Topic 07 (ElectrostaticFieldProblems) Summation

7/31/2019 Topic 07 (ElectrostaticFieldProblems) Summation

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Electrostatic Field Problems:

Spherical Symmetry

EE 141 Lecture Notes

Topic 7

Professor K. E. OughstunSchool of Engineering

College of Engineering & Mathematical Sciences

University of Vermont

2012


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Motivation


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Spherical Coordinates

Coordinate transformations

x = rsin cos , y = rsin sin , z = rcos (1)

r =

x2 + y2 + z2, = arctan

x2 + y2

z

, = arctan

yx

(2)

with 0 r , 0 , and 0 < 2.


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Unit Basis Vectors in Spherical Coordinates

Unit basis vectors

1r = 1r(, ) = 1x sin cos + 1y sin sin + 1z cos ,1 = 1(, ) = 1x cos cos + 1y cos sin 1z sin , (3)1 = 1(, ) = 1x sin + 1y cos ,

where (Orthogonality Relations)

1r 1 = 1,1 1 = 1r, (4)1 1r = 1.

Problem 13. Using the relations in Eq. (3), determine expressions forthe unit basis vectors 1x, 1y, and 1z in terms of the unit basis

vectors 1r, 1, and 1.

Problem 14. Determine the direct coordinate transformation inmatrix notation from c lindrical to s herical coordinates.


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Vectors in Spherical Coordinates

Any vector V may be expressed in spherical polar coordinates as

V =1VV =

1rVr +

1V +

1V, (5)

where

Vr = 1r V= (1x sin cos + 1y sin sin + 1z cos )

(1xVx + 1yVy + 1zVz)

= Vx sin cos + Vy sin sin + Vz cos ,

V = 1 V= (1x cos cos + 1y cos sin + 1z sin )(1xVx + 1yVy + 1zVz)

= Vx cos cos + Vy cos sin + Vz sin , (6)V = 1 V = (1x sin + 1y cos ) (1xVx + 1yVy + 1zVz)

= Vx sin + Vy cos ,with magnitude

V = V = V V = V2 + V2 + V2. 7


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Position Vector in Spherical Coordinates

Position Vector of a point P = P(r, , ) is given by

R = OP = 1r(, )r, (8)

where r =

x2 + y2 + z2.

Notice that this position vector does not have either a 1 or 1component. Nevertheless, the orientation of the radial unit vector1r = 1r(, ) depends upon the and coordinates of the point P.


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Spherical Coordinates - Differential Elements

Differential elements of length along the 1r, 1, & 1-directions:

dr = dr, d = rd, d = rsin d.


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Differential Length, Surface Area, & Volume

The vector differential element of length:

d = 1rdr + 1rd + 1rsin d. (9)

Fundamental quadratic form or metric form

d2 = dr2 + r2d2 + r2 sin2 d2. (10)

Differential elements of surface area:

-Spherical Surface: dsr = 1r (dd) = 1rr2 sin dd.

r-Conical Surface: ds = 1 (drd) = 1rsin drd.

r-Planar Surface: ds = 1 (drd) = 1rdrd.

Differential element of volume:

dV = drdd = r2 sin drdd. (11)


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Distance Between Two Points

Coordinates of P1(x1,y1, z1) = P1(r1, 1, 1):

x1 = r1 sin 1 cos 1, y1 = r1 sin 1 sin 1, z1 = r1 cos 1.

Coordinates of P2(x2,y2, z2) = P2(r2, 2, 2):

x2 = r2 sin 2 cos 2, y2 = r2 sin 2 sin 2, z2 = r1 sin 1 sin 1.

Distance between P1 & P2 is then given by Pythagoreans theorem as

d =

(r2 sin 2 cos 2 r1 sin 1 cos 1)2

+(r2 sin 2 sin 2 r1 sin 1 sin 1)2 + (r2 cos 2 r1 cos 1)21/2=

r21 + r

22 2r1r2

cos 1 cos 2 + sin 1 sin 2 cos(2 1)

1/2

.

(12)

V t Diff ti l O t i S h i l P l


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Vector Differential Operators in Spherical Polar

Coordinates

Gradient Operator

= 1r r

+ 11

r

+ 1

1

rsin

(13)

The gradient of a scalar function of position f(r) = f(r, , ) is thengiven by

f = 1rfr

+ 11

r

f

+ 1

1

rsin

f

. (14)

Laplacian Operator

2 = 1r2

r

r2

r

+

1

r2 sin

sin

+

1

r2 sin2

2

2(15)

V t Diff ti l O t i S h i l P l


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Vector Differential Operators in Spherical Polar

Coordinates

Vector function of position F(r) = F(r, , ) = 1rFr + 1F + 1Fhas divergence

F = 1r2

r(r2Fr) +

1

rsin

(F sin ) +

1

rsin

F

(16)

and curl

F = 1r 1rsin

(F sin ) F

+1

1rsin

Fr

sin r

(rF)

(17)

+11

r

r(rF) Fr

.

Fi ld f P i Ch (S b P i )


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Field of a Point Charge (Symmetry about a Point)

Consider a point charge Q located at the origin of coordinates (ifnot, the origin of coordinates can always be shifted to that point).

The spherical symmetry of the problem then requires that

E(r) = 1rE(r).



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Application of Gauss law to a concentric spherical surfaceS

withradius r > 0 surrounding the point charge Q at the origin O yields

Q

0=

S

E(r)1r 1rda = E(r)S

da = 4r2E(r),

so that

E(r) = 1rE(r) = 1rQ

40r2, r > 0

The absolute potential due to the point charge is then given by

V(r) =

r

1rE(r) 1rdr = Q40r

, r > 0.



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0 1 2 3 4

r

0

Q/40

E ~ 1/r2

V ~ 1/r

Fi ld f U if S h i l Ch Di t ib ti


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Field of a Uniform Spherical Charge Distribution

Consider determining the field due to a uniform spherical chargedistribution of radius r0 > 0 with volume charge density (r) = 0 for

r r0 centered on the origin, where (r) = 0 for r > r0.

The spherical symmetry of the problem then requires that

E(r) =

1rE(r).

Application of Gauss law to a concentric sphere of radius r centeredat the origin then yields

4r2E(r) = 10

V

(r)d3r = 10 43 r3 0, r r04

3r30 0, r r0

Notice that if 0 is positive, then E is directed radially outward fromthe origin, whereas if 0 is negative, then E is directed radially inwardtoward the origin.

Fi ld f U if S h i l Ch Di t ib ti


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The total charge Q contained in the spherical charge distribution is

Q =4

3 r30 0,

so that

0 =3Q

4r30.

The radial component of the electric field vector E(r) = 1rE(r) isthen given by

E(r) = Q

40r30r, r r0

Q4

0r2

, r

r0

Notice that measurements of the electric field due to a sphericallysymmetric charge distribution localized in space (r r0) for r > r0are independent of the radius of the charge distribution.In the limit as r

0 0, the field due to a point charge Q at the origin

is obtained.

Field of a U ifo S he ical Cha ge Dist ib tio


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The absolute potential [Eq. (4.8)] due to the uniform spherical

charge distribution is then given by (with d = 1rdr)

V(r) =

r

E(r)dr.

For r r0,V(r) =

Q

40r

dr

r2 =Q

40r

which is the same as that due to a point charge Q at the origin.For r r0,

V(r) = Q40

1r30r0r

rdr + r0

drr2

=Q

80r30 r20 r2

increase in potential above the surface value+

Q

40r0 potential at the sphere surface.



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0 r0

2r0

r

0

Q/40r

0

Q/40r

0

3Q/80r0

2

E(r)

V(r)



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The potential V(r) may also be determined using Poissons &

Laplaces equations directly, taking advantage of the sphericalsymmetry of the source charge distribution.

Inside the uniform charge distribution (r r0), Poissons equation

2V(r) =

0/0 becomes

1

r2

r

r2

V

r

= 0

0=

r

r2

V

r

= 0

0r2

r2V

r=

0

30r3 + C =

V

r=

0

30r +

C

r2

V(r) = 060

r2 Cr

C = 0+D = V(r) = Q

80r30r2 + D.



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Outside the uniform charge distribution (r r0), Laplaces equation2V(r) = 0 becomes1

r2

rr2 V

r = 0 =

rr2 V

r = 0 r2

V

r= A = V

r=

A

r2

V(r) = Ar

+ B; V() 0 = B = 0

V(r) = Ar

.



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Continuity of V(r) at r = r0 requires that

Ar0

= Q80r0

+ D,

while continuity of V/r [i.e., continuity of E(r)] at r = r0 requires

A

r20= Q

80r20= A = Q

80

so thatD =

Q

80r0 A

r0=

Q

80r0+

Q

40r0.



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The electrostatic potential is then given by

V(r) =Q

80r30

r20 r2

+

Q

40r0; r r0

V(r) =Q

40r

; r

r0

so that the electric field intensity E(r) = V(r) = 1rV/r is

E(r) = 1rQ

40r3

0

r ; r

r0

E(r) = 1rQ

40r2; r r0

in agreement with the result obtained using Gauss law.

Take Home Exam Problem 2


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Take Home Exam Problem 2

A spherical region of radius a > 0 situated in free space contains avolume charge density given by

(r) = 0

1 + r2

; r a,with (r) = 0 for r > a, where 0 and are constants.

1 (20 points) Utilize Gauss law together with the inherent

symmetry of the problem to derive the electrostatic field vectorE(r) both inside and outside the spherical charge region.

2 (40 points) Use both Poissons and Laplaces equations todirectly determine the electrostatic potential V(r) both inside

and outside the spherical region. From this potential function,determine the electrostatic field vector E(r).3 (40 points) Determine the value of the parameter for which

the electrostatic field vanishes everywhere in the region outsidethe spherical charge region (r > a). Plot Er(r) and V(r) as a

function of r for this value of .

Boundary Value Problems Spherical Coordinates


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In spherical coordinates (r, , ) defined by the set of transformation

equations x = rsin cos , y = rsin sin , z = rcos withr [0,), [0, 2), and [0, ], Laplaces equation 2 = 0assumes the form

1

r

2(r)

r2 +1

r2 sin

sin + 1r2 sin2

2

2 = 0. (18)

This equation then admits separated solutions of the form

(r, , ) =1

rU(r)P()Q(), (19)

where the factor r1 is explicitly displayed in order to reflect the formof the electrostatic potential for a spherical charge distribution.



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With this substitution, Laplaces equation (18) becomes

r2 sin2 1Ud2U

dr2 +1

Pr2 sin

d

dsin dPd = 1Qd

2Q

d2 = m2

(20)where m2 is the separation constant. The ode for Q() then has theelementary solutions Q() = eim. In order that Q() be single-

valued when the full azimuthal range = 0 2 is allowed, theseparation constant m must be an integer or zero.The remaining part of Eq. (20) may then be written as

r2

U

d2U

dr2=

1

Psin

d

d sin dP

d +m2

sin2 = ( + 1), (21)

where ( + 1) is another separation constant. The ode for the radialpart of this separated equation then has the elementary solution

U(r) = Ar+1 + Br, (22)

where A and B are constants of inte ration.



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The remaining angular part of Eq. (21) is then

1sin

dd

sin

dP()d

+

( + 1) m2sin2

P() = 0, (23)

where is as yet undetermined. With the change of variable

= cos , for which d = sin()d so that d/d = sin()d/d,this ode assumes the standard form

d

d1 2

dP

d+

( + 1) m

2

1

2

P = 0 (24)

known as the generalized or associated Legendre equation, whosesolutions are the associated Legendre functions. In order that itssolutions represent a physically realizable potential, they must be

single-valued, finite, and continuous on the interval [1, 1].

Boundary Value Problems Spherical Coordinates I


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Case I: Legendres Equation and the Legendre Polynomials:Special case when the problem possesses azimuthal symmetry so that

there is no -dependence and m = 0. The generalized Legendreequation (24) then simplifies to the ordinary Legendre differentialequation

d

d1 2dP

d + ( + 1)P = 0 (25)Solutions of this equation are the Legendre polynomials of order with series representation (obtained by the method of Frobenius)

P() =

[/2]j=0

(1)j (2 2j)!2j!(j)!( 2j)!

2j (26)

which are normalized such that P(1) = 1. Here [/2] denotes thegreatest integer value of /2, where [/2] = /2 if is even and[/2] = ( 1)/2 if is odd.



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Explicit expressions for the first few Legendre polynomials areP0() = 1, P1() = , P2() =

12

(32

1), P3() =

12

(53

3),

P4() = 18 (354 302 + 3), P5() = 18 (635 703 + 15), . . .Series representation of the Legendre polynomials may be written as

P() =1

2!

d

d

[/2]j=0

(1)j !

j!(j)! 2(j)

.

The summation limit of [/2] can now be extended to because thepower of 2(j) is less than after the [/2] term and so its th-order

derivative vanishes. The resulting summation is the expansion of(2 1), so that

P() =1

2!

d

d 2 1

(27)

which is known as Rodrigues formula.



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y p

The Legendre polynomials P() form a complete orthogonal set offunctions on the interval

[

1, 1], satisfying the orthogonality

relation 11

P()P()d =2

2 + 1 (28)

Any sufficiently well-behaved function f() on the interval

[

1, 1]can then be expanded in a Legendre series representation as

f() =

=0AP(), 1 1, (29)

with expansion coefficients

A =2 + 1

2 1

1

f()P()d. (30)


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y p

As an example, suppose that the electrostatic potential is specified asV() on the surface of a sphere of radius a, and it is required to

determine the potential within the spherical region bounded by thatsurface. If there is no charge at the origin, then the potential must befinite there and B = 0 for all . The expansion given in Eq. (31)then becomes

(r, ) =

=0

ArP(cos ), r a. (32)

The coefficients A are then determined by evaluating Eq. (30) onthe surface of the sphere, so that V() = =0 AaP(cos ). Thisis just a Legendre series representation with = cos , so that thecoefficients are given by

A =2 + 1

2a

0

V()P(cos )sin d. (33)



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y p

Suppose next that the electrostatic potential is to be determined in

the region external to the sphere. Because the potential must now befinite at r = , it is required that A = 0 for all . The expansiongiven in Eq. (31) now becomes

(r, ) =

=0

Br+1 P(cos ), r a. (34)

As in the previous case, the coefficients B are determined byevaluating this expression on the surface of the sphere as

V() = =0 Ba+1 P(cos ), so thatB =

2 + 1

2a+1

0

V()P(cos )sin d. (35)



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y

The uniqueness of the potential (r, ) provides a convenient meansby which the solution of some potential problem may be obtained

from a knowledge of the potential along the axis of symmetry.Along the positive z-axis, z = r and cos = 1, so that

(z) =

=0 Ar

+B

r+1 , z 0, (36)while along the negative z-axis, z = r and cos = 1, so that

(z) =

=0(1)

Ar

+B

r+1

, z 0. (37)

If (z) can be determined at any point z on the symmetry axis, andif this potential can be expressed in a power series in z = r of theform given above with known coefficients, then the solution for thepotential at any point in space is obtained simply by multiplying each

power of r and r(+1) by the Legendre polynomial P(cos ).



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An expansion of considerable practical importance is that of thepotential (r, r) =

|r

r

|1 at a field point r due to a unit point

charge at r. Its Legendre series representation is obtained by firstrotating the coordinate axes so that the z-axis lies along the positionvector r to the unit point source. The potential (r, r), whichsatisfies Laplaces equation everywhere except at the point r = r,

then possesses azimuthal symmetry and can therefore be expanded as1

|r r| ==0

Ar

+B

r+1

P(cos ), r = r, (38)

where is the angle between the vectors r and r. If the point r is on

the positive z-axis, then the right-hand side of Eq. (38) reduces tothe form given in Eq. (36) while the left-hand side becomes

1

|r

r

|

=1

(r2 + r2

2rr cos )1/2

1

|r

r

|as 0.



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Let r< denote the smaller of r and r, and let r> denote the larger of

r and r. One then has the expansion

1

|r r| =1

r> r< =

1

r>

1

1 r

=

1

r>

=0

r

.

For points off of the z-axis it is then only necessary to multiply eachterm in this series expansion by the Legendre polynomial P(cos ),resulting in the general Legendre series representation

1|r r| = 1(r2 + r2 2rr cos )1/2 ==0

r

P(cos ) (39)

which is a useful expansion of Greens function (r, r) = |r r|1 inspherical coordinates.

Boundary Value Problems Spherical Coordinates II


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Case II: Associated Legendre Functions & The Spherical Harmonics:The general potential problem in spherical coordinates possesses

azimuthal dependency. In order that the elementary solutionQ() = eim of Laplaces equation (20) be single-valued in thisgeneral situation, m must be an integer. It is then necessary todetermine the solution of the associated Legendre equation

(1 2)d2

Pd2

2dPd

+ ( + 1) m21 2

P = 0, (40)for arbitrary values of and arbitrary integer values of m. For itssolution, one defines

P() (1)m(1 2)m/2 ddm

u(),

in which case the associated Legendre equation becomes

dm

dm (1 2)d2u

d2 2du

d + ( + 1)u = 0. (41)



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The expression appearing inside the square brackets of this equationis precisely the ordinary Legendre differential equation [cf. Eq. (25)]

with a nonnegative integer, and so its solution is the Legendrepolynomial P(). The appropriate solution of the associatedLegendre differential equation (41) is then given by

P

m

() = (1)m

(1 2

)

m/2 dm

dm P() (42)

for positive integer values of m. If Rodrigues formula (27) is used torepresent P(), an expression for the associated Legendre functionsPm () that is valid for both positive and negative integer values of m

is obtained as

Pm () = (1)m(1 2)m/2d+m

d+m(2 1) (43)

from which it is seen that P0 () = P().



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The solutions given by Eq. (43) will be finite on the closed interval

1 1 provided that (1) is either zero or a positive integerand (2) that m can only take on the values

m = , + 1, + 2, . . . ,1, 0, 1, . . . , 2, 1, .

Because the defining differential equation (40) depends only upon m2

and m can only take on positive or negative integer values, it is seenthat Pm () and P

m () are proportional. From Rodrigues formula,

it is found that

Pm () = (1)m( m)!( + m)!

Pm () (44)



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For a given fixed value of m, the associated Legendre functionsPm () form an orthogonal set on the closed interval 1 1,satisfying the orthogonality relation

11

Pm ()Pm ()d = 22 + 1( + m)!( m)! (45)

Because of Eq. (44), this orthogonality relation holds for bothpositive and negative integer values of m. In addition, it reduces tothe orthogonality relation given in Eq. (28) for the Legendrepolynomials when m = 0.



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The solution of Laplaces equation (18) in spherical coordinates by

the separation of variables method assumed a product of singlevariable functions of the three coordinate variables r, , of the formgiven in Eq. (19). It is convenient to combine the angular functionsQm() and P

m () of this solution in such a way so as to construct a

set of orthogonal functions over the unit sphere. Such a set offunctions is called the set of spherical harmonicsThe exponential functions Qm() = e

im form a complete set oforthogonal functions in the index m on the angular interval0

< 2, and the associated Legendre functions Pm () form a

complete set of orthogonal functions in the index for each allowedvalue of m on the interval 0 .Their product Pm ()Qm() thus forms a complete orthogonal set offunctions on the surface of a unit sphere in the two indices and m.



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From the orthogonality relation (45), this normalized set of functionsis

Ym(, ) (2 + 1)( m)!

4( + m)!Pm (cos )e

im (46)

The spherical harmonics satisfy the symmetry relation

Y,m(, ) = (1)m

Y

m(, ). (47)the orthonormalization condition

2

0

d

0

sin d Ym(, )Ym(, ) = mm (48)

and the completeness relation

=0

m=Ym(

, )Ym(, ) = ( )(cos cos ) (49)



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Any sufficiently well-behaved function g(, ) that is defined on theunit sphere can be expanded in terms of spherical harmonics as

g(, ) ==0

m=

AmYm(, ), (50)

where [from the orthonormalization condition (48)]

Am =

g(, )Ym(, )d, (51)

where d = sin dd is the differential element of solid angle, theintegration being taken over the entire surface of the unit sphere.

Notice that all terms in the spherical harmonic expansion (50) withm = 0 vanish at = 0. In that case, this expansion takes on thespecial limiting form g(0, ) = g(, )

=0

=

=0

2+1

4A0 with

A0 = 2+14 g(, )P(cos )d.



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From Eqs. (19), (20), (50), the general solution for a given boundaryvalue problem for Laplaces equation in spherical coordinates is given

by

(r, , ) =

=0

m=

amr

+bmr+1

Ym(, ). (52)

If the potential is specified on a spherical surface, the expansioncoefficients am and bm can then be determined by evaluating thisexpansion on the spherical surface which reduces it to the form givenin Eq. (50).

For the interior problem one typically requires that bm = 0

, m

so that the potential remains finite at the origin, whereas for the exterior problem one requires that am = 0 , mso that the potential vanishes at r = .In either case, the appropriate expansion coefficients are obtained

through direct application of Eq. (51).

Addition Theorem for the Spherical Harmonics


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The addition theorem for the spherical harmonics, which is ofconsiderable mathematical importance, is now briefly considered.To begin, consider two position vectors r and r from the same originO that are separated by an angle , the first having sphericalcoordinates (r, , ) and the second having coordinates (r, , ).Because

cos = cos cos + sin sin cos( ),

it is then of interest to express the Legendre polynomial P(cos ) interms of independent spherical harmonics of the angles , and, . If r is considered to be fixed in space, then P(cos ) is afunction of the angles , with the angles , as parameters. Itmay then be expanded in a series of the form given in Eq. (50) as

P(cos ) =

=0

m= Am(, )Ym(, ).



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However, because P(cos ) is a spherical harmonic of order alone,this expansion simplifies to

P(cos ) =

m=

Am(, )Ym(, ), (53)

with expansion coefficients

Am(, ) =

P(cos )Y

m(, )d =4

2 + 1Ym(

, ). (54)

Combination of these two results then yields the addition theorem forthe spherical harmonics

P(cos ) =4

2 + 1

m=

Ym(, )Ym(, ) (55)

where cos = cos cos

+ sin sin

cos( ).



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If the angle goes to zero so that = and = , then Eq. (55)yields the sum rule

m=

|Ym(, )|2 = 2 + 14

(56)

The addition theorem (55) can be used to express the expansion (39)

of the Greens function (r, r) = |r r|1 into its most general formas

1

|r

r

|= 4

=0

m=1

2 + 1

r

Ym(, )Ym(, ) (57)

where r< denotes the smaller of r and r and r> denotes the larger of

r and r. This expression gives the potential at the point r due to aunit point charge at the point r in a completely factorized form in

terms of the spherical coordinates of r and r

.

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Topic 07 (ElectrostaticFieldProblems) Summation