So I come back to this lecture about -- potentials.
So in the previous lecture, I have just given you the solution to this for General Maxwell's equations
and the solutions are these -- potentials.
But in reality, it's very difficult to find out that these are really the solutions and that this solution
here is true.
So this lecture is optional, and this is because you will not really learn new physics here, because
I already told you that these are the solutions, we just have to verify it.
And furthermore, this proof is rather difficult, mathematically speaking.
So if you're not really that interested in the mathematics, you can go ahead and skip this video and
go to the next video.
But I think it's really worth the effort to go through this video because you'll learn some new cool
tricks about mathematics.
And we will also learn about the so-called Delta distribution, which is a very important concept in
theoretical physics.
So I will only prove one half of this part.
I will only prove that this electrostatic potential with this -- time here solves this Maxwell
equation here.
And then afterwards, if you want, you could verify also that the vector potential solves the other
Maxwell equation.
But the steps on the method in terms of the mathematics, they are very similar.
OK, so let us get started.
So first of all, we must realize that this large class operator acting on the electrostatic potential
is just the divergence of the gradient.
And so we go step by step and first calculate the gradient of fire before we calculate the plaza operator.
Now, since we have here this are ministers, other aren't here as well.
And always we only consider distance might introduce a new variable, which just is Capital R, which
describes this distance here.
And of course, if I don't write it bold, not as a vector, then I mean this absolute value.
So let us calculate the gradient of the electrostatic potential.
So this is a pretty factor and we can also switch around the integration and the differentiation.
So we just have to differentiate this integration to you.
We have to calculate the gradient.
Now, of course, here we have to are dependent terms.
We have to charge density, which depends on R because we have here in this time argument the R and
then we have here another R, so it's a quotient.
Or you could also say a product of two functions that are dependent.
Now another comment, this KNOBLER operator here, this gradient is typically with respect to this position
vector, but since they are linear to each other, it's the same.
If we also write it here with respect to this capital R so we can just differentiate with respect to
this R here.
OK, let's do this.
So we have here this product of these two are dependent functions, so we get a sum of two terms.
The first term here is just this one here times the derivative of one over R, so the derivative of
one of our is of course one of our square then the minus sign.
And we also need to consider the inner derivative, which is the gradient of our.
Then the other term, we know we have to apply the naval operators to discharge density and just have
to multiply by these other functions, so we have to gradient over divided by R.
Now we have to calculate what is the gradient of our I think you know this already, but here I have
written it down once again.
So a comment you could do this much more elegantly if you would use spherical coordinates.
But this in itself is also a bit difficult because then you first have to learn about spherical coordinates.
So in principle, you can also do it in Cartesian coordinates, as you can see here.
But then you need several steps and you need to be very, very careful that you do not miss any any
in a derivative so that you do not miss this term here, for example.
And you will see that this is just are divided by ah, OK.
This means we have now this term here.
So we still need this one, the gradient of the charge density.
And to you we have to gradient with respect to capital R and this coordinate enters this -- time
argument.
So what we have to do is we have to first calculate the outer derivative which is charged density divided
with respect to this argument.
And then we have to multiply and calculate the inner derivative, which is the gradient of this argument.
So we have here robots, there's just a shorthand notation, as you typically use it in physics, and
then for the for the inner derivative, we have the gradient of the argument, which is the gradient
of tea, which gives zero and then the gradient of minus are divided by.
So this in a derivative is the gradient of ah and then times minus one divided by C and the gradient
of are we know already we have it here.
This is the vector are divided by the absolute value of R, so this is our final result for this one
here.
So now let's take these two terms here, this one and that one and put them into our our gradient here.
And this is what we get for the gradient of fire.
We get two terms and they look kind of similar.
But here we have a dependent's one of our square until one of our to the power of three.
And here we get a row until we get the time derivative of row.
So this is what we have, but what we want to calculate is the plant's operator.
So now we need to calculate the divergence of this term.
So we have now these two terms and we need to calculate the divergence of both of these terms.
And now, of course, you can see we have here three functions, Row R and R Square, and here also
three functions that all depend on R.
So we need to use the product rule many times and let us start by calculating these two terms and write
them down.
First of all, I used to the product rule here.
So I write down real dot, for example.
rawData times the divergence of dysfunction here.
So that's done divergence of this one plus the other way around this function, then the scale of products
times the gradient of row dot, and then similarly I do it for this term to get these two terms.
But here we are not finished, I want to go one step further, I have to apply the product rule also
to this term here and to this term, because here we have also the product of two are dependant functions,
so we get even more terms.
So, for example, from this one here, we get these two terms.
And then from this one here, we get these two terms.
Now, this is what we have for our the plant operator and this is what we get for the integrity.
So now we need to simplify all of these six terms.
So let's get started with the first term.
Of course, here the only difficulty is calculating the divergence of R, but actually it's it's not
at all difficult because you can imagine the operator is a vector of the partial derivatives and you
calculate the scale of product with this position vector itself.
So you calculate basically derivative of respect to X of X, which is one plus derivative of Y with
respect to Y, which is one plus the derivative of Z with respect to Z, which is also one.
So we get for this term here three.
So the answer is three times robots divided by C are square.
Now the second term you can see already highlighted that's red.
So we will get a similar functional dependence.
This a bit more difficult.
So here we have to calculate the gradient of one of our square, which is a bit similar to what we previously
did when we already calculated.
So what we what we get here is first we have to construct the outer derivative, which is our to the
power of three.
So one divided by R to the poor three times minus two, which comes from this two here.
But then please do not forget about the inner derivative, which is the gradient of arm.
Now the gradient of are we have also already calculated this was the arbiter divided by R so this is
what we get in total.
And now you can see when we only look at these vectors here, our thoughts are, is of course the absolute
value of our square so we can get rid of the Spectre's here and just right our square and we divide
by our to the power of four.
So we end up with one divided by R to the power of two.
And then all of these three factors, of course, and you can see it's exactly the same dependence here.
But here we have this term three times and here we have it minus two times.
So these two will simplify already.
Now here we get a different turn here.
We have to calculate the gradient of DOT.
And this is, of course, very, very simple because on the previous slide, we have already calculated
the gradient of row.
And so we can just take our solution from the gradient of row.
But instead of getting here what we get now, roadable dot.
But besides that, it's the same thing.
And we need to consider here also this in a derivative, which is, of course, are divided by R and
then we get here also this minus one A.
So if you don't remember how it is done and if you can't do it yourself, just skip a few minutes back
where we have already calculated this gradient of row.
Now here we get the same thing.
We get again, our dot are where we can just ride our square.
And here in the denominator we have our to the power of two times.
Ah.
So we end up with one of our and then these three factors here.
So this is a different dependent's.
So we cannot cancel this term with the other two two terms.
Now, here we get another term here we have to calculate again the divergence of our which is three.
So this one is very simple.
Here we have to calculate the Dow and sorry, the gradient of one of our to the power of three.
So if we know what the gradient of one of our square is, we also know what the gradient of one of our
to the power of three is.
We just get here are to depart four instead of three, and here we get minus three instead of minus
two, which comes, of course, from these three instead of this two.
And then again, we have the derivative, which is the gradient of art, which simplifies to this one
here.
So as a total result here, we get minus three times row divided by R to the point of three.
So you see, it's again very similar here.
We get to minus three instead of two, minus two, and here the three instead of the two.
And see that it's the same functional dependence as this one here.
And in fact, these two terms even cancel out.
So that's really cool, which means we do not have these two terms here anymore because they are exactly
the same for the last term.
We have something here are times the gradient of row and we already know what the gradient of row is.
This is, of course, this minus one of a sea times.
Roedad times are divided by are we have calculated this on the previous slide.
So again, we get our times are, which is our square.
So we get here from this arg to power for only are to the power of two.
And as you can see, this term once again has the same functional dependence as these two terms here.
And furthermore, you can see that all of these terms compensate and that all of these blue terms compensate.
So all of these difficult looking derivation letters to one single term.
So we have that the entire rent is equal to minus roadable dots divided by sea squared times are.
So we have that our the class operator is the integral of this expression here.
Now, what we can do next is we can write down this road double dots as the second order, time to time
derivative, which.
Yeah, second order a time derivative of rope.
And we can then switch around the derivative and integration.
So we write down the second or the time derivative.
Of this integral of roll over are we can do this because these are here is not time dependent and Rose
was the only time dependent thing in his intake around here.
Now, another thing is that I have used this square here in the Integral and have put it outside of
this term here.
And so here you see, I have already introduced these two brackets to highlight this term here.
And if you remember really well what this term was, this was our electrostatic potential from the beginning
to look here.
If you're introduced as are you get the same thing for this term and for this term here.
So what we have here is one of a C Square, the second order time derivative of the electrostatic potential.
So what we wanted to show is that this class operator is equal to one of us square the second or third
time derivative of five, which is this one here.
But there is this term missing here.
Here.
We wanted to also show that we have another term, which is minus one of Epsilon zero.
So I think given the fact that this whole derivation was kind of difficult, it's a good first step
that we have to as a result.
But then again, where is this term?
It cannot cannot be missing.
And the reason why it is missing is that we made a small mistake in the derivation.
So our derivation is only valid when are equal or is only valid when R is not equal to zero.
This is because we divide by art in many different locations.
So whenever we have divided by hour or derivation is in fact wrong, and we need to be much more careful
for the case where ours is equal to zero.
So now, um, what what this means is that this expression here is only true if R is different from
zero.
So this means if R is equal to zero, which is, of course the case, and we also have to integrate
it, this means we get here additional terms that we have been lacking so far.
And so let's go back to this slide here where I have written down all of these different terms.
And I want to tell you which of the terms gives us additional results that we have been missing so far.
And it turns out that all of these four terms here give us nothing new except for this single term here,
which corresponds to these two terms.
So this term for our equals zero gives us another thing and let's look at what it is.
So first of all, we need to, again, remind ourselves that when we straightforwardly calculate this
divergence of this function here, we get these two terms and we get that at zero.
And it's zero for every are different from zero.
But we can also do a different thing, we can just take this term here, so not this role, but this
divergence of are divided by the power of three and we integrate over the whole three dimensional space.
And if we want to calculate this integral, we can use Goussis Theorem as we did before, and transformed
this integration to a surface integral where we integrate not over the divergence, but just over this
function here of our divided by our to the power of three.
And then this one here is this.
Yeah.
Just this element, which is which is a surface and its orientation is normal to the surface.
So it's always in radial direction.
So this one here is actually parallel to our if we consider a sphere.
So what we are doing here is we want to calculate over the whole three dimensional space, which is
infinitely large, and then he on the right hand side, we must integrate over the surface of the three
dimensional space, which is not really possible, of course, because what is the surface of something
infinitely large?
So we must use a trick here.
We must consider a finite sphere at first and what the radius are, and then we must take the limit
of this radius which goes to infinity.
And if we do this, we get an infinitely large sphere, which is of course equal to this whole three
dimensional space.
OK, so we have transformed this one here to this one, which is with just me, which just means that
we take a sphere, we integrate over the surface of the sphere with Radius R and afterwards we make
the radius infinitely large.
And then here, since we have used the sphere, now this surface element here, this orientation is
parallel to our so we get to your unit vector R and then this is given by our square synesthete ADT
to define.
So now let's use this relation here that these two are parallel to each other.
So this just gives us ah, so we get ah to the power of three divided by ah to the power of three.
And that's the whole trick here.
It means that our integration's is not dependent on.
Ah, so this means it doesn't matter how large our sphere is now we can just take the limit ah to infinity
and it will change nothing that will just give us this term here.
And when we calculate this term, this is really easy for us.
First, we can solve this integration with respect to PHI because there is no fi here.
So we just get the function or we just get to the difference of this integration boundary.
So we get two PI times this integration and this integration we can solve by.
Yeah, calculating minus cosine theta from the uh from the boundaries.
So zero to pi.
And so if you do this you get minus minus one, minus minus one, so you get two.
And so our total integral here is equal to four pi.
So what does this mean?
It means that this this term, he had divergence of are divided by R to the power of three is zero everywhere
besides the position are equal to zero.
But if we integrate this term over the whole three dimensional space, which basically means we add
up this expression here at every possible position, it gives us four pi.
So this means at the position R equals zero.
There must be a very large signal or a very large term that gives rise to this integration that gives
for PI.
And so this is something that has bothered physicists for a long time.
And it's really hard to comprehend because it's a bit counterintuitive that you just have a single point
of which you integrate and you get some finite signal.
But this has led to the introduction of the so-called Delta distribution function and this Delta distribution
function you can write here as Delta are, and in very simple terms, which are not really correct.
You could say that the Delta function is zero for everything different from R equals zero and it's infinitely
large for our equals zero.
So this means you really get only a signal from this point.
R equals zero by this definition is not really correct.
The real definition is that when you write it in an integral, you can get rid of the integer and just
take this other function here that you're right as a product.
So you take this function here, which is just for PI and take the argument are at the position zero,
which in this case just gives for PI because there is no argument.
Ah, OK.
This may sound a bit confusing to you maybe, but on the next, on one of the next slides I will show
you a bit more about the Delta distribution.
Then this may become clearer, but the important thing is we can write down that this divergence of
R divided by R to typeof three is given by four PI times the Delta function.
And this also includes our previous results that for a position where R is different from zero, then
this whole divergence is zero.
And it also includes that when we integrate over this term, we do not get zero.
We get a finite term, which is four pi in this case.
So we get our new term here, which is four PI Delta times zero, which you must not forget because
here we have to zero also.
It's now since this Delta distribution is a bit, yeah, difficult to imagine and a bit counterintuitive,
I think here I have prepared one more slide that I want to show you.
And there exists some kind of continuous transformation of this function here that approximates the
Delta distribution very well if you take the limit of eight to zero.
So this function here is a Gorshin function, which is basically x e to the power of minus X square.
And you have here in this argument also some parameter aid and also here you have a perimeter.
And now if you take this parameter eight to zero, you will see that this prefecture becomes infinitely
large and also that this Gorshin function becomes infinitely narrow.
So those are not Desharnais animation on Wikipedia about this.
So you can see here for smaller values of a this peak in the middle becomes larger and larger.
And the yeah, the broadness of the Delta function also becomes more narrow.
And the important thing is that when you take the limit to infinity, you will see that the value of
this function here becomes zero.
It becomes zero everywhere except for the position X equal to zero, and therefore the value will become
infinitely large.
And also, this may be hard to see, but maybe you can imagine this when you integrate over this area
here below this curve, this will give you the same value for every value of a.
So when you have some some final value of A, you can just calculate the integral and it will give you
a number.
Now, if you make a smaller and smaller, this this number stays the same.
And so this means also for the limit of eight to zero, where you could say that this function is zero
everywhere and it's infinitely large at zero, he will still get the same area.
And so this is the whole important thing about this data distribution, that when you integrate a function
times this Delta function or Delta distribution, then you will get a finite value.
And in fact, you will get the value of this function at the position of zero.
And if this data function is not centered around zero, as in this case, where it's centered around
X zero, then you will get the function at the position X zero.
So in our case, this is very similar here, we have an integral role like this and we have shown that
we can transform it to this expression here.
So this means when we integrate over this function, which is just four PI times Delta distribution,
we get this function at the position are equals zero.
And since this does not have any dependence, we just get for PI.
And so this means we can really express this divergence here in terms of this Delta function.
OK, that was really difficult, but this is only one of two occasions where we will use the Delta distribution
throughout this whole course.
So this is why I decided to not spend too much time on the Delta distribution, because for our purposes,
it's not that important.
We could say it's for us just a mathematical trick to write down the things in an elegant way.
But what's important is that, you know, what this Delta function does, and this is when you integrate
over overhead, you get a finite value.
OK, so we have now here over the plant operator and we have transformed our integrations and have found
that it's actually not just a single term, but that there is another term which becomes only important
when R is equal to zero.
So this means when we write it down like this, we get to hear this new term, previously we only had
this term and this term transformed to this term here.
We have shown this already.
Now, the other term here, we have this Delta distribution.
And on the previous slide, we have already learned what the standard distribution does.
It sets the the argument are to this argument are with this dash here, so this means as the integral
here, we get this function, which is for pay roll at or for the case where R is equal to this are
here.
So this means we get no here an hour and we get here a zero, which means we have finally got rid of
the retardation effect.
So this means this integral here is for Pirjo and the arguments are R and T now these four pieces here,
they cancel out and we just get minus one of Epsilon zero zero of our N t and this is exactly this term
here that was missing.
Sort of all this means after this long time, after this difficult lecture, we have finally proven
that indeed this -- scalar potential solves this Maxwell equation.
And now I told you, it is already in the beginning.
You can do a very similar derivation for the vector potential.
So the mathematics is really similar.
But again, it's quite an effort, maybe even more, because this is here a vector and this was only
a scalar.
But I think if you really take your time, you could do it yourself.
If you follow along the steps that I have just shown you for this Caillat potential.