Chapter 2 Coordinate System2

8/9/2019 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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Chapter 2 Coordinate System2

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