tone] one
[tone] kiloHertz, [tone] which is the [tone]
modulated signal, [tone] and it's received by
this antenna, [tone] which is a straight wire
like this, [tone] and this is the emitter.
[tone] And to demonstrate, to you that it really goes from
here to here, if I put my hands in
between, my hands absorb three centimeters.
It's not there anymore. [tone] There it is.
[tone] It's not there anymore. But now, [tone] now I'm going
to [tone] rotate this antenna by ninety degrees,
[tone] so this one is going to be put in this position.
[tone] So now the electric fields -- [tone] so I'm going to
put it now in this position [tone] -- so the electric field,
now, [tone] is going up and down like so.
[tone] But the receiver doesn't like that, because the antenna
of the receiver is like this. [tone] So it cannot [tone]
drive any current. And so you won't have anything
anymore. [tone] So what it means is that
I have changed the direction of polarization by ninety degrees,
[tone] and the receiver cannot see that.
[tone] I'm going to rotate it now ninety degrees.
And it stops. So now, the electric field goes
like this, [shhhht]. The wire here,
the antenna, is like this,
and the wire says, "Sorry, I can't hear you." But
all I have to do now is rotate the receiver ninety
degrees, and so I can rotate this antenna and put it
vertical, and then, of course, it's happy again.
So if I rotate this ninety degrees, [tone] there comes the
signal back. [tone] So now the E field is
like so, [tone] and the antenna receives it like so.
[tone] Now, I have one [tone] -- I have one brainteaser for
you, and I want you to think about that.
We can actually have a vote on that.
I have here a wooden frame with metal bars.
No trick. Metal bars.
And I'm going to put that in between here.
So we have the situation that we are transmitting
electromagnetic waves with the E field going like this,
and the receiver is perfectly happy.
The receiver is also aligned like this.
There you have it. I can put this grid in like so
[wssshhht], so that the bars are in the same
direction as E, but I can also put it in like
so, so that the bars are perpendicular to E.
If I make you a drawing, if you would look in this
direction, then I can do one of two things.
Either the bars are like so, or the bars are like so,
but in both cases, is the electric field coming in
like this? Who thinks that if I have the
bars vertical, that the
electromagnetic radiation will go straight through without any
problem? Who thinks that maybe it will
not get through? Who thinks it will get through?
Who thinks it will not get through?
Now I do this. Who thinks that the
electromagnetic radiation, the oscillating E-field,
can easily go through this? Who thinks it will not get
through? It's every evenly divided,
we have about twenty five, twenty five,
twenty five.
OK, we'll see. [tone] One kiloHertz.
[tone] E field is like this, [tone] receiver is receiving
it. [tone] What do you want me to
do first, this or this? [tone] Let's do this first.
[tone] Nothing. [tone] It's not absorbed.
[tone] Goes straight through here.
[tone] Now I rotate it. [tone] Ninety degrees.
[tone] And I kill it.
[tone] So in this configuration,
[tone] the electromagnetic radiation does not get through.
[tone] I want you to think about it, and I -- if you have
some sleepless nights about it, that's OK, too,
that's very healthy. I have a second demonstration,
and that has also to do with the polarization of
electromagnetic radiation. I have here seventy five
megaHertz transmitter, so that's more like
radio stuff, that is a four-meter
wavelength. And this is the antenna,
this is the transmitter. So let me give you the numbers.
So this is seventy five megaHertz, and that is four
meters wavelength. And I'm going to turn it on
like now. You don't hear anything,
it's not connected to any receiver yet.
Oscillating. Making electromagnetic waves,
going like this, going up.
Not much is going in this direction.
I have a very special receiver. A very special one.
And that very special receiver is this.
It's just a straight copper wire cut in the middle,
and the left side and the right side I connected with a light
bulb. So if there is a strong current
going through here -- that means if it receives a strong signal
-- then the light bulb will indicate that.
I'm going to make it dark so that
we can, perhaps, see that light bulb.
So that is transmitting. And I want to show you several
things. The first thing I want to show
you is that, indeed, if I hold this parallel to the
transmitter, that I can receive it.
The electric field is now going like this, and so this receiving
antenna is very happy. But now look a what happens,
it's same thing I did with the radar receiver,
I rotate it ninety degrees. And now the E-field that comes
in this cannot cause any current like this.
So the light bulb goes out. I can also show you,
which is obvious, that if you go farther away
from the transmitter, that the strength of the signal
is less. Remember, the pointing vector
goes as one over R squared. So the light in the light bulb
should go down by one over R squared.
So I go a little closer -- cannot too close,
because I can really burn out the bulb if I come too close.
But you see it's quite bright now,
and when I walk away, by holding it still in the
right direction, you see, that,
indeed, the light bulb gets fainter.
One over R square relationship. What is also interesting,
we can explore how much radiation is sent out by this
transmitter in the direction of the antenna, which is the
direction in which we argued that no radiation should go out.
So let me walk over here -- or let me walk to the other side,
because it's darker here. So right here,
I am now standing such that theta
is zero. No matter what I do,
whether I hold my antenna like this, or like this,
or like this, there's nothing I can do.
I can even come a little closer, you see no light,
because there is no energy flowing in this direction,
no electromagnetic wave. Largely going out in the plane
perpendicular to the antenna, but of course,
if I'm here, which is not exactly
perpendicular, then of course,
I do pick up also, some radiation.
OK, see you Friday.