So let's take a look at

the equations that we summarized in

terms of electrostatic potential

and magnetic vector potential.

So this is electric scalar potential.

So you see Laplacian of

electric potential minus one

over c squared times Rho square.

Electric potential over Rho t square

is equal to minus Rho over Epsilon naught,

and the same form is taken by

the vector potential. Go ahead.

I just wanted to say that reminds students

there's three of those for A.

For three. Yeah. We have three components for A.

So in fact, it seems like two equations,

but in fact you need four equations.

Correct.

Very good. So here,

for the right term,

Rho and j, because we are far away from sources.

We can say they are zeros.

So that's why we put zero to the right term,

does in free space to scalar potential

Phi and each component of

the vector potential A all satisfy

the same mathematical equation as you can see.

So, once you understand

solution to one of these equations,

you can solve for all of them.

Okay. So, universal solution

for Maxwell equations is

really to solve the one of these equations.

Suppose we let Psi stands for

any one of the four quantities Phi,

A sub x , A sub y,

A sub z as melody just mentioned.

Then you can see this will be

the universal equation and

universal solution will be derived from this equation.

You can see they're neatly

arranged in a linear combination

and you can see it's symmetric.

Only the time is different because it

has different dimensions with the space.

So let's take a look at

the field as a three-dimensional wave,

so we will revisit this equation,

the Laplacian of A minus one over c square,

Rho square A over Rho t square is equal to zero.

Now we are going to use the equivalence

of changing the sequence of the operator.

So instead of first doing Laplacian we're

going to take the kernel and then do Laplacian,

and from the fact that B field is the curl of A,

we can replace this part by B,

so it becomes the Laplacian of B.

For the second term here,

we will do the similar trick,

we'll use a similar trick.

So, instead of first doing

the curl and then do the time derivatives,

we're going to first do the time derivative here,

the left side, and put curl here.

So, in fact, I told you to put the first here,

but in fact the sequences that

you do the time derivative first and then do curl here.

But in this case, you do curl

first and then do time derivative.

Now, from the same definition,

we can see that we can replace this del cross A by B,

so it becomes one over c

square Rho square B over Rho t squared.

So if we do this transformation we can understand that

the Laplacian of B minus one over c squared

times Rho squared B or Rho t square is equal to zero.

So you can see they have

the same form and so each component of

the magnetic field B

satisfies the three-dimensional wave equation.

Now likewise, let's take a look at

electric field as a three-dimensional wave as well.

You can probably think oh,

because we know they are coupled and they have

orthogonal relationship that should

be something that we expect to see.

But the mathematical derivation

might be a little bit more complicated.

So let's take a look here.

Look at this equation.

So, electric potential, Laplacian minus one over c

squared times Rho squared Phi over Rho t squared,

we can use the fact that the electric field is

the combination of the electrostatic potential gradient

and the time derivative of magnetostatic potential.

Using the fact, we can now expand the curl of E,

and if I took this here,

the curl be E becomes minus del

cross del Phi minus del cross d over dt.

We know from our earlier lecture,

gradient of a scalar function has no curl.

There's no rotation, so this should be zero.

So what is left is only the latter part.

Now this latter part we can do

the same trick to change a sequence of the operations.

So instead of del cross d over dt,

we can do the d over dt del cross A,

and del cross A again is B. All right.

So then you can see it's minus Rho B over Rho T. In fact,

this is the second equation of Maxwell equations.

Then we can also use the fact del cross,

del cross E from the backup rule is del,

del.E minus Laplacian of E field,

and the divergence of E is

the charge density at the point of

interest divided by permittivity.

But because we are free of source, this should be zero.

So, what is left is only this Laplacian of E field.

Laplacian E field from here as you can see,

this one goes into here.

So, it is equal to minus del cross

Rho B over Rho t. We can change the sequence again,

which becomes minus Rho over Rho t del cross

B. Del cross B we know from the equation,

that it is minus one over c squared times Rho squared

E over Rho t because the source J is zero.

So, putting this one and putting

this one equal we come up with the equation here,

which has the same form as three-dimensional waves.

So, again, it follows

that the electric field E in free space

also satisfies the three-dimensional wave equation.

So then I know magnetic potential

that the potential electric potential,

scalar potential, magnetic field,

electric field they all follow the wave equation.

So, now, it is the time to solve for the wave equation.

So all of our electromagnet fields

satisfy the same wave equation. That is to say.

The Laplacian of Psi minus one over c square,

Rho square Psi over Rho t square is equal to zero.

So, what is the most general solution to this equation?

So Melody, do you have any clue what

the form of the solution will be for wave equation?

I'm a little bit rusty on my

physics I didn't memorize that.

Okay. That's okay.

So here's the thing.

If you don't memorize this

as our teaching assistants does,

there's no worry to do.

You can try to follow us

and repeat it until it becomes second-hand nature.

So for here, we will see the general solution

will take a form of a sinusoidal or cosine function,

and inside the cosine or sinusoidal function,

you will have an argument that is

a function of the position and time,

and then you will have defined

the wave vector and the angular velocity of the wave.

But that is okay. So rather

than tackling this difficult question right away,

we will look first at what can be said general

about the solution where nothing varies in y and z.

So we'll go step-by-step.

So let's suppose that the magnitudes of

the fields depend only upon x,

that there are no variations of field with y and z.

So we are considering plain waves again.

So why do we do it all over again?

We didn't show the waves,

we found what most general solutions for plane waves.

We found the fields only from

a very particular current source,

which was all of a sudden a plain current was turned on.

That was a particular example,

but now we want to expand our horizon.