All right, so let us continue.
So in the previous lecture, we have introduced the Hetson dipole and we have calculated the electrostatic
potential fire and the vector potential, eh?
So now we can go ahead and use these two equations that correspond to two equations and we can calculate
our fields, E and B..
So this is what we finally want.
So we will start with this.
B is equal to the rotation of E that is highlighted here because it is the more simple operation.
So we just take our vector potential here with this -- time argument and we calculate the rotation.
So basically we just have to calculate the rotation of p dot vector divided by R, but here it's again
really important to remember that this time argument is also R dependent.
So this derivative here also acts on the P.
So what we get here is we have to use yet this identity here for the rotation, so if we calculate the
rotation of a scalar function times a vector.
So in our case, this would be one of our times the pitot vector.
We get these two terms here and as I have written down here, so we get to scale our function and then
the rotation of the vector minus the vector and then the the vector product with the gradient of the
scale of function.
So now we have to calculate the gradient of one of our and we have to calculate the rotation of pitot.
So here this gives us these two terms, the gradient of one of our this we have already calculated a
couple of times.
This is minus our vector divided by our to the power of three.
And then we get here for the rotation of P, we get first the outer derivative, which is the time derivative
of P dot, which is P double dots.
And then we also have to consider the inner derivative which is um yeah.
Here this t does not give us anything because it's, it's not dependent on R but this term here again
gives us something.
So we have to calculate minus one of the C times the gradient of our gradient of are we already had.
This is our vector divided by R so we get this expression here and once again consider at least the
time argument with the retardation.
So now we can simplify, we can.
Holders are divided by r out of the brackets because we have it here and here.
And so our final result for the magnetic field looks like this looks kind of simple.
And we have here a second all the time derivative of P and the first all the time, the time derivative
of P, which is our electric dipole moment.
So while this calculation of B was kind of kind of simple, just like four or five steps, the calculation
of E will be a bit more difficult because you can see we need here the time derivative of A and we need
the gradient of.
So first we have to calculate these two things.
So the time derivative of A is actually really, really simple because the only time dependent thing
in A is P dot.
So a dot is just dependent on pedulla dot.
So that's really all you have to know.
It's just straightforward writing here.
Another dot.
OK, but the calculation of the gradient, a fight is more difficult.
So here this is because we already had two terms and is fine and now we have to calculate the gradient
of this whole expression here.
So what we have to calculate is first we have to calculate the gradient of our dot dots divided by our
square.
And then we have to calculate the gradient of our dot p divided by R to the power of three.
So you can see already we have to gradients of functions that have very similar shapes.
So it turns out when we calculate this one here, it's done really simple to calculate only also this
one because we have to apply exactly, exactly the same steps.
We just have to be careful because here we have P dots.
Here we have only P and here we have a path to and here we have a panel of three.
But still now what we do is we use this identity for the gradient, so if we calculate the gradient
of such a scale product of two vectors, we actually get four terms.
And since we have here to two terms already, we get in total eight terms.
So I have used now this law here to write down all of these eight terms.
And here I have used that our AI is P dot or P P Dot and Pete and our B is our vector divided by our
square or our vector divided by R to the power of three.
All right.
So now we have to go ahead and calculate all of these eight terms and we will.
So I will I will start with the very first term here.
And we calculate it in Cartesian coordinates.
And you will see that this takes a lot of effort and is quite difficult.
So for all of the other terms, we will then turn to a circle coordinates and make our lives very much
easier.
So this will then take not so much time.
But I think for this term, it's it's just a good practice to just tried to calculate it once and Cartesian
coordinates to really see what is happening here, because this is quite an unusual term that you don't
see so often, because if you would not have these brackets here, you would have the gradient of a
vector.
And that's a bit a bit strange because the gradient is a vector and then you calculate a vector vector.
So that's really a bit difficult.
So you need to be careful that you first calculate this DOT product here.
So it means we get P extorts times, this partial derivative with respect to X, and then we get these
two other terms where we have replaced the X with A, Y and Z.
And now we have to sum here because the DOT product is of course a sum.
And these three terms all have to act on this vector here, which is the vector X, Y, Z divided by
our square.
OK, so we must calculate now these three operations on all of these three terms.
So this will give us nine independent calculations.
But also you need to be careful because this hour vector is actually so our square is actually X squared
plus Y squared plus C squared, because you need to be careful because this is also X, Y and Z dependent.
So the derivatives will not be zero here.
OK, so you get four.
Let's maybe let's start with the first was the first operation where we have to calculate P, dot,
x and then the X derivative of X divided by our square.
So as I already mentioned, we have here in art that is also X dependent.
So we have a product or quotient of two X dependant functions.
So we need to once again apply the product rule.
So the first term is I write down minus one of our square times to the X derivative of X, which is
of course one.
So this gives us one of our square and we have to multiply also by accident.
That's this term here.
And then we get the other term where we write X times the X dependence of one overall square.
So now we have to calculate the exact derivative of one of our square and one of our square is one divided
by X squared plus Y squared plus Z square.
So we get then in the denominator, we have to increase the the power of the whole thing, which is
which is now minus one.
We have to decrease it by one.
So we get minus two.
So this gives us now an our vector here to the power of four because the vector is actually the square
root.
OK, and then we need to write this minus one also as a prefect here and we have to calculate the inner
derivative, which is four X square, two X.
This is where this one comes from here.
So this may sound a bit confusing, but please go ahead, take a sheet of paper, write down right on
this one here as X divided by X squared plus Y square, Plessey Square, and then do the chain rule
yourself and you will see you will get this term.
And don't forget the dot from here.
Now for the other terms where this derivative X on this one and this one on this one, we do not have
to apply, have to apply the product rule because X is of course not Y dependent.
So we just get a single term, which is essentially the same one as this one.
But here we get the Y because the inner derivative is here because of this y it's two Y, so we get
to hear multiple terms that are very similar.
Here we have a Z instead of the Y and the X, and then for the four, the second component, we get
essentially the same thing for yapped for these three terms.
But you have to, of course, consider the second term of the product rule by using disintegrate.
This differentiation here.
Yeah.
So in total, we get four terms for all of these components and now we can simplify what is written
down here.
So these first things here, this is actually the p dot vector.
So because we have pipsy, dot, dot, dot, dot, dot, dot vector divided by our square and then these
things here, these are minus two in all cases, minus two times.
And then we have here the OK, here we have.
X times X, plus Y times P, Y dot, plus C times dot, and we have to send every single component and
this X times dot plus Y Y plus this is actually just ah dot product P dot.
And then we just have here is a pre vector x, y, z, x, y, z, x, y, z, x, y, z.
And don't forget the denominator are two the power of four.
So this is what we have and this is of course here the position vector.
So we get as a final result, prad vector divided by our square minus two times our dot plot and our
vector divided by our to the power of four.
OK, wow, that was really difficult, but I think it's really purposeful practice if you try yourself
and sort of recommended that you follow along these steps to see that this is really true.
But of course we have seven other terms in front of us and we do not want to do such a difficult calculation
seven more times.
So for the only time in this course, I will turn to spherical coordinates and I will show you a nice
trick.
So if you really know about spherical coordinates, you can derive that enable operate and spherical
coordinates.
Looks like this one here.
You in spherical coordinates.
You do not have the X, Y and Z components, but you have the unit vectors along the radial direction
along the as you move that direction and along the polar direction.
And all of these three components will have of course, individual derivatives for this KNOBLER operator.
However, now it's really important to realize that all of our terms here depend only on R and we have
no theta and no fi dependent's.
So this means when we let this novela operate to act on on our terms here, these will always give zero.
So this means we can just consider our Nabala operator to be e r times this partial derivative or respect
to this radius r and you can write of course E are also as the correct divided by R.
All right, so now we can use this and go ahead and calculate the other terms.
For example, we can calculate now this one here.
Yeah, this one here.
So we write down instead of Arnab operator are divided by our times, this derivative here.
So we have now our times, our vector divided by R-squared, divided by R and R, the times our vector
is just R squared.
So we get one of our in total then this are derivative of p dot.
And this is, of course, yeah, because this is now not not the gradient here, but it's really just
a partial derivative with respect to our we get the outer derivative, which is our double dots, and
the inner derivative is just minus one over three times partial derivative of R with respect to our
which is one to our total result.
Is that one here.
Now for the next term, we need to apply a bit, a bit more thought.
So here we have here we have are divided by our square then the vector product of the operator and vector
product of P Dot.
And so once again we just take this one here instead of an apple operator and put it in so we can pull
out the one of ours.
We get here one of our three, then we get our vector product with our vector product with this our
derivative of P Dot and this.
In fact, we have already calculated here in this step there, we have found out that and this derivative
of pi dot dot is just minus one of the C times P double dot.
So this is what we get.
And here.
Here, you know, we can just use another law for this double vector product where we can write it down
in terms of B, A, C minus CAIB.
This is one of the famous rules.
And since we have here actually A equal to B, we can also write it down as a times dot product of A
C minus C times dot product of A, which is just the absolute value of a square.
And since here are a is are we get for this one the absolute value of our square.
So this is why in this term here we have all squared divided by R to the power of three, which is just
one of our.
OK, so this is what we get for this term and know the last term, that is the most easy one we use
again, this one here in terms of our KNOBLER operator.
And now you can see we have here the vector product of our and some things and are and some vectors.
And so we have a vector product of R and R, and since these two are of course parallel.
So they because they are the same today, of course parallel, then the vector product will always be
zero.
And so this whole thing will be zero.
OK, so this is what we get, we have our four terms now and now we can calculate the other four terms
in the exact same way and really not much changes here.
Of course, for all the peace, you will have to remove one of the dots.
So you will get here p p dot, dot, dot, where you have here, p p double, double, double that.
And you also have to take care of this different exponent here in this R, so here you have three three,
three three and here you had two everywhere.
So this means you will typically have to increase here the power of all of these one of ours.
So yeah, it's basically everywhere you can you can see.
And then here for this term, you need to be really careful.
Maybe you could you you write it down yourself and take your time.
But of course you will get to hear different prefecture because here you have now two, three instead
of the two instead of this two.
I'm sorry.
All right.
That was quite an effort.
But we have finally done it.
We have written down we have calculated all of the eight terms so we can now write down what our gradient
of the electrostatic potential looks like.
So we have these six terms because these two give us zero.
And you can also see that some of the terms look exactly the same.
So they will.
So you can simplify them.
For example, you have this term year and this term here.
Yeah, because here you see you also have a one of a C, so this term and this one can be transformed
to.
Yeah.
Basically this term with a three.
And when you do this really carefully, you can write down all of these terms and you will see that
you have five terms that have different functional dependencies.
OK, so here we are, we have the electrostatic potential, we have developed a potential, we have
the gradient of fire, we have a dot, and now we have already cut what we had already calculated be
the magnetic field.
And now we can finally calculate the electric field.
And this is basically just this expression here with a minus sign.
And then we have to take into account also this single term because we have also minus a dot.
So this means now, instead of five terms for gradient five, we get six terms.
And here I have written all of them down and I have ordered them with respect to p p p dot, dot, dot,
dot, dot, dot.
And now in the next session, we will discuss the shapes and the dependencies of these two fields,
B and E, and we'll look at the fields and I will show you what they look like.