## 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?

Seattle.

That’s the story of my life.

So nothing else?

Yes?

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?

No.

With the exam?

Yeah.

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?

Yes?

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?

Operationally,

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?

Yes?

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?

Student:

>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.

Consequently,

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.

Therefore

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 9ï¿½10^(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.

dE_z/dy,

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

ω^(2)=k^(2)c^(2)ω

=kc or -. Or if you like,

k is ï¿½ω/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.