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14. Maxwell’s Equations and Electromagnetic Waves I

14. Maxwell’s Equations and Electromagnetic Waves I

Prof: Before we start
today’s lecture, I just wanted to explain to you
that at the end of the class, I have to run out and catch a
plane. So if you have to discuss any
other administrative matters, you should do it now,
because you won’t find me after class.
I couldn’t take this weather.
You know where I’m going?
That’s the story of my life.
So nothing else?
Student: Is there going
to be an equation sheet for the midterm and if so,
is it possible to get it? Prof: You will get the
equation sheet. Do you want it now?
With the exam?
I don’t give it out before,
but you will have all the reasonable equations on that
sheet. You know the reason for that,
right? The reason is that some of you
may be going into medicine and you control the anesthesia on
me, one of these days in the future.
If I held back the equation
sheet, I know you will hold back the painkillers.
It’s just making a deal,
the way things are done nowadays.
I made a deal with you guys.
I couldn’t sleep last night,
because today is when we’re going to solve Maxwell’s
equations and get the waves. I just couldn’t wait to get to
work. It’s great.
No matter how many times I talk
about it, I just find it so amazing.
So here are these great Maxwell
equations. I’m going to write them down.

Those are the equations.
The first one tells you that if
you draw any surface and integrate
B over that surface, namely, you’re counting the net
number of lines coming out, you’re going to get 0.
That’s because lines begin and
end with charges and there are no magnetic charges.
If you look at any magnetic
field problem, the lines don’t have a
beginning or an end. They end on themselves.
So if you put any surface
there, whatever goes in has to come out.
That’s the statement of no
magnetic charges. This one says,
yes, there are things that manufacture electric field lines
called charges. They make them and they eat
them. Depending on how many got in
the volume, the net will decide the flux out of that volume.
This one, without that,
used to say electric field is a conservative field.
It defines a potential and so
on. But then we found in the time
dependent problem, E⋅dl,
we said rate of change of the magnetic field.
So changing magnetic field
sustains an electric field. And this one says a changing
electric field can produce a magnetic field.
In addition,
a current can also produce a magnetic field.
This is what we have to begin
with. Now I’m going to eventually
focus on free space. Free space means we’re nowhere
near charges and currents. And the equations become more
symmetric. This has got no surface
integral, that has got no surface integral.
The line integral of this guy
is essentially the rate of change of flux of that.
And the line integral of that
guy is – the rate of change of flux of this.
If I’m going to get a wave
equation, the triumphant moment is when I do this and do that
and out comes the wave equation. But if at that moment you have
not seen the wave equation, you’re not going to get the
thrill. So I have to know,
how many people remember the wave equation from the first
part of the course? Okay, so I think I have to
remind you of what that is. It’s a bit like explaining a
joke, but still, I have to do that,
because otherwise we won’t have that climactic moment.
So let me tell you one example
of a wave equation. People have seen them before.
Here is the simplest example.
You take a string,
you tie it between two fixed points and it has a tension
T. Tension is the force with which
the two ends are being pulled. Then you distort it in some
form and you release it, and you see what happens.
Now let’s measure x as
the distance from the left end and y as the height at
that time t. The wave equation tells you an
equation governing the behavior of y.
It tells you how y
varies. Now if you want to do that,
you may think you have to invent new stuff,
but you don’t. Everything comes from Newton’s
laws. That’s to me the amazing part,
that you have these equations, then of course the Lorentz
force equations. Then you’ve got Newton’s laws
and that’s all we have up to this point in the course.
So there should be no need to
invoke anything beyond Newton’s laws to find out the fate of
this string. And namely, we are going to
apply F=ma, but not to the whole string at
one time, because a is the
acceleration and whose acceleration are we talking
about? Piece of string here,
piece of string there? It can all be moving at
different rates. So you pick a tiny portion,
which I’m going to blow up for your benefit.
That’s a piece of string.
Its interval,
it lies under region of length dx.
In other words,
you take a dx here and look at that string.
It is being pulled like that
with a tension T and it’s being pulled like this with a
tension T here. That’s the meaning of tension,
pulled at both ends. If this was a perfectly
straight segment, there’ll be no net force on it,
because these forces will cancel.
But if it’s slightly curved,
then this line and that line are not quite in the same
direction and the forces don’t have to cancel.
So let’s take that force and
write its vertical part. If this angle is
θ, T times
sinθ is the force here.
-T times (because
T is pointing the opposite way)–
let me call this θ_1 and
θ_2 is the angle you have here.
Do you understand?
The vertical force is restoring
the strength to where it was is what I’m focused on.
So you resolve that force into
a vertical part, up and this one has the
vertical part down. Now if θ_1=
θ_2 they cancel and that’s what I just
said. But θ_1
and θ_2 are generally not equal,
because if I go to the top of the string,
for example, this end is pulling this way,
that end is pulling that way. There’s a net force down.
So you have a net force because
this angle or the tangent to the string is changing with
position. That’s why they don’t cancel.
Now the approximation I’m going
to make is that theta is small, measured in radians,
and I remind you that sinθ is essentially
θ – very small corrections.
And cosine θ is 1
− θ^(2)/2 corrections.
And tanθ,
which is the ratio of these guys, is θ other
corrections. For a very small
θ, you forget that,
you forget that. You keep things of order theta
in every equation. So for small angles,
cosine is just 1, and sinθ is
θ. If you do that,
and sinθ and tanθ being equal,
I can write it as tanθ_1 –
tanθ_2. But that’s in the
approximation, where tan and sine are all
roughly equal. That is really T times
dy/dx at the point x dx −
dy/dx at the point x.
Are you with me?
dy/dx from calculus,
you remember, is the slope.
It’s a partial derivative.
Why is it a partial derivative
now? Can you tell me?
Student: It’s all
dependent on time. Prof: It can depend on
time and I’m doing F=ma at one instant in time.
So I don’t want to take this
guy’s slope now and that guy’s slope an hour later.
I want them at the same time,
at a fixed time. Now what do I do with this?
Have I taught you guys how to
handle a difference like this? There’s one legacy I want to
leave behind is what to do when confronted with a difference
like this. You have any idea what to do?
Pardon me?
Student: Multiply by
one. Prof: Multiply by one.
Then what do you do with that?
I’m asking you to evaluate the
difference between this guy at x dx – the same thing at
x. What is it equal to
approximately for small dx?
Let me ask you another question.
Suppose I have f of x
dx and I subtract from it f of x,
what is it equal to? Student:
>Prof: If I divide by
Δx and take the limit, it becomes a derivative.
If I don’t divide by
Δx, this is then df/dx
times Δx. There may be other corrections,
but when Δx goes to 0, you may make that
approximation. That is the meaning of the rate
of change. If I tell you,
here’s a function and that’s the same function somewhere to
the right, what’s the difference?
That’s what the role of the
derivative is, to tell you how much it
changes. This slope changes from here to
there, therefore it’s a derivative of
something which is already a derivative,
so it’s d^(2)y/dx^(2) times Δx.
In other words,
the f I’m using here is itself dy/dx and the
df/dx is then the second derivative.
So that’s the force.
Now that force we must equate
to ma. That’s the easy part.
What is the mass of this
section, this portion of string? If the mass per unit length is
μ, that’s the mass of that section.
Then what’s the acceleration.
Acceleration is the rate at
which y is going up and down. That is d^(2)y/dt^(2).
Canceling the Δx,
we get this wave equation which says d^(2)y/dx^(2) –
μ over T d^(2)y/dt^(2) is 0,
and that’s the origin of the wave equation.
Any time you get something like
this, you call it the wave equation.
I’m going to write this as
d^(2)y/dx^(2) – 1 over v^(2) d^(2)y/dt^(2 )=0
where v is √(T/μ).
I use the symbol v,
because this will turn out to be the velocity of waves on that
string. Namely, if you pluck the string
and make a little bump here, let it go, the bump will move
and v is going to be the speed at which it moves.
The speed of waves is
determined by the tension and the mass per unit length.
Now I want to show you that
this really is the velocity of the wave.
How do we know that’s the
velocity? It’s got dimensions of
velocity, but maybe it’s not the velocity of this wave.
Maybe it’s something else.
I want to show you that
v stands for the velocity of the waves in this medium.
The way you do that is you ask
yourself, what is the nature of the solutions to this equation?
If y(x,t) satisfied the
equation, what can you say about its functional form?
What functions do you think
will play a role? Does anybody have any idea what
kind of functions may play a role?
Student: Sine function?
Prof: Sines maybe.
With single oscillators,
when you have something something is d/dt
squared, you had cosine ωt.
Whatever you can do for t
you can do for x. You can have a cosine in
x. So you might think it is all
sines and cosines and exponentials.
All that is true,
but it turns out the range of solutions is much bigger than
that. I’m going to write down for you
the most generation solution to the wave equation,
namely what’s the constraint it imposes on the function?
The answer is this – y
can be any function you want of x – vt.
I don’t care what function it
is. So if I call z as a
single variable x – vt, y can be a function of
this single combination, x – vt.
In other words,
y can depend on x and y can depend on
t. If it depends on x and t in an
arbitrary way, of course it won’t satisfy this
equation. But I’m telling you that if it
depends on x and t only through this combination,
x – vt, it will satisfy the wave
equation. In other words,
here’s the function. You don’t even have to
think, x – vt squared over some number
x_0^(2). That’s a completely good
solution to the wave equation. I guarantee if you took this,
it would satisfy the wave equation, because it’s a
function only of x – vt. A function that’s not a good
function is y=e^( −x2) −
v^(2)t^(2). That’s not a function of x –
vt. That won’t satisfy the wave
equation. So why is that true is what I
want to show you. Let us take dy/dx.
Remember, y depends only
on this combination z, therefore it equals df/dz
times dz/dx. But z is x – vt,
dz/dx is 1, so it’s just df/dz.
Now let me rush through this
and do it one more time. Everything will go the same way.
You will get this.
But how about dy/dt?
If you want dy/dt,
again you take df/dz times dz/dt.
But that gives you–this should
be written as a partial–dz/dt is
df/dz times -v. Can you see that?
By the chain rule,
you take any function you have, first differentiate with
respect to z. Then differentiate z
with respect to x and with respect to t.
If you do an x
derivative, you get 1. If you do a t
derivative, you get -v.
Now if I do it twice–again,
I don’t want to spend too much time–if I do it twice,
I’m going to skip the intermediate step.
I’ll get a -v and a
−v^(2). It follows then that if I took
1/v^(2) times that, can you see that now?
1/v^(2) d^(2)y/dt^(2) is
the same as v^(2)y dx^(2).
Therefore to satisfy the wave
equation, it can be any function at all,
as long as it depends on x and t in the
combination x – vt. That is just a mathematical
fact. You just have to do the chain
rule, you will find that it’s true.
But more importantly,
I want you to understand what it tells you about the wave
propagation. Let me pick x_0
=1 for convenience. Let me plot this guy at time
t=0. This solution looks like
e to the -x squared, which is a function
like that, at t=0. What does it look like at t
=1 second? Think about it.
t=1 second looks like
e to the -x – v whole squared.
That means it’s the same bump,
shifted by an amount v. See this function x – v
squared behaves with respect to x=v the same as the
original function around x=0.
Whatever happens to this guy
here happens to that guy at distance v to the right.
You can see that if you wait
t seconds, it would have moved a distance
vt. That’s why you understand that
vt, v is the velocity of
propagation, because what this is telling
you is that if you start the system out with some
configuration at time t, t=0,
later on the function looks like the same function
translated to the right at a velocity v.
That means its profile,
whatever profile you have, is just moved to the right.
Now can you think of another
family of solutions besides this one?
Why should the wave move to the
right? Yes?
Student: It’s x
vt. Prof: Do you understand
that if it was x vt, you go through this derivation,
you’d get a v^(2) instead of a
−v^(2) and that doesn’t matter.
So the answer is,
the most general solution to the wave equation is any
function you like of x – vt any function you like of
x vt. This will describe waves going
to the right; that will describe waves going
to the left, and you can superpose them.
Because it’s a linear equation,
you can add solutions. This is what I want you to know
about the wave equation. This is one example of how the
wave equation comes and what the meaning of the symbol v is.
Now let me go to the old
Maxwell equations. Now I want to do them and ask
myself. I’m going to write down a
solution and I’m going to see if it obeys these four Maxwell
equations. If I write down a solution for
the electric field and the magnetic field,
they have to obey all those conditions, those four
equations. That means, let’s take first
the case of working in a vacuum, when there is no current,
when there is no charge. So you want a surface integral
of E on any surface to be 0
and the surface integral of B on any surface to be 0.
Look at the surface integral
first. Do you understand?
This was never going to be
non-zero. This will be non-zero near
matter but we are far from any charges, so there is nothing to
enclose in the volume. There are no charges there.
Look at those equations.
I give you a function and I
say, “Tell me if it satisfies Maxwell’s
equations,” what do you have to do?
what is it that you have to do? Any ideas on what you must do
at this point to test? Any ideas?
What do you have to see of a
potential solution? Come on, this is not the New
York subway. You can make eye contact.
What is this?
You have no idea what you may
have to do? What if I give you a function
and say, “See if it satisfies the wave
equation,” what will you do then?
Take the function and then what?
Student: Differentiate
it and see– Prof: Yes!
Take x derivative,
y derivative and all that, second derivative.
If it vanishes,
you’ll say it satisfies it, right?
That’s what it means to satisfy
it. If I give you fields
E and B that depends on space and time and I
want you to verify if it satisfies this,
what do you have to do, is what I’m asking?
Student: Come up with a
situation? Prof: No,
it must be true in every situation.
In other words,
what’s the surface on which this integral is done?
On what surface do we do the
integral? Pardon me?
Student: Closed.
Prof: Closed.
But beyond closed,
anything else? Where is it located?
How big is it?
Pardon me?
>Prof: I didn’t hear that.
Student: It’s
everywhere. Prof: It could be
anywhere and it could have any size.
It could have any shape.
And it must be true for all
those surfaces. So if I just give you a field,
E (x,y,z and t), it’s a lot of work,
right? Draw all possible surfaces.
On them, you can take the
surface, divide it into patches, do the surface integral patch
by patch, and you better keep getting 0.
When you’re done with that,
then you take the magnetic field.
Then the other two Maxwells
tell you, take a loop, any loop, go round that and do
the line integral of B. That should be the flux
crossing it, the electric flux or magnetic flux.
And it should be done for every
loop and for every surface. You realize that looks
impossible. But there is a quicker way to
verify all of these equations, and the quick way is the
following – if these equations are true for a tiny loop,
in this case, true for a tiny volume or a
tiny surface enclosing a tiny volume,
then it’s true for big ones. And likewise,
if true for a tiny loop, it is true for any arbitrary
loop. In other words,
if I can show that at any arbitrary point,
if I pick an infinitesimal loop or infinitesimal surface,
and these guys satisfy these equations on an infinitesimal
thing, it’s going to satisfy it on a
macroscopic thing. That’s what we want to
understand. How does that come about?
It comes about due to a very
beautiful way in which we define the sum of two surfaces.
So here is a surface.
I’m going to take everything to
be a cube, but it’s not important.
Here’s a surface.
Let’s call it
S_1. And it’s got its outward normal.
That’s the definition of
surface integral is you draw the normals pointing outwards and
you take E⋅dA
on every face of the cube and add it.
The area vector is defined to
be outward pointing normal to every face.
So you take the little cube.
E may be pointing this
way there or that way there, but you take the dot product of
the area and you add them. Then I take a second cube here.
Let’s see.
It’s got its own outward normal.
This is surface 2.
Now imagine gluing this guy to
this guy, so it looks like this. You glue this common face.
Bring that surface over and
glue them with that face common to both.
Then I claim that the surface
integral on this one the surface integral on that one is the
surface integral on the union of the two.
In other words,
let’s see, take one cube, take another cube,
and that surface integral will be the same as on a longer
object. I don’t know what this is
called, something over that. The two cubes used to be
sharing this. Do you understand why?
Let’s understand why the
integral on this guy, which is, if you want,
two cubes joined to form a rectangular solid.
That surface integral is the
sum of these two surface integrals.
Think for a minute about why
that is true. If you compare them,
piece by piece, this face matches that face.
This face matches that half.
This one matches that half.
But something is missing in
this that’s present here, do you agree?
What is it we don’t have?
What’s missing in the equal
union? Yes?
Student: Two sides
equal _________. Prof: The two common
faces are missing in the big surface, but it does not matter.
The two common faces do not
matter because when you do the surface on the first one,
the outward normal will point out of that,
whereas if you do it on the second one,
the outward normal will point into the other one.
When you glue them together,
the common face is opposite orientations.
the surface integral on the big solid is equal to the sum of the
surface in integrals in the two small solids.
Do you follow that?
That’s the way it works.
Then you can go on adding more
and more pieces to this, right, and you can build like
Legos and arbitrarily complicated blob.
Remember, these are all very,
very tiny volumes, so you can build them up to
look like a big thing. It may look like a pyramid.
If you look very closely,
you’ll have the steps like in a real pyramid,
but the steps here can be made arbitrarily small.
Therefore if the surface
integral was 0 on every little piece that made up the big
object, then it’s going to be 0 on the
big object, because the integral on the big
object is the integral of the tiny pieces that make up the big
object. This is a very profound idea,
because the big object has fewer surfaces than the small
ones, because when you cut it,
you create two new surfaces, but they don’t contribute
between the two of them, because they cancel.
And in particular,
if the surface integral had not been 0,
but because of matter it was equal to charge inside volume 1,
and this right hand side of this one was the charge inside
volume 2, you can see that the surface
integral on the bigger surface will be the sum of the two
charges. So anything you’re trying to
prove about surface integrals, even if the right hand side is
not 0, if it is true in a tiny cube,
true in a union of two cubes and then in a union of any
number of these tiny guys, therefore on any surface.
That’s the thing to remember.
What we will do then is not to
take arbitrary surfaces, but infinitesimal surfaces and
we’ll prove for them that the Maxwell equations are satisfied
for the solution that I come up with.
Then you’re guaranteed that it
will work for an arbitrary surface.
When you do line
integrals–this is for surface integral–line integral is even
easier. Suppose I’m taking the line
integral of some function around this loop.
This is loop
L_1. I take a line integral of some
field on L_1. Then somebody wants to do a
line integral on L_2 that looks
like this, E⋅dl
on line 2. You agree that this that really
is the same as the line integral on the union of the two loops
where you delete the common edge.
You have to understand this.
I will not leave any child
behind. You have to know why this is
true. I’m taking a lot of time so you
know where it’s coming from. If you compare an integral of
any field on the big rectangle compared to the two squares,
the only difference is that the two squares had these two sides,
but they were doing them in opposite directions,
so it does not matter. So if the line integral of
this, for example, was some flux coming out of
this one, and the line integral of that
was the rate of change of flux coming out of that one,
if it was true, then it’s guaranteed that the
line integral on the big rectangle will be the sum of the
fluxes or the rate of the change of fluxes coming out of both.
So any Maxwell equation,
if it’s true for a tiny square, infinitesimal square,
will be also true for anything bigger.
There’s only one subtlety when
you do loops and that’s the following – if you’re living in
a plane, you can prove the result for
the plane in the plane of the blackboard,
because there no loop I cannot form by joining these guys.
But we live in three dimensions
so that we may have a loop like this, floating in three
dimensions. Then what you have to do is,
you’ve got to find any surface with a loop as a boundary,
maybe this dome. Then the integral on–let me
hide this for you. This is the part behind;
this is the part you can see. That’s the same as tiling it
into little squares and doing an integral on each one of them
like this. You have a dome of some big
building, you take little tiles and you tile the building.
Then all the interior edges
cancel, and all that remains is the edge that is the edge of the
original surface. So even in 3D it’s going to
work, but now you should be ready for loops that are not
lying in the xy plane. So you will have to prove it
for the three independent loops. You will have to prove it for
an infinitesimal loop in that plane, infinitesimal in that
plane, infinitesimal in that plane.
If it’s true for three
independent directions, then by combining those little
pieces, you can make yourself any surface you want.
In other words,
I’m saying, given a rim, you can build any surface with
the rim as the boundary. If you can take the little flat
pieces in any orientation, and it’s enough to have them in
the xy, yz, and zx planes.
So the strategy that I’m going
to follow, this is something one can skip,
but I wanted you to know the details,
if you really want to know where everything comes from.
I’m going to write down or
search for a solution to Maxwell equations.
I’m going to make it have a
certain form. You remember how we do this
with equations. We assume a certain form,
stick it into the equation, play with some parameters till
it works. I’m going to make it work on
infinitesimal loops and infinitesimal cubes and that’s
going to be enough, because if it works on this
tiny loop, it works in a big loop,
works in a tiny cube, works on a big cube or
arbitrary surface. So let us write down the
functional form that I’m going to use.
The functional form that I’m
going to use looks like this. I’m going to take an electric
field that is entirely in the z direction and the
z field will depend on y and t.
And I’m going to take a
magnetic field which is going to be in the x direction and
the B field is going to depend on the y and
t. So let me tell you how my axes are defined here.
This is x, this is
y, this is z. The electric and magnetic
fields, the electric field will always point along the z
axis. The magnetic field will always
point along the x axis. They will not vary,
as you vary x or z.
They’ll vary only if you vary
y by assumption. In general, it can vary with
everything, but I’m trying as a modest goal
to find a simple solution which has a dependence on only two of
the four possible coordinates. It depends on only y and
t, rather than x, y, z and t.
Let’s get any solution.
We cannot get every possible
one; let’s get something and
something, I assume, has this form.
So this is called a plane wave,
because if you take the plane y=0,
E and B are constant on that plane.
Because when y is fixed
or some value, E and B are not
changing as you vary x and z.
So you should think of it as
plane after plane and on each plane, the field is doing
something. It’s doing the same thing.
If I draw a plane here,
that field is a constant. E is a constant and that
plane and B is a constant on that plane.
On another plane,
it could be a different constant, but within the plane,
it varies from plane to plane. So that is not an axiom,
that’s not a law. That’s an assumed simplicity in
the function I’m looking at. There’s no theorem that says
that every solution to Maxwell equations much obey this
condition. In general, these functions
will be functions of x, y, z and t.
The trick will be superposition.
If I can get a solution that
depends only y and t,
and you can get a solution that depends on z and
t, I can add them up and they will
still be a solution, because the wave equation is
linear. You can add solutions.
Then our sum of the two
solutions will be a function of y and z.
You can bring in the x
and so on. So you do the simplest one,
then you can add them. So here is my function and I
have to know what I can say about these two functions.
You understand?
In general, the electric field
has three components, the magnetic field has three
components. Each of them depends on four
quantities, x, y, z and t.
Big mess.
In our simplified solution,
the only unknown component of E is
E_z. I’m assuming there is no
E_x and there is no E_y.
And there is no
B_y and there is no B_z.
Now you might say,
“Why don’t you go a little further?
Kill the B also.”
You can try that.
If you kill the B,
you will find the only solution is to get everything equals 0.
So by trial and error,
we know this is the first time you can get something
interesting going. If you make it simpler than
that, you get nothing. You can make it more
complicated, but not simpler. So now, will this satisfy the
surface integral condition? Let’s check that?
So what do I need to do for
that? I have to take a cube, right?
Let me take the cube.
It’s infinitesimal,
but I’m just keeping it near the origin.
It can be anywhere you want,
but I’m drawing it near the origin.
It’s a surface,
and it’s got these outward going normals.
And I must take E ⋅
surface area for every face. And I’ve got to get 0.
That’s the condition.
So is that going to work or not?
Let us see.
There are six faces in this
cube, so I’m going to draw 1,2 and 3 that I can see,
and -1, -2 and -3 refer to the faces on the opposite side.
This is 3, that’s -3.
So I can only show you 1,2 and
3. Let’s look at surface 1 and ask
what I get from the surface integral of E.
Do you agree that E
points this way? So
E⋅dA is a non-zero contribution on
surface 1. But on surface -1,
E still points up, but dA points down,
because E doesn’t vary from the upper face to the lower
face, because in going from upper to
lower, I’m varying the coordinate
z, but nothing depends on z.
The same electric field is
sitting on the upper plane of this cube as on the lower plane
of the cube. Therefore the surface integrals
will cancel and give you 0, because the area vectors are
opposite. That is a simple statement,
that if you’ve got a constant electric field going through a
cube, the net flux will be 0 because
what’s coming in on one side goes out of the other side.
So that’s the cancelation of 1
and -1 giving me 0. But there are other faces,
like 3 and -3. What surface integral will I
get from 3? The area vector looks like
this, the electric field looks like that, the dot product is 0.
In other words,
the field lines are parallel to this face, so they are not going
to penetrate it. You’re not going to get any
flux. Or the area vector is
perpendicular to the field vector.
So on this face,
E is 0, on the opposite face is also 0.
The same thing goes for 2 and
-2. If you go to face number 2
here, the electric field is pointing like that,
but there is no flux and there is no flux on the opposite face.
So if you get 0,
either because the field is parallel to the face,
or if it’s perpendicular, it has the same value on
opposite faces, with opposite pointing normals,
or opposite pointing area vectors and you get 0.
That’s how you get the surface
integral of E to be 0 on this tiny cube.
But if it’s 0 on a tiny cube,
it’s 0 on anything you can build out of tiny cubes.
That means 0 on any surface.
If you repeat the calculation
for B, you’ll get pretty much the same
logic, except that B now points like this.
So on the top face,
B will have no flux because it’s running along the
face. There is no flow through that.
Top and bottom are 0 and 0.
The only faces that matter are
2 and -2, because 2 is coming out of the board and B is
coming out of the board. But on the other face,
which you cannot see, the -2, B is still going
this way, but E, the area vector is going the
opposite way. The key to this is that
B does not vary. You see, if the flux was not
constant, if the field was not constant,
the fact that you’ve got two faces with opposite pointing
area vectors doesn’t mean the answer is 0.
Even though the area vectors
are opposite, the field strength would be
bigger on one face, smaller on the opposite face,
in which case they won’t cancel.
But the field is not varying in
the coordinate in which I’ve displaced the planes.
That’s the reason you get 0 for
both of those. All right.
So far what I’ve shown you is
that the solution I have automatically satisfies 0
surface integral without any further assumptions.
Of course, it’s very important
that E did not vary with x and z.
But with the assumed form,
I don’t have to worry. I still have to only worry
about the other two Maxwell equations involving line
integrals. I’ve got to make sure that
works out. So let’s see.
So here I have to take loops
and I told you, when you take loops now,
you’ve got to take loops in that plane,
that plane, and that plane, because it takes three kinds of
little flat Lego pieces to form a curved surface in 3D.
So let me take this loop first.
I want to take an infinitesimal
loop that looks like this. I choose the orientation of
this so that if I do the right hand rule, the area vector is
pointing in the positive y direction.
In other words,
this is a tiny loop of size Δy–let’s see.
Δx this way and
Δz that way. The area vector is coming out
like that. So I have to now look at the
line integral of E⋅dl
and equate it to -dΦ
_B /dt. I have to see,
is that true or false? Well, take this loop and look
at the line integral of E and see what happens.
There is an E going up
this face. E is perpendicular to
that face , and E is perpendicular to that edge.
E is anti-parallel to
this edge. You see that?
If E is pointing up,
it cancels between these two. These two have no contribution,
because E is perpendicular to dl.
But these two cancel.
They cancel again because the
E that you have here going up is the same E
that you have here also going up,
but the dl is going in opposite directions.
E⋅dl around this tiny loop is
actually 0. You understand why it is 0?
Cancelation between opposite
edges, and two edges that don’t give you anything.
So what we hope for is that on
the right hand side, there better not be any
magnetic flux coming out of this thing,
because otherwise right hand side will give you a non-zero
contribution. But luckily,
that is the case, because the magnetic field
looks like this. It’s in the plane of this loop.
The dot product of the area
vector, which is perpendicular to the loop and the B
field is 0, so that’s also 0. So this is identically
satisfied on that loop. Now if you take line integral
of B⋅dl,
and that’s supposed to be μ_0
ε_0 dΦ/dt of
the electric flux, again, you should try to do the
exercise with me. Magnetic field,
I said, is going in the x direction,
so it has nothing to do with those two edges,
because they are perpendicular to it,
but it cancels between these two.
It cancels because the edges
are going in the opposite direction, but B doesn’t
vary from this edge to that edge.
It varies only with y,
so again you get 0. And there is no electric flux
coming out of this surface, because electric field lines
also lie in the plane of that loop.
They don’t cross it.
So those equations are also
satisfied. But I’m not done,
because I still have to consider loops in this plane and
loops in that plane. So far I’ve gotten no
conditions at all. What this means so far is that
there are no further restrictions on E_z
and B_x. They can be anything you like.
But I’m going to get
restrictions by finally considering loops in this plane
and loops in that plane. So let’s see how you get
conditions on one of them. This is x, y, z.
So let’s take a loop that looks
like this. I’ve chosen it so that with the
right hand rule, the area vector is
perpendicular and coming out the x axis.
I remind you once again,
E looks like that, and B looks like this.
So let me give the edges a
name, 1,2, 3,4. And let’s take the condition
E⋅dl=- d/dt of the magnetic
flux. Now this loop has dimension
dy in this direction, dz in that direction.
This is an infinitesimal loop.
I’ve drawn it so you can see
it, but it’s infinitesimal. Okay, so what do I get on the
right hand side? Right hand side says
−d/dt of the magnetic flux coming out of the
board. The magnetic flux is coming out
like this, right? It’s coming out of the
blackboard, so there really is a magnetic flux.
That magnetic flux is the
number I wrote down, B_x times the
loop area which is dy/dz. That gives me –
ΔyΔz B_x dt.
That’s the right hand side.
This loop is so tiny,
I am approximating B by the value at the center if you
like. You might say,
“Look, I don’t think B is a constant on the
loop. B is varying.
Why do you take the value of
the center?” Well, if it varies,
the variation is proportional to Δx .
I’m to Δy or
Δz, therefore that will be a term
proportional to Δy squared or Δz
squared. But we are going to keep things
to first order in Δy and Δz,
so it doesn’t matter. Now how about the left hand
side? If you look at the left hand
side, I hope you’ll try to do this one.
If you go like that on edge 2,
you will get E_z times
Δz –I’m sorry. Let me write it as
Δz times E_z at the
point y Δy. And on this edge,
I will get – Δz times E_z
at the point y. In other words,
the electric field is everywhere, going up,
so it goes up here. E is parallel to it,
so that should be that times Δz.
These two don’t contribute,
because E and ΔL are
perpendicular, so forget that.
But this one,
it goes in the opposite direction,
so I subtract it, but I bear in mind that this is
y Δy, but that’s only y.
So here I hope you guys will
know what to do. You will say that is then
roughly equal to dE_z/dy times
Δy. Therefore the line integral of
the electric field is proportional to the derivative
of dz with respect to y, times the area of the
loop. And the surface integral of the
flux change is also proportional to the area of the loop,
and you get dB_x/dt.
So this is the final equation
you get by considering that loop.
So what is remarkable is the
line integral and the surface integral are both proportional
to the area. The surface integral being
proportional to the area of the loop is obvious,
because it’s the surface integral.
Why is the line integral
proportional to the area? Because one part of the line
integral is the width of the loop,
Δz, other one comes in because the
extent to which they don’t cancel is due to the derivative
of the field in the transverse direction.
That brings you a
Δy. So if you cancel all of this,
you get the equation that I’m interested in,
which is very important – dE_z/dy=
-dB_x/dt. This came from looking at this
equation. The last one,
another loop equation, I’m going to go through
somewhat quickly, because it’s going to be 0=0.
Suppose I take the integral of
the magnetic field around this loop.
Do you understand why it is 0?
The magnetic field is coming
out of the blackboard. dl’s are all lying in
the plane, so line integral of B is then 0.
And that better be equal to
-μ_0 ε_0
dΦ _electric
/dt. That is the case,
because there is no electric flux coming out of this face.
The electric lines are in the
plane of the loop. There is no electric flux,
there is no rate of change, so it’s 0=0.
So from this loop,
I’ve managed to get one equation.
So my plan now is,
I don’t want to do yet another loop, because you’ve got the
idea, or it’s not going to help to draw one more loop.
But I would just say that if
you draw the last loop, which lies in what plane?
Which lies in this plane.
You will get one more equation.
I’m just going to tell you what
it is. I got dE_z/dy
is -dB_x/dt. You’ll get another equation,
dB_x/dy=- μ_0ε
_0dE _z/dt.
So I confess,
that I’ve not derived this one, but it’s going to involve just
drawing one more loop and doing it.
Now it’s really up to you guys
how much you want to do this, but you should at least have
some idea what we did. It turns out these are the only
restrictions I have to satisfy. If this is true of my field,
I’m done, because I’ve satisfied every Maxwell
equation. Every surface integral was 0 on
every tiny cube and therefore 0 everywhere for E and
B. The line integrals,
some were identically satisfied, some were 0=0.
There were some that led to non
trivial conditions and it’s these two.
Therefore I’m told that if the
fields that I pick, that depend on y and t,
have the property that the y derivative of this guy is the t
derivative of that guy, and the y derivative of
B_x is up to some constant,
the -t derivative of the other guy,
you’re done. That’s all you require of that
function. So let’s do the following –
take this equation, take its y derivative on both
sides. IN other words,
I want to take y derivative of this and I want to take y
derivative of that. Then here you will get
d^(2)E _z/dy ^(2)=-dB_x/dy
dt. Take the d/dz of this.
Now you know partial
derivatives, you can take the derivatives in any order you
like, so let’s write it as d/dt, d/dy.
But dB_x/dy is
μ_0dE _z/dt.
Another dt makes it
μ_0ε _0d^(2)E
_z/dt^(2). Take first the y derivative,
dB_x/dy, is that one single time
derivative. Take one more time derivative,
you get this. Now we get this wonderful–we
really have the wave equation now,
because I get d^(2)E _z/dy ^(2) –
μ_0ε _0d^(2)E
_z/dt^(2)=0, which you recognize to be the
wave equation. That’s what the wave equation
looks like in the top of the blackboard.
But the things that are
oscillating now are not some string.
It is really the electric field
oscillating. It’s not a medium that’s
oscillating. There’s nothing there.
This is all in vacuum.
So let’s find the velocity.
v^(2) will be
1/μ_0ε _0,
right? Because this thing that comes
with our equation is the 1/v^(2) term.
Then you go back to your
electrostatics and magnetostatics and find out what
these numbers are. And I remind you that
¼Πε _0 is 9�10^(9)
and μ_0/4Π is
10^(-7). So let’s write this as
4Π/μ_0 times ¼Πε
_0, because we know what those guys
are. ¼Πε
_0 is 9�10^(9). And 4Π/μ
_0_ is 10^(7), so you get 9�10^(16).
Or the velocity is 3�10^(8)
meters per second. So this was the big moment in
physics, when you suddenly realize that these things are
propagating at the velocity of light.
So people knew the velocity of
light from other measurements. They had a fairly good idea
what it was. They know μ_0 by
doing experiments with currents. They knew ε_0
from Coulomb’s law. It’s one of the greatest
syntheses that you put them together and out comes an
explanation. Now this doesn’t mean that
electromagnetic waves are the same as light.
You agree that if you run next
to a buffalo at the same speed, you are not a buffalo.
You just happen to have the
same speed. So that was a bit of a leap to
say it is really light. But it was also known that
whenever you have sparks in electrical circuits and so on,
you see light. So it took a little more than
that, but it’s quite amazing, because gravity waves also
travel at the speed of light. You cannot assume that the
speed means the same phenomenon. But they were really right this
time. It really was the speed of
light. So we now have a new
understanding of what light it. Light is simply electromagnetic
waves traveling at the speed. It consists of electric and
magnetic fields. What we have is an example of a
simple wave, but one can show in general
that if you took the most general E and B
you can have, you will get similar wave
equations. You will get it for every
component of E and every component of B.
So waves in three dimensions
will satisfy the general wave equation.
But I’m not interested in the
most general case, because this is enough to show
you where everything comes from. So you have to think about how
wonderful this is, because you do experiments with
charges, with currents,
and you describe the phenomenology as best as you
can. Then Maxwell added that extra
term from logical consistency, by taking this capacitor and
drawing different surfaces, and realizing that unless you
added the second term, dΦ
_E /dt it didn’t work.
And without the second term,
you don’t get the wave. It’s only by adding the second
term and then fiddling with equations, he was able to
determine that waves can travel. So the reason that
electromagnetic waves travel in space without any charges is,
once you’ve got an E field or B field
somewhere, it cannot just disappear.
It’s like the LC circuit.
If your capacitor is charged,
by the time it discharges, it has driven a current in the
inductor. Inductor is like a mass and
it’s moving with velocity. Current is like velocity,
so it won’t stop. So the current keeps going till
it charges the capacitor the opposite way.
Then they go back and forth.
They are the only 1 degree of
freedom, which is the charge in the capacitor or the current in
the circuit, which are related. Here E and B are
variables defined everywhere, but you cannot kill E
because the minute you try to destroy E,
you will produce a B. The minute you try to destroy
B, you’ll produce an E, so they keep on
swapping energy and going back and forth.
So it is self sustaining and
it’s an oscillation if you like, but it’s an oscillation over
all of space, and not over 1 or 2 degrees of
freedom. So I’m going to now write down
a specific form. The specific form I’m going to
write down looks like E=k times E_0.
Now I’m going to pick a
particular function, ky – ωt,
and I’m going to take B to be I times B_0 sine ky –
ωt. Now I’m making a very special
function of y and t. Till now, I just said it’s a
function of y and t. All I needed was it’s a
function of y and t. That was enough to get me all
this. I’m going to find a relation
between these constants, E_0 and
B_0, by putting them into these two
Maxwell equations I had. That’s all I want to do now.
So one condition I had was
dE_z/dy was -dB_x/dt.
Then I had
dB_x/dy is -μ_0ε
_0dE _z/dt.
These are the two equations.
I’m going to demand that these
particular functions obey these two equations.
If they obey these two
equations, they will obey the wave equation,
because when I combine this equation with that,
I got the wave equation. Do you understand?
I have an equation A and an
equation B, if I put one into the other, I got the wave
equation. But you really should satisfy
separately equation A and equation B, because it’s not
enough to satisfy the one you get by shoving one in the other.
They should be true
independently. So let’s demand that this be
true and demand that be true. So what is
dE_z/dy? If I take the d/dy of
this, this is what I call E_z and this is
what I call B_x.
if you take the d/dy of that,
will give me kE_0cos(ky –
ωt). And we want that to be=
-dB_x/dt. So take the d/dt of this
guy. d/dt of this guy is
B_0 – ωcos(ky –
ωt). So these things cancel out,
then I get B_0ω.
So this tells you that
E_0=ω/k B_0.
It tells you about the
magnitude of the E and B vectors,
what size they should bear in relation to each other.
This is one condition.
I’m almost done.
So I want to take this
condition now. So dB_x/dy,
what is that equal to? B_0kcos(ky
– ωt)=- μ_0
ε_0 dE_z/dt which will
be – ωE_0 cos(ky –
ωt). Cosines cancel,
then I get a condition B_0k=
μ_0ε _0ω
times E_0. But that equals ω/c
squared times E_0,
because μ_0 ε_0 is
1/c^(2). So I get a second condition
which I can write as E_0=
c^(2)/ω times B_0.
So these are the two conditions
in the end. If you want the function to
look like this traveling wave, with sine waves in it,
then the amplitude for E and the amplitude for B
have to satisfy these two conditions.
But look at these two equations.
They’re both telling you
something about E_0.
One says E_0
should be ω/k times B_0.
The other says
E_0 should be c^(2)/ω times
B_0. That means ω/k
better equal c^(2)–I bet I dropped a k
somewhere. Student: You had
B_0k in the– Prof: Thank you, yes.
So c^(2)k/ω.
That tells me that
=kc or -. Or if you like,
k is �ω/c.
It doesn’t matter how you write
it, but you can see what that means.
What this is telling you is
that if you take the function sine ky – ωt,
and if ω=kc, it becomes sine of ky –
kct. That becomes sine of k
times y – ct which is a function of y – ct.
In other words,
we knew this was going to happen,
because the wave equation says the function should be a
function only of y – ct. That will happen if ω
=kc. You see that?
If ω
=kc, you pull the k out, you get y – ct.
Or you can have y ct,
but my solution was y – ct.
I’m almost done.
So the last thing I want to get
from this same equation, now that that condition is
satisfied, if ω=kc, I get
E_0=c times B_0.
So now I’m going to summarize
this. Don’t worry about the details.
I will tell you the part you
should know all the time. So I know this is somewhat
heavy. I had to do this calculation at
home to make sure I got all the – signs right.
There’s an orgy of – signs.
I’m not that interested in that.
I know that you guys,
unless you’re going to major in physics and want to do it for a
lifetime, don’t want to know all that.
So what should you know?
What you should know is that by
doing a few experiments, one wrote down these Maxwell
equations. You’ve got to understand a
little bit where everything came from.
Then the whole class was
demonstrating that they implied some waves and the demonstration
was shown for proving it for infinitesimal loops and
infinitesimal cubes and then seeing what conditions I had.
And I found that I could get a
set of functions that are dependent on y and t,
and that they really travel at the speed of light.
So that was the bottom line.
Actual derivatives and how
everything happened is really not something I expect you to
carry in your head, so don’t let the exams ruin
that for you. I don’t care.
I will not quiz you on that
part of this derivation. But you must understand that
the structure of physics was such that it was an interplay
between some experiments and some purely theoretical
reasoning on the nature of equations.
And you put them together and
what makes it worthwhile in the end is to get fantastic
productions like this that unify electricity and magnetism and
light into one shot. So the picture I get now,
if you look at all of this, the final answer is E
looks like k times some number E_0 sine
ky – ωt. But I want you to know that
ω=kc there. Then I get B=
IB_0 also the same sine,
with the extra restriction that B_0 times
c=E_0. So here is what the
electromagnetic wave looks like. One guy, the electric field,
is always living in this plane. This is the E field.
At one instant,
if you take a snapshot, that’s what it will be doing,
going up there, coming down here,
going up there. That’s E.
B field looks like this,
lies horizontal here, goes like hat and comes out.
So this is in that plane and
that is in the vertical plane. That’s horizontal plane,
that’s the vertical plane. So that is E and this is
B. And the point is,
E over B=C.
In other words,
E is much bigger than B.
The ratio of them is the
velocity of light. This is interesting,
because if you took the force on a charge, you remember is
q times E v x B.
That means if you took an
electron and you left it in the electromagnetic field,
the field comes and hits you. If E field is
oscillating one way, the B field is
oscillating perpendicularly. The wave is traveling in a
direction perpendicular to both. If you want,
it’s in the direction of E x B,
is the direction of propagation of the wave.
If it hits an electron there,
the oscillating up and down electric field will make the
electron move up and down. It will also be feeling a
magnetic force, V x B.
But notice that the size of
B is the size of E divided by c.
So the electric force to
magnetic force ratio, or magnetic electric will be
the ratio of v over c.
So for most velocities,
for electrons and circuits and so on, the velocity is much
tinier than the velocity of light.
So when a radio wave hits your
antenna, gets the electric charges in
action, it’s the electric field that
does most of the forcing, not the magnetic field.
But in astrophysics and cosmic
ray physics, where particles can travel at
velocities comparable to that of light,
then these forces can become comparable.

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