There is energy in an electric field and there is energy in a
magnetic field. You remember that from 8.02.
And the energy density, mind the units in terms of
joules per cubic meter, for the electric field we write
with that u equals one-half epsilon zero E2.
There is no such thing as a free lunch.
You have to do work to create an electric field.
You have to assemble charges and bring them together.
That means work that creates an electric field.
And the same is true for magnetic fields.
When you have a solenoid and you create the magnetic field
inside the solenoid that caused energy.
And the energy density, mind the word density,
it is per cubic meter, for a magnetic field is B2
divided by 2mu zero. Now traveling EM wave,
traveling electromagnetic wave, so this is now a traveling
wave, we know that the magnitude of B is the magnitude of E at
any moment in time divided by c. I can write this as E2 divided
by 2mu zero times c2. But c2 is one over epsilon zero
mu zero. This is also one-half epsilon
zero E2. And what you see now is so
wonderful, so beautifully symmetric in electromagnetic
waves. One is the same as the other.
One cannot exist without the other.
And look at the energy density in the electric field of a
traveling wave. It is exactly the same as the
energy density in the magnetic field.
And so the total energy density is the sum of the two,
is epsilon zero times E2. But you can also write for that
epsilon zero times E times Bc, if you prefer that.
This, of course, is only true in vacuum when the
speed of propagation is c. Now it is a traveling wave.
And this traveling wave moves, and so it carries energy with
it. Now the question is how much
energy flows through an area which is, say,
one square meter area perpendicular to the direction
of propagation? Suppose I have a box here and
this side is one square meter, and I want to know how much
radiation comes out of there in one second.
The radiation is flowing in this direction with speed c.
In one second this box, it's quite a large box,
3 times 10 to the 8 meters. And all the energy that is in
there will flow through this one square meter in the time of one
second. And so the dimensions that we
are talking about now are joules per second per square meter,
which is also watts per square meter.
And that now is, of course, the total energy
density times c, the speed of light for which
you can write epsilon zero times E times B times c2,
if you like that. But you can also write that EB
divided by mu zero because c2 is one over epsilon zero mu.
And this should remind you of something that is in your far
distant past which is what we earlier have called in 8.02 the
poynting vector. And the poynting vector S was E
cross B divided by mu zero, and the units were watts per
square meter, which is exactly what this is.
And the reason why the cross disappears here is that with
electromagnetic traveling waves, E is always perpendicular to B.
That takes care of the cross. Now, both E and B are time
variable. And so the poynting vector will
obviously also be time variable. E is going to be proportional,
or you can write E is some E0 times cosine omega t and B is
some B0 times cosine omega t. In the poynting vector,
you get the cosine square of omega t.
But, since we are never interested in the poynting
vector on a timescale smaller than the period of oscillations,
we want to know the average over many oscillations.
What matters there is the average value of cosine square
omega t, one from the E and the other one from the B.
And that is one-half. And so we can conclude then
that the average value of the poynting vector,
time averaged, is one-half that value,
one-half EB divided mu zero for which you can also write
one-half, if you want to kill completely --
By the way, this is one-half E0B0.
It is important that you have the E0 because the cosine square
average is one-half. Here are the amplitudes.
You can also write for that one-half E02.
And so you write down for BE divided by c,
and then you get downstairs mu zero divided by c.
And the reason why I write it in this form is that it tells
you that if you know what the strength of the E field is,
that alone tells you then what the poynting vector is.
Because B is coupled to E through Maxwell's equations.
That is where B is E over c. And so all that matters then,
if you want to calculate what the poynting vector is,
is E0. Of course, B0 alone would also
be fine. Let's take an example.
Suppose we have an electromagnetic wave whereby E0
were 100 volts per meter, then I can calculate now --
And it is a traveling electromagnetic wave.
I can calculate what the average value of the poynting
vector is. That would become one-half
times 100 squared divided by mu zero times c.
And, if you do your homework on that, you will find that it is
13 watts per square meter. Now, if you exposed yourself to
13 watts per square meter, visible light and infrared
light, you take all your clothes off and expose yourself to that,
your body will absorb that. It will not go through your
body. X-rays may go through your
body, gamma rays certainly, but infrared radiation
invisible light gets absorbed by your body.
And so the question now is will that harm you?
And then the answer is no. You would hardly notice 13
watts per square meter. Your body itself radiates about
100 joules per second because of the body heat that you have.
And so the 13 watts per square meter that you absorb,
let's say your cross-sectional area is about one square meter,
just to round it off, so that means 13 joules per
second would be absorbed by your body.
And that will not affect you in any way.
But let's now take a situation that we have E0 equals 10 times
higher, which is 1000 volts per meter.
Now the poynting vector goes up by the square of E,
so now you are going to get that the poynting vector is 1.3
kilowatts per square meter. And that will fry you.
If you walk naked in a field of infrared or optical,
whereby the absorption 1.3 kilowatts per square meter,
that is very dangerous. You get skin cancer and worse.
Now, why do I mention that and why do I focus on that 1.3
kilowatts per square meter? Because that is exactly what
the sun does. Here is the sun,
and the sun is a rather powerful light bulb,
about 3.9 times 10 to the 26 watts.
And this is where you are on earth, and the distance to earth
is 150 million kilometers. I can calculate now how many
joules per second go through one square meter here,
and this one square meter is held perpendicular to the
direction to the sun. That value for S is,
of course, the radiation that leaves here.
That is my 3.9 times 10 to the 26.
And now I have to divide it by the entire surface area of this
sphere that goes in all directions that has this radius.
And that surface area is four pi times that radius square,
so that is 150 times 10 to the 9, because I have to go to MKS
units squared. And that is now the number of
watts per square meter that we receive at earth from the sun.
And that number, a very famous number,
is 1.4 kilowatts per square meter.
And that is why I picked the 1.3 to show you that if you walk
around on the beach and do not take care of yourself,
if you do that too long that is very dangerous because your body
absorbs that. This number,
called the solar constant, has major implications,
of course, for people who want to harvest solar energy.
The maximum that you can ever harvest, for every square meter
that you dedicate to solar energy, you only get 1400 joules
per second. You can never get more.
The electric power capacity of the United States is 700,000
megawatts for which you need 700 power plants.
A full size power plant is about 1000 megawatt power plant.
You need 700 of these power plants to have the capacity for
the United States. We are energy hungry.
We consume more than one-quarter of all the energy in
the world. If you want to replace this by
solar energy that is a major problem because you can only get
1.4 kilojoules per second for every square meter.
You can calculate how many hundreds of square miles in the
desert you would have to commit with solar cells,
which are extremely expensive, in order to get electricity.
And the efficiency, of course, is never 100%.
And also, when the sun is low in the horizon,
then you don't have this one square meter perpendicular to
the direction of the sun. All of that has to be taken
into account. Solar energy is not very
important in our lives, unfortunately.
And this is the number that is the ultimate limit.
Now comes the question, is there such a thing as an
electric field over 1000 volts per meter in the solar
radiation? Could we actually measure the
electric field just from the radiation of the sun?
And the answer is no. And the reason is that the
radiation is not really in the form of our idealized plane
waves, but more important than anything else is that there is
no such thing as just one wave from the sun that has the
amplitude of 1000 volts per meter.
In fact, the radiation reaches us in small packages.
It is broken up in pieces, so to speak.
However, since the energy flow is 1.4 kilowatts per square
meter, it is perfectly OK with me that you refer to this number
as the poynting vector. I have no problem with that.
But it is a little bit naive to associate with that an electric
field that can be measured that has an amplitude then of 1000
volts per meter. And so this now is the right
time to take a close look at how electromagnetic waves are
produced. In a nutshell it comes down to
this. You can create electromagnetic
waves if you accelerate charges. Charges that are stationary or
moving at constant velocity are surrounded by a radial field
poynting away or poynting inwards, depending upon whether
the charge is positive or negative, and there is no kink
anywhere in these fields. Whether it has a constant
velocity or whether it stands still they are radially electric
field lines. The moment, however,
that you accelerate it, as you will see today,
you introduce a kink in those field lines.
And that kink is responsible for the electromagnetic
radiation. It manifests itself as
electromagnetic radiation. I will follow a classic
derivation that is verbatim given that way in Bekefi and
Barrett. I therefore advise you
strongly, for the next 30 minutes, not to take any notes
but try to follow my arguments. That will help you way more
than if you try to also takes notes, because it is really
verbatim from Bekefi and Barrett.
It is a classical derivation with many simplifying
assumptions, but it gives a very nice result which has great
practical applications. Suppose I have here a charge q
which is located at position A. We have a charge q.
It is at 0 and it is at rest. And I am going to accelerate
that in this direction. I am going to accelerate it
with an acceleration which is in this direction.
I will not put the vector in there now.
I will do that later. Otherwise, the picture becomes
too complicated. And I do that for delta t
seconds, only very brief. And then it ends up at location
0 prime which is here. Now it has a velocity in this
direction, and that velocity u in this direction is now A times
delta t. That is 8.01.
And so it is cruising now with constant velocity,
and we just let it cruise all the way.
It is now, again, a charge with constant
velocity. And I look at it t second
later. And so at time t we will find
it in 0 double prime. And it is still cruising,
and we just let it cruise. We are not going to interfere
with it anymore. And here is then 0 double
prime.
The whole exercise from 0 to 0 double prime took so many
seconds. That means there is a sphere
around point zero and that sphere has a radius which is c
time t plus delta t. Outside that sphere the world
has no knowledge that this charge was accelerated because
that message has to travel with the speed of light.
I am going to draw a circle, but in reality it is a sphere
in all directions, which has a radius c,
t plus delta t. Delta t, by the way,
is way much smaller than t. And outside that sphere there
is no knowledge, the world does not have any
clue about the fact that this object is being accelerated.
I will mark again to make sure that you can make a connection.
This one has its center at 0 and the radius is c times t plus
delta t. And if you ask me what is the
electric field right there that is radially pointing outwards if
q is positive, this world does not know yet
that there was an acceleration. And if you ask me what is the
electric field here, the field line is line so.
And so the electric field, if it is a positive charge,
is pointing radially outward and has no knowledge that there
was any acceleration. And so I call this my position
vector r, but you can take that r in any direction that you
want.
Now let's look at the world that does know that there was a
change. When the object was at 0 prime
the acceleration stopped, and so it started cruising with
a constant velocity. And so from that moment on the
field lines are again nicely radially outwards.
And so by the time that it reaches point 0 double prime,
I can draw these field lines radially outwards.
And I think, of this original field line,
as a stick that was connected to the charge in that direction.
I could have picked another direction.
And so that field line is now here outwards.
And the world that knows about this is the world which has a
sphere around 0 prime with a radius c times t.
And so I am going to draw another circle,
which in reality is a sphere about point 0 prime.
And this sphere here has now radius ct.
And so this, to remind you what it is,
has the origin at 0 prime. And the radius is ct.
And so everything inside that sphere recognizes that the field
is radially outwards from 0 point 0 double prime and
everything out side here still thinks that the electric field
is like this. But this field line was this
one, the stick that I attached to it, and so there must be a
connection between here and there.
And that connection is only in that very thin shell which has a
thickness c delta t. Let me first make a drawing of
a triangle. That is going to be important.
It is this triangle. This length here has a length u
perpendicular times t. And you can easily see that why
that is the case because this distance here is ut.
Now, you may say, well, it wasn't going with
velocity u here. I grant you that.
But, keep in mind, in whatever follows that u is
way, way smaller than c, delta t is way,
way smaller than t, and so r which is c times t is
way, way larger than ut. When you see a little
distortion in this picture, I cannot, of course,
make delta t way, way smaller than t.
You wouldn't see any decent picture.
Let me first go to u perpendicular t.
There is here a velocity vector u.
And the component perpendicular to r is a component in this
direction. It cruised for t seconds,
so this length is u perpendicular t,
to a very good approximation. Forgetting about this teeny
weenie little distance. We know what this is.
U perpendicular is the component of the vector u
perpendicular to my position vector r.
And this little section here must, of course,
have thickness c delta t. That is the difference in this
radius and this radius, c delta t.
Look at this green triangle. Now comes a key thing.
This electric field line must connect with that electric field
line. It was one in the same.
It has been broken now. And the connection must go
through this thin layer. I will draw here a connection.
And then here you have to round it off a little.
It cannot be sharp. And this rounds off a little.
This point here represents the beginning of the acceleration,
and this represents the end of the acceleration.
And so then the field line goes swoosh.
And now comes my task, and that is to calculate the
electric field inside that shell because that,
and only that, is responsible for the
electromagnetic radiation. And so the field line goes like
this. There must be a component of
the electric field which is in this direction and there must be
a component which is in this direction.
I just decompose it. And I call this one E parallel.
My notation parallel always means parallel to r,
and my notation perpendicular always means perpendicular to r.
And so this then would be called E perpendicular.
And this triangle that you see here is congruent with this
green one. That is why I gave you the
dimension of the green one. And so the question now is what
is this component E perpendicular?
You can just feel in your stomach that that is the one
that is responsible, that is electric field in the
traveling wave that moves outwards because this whole
shell moves outwards with a speed c.
And notice this one is perpendicular to the direction
that I have chosen of propagation you are watching
here. And this shell comes over you.
And it is always the E field that is perpendicular to your
line of sight. That is the E field in the
traveling wave. Our task is now to calculate
the E vector perpendicular here. Well, if you look at the fact
that this triangle is congruent with this one then you see that
E perpendicular divided by E parallel must be u perpendicular
t divided by c delta t. U perpendicular t,
we know that u is at, so u perpendicular is a
perpendicular time t. A perpendicular is now the
component of the acceleration perpendicular to r.
I don't want to put it in here because it becomes too
cluttered. I will make a new drawing
shortly. I will put the a in there.
Acceleration is in this direction.
And so a perpendicular is like this.
And so I can replace u perpendicular t.
I can replace it by a perpendicular.
Sorry, this is delta t. You guys should have screamed
because the acceleration only lasts for delta t seconds.
Now I get a perpendicular times delta t times t divided by c
delta t. And so that is a perpendicular
times t divided by c. If now I can calculate what E
parallel is, I am done because then I know what E perpendicular
is. I put E parallel here and I am
in business. Before I do that,
I want to eliminate t. And I am going to write for
this t, r divided by c. It is also a perpendicular
times r divided by c2. And so I can write down now
that E perpendicular is this time E parallel.
How do we find E parallel? 8.02 Gauss' Law.
I make a pill box, and the pill box is going to be
like this. I will make a drawing of it.
And this surface is here in the world that doesn't know yet what
happens. And this surface here is in the
world which is in turmoil, which is the transition.
And so I am going to make that pill box for you.
And so I draw here again this. This is a line that goes
somewhere through this point here.
It is not this one. It is somewhere there.
And here is my pill box.
That is my pill box. And so, in that pill box,
I have in the outside world, the outside world here I have
that E field. I will put that in as an E.
It is in the world that doesn't know yet what happens.
That is radially go through point 0.
Inside the box, I have this E parallel coming
in like this. And then going straight through
the sides of the box is this one E perpendicular.
This comes in E perpendicular, and that E perpendicular goes
out. Now, since this is in vacuum,
there is no charge density inside this box.
The divergence of E must be zero.
And that means that the E vector here must be exactly the
same as the E vector here, because the contribution these
two is zero so this must also be zero.
There is no charge inside the box.
But I do know what this E field is.
That is 8.02. If someone tells you that I
have a charge sitting there. What is the electric field at a
distance r that is Coulomb's Law?
This is simple. It is not Coulomb's Law.
It is Gauss's Law. But, in any case,
you will find very easy that his E vector is q divided by 4pi
epsilon zero r2. It falls of as one over r2.
You people should remember that from 8.02.
And so now we have accomplished,
the 13 minutes are almost up. E perpendicular now is,
therefore, A perpendicular times r divided by c2 times q
divided by 4pi epsilon zero times r2.
And you lose one r. And so this is the classic
derivation also already known in the late 19th century which is
now the electric vector, the strength of the vector
which clearly is responsible for the electromagnetic traveling
wave because it is perpendicular to my direction r that I have
chosen. It is inversely proportional to
r. I will come back to that.
That is a natural consequence of the conservation of energy.
And this field is very different from static electric
fields which fall off as one over r2.
This is a travel wave that has an E field that falls off as one