chapter 2 coordinate system2
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Gradient of a scalar Field
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1 1
2
2
Suppose that ( , , ) is the temperature at point P
( , , ) in some region of space, and T ( , , )
is the temperature at a nearby point P as shown in Figure 2.19.
the differential di
T x y z
x y z x dx y dy z dz+ + +
stances dx, dy, and dz are the components of the
differential distance vector l. That is:
l=x y z (2.69)
d
d dx dy dz+ +
2 1
From differential calculus, the differential temperature
- is given by
T T T= (2.70)
By definition =x. l and =z. l, eq 2.70 can be rewritten as
=x
dT T T
dT dx dy dzx y z
dx d dz d
dT
=
+ +
T T T . l y . l z . l
T T T = x y z . l (2.71)
d d dx y z
dx y z
+ +
+ +
This is called as gradientof
T or grad T and usuallywritten symbolically as T
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T T T grad T x y z (2.72)
= . l (2.73)
The symbol is called the del or gradient operator and
is defined as
x y z (c
Tx y z
dT T d
T
x y z
= + +
+ +
artesian) (2.74)
With l=a where al is the unit vector of l,
the directional derivative of T along the direction a
is given by:
dT.a . (2.75)
dl
l
l
l
d dl d
T=
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If T is known function of the coordinate variablesof a given coordinate system, we can find the difference
(T2-T1) where T1 and T2 are the values of T at
points P1 and P2, respectively, by intergrating
2
1
P
2 1
both sides of Eq 2.73
T -T = . l (2.76)P
T d
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Example: Directional Derivative
Find the directional derivative ofalong the directionand evaluate it at (1,-1,3) x2 y3 z2+
2 2T x y z= +
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Gradient Operator in Cylindrical and
Spherical CoordinatesTo convert equation 2.72 (Cartesian coordinate) into cylindrical,we start with coordinate relationship
2 2 tan
From differential calculus,
T T T T z
(2.78)
yr x y
x
r
x r x x z x
= + =
= + +
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2 2
Since z is orthogonal to x, the last term is equal to zero
because z/ x=0. Using the coordinate relations given byeq 2.77, it easy to show that
cos (2.79a)
1sin
r x
x x y
x r
= =
+
=
(2.79b)
Hence,
sincos (2.80)
T T T
x r r
=
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In addition, we use the relation x=rcos - sin
and y=rsin + cos
From eqs (2.57a) and (2.57b), eq (2.72) becomes
T 1 T T T=r (2.81)r
Therefore the gradient operator in
coordin
cylindrical
zr
+ +
ates can be defined as
1 =r (2.82)r
The gradient operator incoordinates can be defined as
1 1 =R (2.83)R
( )
spher
sin
ical
( )
z cyr
R
lindrical
spheric lR a
+ +
+ +
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Properties of Gradient Operator
For two scalar functions U and V, thefollowing relations apply;
1
1. ( ) (2.84 )
2. ( ) (2.84 )3. (2.84 )n
U V U V a
UV U V V U bVn nV V for any n c
+ = +
= +
=
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Divergence of Vector field
Flux lines of the electric field E due to a positive charge q.
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The field line of flux is called as flux
linesFlux density is defined as the amount ofoutward flux crossing a unit surface ds:
. E.ndsFlux density of =
| s| |ds
where n is the outward surface normal of s
The total of flux crossing the surface S,
such as the enclosed surface of theimiginary sphere outlined in Fig 2.20
Total
E dsE
d
d
=
S
flux= E. s (2.86)d
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Flux lines of a vector field E passing through a differential rectangular parallelepiped of volume v= x y z.
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Consider Figure 2.21
A vector E (x, y, z) exist in the region ofspace containing the parallelepiped.
The area of the face marked 1 in Figure 2.21is and its unit vector Hence,the outward flux F1 through face 1
Let E defined as E=x +y +z (2.87)
x y zE E E
y z 1 n x.=
1
1
1
(x +y +z ).(-x)
(1) (2.88)
Face
x y z
Face
x
E E E dydz
E y z
=
=
1 E.nF ds=
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where (1) is the value of at the center of face1.
Approximating over face 1 by its value at the center is
justified by the assumption that the differential volume
under consideration is very sam
x x
x
E E
E
1
2
ll.
Similarly, the flux out of face 2 (with n x) is
(2) (2.89)
Where (2) is the value of at the center of face 2. Over
a differential separation , (2) is related to (1) by
(2)
x
x x
x x
x x
F E y z
E E
x E E
E E
=
=
= (1) (2.90)xE
xx
+
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2
where we have ignored higher-order terms involving
and higher powers because their contributions
are neligibly small when is very small. Substituting
Eq (2.90) into Eq(2.89) gives
(1) xx
x
x
EF E xx
= +
(2.91)y z
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The sum of the fluxes out faces 1 and 2 isobtained by adding eqs (2.88) and (2.91)
Repeating the same procedure to each of the
other pairs of faces leads to
1 2 (2.92 )xEF F x y z a
x
+ =
3 4 (2.92 )yE
F F x y z b
y
+ =
5 6 (2.92 )zEF F x y z c
z
+ =
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The sum of fluxes F1 through F6 gives thetotal flux through surface S of theparallelepiped:
E. s
(divE) (2.93)
where is and div E is a differential function
called the divergence of E is defined in Cartesian Coordinates as
divE=
yx z
s
yx z
EE Ed x y z
x y z
v
v x y z
EE E
x y
= + +
=
+ +
z
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Divergence TheoremThe result given by Eq (2.93) for a differentialvolume can be extended to relate the volume
integral of over any volume v to the flux of Ethrough the closed surface A that bounds v. That is
The relationship, known as the divergence theorem,
is used extensively in EM
v.E
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Calculating Divergence
2 2
3 2 3 2
Determine the divergence of each of the following vector fields
and then evaluate it at the indicated point:
(a) E=x3x +y2z+zx z at (3,-2,0)
(b) E=R( cos / )- ( sin / ) at ( / 2,0, )a R a R a
Solution
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(b)
( / 2,0, )
at R= / 2, 0,
. | 16a
a and
E
=
=
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Curl of Vector fieldThe curl of a vector field B describes the
rotational property, or the circulation of B
For closed contour C, the circulation of B is
defined as the line integral of B around c.That is
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For uniform field B=xB0, who field line
is depicted in Figure 2.22(a). For rectangular contour
abcd shown in figure, we have
Circulation is zero for the uniform field in (a), but it is not zerofor the azimuthal field in (b).
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The magnetic field B induced by an infinite wirecarrying a d-c current I. If the current is free space
and it is oriented along z-direction, then
0 permeability of free space
radial distance from x-y planeFor a circular contour of radius r, the differential
length vector l= and circulation of B around C is
r
d rd
=
=
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The curl of vector A is:
For a vector B given in Cartesian coordinates as
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Vector Identities involving curlFor any two vectors A and B,
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Stokess Theorem Stokess Theorem convert the surface integral of the
curl of a vector over an open surface S into a line
integral of the vector along the contour C boundingthe surface S
The geometry is shown in figure 2.23.
The direction of the unit vector n-hat is along the thumbwhen the other four fingers of the right hand follow dl.
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Mathematically, Stokess Theorem is given
by:
If curl b is zero, the field B is said to be
conservative or irrotational because itscirculation, represented by the right-handside of eq 2.107 is zero
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Example: Verification of Stokess Theorem
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Solution
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Right hand side: The surface S is bounded bycontour C=abcd in Fig 2.24.
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abOver segment ab, the dot product of B =z(cos )/2 and
l= is zero, and the same is true for segment cd.Over segment bc, / 2
d r d
=
bc
da
Hence B =z(cos / 2)/2=0For the last segment, B =z(cos / 3)/2= / 4
and l=z . Hence
z
d dz
0
3
1 1 3 B. l z .z
4 4 4
a
C b
d dz dz
= = =
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Laplacian OperatorFor a scalar function V defined in Cartesiancoordinates, its gradient is
Where we defined a vector A with components
The divergence of
/ , / / .x y zV x A V y and A V z= = =
V
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Laplacian of V:
The Laplacian of scalar can be used to definethe Laplacian of a vector
For a vector E in Cartesian coordinates by:
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The Laplacian of E is defined as
Thus, in Cartesian coordinates the Laplacian of a vector is avector whose components are equal to the Laplacians of avector components. Through direct substitution, it can be shown
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