gradient diver curl
DESCRIPTION
gradient divergence and curlTRANSCRIPT
Gradient, Divergence & Curl
Chapter 1
2
Tc
Td
Scalar and vector fields
va
vb
Fluid
Imagine a cooling system of a reactor which is using fluid as the cooler medium
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FIELD is a description of how a physical quantity varies from one point to another in the region of the field (and with time).
(a) Scalar fields
Ex: Depth of a lake, d(x, y)
Temperature in a room, T(x, y, z)
Depicted graphically by constant magnitude contours or surfaces.
y
x
d1
d2d3
4
At any point P, we can measure the temperature T. The temperature will depend upon whereabouts in the
reactor we take the measurement. Of course, the temperature will be higher close to the radiator than the opening valve.
Clearly the temperature T is a function of the position of the point. If we label the point by its Cartesian coordinates , then T will be a function of x, y and z, i.e.
. This is an example of a scalar field since temperature is
a scalar. ),,( zyx
),,( zyxTT
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1. A field is a quantity which can be specified everywhere in space as a function of position.
2. The quantity that is specified may be a scalar or a vector.
3. For instance, we can specify the temperature at every point in a room.
4. The room may, therefore, be said to be a region of “temperature field” which is a scalar field because the temperature T (x, y, z) is a scalar function of the position.
5. An example of a scalar field in electromagnetism is the electric potential.
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Meanwhile, at each point, the fluid will be moving with a certain speed in a certain direction
That is, each small fluid element has a particular velocity and direction, depending upon whereabouts in the fluid it is.
This is an example of a vector field since velocity is a vector. The velocity can be expressed as a vector function, i.e.
where will each be scalar functions.
kjivv ),,(),,(),,(),,( 321 zyxvzyxvzyxvzyx
321 , vvv and
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1-7Example: Linear velocity vector field of points on a rotating disk
8
Physical examples of scalar fields:
Electric potential around a charge
Temperature near a heated wall
+
(The darker region representing higher values )
9The flow field around an airplane
Electric field surrounding a positive and a negative charge.
Hurricane
+ −
Magnetic field lines shown by iron filings
Physical examples of vector fields:
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Gradient of Scalar Fields
The gradient of a scalar field is a vector field,
which points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change.
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grad
x, y, z ctt
dr��������������
Physical meaning: is the local variation of Φ along dr. Particularly, grad Φ is perpendicular to the line Φ = ctt.
grad dr��������������
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Gradient operator
Suppose , we have a function of three variables- say, the temperature T(x, y, z) in a room.
For the temperature distribution we see how a scalar would vary as we moved off in an arbitrary direction.
Now a derivative is supposed to tell us how fast the function varies, if we move a little distance.
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kz
Tj
y
Ti
x
TTgrad
z
ky
jx
i TTgrad
If T(r) is a scalar field, its gradient is defined in Cartesian coordinates by
It is usual to define the vector operator
,
which is called “del”. We can write
.
Without thinking too hard, notice that grad T tends to point in the direction of greatest change of the scalar field T.
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dzz
Tdy
y
Tdx
x
TdT
ldT
kdzjdyidxkz
Tj
y
Ti
x
TdT
The significance of the gradient:
A theorem on partial derivatives states that
This rule tells us how T changes when we alter all three variables by the infinitesimal amounts dx, dy, dz. Change in T can be written as,
The conclusion is that, the RHS of above equation is the small change in temperature T when we move by dl.
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.
ldofdirectiontheinvectorunitaisdl
ldbut
dl
ldT
dl
dT
getWe
^^
aTaofdirectionindl
dT
If we divide the above eq. by dl
So , we can conclude that, grad T has the property that the rate of change of T w.r.t. distance in any direction â is the projection of grad T onto that direction â.That is
In general,
• a directional derivative had a different value for each direction,
• has no meaning untill you specify the direction
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dl
dT is called a directional derivative.
the quantity
Gradient Perpendicular to T constant surfaces
If we move a tiny amount within the surface, that is in any tangential direction, there is no change in T , so
Surface of constant T,
These are called level surfaces. Surfaces of constant T
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.0dl
dT
0
dl
ldTsurfacethein
dl
dl
Conclusion is that; grad T is normal to a surface of constant T.
So for any
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Geometrical Interpretation of the Gradient
Like any vector, a gradient has magnitude and direction. To determine its geometrical meaning, lets rewrite the dot product In its abstract form:
cosTdldlTdT
(1.5)
where θ is the angle between and dl. Now, if we fix the magnitude dl and search around in various directions (that is, vary θ), the maximum change in T evidently occurs when θ =0 (for then cos θ = 1). That is for a fixed distance dl, dT is greatest when I move in the same direction as .
T
T
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In the above two images, the scalar field is in black and white, black representing higher values, and its corresponding gradient is represented by blue arrows.
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Example 1
If (x,y,z) = 3x2y– y2z2, find grad and at the point (1,2,−1).
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Solution
kji
kji
)2()23(6
grad
222 zyyzxxy
zyx
At the point (1,2,−1),
kji
kji
812
)1()2(2])1)(2(2)1(3[)2)(1(6 222
2098)1(12812 222
)1,2,1(
kji
Example 2
If
2
^
2
^^^
222
1)(
2)()(
.
?
r
r
rb
rra
thatShowQ
kzjyixrhere
routFind
zyxr
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Example 3
Find , if
Given
ji )(sin cos yexxy
),( yx
.0)0,0(
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SolutionSince , we haveji )(sin cos yexxy
.....(1) cos xyx
.....(2) sin yex
y
Integrating (1) and (2) w.r.t. x and y respectively, we obtain
.....(3) )(sincos yfxyxdxy
.....(4) )(sin)(sin xgexydyex yy
25♣
Comparing (3) and (4), we can conclude that
Ceyf y )(
Hence, Cexyyx y sin),(
To find constant C, use .0)0,0(
1
01
00sin0)0,0( 0
C
C
Ce
Therefore, 1sin),( yexyyx
and Cxg )(
where C is an arbitrary constant of integration
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Example 4
Find if
and .
),,( zyx
kji )34()23()2( 2233232 yzxzzxxyxyzy 2)0,0,0(
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Integrating (1), (2) and (3) w.r.t. x, y and z respectively, we obtain
SolutionWe have .....(1) 2 32 xyzy
x
.....(2) 23 32zxxyy
.....(4) ),()2( 32232 zyfyzxxydxxyzy
.....(3) 34 223 yzxzz
.....(5) ),(3)23( 32232 zxgyzxxyydyzxxy
.....(6) ),()34( 324223 yxhyzxzdzyzxz
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Comparing (4) with (5) and (6) we get
Czyzyf 43),(
Czyyzxxy 4322 3
To find constant C, use 2)0,0,0(
23 4322 zyyzxxy
Therefore
Problem 5
29
0.
),,(
.),,(
^
rd
kdzjdyidxz
ky
jx
i
odzz
dyy
dxx
d
kzyx
constiskwherekzyxsurfacetheto
larperpendicuvectoraisthatShow
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Application of gradient: Surface normal vector
A normal, n to a flat surface is a vector which is perpendicular to that surface.
A normal, n to a non-flat surface at a point P on the surface is a vector perpendicular to the tangent plane to that surface at P.
n
),( yxfz
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Therefore, for a non-flat surface, the normal vector is different, depending at the point P where the normal vector is located.
Unit vector normal is defined as
n
),( yxfz
n
nn ˆ
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To find the unit vector normal to the surface
we follow the following steps:
(i) Rewrite the function as
(ii) Find the normal vector that is
(iii) Then, the unit normal vector is
(iv) Hence, the unit normal vector at a point
is
n
nn̂
),( yxfz ),( yxfz
constantany ,),,( kkzyx n
),,(
),,(ˆ
000
000
zyx
zyx
n),,( 000 zyxP
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Example 5
Find the unit normal vector of the surface at the indicated point.
(a) at
(b) at
226 yxz 23 zyxe y
)2,3,1(
)1,0,1(
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Solution
(a) Rewrite as2214 yxz 14222 zyx
Thus, we obtain 222),,( zyxzyx
Then, )(2222 kjikji zyxzyx
At the point (−1,3,2),
The unit normal vector is14
23ˆ
kjin
14223)1(2 222 and)23(2 kji
35
(b) Rewrite as
Thus, we obtain
Then,
At the point (1,0,−1),
The unit normal vector is6
2ˆ
kjin
6211 222 and
23),,( zyxezyx y
kji zyxee yy 2)3( 2
23 zyxe y 023 zyxe y
kjikji 2)1(2])0(31[ 200 ee
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Divergence of Vector Fields
The divergence is an operator that measures the magnitude of a vector field's source or sink at a given point
The divergence of a vector field is a scalar
(x, y, z)V��������������
(x dx, y, z)V��������������
x x+dx
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The divergence of a vector field
is defined as
z
F
y
F
x
F
FFFzyx
321
321 )(
div
kjikji
FF
kjiF ),,(),,(),,(),,( 321 zyxFzyxFzyxFzyx
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Divergence
y ux zu
vv vdiv = v
x y z
v v����������������������������
2 – Physical meaning
(x, y, z)v��������������
is a differentiable vector field
div v��������������
is associated to local conservation laws: for example, we will show how that if the mass of fluid (or of charge) outcoming from a domain is equal to the mass entering, then is the fluid velocity (or the current) vectorfield
div 0v��������������
v��������������
(x, y, z)V��������������
(x dx, y, z)V��������������
x x+dx
Geometrical Interpretation.
The name divergence is well chosen, for . F is a measure of how much the vector F spreads out (diverges) from the point in question.
The vector function has a large (positive) divergence at the point P; it is spreading out. (If the arrows pointed in, it would be a large negative divergence.)
P
39
NOTE: P=electric field due to charge (+ ve or – ve)
On the other hand, the function has zero divergence at P; it is not spreading out at all.
40
P
So, for example, if the divergence is positive at a point, it means that, overall, that the tendency is for fluid to move away from that point (expansion); if the divergence is negative, then the fluid is tending to move towards that point (compression).
Fundamental theorem of divergence
The fundamental theorem for divergences states that:
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surfacevolume
daFdF
This theorem has at least three special names: Gauss’s theorem, Green’s theorem, or, simply, the divergence theorem.
is function at the boundary element of volume (in Cartesian coordinates, = dx, dy, dz), and The volume integration is really a triple integral.
d
d
d
da represents an infinitesimal element of area; it is a vector , whose magnitude is the area of the element and whose direction is perpendicular ( normal ) to the surfaces, pointing outward.
42
On the front face of the cube, a surface element is idzdyda ˆ
1
43
jdxdzda ˆ2
on the right face, it would be
whereas for the bottom it is
kdydxda ˆ3
Problem 1
Q. Calculate the divergence of the following vector functions?
zxkyzjxyivb
xzkjxzixva
32)(
23)(
2
221
Problem 2
Check the divergence theorem using the function
45
kyzjzxyiyv )2()2( 22
And the unit cube is situated at the origin.
Z
Y
X
46
Curl of Vector Fields
Curl is a vector operator that shows a vector field's rate of rotation, i.e. the direction of the axis of rotation and the magnitude of the rotation.
0curl v��������������
0 v
47
48
The curl of a vector field
is defined as
kjiF ),,(),,(),,(),,( 321 zyxFzyxFzyxFzyx
y
F
x
Fk
z
F
x
Fj
z
F
y
Fi
FFFzyx
121323
321
curl
kji
FF
Problem 1
Find the curl of
0
3
xyzyx
kji
v
xjyiv 3
The curl of v3 points in the z-direction
Z
y
X
Curl To find a possible interpretation of the curl, let us consider a body rotating with uniform angular speed about and axis l. Let us define the vector angular velocity to be a vector of length extending along l in the direction Take the point O as the origin of coordinates we can write R = xi + yj + zk
the radius at which P rotates is |R||sin| Hence, the linear speed if P isv = |R||sin| = |R||sin|
If we take the curl of V, we therefore have
that is
Expanding this, remembering that Ω is a vector, we find
Conclusion: The angular velocity of a uniform rotating body is thus equal to one-half the curl of the linear velocity of any point of the body.
^^
cossin jyzexiyzexu xx
wvuzyx
kji
u
^^^
^^
cossin jyzexiyzexu xx
^^^
^^^
^^^
coscos1cossin
sincossincos
0cossin
kyzzeyzejyzyeiyzye
kyzexy
yzexx
jyzexz
iyzexz
yzexyzexzyx
kji
u
xxxx
xxxx
xx
Example: For velocity field,
, find the angular velocity ω.
For the field,
, we obtain:
Fundamental theorem of curl
The fundamental theorem for curls, which goes by the special name of Stokes’ theorem, states that
lineboundarysurface
dlvdav .
The integral of a curl over a region (a patch of surface) is equal to the value of the function at the boundary (the perimeter of the patch).
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55
Example 6
Find both div F and curl F at the point (2,0,3) if
kjiF )2(cos2),,( 2 yxyxzzezyx xy
56
Solution
yxzyze
yxz
yxzy
zex
z
F
y
F
x
F
xy
xy
sin22
)2()cos2()(
div
2
2
321
FF
At the point (2,0,3),
00sin)3)(2(2)3)(0(2 )0)(2(2 eF
[Notice that div F is a scalar!]
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kji
k
j
i
kji
FF
)2cos2()1()cos22(
)()cos2(
)()2(
)cos2()2(
2cos2
curl
22
2
2
2
xyxy
xy
xy
xy
xzeyzeyx
zey
yxzx
zez
yxx
yxzz
yxy
yxyxzzezyx
58
At the point (2,0,3),
ki
k
jiF
62
])3)(2(20cos)3(2[
]1[]0cos)2(22[)0)(2(2
)3)(0(2
e
e
[Notice that curl F is also a vector]
59
Properties of Del
If F(x,y,z) and G(x,y,z) are differentiable vector functions (x,y,z) and (x,y,z) are differentiable scalar functions, then
(i)
(ii)
(iii)
)(
)(
2
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(iv)
(v)
(vi)
(vii)
(viii)
GFGF )(
00 grad curlor )(
GFGF )(
2
2
2
2
2
2
2
2
2
2
2
22
)(
zyx
zyx
0 curl divor 0)( FF
*Notes: In (vi), is called the Laplacian operator2