In the last lecture, we had introduced you
with the ideas of divergence of a vector field.
We had seen that the definition of divergence
is given. Divergence is also written as del
dot V or div V, if you like. We had seen that
this is given by d V x by d x plus d V y by
d
y plus d V z by d z. This is the Cartesian
expression for that. One of the things that
we
talked about is, what is known as the divergence
theorem, which connects the surface
integral of a vector field written as, for
instance, if F is a vector field, then F dot
d S is
the same as the volume integral of divergence
of F over the volume, which is described
by this surface. Now, this is the theorem
which has many many applications in subjects
such as fluid mechanics and as we will see
in electricity magnetism, which we are
discussing as well.
What I wish to do today, is to take this concept
of divergence a little further and make
you more familiar with, how to use the divergence
theorem and what is the physical
meaning of the word divergence. I would do
that and subsequent to this, I will also
.introduce to you what is known as curl of
a vector field. As we can see that the
divergence of a vector field is a scalar field,
because del dot is there. del dot F is a scalar
field.
.
So, let us proceed with that. So, as the name
divergence suggests, the divergence of a
vector field essentially is a measure of the
amount of spread that a vector field has got
at
a particular point. Now, let us, for instance,
if you could see these pictures you will find
that, in this case, let us take the origin
and you find that the fields are spreading
out from
the origin. On the other hand, this is the
type of vector field that you would expect.
For
example, the electrostatic field due to a
positive charge. Of course, it will not be
exactly
this, but essentially it will be spreading
out. This, on the other hand, you notice that
the
fields are converging to the center. So, these
are examples of positive and the negative
divergence.
..
Now, let us look at what it actually means.
So, one of the things that I would like to
point
out is that, the concept of divergence curl
etcetera, they all came because of, they were
first used in the field of fluid dynamics.
So, let me try to illustrate the concept of
a vector
field using fluid dynamics as example. So,
let us look at an elemental volume at the
point
x y z, having a length, breadth and width
dimension of dx, dy and dz. So, that is what
we
are doing and what we are saying is that,
at the point x, y, and z, the density of the
fluid
is given by rho x y z and the fluid velocity
at that point is given by v of x y z.
So, this is, I am simply showing what happens
to the y component of the velocity. For
convenience, I define a vector capital v at
the point x, y, z as the velocity vector at
that
point, small v x y z multiplied by the density
rho at that point. This sort of tells you
that
this is essentially, this quantity entering
an elemental volume dx dy dz and here, from
the
other phase it is leaving. This is just the
y component of that coming in here and this
is
the y component at the point y plus dy. So,
this is the y component is y here, and it
is y
plus dy.
..
Now, let us let us look at this elemental
volume and ask the question, what is the mass
of
the fluid flowing in. Now notice, this is
the phase, the perpendicular to which is along
the
minus y direction or minus j direction. This
is the phase for which the perpendicular is
along the plus y direction, but the y coordinate
of this is at y and the y coordinate of this
phase at y plus d y. So, how much fluid is
flowing into this volume though this phase?
Now, obviously, we are talking about a mass.
So therefore, the rho times V y, that is the
distance moved per unit time along this direction
and of course, you multiply it with dx
dz, which is the perpendicular phase there.
So, this is the amount of fluid that is flowing
in through the phase n is equal to minus j,
namely capital V y, which as I told you is
a
product of rho multiplied by small V y that
is the velocity times dx dz.
Now, so that is the amount of fluid that is
getting in and how much is the amount that
is
getting out. Now, the difference between this
phase and this phases that their areas are
the same, but this, its y coordinate is y,
this as the y coordinate, y plus dy. So, what
we
do is this. We assume that this elemental
volume dx dy dz are small. So that, I need
to
only retain the first order change in quantities
to calculate how much is the mass of the
flowing that is flowing out. The amount of
fluid that is flowing out is given by y
component of the velocity here. What is the
y component? That is equal to V y, that is
the velocity on this phase plus the rate of
change of the velocity with distance, namely
d
V y by dy times dy, because that is the distance
though which it has moved and of
course, multiplied by the area. So, that is
the amount of mass that is flowing out.
.Now, so therefore, if this is the mass that
is flowing in and this is the mass that is
flowing
out, the net amount of mass that is accumulating
inside this mass, namely the net
increase in the mass of the fluid is this
minus this, which is simply minus d V y by
dy
into dx dy dz. You recall, dx dy dz is the
total volume of this element. Now, mind you,
this is just the increase only from the flow
along the y direction.
.
Now, what I am going to do now is to, by symmetry,
I can write down an identical
expression for the flow from the x direction
and the y direction.
.
.So, since that term was minus d V by dy,
so what I get is, the result would be the
net
flow, net increase in mass If you like, there
is a minus sign in front of that, multiplied
by
d V x by d x plus d V y by d y plus d V z
by d z multiplied by the volume, which was
d x
d y d z. This is of course, your volume of
the element, which if you like, I will not
write
it as d V, so that, you do not get confused
with this velocity field V. So, this is, let
us say
d tau.
So therefore, now there is another way in
which I can talk about the rate of increase
of
mass. So, what I can do is this. I know mass
is nothing but the volume times the density.
Now obviously, the volume here is fixed. Now,
since volume is fixed, the rate of change
of the mass is simply given by time rate of
change of the density times dx dy dz. These
two must be identical. This two must be identical.
This is one way of doing it and this is
from the definition. That tells me that I
have del dot V plus d rho by d t is equal
to 0.
Now, in fluid dynamics, this is known as the
equation of continuity. Now, I have drawn
some pictures.
.
So, let us look at what are these. This will
give you an appreciation for the name
divergence as well.
..
But, let me return by to the equation of continuity.
I have got del dot v plus d rho by d t
is equal to 0. That is my equation of continuity.
I am giving an illustration. Look at this
picture. Now, this is the field which is written
on the top. So, this field, the left hand
side
field is given by x s square y times i, the
unit vector along the x direction plus x y
square
times j. So, the field is, let us say, let
us called field as F and this is equal to
x s square y
times the unit vector i plus x y square times
unit vector j.
Now, this is the field that has been plotted
in this graph. You remember that, we had said
that, this we have used mathematic to plot
it and the way we plotted is that the vector,
the
size of the arrows are proportional to the
magnitudes of the vectors and the direction
is
represented by the direction on the arrows.
Now, if I look at the first quadrant of this,
I
am looking at the circular region to set two-dimensional
plot. You notice that the, let us
suppose this represents a fluid field, the
velocity field of a fluid. So, this is, so
here you
notice that, if you call that these small
arrows are the velocity vectors for the fluid,
you
notice that the fluid velocities which are
entering into this circle, they are of smaller
magnitude than those which are going out.
In other words, from the circular region,
more
fluid is going out than it is coming in. Thus,
more outflow. There is more outflow. Now,
obviously, such a thing can happen if the
density is decreasing with time. So, if d
rho by
d t is negative, then the divergence of the
field del dot V will be positive and this
represents a case of positive divergence.
So, this is and the identical statement would
be
.true, if we look at the third quadrant as
well. I have not shown it here, but you can
sort of
check that if you draw a circle here, this
same identical argument would be true.
Look on the other hand. In the second quadrant
here, now in the second quadrant here or
the picture is given on the forth quadrant,
identical story would be true of the second
quadrant. If you look at the forth quadrant
here, you notice that the arrows which are
pointing in into this circle are much bigger
in magnitude than those which are going out.
In other words, this is a case of net inflow.
More fluid is coming in than going out. Now,
such a thing can happen if there is an increase
in the density of the fluid with time.
So, this is the field is diverging and divergence
is positive and this field is divergence is
negative. Let us look at this in a slightly
different field, for which, d rho by d t is
0. It is
an incompressible fluid. Now, if d rho by
d t is 0, then del dot of V will also be 0.
In
other words, the velocity field has 0 divergence.
Look at this picture here. You notice
that, if you take a circle or circular region
here, as much fluid is getting in as is going
out. Now, in such a situation, where the divergence
of the field is 0, for example, is a
field which is x i minus y j which we had
seen earlier. Such a field for which the
divergence of the vector field is 0 is known
as Solenoidal vector field.
I have already introduced you with what we
call as divergence theorem. So, let us recall,
what is the divergence theorem. The divergence
theorem, if you recall, connects a
surface integral with the volume of the surface,
volume of the body bounded by the
surface.
..
So, suppose I have a vector field, which I
represent by F. Then, the surface integral
of F,
F dot dS over whatever surface you are talking
about. Now remember, we had said that,
only some special types of surfaces are permitted
or rather special type of surfaces like
mobia strip are not permitted. This F dot
dS, the direction of S is according to
convention, the outward normal and this is
equal to the volume integral of the divergence
of F.
.
.Now, what I am going to do is to illustrate
the use of such a thing. So, here what I am
doing is that, I am trying to evaluate, this
is a cylinder, the base here x y and the height
of
the cylinder is along the z direction. I want
to integrate, find the surface integral of
the
vector r, r dot n dS. ‘n’, if you recall
is along the outward normal to any element
of
surface. Now first, let me do it the easiest
way. The easier one is this; to use the
divergence theorem.
So, in this case, the vector field is my position,
vector r. Now, I am interested in finding
out, what is r dot dS over the surface of
the cylinder that has been shown in this picture.
So, let us look at this. So, in other words,
this surface integral is same as integral
of
divergence of the vector r over the volume
of the cylinder.
So, this is over volume of the cylinder. Incidentally,
divergence of the position vector
which keeps on coming in various applications
is a good thing to remember. It is trivial
to calculate because, you recall that vector
r is given by i x plus j y plus k z. So therefore,
del dot r, which is d by d x of the x component
of the vector, namely x, plus d by d y of
the y component of the vector, which is y.
Similarly, d by d z of the z component of
the
vector which is z, which is simply 1 plus
1 plus 1 which is equal to 3.
So, divergence of r is 3. So therefore, if
I write it here, this is over the volume of
the
cylinder, 3 which is the number of times d
V and how much? Because, there is nothing
to
integrate, it is integral d V, which is the
volume. We all know the volume of the cylinder
is pi, a is the radius, so a square times
the height h. So, this is the result. So,
this surface
integral, which we calculated in an indirect
fashion, namely calculating through the
divergence, works out to 3 pi a square h.
Now, what I am going to do now is, to show
that this is exactly the result that you would
get, if you calculated the surface integral
directly.
..
So, let us do that. So, let us look at what,
so, let me redraw this picture here.
.
This is x y z and I have a cylinder, whose
base is at the origin. First thing to know
is this.
The cylinder has three surfaces. It is a closed
cylinder because, we have said it is over
the
surface of the cylinder, closed cylinder.
It has a top surface, which is this one and
the
outward normal to the top surface is just
the unit vector k. It has the bottom surface,
which is, because it is a outward normal,
so, it is minus k. We will come back to what
happens to this side surfaces. But let us
first compute, how much is the contribution
to
.the surface integral from the top and the
bottom surface first. Let us recall my field
r is i
x plus j y plus k z.
I am interested, let us say, first calculating
the top surface. So, the top surface is surface
integral of r dot k. Now, r dot k d s; d s
is an element of the surface. Now, so, i dot
k is 0,
and j dot k is 0. I am only left with k dot
k. So therefore, k dot k is 1. I am left with
z and
of course, dS, the element of the surface
and it is only the top surface. But notice
that
height of the cylinder is h and this is z
equal to 0. So, the z value on the top cap
of the
cylinder is fixed and is equals to h. So,
this is nothing but h times dS on the top
surface
and h is constant, so it comes out.
So, I am left with simply the area of the
top surface, which is pi a square. Now, the
calculation of the lower one is equally straight
forward. So, let us look at the bottom
surface. The only difference now is the unit
vector outward normal is along minus k. So,
this will not be z dS, but will be minus z
dS. So, let us write down, integral bottom
surface of z dS. Now, this is actually even
simpler because, the value of z for this
surface, the lower surface is 0. z is equal
to 0 for bottom surface. So therefore, this
integral is 0. So, from the top and the bottom,
the surface integral gives me pi a square
h.
I will now calculate this for the curved surface,
which is the only surface remaining now.
.
Now, what I have to do now is this. I have
to find out what 
is the unit vector on the
curved surface. So, this will be some something
like this. Now, let us see what it is. So,
.notice that this is parallel to the x y plane,
a perpendicular to the curved surface is
parallel to the x y plane and it is on the
surface of the cylinder. So therefore, the
unit
vector n is nothing but i x plus j y, which
is just the radial vector in the x y plane.
But I
have to find a unit vector.
So therefore, I have to divide it by square
root of x square plus y square, where x and
y
are on the curved surface of the cylinder.
But remember that if the radius of the cylinder
is a, the square root of x square plus y square
is nothing but the radius a itself, because
it
has to be on that circle. So, this is i x
plus j y divide by a. x and y are arbitrary,
but
because it is on the curved surface, this
relationship is there.
So now, let us compute F dot m. F is i x plus
j y plus k z dotted with i x plus j y divided
by a. So, let us look at what it gives me.
So, x into x i dot i is 1 x into x, I get
x square. j
dot j is 1. I get y square. k dot i and j
are both 0. So therefore, this is x square
plus y
square by a, which is nothing but a square
divided by a. So, which is equal to a itself.
So, what do I have? I have here, I have to
calculate r dot m dS and r dot n is a times
integral of dS. Now, how much is the area
of the curved surface? The area of the curved
surface of height h is nothing but the circumference
of any of these circles multiplied by
the height h, which is 2 pi a times the height,
which gives me 2 pi a square h. If you
recall, from the top and the bottom surface,
we had pi a square h. From the curved
surface, I had 2 pi a square h. So, the net
result I get 3 pi a square h, which is the
result I
had obtained from the divergence theorem.
So, divergence theorem makes it easy to
compute certain surface integrals. That is
one of the major applications of the divergence
theorem.
..
As I further example, let me show you a rather
nasty looking field. As you can see it, the
field F is i times 2 x plus z to the power
5 j times y square minus sine square k z k
times
x z plus y cube e to the power minus x square.
I want surface integral over a cubical box
of, you know, 1 by 1 by 1 cubical box x from
0 to 1, y from 0 to 1 and z from 0 to 1.
Now, you realize that if I am trying to attempt
to calculate this directly, it is going to
be a
mess because, I have to worry about how to
integrate many of these things. However,
this situation is not as bad as it looks.
.
.The reason is the following. If you look
at what is the divergence of this vector,
remember the divergence is d F x. Only x derivative
of the x component plus d F y by d
y plus d F z by d z. Now, x component? Notice
this z to the power 5, which was
somewhat nasty, it has its derivative with
respect to x is 0. So therefore, my derivative
is
simply 2. y component again, the sine square
x z d by d y is 0. So therefore, I need only
d
by d y of y square, which is 2 y. Finally,
d by d z of F z, again this is a function
of x and
y. So therefore, x z has to be differentiated
with respect to z and I simply get x. So,
it is 2
plus 2 y plus x.
Now, I need to integrate this, but fortunately,
over the volume of the cylinder. So, I need
to calculate, if you like, I will write it
explicitly as triple integral d x d y d z
of 2 plus 2 y
plus x. First thing to notice in this integral
is there is no z dependence. Now, since there
is no z dependence, I can integrate z out
from 0 to 1 and it simply gives me 1. So
therefore, I am left with a double integral
d x d y of 2 plus 2 y plus x. Each one of
them
is rather simple to work out. First, so, this
is all are from 0 to 1.
First this 2, so, two times integral d x d
y. So, it is 2 into 1 into 1, and that is
2 plus 2
times. Now, integral d x over 0 to 1, since
there is no x dependence gives me 1 and I
have got y square by 2, which is 0 to 1 is
1 by 2 plus; this has only x. So, y integral
is
done, which gives me 1 into x s square by
2 which when, so x s square by 2 from 0 to
1
is 1 by 2. So, what I do? What do I get? I
get 2 plus 1 plus a half, which is 7 by 2.
So,
this is the result of this surface integral.
Now, because the function is so nasty, I will
not
be attempting direct evaluation on the surface.
So much about divergence.
For vector fields, so, divergence of a vector
field is a scalar. So, it is a scalar field.
Now,
for a vector field, it is possible to have
an operation, which results in another vector
field
and this is called the curl of a vector. The
curl of a vector came from the word
circulation.
..
Now, we will see as the name suggests, the
meaning of the CURL is associated with how
much a vector field is curling about that
point.
.
So, let us, so, what I have done here is to
draw a picture of a. This is an open surface,
something like inverted pot and this open
surface has a circular boundary. So, this
is a
boundary. Now, let us look at how does one
calculate the surface integral of a vector
field over this surface. Now, let me illustrate
that problem a little bit. So, I have this
surface. This is an open; I have given it
a sense of direction. Now, if I make segments
of
.the surface, so, what I do is this. Just
draw these segments as has been shown there.
Make elements of segments.
Now, let me try to calculate the surface integral
over, for instance, the area bounded by
this. Now, let us look at first, instead of
going to the area, which I am coming to in
a
second, let me concentrate on this element
of area and let me give it a direction. So,
the
direction that I will give is this. That is
an anticlockwise direction and the surface
corresponding to this has an outward normal.
So, I will call this element as, supposing
this is the i th segment, let me called it
n times d S i and this curve, which is the
boundary of this d S i, I will called it d
C i.
Now, the thing that I want you to notice is
this. If I go in each of these segments, if
I take
the line integral in the same sense all the
time, in this case I am taking in the
anticlockwise fashion, then you notice from
an adjacent circuit or curve, my result will
be something like this. This will go like
this, this will go like this and this will
be exactly
in the opposite direction to the previous
one. Let me let me illustrate this by making,
amplifying these elements. Supposing the same
two adjacent elements, I am amplifying.
These are two adjacent elements.
So, this is my d S 1. Let us say this is d
S 2 and I am going on this, the upper one
in an
anticlockwise fashion. The line integral will
be over this, over this, over this and over
that. Now, when I come to number 2 and I still
go in the anticlockwise fashion. Just to
make it clear, let me give these arrows in
a slightly different manner.
Supposing this is the anticlockwise arrow.
You notice that this common line for the top
one is traversed this way, and for the bottom
one is traversed in the exactly opposite
direction. So, if I am to now add up, supposing
I want to find out how much is the
surface integral over this plus that. What
remains are only the contribution from the
outside boundaries. This will happen that,
supposing I am now add up, add another one,
it will be like this. You notice again, this
has cancelled and this has canceled.
So, what will I be left with? So, if I split
up this into such elemental curves on the
surface, then I will be left with only the
outside boundary, which is nothing but this
edge
of this object. This is made clear in the
next picture.
..
So, you notice here that I have shown a little
stretched out thing and everywhere I have
gone with the same sense, the anticlockwise
fashion. So, if you look at this one and that
one, the common areas cancel out and you will
be always left with only the outside
things.
.
Now, so therefore, I can write down that the
net contribution from this, so, let me write
it
down clearly, is, supposing I am talking about
the entire curve F dot d l.
..
Over all these little close curves that I
showed you, then this can be written as sum
over
i, which is summing over all those little
curves and this integral is to be taken over
the i
th curve of F dot d l. Now, what I do is this.
Now, this quantity here is different for
different curves. Let me divide this by the
area of the surface enclosed by the i th curve
and multiply this with the same number. Now,
what is this quantity? So, this quantity,
which I have written as c i F dot d l over
the i th closed curve, divided by delta s
i. Now,
this is the line integral of the boundary
of the i th surface and the area of that i
th,
enclosed by i th curve is delta s i. The direction
associated with this area is the outward
normal, that I will call as n i. This quantity
is defined as 
the curl of the vector F at that
point i. It is a point relationship because,
this relationship is true only in the limit
delta s i
going to 0. So, if I take the limit of this
delta s i going to 0, so, this is a point
relationship
at the point i.
..
Now, this definition, just as when we define
divergence, we obtained a relationship
between the surface integral of a vector field
with the volume integral of the divergence
of that vector field. This definition of the
curve in a very similar way gives me or gives
us a relationship between line integral of
a vector field with the surface integral of
the
curl of the vector. This relationship is known
as the Stoke’s theorems.
Let us look at how does this come. So remember,
the line integral of this curve, now this
is the curve bounding the surface in the picture
of the inverted pot that I have showed
you. It was the rim, the circular rim that
I had showed you. So, this integral is nothing
but sum over the integrals; the line integrals
of little constituents on the surface. So,
sum
over i integral over c i. Now, what do is,
divided by delta c i multiplied by delta c
i and
suppose, I take its limit, that is making
the circuits smaller and smaller. Then, since
this
quantity is defined to be the curl of the
vector, I get integral F dot d m is nothing
but the
surface integral of the curl of the vector.
That is called Stoke’s theorem. It is an
extremely important theorem.
..
So, what is the c? You take a surface. You
take an open surface. It is very important
to
realize. It is an open surface and not a closed
surface, like that pot that I showed you.
So,
line integral on the boundary of the open
surface. So, for example, going back to that
picture, if I am interested in surface integral
over this surface, I am relating it only to
the
line integral on this boundary. So, F dot
d L is equal to, it is the surface that is
bounded
by this curve and the curl of that I have
to take, so curl of F dotted with d s over
the
surface, which is described by this. This
is called Stoke’s theorem. What I will do
next
time would be to obtain an expression for
the curl in the Cartesian coordinate system.
This will be, the essential will be following
the same technique as we followed for
obtaining an expression for the divergence.
To summarize what we have done today is to
look at two things. We started with an
interpretation of the divergence of a vector
field. I repeat, the divergence of a vector
field
is a scalar field and divergence as the name
suggests, is a measure of, it is a point
relationship, it is a measure of how much
the vector field is diverging or of course,
it
could be converging at that point. This will
be extremely important when we look at the
electrostatic phenomenon. Look at for example,
the electrostatic field due to the positive
charges or negative charges. We will be returning
back to the divergence of a vector
field. The divergence of a vector field gives
us a handle for computing surface integral
of
a vector in terms of the volume integral of
the divergence.
.The next thing that we did, which we will
take up in greater detail in the next lecture
is to
define the curl of a vector field. The curl
of vector field is itself another vector field.
As
we will see next time, the curl gives a measure
if you like of how much a vector is
curling around, as the name suggests, at a
given point. What we have done is similar
to
the divergence theorem. We have obtained a
theorem, which relates the line integral of
a
vector field with the surface integral of
the curl of the vector field. In the next
lecture,
when I have a mathematical expression for
the curl of a vector field, we will also give
a
few examples of the application of Stoke’s
theorem.
.
