Now that we have good qualitative understanding where the Maxwell's equations come from, I want to
show you something different.
So far we have introduced to Maxwell's equations in the differential formulation.
So as you can see on the right hand side of these equations, there are quantities like the charged
density or the magnetic fields or also E and J and all of these depend not only on the time, but also
on the position vector.
So what this means is if we look at one particular position vector, we can calculate also the left
hand side of these equations and we can determine E and B, and so we know what E and B look like at
this particular position vector.
So it's something like a local formulation.
However, we can also integrate over these equations and so we are not restricted to one particular
position vector, but it's more like an integral formulation, like it's the same equation, but from
a different point of view, so to say.
So in this lecture, I want to show you how we can derive these four integral formulations of the Maxwell
equations.
And for this, we need two laws from mathematics.
The first is divorce law.
And this is actually quite difficult to derive.
So I hope that it's sufficient for you that you just believe me that this is true from the mathematical
side.
And if not, you can look up proofs for this law on the Internet or in a textbook.
But it is quite difficult.
But here, I want to just just show you the result and want to explain to you what this equation means.
So Gaza's law tells us that when we integrate over the divergence of some vector.
So this is here, the integral over in Cartesian coordinates.
This would be the X divided.
So I just tried D.V., which is integral over the volume, and we integrate over some volume that we
have defined.
For example, it could be such a shape, but it could also be any other smooth shape.
So that's a three dimensional integral.
So did that Virgin's of this vector A is equal to the Vector eight itself, but now it's not a volume
integral anymore, but it's an integral over the surface of this volume.
So this is what I have noted here with this.
And here the we integrate over this area element.
So this D.
S means or is characterized by such an element of the surface, and it is characterized, of course,
by the absolute value of this element and a normal vector, which is always perpendicular.
So you could say you have your your your vector field, which depends, of course, on on our on the
position vector.
So it can be different inside of this volume and on the surface of this volume.
Now you calculate here the the scale of product with this vector A and this surface, which is essentially
just the normal of the surface, then you integrate over the whole surface and it turns out that the
left hand side is equal to the right hand side of this equation.
So now that has used his theorem and apply it to our Maxwell's equations, because maybe you remember
this, the first two Maxwell equations were divergences of E and B, so let us first take the divergence
of E, which was equal to a constant times, the charge density row of R so we can rewrite this.
We just multiply by Epsilon zero.
So nothing really happened here.
And then of course, since this is true for every position vector, we can just we can just integrate
over our.
So basically we integrate over the whole volume.
So again, this could be any volume.
As long as it's smooth, it could be a sphere.
It could be a cube.
Doesn't really matter here.
It's true for any volume.
And everything that is done here is just adding the integral.
And I think you you know, this an integral.
You could also think of some of this integral here at several different positions inside of the volume.
And if you make these divi volumes very, very small, then this becomes the integral.
So this is, I think, quite straightforward to understand what happened here.
I have just introduced the integral because it is true for every position.
So it must also be true for the integral.
And now we apply the Gauss Theorem where we say, OK, let the right hand side does not really change,
but the left hand side here, we change the divergence of E to E itself and we change the volume integral
to the surface integral.
And the right hand side, um, is the integral over the charged density.
This is the same as here.
And if you think about it, if we integrate over the density of our whole volume, then this is just
the total charge in our volume.
So we know that a surface integral of the electric field is equal to the absolute charged in our volume.
B is the integral version of this differential maximal equation.
Now for the electric field, it is very similar, but it's a bit more easy, actually, because the
right hand side is zero in this case.
So we do the same thing.
We know that this is true for every position vector.
So it must also be true for the integral.
And now we can apply Goussis theorem here.
So the divergence of view becomes just beat and the volume integral becomes integral over the surface
area.
So we have already derived two of the integral versions of the Maxwell equations.
Now that is drive the other two.
And for this we need a different mathematical law.
And this is the Stokes law.
So it is kind of similar to Kansas law, but also a bit different because here we do not have the diversions
of a vector, but we have the rotation of a vector.
And this whole integral is not a three dimensional integral of the volume, but it's a two dimensional
integral of a some surface eight, for example, could be this one here.
And so the two dimensional surface integral of this rotation of A becomes equal to the vector itself,
integrated along the confining line.
So does this align integral?
You have this vector DL, which is always tangential to this area, eight so tangential to the confinement
and you calculate here the the product of these two vectors and then you integrate over the closed loop.
And if you do this, you will find that the left hand side is equal to the right hand side.
So now we see here we have two dimensional integral over the rotation of a vector so that we can use
this to formulate the integral version of the other two maximal equations, because there we have rotations
of B and rotations of each.
So first of all, and we'll just rewrite this a tiny bit.
So I will just divide by Muzio.
So nothing special happened here.
And then we know that this is again true for every position.
Makeda So it must also be true for an integral over these quantities.
So we integrate over this one for this one and over this one.
And in every case we integrate over the same area.
And now we leave to the right hand side as it is.
And we just use here stokes law because we have here a surface integral over the rotation of a vector.
So this means it must be equal to the vector itself and the surface integral becomes an integral over
the surrounding of the surface.
So it's just a line integral.
And the only other thing that I have done here is I have considered this integral over to current density.
So it is just the total current in the area.
A Those are our integral formulation for this maximal equation and for the electric field.
It's very similar.
We know that this is true for every position vector so we can integrate and this will remain true.
And so now here we can apply Stokes law.
So the rotation of E becomes E and this becomes a line integral.
So this means we have derived the integral versions of these four different Maxwell's equations and
we have come from a local definition of these relations to a global definition.
And in the following lecture, I want to discuss both versions of these Maxwell's equations in more
detail.
And I want to show you how we can reestablish our well known results from the very first section of
the scores, for example, the induction law, or also the umpire's law.