Have a wonderful good day, ladies and gentlemen.
Today we want to use this little video from the Electromagnetic Theory series to
think about the behavior of the fields at the interfaces, as it is
obtained directly from Maxwell's equations.
My name is Hans Georg Krauthäuser, I am the holder of the professorship for
Electromagnetic Theory and Compatibility at the TU Dresden.
Well, when we start we can start right away with the Coulomb-Gaussian
Law that is written down here again.
Divergence D equals rho_v.
And we apply these laws to a geometry that is also popular in the
Literature is referred to as Gaussian box or Gaussian box or also as
Gaussian box or something like that and yes the geometry is very simple.
We have here an interface between two media.
Medium 1 is characterized by an epsilon_1 and a mu_1.
Medium 2 is characterized by an epsilon_2 and a mu_2.
And now we imagine a volume that is just such a small box or can
that begins in one volume on one side and extends down to the other
Page extends and for the sake of simplicity we take one here
cylindrical area.
The interfaces are parallel to the interface.
The interface itself is characterized by its normal vector n and the
The front sides, i.e. the top and bottom of the box, so to speak, are characterized by the
Normal vectors n_1 on the one hand and n_2 on the other.
Yes, and we actually look at dielectric displacement first
D, which now has some values ​​on both sides.
And we will assume that these surfaces of the cylinder - top and bottom
of the cylinder - are small enough to view D as constant over this area.
Well so what can we do?
We have a volume, we have the differential relationship - the local one
Relationship - and yes then I think it's obvious that we should just get on with it
skip an integral relationship.
We do that on both sides and apply the volume integral as we already do
know, we can use Gauss’s integral theorem from the volume integral over the
Make divergence a surface integral.
And now we make a limit h equal to 0, h is the height of this cylinder.
That means we let this whole structure shrink , the volume is then
no longer a real volume, of course.
That means at the border crossing the volume to 0 and then it is clear - there we even have
no more lateral surface of the cylinder - then becomes the surface integral
actually only the top and bottom of this cylinder contribute.
Because we said that D should be constant over this area A, that is
Then quickly written down integral.
That is simply the contribution of one area - the first area - n_1 times D_1
scalar multiplied and the whole thing then multiplied by the area A itself.
And to this comes the contribution of the soil, area 2, n_2 times D_2 as well
again multiplied by the size of the area.
Very simple surface integral in the case.
On the other hand we have the volume integral where we are now
must take into account that the volume is compressed onto the surface at
The limit transition h towards 0 and, accordingly , the volume charge density is also none
real volume charge density more, but something that only has one when x equals 0
Has value. We can express that by being here
actually replaced the bulk charge density with a surface charge density
rho_F, which then has the unit coulomb per square meter multiplied by the
Delta function of x.
Here you have to note that x is a dimensional quantity, a length,
accordingly, the delta of x is also dimensional and has the unit 1
by length, 1 by meter.
If you don't remember the relationship, please look again
Video with the previous knowledge.
That's explained in there.
Yes, and in that case this volume integral will also be relatively simple, we can
execute the x integration immediately, there then only remains an area integral
left. And with the area integral we again assume that the area is like this
small is - which we can of course achieve again and again - that roh_F on the
Area is also constant again, then it is simply rho_F times A.
Here we can of course shorten the A on both sides and n times
an n times D - that is now the normal component D.
We just have to be a little careful in which direction we do it.
We actually had n_1 and n_2 here before , we have now replaced,
n_2 is directly n, but n_1 is equal to minus n, so here comes straight
also this sign and therefore there is minus n times D_1 plus n times D_2.
These are normal components, so that we can summarize them now in
this clear form: normal component D_2 minus normal component
D_1 is equal to rho_F, i.e. the normal component of D is at the transition
from medium 1 to medium 2 discontinuously and we can also specify how big this is
Discontinuity, how big the jump of the normal component is.
Attention: even for simple media, linear isotropic homogeneous media, this can
cannot be transferred to the E-field because they are one on both sides
have different epsilons , which of course have to be factored in here
so that they get a slightly different formulation here .
Similarly, we can of course do the whole thing for B, for the B-field, for them
magnetic induction.
Here it becomes even easier because the divergence B is equal to 0, that is, if
we carry out the same calculation, the 0 and are always on the right-hand side
accordingly we get here that the normal component of B is continuous,
this means that the normal component does not jump at the transition from medium 1 to medium 2.
Here again, attention: what we said here for the B field
does not apply to the H field.
With the H-field you would still have different mus on both sides , so that
they would actually get them in one go.
But first of all, these two relationships are important.
Well, these are the statements for the normal components of D and B.
And in fact we can make statements for them in a similar way
Tangential components of E and H.
And for that we consider a slightly different geometry and we do that first
here on this slide actually only for the E-field.
What we are looking at is a so-called Stokes surface.
So, here we have our interface between
the two media 1 and 2.
n is again the normal vector on this interface.
And now we put a loop, a loop, perpendicular to this interface
rectangular loop as shown here
which spans the area A.
The contour of A would therefore run around accordingly if the
The normal vector of this surface would now come out of the film plane.
Otherwise it has the length l on one side and the width h on the other side.
The E field on both sides is again labeled E_1 and E_2.
Here, too, we would first do an integration.
Now, however, an area integration, namely the integration via this right now
pasted surface A.
On the one hand we get an area integral over the rotation of E, da
we know again, we can convert that using Stokes' integral theorem
into a contour integral, i.e. a line integral that is closed
Way around the area. That's this integral here E ds.
And on the other hand there is simply the area integral, that is
minus integral A dB after dt times dA.
What is written here is ultimately the change in flow through this area.
If we now let the border crossing point h go towards 0, then it is clear that the
Flux change through this area, when this area becomes 0, will also be the
Flux changes through this area disappear, so that on this side on
in any case, a 0 comes up.
And with the orbital integral it is so that in the border crossing h against 0, of course, this
Cross pieces of length h no longer play a role, only these two
Long sides still make a contribution.
This post is written here.
The whole thing is now written down in terms of the tangent vector
t introduced here above.
And we see that this one path here in medium 1 runs in the direction of minus t,
therefore this minus sign appears here.
The other way is in the direction of t, accordingly there is a positive one here
Character. I can then shorten the length l so that I can see immediately that
the tangential component of E is continuous at this transition.
Again, please be careful again: you could use it for simple media now
of course replace E by 1 by epsilon times D, but there the epsilon on both
Pages were different for simple media , they would get a discontinuity.
So you can not refer to the continuity of the tangential component of E
Infer the continuity of the tangential component of D.
The tangential vector t is of course not unique.
You can access this area, which is characterized by the
Normal vector n define any number of tangent vectors in which you
simply rotate this tangential vector.
And you could of course make a corresponding consideration again.
You can then introduce a surface again, introduce a Stoke surface,
which is also spanned again by t and n, so that ultimately it is this
Continuity condition for any tangential vector of this boundary surface
to get. Yes, and we can now also consider that in a very analogous way
Make the H-field, but there is the equation, our Maxwell equation, that
Flooding law a little more complicated.
Therefore, the result is also a little more complicated.
But first of all we would lay a surface just as flat again , vertically
for the interface between media 1 and 2 and the steps
are initially the same.
First of all I do an area integral, use the again
Stokes' integral theorem um from the area integral over the rotation
To make circulation integral.
And look at my orientation again and see again that one way goes
toward minus t and the other way is toward t, so I'm on side 1
get a negative sign again.
The positive sign remains on page 2.
And the change in the displacement current will also disappear again when the
The area becomes small, so that this term, the integral over this term, falls into the limit
h also disappears towards 0.
What remains is then the integral over the current density J dA.
And that is a volume flow density, first of all, that in this one
Area would prevail.
If I let the area shrink, let h go towards 0, then I have
of course, again only the contribution on the interface itself.
I do that in a very similar way to what I have just done by introducing a delta function and
now have a surface current density distribution J_A, i.e. J_A times the delta of x
my normal current density again.
And so these are the same integrals, only I can now do the integral back here
easy to evaluate, of course, because the delta function is included.
And if we again assume that the loop is small enough that this
Surface current density is the same over the whole area , then I get it here
accordingly, simply just multiply the tangential component
out with the length.
Now I have to be careful which tangential component.
The current density would be the current density that penetrates this area.
And if I let that shrink to the interface now , that's it
Tangential component which is given here by this tangential vector t_2, which is here
plays a role. t_2 is nothing more than the normal vector on this loop that i
I've defined here and well, obviously these vectors also form
a legal system and that is how I get this simple relationship.
And here too, of course, you can immediately see that the length l is shortened out again
can, so that overall I get a simple statement again .
Namely the one that the tangential component of H is discontinuous and the discontinuity of the
Tangential component is given by a tangential component of
Surface current density. But we have to be careful, t and t_2 are different
Tangential components. So if you have the discontinuity in relation to the
Want to write down the tangent vector t, then you have to see that you get from the
Surface current density get the tangential component with respect to t_2.
So they are perpendicular to each other.
First of all, of course, this applies again to any tangential components t that would be
yes again freely selectable.
But I also have to turn t_2 accordingly so that it works, so
t_2 must be the normal vector on this loop.
Or to put it another way: there is a fixed relationship between t_2, n and t that always
must be adhered to. Otherwise t is arbitrary.
Well. Yes, that's what we say about the interfaces, the components of the fields
wanted to testify at the interfaces.
The relationships are fundamental.
You can see that we can get that from the Maxwell equations only with pure
Have derived mathematics, did not need anything else.
In other words, you can actually always fall back on these border relationships,
these are not approximations or approximations, they actually always apply.
Well. Yes, thank you for your attention.
As always, you can find more information on the website.
Until next time.