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Welcome to this lecture number 6 in our course
on fundamentals of transport processes. We
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are going through right now, some
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preliminary material on calculus of vectors.
So, that our future development becomes easier.
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So, to review briefly, what we had done in
the previous lecture, previous few lectures,
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we will consider vectors and tensors as
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objects in themselves, with some one or more
directions associated with them in space.
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For example, the velocity vector typically
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it is written as summation i is equal to 1
to 3 u i times e i. If you recall we using
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a cartesian coordinate system in which the
three
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coordinates are x 1, x 2, x 3 and we have
unit vectors e 1, e 2, e 3. So, this is a
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vector it has three components, but we will
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consider the vector as an object in itself.
Similarly, I had defined for you the stress
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tensor equal to 1 to 3 T i j e i e j. So,
there is a stress tensor it has two directions
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associated with it, direction of the force
acting at the surface. The direction of the
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unit normal to the surface and the dot product
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of two vectors is written as A dot B is equal
to A 1 B 1 plus A 2 B 2 plus A 3 B 3. And
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I had written this for you as summation i
is
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equal to 1 to 3 of A i B i dot product there
are no unit vectors because once, I take the
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dot product of two vectors it becomes a
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scalar.
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And we had made a notational simplification,
we got loss of generality one can always remove
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the summations and the unit
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vectors in all of these. Whenever there is
an index that is not repeated, it implies
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that there is a summation and a unit vector.
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When an index is repeated two times there
is a summation, but there is no unit vector.
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So, there is no direction associated, with
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that index that is repeated two times because
it has already become a dot product.
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Cross products we had written in a manner
similar to dot products A cross B is equal
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to epsilon i j k A j B k where epsilon i j
k
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was the anti symmetric tensor is equal to
1. If i j k is 1 2 3, 3 1 2 or 2 3 1 it is
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minus 1, if it is the other way 1 3 2, 3 2
1 or 2 1 3
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and its equal to 0, if any two indices are
repeating. So, this anti symmetric tensor
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third order, it has three indices that is
one of the
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special tensors.
The other one that we had seen earlier was
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the isotropic tensor delta i j is equal to
1 for i is equal to j i is equal to 0 for
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i is not
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equal to j. So, delta 1 1 is 1 delta 1 2 is
0 and so on. Delta i j is also the dot product
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of the unit vectors e i dot e j. So, this
can also
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be written as e i dot e j. So, if i and j
are the same then this dot product is equal
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to 1, if i and j are different it is equal
to 0. The dot
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product can be written equivalently as A i
B j delta i j. So, that is another way of
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writing dot products and then we had gone
on
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just to briefly review the rules that we will
use here, one free index means there is a
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unit vector and the summation. So, it
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represents one direction.
So, the order of a tensor is the number of
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unrepeated indices that are there, when an
index is repeated it is a dot product. So,
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there is no unit vector associated with that
index and in general, the order of the tensor
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of all terms in an equation have to be the
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same because one cannot equate a scalar to
a vector or a vector to a tensor, a second
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order tensor you do not have the same
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number of components.
So, the order of all terms has to be the same
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and the when there is one cross product between
two real vectors, what one gets is a
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pseudo vector something that changes sign,
when the coordinate system changes from right
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to left handed coordinate system. So,
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if term in equation is a pseudo vector then
all other terms should also be pseudo vectors.
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A couple of other things before we go on
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to the next, subject the first thing is that.
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So, the first thing is that we are also defined
for you the elements of vector calculus, vector
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calculus the gradient of phi of a scalar
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function phi. Let us just assume that this
is a temperature field the gradient of a temperature
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field is defined such that, this thing
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dotted with delta x is equal to T at x plus
x minus T at x. Physically, what this represents
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if I am sitting in a room with some
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temperature variations, I am sitting at some
location x that travel a small distance to
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some other location x plus delta x, the
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difference in temperature between these two
locations T at x plus delta x minus T at x
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is related is equal to distance travelled
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dotted with the gradient of T.
The gradient of T is a quantity, which is
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defined uniquely a vector quantity defined
uniquely at each position in space and I have
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derived the equation for grad T, which is
in cartesian coordinates e 1 partial t by
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partial x 1 plus T 2 partial T by partial
x 2 plus e
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3 partial T by partial x 3. So, this is one
type of derivative acting on this scalar field,
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the inverse of this derivative is the integral
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and that integral relation basically, relates
the integral of grad T dotted with distance
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travelled longer path to the difference in
the
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temperature between the two end parts.
So, the integral equivalent of this is that
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if I have two points here A and B, if I take
a path between these two points then integral
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along that path. The vector distance, the
vector displacement dotted with grad T between
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these two end points is equal to the
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difference in temperature between those two
end points. Corollaries, the difference in
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temperature is only a function of the n
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points. Therefore, along any path I should
get the same integral of d x dot grad T and
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if I go around and come back to the same
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location, this has to be equal to the 0. So,
those that was one element of vector calculus
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and that was the gradient.
The second is the divergence, the divergence
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of a vector A is defined as if I want to know
what is the divergence at some location
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x, at some location x. I want to know what
is the divergence of A I construct a small
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volume delta v with a surface S I construct
a
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small little volume delta v around this point
with a surface S, that surface has the unit
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normal, outward unit normal at various
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points along that surface.
So, what I do is I take integral over that
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surface of a vector dotted with the unit normal
at that location on the surface at each
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location of the surface I take A. And dotted
with a unit normal and divided by the volume
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delta v in the limit as delta v goes to 0.
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That is the definition of the divergence,
we had derived this for a cubic differential
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volume and for that we got divergence of A
is
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equal to partial A 1 by partial x 1 plus partial
A 2 by partial x 2 plus partial A 3 by partial
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x 3.
This can also be written as del dotted with
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A, where the del operator of course, is e
1 d by d x 1 plus e 2 d by d x 2 plus e 3
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d by d
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x 3. So, there is a definition of divergence,
once again it is a derivative, it contains
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partial derivatives of components of a with
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respect to the different coordinates. The
integral equivalent of this is for any macroscopic
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large differential volume, you can show
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that integral over the volume d v of divergence
of A is equal to integral over the surface,
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surrounding that volume of n dot A is
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called the divergence theorem, the greens
theorem.
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And for this rather than taking a differential
volume for which in the limit as this volume
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goes to 0, rather than taking differential
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volume in the limit as the volume goes to
0, we actually take a large volume macroscopic
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volume v with a surface S and with unit
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normal defined at each point on the surface.
And if I take integral d s of n dot a over
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this surface, this is equal to integral over
the
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volume of divergence of A. So, what this means
is that the divergence of a integration over
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the entire volume depends only on the
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values of A on the surface. And we will see
some physical examples of where this might
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be applicable we will see at a little later
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also, but briefly if you.
For example, A over a heat flux, if A over
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a heat flux, then you know that integral ds
n dot q, q is the energy transported per unit
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area per unit time, q dot n is the energy
transported perpendicular to the surface,
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integrating that over the entire surface that
is
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equal to the total energy that comes out of
this surface, and that is equal to integral
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over the volume d v of divergence of q. So,
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only if there is a net amount of energy coming
out of this surface with the divergence of
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q integrated over the volume be non
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zero.
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The third element that we had considered in
the last lecture was curl of a vector. Once,
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again you have our coordinate system x 1,
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x 2, x 3 and we construct once again a small
volume v over the surface S, there is a unit
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normal at each point on the surface. And
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the curl of A is defined as integral over
the surface of now, the unit normal crossed
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with A, the cross product of the unit normal
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length A divided by the volume, and limit
as delta v goes to 0.
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So, that is the third class of derivatives
of vectors, note that all of these are defined
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independent of the coordinate systems, they
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are not derivatives of components, their derivates
are integrals of the vectors themselves. So,
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I got an expression for curl of A in
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the last lecture is equal to epsilon i j k
partial by partial x j of A k and it is also
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equal to I can write it in matrix form e 1,
e 2, e 3
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partial by partial x 1, A 1, A 2, A 3 also
del cross A. So, this is the definition of
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curl the gradient operator cross product of
the
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gradient operator acting on this vector A.
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The equivalent integral theorem for this is,
as I said for the gradient the integral theorem
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relates the line integral to the difference
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in a function between its end points, line
integral to the difference in the function
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between its end parts. The divergence theorem
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relates the volume integral to a surface integral.
So, that is the divergence theorem.
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The integral theorem for the curl relates
the surface integral to a line integral. So,
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if you have some surface like this with some
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perimeter c. So, this is the surface S this
is the perimeter of the surface c and this
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surface of course, has its own unit normal
and
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the integral theorem for the curl states,
that integral d s of n dot curl A is equal
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to integral over the perimeter c over the
closed
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perimeter c of d x dotted with A. So, this
is the integral theorem for curl, it is also
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called the stocks theorem.
This relates the surface integral over an
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open surface to the integral over the perimeter,
the contour that is the perimeter of this
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surface has some physical implications. Integral
over the perimeter is the same for all surfaces
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that are bound by the same
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perimeter. So, you can have different surfaces,
which have exactly the same perimeter this
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integral is the same for all of those
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surfaces. And for a closed surface it has
no perimeter. So, the integral of n dot curl
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a has to be equal to 0, over a closed surface.
Now, you have taught in mathematics courses
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that these the gradient acts on the scalar
that divergence acts on a vector, the curl
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acts on the vector it does not have to be.
The divergence could act on a vector, a tensor
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the gradient could act on a scalar, a
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vector, a tensor and so on. The only distinction
between these is whether, there is a dot product
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or not. So, I could take the
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gradient of a vector if I write this as partial
u i by partial x j.
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Now, let me just leave some space for myself,
if I write this as partial u i by partial
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x j. Now, there is no repeated index i appears
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only once, j appears only once. That means,
there is one unit vector associated with i
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and one summation associated with i, there
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is one unit vector associated with j and one
summation associated with j. So, this is a
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second order tensor, the gradient of the
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velocity is a second order tensor. It has
two directions one is the direction of the
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velocity i, in this case is the direction
of the
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velocity the direction in which the fluid
is moving, j is the direction of the gradient,
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the direction in which you are moving to find
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out what is the variation in velocity.
So, this give me a second order tensor. On
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the other hand if I have del dot u I will
write this as partial u i by partial x i del
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dot u is
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partial u i by partial x i. In this case i
is repeated so, there is no unit vector and
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there is only one summation. So, this is now
a
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scalar it is the divergence
I could also write the divergence of the second
order tensor. For example, the stress tensor
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that I have
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del dot tin this case there is repeated index,
involving this gradient operator and one of
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the two directions of T, it is not clear from
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this notation what the direction should be,
but let us just define this for ambiguity
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without ambiguity. T has two components i
j
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and there is a dot product.
So, that means that the dot product can be
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with respect to either i or with respect to
j. So, let me just write this without ambiguity
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as T partial by partial x j, this is the divergence
of the second order stress tensor, it has
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one unrepeated index that is j that is i and
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therefore, there is one unit vector and a
summation, i is equal to 1, 2, 3. There is
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one repeated index j and there is no summation
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over that I am saying, there is no unit vector
for that, but there is still a summation for
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j.
So, this represents the divergence of the
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stress tensor, one dot product a gradient
and a second order tensor because of one dot
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product, it reduces the order by 2 and I get
back a vector, when I take a divergence you
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take a divergence of velocity vector you
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get a scalar. If you take the divergence of
a second order tensor, you get a first order
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tensor or a vector. You could as well take
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the gradient of the stress this is also possible.
So, this is now a third order tensor because
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this is partial by partial x k of d i j. So,
I am taking the derivative with respect to
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index
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k of T, which has indices i and j three unrepeated
indices. So, you take a gradient of a second
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order tensor, you get a third order
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tensor. So, taking the gradient increases
the order of tensor by one, divergence decreases
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it by one and you can do it for any
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scalar vector tensors. Of course, divergence
and curl can be taken only for vectors and
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tensors gradient can be taken for all
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scalar, vector and tensor.
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A couple of other things the first is that
if I take the curl of the gradient of something,
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this has to be equal to 0 it is because this
is
215
00:22:06,320 --> 00:22:13,320
if I write it on in in long hand notation,
if this is the matrix e 1, e 2, e 3 d by d
216
00:22:15,740 --> 00:22:22,740
x 1, you can evaluate this determinant into
integral
217
00:22:34,380 --> 00:22:40,690
c is equal to 0. So, the curl of the gradient
of anything is equal to 0, I could have told
218
00:22:40,690 --> 00:22:42,980
you this quiet easily because this is equal
to
219
00:22:42,980 --> 00:22:49,980
epsilon i j k partial by partial x j of partial
T by partial x k.
220
00:22:51,020 --> 00:22:58,020
Now, in this one if I interchange two indices
I will get this is equal to minus epsilon
221
00:23:00,679 --> 00:23:07,330
i k j partial by partial x k, partial by partial
x j
222
00:23:07,330 --> 00:23:13,650
of T. Let us interchange two indices and I
get minus epsilon i j k because I interchange
223
00:23:13,650 --> 00:23:16,360
two indices, I get the negative sign. So,
224
00:23:16,360 --> 00:23:23,360
this is equal to minus curl of grad T, this
is also equal to minus curl of grad T, if
225
00:23:25,340 --> 00:23:27,539
a number is equal to the negative of itself
it can
226
00:23:27,539 --> 00:23:34,539
only be 0. So, the curl of the gradient of
anything is equal to 0. The converse is also
227
00:23:35,820 --> 00:23:42,820
true, the converse is also true.
228
00:23:43,549 --> 00:23:50,549
If del cross a is equal to 0 then A can be
written as the gradient of some scalar function,
229
00:24:05,960 --> 00:24:08,159
if the curl of A is equal to 0 then A can
be
230
00:24:08,159 --> 00:24:15,159
written as the gradient of some scalar function.
The other thing, if A is a vector then I also
231
00:24:17,980 --> 00:24:21,860
have the divergence of del cross A is
232
00:24:21,860 --> 00:24:28,860
equal to 0. So, this as you recall is a triple
product the
divergence of the curl of A is also equal
233
00:24:35,580 --> 00:24:39,610
to 0, this you can easily see is
234
00:24:39,610 --> 00:24:46,610
equal to d by d x 1 d by d x 2 d by d x 3,
two rows of this matrix are identical. Therefore,
235
00:25:01,710 --> 00:25:08,710
the determinant has to be equal to 0.
236
00:25:09,440 --> 00:25:16,440
And corollary of this is that if del dot any
vector B is equal to 0, then B can be expressed
237
00:25:29,309 --> 00:25:35,039
as del cross some vector A. If del dot B
238
00:25:35,039 --> 00:25:42,039
is equal to 0 then B can be expressed as some
vector as the curl of some vector A. Finally,
239
00:25:42,120 --> 00:25:46,970
often you will come A cross, A double
240
00:25:46,970 --> 00:25:53,970
cross product A cross B, cross C. If there
is one cross product it becomes a pseudo vector
241
00:25:57,929 --> 00:26:02,450
because the direction changes sign.
When you go from a right to a left handed
242
00:26:02,450 --> 00:26:08,260
coordinate system, if there are two cross
products both of those change sign and so,
243
00:26:08,260 --> 00:26:14,809
you get back a real vector. If I have to write
this in long hand notation I would write B
244
00:26:14,809 --> 00:26:21,809
cross C as epsilon k l m B l C m. So, this
is
245
00:26:22,830 --> 00:26:29,799
B cross C it is a vector, which has direction
k and I am taking a crossed with this vector.
246
00:26:29,799 --> 00:26:32,929
So, I take A cross with this vector I get
247
00:26:32,929 --> 00:26:39,929
epsilon i j k, A j times epsilon k l m, B
l and C m.
248
00:26:41,039 --> 00:26:48,039
Now, this involves the product, this involves
product of two epsilons, this involves product
249
00:26:49,179 --> 00:26:52,049
of two epsilons. And there is a short
250
00:26:52,049 --> 00:26:59,010
hand there is an identity, which relates this
to the delta functions epsilon i j k epsilon
251
00:26:59,010 --> 00:27:04,760
k l m is equal to delta i l delta j m minus
252
00:27:04,760 --> 00:27:11,760
delta i m delta j l. So, this identity you
can derive for yourself verify that if any
253
00:27:15,029 --> 00:27:19,179
two of epsilon i j k and epsilon k l m are
equal
254
00:27:19,179 --> 00:27:23,020
then it should be 0.
You can verify that the right hand is 0, where
255
00:27:23,020 --> 00:27:28,370
deltas are the identity tensors and if all
three are different you can verify you
256
00:27:28,370 --> 00:27:35,350
actually get this result for epsilon i j k
times epsilon k l m. So, these are the kinds
257
00:27:35,350 --> 00:27:37,600
of identities that we will use as we progress
in
258
00:27:37,600 --> 00:27:44,600
the course. So, everything I have done for
you so, far is for a cartesian coordinate
259
00:27:44,940 --> 00:27:48,230
system. How does all this carry over to a
260
00:27:48,230 --> 00:27:53,700
curvilinear co-ordinate system? If you recall
when we did fundamentals of transport processes
261
00:27:53,700 --> 00:27:56,700
one, we actually did shell balances
262
00:27:56,700 --> 00:28:03,700
in those coordinate systems, where the surfaces
of the control volume were chosen to be surfaces
263
00:28:04,110 --> 00:28:07,730
of constant coordinate.
And when we looked at the fluxes going through
264
00:28:07,730 --> 00:28:12,270
those volumes took the difference to find
out, what rate of change of the
265
00:28:12,270 --> 00:28:17,919
temperature or concentration within that volume
was. And once, we got the equations then we
266
00:28:17,919 --> 00:28:21,289
identified the divergences the
267
00:28:21,289 --> 00:28:26,600
gradients etcetera, within those equations.
There is a another way to do it, and that
268
00:28:26,600 --> 00:28:29,929
is to look at the transformation of unit vectors
269
00:28:29,929 --> 00:28:35,340
themselves. So, I will just briefly go through
that as to how you transform unit vectors,
270
00:28:35,340 --> 00:28:37,370
in order to get relations between unit
271
00:28:37,370 --> 00:28:37,990
vectors.
272
00:28:37,990 --> 00:28:44,990
So, let us take the spherical coordinate system
for as an example
x 1, x 2, x 3 and this spherical coordinate
273
00:28:56,380 --> 00:28:57,480
system, the three
274
00:28:57,480 --> 00:29:04,480
coordinates are one is r the distance from
the origin. So, r theta and phi, r is the
275
00:29:09,110 --> 00:29:14,220
distance from the origin, theta is the angle
that
276
00:29:14,220 --> 00:29:20,559
this makes with the x 3 coordinate, theta
is the angle it is called the azimuthal angle.
277
00:29:20,559 --> 00:29:22,730
The angle that the radius vector makes with
278
00:29:22,730 --> 00:29:29,730
the x 3 coordinate.
And phi is the angle in the x-y plane made
279
00:29:30,260 --> 00:29:37,260
by the projection, in the projection of the
radius vector on to the x-y plane. So, those
280
00:29:39,179 --> 00:29:44,380
are the three angles r theta I am sorry three
coordinates r theta and phi note that r is
281
00:29:44,380 --> 00:29:48,899
a distance theta, and phi are angles. So,
they
282
00:29:48,899 --> 00:29:55,899
are dimensional less. Now, at each point you
have, a unit vector in the r direction e r
283
00:29:56,320 --> 00:29:59,080
unit vector in the theta direction, the unit
284
00:29:59,080 --> 00:30:04,289
vector in the theta direction goes in the
direction of increasing theta, theta increases
285
00:30:04,289 --> 00:30:08,640
downwards.
So, the unit vector in the theta direction
286
00:30:08,640 --> 00:30:15,580
will go in the direction of increasing theta,
and the unit vector at the phi direction will
287
00:30:15,580 --> 00:30:16,000
go
288
00:30:16,000 --> 00:30:23,000
in the direction of increasing phi. And obviously,
these unit vectors now depend up on position.
289
00:30:24,440 --> 00:30:27,899
So, if I go to some other location
290
00:30:27,899 --> 00:30:34,899
here the unit vectors are in different directions,
the unit vectors are in different directions.
291
00:30:37,720 --> 00:30:39,809
So, the unit vectors do depend up on
292
00:30:39,809 --> 00:30:46,809
location. Now, if I go a small distance delta
r in the, if I change r coordinate from r
293
00:30:51,340 --> 00:30:58,340
to delta r, the distance travelled
if I go from r
294
00:31:02,659 --> 00:31:09,659
to r plus delta r. The distance travelled
is delta r itself because if I go a small
295
00:31:15,809 --> 00:31:19,380
distance in the r direction along that along
the r
296
00:31:19,380 --> 00:31:25,140
direction. If I go a small distance, the distance
travelled is delta r itself. On the other
297
00:31:25,140 --> 00:31:27,169
hand if I sit at this point and then move
a
298
00:31:27,169 --> 00:31:34,169
small change theta to theta plus delta theta,
the distance travelled is not delta theta
299
00:31:36,850 --> 00:31:41,169
because the distance travelled is actually,
it
300
00:31:41,169 --> 00:31:46,600
has to have dimensions of length.
Since, the distance from the centre the radius
301
00:31:46,600 --> 00:31:53,600
vector is r the distance travelled is actually,
r times delta theta and if I change phi to
302
00:31:58,399 --> 00:32:05,399
phi plus delta phi, the distance travelled
is not delta phi. The projection here, the
303
00:32:06,860 --> 00:32:12,679
projection on to the x y plane this length,
this
304
00:32:12,679 --> 00:32:19,679
length is equal to r sign theta because the
projection on to the z axis is r cos theta,
305
00:32:22,100 --> 00:32:25,029
the projection on to the x y plane is r sine
theta.
306
00:32:25,029 --> 00:32:32,029
So, the distance travelled is going to be
actually r sine theta delta phi. That means,
307
00:32:34,000 --> 00:32:36,450
that when I take derivatives with respect
to
308
00:32:36,450 --> 00:32:43,450
coordinates to define a gradient, I should
also use these scale factors which transform
309
00:32:44,649 --> 00:32:48,000
the change in coordinate to a distance. That
310
00:32:48,000 --> 00:32:53,440
means, I should also use a scale factor, which
transforms the change in coordinate to a distance.
311
00:32:53,440 --> 00:32:57,590
For example, you know that for
312
00:32:57,590 --> 00:33:04,590
the definition of gradient delta T is equal
to delta x dot grad T. In a cartesian coordinate
313
00:33:06,529 --> 00:33:11,700
system I had written grad T as e 1 partial
314
00:33:11,700 --> 00:33:18,700
T by partial x 1 plus e 2, partial T by partial
x 2 plus e 3, partial T by partial x 3, but
315
00:33:23,360 --> 00:33:25,549
in this coordinate system I have to use scale
316
00:33:25,549 --> 00:33:31,190
factors because if I write in terms of r theta,
and phi these are not distances.
317
00:33:31,190 --> 00:33:38,190
So, what I have to define this is as e r partial
T by partial r plus e theta divided by r partial
318
00:33:40,480 --> 00:33:45,100
T by partial theta plus e phi by r sine
319
00:33:45,100 --> 00:33:52,100
theta partial T by partial phi because when,
I move a small distance delta phi, the distance
320
00:33:52,940 --> 00:33:55,330
travelled is actually r sine theta times
321
00:33:55,330 --> 00:34:02,330
delta phi. And that infinitesimal displacement
delta x has to be written as delta r e r plus
322
00:34:04,950 --> 00:34:11,950
r delta theta e theta plus r sine theta delta
323
00:34:15,000 --> 00:34:22,000
phi delta phi times, times e phi. So, these
things are what are called the scale factors,
324
00:34:33,740 --> 00:34:36,869
in the case of r it is just 1, but for theta
I
325
00:34:36,869 --> 00:34:43,820
have r and for phi I have r sine theta. So,
these are what are called as scale factors
326
00:34:43,820 --> 00:34:46,530
in any general coordinate systems, where the
327
00:34:46,530 --> 00:34:52,579
coordinates themselves may not have dimensions
of length, this can in general be written
328
00:34:52,579 --> 00:34:59,579
as S a e a delta x a plus S b e b delta x
b
329
00:35:05,400 --> 00:35:12,400
plus S c e c delta x c. In this case for the
cartesian coordinate systems a was equal to
330
00:35:15,500 --> 00:35:18,320
one because the pre factor in delta r is equal
331
00:35:18,320 --> 00:35:25,320
to ones, b is equal to just r and S c is equal
to r sine theta.
332
00:35:26,130 --> 00:35:30,690
When you have these scale factors which are
non 0, there is also a variation in the unit
333
00:35:30,690 --> 00:35:36,070
vectors with position. And one can derive
334
00:35:36,070 --> 00:35:40,510
the derivation the variation in unit vectors
with respect to position quiet easily, for
335
00:35:40,510 --> 00:35:42,210
the general case I will write down the
336
00:35:42,210 --> 00:35:48,790
derivation and then I will apply to the specific
case, but the reason I am doing this is because
337
00:35:48,790 --> 00:35:50,609
I have defined here the gradient for
338
00:35:50,609 --> 00:35:57,609
you. I have defined here the gradient for
you it is e r d by d t by d r plus e theta
339
00:35:57,710 --> 00:36:01,070
by r d t by d theta plus e phi by r sine theta
d t by
340
00:36:01,070 --> 00:36:01,650
d phi.
341
00:36:01,650 --> 00:36:08,650
So, I want to go to a general coordinate system,
the gradient of T will be equal to 1 by
e a by S a partial T by partial x a plus e
342
00:36:20,829 --> 00:36:22,190
b
343
00:36:22,190 --> 00:36:29,190
by S b partial T by partial x b plus e c by
s c partial T by partial x c. Where S a, S
344
00:36:40,200 --> 00:36:44,260
b and S c are the scale factors in the a,
b and c
345
00:36:44,260 --> 00:36:49,310
directions. This is for a general orthogonal
coordinate system. However, when it comes
346
00:36:49,310 --> 00:36:51,460
to defining divergences it gets a little
347
00:36:51,460 --> 00:36:58,460
more complicated, we know that del dot a is
equal to
partial by partial x a plus e b by S b
348
00:37:18,869 --> 00:37:25,869
dotted with. So, when I am taking this
349
00:37:31,410 --> 00:37:38,410
divergence I have to take derivatives, I have
to take derivates of the components as we
350
00:37:43,210 --> 00:37:45,140
did usually in cartesian coordinate
351
00:37:45,140 --> 00:37:50,770
system, as well as of the unit vectors here,
even you take derivatives of the unit vectors
352
00:37:50,770 --> 00:37:54,380
as well.
353
00:37:54,380 --> 00:38:00,440
Because as I showed you in the spherical coordinate
system, the unit vectors do depend up on position
354
00:38:00,440 --> 00:38:00,970
and you have to take
355
00:38:00,970 --> 00:38:05,190
derivatives with respect to the unit vector
as well.
356
00:38:05,190 --> 00:38:10,290
So, for this reason it becomes important to
be able to define derivatives of unit vectors,
357
00:38:10,290 --> 00:38:11,960
and that is the reason I will spend a little
358
00:38:11,960 --> 00:38:18,180
bit of time on that. So, how do we get the
derivative of a unit vector with respect to
359
00:38:18,180 --> 00:38:19,680
position.
360
00:38:19,680 --> 00:38:26,680
If I take a small displacement delta x, this
can be written as e a, S a, delta x a plus
361
00:38:33,030 --> 00:38:40,030
e b, S b, delta x b plus e c, S c delta x
c. Where
362
00:38:42,410 --> 00:38:48,020
the e’s are the unit vectors, S are the
scale factors and delta x’s are the small
363
00:38:48,020 --> 00:38:52,579
change in coordinate. If I take the derivative
of this
364
00:38:52,579 --> 00:38:59,579
with respect to x a for example, I will get
e a times S a, if I take the derivative with
365
00:39:05,650 --> 00:39:12,650
respect to x b is equal to e b and S b. Now,
the
366
00:39:15,810 --> 00:39:20,119
derivative with respect to x a that I had
taken, I can take one more derivative with
367
00:39:20,119 --> 00:39:26,200
respect to x b. If I take partial by partial
x b of
368
00:39:26,200 --> 00:39:33,200
partial x by partial x a is equal to S a,
partial e a by partial x b plus e a partial
369
00:39:46,480 --> 00:39:53,480
S a by partial x b.
Now, I can take the derivative of this with
370
00:40:07,650 --> 00:40:14,650
respect to a d by d x a of partial x by partial
x b. And write it quiet easily once again
371
00:40:20,230 --> 00:40:27,230
so, this is equal to partial e b by partial
x a, S b plus e b partial S b by partial x
372
00:40:35,550 --> 00:40:42,550
a. Now, if we look at this relation of course,
when I
373
00:40:44,619 --> 00:40:49,270
am taking partial derivatives, the order of
the derivative should not matter. So, obviously
374
00:40:49,270 --> 00:40:51,290
this one in which I am taking with
375
00:40:51,290 --> 00:40:56,079
respect to a and then b should be the same,
as this one where I take first with respect
376
00:40:56,079 --> 00:40:59,630
to b and then with respect to a both of these
377
00:40:59,630 --> 00:41:04,099
are vectors.
In this case the first one is S a partial
378
00:41:04,099 --> 00:41:10,790
e a by partial x b plus e a partial s a by
partial x b. The second is equal to partial
379
00:41:10,790 --> 00:41:11,300
e b by
380
00:41:11,300 --> 00:41:17,890
partial x a S b plus e b partial S b by partial
x a. The coordinate systems that we are considering
381
00:41:17,890 --> 00:41:22,140
are orthogonal. So, e a and e b
382
00:41:22,140 --> 00:41:25,900
are perpendicular to each other, you will
restrict attention to orthogonal coordinate
383
00:41:25,900 --> 00:41:30,050
systems. Since, e a and e b are perpendicular
384
00:41:30,050 --> 00:41:34,920
to each other and these two vectors, these
two vectors have to be the same.
385
00:41:34,920 --> 00:41:41,920
Whereas, a and e b are perpendicular to each
other, you require that the e a the term proportional
386
00:41:44,160 --> 00:41:45,300
to e a here has to be
387
00:41:45,300 --> 00:41:51,579
proportional to has to be equal to this term
because obviously, e is e a is orthogonal
388
00:41:51,579 --> 00:41:54,460
to e b. So, it has no component along e b
that
389
00:41:54,460 --> 00:42:00,750
means, that e a has to be equal to this term.
Similarly, e b times this has to be equal
390
00:42:00,750 --> 00:42:04,480
to this. This gives us the relation for the
391
00:42:04,480 --> 00:42:09,190
derivatives of unit vectors in terms of the
scale factors.
392
00:42:09,190 --> 00:42:16,190
For example, partial e a by partial x b is
equal to e b by s a partial s b by partial
393
00:42:23,040 --> 00:42:28,290
x a. So, if I know what the scale factors
are, then
394
00:42:28,290 --> 00:42:32,910
I automatically, know what the derivatives
of the unit vectors are. So, this one I got
395
00:42:32,910 --> 00:42:39,910
by equating this term with this term because
I
396
00:42:41,210 --> 00:42:46,980
know that this term cannot be equal to the
first term on the right hand side because
397
00:42:46,980 --> 00:42:50,099
e a, and e b are perpendicular to each other.
398
00:42:50,099 --> 00:42:57,099
So, this gives me a relation for the derivatives
of the unit vectors with respect to position.
399
00:42:58,510 --> 00:43:00,780
And how do I use that to advantage, in
400
00:43:00,780 --> 00:43:03,619
order to derive the divergence.
401
00:43:03,619 --> 00:43:10,619
So, this case we the derivative is a unit
vectors for a spherical coordinate system.
402
00:43:14,609 --> 00:43:21,609
I know that S r is equal to 1 S theta is equal
to
is
403
00:43:30,170 --> 00:43:37,170
equal to r and S phi is equal to r sine theta.
I also I know that partial e a by partial
404
00:43:40,339 --> 00:43:46,819
x b is equal to e b by S a partial S b by
partial x
405
00:43:46,819 --> 00:43:53,500
a and from this you can derive the derivatives
of the all unit vectors. For example, partial
406
00:43:53,500 --> 00:43:58,500
e r by partial theta is equal to e theta by
407
00:43:58,500 --> 00:44:05,500
s r partial S theta by partial r.
And partial S theta by partial r is just one
408
00:44:07,510 --> 00:44:14,510
because S theta is equal to r and S r is 1,
this just becomes equal to e theta. Similarly,
409
00:44:17,369 --> 00:44:24,369
partial e r by partial phi is equal to e phi
by S r partial S phi by partial r S r is 1
410
00:44:31,630 --> 00:44:34,670
S phi is r sine theta. Therefore, partial
S phi by
411
00:44:34,670 --> 00:44:41,670
partial r is just sine theta. So, this just
gives me sine theta e phi, what about the
412
00:44:45,150 --> 00:44:47,940
diagonal terms this just gives you the derivatives
413
00:44:47,940 --> 00:44:53,200
of unit vector in one direction, with respect
to some other direction.
414
00:44:53,200 --> 00:44:56,970
You can also derive the unit vectors in one
direction with respect to that same direction
415
00:44:56,970 --> 00:45:01,200
itself, as follows. You know that partial
e
416
00:45:01,200 --> 00:45:08,200
a by partial x a, if I want to evaluate this
one, what I do is to express e a in terms
417
00:45:09,720 --> 00:45:16,720
of e b and e c. So, I express d by d x a of
e b
418
00:45:17,300 --> 00:45:24,300
cross e c. So, this is equal to partial e
b by partial x a cross e c plus e b cross
419
00:45:33,220 --> 00:45:38,099
partial e c by partial x a. One has to be
careful here
420
00:45:38,099 --> 00:45:42,400
because the order of the cross product has
to remain the same because if you interchange,
421
00:45:42,400 --> 00:45:44,210
the order of the cross product it
422
00:45:44,210 --> 00:45:51,210
becomes the negative of itself. So, this first
term here partial e b by partial x a is equal
423
00:45:52,770 --> 00:45:59,770
to e a by S b partial S a by partial x b cross
424
00:45:59,900 --> 00:46:06,900
e c plus e b cross e c by S e b cross e a
by S c partial S a by partial x c just using
425
00:46:31,680 --> 00:46:36,609
the formula that we just derived, and e a
cross e
426
00:46:36,609 --> 00:46:43,609
c is minus e b e a cross e b is minus e c
is minus e b. So, I get minus e b by S b partial
427
00:46:45,930 --> 00:46:50,619
S a by partial x b and e b cross e a is minus
428
00:46:50,619 --> 00:46:57,619
e c. So, I get minus e c by s c partial s
a by partial x c. So, this gives me the derivative
429
00:47:05,079 --> 00:47:07,950
with respect to a of the unit vector in the
a
430
00:47:07,950 --> 00:47:14,490
direction itself. So, in that way you can
derive all of the unit vectors good. Now,
431
00:47:14,490 --> 00:47:18,240
how do we use this in order to calculate the
432
00:47:18,240 --> 00:47:20,800
divergence.
433
00:47:20,800 --> 00:47:27,800
You know that divergence of A is equal to
e a by S a partial by partial x a plus e b
434
00:47:32,660 --> 00:47:39,660
by S b d by d x b plus e c by s c partial
by
435
00:47:40,369 --> 00:47:47,369
partial x c dotted with. Let us just take
the first derivative alone, in order to illustrate
436
00:47:58,760 --> 00:48:01,190
the case, we will just calculate the first
437
00:48:01,190 --> 00:48:08,190
derivative alone e a by S a d by d x a of
we just take the first term alone. And here
438
00:48:24,829 --> 00:48:27,890
I have to take the derivative of both the
439
00:48:27,890 --> 00:48:34,890
component and the unit vector. So, I will
have e a by S a d by d x a
dotted with so, the first term will be the
440
00:48:43,020 --> 00:48:43,270
derivative with
441
00:48:43,240 --> 00:48:50,240
respect to the component. So, this will be
e a by s a dotted with e a partial A a by
442
00:48:52,450 --> 00:48:59,450
partial x a plus A a partial e a by partial
x a plus
443
00:49:05,260 --> 00:49:12,260
e b partial a b by partial x a plus a b
partial e b by partial x a plus e c partial
a c plus a c partial e c by partial x a. So,
444
00:49:42,390 --> 00:49:43,589
I take the
445
00:49:43,589 --> 00:49:50,300
derivative of both the component as well as
the unit vector point to note here partial
446
00:49:50,300 --> 00:49:52,380
e a by partial.
447
00:49:52,380 --> 00:49:57,599
Here I have the derivative of partial e a
with respect to x a is equal to minus e b
448
00:49:57,599 --> 00:50:01,920
by S b partial S a by x b minus e c by S c
partial x
449
00:50:01,920 --> 00:50:07,130
S a by partial x c. That means, this derivative
is perpendicular to the unit vector itself,
450
00:50:07,130 --> 00:50:09,329
it has no components along a it has only
451
00:50:09,329 --> 00:50:10,280
components along b and c.
452
00:50:10,280 --> 00:50:17,280
Similarly, partial e a by partial x b is equal
to is in the direction of e b that means,
453
00:50:17,839 --> 00:50:19,809
the derivative of unit vector with respect
to
454
00:50:19,809 --> 00:50:25,150
some direction is perpendicular to the unit
vector itself, and that I can use to advantage
455
00:50:25,150 --> 00:50:27,520
here.
456
00:50:27,520 --> 00:50:34,520
The first term is just e a dotted with e a
that gives me 1. So, I just get one by S a
457
00:50:34,640 --> 00:50:40,000
partial A a by partial x a. The second is
e a dotted
458
00:50:40,000 --> 00:50:47,000
with partial e a by partial x a partial e
a with respect to x a is perpendicular to
459
00:50:47,030 --> 00:50:50,839
e a itself. So, when I take the dot product,
this term
460
00:50:50,839 --> 00:50:57,609
when I dotted with e a becomes 0. So, I get
only due to the variation of component in
461
00:50:57,609 --> 00:51:01,859
the direction the second and third terms,
I
462
00:51:01,859 --> 00:51:07,160
am sorry the third and fourth terms the third
term is the component derivative with respect
463
00:51:07,160 --> 00:51:12,089
to a times e b, but e b dotted with e a
464
00:51:12,089 --> 00:51:16,050
is equal to 0.
So, this term since e b is perpendicular to
465
00:51:16,050 --> 00:51:22,890
e a this becomes 0, the next term of course,
is not zero because partial e b by partial
466
00:51:22,890 --> 00:51:24,700
x a
467
00:51:24,700 --> 00:51:31,700
becomes e a by S a dotted with a b partial
e b by partial x a is e a by s b partial s
468
00:51:41,250 --> 00:51:46,329
a by partial x b. So, that is the expression
for
469
00:51:46,329 --> 00:51:53,190
partial e b by partial x a. Similarly, in
the fifth term e c dot e a is equal to 0,
470
00:51:53,190 --> 00:51:57,020
the fifth term e c dot e a is equal to 0.
And the sixth
471
00:51:57,020 --> 00:52:04,020
term just gives me e a by S a into A c times
e a by S c partial S a by partial x c. So,
472
00:52:13,750 --> 00:52:15,740
this first term alone if I write it out in
473
00:52:15,740 --> 00:52:21,980
expanded form because one over S a partial
A a by partial x a plus.
474
00:52:21,980 --> 00:52:28,980
This e a dotted with e a is equal to 1. So,
I get a b by s a s b partial s a by partial
475
00:52:30,160 --> 00:52:37,160
x b plus A c by S a S c partial x a. I am
sorry
476
00:52:42,030 --> 00:52:49,030
partial by partial x c that was only for this
first term here. And now, I need to add up
477
00:52:51,569 --> 00:52:54,010
for the second and third term, I will just
478
00:52:54,010 --> 00:53:00,099
write it down, I would not go into the details
it is just an it is just algebra of the form
479
00:53:00,099 --> 00:53:07,099
1 by S b partial A b by partial x b plus A
a by
480
00:53:09,030 --> 00:53:16,030
S a S b partial S b by partial x a plus A
a plus A c. A c by S b S c partial S b by
481
00:53:29,790 --> 00:53:36,680
partial x c plus 1 by s c partial A c by partial
x c
482
00:53:36,680 --> 00:53:43,680
plus partial S c by partial x a plus A b by
S b S c partial S c by partial x b. So, that
483
00:53:57,069 --> 00:53:58,819
is the final expression that you get after
484
00:53:58,819 --> 00:54:05,819
incorporating all the derivatives. I can write
this, I can just collect terms here I can
485
00:54:07,020 --> 00:54:11,200
just collects terms in this equation.
486
00:54:11,200 --> 00:54:18,200
You can write it compactly as divergence of
A is equal to 1 by S a S b S c into partial
487
00:54:27,359 --> 00:54:34,359
by partial x a of S b S c A a plus partial
by
488
00:54:35,059 --> 00:54:42,059
partial x b of S c S a A b plus partial by
partial x c of S a S b A c. This is the divergence
489
00:54:48,940 --> 00:54:53,640
in terms of the scale factors, what is it
for
490
00:54:53,640 --> 00:55:00,640
a cartesian coordinate system, where I have
S r is equal to 1 S theta is equal to r and
491
00:55:01,630 --> 00:55:08,040
S phi is equal to r sine theta. This is the
492
00:55:08,040 --> 00:55:15,040
divergence of A is equal to 1 over S a S b
S c that is r square sine theta of A is r.
493
00:55:18,079 --> 00:55:25,079
So, d by d r of S b times S c r square sine
theta. I
494
00:55:26,270 --> 00:55:33,270
am sorry S c S a r plus d by d theta because
this is S theta times S phi that is r into
495
00:55:42,390 --> 00:55:49,390
r sine theta.
Second one is S phi into S r, this becomes
496
00:55:50,190 --> 00:55:57,190
r sine theta A theta plus d by d phi of r
times A phi because d by d phi of S r into
497
00:56:05,210 --> 00:56:05,579
S phi.
498
00:56:05,579 --> 00:56:12,579
So, this becomes r times a phi. So, this in
the first one I can cancel out r and I will
499
00:56:14,230 --> 00:56:18,109
get in the first one. Since, I am taking the
500
00:56:18,109 --> 00:56:21,630
derivative with respect to theta here, I am
sorry derivative with respect to r of sine
501
00:56:21,630 --> 00:56:23,829
theta I can cancel out sine theta, and I will
get
502
00:56:23,829 --> 00:56:30,829
1 by r square d by d r of r square A r. The
second one I can cancel out one r. So, I will
503
00:56:32,190 --> 00:56:38,319
get 1 by r sine theta d by d theta of sine
504
00:56:38,319 --> 00:56:45,319
theta A theta plus 1 by r sine theta partial
A phi by partial phi.
505
00:56:49,500 --> 00:56:55,500
Go back to the lectures, were we calculated
shell balances in a spherical coordinate system,
506
00:56:55,500 --> 00:56:57,109
and you will find that the divergence
507
00:56:57,109 --> 00:57:04,109
of q that I calculated del dot q the expression
was exactly the same in the spherical coordinate
508
00:57:05,369 --> 00:57:07,140
system, we got the same thing by
509
00:57:07,140 --> 00:57:14,140
taking into account the variation of unit
vector with respect to position. Similarly,
510
00:57:14,480 --> 00:57:16,740
one can get the simplified expression for
the
511
00:57:16,740 --> 00:57:20,619
curl, I would not go through the details here,
it is just algebra of the same kind that we
512
00:57:20,619 --> 00:57:26,869
done before. The curl of A can be
513
00:57:26,869 --> 00:57:33,869
calculated as 1 by S a S b S c times the determinant
of S a e a S b e b S c e c d by d x a d by
514
00:57:47,440 --> 00:57:54,440
d x b d by d x c times S a A a S b A b
515
00:57:57,950 --> 00:58:04,950
S c A c. Good exercise to undertake, calculate
the value of curl in this coordinate system.
516
00:58:08,030 --> 00:58:10,710
I mean the spherical coordinate system
517
00:58:10,710 --> 00:58:17,180
use this formula here to calculate the value
of the curl.
518
00:58:17,180 --> 00:58:22,640
And the other thing that we saw in the course
on fundamentals of transport processes was
519
00:58:22,640 --> 00:58:27,460
the Laplace del dot grad of T just l
520
00:58:27,460 --> 00:58:34,460
square T. So, this I can write simply as 1
by S a S b S c into partial by partial x a
521
00:58:38,210 --> 00:58:41,150
of S b S c.
522
00:58:41,150 --> 00:58:48,150
Into A a here the component A a is equal to
the gradient of T in that direction, which
523
00:58:49,609 --> 00:58:55,230
is 1 over S a partial T by partial x a plus
d by
524
00:58:55,230 --> 00:59:02,230
d x b of S a S c 1 by S b partial d by partial
x b plus d by d x c of S a S b by S c partial
525
00:59:11,730 --> 00:59:15,730
d by partial x c.
526
00:59:15,730 --> 00:59:20,680
You can calculate this for a spherical coordinate
system, this becomes 1 by r square d by d
527
00:59:20,680 --> 00:59:24,530
r of r square partial d by partial r plus
528
00:59:24,530 --> 00:59:31,530
1 by r square sine theta, exact same expression
that we got when we did shell balances. So,
529
00:59:47,329 --> 00:59:48,780
go through this the way we have done
530
00:59:48,780 --> 00:59:53,410
it in this class compare it with what we have
done in the previous fundamentals of transport
531
00:59:53,410 --> 00:59:56,040
processes 1, where we did the same
532
00:59:56,040 --> 01:00:00,059
thing on the basis of shell balances.
We managed to get conservation equation sine
533
01:00:00,059 --> 01:00:04,710
spherical coordinate system, you will find
that the divergence operator, the
534
01:00:04,710 --> 01:00:09,319
Laplace an operator in both of these cases
are identical. If I able to compare those
535
01:00:09,319 --> 01:00:11,030
you get a better understanding of how we
536
01:00:11,030 --> 01:00:17,000
derived these in this of course, we made no
reference to the basic to the underlying reference
537
01:00:17,000 --> 01:00:19,530
coordinate system. And in this case
538
01:00:19,530 --> 01:00:23,859
,we have managed to derive these operators
independent of coordinate systems depending
539
01:00:23,859 --> 01:00:25,780
only up on the scale factor, in the
540
01:00:25,780 --> 01:00:30,460
variation of unit vectors with respect to
position. We will continue this in the next
541
01:00:30,460 --> 01:00:32,460
class, while we will try and apply some of
these
542
01:00:32,460 --> 01:00:38,210
concepts to the velocity field in fluids,
we will start fluid mechanics in the next
543
01:00:38,210 --> 01:00:38,530
lecture.