Today we are going to talk about the interactions of
electromagnetic waves with conductors.
No electric fields can exist inside an ideal conductor,
so when electromagnetic waves are incident to an idea
conductor somehow that electromagnetic wave must be
reflected. And so we are going to look
this morning at the boundary conditions and at the
consequences of those boundary conditions.
We have four Maxwell's equations so we expect four
boundary conditions. The key is that inside an ideal
conductor the electric field must be zero.
The magnetic field does not have to be zero.
A static magnetic field is possible inside a conductor.
However, you cannot have a change in magnetic field inside
a conductor because the curl of E is minus dB/dt.
And so, if E is zero inside, dB/dt is also zero inside.
Now, before we work this out, the consequences,
I have to be fair to you and tell you that in reality no
conductors are ideal. And this is what Bekefi and
Barrett discuss on page 442 of the book.
The electric field falls off exponentially in a very thin
layer which we call the skin depth.
And that skin depth's delta is defined in such a way that the
electric field falls off by a factor of E.
And that skin depth, which I will not derive here,
is the square root of two divided by the frequency omega
times mu zero times the conductivity sigma.
And the conductivity for an ideal conductor would be
infinitely high so that the skin depth is zero.
But even if you take something like copper, which is an
extremely good conductor, then sigma, of course,
is not infinitely high, but sigma is something like 5.8
times 10 to the 7. And the units of conductivity
are ohmic meters to the power minus one is SI units.
If you use this sigma and you take a thousand megahertz
radiation, which has a wavelength of 30 centimeters,
then the skin depth is only two microns.
It is very, very small. If you take optical light,
which has a frequency of 5 times 10 to the 14 hertz and you
put that in here then you will find the skin depth is only
three nanometers, but it is 30 atomic distances
that the E field decays into the layer and reduces them by a
factor of E. The wavelength of visible light
is 500 nanometers roughly, and so the skin depth is only
something like three nanometers. I want to start now deriving
the boundary conditions, and I will start with the first
Maxwell's equation which is the divergence of E equals rho
divided by epsilon zero, or this rho is the volume
charge density, so this rho is in Coulombs per
cubic meter. You should remember this from
your 8.02 days. I will write it in integral
form. If I have the integral of E dot
dA, I have a closed surface. And you will see it at work
very shortly. A close surface is very
essential. This is the electric flux that
emerges from that surface. It is a dot product.
There is a dot here. This is one over epsilon zero
times the volume integral of all the charge that is inside.
You have called that earlier q inside divided by epsilon zero,
but this is a more mature way of writing it.
So you integrate it over that whole volume.
05:00 I am decomposing this now in a
component which is tangential, that is in the plane of the
surface, and I call that E tangential.
And a component which is normal to the surface,
and so I give that an n which stands for normal.
And so now I am going to make my pill box.
Here is my pill box. 06:00
That is the flux leaving that surface.
Then I have an electric flux which I will call just phi of E.
This is also an electric flux, but I call it phi of E
shorthand notation for everything that goes through the
cylindrical surface. It is what I call the curved
surface. Let's call it the cylinder.
Well, when I am going to make dl zero, that cylindrical
surface goes to zero so no electric flux can escape there.
And so this one then equals zero.
Now, there is no electric field here, so this is the entire
electric flux as it escapes from this surface,
from this box. And so that is now one divided
by epsilon zero times the volume charge density that I have in
here. And so that is rho times the
volume which is dl times dA. This is now Coulombs per cubic
meter. You will say now,
when dl goes to zero, that that goes to zero.
But that is not true because there can be,
and there will be charge right at the surface of a conductor
which is not Coulomb's per cubic meter but it is the surface
density in Coulombs per square meter.
Totally different dimension. And that is rho of S,
how much charge there is per square meter on that surface.
And so in the case when dl goes to zero this does not go to zero
but this becomes rho s divided by epsilon zero times that dA.
And this rho s now is Coulombs per square meter.
That is the surface charge density.
And so you see that we have derived now our first boundary
condition which I will write down on the blackboard there.
And that is that the normal component of any changing
electric field that is due to the electromagnetic radiation
equals the surface charge density divided by epsilon zero.
Now when you see this result you will say that is nothing
new. We have seen that also in 8.02
for static charges. That is true.
It is also true for static charges.
But here there is something special.
And that is that this E vector is changing all the time with
the frequency of the incident wave and, therefore,
rho S must also change all the time.
That is all the time rearrangements of that surface
charge density to make sure that this boundary condition is met.
But, yes, it also holds for static charges.
The next one is the divergence of B.
You are going to do that. You are going to make a pill
box and you are going to demonstrate to me that if you
use the divergence of B, which is the second equation
Maxwell, that now you find that the normal component of the
changing magnetic field at the surface is zero.
Keep in mind it is time variable, the magnetic field,
as it comes in. And it is the time variable
component, the normal one that has to be zero.
The next one in line is the curl of E minus dB/dt,
the famous Faraday's law. And then we have the curl of B.
I will do the curl of E, and I will leave the curl of B
up to you. The curl of E is minus dB/dt.
This equation runs our economy. This is why we have light.
This is why we have energy. This is Faraday's law.
And I am going to write it in integral form.
Of course, you are probably used to that.
This is a closed loop now. Any closed loop that you may
choose of E dot dl, it is a dot product,
it is a line integral equals minus d phi B/dt.
And this is a magnetic flux that goes through the surface
that you attached to that closed loop.
It is an open surface. Any surface that you may choose
is fine. It is an open surface.
I cannot stress that enough. You choose a closed loop.
You can do it right here in space.
You can attest to that an open surface, which could bulge out
into the audience any moment in time, this will hold.
And so whenever you deal with curls you do line integrals.
And so let's do a line integral and then see what boundary
condition that leads to. We do the same thing that we
did there. We have here the conductor and
here we have vacuum. And so the E field here is
everywhere zero. And here is your electric field
vector coming in from your electromagnetic wave.
Here is your Et and here is your En exactly in the way that
I defined them. This is the normal component to
the surface. Now I am going to make a closed
loop. This is my closed loop.
It shouldn't surprise you that I choose it this way.
And I am going to march around this closed loop in any
direction that I want to, but I chose this direction.
It makes no difference which direction you choose.
And let this be dl and let this be dB.
And I am going to do exactly the same thing.
I am going to go in the limiting case when dl goes to
zero so that I get the boundary condition at the surface.
If you are ready for this, I start here and I move in this
direction. The E of t has no impact
because it is perpendicular to my direction of motion.
The dot product is zero. All I worry about is this one.
And so I get E of n times one-half of dl.
Let's suppose that this is one-half of dl and this is
one-half. That is when I go from here to
here. And now I go from here to here.
And now, of course, E of n has no effect because it
is perpendicular to dl. dl is a vector in the direction
that I am moving. Now I only deal with E of t.
Now I get plus E of t times dB. Now I am here and I go down.
Now E of t is up, but I go down so the dot
product creates a minus sign. Now I get minus E of n times
one-half dl. I said E of t,
but I meant E of n. E of n is in this direction,
and you are moving in this direction.
And so that is why you get a minus sign.
And E of t has no effect here because it is perpendicular to
my direction of motion. And the electric field here is
zero. This now becomes minus the
change of the magnetic flux through that surface,
this surface if you wish, or a surface that bulges out in
the audience, I don't care.
It is minus d phi B/dt. And that goes to zero because I
am going to make dl zero. There is no surface left,
and so that one goes to zero. And notice that this one
cancels out against this one. And so now we find the third
condition, and that is that at the boundary condition E
tangential is zero. That is the third one.
I want to remind you, though, there is the skin
depth. In reality, there is an
exponential decay of this E vector that is in the surface
that decays exponentially over that skin depth delta that we
discussed earlier. That is that skin depth.
Now there is the curl of B, which is your turn.
You are going to do a closed lope integral of B dot dl.
And you work your way through that.
And then you will find that the magnitude of B of t,
that is the component of the magnetic field in the plane of
the conductor, the tangential component,
that that magnitude is mu zero times what we call surface
current density. J of s is a surface current
density, and it has units. MP is per meter.
And so there are actually oscillatory currents on the
surface that occur when the electromagnetic wave comes in.
These currents change with the frequency, of course,
of the incident radiation. Not only is there all the time
a readjustment of rho of s which is the surface charge density,
but all the time as that electromagnetic wave interacts
with the conductor is there going to be a current
oscillating in order to make sure that we meet this boundary
condition. Real things are really
happening. We can use these four boundary
conditions now to start looking into some results,
and I will start with a very simple example because I can
demonstrate that. I take linearly polarized
radiation. This is my z direction.
It is going to make a traveling wave.
This is my x direction and this is my y direction.
I have linearly polarized radiation.
It only has an x component. I am going to make it move in
the z direction. 18:00
It is propagating in this direction.
This is the k vector. And that is the incident wave,
and it is in the x direction. They are plane waves
perpendicular to z infinite in all directions.
And the E vector is then everywhere in those planes.
This value changes with z and it changes with time.
Now what I am going to do is put here a perfect conductor.
And so the E vector arrives there.
There is no normal component to the E vector so I don't worry
about E of n. There is no normal component.
I only have E of t, and that E of t must be called
zero at the conductor. And so this reminds you of the
string that we had when we were wiggling the string at one side
and we fixed that side of the string.
That in order to make sure that there was no motion of the end
of the string, the reflectivity was minus one,
a mountain came back as a valley to make sure that this
point never moved. And so exactly the same thing
is happening now. We must make sure that right
here the electric vector is always, at all moment in time,
zero. If I just call this for
convenience z equals zero at this location then the reflected
wave must be minus E zero pi. That is the mountain becomes a
valley. And then it moves in the
opposite direction so you get kz plus omega t in the x direction.
Now, you will see that the sum of these two,
at any moment in time, will make the E vector right
here in this entire plane, which is infinitely large,
zero. That is the boundary condition
that we have to meet. The total electric field,
which is the sum of the two, one moving in this direction
and this one moving backwards is, of course,
the factorial sum. I have the cosine of alpha.
I call that alpha, minus the cosine of beta.
Your high school days tell you that it is twice the sine of
half the sum times the sine of half the difference.
Although, I always have to look that up because I forget that
just as well as you do. And when I add them up then I
get that the total electric field, which is the sum of the
two, is going to be 2E0i, and then I get the sine of kz
and I get the sine of omega t. And that is a standing wave.
All the spatial information is in here and all the time
information is in there. And, indeed,
if you substitute in there z equals zero, notice that the E
field is always zero. This, by the way,
is in the x direction. And so we have a standing wave
now. And what that means is that
there are surfaces perpendicular to the direction of z where the
E vector is always zero. An entire surface.
We have now a nodal surface like this, and they are
separated by distance one-half lambda.
This is a nodal surface, but this is also a nodal
surface. I am trying to draw the curve
of the E vector, sinusoidal curve,
so the E vector may be up here. And then the E vector would be
down here. But it is all in the plane
perpendicular to the blackboard to same.
So this is a nodal surface, this is a nodal surface,
this is the wavelength and the
E vectors go like this. Nothing is moving anymore in
this direction. It goes like this,
a standing wave with nodal surfaces.
This standing electromagnetic wave has an electric field which
is standing. There is an associated magnetic
field with it, which is a standing magnetic
field. And I would like you to work
that out. I may even have put that in one
of my problem sets. And what you find out,
much to your surprise, certainly much to my surprise
when I saw this first, that the magnetic field is now
90 degrees out of phase with the electric field both in space and
in time. That means where you have the
nodal surfaces in E, you have the antinodal surfaces
in B. And where the E fields reach
the maximum you will have that the magnetic field is zero.
Unlike a traveling wave where the magnetic fields and the
electric fields are in phase both in space and in time.
They are now 90 degrees out of phase.
When you think about it that is perhaps not so weird because
there cannot be any net energy transport in one direction with
a standing wave. And so the mean value of the
pointing vector must be zero. And, if you make E and B 90
degrees out of phase then there is always mass.
The mean vector of the pointing vector then becomes zero.
But, in any case, if you use Maxwell's equations
you will see that the magnetic fields and the E fields are 90
degrees out of phase and 90 degrees in space and in time.
I have here our famous 88 megahertz transmitter.
This is an antenna where we accelerate charges back and
forth. Currents are running at a
frequency of 88 megahertz, 88 million oscillations per
second. We are creating,
because we went through this, that accelerated charges give
rise to electromagnetic radiation.
We are creating electromagnetic waves.
Lambda equals c divided by the frequency.
You will find that the wavelength is about 3.4 meters.
And I have here a receiving antenna, which are two cooper
wires that are not connected to each other.
Well, they are. They are already connected
through the filament of a light bulb.
And so when the electric field reaches this receiving antenna,
current will start to flow in this direction and then the
light will go on. I have demonstrated this
before, but I want to do it again, that if I hold up this
receiving antenna like this then you will see the light bulb go
on. If I do it like this then it
will not go on because not currents can flow in this
direction. But that is something that I
have already shown you before. That is not the reason why I
want to do this again. The reason why I want to do
this again is that the blackboard here,
behind the blackboard is metal. And so the electromagnetic
waves that go out in your direction and to the blackboard
are going to reflect off here and they are going to reflect
off the walls. It is a very chaotic way that I
could not predict nor calculate. But there will be,
in this lecture hall, locations of nodal surfaces
where the E field is very low or near zero.
And certainly that has to be the case near the blackboard
because this whole plane behind here is a conductor.
And so the sum of the incoming wave and the reflecting wave
must be zero here. So I don't want to see that
light at all. And then I will walk out in the
lecture hall and see whether there are locations in the
lecture hall where we actually receive electromagnetic
radiation, we will see the light, and where we will not see
the light. Make sure this is not too close
to the transmitter. Otherwise, we will blow the
bulb. Let me first show you then that
when I stand here that I am receiving this electromagnetic
radiation. Remember that the E field is
proportional to the sign of theta.
And so I am here in an excellent position so I get a
maximum electric field. If I rotate it like this,
it does nothing, and that has to do with the
polarization. It has nothing to do with
standing waves, simply polarization.
Now look. I claim that there must be here
a nodal surface. There are no electric fields.
There it is. Isn't that amazing?
This is a standing wave that I am building up.
I am now in that surface that I pointed out to you on the
blackboard. Now I can also try to walk
through the audience and see whether there are locations
where I see light. Right here I see light so I am
receiving. Now, how it bounces off these
walls, I have no idea. That is probably one of the
main contributors. But the walls also have metal
in them. I walk further in and the light
goes out. You may say,
yeah, it goes out because you are further away from the
transmitter. That is a good argument,
but if there are standing waves in this room perhaps it will go
back on again. And when I am here,
I would like to see it go back on again.
It doesn't want to do that. There it is.
Because I tested it out, of course, before you came.
You see the light again? Do you see it?
Just tell me that you see it. It is not very strong,
we are very far away, but it is clear this light is
substantially brighter than it is here.
Nothing here, but here it is stronger.
There it is. And so there is a huge
incredibly difficult pattern in this room of locations where the
E fields are very weak and where the E fields are very strong
like here. And here is even weaker.
I think the most spectacular one is what I did first,
and that is to show you that it becomes zero at the blackboard
which is really a very well controlled plane of conductor.
I know would like to discuss with you another example which
we give the name transmission lines.
We are going to try to have traveling waves,
just like we have with this system.
The reason why I chose this example is because I can also
demonstrate this. Here we have a copper wire.
You see it right in front of you here.
It is 425 centimeters long. And there are two.
And here I have a power supply, so I am driving this with a
variable voltage, v0 times the cosine omega t.
And I am going to use my same transmitter for that.
It is going to be the 88 megahertz transmitter.
I call this y a one and I call this y a two.
We are getting a wave of voltage cosinusoidal that is
going to propagate in these wires.
And what is going to happen is not what you think.
First of all, there cannot be any electric
field in the wires because it is a conductor.
The electric field in this direction does not exist,
so there is no potential difference from here to there.
But, if there is any potential difference, it is going to be
between here and there. If I take this wire and I make
a cross-section of that wire, this is wire one and this is
wire two, then at any moment in time, as the voltage wave passes
by this may be positive. That is that row of s.
And this will be negative. That is that row of s.
And the electric field lines then go like this.
That means if this is, at this moment in time,
positive and this is negative then there is an electric field
in this direction between these two wires.
Not along the wires but between the wires.
But then, of course, there are also going to be
locations where this is minus and this is plus and then where
this is plus and this is minus. And so then the electric field
is in this direction and the electric field is in this
direction. The potential difference
between here and here is positive.
The potential difference between here and here is
negative because it is integral E dot dl.
The potential difference v1 minus v2 is going from 1 to 2
from E dot dl. And so if the electric field is
in this direction, this has a higher potential
than this. If it is in this direction,
this has a lower potential than this.
This is how this voltage wave propagates.
The E field is between the wires, not along the wires,
so there is only an electric field potential between the two
wires. There are oscillating surface
charge densities. In order to meet those boundary
conditions as the voltage flow comes by.
It is positive at this moment, but a little later in time it
will be negative and this will be positive.
And of that moves with the speed of light.
And then there are also oscillating surface current
densities, this j of s. Let us assume now,
I call this z equals zero, that I am going to short this
out. I am simply putting there a
wire to totally short it out. 33:37
And so here is the situation that the end of the lines which,
by the way, is there, is shorted out.
If it is shorted out, it acts like a conducting wall,
just like what we discussed before.
There cannot be any potential difference between here and here
because the E field in this direction must be zero.
And so you get a similar situation that the reflectivity
is minus one. That a mountain comes back as a
valley. So that the E field of the
incident one and the E field of the reflected one kill each
other exactly at this location. That means you get a standing
wave. So that means this here is a
nodal line. Everywhere in this line the
electric vector will always be zero at all moments in time.
But, since I get a standing wave, incident wave and
reflective wave, right here there is another
nodal line. And this distance then is
one-half lambda. And right here there is another
nodal line. And the electric field here,
along this line, is everywhere zero.
There is no potential difference between y and one and
two here. There is no potential
difference here and no potential difference here at all moments
in time, because it is a standing wave.
And in between you see the electric field.
Let's assume it is in a down direction here,
so it is smaller here, smaller here,
smaller here and then zero here and zero here.
And here the electric field would be like this.
And that whole pattern then is a standing wave,
and so this is down and this is up and then it oscillates like
that. And these are always nodal
lines. Imagine now that this end is
open. So I just remove the short
circuit and I leave it open. Well, I hope you remember the
experiment we did with strings. This is really the equivalent
of the string whereby the end is attached, so you get a node
there, reflectivity is minus one, mountain comes back as a
valley. But this now,
when the end is open so there is completely nothing at the
end, that is the situation that it is right now,
by the way, this is completely open.
This wire here and this wire are not connected.
When this is open you expect a reflectivity of plus one,
just like with the string, that the mountain comes back as
a mountain. And, if that is the case,
then here you get the maximum of the E field.
You get an antinode here. But right here in the middle
you get your nodal lines and right here in the middle you get
your nodal lines, so you get a shift of
one-quarter wavelength. From where you earlier had
nodal lines, you get now antinodes.
And where you had earlier antinodes you now get nodes.
If this is open then these are the nodal lines.
And so those add locations across the wire here where there
is no potential difference at all moments in time.
And so this is then what we would refer to as closed.
And I want you to draw the parallel with our strings.
And I even think that in my problem sets I give you a chance
to work a little bit on this. Remember with strings,
we can fix the lengths of the string in such a way that we get
resonance. We can drive a string at one
end, just moving it a teeny weenie little bit and the end is
fixed. And we can set that system at
resonance. The wave length that we then
get, lambda of n, n being an integer,
so n equals one, two, three, etc.
That wavelength then is 2l divided by n.
You can also wiggle it here and leave this end open.
We even discussed that and I even demonstrated that.
In which case you get resonance for lambda n equals 4l divided
by 2n minus one. And, of course,
the frequency in which you reach that resonance omega of n
is always k of n times the speed of propagation which,
in this case, is the speed of light.
And k of n is then 2pi divided by the wavelength.
And so the way I am going to do this demonstration for you,
we have chosen the length of these wires such that the system
is at resonance when it is open. We have a length which is 4.25
meters, the wavelength is 3.4 meters, so the length is 1.25
wavelength.
And so when it is open, which it is here,
and this is one-quarter wavelength and this is one-half
wavelength and so on. If I leave this open,
and that is what I am going to do, and I have here a light that
is connected between these two wires that I can slide along,
so I have a light here, a light source,
this is a very special light source, has an enormously high
ohmic resistance. So that when I put it here that
I am not disturbing the potential difference between the
wires. I am not changing the boundary
conditions, really. Because it has an enormously
high ohmic resistance. And that light bulb lights up,
even with a minutely small current.
And so the first thing that I am going to show you is that if
I turn on the transmitter, which I will use the same one I
have here, and make sure I don't get electrocuted now.
We are going to take this off.
And it may help, actually, if we give you a nice
light situation so that it is not completely dark.
Now I am going to move the light across the two wires,
and you are going to see that there are locations were the
potential difference is very large because we are at
resonance. And, therefore,
you will see a lot of lights. And there are locations where
there are nodal lines, and you will see no light.
Let's first go here. That is the open end.
The open end is clearly where the electric field is very high.
It is like the situation of a string with an open end.
You have an antinode there. You have an antinode right
here. But, look, when I move the
light a quarter wavelength it goes out.
I must be now near a nodal. Maybe not exactly on the nodal
line. And now I move it another
quarter wavelength and the light goes on again.
Now I am going to move it back to the open end.
And you see the light is on. I move it quarter wavelength,
the light goes off. I move it another quarter
wavelength, light goes on. I move it another quarter
wavelength, light goes off. Isn't this amazing?
Now I think I am going to move it another quarter wavelength.
And the light goes on again. And so it is an amazing thing
that you see in front of your eyes that there is a standing
voltage wave between these two wires, and it is across the
wires, not along the wires. Not what I am going to do is
short this out here. When I put the light bulb right
in the middle here, I put the light bulb at quarter
wavelength. Now, when I short it out at the
end, I am no longer at resonance.
You no longer meet the resonance condition which I meet
now at an open end, something that you can very
easily verify. Now, the moment that I am no
longer at resonance. I no longer have a very high
potential difference at the antinodes, which I had before
when I was at resonance. But, for sure,
the moment that I short out these lines at the end,
the locations which are now nodal lines must become
antinodes. Now, maybe not super duper
antinodes, which are a huge potential difference,
but probably high enough for that light to go on.
Are you ready for this? Now I am going to short this
out. And, as you can see,
the light goes on. Now, it may not be as bright as
you have seen it before. But this is,
again, an antinode and the light is on.
And so you see a remarkable demonstration here where you see
the standing waves, you see the nodal lines,
you see the antinodal lines and you also see that it depends on
the boundary conditions that if you make this open then
one-quarter wavelength from the open end is an antinode.
If you have it open one-quarter wavelength from the open end is
a nodal line. And if you short cut it then it
becomes an antinodal line. I have a second demonstration
which is also very interesting. Let me turn this off before
some of you touch those wires and we have some problems.
I have another very nice demonstration,
which is a cable that is 127 meters long.
You see it here. It starts there,
goes all the way here to make you see that it is really big,
and then it goes back there. And, in this cable,
I am going to send in a voltage pulse.
It is a coaxial cable. And a coaxial cable has a wire
which is at the core, metal, and then there is a
cylinder around it. That is why it is called
coaxial. This is a conductor and this is
a conductor, and it is filled here with dialectic material to
insulate the two conducting surfaces.
And what are we going to do? We are going to send in here
voltage pulses which have a length of 100 nanoseconds,
100 times 10 to the 9 seconds, 100 nanosecond length of the
pulse. In other words,
if you look at the wire from the side, here is the wire and
here is the shell, just take a cross-section here
so you see only that shell. As the voltage pulse moves in
this direction then you will see locally, between the outer
conductor and the inner conductor, that E field.
And it goes to zero here and here.
And that marches then with a certain speed along the wire.
And so, as that pulse passes, there will be an E field like
this. But then there is no pulse.
There is no E field between this and that.
In that sense, it is sort of similar to this
transmission line, except the geometry is
different here. It is a coaxial cable.
Now comes the punch line. I have an option to keep this
end open so that the line at the very end is completely open,
and I have the option to short cut it.
And, when I short it out, a mountain comes back as a
valley. When I leave it open,
a mountain comes back as a mountain.
We are going to show you this pulse at three locations.
Where the pulse enters the system right at the beginning,
which is here. Then we will show you the pulse
at the end of the line here. And then we show it to you
again as it comes back. You are going to see it three
times, and you do. And the system now is closed at
the end. Here is the pulse that goes in.
And 1.3 microseconds later between this line and this line,
there is a marker here which says 1.3 microseconds,
the mountain has returned as a valley.
This is the pulse that enters and then when it has come back.
But now I am going to make it an open end.
Here we see the incoming pulse. And you see the one that comes
back is now 1.29 microseconds. Big deal.
A mountain comes back as a mountain, but now there is
something interesting. At the end of the cable where
it is open you see twice the voltage.
Do we understand that? I see someone say yes.
Why do we see twice? Remember the open string?
Now the reflecting wave and the incident wave have the same
polarity, so at the end they add up together and they give you
double the amplitude. That was exactly the same
situation with the string. When we drove in a pulse on the
string, when the end was open, remember the end moved up by
twice the amplitude of the individual pulse.
And you see that, too, here.
It has twice the amplitude. This is right at the end.
But now there is even more which I discovered purely by
accident. Well, not so much of an
accident. I always do some consistency
checks to see where physics works.
And so I said to myself, OK, 1.3 microseconds,
that is what it takes to make a roundtrip through the wire.
With all the confidence that I have in physics I said 2l
divided by c, l being 127 meters,
should therefore be 1.3 microseconds because that goes
with the speed of light. But when I calculated this,
I found that this is only 0.84 microseconds.
But then I said to myself why is this speed slower than the
speed of light? And I decided to ask you.
Why is the speed substantially lower?
Remember, we observe 1.3 microseconds.
That is what we observe. This is only 65% of the speed
of light. She knows the answer.
Who else knows the answer? I don't blame it.
It took me also two seconds before I realized it.
Nice try but it is the wrong answer.
But it is better that you try than that you don't try.
I think you mean well. You may not express it in the
most effective way. Do you remember,
or do you not, that we discussed that the
speed of propagation v is c divided by the square root of
the index of refraction, of the dielectric constant?
Remember that? I did at one point.
I even showed you how this dielectric constant is a strong
function of frequency for water. Well, when I looked up the
dielectric constant of the material that is in here,
I found that the dielectric constant kappa E is about 2.3.
And if you calculate one divided by the square root of
2.3 that happens to be 0.56. And so that's why the speed is
65% of the speed of light. So you see also the effect of
the dielectric constant here which is quite wonderful.
Remember the square root of kappa E was also called the
index of refraction. This is a nice moment for a
break. We will reconvene in four
minutes.
I would like to bring to a test now what we have learned,
and so you get a chance to do some thinking for a change.
I have here an electromagnetic wave linearly polarized E vector
in this direction, going to propagate in this
direction. And it is set up here 10
gigahertz radiation, radar, so lambda is about three
centimeters, and it is going to be polarized in this direction.
You have to take my word for it.
And, as it arrives here, this receiver is set so that it
receives a radiation in this direction.
If I rotate it 90 degrees then it doesn't receive it,
but that is a different demonstration that we did
before. Now I have a peculiar comb that
is metal. These bars are not connected in
here. They are just metal bars.
And I am going to put this here at the end, and I am going to do
it in two different ways. In one situation,
the comb is like this. And in the other situation the
comb is going to be like this. In both cases,
will the electric field of the wave that comes in be in this
direction? Here it will be in this
direction. In one of those two cases,
the radiation will go straight through.
I will hold it in there and the receiver is happy.
In the other situation, the radiation will not go
through. It will be reflected.
And now I am asking you to make a prediction.
Do you think that if I hold the comb like this in the beam,
that means like this, that the radiation will go
through? Or, will the radiation be
reflected? Or, do you think that in this
case the radiation will go through or will it be reflected?
Let's have a vote on that. First I want to ask the class,
if I hold the comb in the beam like this, do you think that the
wave will be reflected? Who thinks that the wave will
be reflected? Who thinks that the wave will
go straight through? Who thinks that in this case
the wave will go straight through?
Who thinks that it will be reflected in this case?
I am actually quite impressed by your answers.
Most of the time I get more wrong answers than correct
answers. Indeed, it is in this case that
the wave is practically 100% reflected.
And, it is in this case, that the wave will go through.
And, if you have a good understanding about the currents
that are running at these surfaces to make sure you make
the boundary conditions, you will be able to give the
answer for yourself. If you cannot then see me or
write me an email, but I want you to think about
it a little bit for yourself. I will demonstrate it now.
We have this 10 gigahertz signal modulated with 550 hertz
sound so you can hear it. That is the only reason why we
modulate it. So you can hear it.
It is a nasty sound. Here it is.
This is the transmitter and this is the receiver.
Here is the comb. I am going to first stick it in
like this, which is this situation.
And I kill it. Now I rotate it and now it goes
through. It is in this situation that it
goes through.
My hand is not a bad conductor. All of your bodies are very
good conductors. Look at my fingers.
I kill it. Look now at my fingers.
It goes through. You don't even need a comb,
which is a conductor. You can just do it with blood
and flesh. It goes through.
This experiment should give you some insight into the secret
behind Edwin Land's linear polarizers for light,
which we never discussed. Imagine that the radiation that
hits this plate is not linearly polarized but is completely
unpolarized. Then the component in this
direction will be reflected. But only the component in this
direction will be allowed through.
That becomes a linear polarizer.
That is the whole idea behind the linear polarizer.
So this, in a way, is for radar a linear
polarizer. Shine on here unpolarized radar
and all components in this direction will be gone,
only this component will go through.
Linearly radar light will come out.
And so it may give you some thoughts about how the optical
polarizers work. What they do is they align
strings of molecules in such a way that you get a behavior not
unlike these metal bars. Now I want to change gears,
and I want to talk for the remaining time about radiation
pressure. In modern physics,
we don't think of electromagnetic radiation as
plane parallel waves, which are infinite in all
directions. But we think of them as
bullets, as packages of waves. We call them photons.
Photons can be produced by, for instance,
atoms or molecules if they are in an excited state.
And then they decay to a low energy state,
they can radiate just one photon.
But photons, in our modern physics,
have momentum. If you throw a tomato at me,
and it is a rotten tomato, it will go splat and I feel a
push. Another tomato,
another push. Another tomato,
another push. So I feel a force.
And, in that same way, the light, when it hits me also
pushes on me. And that is what I want to
discuss with you. First of all,
the energy, and we have so many Es today, we even had an E of n,
and we had an E which was the E vector, and now we have E for
energy, so I will write the n here.
This stands now for energy. The energy in a photon is Max
Planck's constant times the frequency of the photon.
And Max Planck's constant which, of course,
plays a key role in quantum physics, is 6.63 times 10 to the
minus 34 joule seconds. If you tell me what the
frequency is, I will tell you what the energy
is, which is very different from the classic idea that we always
said, well, the energy is proportional to the electric
field strength squared. And we had plane waves with
infinity in this direction and infinity in this direction.
So we always had infinite energy into a wave of a plane
wave. Not for photons.
Take the example radial waves. I will do only one example for
you. We take 10 megahertz,
10 to the 7 oscillations per second, so that would be a
wavelength of about 30 meters. One photon contains energy,
and each radial photon with this frequency has exactly the
same energy. It is a quantized amount of
energy. And if you substitute for this
frequency, you multiply it by eight, then you get that the
energy is about 6.6 times 10 to the minus 27 joules.
In physics, we don't like to work with such very small
numbers so we often convert them to what we call electron volts.
One electron volt is 1.6 times 10 to the minus 19 joules.
And so that energy then becomes a more digestible number.
It is still small. It is about 4 times 10 to the
minus electron volts. I did this exercise only for
radial waves, but you can do it for any other
frequency that you choose. And we call that then a photon.
In modern physics, photons have momentum.
And the momentum of a photon p is the energy divided by c.
In other words, it is also hf divided by c.
But you remember from 8.01 that the force is db/dt.
Let's return to our tomatoes first so that you feel
comfortable that you know what we are talking about.
I am throwing rotten tomatoes at you.
And those tomatoes come in like this.
And then they lose all their momentum in this direction.
Of course they go plop. Momentum is conserved.
So the momentum that they lose must become my momentum because
momentum is conserved. Suppose I throw at you ten
tomatoes per second, that the mass of each tomato is
one-quarter of a kilogram. And I throw them at you at 40
meters per second, which I can do because I hire a
baseball pitcher. And the baseball pitcher can
throw them at 90 miles per hour, which is 40 meters per second.
And now I can calculate what the average force you will
experience is when all those tomatoes hit you.
You feel a force because the momentum of a tomato is mv and
the force that you will feel in the x direction.
Momentum transfer is now going to be 10 times one-quarter times
40 which is about 100 newtons. That is a substantial force.
It may actually throw you over. It is a huge force.
Let's now turn to photons. And so we have a beam of light
that comes in. And let's give the beam of
light cross-sectional area A. The light comes in this
direction. And this light strikes you and
you absorb it. It hits your face and it is
absorbed. And so the momentum in this
direction is destroyed. Momentum is conserved so you
must have a momentum in this direction.
You must feel a force. The pressure that you feel is
obviously the force that you experience divided by the area.
The force is db/dt divided by the area.
But that is also, if this is the energy divided
by c, I take db/dt, so I get one over c,
then I get my one over A and then I get dE/dt.
E being the energy now. I get here d of energy over dt.
This is joules per second. How many joules per second hit
me? That is what that means.
And this is the area of the cross-section of B.
And this is what we earlier called in this course the
pointing vector. We called that the mean value
of S. This is joules per second.
So this is joules per second per square meter,
which is exactly what the pointing vector was.
And so we can write now that the average pressure that you
will experience, if you absorb that light or the
radial emissions or radars or whatever that is,
is given by this relationship, namely that average pressure is
the average pointing vector divided by c.
You no longer have to think of S being E cross B divided by mu
zero. In fact, the whole idea of
individual photons, which is perfectly fine,
that is where it all came from, can simply be replaced now by
how much energy per second hits you.
And whether this is uv or infrared or radial or radar,
as long as you absorb it, if it goes straight through,
of course, then nothing happens.
But as long as you absorb it that is then the pressure that
you will experience. Remember that we calculated the
pointing vector at earth due to the radiation from the sun.
And I still remember that number because it's the solar
constant. It's the famous 1.4 kilowatts
per square meter. That is the pointing vector
from the solar radiation as it reaches earth,
joules per second per square meter.
I can calculate now, when I expose myself to the
sun, what the pressure is, the radiation pressure,
because I absorb it. You better believe it.
None of that radiation goes through me.
I absorb it. And I feel a pressure.
And I can calculate that pressure by taking this number
and dividing by c. That pressure is very,
very low. The c is huge.
In fact, I can even calculate, if I hold my hand up in the
direction of the sun, what the force will be on my
hand. Well, the force is the pressure
times the area. And, when I do that,
then I find that the force on my hand is only 5 times 10 to
the minus 8. So I hold my hand here.
There is the sun and there is this force on my hand of 5 times
10 to the minus. It is not going to knock me
over. It is not going to be like the
tomatoes. In fact, I don't even notice
it. It is completely insignificant.
The force, due to the radiation pressure of the sun,
is for you and me insignificant.
If I hold a mirror in my hands and I reflect the light back to
the sun then, of course, the momentum change
of the light doubles. It comes in like this and it
goes back like this. You have twice the momentum
transfer to you. And so then the force on you
and the radiation pressure on you doubles.
It is 100% reflection. The radiation pressure is twice
what it is for 100% absorption. It is clear that you and I will
never experience in our lives radiation pressure.
For all the light sources that we deal with and the distances
involved, the pointing vectors are always so small that it will
not be noticeable for you and me.
However, in astronomy that is different.
There are stars 20, 30, 40 times more massive than
the sun, which radiates so much energy that the atmosphere of
the stars is floating on the radiation pressure.
If the star would radiate more, it would blow its atmosphere
off. And, if it radiates less,
the atmosphere would come down. So it is very important in
astronomy. Comets have two tails.
I will show you shortly a picture of a comment.
It has two tails. In general, you and I can only
see one of the tails, which is the white tail,
but there is also a blue tail. The blue tail is the result of
solar winds. The sun, apart from emitting UV
and electromagnetic radiation, also loses hydrogen.
Protons and electrons. And, when they reach the earth
in large quantities, we see Northern Light at night
because they interact with the atmosphere of the earth.
But they also interact with carbon monoxide and the carbon
dioxide of the comet tail. And when they excite those
molecules, and the molecules de-excite, they emit blue light.
That is where the blue tail comes from.
But independently a comet has dust.
And it is the radiation pressure on these dust particles
that blows the particles away from the sun,
purely radiation pressure. These particles are a few
microns in size. And so you get a huge tail of
dust and now you get reflected sunlight off that tail of dust.
The dust tail is formed by radiation pressure.
And the reason why that light is not very bluish,
which you might expect if you accept the scattering,
is probably that those particles are a few microns in
size whereby the preferential scattering for blue light is no
longer very effective. That is why these tails,
which can be 100 million kilometers in length,
are white. Bright white ones are due to
the radiation pressure. And the blue one,
which is very faint, is due to the solar winds.
And I would like to show you Hale-Bop which was very
prominent in the sky in 1997. I hope that many of you have
seen it. It was extremely easy to see.
And I saw it 20, 30, 40 nights in a row.
I couldn't get enough of it. Every night I was staring there
for hours at the sky to see these gorgeous comets,
but I only saw one tail. That was my problem,
of course. Let's take a look at this
beautiful comment Hale-Bop. Who had seen it in 1997?
You were all born. Good.
There you see Hale-Bop. You see the blue tail.
That is the result of the solar winds, which is very faint and
not easy to see, unless you have a long exposure
like this picture and then you see the white tail which is the
result of radiation pressure. The sun is pushing onto the
small particles. And then you see the reflected
sunlight. You would expect that if the
dust particles are small enough, and if you see it at the right
angle that you would actually sometimes see this tail which is
mostly white, that you might see it actually
a little bluish, which we understand why that
would have to be bluish. Now I want you to test your
knowledge. Call it the brain teaser.
I have here a radiometer. Many of you may have a
radiometer in your rooms. They are very cheap.
It is a little glass bulb with low pressure inside.
It is not vacuum but it is low pressure.
And it has an arm here and an arm perpendicular to that like
windmills like this. And they can rotate with very
low friction in this direction. There are two,
one perpendicular and one like this.
And here is a little surface no larger than, which you will see,
a square centimeter or so. And this side is white and the
backside is black. And this side is black and the
backside is white. And then there is another one
like this. And we are going to radiate
light on this. The light will interact with
this whole thing. Now I am testing you.
Not very fair what I am doing because I know the answer.
Do you think, as the light strikes,
that this white surface will start to move in the board?
Because it is going to rotate. You are going to see that.
Will this go in the board and this come out of the board or
will it be reversed? Let's have a vote.
The first one to know who thinks that this white surface,
as it rotates, goes in the board and that this
will come out of the board? Who thinks that,
in fact, this white surface will come to you and this will
go in the board? It is about 50/50.
It is interesting. I can demonstrate it to you.
The one thing I can tell you, though, no, let me first
demonstrate it. Then we can test it.
It is nicer. There it is.
We will make it a little darker.
Try to remember what your prediction was.
Now I am shining onto it. You can clearly see the white.
This part is white and this is black.
And so the white surface is coming to you and the black
surface is going away from you. Have of you had it right and
half of you had it wrong. And the other half were not
even thinking so they had no answer.
How can it be that the white is coming to us?
Because radiation pressure would predict that the force on
the white surface would be about twice as high as the force on
the black surface because the white surface reflects and the
black surface absorbs. Well, the answer is very clear.
This has nothing to do with radiation pressure.
If you calculated the force on these surfaces due to radiation
pressure, it is insignificant. It is nothing.
It is even less than the force on my hand from the sun.
What makes this rotate has nothing to do with radiation
pressure. That is why it was not quite
fair when I said I am going to test you.
I like the idea that you may have sleepless nights,
you know that, and so I would leave you with
this. Think about why it is rotating.
And, if you cannot sleep, send me an email and we will
discuss it. Have a good weekend and see you 1066 1:16:01,984 next Tuesday.