Good day, ladies and gentlemen.
I warmly welcome you to another contribution in the Electromagnetic Theory series.
My name is Hans Georg Krauthäuser, I am the holder of the professorship
for Electromagnetic Theory and Compatibility at TU Dresden.
We've got electrostatics ourselves in the last few events
and are there until the formal solution of the problem with the help of the
Greenschen function, thus penetrating the fundamental solutions.
Today we are going to get to know a very extremely important method for the
Green's function can actually be found with given boundary conditions .
Because it depends of course, the formal solution should only be used when
you also know the Greensche function.
The method we're going to go through today is the so-called
Mirroring method or image loading method.
In principle, it is based directly on what we used to derive the formal solution
have already seen.
The basic problem is therefore the solution to the Poisson equation, but now with
Boundary conditions in the finite.
We saw in the employment of the Green function that this
Superposition principle is really fundamental.
Because we can get out of the solution for a point charge - and nothing else
is ultimately the Greensche function - immediately also solutions for any, not
Open up delta-shaped inhomogeneities.
In the form we have already seen several times: as an integral
via the Greensche function multiplied by the volume charge density rho_v.
If we have this, then this part of the potential is ultimately - as you can see here -
completely determined by the solutions in the solution volume.
There are no interfaces yet.
In order for the overall solution to meet the intended boundary conditions, it must
Green's function of open space - that is ultimately nothing else than that
Potential of a point charge, where we now form the charge as one
that is exactly what Green’s function is.
We have to change that.
An additional term is added here, the solution of the Laplace equation
-so the associated homogeneous equation- must be.
And it's clear when I find a solution to Laplace's equation - the homogeneous equation-
add to G, then G will still be a solution to Poisson's equation.
This follows from the linearity of the Laplace operator and only for linear ones
Differential operators choose Green's formalism.
The idea now is as follows: So if I use the greens function of open space
know and now introduce boundary conditions, then these boundary conditions will be included
lead to the fact that the field of point charges - here in the case or this Greensche function -
is no longer the Green function of the changed problem.
Because the boundary conditions are no longer met.
We will actually see this again in more detail using the example .
Now of course I cannot place any additional charges in the solution volume,
because with that I would have a completely different problem.
But I have the option of modeling the influence of the edge by using
introduce charges outside of the solution volume that are not there at all, I
only take as model charges.
And do it so that the original load plus the model loads that
associated field or the
associated potential then
the boundary conditions are met.
If that's the case, then because of the uniqueness , I'm actually done.
I then found a potential that is the solution in V and at the same time
the specified boundary conditions are met.
And then we already know: if we have found that, then it is also clear,
until it then goes to any additive constants that give us
no further interest.
These charges, which I place outside of the solution volume, have none
direct influence, but of course they influence the solution, otherwise we would
don't do the whole thing.
You can imagine it that way, that the surface itself has an influence
on the solution.
Because, on a surface, charges are present in the volume itself
Surface charges influenced.
And we try to investigate the influence of these influenced surface charges
simulate modeling by going outside the solution volume instead
Place charges.
I think this becomes clear best if you look at a specific example
performs. Of course we start with a particularly simple one, but also definitely
with an important example.
Namely, we consider a single point charge Q that precedes a
should be infinitely extensive plane.
We assume that this level is also arbitrarily good conductive, that is
so a perfect ladder extended infinitely.
Regarding the nomenclature: I already said that we have charge Q, which is in one
Distance a to the surface and if we now observe this at a point r,
then of course we have different sizes that we look at.
We have the position vector r, which is in cylindrical coordinates and that is
a cheap choice of coordinates, of course.
In cylindrical coordinates it will have the z-component on the one hand and on the other hand
a rho component, the phi component didn't play any here at first
Role, because the whole thing can also be thought of as rotationally symmetrical
the z-axis around. So, coordinate rho, coordinate z are clear and as always at
Of course, the distance between the cargo and the load also plays a role in such arrangements
Observation point a role.
This is the vector R here and of course that is nothing other than r minus r 'and
Here r 'would be a times the unit vector in the z-direction.
The length of R which is obviously rho square root of rho square
plus z minus a square.
z minus a is the piece in front here.
Yes, a normal vector is already written on it, that's because we
now the boundary conditions will also be brought into play.
As always, we will arrange the normal vector so that it comes out of the volume
shows and we used to designate the materials in such a way that one
Side - this side here - is one and this side into the n
shows, then side is two.
The boundary conditions for the electric field or
I wrote down the dielectric shift here again.
It then applies to the normal components that D_2 normal minus D_1 normal just that
The surface charge density is rho_F and it applies to the tangential component
continuous E is tangential 2 minus E tangential 1 is equal to 0.
If we now assume a vacuum here, then D 1 would of course be epsilon_1 times E_1
-We will then use that right away- and for the metal, what is on this page
is, we already know that there will be no field on this side, that is
both E_2 and also D_2 will be 0, so this equation works accordingly here
simplify. E tangentially equals 0, which is what it says here, in particular, means that
the E-field is perpendicular to this surface of this perfect conductor.
We have already derived that in the past and have already done so
used, so not surprising.
Well, that is the arrangement and we can now briefly recollect it
shout what's going to happen now.
If I didn't have this border here, then I would have radial around Q here
symmetrical electric field, with corresponding circular potential lines.
I won't have that here, but this electrical charge that is here in front of the
conductive wall is, of course, will have an impact on the charge carriers, which yes
are freely movable in the perfect ladder.
Specifically, this means, let's assume that there is a positive charge here, then
negative charge will accumulate here and it is also clear that
it then drops and the further we are from this point rho equal to 0,
the less influence will of course be.
This surface charge that results here is of course on again
Field - a potential - connected and what ultimately interests us is that
Overlay here in the solution volume.
If we wanted to solve this directly, it would lead to an integral equation for the
Surface charge density and that is relatively complicated.
With such simple geometries, however, we can very quickly consider how
the problem is easier to solve.
But we draw it a little differently.
So on the one hand we still have our solution volume with us
our cargo. And I have now indicated what the field would look like
if this conductive plane weren't there.
Then we would have this radially symmetrical field and it is already
of course, if you look at that, it would of course be an E-field, the one up here
and is of course not automatically perpendicular to the surface down here .
In the middle it cuts it, but outside the middle we become
always have a certain angle.
But we are not allowed to have that.
This tangential component here must disappear because the tangential component
is continuous, and we know it's 0 on one side, so it must be on the
on the other side also be 0.
The tangential component would be a small component here
is directed from the bottom up.
So we can only compensate for it by adding another field
have the one corresponding
Has tangential component from top to bottom.
As one can easily imagine, this will be fulfilled when this is this one
small, this blue arrow would be, then namely the superposition of these two
Field this brown arrow, which is now longer overall, but vertical
stands on the surface.
It is easy to think that if I do it like that, then it will also work in the
Center. In the middle it will be twice as strong, but also stand upright and
down here, of course, it also works with the corresponding construction instructions.
Again with a larger resulting field, but above all
with vanishing tangential components in the resulting field strength.
Yes, and now of course it's easy to say: Okay, where does this field come from?
And if I trace that back, of course , I easily come to the conclusion that I am here
can introduce an image charge.
And if I introduce this image load without this layer being there now, then
would be the tangential components in the combination of the charge and the image charge
along this surface - which is now for the
Moment is no longer there - actually gone.
For this it has to be a real image load and that means the distance on the
one side and the distance on the other side, must be the same.
And if I have a positive charge here, then I need one on the side
negative charge of the same size.
Otherwise it will not work with the compensation of the tangential component.
By the way , we already know the picture that comes out of it, of course, we have it
Already looked at the dipole, here again called to mind: If we
draw a dipole, then exactly this condition is here at the center line,
that the E-field is always perpendicular to the line is fulfilled.
That is, the dipole does just that.
From this point of view, for this simple example, we can already say: One half
this dipole field, that will also be the solution and we calculate exactly that now
but actually from one more time.
Boundary conditions are very fundamental, I have E-tangential when z equals 0 is 0 resp.
phi when z equals 0 is 0 because this is a perfect conductor and it is grounded
is, then I have potential 0.
I specified that.
I can now use the superposition of the two potentials, i.e. one of them
write down the original charge and once the picture charge directly.
Geometrically, it's easy to make clear: this was the capital R that we did before
had introduced.
And the distance from the image charge to the observation point is simply the root
from rho square plus z plus a to the square.
If I am here now, of course, the potential in the appropriate way
then of course I can immediately E equal minus gradient phi that too
Calculate the E-field. We're doing this in cylinder coordinates.
We don't have a phi dependency, so I need an education for the gradient
not to worry about it either.
There is then only one rho and one z component for the E field, which I now have
just wrote down here.
The interesting thing now is the level z equal to 0, where we look at that again
can see how is the E-field at z equal to 0.
Now you simply set z equal to 0 everywhere, and you find that the rho component
disappears and actually just a z-component remains, which is here
is written accordingly.
If I only have one z component, that's exactly what I wanted.
I obviously have no tangential component of the electrical
Field, but just a normal component.
If we have the electric field, then we can actually do a lot now
quickly calculate the surface charge density because we have that
Continuity component for the normal component of the dielectric displacement.
And in that case I can just write it as epsilon_0 times E.
That is only the one side that we called the second side.
On the second page is the E field or
the dielectric shift is 0 anyway, so that only this one term remains.
The normal component - the normal vector was directed outwards - so it is
Normal component just the negative z component, so this sign
then just disappears here.
And this discontinuity is just given by the surface charge, so that we
simply, when we have the field, we get the rho_F by still having the field
multiply once with ebsilon_0 -at the place of the surface of course- and that gives
then this expression, which we then wrote down here again.
Which, by the way, we have found so much easier than when we found a corresponding one
Integral equation should have solved.
If we know the rho_F, then of course we can also calculate the total charge of the area
calculate. We do this to check the plausibility of the whole thing again.
For this we have to do the rho integration from 0 to infinity and the
phi integration from 0 to 2 pi.
Since we have no pi dependency, there is of course just a factor of 2 pi
who hereby compensates.
And the rest of the integral is then quickly solved.
In fact, if you do that, minus Q comes out here .
That is very plausible, because the minus Q is also exactly the charge that we call
Image charge have started and the overall charge of the arrangement is allowed to increase
of course through this modeling step i did that i instead of the
Surface charge density inserted an image charge through this
Modulation step, of course, the overall charge must not affect the arrangement
change. All in all, the arrangement has no overall charge, and therefore none
Monopoly term in multipole expansion.
So this is such a pure dipole, as we already suspected.
I had just shown the picture of the dipole as the solution, which confirms that
just here again.
Please note: you always have to be careful with the image loading method
The solution that we have now calculated are still only solutions in this case
just for z greater than 0, i.e. in the original solution volume.
On the other side of the conductive surface, we also know the solution
beautiful. We know that E is equal to 0 and phi is equal to 0, in other words, there we have
absolutely no reason to calculate the E-field and potential.
Only sometimes is it like this, if you calculate this in the context of the image loading method,
you have almost forgotten the original problem when you find the solution
found and then also writes this E-field for z less than 0
as a solution. Of course you can't do that.
We found the potential.
I wrote it down here again.
And if we have the potential, we know we have that too, in the end
Greensche function found.
That means that we can now use this potential directly to do the Greensche function
the half-space - the space with any highly conductive surface
is complete- calculate.
As if the half-space z were larger than 0 and I just need to write that down now
by simply setting Q = 1.
Then that is immediately the Green function.
And of course we wrote here without coordinates, but the
Stepping from here to here shouldn't be too difficult.
The key point is that I am using the Greenschen feature that I am at
found an example, namely in which I said this area is
grounded and has the potential 0 everywhere I now also use these greens
can take to solve any other boundary value problems of the same geometry.
The same geometry means that this conductive plane must of course stay where it is.
Otherwise I would have a completely different new problem.
We have seen that the boundary conditions for the E-field are also very
were elementary. I took advantage of it, so it has to continue indefinitely
be conductive so that E is perpendicular to the surface , but we could too
any other values ​​for the potential or for the normal derivatives of the
Now accept potential on this surface and then get out of the formal solution
immediately write down the solution for the potential.
Because we now know the Greensche function.
And if we know the Green's function , we also know the normal derivatives
the Greenschen feature, which I'll just repeat here just in case
have written down.
That means, as soon as now at the places of the surface, either phi or d phi after dn
is given, you can actually use the well-known Greensche function and the
known normal derivation of the Greenschen function formally solve this expression here.
Formally solving means that everything that is written here is known.
The integrals may then not be analytical in the last step
but I have at least one closed term that I am
in case of doubt can solve numerically.
We will be here in this case because that is when you see this for the first time
It is amazing to actually do the math for a simple case.
We now assume that the potential on the surface is no longer 0
should be, as we have assumed so far, but have a finite value phi_0.
What we expect is of course - since the potential is only up to there anyway
a constant is determined - that I get the same expression as for that
Problem where the potential at the surface was 0 and just the phi_0 as additive
Constant added. And we are now trying to understand that this is actually the case.
For the concrete calculation we are going to use cylindrical coordinates again .
That is the familiar expression that we had just seen and
in general - also wrote that down again - I would have just that now
Volume integral with the charge distribution, the Greens function and the two
Surface terms. Specifically, we have Dirichlet boundary conditions here, that is, we
in this case would take a look, where we find the normal derivatives of the Greens function
need. This term would not matter because we have Dirichlet boundary conditions everywhere
have and yes we know that we do not have both boundary conditions
must have a point.
That means, we can safely drop this term here, we need it
not to be considered further. The surface integral for the second term applies
ultimately to calculate.
But for this we first need the normal derivatives at the location of the surface,
so d G after dn evaluated at the surface of V, that is by definition
Gradient G times the normal vector and in the special case - the normal vector is yes
the vector minus d z- so the normal component of the gradient would now be in that
Case the negative z-component, i.e. minus d G evaluated according to dz when z equals 0.
The derivation of the Greens function here with respect to z, of course, happened quickly.
I wrote down the intermediate step again , you are welcome to do it again
comprehend. And if you put in z equal to 0 here , you get first
this expression and then the final expression as it is here in front.
Now I know the normal derivatives of the Greens function.
Now all terms are known and I only need to insert for them
Dirichlet boundary conditions.
So now the volume integral - that's highlighted here as a green term - and
the surface term that's this integral here in red and we need it now
of course only to use the concrete derivation.
We want to work out the red term.
Here in the first line I just started.
So then you pull the constants forward and pull them together.
With the integral that is still here, you can see very quickly that it is
is appropriate to substitute, for example, as follows.
We put zeta equal to rho square plus a square, then d zeta is equal to 2_rho d rho.
In principle, this is the correct expression up here.
We just have to consider the 2.
That gives the factor half up here.
We extract the a from the integral.
And the integral, what is left - with which of course the limits are adjusted
have to be - this is trivial to solve.
And if we do that now, it just ends up being phi_0.
That's exactly what we were expecting.
That is, as expected, in that case the total potential would simply be the potential
what we have calculated, in the event that we have potential 0 on the surface
have, plus the potential that we are actually in
the case on the surface.
That is very satisfying, the result.
This is exactly what we expected and shows us again that we are
Didn't make any obvious mistakes in between .
That was the first example, which is relatively simple, but actually very
is important because conductive surfaces are actually a very important part of electrical engineering
play a fundamental role.
A second example where we can look at the situation again is that
Reflection on the ball.
Spheres are also geometries that actually appear again and again.
So now we consider the following geometry as the second.
We have a sphere with a radius R -also that should again be arbitrarily conductive-
and it too should first be grounded.
And at a distance s_1 from the center point we have a charge Q.
We are interested in the potential or field at observation point P.
And this point P is initially at a distance R_1 from the original charge Q, the
we can of course simply specify.
One already suspects that one can cope with reflections again here , but it is
not so obvious how big the picture load must be now.
That is why we write something again here in general that it is minus alpha times Q.
The alpha is still unknown and you do n't know directly either
where to put them.
And so we say we put it at a distance s_2 from the center.
The potential for these two charges, we can write that down again by
we make it clear what the corresponding distances are.
The distance vector R_1 is simply actually the vector R - and that is yes
nothing else than R times e_r- minus the vector r 'and that is the vector s_1 times
he'. For the other charge which has the charge minus alpha times Q accordingly
written down. We are now demanding that the potential be on the surface of the sphere
should vanish, i.e. phi of r equals R times e_r, so when the point is on
the surface lies, that should be 0.
If you use that and demand that this is 0, it leads to this relationship here.
This relationship we met, you can easily see that this relationship
is fulfilled if s_2 -the unknown s_2- is replaced by the known R and
s_1 and if the alpha is again expressed by the well-known R and the
known s_1 is expressed in this way and whiteness.
If you use this accordingly, you will see that the condition is met.
That is, we have determined the two unknown quantities s_2 and alpha,
by considering the boundary conditions for the potential -the given
Boundary conditions for the potential - namely the potential here
should be equal to 0.
In this way, of course, we can now also use the Green's function of the sphere
determine. We know the potential.
And when we know the potential, we already know that we only need the charge
formally to be set as 1 -Q is now set to 1 -the other sizes are used.
That's the way we get that expression for them
Green's function of the sphere.
You can also write it down differently by clicking here accordingly
the amount and this is how it looks, not as handy as in the example what
we just had, but still expressions that can be handled.
That is obviously symmetrical, that’s another check.
Because we know that the G of r and r 'must be the same as G of r' and r, so
interchanging r and r 'must not play a role.
You quickly convince yourself that it is fulfilled here.
The boundary conditions of the potential are fulfilled.
The Poisson equation in the solution volume is fulfilled and we then already know
that we have the uniqueness of the solution and that means we know at that point
already, that we are already finished by then.
Again you can get here by solving this one problem with a point charge
of a grounded sphere can actually solve a whole class of problems by one
this Greens function that we have now found , then just use and then
the potential - no matter how the boundary conditions are given - whether as
von Neumann boundary condition or as Dirchlet boundary condition or as
mixed boundary conditions do not matter.
Formally, we can calculate this because we now know the Greens function.
Of course, we still have to know the normal derivatives.
In that case it can be calculated quickly.
We don't have to go through this term by term either , I just have it
written down again.
Well, now you can see how it works and honestly, it is not
always easy to find the greens function.
But for certain, simple geometries, you can actually use this
Image loading method - so with this mirroring method - the Greens function
very easy and very elegant to find.
We have not yet considered an important case.
We have now only ever looked at one surface, that for reflection
is used. But it is often the case that you have several surfaces.
And even in this case you can actually sometimes use the
Mirroring method to get ahead. We'll look at an example very, very briefly
look at. Namely the case - we will see in a moment - that we two parallel
have infinitely good governance levels.
Whenever you have several mirror planes or have several edges, then you have to
note that you don't just have to mirror the original charge on each of the edges,
but you actually have to re-apply the mirror charge at the edges
mirror on which they have not yet been mirrored.
So we have to take into account reflections of the reflection.
It's really like standing between two mirrors
-two real, optical mirrors- stands.
Then it is also the case that you see very, very many reflections and because of yourself
reflect the mirror images again.
This is actually the case here.
This is because we actually only solve this for certain geometries
because that then automatically leads to a series development.
Whether this series then converges or not , that actually depends on the
concrete geometry.
A simple example: in which it will also converge, but we will not
write down to the last point.
Let's assume that we actually have two conductive plane-parallel edges.
And in between we have a load distribution.
I've now written them down in triangular shapes, that’s the same for them
To see reflections a little better.
What the mirror image of whom is seen a little better with the triangle than when
I would only make one point.
What has to be done first, of course, is that we use this charge distribution to the
Mirror the example at this first level.
That would result in a corresponding mirror image, as I have drawn in here.
With a corresponding charge distribution with the opposite sign.
Of course, I also have to mirror this at the right interface and that gives it.
So now it goes on.
I need the charge that I have now mirrored at this interface, this one
Image charge, I actually have to reflect it on the other surface as well,
just as I still have to mirror this image charge on the left surface.
Let's start with that.
This image charge, mirrored on the left surface, obviously gives a charge
who sits here and which I could also imagine as a mirror image of the first image charge,
if I simply put the structure in front of it periodically, the red lines should indicate that here.
It goes in the other direction accordingly.
This charge, now mirrored at the right-hand interface, is again a
Image charge - second order, so to speak - and I could also imagine that as
the image of the first image mirrored on a periodic continuation of the structure.
Yes, and now I think it's clear how this will continue.
You would have to put this whole chain in front of infinity, you have to
write down. Look at all the terms and then just see that you
writes down the series expansion and shows the convergence and the corresponding
Write down the end value of the series - the sum of the series.
In this way you can also use the greens function of this room in the
Calculate the gap between two infinitely extended plates.
Once we have the Greens function, we know that, then we can use it
Geometry solve all problems.
Well. Yes, it should actually have been at the point and you can see how
The Greens functions are actually powerful and how we actually do them for them
certain geometries are very easy with the help of the mirroring method
can find. We will actually also be in the next events
get to know other methods.
But the bottom line is that the electrostatics
The method of mirror charges is one of the most important of all for the
Solve problems to find the Greenschen functions.
As always, thank you for your attention and as usual you will find
there is also more information on the website. Thank you, see you next time.