19403003 Arfken Mathematical Methods CH 6 HW

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r e 1")IT Y1

r

av,1 Ptf. 1 I,ve.

Math

15 .Seer 199 7

.I5c\

FkiAct GPI tke_ .2 e 6)3 0 1 .rt -

= L Err j

-e = LALF4i

r1 •

421 e =

1 Trk L.4 rl

kvi

w X 2c4

x Zz rt ni

A

- r 0

A A

0 rcQse

Jr c.c .. c' r 5, C.

(Ai

t 0

6.;

cat

rs —j

``:c t w

bcw,J PerkA;e„11

kIC ;110f-re7 c t c. cu 4-7

re/ f-c., a (c+ct+ i o coord.;,ic,.f-e s tj Ltefryn a s

YYI j2 171" i41 Gt x(-(:)x);') - x x

Gtt5

,

.C. oo .“ de toe C -1.‘r X + c,) .7- (A)

t-ck, S c1/41- the rt ig icicew,-e4 t

r X X 4-3 ,j

A rt. L a 1Paas

z , t Louk' -

r - r /2. ckt •

c"\v 11 od. e t

ct •/_ t-:1„x(L, t al

C

q., I z

(J LA:, , 0 r cc,e ri ..,0 0

R

--> (.....% ., ....\0

LAJ y, k.i....; A z j __

11 A

- r S; 0 0 L; e Cc,,, e I

r.:(A) r'. n 0 + 0

Y'S., 6'

r •e• r

- -r ,,(6t).1tn) ei = re

L 61( a /-2..1iK)

re,1(0-arm)

I

Dcw. , (A fie.vu,t)e)l

;vi t•ut1;(.1,‘ t coo +01,, r C is A C i rcie ciie -4,0-2c1 /,_J

1 zl . 1Z >1, IL,,d-: 1 , y- fc ..}- U(St 047 1- he Cclt,i cit,j

vl 'te@r A t flA eort b-) n s me j .-ti. Ti,1, ; 0 t.e 5,-,:, 1 AA ctj ile ev,..1'1,46,1.4 c -1- Ycle16--ccr AA; A 6) . -ii..) p, I , r Ce;',.;rit, "1.:A re' .3' ot c.1 ,t ,,,. tA <,.) 1

.i,,,ie.>-

(6,-) +2 .rt ' da L ct(0-t.)

z 2 LA (64.i.11,/t) r c e

r e ire

here r 7-: 1 coo G (3 z- e;

e gr 21r'7)) = L(e2-fro)

e - th Ee' 1J 1

5/4 .2/1 0

..r--—• ir,

C Z t 11-

e +11] --.) -../.,,, L23 -e , 1 v. - O

r-7 t - ,4

2

e - . - I .1

J

Tr/i.r 3 3,j3:1 r

e 1.41_ e e / efi j C LA L'-e +(J

r

e Li Le

d

+-

O

3 i (.40 c ) t.`2 .... 0

ZL1-2. .i

bay: eert,,Je

crAlc,k,s f re.s4Les us,

1; ( z -a)/ ' I 1

2(tri

1.3

; 5 1-;

•C

Zti

c 2(es tosed) -1.1= cR (L) ----- 0

[(-2-L)Af.() , Zo (v1 -‘) ) ckz 4

Fern e e 9ra.cl bras p,tes e-IG•( 0 c' f2 or,:ter S:r1Ce

CA.•n1)(.1

A 4.•• 4- -r -

A r re

ex

„evl (Lcinti #it

.2 I,.1

2

3 • 4_ ( 6 0.)

3 !Li -= A

- •

I _bevi (GA)

I'Lle1C.:+;04 C:(a .) ccw, eApclAded, c‘

1—.0,revd- se r,e,s (.1,:u} e or 1/1 tin Vie co tsctrecA1,

S rk t e c- C (-7 n-n c- f e c 1-11, S fLoic+0).1 2 ; 5 the aim e c 1- e c 3 re t :s-

4 0,4 t-e 4E r

f(i) r"_ .

I + 66+ 1,1 (:(04) LI (eh) t a ! I

(.,c44) ((' 0 0)4 ) 5 -t- - i, Y,1 S

prt

ill

=0 m.c.,•_17 •

1),c\v,J Pert e 11

c c4

Z = F -fr = -Rea I

f(Z) S'l IRLC3,T- t* CO S ft\) 5,4Z

-F() = -P \ 1 - I r 1, I . • •

5 i 71 /\ i 6i

-3

(-7 — T r- t•• — e e

'117!

f zI

R — .., ,Li

1 +31 ,_ IR

,v, -1 )

4_ R,

,1

(/ RZ

R_3

-1-R.5-

4- I j- + • •

.1 fIC,-, 3 -7= 3

I-4er e z ref)fesev, ij 1- 14 474 c-1- ci ; Al " `I a r.1)

r%t yA ln e C‘.

;n

. A LA to

el I

t-s t

fc the ti 3.1,e

C 5

47

7 • • •

7;

fi

-4--

(5'7

K - IR -I- th - R .4.

3; 5i '7!\'1 A

c1zk

3_3 55 . 7

E l 71

• •

)!, iC ; t _ r \

n tic

= c„U

, Ja v d i'oiL,;e:11

,

thus:

(,..c.„-1-)

av; Ci ?e01,,,e,"

, (i-) -i)

r L e -0)

. , ,

1- 4. rue,. r &Pt / A

r (0)

to ire (r / 2 I

( )3 + r

P1

F(7) - (a,=

Le re

tL) it - re - mod

-r re IL . 4

)).

l 2) = i. r- -re -k- L -r(.92

r 3 —

ar

2 , 3 r r0 r0

4-re -L:r___QZ-r13

V

=

AtL I (N.

V X . X

1 1. '"i L

l. LLViCt.";

ir.. (A ce f(° 3 e O 1 -ke r ci 4 tke

z -pitto 041.3 IC: isn -EC r

nA t-Cec=61t c k— the n-sr ∎ ; ...4 tvtr 7-p'ciiie

L7-1 2 X 1 LI L

-2. \ = ÷ I X V =

\ /

( 44C?1 ---: rlri

t'l

c) • te f rAal- -co( r IA > X

r v <

r < I vj > I k. iN 4 z

11

As 1 . 71 c W, (2) .= x o .

t- c; rc (c K tile 1 - Ltie 'r;t h.") ,C`ckNete

4-1-tct S tr,A. j .1, 1.! CA th -k,;( .5 vt • J ‘ E t-

ieA ) th a

Jt'

2. • 7.

• A r— I X

• (-I

v=( (2+1.

\

HQ d0 (2.4-cles _ i-e(e( i _ (Jr,

}-rcios

2t, 1L j

sA/2_ (2) =

zz

(A -,:n)

X 4-

L

• c ce,,f erect t.tt the ;

r2 Liz - -

r r vki tkta

1 U < X r-eci;LIn

r) we_ , e.t 0 fri.,>2

ilA6j c,,te- kA t fsL ti4e 2..),

• r LA, e ct CI. Sfr. 5 kt IA The

• kl 3 T

-t-ke fraviSfc r

1))Ma Ft (1,...;e11

e wet -a

iir

e

(ezi.() = (e

iv =

-1

C' L a

to zn

(e(3-e3)

eZx

az

ZW• A A (222e:j 1 4- e a

- e2x

-)cex 5:4

4.) ex cc-

-e -

teA ) co,

/ x X :01 e -e - LS,A

(e x 4-e-')•

2

k o ..; ccorJ,A.de +lie p4tae

let ave oa co As-+rcA ci- ed .

/ e (at (.0) - (A)

(A) - e x e'')

( e'e

p

2.x i (-t e x'

(A) - e2x e x -e `y)

e E (e + e

z„ i • e -e e

I L-1 t fi x/

e' i-e")

w-:. e x S A y

_ t -tez-tJe4ca.;i

-x x

-e (. -X A

e vs y

Ck 5, Vdel

CC -A A --

CI t2

a- 2

• C L.:n L rt.

X = a S n II 'Z

CC :; C0t)

7 . 3

CA1

- 5,4k X — 5,.1

t• COS ‘..j

— a A s, k

n„t e Cc4t1 Z

co, A t X 4 L. el

• A 3 s n y a s,

co. in C-CS — cost.X ."C"3 C. t"..

We ,1 (..\.ve

re (2,4- eck ck 8. (Oa r C c,:rct; nate _ S^Stew1, .

rea( 'tit'

Cvre Kit c,..‘,4 ti,e cccr,-t,/tAke. -+k€

re) ? ct itA e CA (

. 1-11 e trail ryK at 611

eii4c4; OA 5 (Are

eL

V `-LA 2

( a 1" IA ) V

\ V

-r,.)1-#%/1

Re( e eX

- v1 2

_

(at.)t-rvi

1.ne ,e. X=G}

P (..\v, a tev, .e.t1

(2, '7 3 c,,t)IA Grjer -1-t See ko,.. cccr-cl ∎ ‘ -a-te. AeS -Fr-A A.> -f-vri,1

t:Ae X Co:rd. t, 61 =G) cl,.1(4.-e4t13

e =

CA +v.-61 v

(citu-

(Oci- LA -iv

. 2.• 2. (..% V - LA - -V -

Z

C .n -tau tau A- LA -A-

• L

2.

(Cyr,.t1

. 2 1. Z ; 1. /ct i4 1- L4 *V = CA - v LA.

2. 1 Aco4 -4Dv 2 - 0

2Ain

\ z / =

(^1;.ci) iS fife efctqfic63 C; -c e C eil.+e re61 en 4- i1 e 4 -CXX n S

V

b av, 61 Pet„,A;e0

CIA LA

Livc --'— cos t:

x .11 't s.-n 17.5-r,-h4fe/

el...,;4,4+;x9 1.

1 3c S:

k -

5,4

\ 2

x C1Cc4 (-12- CS cl

Lk):ckt cosAr6e are circles cevi ffer-e61 0.4 t Vcvxrs.

C1 rl fe VA A+

, 2

(x ccik r1-) kj- ct Z csch

• - .

1) a 4 (.1 ?en well

e LCAtj \400.1e fOia•r". rikelS

r

1-14e 01-ec,A:t;c4 .

LC LA, lo5 Ct-t- Tkey, e csve 11' P444-r ilnayt,c-cSrk6v1Sn:A.

a- 2

C-,Ar 42e6..,,it, CA .s "forw%

1 0.5 I C tAi

4i4)

ktsptEct-; 0 ,, we sec --1-ke (A; reLt;c4A ,App; , Aver,k,4

0A-1" fr"I° p•45 ver ( rl.ck g,ec.wr

T WC. Ss Ac IN ; e GeS

aus •

,r2 ,&s rc,5;0AAik ot.

•2.9 BIPOLAR COORDINATES (,5, z) 87

v< EXERCISES

2.7.1 Let cosh u iii , cos v ----- (12, -.: q3. Find the new scale factors hq, and h".

., .,' q i — 11 72 '

It" - a , ) (qi - — 1

h„., -- a or -- 11,

2.8 Parabolic Cylindrical Coordinates (,:=, q,

The transformation equations,

X =

= )0/2 (2.79)

Z =

generate two sets of orthogonal parabolic cylinders (Fig. 2.6). By solvin g, Eq. 2.79 for and ti we obtain the following:

I. Parabolic cylinders, ‘; = constant.' — Co < <

(2.74)2. Parabolic cylinders, iI = constant, 0 tl <

(2.75)3. Planes parallel to the .vi-plane, = constant, — Co < < CO.

(2.76) From Eq. 2.6 the scale factors are

= h 4 + /12)12,

(2.77) h

+ 11 2 ) 1 2 (2.80)

the major one. II, It,. -=

li \ -/ 1 \

V=

=

(

(2.78)

hapter 6 ∎s hen

41)

2.9 Bipolar Coordinates

This is an oddball coordinate system. It is not a degenerate case of the confocal ellipsoidal coordinates. Equation 2.1 is not completely separable in this system even for k 2 = 0 (cf. Exercise 2.9.2). It is included here as an example or how an unusual coordinate system may be chosen to lit a problem.

The parabolic e N lindcr c.); constant is invariant to the sign of We must let ,; (or id go

negati ve to cover negative values of .v.

88

2 COORDINATE SYSTEMS

FIG. 2.6 Parabolic cylindrical coordinates. (Top) Cross section

The transf(

Dividing Eq.

Using Eq. 2.

Usin g Eq. 2.

From Eqs

I. Circula

2. CircuL

3. Planes

When 17 -

y = 0. Simil to a point, t Eq. 2.84 pa y = 0 are s(

The scale

To see h (a, 0), (—a

= constant,

3. Planes parallel to .vy-plane,

= constant,

- CC < 1/ < CC .

- SO < < Cf-

2.9 BIPOLAR COORDINATES (5, 77 , z) 89

The transformation equations are

a sinh(2.81a)

(2.81b)

(2.81c)

x =cosh — cos 'C.

a sin 'C = cosh q — cos

z = z.

Dividing Eq. 2.81a by 2.81h, we obtain

x sinh(2.82)

y sin;

Usin g Eq. 2.82 to eliminate C from Eq. 2.81a, we have

) 2 + y2 = C/ 2 CSC112 (x — a coth (2.83)

• Eq. 2.82 to eliminate q from Eq. 2.81h, we have

X 2 + (3' - a cot ,;) 2 = a 2 csc2 (2.84)

•From Eqs. 2.83 and 2.84 we may identify the coordinate surfaces as follows:

I. Circular cylinders, center at i = a cot *C,

= constant, 0 5 C 27t.

2. Circular cylinders, center at .v = a coth

when coth q —* I and csch q 0. Equation 2.83 has a solution x = a,

y = 0. Similarly, when q — cr.>, a solution is .v = —a, r = 0, the circle degenerating to a point, the cylinder to a line. The family of circles (in the .vy-plane) described by Eq. 2.84 passes through both of these points. This follows from noting that .v = +a,

y = 0 are solutions of Eq. 2.84 for any value of C. The scale factors for the bipolar system are

a h , h: =

cosh q — cos c

11, =a

= (2.85) cosh q — cos c

11 3 = -= 1.

To see how the bipolar system may be useful let us start with the three points

(a, 0), (—a, 0), and (.v, y) and the two distance vectors and i )„,_ at angles of 0, •

90

2 COORDINATE SYSTEMS

FIG. 2.7 Bipolar coordinates

FIG. 2.8

We define'

and 0 2 from the positive .v-axis. From Fig.2.8

Pi = (x — a) 2 + y2,

2.86 = (x + a) 2 + y2( )

and

tan 0, = x —

(2.87) tan 0 2 =

x + a

012 = inP2

Pi

= Ol 02.

By taking tan 5 I2 and Eq. 2.87

tan 0, — tan 02 tan C, 2 =

+ tan 0 1 tan 02

1 , /( x — a) — y/(V a)

I + y 2 /(X (1 )( x + ( 1)

1 The notation In is used to indicate loge.

(2.88a)

(2.88b)

(2.89)

2.9 BIPOLAR COORDINATES (6, 7) , z) 91

From Eq. 2.89, Eq. 2.84 follows directly. This identifies ,; as 5 12 = 0, — 0 2 . Solving

Eq. 2.88a for p 2/p, and combining this with Eq. 2.86, we get o z, + + y 2 v -pi (x + y2

Multiplication by e - " 2 and use of the definitions of hyperbolic sine and cosine produces Eq. 2.83, which identifies q as th, = In (0,1p,). The following example

exploits this identification.

EXAMPLE 2.9.1

An infinitely long strai g ht wire carries a

current I in the negative :-direction. A second wire, parallel to the first, carries a current / in

the positive z-direction. Using

dA =dk (2.91)

47r r

find A, the magnetic vector potential, and B, the magnetic inductance.

From Eq. 2.91 A has only a z-component. Integrating over each wire from 0 to P and

taking the limit as P oo, we obtain

clz A_ =--11 ° 1 lim( j.

47r p—. (),//); + z2

.8

(2.90)

FIG. rents

2 P CI= \

J o \/ + Z2

2.9 Antiparallel electric Cur-

(2.92)

/ 2 P + + P p, —Poi

( lirn 2 In 2 In . 47 p— P + p2 Pi.

This reduces toP0 I , 1 1 2 /101

.-1_ = — i 11 — = — — il. (2.94)

1_7 p i27

So far there has been no need for bipolar coordinates. Now. however. let us

calculate the magnetic inductance B from B = V x A. From Eqs. 2.22 and 2.85

1l440 Ito° k

B=

(cosh ii — cos (;) 2 c, il -- 7.—

u 22 (IL,'" (._.

11 ('.7

(2.93)

0 0

(cosh — cos = %,o a )TE

—I1„1

(2.95)

Rol A - —

" 47t 1"--•

lim + \/1) :1, + z 2 )1 1

n-1 0' — In(z + +

92

2 COORDINATE SYSTEMS

The magnetic field has only a 4 0 -component..The reader is ur ged to try to compute B in some other coordinate system.

We shall return to bipolar coordinates in Sections 2.13 and 2.14 to derive the toroidal and bispherical coordinate systems.

EXERCISES

2.9.1 Verify that the surfaces 6 h i and c; are orthogonal by the following methods: (a) Show that the slope of one surface (the intersection with a = constant)-plane) is the

ne gative reciprocal of the slope of the other surface. (b) Calculate q„.

2.9.2 (a) Show that Laplace's equation, V 0(6, 7), — 0 is not completely separable in bipolar coordinates.

(b) Show that a complete separation is possible if we require that tb — 0(6, 7)), that is, if we restrict ourselves to a two-dimensional system.

2.9.3 Find the capacitance per unit length of two conducting cylinders of radii b and c and of infinite length, w ith axes parallel and a distance apart.

277E0

C- - 7/2

2.9.4 As a limiting case of Exercise 2.9.3, find the capacitance per unit length between a conduct-ing cylinder and a conductin g infinite plane parallel to the axis of the cylinder.

27reo

n

2.10 Prolate Spheroidal Coordinates (a, 1 , , (p)

Let us start with the elliptic coordinates of Section 2.7 as a two-dimensional system. We can generate a three-dimensional system by rotating about the major or minor elliptic axes and introducing (I) as an azimuth angle (Fig. 2.10). Rotating first about the major axis gives us prolate spheroidal coordinates with the following coordinate nate surfaces :

1. Prolate spheroids,

= constant, 0 u < v4.

2. Hyperboloids of two sheets,

c= constant, 0

3. Half planes through the .7-axis,

= constant, 0 2n.

top related