So let's go ahead and talk about the Nabila operator and the previous lecture, we have already introduced
the operator as being a vector of the partial derivatives.
And alternatively, you could also write it down like this to save some space.
So this t just means you transpose this line vector in such a column vector.
And of course, if you just take this KNOBLER operator by itself, it doesn't mean anything, you always
had to let this operator act on the function.
So that makes sense then.
Other text books or on websites, you may also find other notations.
For example, you could write down does not an operator as partial derivative with respect to a vector,
which I don't really like because it would somehow indicate that you are dividing by a vector, which
doesn't make any sense.
So I prefer this KNOBLER operator.
But the downside of this Knapp's operator, is that you don't really know with respect to which variables
you differentiate.
So typically it's just X, Y and Z.
But if you have some more difficult problem where you have multiple vectors included in your function
and you maybe also want to derive with respect to different variables.
For example, when you have a function that depends on the vector, which is X, Y, Z, but also on
the vector M, which is M, X and Y and Z, and you want to differentiate with respect to M, then you
could write it down like this.
You write down the block operator and you write down M on the bottom.
And this indicates that you don't want to derive with respect to X, Y, Z, but with respect to different
variables.
And if you write down nothing here, then it's typically assumed that you derive with respect to X,
Y and Z.
Now, we have also learned already about the gradient, which is when you take your Anabella operator
and you let this operator act on the scalar function.
So this function ideally depends on all of these variables X, Y and Z.
And it could also depend on other variables that are not included in this KNOBLER operator.
And then this is what we get.
We get a vector with three components.
So now let's see what this looks like, for our example, so let's take again our mountain.
And this time I want to show you the top down view of our mountain.
And what we get is something like this.
So we look from the top and you can see here the color in this 3D plot corresponds to this color here.
And so we have several lines to go around in such oval shapes, so it goes around the top of the mountain
here, for example, and on these lines, the height of a mountain is equal.
So on every point of this line and then we go to a different line.
So the other line is different.
But on this line, the height is again constant.
So this is something you would, for example, see on maps where you have a mountain and then this is
typically indicated with such a plot because it's not really sensible to plot a map in 3D.
So now what we can do is we can calculate our gradient.
So we have already calculated the partial derivatives in the previous lecture.
So I will just show you the results.
Of course, here we would write partial derivative with respect to Z of F, but since our first two
dimensional and does not depend on Z, this just gives us zero.
So we get this one here as our gradient.
And now you see this is a field of vectors.
So at every position, X and Y.
So for every position on this map, we get a vector which can be indicated by an arrow.
And now if we plot this vector field here, this gradient, it looks like this.
And here we see a special feature of the gradient.
All of these arrows are always pointing perpendicular to such and such a line of equal height.
And that's something that's generally true.
The gradient is always perpendicular to these lines.
And what this means is the gradient always tells us the direction of the steepest slope.
For example, imagine you're standing here on the mountain at this point and you want to know which
direction is the steepest.
Then you just calculate the gradient and it gives you the vector along which you have to go so that
you have the steepest slope.
So the gradient points always along the direction of the steepest slope.
Now, besides the gradient, we can also define different operations with the naval operator.
One of them is the divergence.
So far, we have considered scalar functions, which are no vectors themselves, but just Skylar's.
But now let's consider a different function, let's call it H and H is a vector.
So it's it has multiple components.
And here we want to go to three dimensions.
So it has H, X, X, Y and Z.
Now, if we calculate the divergence of H, we have to calculate the dot product of our Nabala vector
with this function H.
So this means we have to calculate this DOT product, which of course means we have to calculate the
partial derivative of H X with respect to X plus these other two terms.
And you can see here and now we go ahead and do an exercise.
So let's go ahead and calculate, calculate these two divergences.
So we start with a very simple one, the divergence of the position vector R, where our vector R is
just to vector X, Y and Z.
So of course then the divergence of R is the sum of partial derivative with respect to X of X plus the
other two terms.
Here we have Y and here we have C.
And of course you notice the derivative of X with respect to X is just one.
So we have one plus here we have also one plus one which is three.
So the divergence of the position Vector R is three.
Now we can use this result to calculate divergence of a more difficult function we have here.
The position vector are divided by R, so R is the absolute value of the position vector, which is
the square root of X square, y square, the square.
And we could also write this down as this brackets here, which is on the and this square root to the
power of one half.
OK, so here we have a quotient or a product.
We could also say and we can apply the product rule in this case.
So we write down the product run one of our times, the derivative of of.
Ah.
So this is because our first function is our and our other function is one of our.
So this is our first term and then we get the other term, which is the vector R, so we leave that
invariant and we derive with respect to the other function.
So now we write down gradient of one over R, so this one here, we know already this is all three,
so let's write it down.
We get three of our class, our times, the gradient of one over R so let's calculate the gradient of
one of our artists a bit more difficult.
So gradient of one of our is the gradient of X squared plus Y squared plus Z square to the power of
one half and now we get of course a vector.
And I want to start with calculating just the X component.
So we have to calculate the partial derivative with respect to X of this whole expression here.
So we first have to calculate the outer derivative, which is one 1/2 times.
Sorry I made a mistake here.
This is, of course, one of ours, so this is not one half year, it's minus one half.
So we get the outer derivative, which is minus one half times this bracket.
And then we have to decrease the exponential by one, so we get minus three over two, and then we also
have to consider the inner derivative and the inner derivative of X squared plus Y square, blowsy square
is two X, so we get to X and here for Y, we get the same thing for the outer derivative and we get
to Y for the inner derivative T and the same thing times to Z.
So what this is, is actually we can take here just one 1/2 and this too, so they cancel out.
So we get minus, um, in all cases we have this expression here.
So this is ah.
To the power of minus.
Three, because our is this expression in the brackets to the power of one half, and then we have the
vector X, Y, Z.
So we can also write this down as being, um.
Minors are to the past three, and then here we have our Victor R..
OK, yeah, so here, of course, make sure that this is also the victor, so now we can go ahead and
continue calculating this divergence of this whole term where we have to make sure that this one that
we calculate this one here.
So let's go back.
Let's calculate the divergence of our divided by hour.
So we already had our first term, three of our plus, and now we have our vector dot products with
minus factor are divided by our to the power of three.
And so this one here, our times are is because these two are parallel because they are the same vectors,
just our square.
So we get our square divided by to the power of three which is one over.
So we get three of our minus one over our.
So this is.
Two of our and this is our result.
OK, so let's go ahead and plot this.
So here's our function age, which we have just calculators, is the vector are divided by the absolute
value of R, so that's a vector.
That's a scalar.
So this function depends on every every point in the three dimensional space, it's dependent on X,
Y and Z because we have learned what our vector and the absolute value of our are.
So this means we get for every position in the three dimensional space, a value of our function.
And our function itself is again a vector.
So it has three components.
And a way to plot this is to position an arrow at every point in history, dimensional space and its
direction and length corresponds to the value of this function.
So you can see what this looks like.
It is vector field that points along the radio direction.
So here we have the center point and every arrow points outwards along the radial direction, which
is of course, because we have here is dependent on the position Vector R.
And also you can see, since we always divide by the absolute value of our all of these arrows have
the same length and the only point where this field is undefined is the position of zero.
OK, so now we can this is a bit looks a bit confusing, this is because that's a three dimensional
plot with another three dimensional field.
Yeah.
Included in this plot.
So it's too much information to really be recognizable easily.
So what's more easy is to take a cut of this field.
So that's a special case for setting Dessy Component zero.
So in here, you can see already I have included this plane for Z equals zero.
So this is what it looks like when we look from the top.
This is what our field looks like for Z equals zero.
So here you can see again this radial profile of our vector fields.
All of these arrows point along the radial direction outwards and all of them have the same length.
Now, if we calculate the divergence of this field, as we have done in the previous lecture, we get
to Örvar and if we plot to over, we get this function here.
So far, positions far away from the center, we get zero because then this R becomes large and this
expression here becomes zero.
But when we approach the zero point here, then this function becomes larger and larger and in fact
it even goes to infinity.
This is because this is why.
This is why here I have just cut off the plot at some point because I cannot plot infinity.
And what this means is that our vector fields are divided by our has a very strong divergence in the
center.
And this in terms of physics, for example, would be the electric field or the electric field of a
charge.
So if you have a charge and charge, it sits at some point, then the electric field will have a radio
shape and it will have a divergence because the charge is the divergence of the electric field.
And there are other examples in physics and also in other fields of science where the divergence plays
an important role and it's a very useful feature to characterize vector fields that look difficult.
But you can really tell, OK, the zero position seems to be a special point by just looking at these
vectors.
And in fact, if we look at the divergence, then it's very large at this particular position.
Now, the other thing that we can do with the operator and a vector function is to calculate the kernel
instead of the divergence.
So let us once again consider our function age, which is still a three dimensional function, and let's
calculate the kernel so the kernel of our function age is just the vector product of the knuckler operator
with the function age.
So you can write it down like this.
You have this vector product.
Now you just have to know how to calculate a vector product.
So you get as a result, a three component vector, which looks like this.
For example, for the first component, you have to take the partial derivative with respect to Y and
let it act on HHC and then subtract the partial derivative, which is big to see acting on H.
Y and then the other terms you get in a very similar fashion.
So let's go ahead and calculate an example.
So here we have an example in cylindrical coordinates.
So that's something that should not confuse you.
Does this term role here is just the the length in the plane.
So this is the square root of X squared plus Y square.
So it's a bit different to R where we would have the same expression, but another plus C square here.
So this is just the distance in the X Y plane.
So now we can write this down as the rotation of our vector is minus Y divided by X squared plus Y square.
And here we have X divided by X squared plus Y square.
And here we have zero.
All right, so if we calculate this, then we can look at our equation and here we have only X and Y
components.
So we do not have this one here and we do not have this one here.
Furthermore, we have only a wider pendent X component and an X X dependent Y component.
So this means the only terms that give results different from zeros are partial derivative of the X
component with respect to Y.
So let's search for this year.
We have it X component which was packed.
Why does one and the other term that gives non-zero is the X derivative of the Y component.
So we have to search this is this one here.
OK, and the other ones are also zero.
So we have only a Z components, so instead of writing this down as a vector, I just write here we
have our unit vector along Z, which is so easy is zero zero one and times this this expression here.
So we have partial derivative acts of the of the Y component, which is X divided by X square, y square
square root minus um the Y derivative of the X component, which is this one here.
So we have another minus sign, so we get a plus and then we have here Y divided by X squared plus Y
square.
OK, and then we make such a racket here and that's it.
OK, now we just have to calculate these partial derivatives here so we get easy.
And now as the first derivative we get, we have to apply the the product rule and then later also the
chain rule.
So we get a first term where we have to where we write one over square root, X squared plus Y square
times.
The derivative of this term here, which is one.
So that's it.
And then we have plus.
And this time we leave the X here and we have to differentiate one over the square root of X squared
plus Y square.
So we get as the outer derivative in this case, minus one half and then one divided by X squared plus
Y square to the power of three because he the exponent is minus one half.
We have to decrease the exponent by one.
So we get minus three 1/2, which is exactly this one here.
And as a prefect we get minus one half and then we have to consider the inner derivative of X squared
plus Y square, which is to X.
So we have to multiply two X and then here we get very similar terms.
That's just copy them.
So we get for the first time exactly the same.
This is this one here.
And for the other term we get, uh, y then minus one half.
This one is the same and here we get four in a derivative to Y and here we get X squared plus Y square
to the power of three.
So what we have is easy times.
Then let's, let's uh, let's look at these individual terms.
So we have twice this first term here.
So this is two times and this one here is are two of our and now we only have the second and the fourth
term left.
So let's write them down.
We have a minus sign in both cases.
We have in both cases one half and then two later.
So these cancel out and then we have here X Square and we we divide by, um, sorry, I made a mistake
here.
This is of course not r this is real.
So this is what we introduced here.
So this is not r this is real.
And here we get to row again, but wrote to the power of three and here we have then the fourth term
which looks exactly the same, but instead of X Square we get Y squared so we can write it down like
this.
As you can see here, we have X squared plus Y square, which is row square, so we have zero square
divided by row to the power of three, which is one overall.
The result is two of a row, minus one overall is one of a row times easy.
This is the rotation of this function here.
Now let us look what this whole field looks like.
So, first of all, let us plot our function H which was this vector here with minus Y and the X component
and X and the Y component, and then we divide by the distance in the plane.
And so our vector field looks like this.
As you can see, we have no Z dependent's.
So this means the field looks the same for every value of C.
So we get something like a cube here.
We can cut through this tube and the profile in every cut looks exactly the same.
And if we look at such a cut here, you can see that it looks like such a rotating, rotating vector
plot.
So once again, at the position zero, it's not defined because we then would divide by zero.
But besides that, we we get such a plot here where we have these arrows, which all have the same length,
and they go around the center in circles.
So we have calculated the curl of this of this thing sometimes I also call it the rotation, because
this is what we call it in my native language.
And I think it also makes sense to call it the rotation, because when we have such a rotating vector
fields, you see that everything rotates around the center.
And if we plot the rotation, if we plot the Z component that we have just calculated, we get again
something that we had previously where we have a very large signal of this curl here of of this rotation
in the center for the position zero.
So this means this point here is special.
It's a special point of the curl of this vector field.
And this shows also in this plot here where we have plotted the kernel.
So we have learned now about the divergence and the curl.
And both of these are very useful features to characterize difficult looking vector fields.
And for example, the kernel can be used to characterize the magnetic field that arises from here from
from a current.
For example, if you have a wire and you let the current flow through this wire, then you will get
a magnetic field that rotates around the wire and then the to the current or the wire itself then acts
like the curl of this magnetic field.
Now, the very last thing that I would just want to mention briefly, because we will not really use
it that often is to lab class operator Plata operator is a second order differential operator.
And it is basically just as some of the second order partial derivatives.
So you can let it act on a scale or function and then you have to calculate the second order, partial
derivatives.
And when you look at this, you will see that this is nothing different than the divergence of the gradient
of.
So when you want to calculate the A plus operator acting on an F, you can do it like this.
Or you can first calculate the gradient and then calculate the divergence of this gradient.
And the cool thing is that this lab class operator can also act on vector functions because this is
just some of the scale of operation.
So it doesn't matter if you have a scalar here or a vector here.
So you can also calculate this lab class operator acting on this function here.
And the operator is used, for example, in the wave equation or even in the equation in quantum mechanics.
So now you know about the operator and you know about gradients, divergences and curls, and we are
now ready to apply them to our problems.