Chapter 02 TRƯỜNG ĐIỆN TỪ

Chapter 2 COORDINATE SYSTEMS AND TRANSFORMATION Education makes a people easy to lead, but difficult to drive; easy to govern but impossible to enslave —HENRY P BROUGHAM 2 1 INTRODUCTION In general, t[.]

Chapter 2

COORDINATE SYSTEMS AND TRANSFORMATION

Education makes a people easy to lead, but difficult to drive; easy to govern but impossible to enslave.

—HENRY P BROUGHAM

2.1 INTRODUCTION

In general, the physical quantities we shall be dealing with in EM are functions of spaceand time In order to describe the spatial variations of the quantities, we must be able todefine all points uniquely in space in a suitable manner This requires using an appropriatecoordinate system

A point or vector can be represented in any curvilinear coordinate system, which may

be orthogonal or nonorthogonal

An orthogonal system is one in which the coordinates arc mutually perpendicular.

Nonorthogonal systems are hard to work with and they are of little or no practical use.Examples of orthogonal coordinate systems include the Cartesian (or rectangular), the cir-cular cylindrical, the spherical, the elliptic cylindrical, the parabolic cylindrical, theconical, the prolate spheroidal, the oblate spheroidal, and the ellipsoidal.1 A considerableamount of work and time may be saved by choosing a coordinate system that best fits agiven problem A hard problem in one coordi nate system may turn out to be easy inanother system

In this text, we shall restrict ourselves to the three best-known coordinate systems: theCartesian, the circular cylindrical, and the spherical Although we have considered theCartesian system in Chapter 1, we shall consider it in detail in this chapter We should bear

in mind that the concepts covered in Chapter 1 and demonstrated in Cartesian coordinatesare equally applicable to other systems of coordinates For example, the procedure for

'For an introductory treatment of these coordinate systems, see M R Spigel, Mathematical

Hand-book of Formulas and Tables New York: McGraw-Hill, 1968, pp 124-130.

28

2.3 CIRCULAR CYLINDRICAL COORDINATES (R, F, Z) 29finding dot or cross product of two vectors in a cylindrical system is the same as that used

in the Cartesian system in Chapter 1

Sometimes, it is necessary to transform points and vectors from one coordinate system

to another The techniques for doing this will be presented and illustrated with examples

2.2 CARTESIAN COORDINATES (X, Y, Z)

As mentioned in Chapter 1, a point P can be represented as (x, y, z) as illustrated in Figure 1.1 The ranges of the coordinate variables x, y, and z are

- 0 0 < X < 00 -00<-y<o> (2.1)

2.3 CIRCULAR CYLINDRICAL COORDINATES (p, cj>, z)

The circular cylindrical coordinate system is very convenient whenever we are dealingwith problems having cylindrical symmetry

A point P in cylindrical coordinates is represented as (p, <j>, z) and is as shown in Figure 2.1 Observe Figure 2.1 closely and note how we define each space variable: p is the radius of the cylinder passing through P or the radial distance from the z-axis: <f>, called the

Figure 2.1 Point P and unit vectors in the cylindrical

coordinate system

30 Coordinate Systems and Transformation

azimuthal angle, is measured from the x-axis in the xy-plane; and z is the same as in the

Cartesian system The ranges of the variables are

where ap> a^, and az are unit vectors in the p-, <£-, and ^-directions as illustrated in

Figure 2.1 Note that a^ is not in degrees; it assumes the unit vector of A For example, if aforce of 10 N acts on a particle in a circular motion, the force may be represented as

F = lOa^, N In this case, a0 is in newtons

The magnitude of A is

= (A lp ,2x1/2 (2.5)Notice that the unit vectors ap, a^, and az are mutually perpendicular because our co-ordinate system is orthogonal; ap points in the direction of increasing p, a$ in the direction

of increasing 0, and az in the positive z-direction Thus,

a^ = az • az = 1

a = a7 • a = 0

n p X a<j> = a ,a^ X az = a,

a z X a p = a*

(2.6a)(2.6b)(2.6c)(2.6d)(2.6e)where eqs (2.6c) to (2.6e) are obtained in cyclic permutation (see Figure 1.9)

The relationships between the variables (x, y, z) of the Cartesian coordinate system and those of the cylindrical system (p, <j>, z) are easily obtained from Figure 2.2 as

Figure 2.2 Relationship between (x, y, z) and (P, * z).

The relationships between (a x , a y , a z ) and (a p , a^, a 2 ) are obtained geometrically from Figure 2.3:

32 Coordinate Systems and Transformation

Finally, the relationships between (A x , A y , A z ) and (A p , A0, A z ) are obtained by simply

substituting eq (2.9) into eq (2.2) and collecting terms Thus

A = (A x cos <j> + A y sin <j>)a p + (~A X sin <j> + A y cos 0 ) a 0 + A z a z (2.11)or

A p = A x cos <t> + Ay sin <f>

A,/, = ~A X sin <f> + A y cos tj> (2.12)

In matrix form, we have the transformation of vector A from (A x ,A y ,A z ) to (A p , A0, A,) as

(2.13)A,

The derivation of this is left as an exercise

2.4 SPHERICAL COORDINATES (r, 0, (/>)

The spherical coordinate system is most appropriate when dealing with problems having a

degree of spherical symmetry A point P can be represented as (r, 6, 4>) and is illustrated in

Figure 2.4 From Figure 2.4, we notice that r is defined as the distance from the origin to

2.4 SPHERICAL COORDINATES (r, e, 33

point P or the radius of a sphere centered at the origin and passing through P; 6 (called the

colatitude) is the angle between the z-axis and the position vector of P; and 4> is measured

from the x-axis (the same azimuthal angle in cylindrical coordinates) According to thesedefinitions, the ranges of the variables are

O < 0 < i r (2.17)

0 < <f> < 2TT

A vector A in spherical coordinates may be written as

(A r ,A e ,A^) or A& r + Aga e + A^ (2.18)

where a n a e , and 3A are unit vectors along the r-, B-, and ^-directions The magnitude of A is

|A| = (A 2r +A 2e + Aj) 112 (2.19)The unit vectors an a^, and a^ are mutually orthogonal; ar being directed along the

radius or in the direction of increasing r, a e in the direction of increasing 6, and a0 in the

di-rection of increasing <f> Thus,

34 • Coordinate Systems and Transformation

The space variables (x, y, z) in Cartesian coordinates can be related to variables (r, 0, <p) of a spherical coordinate system From Figure 2.5 it is easy to notice that

a x = sin 0 cos 4> ar + cos 0 cos <£ a s - sin

83, = sin 6 sin <£ ar + cos 6 sin 0 ae + cos <j

a z = cos 6 a r — sin 0 as

a r = sin 0 cos 0 a* + sin d sin <£ ay + c o s

a^ = cos 0 cos <t> a x + cos 0 sin </> a y — sin

2.4 SPHERICAL C O O R D I N A T E S (r, e, <t>) 35

The components of vector A = (A x , A y , A z ) and A = (Ar, A e , A^) are related by

substitut-ing eq (2.23) into eq (2.2) and collectsubstitut-ing terms Thus,

A = (A x sin 0 cos 4> + A y sin 0 sin 0 + A z cos 0)ar + (A x cos 0 cos 0

+ Ay cos 0 sin 0 — A z sin d)a e + {—A x sin 0 + A y cos <A)a^, (2.25)and from this, we obtain

A r = A^ sin 0 cos <t> + A y sin 0 sin <j> + A z cos 0

A e = A x cos 0 cos 4> + A y cos 0 sin <f> — A z sin

000

trans-and 2.6 (where <f> is held constant since it is common to both systems) This will be left as

an exercise (see Problem 2.9) Note that in point or vector transformation the point orvector has not changed; it is only expressed differently Thus, for example, the magnitude

of a vector will remain the same after the transformation and this may serve as a way ofchecking the result of the transformation

The distance between two points is usually necessary in EM theory The distance d between two points with position vectors r l and r2 is generally given by

36 • Coordinate Systems and Transformation

Figure 2.6 Unit vector transformations for cal and spherical coordinates.

cylindri-or

d 2 = (x2 - x , f + (y 2 - y x f + (z2 - zif (Cartesian)

d 2 = p\ + p 2 - 2p,p2 cos((^2 - 0 0 + (z2 - Z\f (cylindrical)

- 2r^r 2 cos d 2 cos 0jsin 02 sin d x cos(</>2 - 0i) (spherical)

d 2 = r\ + r\ - 2r^r 2 cos d 2 cos 0j

(2.31)(2.32)

(2.33)

EXAMPLE 2.1 Given point P(—2, 6, 3) and vector A = ya x + (x + z)a y , express P and A in cylindrical

and spherical coordinates Evaluate A at P in the Cartesian, cylindrical, and spherical

systems

Solution:

At point P: x = - 2 , y = 6, z = 3 Hence,

p = V x2 + y2 = V 4 + 36 = 6.324> = tan"1 - = tan"1 = 108.43°

2.4 SPHERICAL COORDINATES (r, e, • 37

For vector A, A x = y, A y = x + z, A z = 0 Hence, in the cylindrical system

A p A*

00

0

001

But x = p cos <j>,y = p sin 0, and substituting these yields

A = (A p , A$, A z ) = [p cos 0 sin 0 + (p cos 0 + z) sin 0]ap

+ [ - p sin 0 + (p cos 0 + z) cos

V 4 0 / V 4 0

-= -0.9487a,, - 6.OO8a0

A r

A e A*

or

sin 0 cos </> sin 9 sin 0 cos 0 cos 6 cos 0 cos 6 sin 0 — sin 6

—sin 0 cos 0 0

A r = y sin 0 cos <j> + (x + z)sin 6 sin 0

A9 = y cos 0 cos 0 + (x + z)cos 0 sin

A4, = ~y sin (j> + (x + z)cos 0

x + z

38 I f Coordinate Systems and Transformation

But x = r sin 6 cos (j>, y = r sin 6 sin </>, and z = r cos 0 Substituting these yields

, z ) | = |A(r, 0, <^)| = 6.083

PRACTICE EXERCISE 2.1

(a) Convert points P(\, 3, 5), 7X0, - 4 , 3), and S ( - 3 , - 4 , - 1 0 ) from Cartesian to

cylindrical and spherical coordinates

(b) Transform vector

Q =

Vx2

to cylindrical and spherical coordinates

(c) Evaluate Q at T in the three coordinate systems.

2.4 SPHERICAL COORDINATES (r, e, 39

Answer: (a) P(3.162, 71.56°, 5), P(5.916, 32.31°, 71.56°), T(4, 270°, 3),

T(5, 53.13°, 270°), 5(5, 233.1°, - 1 0 ) , 5(11.18, 153.43°, 233.1°)

(b) : (cos 4> a p — sin <j> a^, — z sin <j> az), sin 9 (sin 0 cos <j> —

r cos2 0 sin <j>)a r + sin 0 cos 0 (cos 0 + r sin 0 sin </>)ag — sin 0 sin

(c) 0 2.4az, 0 2.4az, 1.44ar - 1.92a,,

EXAMPLE 2.2 Express vector

cos 0 cos <$> -sin</>

cos 0 sin 0 cos <t>

- s i n 0 0

K) r

40 H Coordinate Systems and Transformation

Substituting all these gives

sin ^ cos 6 0

0 0 1or

Express the following vectors in Cartesian coordinates:

(a) A = pz sin 0 a p + 3p cos 0 a^, + p cos 0 sin 0 a.

42 M Coordinate Systems and Transformation

Cartesian system, if we keep x constant and allow y and z to vary, an infinite plane is

gen-erated Thus we could have infinite planes

x = constant

y = constant

z = constant

(2.34)

which are perpendicular to the x-, y-, and z-axes, respectively, as shown in Figure 2.7 The

intersection of two planes is a line For example,

is the line RPQ parallel to the z-axis The intersection of three planes is a point For

example,

x = constant, y = constant, z = constant (2.36)

is the point P(x, y, z) Thus we may define point P as the intersection of three orthogonal infinite planes If P is (1, - 5 , 3), then P is the intersection of planes x = 1, y = - 5 , and

are illustrated in Figure 2.8, where it is easy to observe that p = constant is a circular

cylinder, <f> = constant is a semiinfinite plane with its edge along the z-axis, and

z = constant is the same infinite plane as in a Cartesian system Where two surfaces meet

is either a line or a circle Thus,

z = constant

x = constant Figure 2.7 Constant x, y, and z surfaces.

is a circle QPR of radius p, whereas z = constant, <j> = constant is a semiinfinite line A

point is an intersection of the three surfaces in eq (2.37) Thus,

44 Coordinate Systems and Transformation

is a semicircle passing through Q and P The intersection of three surfaces gives a point.

Thus,

is the point P(5, 30°, 60°) We notice that in general, a point in three-dimensional space can

be identified as the intersection of three mutually orthogonal surfaces Also, a unit normal

vector to the surface n = constant is ± a n , where n is x, y, z, p, </>, r, or 6 For example, to

plane* = 5, a unit normal vector is ±a x and to planed = 20°, a unit normal vector is ậ

EXAMPLE 2.3 Two uniform vector fields are given by E = - 5 ap +

(c) Utilizing the fact that the z-axis is normal to the surface z = 3, the angle between the

z-axis and E, as shown in Figure 2.10, can be found using the dot product:

E a z = |E|(1) cos 6 Ez -> 3 = V l 3 4 cos $ Ez

(b) H X a ,

(c) The vector component of H normal to surface p = 1 (d) The scalar component of H tangential to the plane z = 0

Answer: (a) -0.433, (b) - 0 5 ap, (c) 0 ap, (d) 0.5

EXAMPLE 2.4 Given a vector field

D = r sin 0 ar sin 6 cos 0 a e + r 2 a^

determine(a) DatPQO, 150°, 330°)

(b) The component of D tangential to the spherical surface r = 10 at P (c) A unit vector at P perpendicular to D and tangential to the cone d = 150°

Solution:

(a) At P, r = 10, 6 = 150°, and 0 = 330° Hence,

D = 10 sin 330° a r - ~ sin 150° cos 330° a e + 100 a0 = ( - 5 , 0.043, 100)

46 SB Coordinate Systems and Transformation

(b) Any vector D can always be resolved into two orthogonal components:

D = D, + D n where D t is tangential to a given surface and D n is normal to it In our case, since a r is

normal to the surface r = 10,

(c) The vector component of A along a z at (1, TT/3, 5ir/4)

Answer: (a) - 6 , (b) 34.48, (c) -0.116ar + 0.201a,,

SUMMARY 1 The three common coordinate systems we shall use throughout the text are the

Carte-sian (or rectangular), the circular cylindrical, and the spherical

2 A point P is represented as P(x, y, z), P(p, <j>, z), and P(r, 6, 4>) in the Cartesian, drical, and spherical systems respectively A vector field A is represented as (A x , A y , A z )

cylin-or A^n x + A y a y + A z a z in the Cartesian system, as (A p , A$, A z ) or A p a p + A^a^ + A z a z

in the cylindrical system, and as (A n A e , A^) or A^a r + A e a e + A^a^ in the sphericalsystem It is preferable that mathematical operations (addition, subtraction, product,etc.) be performed in the same coordinate system Thus, point and vector transforma-tions should be performed whenever necessary

3 Fixing one space variable defines a surface; fixing two defines a line; fixing threedefines a point

4 A unit normal vector to surface n = constant is ± a n

( f ) - 7 T < 0 < 7 r , - X < 0 < 7 T

2.2 At Cartesian point ( — 3, 4, — 1), which of these is incorrect?

(a) p = - 5 _

(b) r = \Jlb (c) 6 = tan"1 —

(d) <t> = t a n "1 ^ "

2.3 Which of these is not valid at point (0, 4, 0)?

(a) (b) (c) (d)

a* = a« =

a r a»

a 0 none of the above 2.5 At every point in space, a 0 • a# = 1.

(a) True (b) False

2.6 If H = 4a fi - 3a0 + 5a z , at (1, x/2, 0) the component of H parallel to surface p = 1 is

(a) 4a p

(b) 5a z (c) - 3 a * (d) -3a,,, + 5a 2 (e) 5arf, + 3a

48 B Coordinate Systems and Transformation

2.7 Given G = 20ar + 50as + 4Oa0, at (1, T/2, TT/6) the component of G perpendicular to surface 6 = TT/2 is

(a) 20ar(b)(c) 0(d) 20ar +(e) - 4 0 ar

2.8 Where surfaces p = 2 and z = 1 intersect is

(a) an infinite plane(b) a semiinfinite plane(c) a circle

(d) a cylinder(e) a cone

2.9 Match the items in the left list with those in the right list Each answer can be used once,more than once, or not at all

x / 3 , <f> = w/2

5n73

= 1

TT/6 TT/3

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix)(x)

infinite plane semiinfinite plane circle

semicircle straight line cone cylinder sphere cube point

2.10 A wedge is described by z = 0, 30° < <t> < 60° Which of the following is incorrect:

(a) The wedge lies in the x — y plane.

(b) It is infinitely long

(c) On the wedge, 0 < p < <*>

(d) A unit normal to the wedge is ± az(e) The wedge includes neither the x-axis nor the y-axis

Answers: 2.1b,f, 2.2a, 2.3c, 2.4b, 2.5b, 2.6d, 2.7b, 2.8c, 2.9a-(vi), b-(ii), c-(i), d-(x),

e-(vii), f-(v), g-(iii), h-(iv), i-(iii), 2.10b