Chapter 19 Laplace's equation in spherical coordinate: Green's function Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: November 11, 2010) Addition theorem Green's function in the spherical coordinate Dirac delta function in the spherical coordinate See Chapter 22 for the detail of the Legendre function. Some content in Chapter 22 is the same as that in this Chapter. 19.1 Formal solution of Laplace's equation We consider the solution of Laplace's equation 0 ) ( 2 r . where ) (r is a scalar electric potential. ) ( 1 1 ) ( 1 1 ) ( 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 r r r r r r r r r r r r r r L L L 0 ) ( 1 )) ( ( 1 2 2 2 2 2 r L r r r r r . Here we assume that ) , ( ) ( ) ( m l Y r U r . (separation variable) 0 ) , ( ) ( 1 )) ( ( ) , ( 1 2 2 2 2 2 m l m l Y r U r r rU r Y r L . We use the relation
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Chapter 19 Laplace's equation in spherical coordinate: Green's function Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton (Date: November 11, 2010)
Addition theorem Green's function in the spherical coordinate Dirac delta function in the spherical coordinate See Chapter 22 for the detail of the Legendre function. Some content in Chapter 22 is the same as that in this Chapter. 19.1 Formal solution of Laplace's equation
We consider the solution of Laplace's equation
0)(2 r .
where )(r is a scalar electric potential.
)(11
)(11
)(11
2
22
22
22
222
22
222
2
rrrr
rr
rrr
rr
rrr
L
L
L
0)(1
))((1 2
222
2
rLrr
rrr
.
Here we assume that
),()()( mlYrU r .
(separation variable)
0),()(1
))((),(1 2
222
2
m
lm
l YrUr
rrUr
Yr
L
.
We use the relation
),()1(),( 22 ml
ml YllY L .
Then we have
0)()1(
)]([1
22
2
rUr
llrrU
rr.
The solution of U(r) is given by
)1()( ll BrArrU .
The general solution is
0
)1( ),(][),,(l
l
lm
ml
llm
llm YrBrAr .
19.2. Dirac delta function in the spherical co-ordinate
We define the Dirac delta function as
1')'( 3 rrr d .
Suppose that
)'()'()'()'()'( rrrArr ,
with = cos and ' = cos. From the property of the delta function, we have
)'('sin
1)]'('sin[)'(
,
where and ' are in the range between 0 and . Then we have
1)()'('')'(
)'()'()'('sin
)'(''sin''')'(
2
0
2
2
000
23
rrArrdrrrA
rrrA
ddrrd
rrr
or
2
1)(
rrA
In summary,
)'()'()'('
1)'( 2 rr
rrr
.
19.3 Green's function
We consider the Green's function given by
)'()',(2 rrrr G ,
with the boundary surfaces which are concentric spheres at r = a and r = b (b>a). Note that
0)',( rrG for r = a and for r = b.
where r is the variable and r' is fixed.
Within each region (region I (a<r<r'<b) and region II (b>r>r'>a), we have the simpler equation
0)',( rrG .
The solution of the Green's function is given by the form
0
),()',',',()',(l
l
lm
mllm YrrAG rr .
Then the differential equation of the Green's function is given by
)'()'()'(
),(])1'('
)(1
2','
''''2''2
2
r
rrYA
r
llrA
rrml
mlmlml .
Note that
),(),(sin,','*'
'',', ml
mlmmll YYddmlmld nn ,
where
ddd sin .
Then
)'()'()'(
),(])1'('
)(1
)[,(),(2
*
','''2'.'2
2*
r
rrYdA
r
llrA
rrYYd m
lml
mlmlm
lm
l
or
)'()'()'(
),(])1'('
)(1
[2
*
'',',',''2''2
2
r
rrYdA
r
llrA
rrm
lml
mmllmlml
or
)','()'(
)'()'(sin)','()'(
)'()'(),()'()1(
)(1
*
2
*
2
*
222
2
ml
ml
mllmlm
Yr
rr
ddYr
rr
Ydr
rrA
r
llrA
rr
Since )','(*
mlY is constant, we put
)','(
)',',',()',( *
m
l
lml
Y
rrArrg .
Then we get
222
2 )'()1()(
1
r
rrg
r
llrg
rr ll
,
with the boundary condition
0)',( ragl , 0)',( rbgl
(i) )',( rrgl is continuous at r = r'.
(ii) 20'0' '
1|
)',(|
)',(
rr
rrg
r
rrgrr
lrr
l
Using Mathematica we get the Green's function
)12)((
)'(')(1212
1212)1(1212)1(
lab
rbrarrg ll
llllll
Il for a<r<r'
)12)((
)()'('1212
1212)1(1212)1(
lab
rbrarrg ll
llllll
IIl for b>r>r'
or
0
*
12
1211
12
0
*
121211
12121212
)','(),(
]1)[12(
)'
'
1)((
)','(),()('
)')(()
12
1()',(
l
l
lm
ml
mll
l
l
ll
ll
l
l
lm
ml
mlllll
llll
I
YY
b
al
b
r
rr
ar
YYabrr
rbar
lG
rr
0
*
12
1211
12
0
*
121211
12121212
)','(),(
]1)[12(
)1
)('
'(
)','(),()('
))('()
12
1()',(
l
l
lm
ml
mll
l
l
ll
ll
l
l
lm
ml
mlllll
llll
II
YY
b
al
b
r
rr
ar
YYabrr
rbar
lG
rr
Or more simply, we have
0
*
12
1211
12
)','(),(
]1)[12(
)1
)((
)',(l
l
lm
ml
mll
l
l
ll
ll
YY
b
al
b
r
rr
ar
G rr.
This means that
'rr
rr
in the region I (a<r<r'<b)
'rr
rr
in the region II (a<r'<r<b)
Fig. Plot of the Green's function gl(r, r') as a function of r. a = 1. b = 2. l = 3. r' is changed as a
parameter; r' = 1.2, 1.4, 1.6, and 1.8. When a→0,
0
*121
)','(),()12(
)1
(
)',(l
l
lm
ml
ml
l
l
l
l
YYl
b
r
rr
G rr
___________________________________________________________________________
19.4 Special case (b∞, )0a )
In the limit of b∞ (but )0a , we have
0
*
11
1212
)','(),('
)()
12
1()',(
l
l
lm
ml
mlll
ll
I YYrr
ar
lG rr ,
for the region I (a<r<r'), and
0
*
11
1212
)','(),('
)'()
12
1()',(
l
l
lm
ml
mlll
ll
II YYrr
ar
lG rr ,
for the region II (r>r'>a). Or more simply, we have
r'=1.2r'=1.4
r'=1.6
r'=1.8
1.2 1.4 1.6 1.8 2.0r
0.02
0.04
0.06
0.08
0.10
Gr,r'
0
*
11
1212
)','(),()12(
)()',(
l
l
lm
ml
mlll
ll
YYrrl
arG rr .
Further we assume that a0.
0
*
1 )','(),(12
1)',(
l
l
lm
ml
mll
l
YYr
r
lG rr .
____________________________________________________________________ 19.5 Mathematica
Find the Green's function for
222
2 )'()1()(
1
r
rrg
r
llrg
rr ll
,
with the boundary condition
0)',( ragl , 0)',( rbgl .
____________________________________________________________________________ 19.6 Addition theorem for spherical harmonics
The Green's function G(r, r') for the differential equation
)'()',(2 rrrr G
is given by
Clear"Global`";
eq1 Dr Gr, r, 2 L L 1r
Gr 1
rDiracDeltar r1;
eq2 DSolveeq1, Gr, r FullSimplify, L 0, 0 a r1 b &
Gr r r1L r1L r12 L C1 C2 r12 Lr112 L HeavisideThetarr1
r12 L r1
r
GIr_ Gr . eq21 Simplify, 0 a r r1 b &
rL C1 r1L C2
GIIr_ Gr . eq21 Simplify, b r r1 a &
r r1L r12 Lr112 L
r12 L r1 r1L r12 L C1 C2r
eq3 SolveGIIb 0, GIa 0, C1, C2 Simplify
C1 r11L b12 L r112 La12 L b12 L 1 2 L, C2
a12 L r11L b12 L r112 La12 L b12 L 1 2 L
GI1r_ GIr . eq31 Simplify
r1L a12 L r12 L r11L b12 L r112 La12 L b12 L 1 2 L
GII1r_ GIIr . eq31 Simplify
b12 L r12 L r r11L a12 L r112 La12 L b12 L 1 2 L
|'|4
1)',(
rrrr
G
This leads to the relation
0
*
1 )','(),(12
1
|'|4
1
l
l
lm
ml
mll
l
YYr
r
l
rr
or
0
*
1 )','(),(12
4
|'|
1
l
l
lm
ml
mll
l
YYr
r
l
rr
Here we note that
0
1 )(cos|'|
1
lll
l
Pr
r rr
where is the angle between r and r'.
Fig. The angle is the angle between the vectors r and r'.
)cos,sinsin,cos(sin rr ,
)'cos,'sin'sin,'cos'(sin'' rr
]'coscos)'cos('sin[sin'
'coscos'sin'sinsinsin'cos'sincos(sin'cos''
rr
rrrrrr
or
'coscos)'cos('sinsincos
x
y
z
r
r'
q q'
ff'
From
0
*
10
1 )','(),(12
4)(cos
|'|
1
l
l
lm
ml
mll
l
lll
l
YYr
r
lP
r
r rr
we have
l
lm
ml
mll YY
lP )','(),(
12
4)(cos
*
(addition theorem)
____________________________________________________________________________ 19.7 Electric potential - I Jackson: Classical Electrodynamics
We consider a hollow grounded sphere of radius b with a concentric ring of charge of radius
a and the uniform charge density (= Q/(2a). Q is the total charge. The ring of charge is
located in the x-y plane. We discuss the distribution of the electric potential .
The electric potential is described by a Poisson equation,
0
2
Using the Green's function, the electric potential can be obtained as
V
dG ')'()',(1
)( 3
0
rrrrr
where
)'()'()'( arA r .
Note that the constant A is determined as
2
2
0
1
10
23
2
'')'('')'(')'(2
Aa
dddrrarAdaV
rr
or
22 a
Q
aA
,
since the total charge Q is )2( a . Here we use the Green's function (a0);
0
*121
)','(),()12(
)1
(
)',(l
l
lm
ml
ml
l
l
l
l
YYl
b
r
rr
G rr
Then the electric potential )(r is obtained as
0
2*121
0
'''')','(),()12(
)1
(
)'()'(1
)(l
l
lm
ml
ml
l
l
l
l
V
dddrrYYl
b
r
rr
ara
r
Here ),( mlY can be also expressed by
)(cos)!(
)!(
4
)12()1(),(
m
limmm
l Peml
mllY
,
where )(cosmlP is the associated Legendre function. We note that
0,
0,
2
0
'*
)0(4
)12(2
)0()!(
)!(
4
)12()1(2
'')'()'()!(
)!(
4
)12()1('')','()'(
ml
mm
lm
imml
mml
Pl
Pml
mll
dedPml
mllddY
0121
0
0121
0
2
0
0
121
0
2
0
)(cos)1
()0(4
)(cos)1
()0()'('2
1
)0(4
12)(cos
4
12
12
)1
(
)'(''2
)(
lll
l
l
ll
lll
l
l
ll
lll
l
l
l
l
Pb
r
rrP
Q
Pb
r
rrPardrr
a
Pl
Pl
l
b
r
rr
arra
dr
r
where
ar
rr
for r>a
rr
ar
for r<a
For r<a, Lmax = 5 (the highest term in the summation),
)(r =
For a<r<b, Lmax = 5 (the highest term in the summation),
)(r =
1
4 0Q
1
a
1
b
1
4
1
a3
a2
b5r2 1 3 Cos2
3
64
1
a5
a4
b9r4 3 30 Cos2 35 Cos4
1
4 0Q
1
b
1
r
1
4a2 1
r3
r2
b51 3 Cos2
3
64a4 1
r5
r4
b93 30 Cos2 35 Cos4
Fig. Plot of the electric potential as a function of r. red ( = 0) and blue ( = /2). a = 1. b =
2. Q = 1. 0 = 1. Lmax = 5.
Fig. TheContourPlot of the electric potential and the StreamPlot of the electric field. a = 1.,
b = 2, 0 = 1. Q = 1. We use Lmax = 5.
0.5 1.0 1.5 2.0r
0.01
0.02
0.03
0.04
0.05
0.06
0.07
F
x
z
-2 -1 0 1 2
-2
-1
0
1
2
19.8 Electric potential - II Jackson: Classical Electrodynamics
We consider a hollow grounded sphere of radius b with a uniform line charge of total charge
Q, located on the z axis between the north and south poles of the sphere. We discuss the
distribution of the electric potential . The volume charge density is given by
)]1'(cos)1'(cos['4
)'( 2
br
Qr .
Note that
Qbb
Q
dddrb
Q
br
Qdddrrdddrr
b
bb
)2(4
2
)]1'()1'([''''4
)]1'(cos)1'(cos['4
''sin''')'(''sin'''
0
2
0
1
1
20
2
0 0
2
0
2
0 0
2
r
The electric potential )(r can be described by
0
2*121
20
'''')','(),()12(
)1
(
)]1'()1'(['4
1)(
l
l
lm
ml
ml
l
l
l
l
V
dddrrYYl
b
r
rr
br
Q
r
Here we note
0,
0,
2
0
'*
)]1()1([4
122
)]1()1([)!(
)!(
4
)12()1(2
'')'()]1'()1'([)!(
)!(
4
)12()1('')]1'()1'()[','(
mll
mm
lm
lm
imml
mml
PPl
PPml
mll
dedPml
mllddY
)(cos4
12),(0
ll P
lY
and
1)1( lP , llP )1()1(
.
Then we have
b
l
l
l
l
llll b
r
rrdrPPP
b
Q
0121
00
)1
(')]1()1()[(cos8
)(
r
For the integrand,
'rr
rr
for 0<r<r'<b
rr
rr
'
for 0<r'<r< <b
Then we get
]1[)1(
12
)'
'
1(''')
1(
)'
'
1('')
1()
1('
1210
121
1210
1210
121
l
b
rl
l
ll
rl
l
l
l
b
rl
l
ll
r
l
l
l
b
l
l
l
l
b
r
ll
l
b
r
rdrrdrr
b
r
r
b
r
rrdrdr
b
r
rb
r
rrdrI
For l = 0,
r
bI
llnlim
0
Thus we get
10
]}1[)1(
12
2
])1(1[)(cos){ln(
4)(
l
ll
l b
r
ll
lP
r
b
b
Q
r
where
1)(cos0 P .
19.9 Electric potential due to charge distribution Susan M. Lea: Mathematics for Physicists
We consider a grounded sphere of radius a and a ring of charge of radius b with the uniform
charge density . The ring of charge is located in the x-y plane. We discuss the distribution of the
electric potential . The electric potential is described by a Poisson equation,
0
2
Using the Green's function, the electric potential can be obtained as
V
dG ')'()',(1
)( 3
0
rrrrr
where
)'()'()'( brA r .
with
bA
.
Note that the constant A is determined as
2
2
0
1
10
23
2
'')'('')'(')'(2
Ab
dddrrbrAdbQV
rr
or
bA
,
since the total charge is )2( b .
0
2*
11
1212
0
'''')','(),()12(
)()'()'(
1)(
l
l
lm
ml
mlll
ll
V
dddrrYYrrl
arbr
b
r
Here
),( mlY can be also expressed by
)(cos)!(
)!(
4
)12()1(),(
m
limmm
l Peml
mllY
where )(cosmlP is the associated Legendre function.
Note that
0,
2
0
'*
)0(4
122
'')'()'()!(
)!(
4
)12()1('')','()'(
ml
imml
mml
Pl
dedPml
mllddY
011
1212
0
011
1212
0
2
0
011
1212
0
2
0
)0()(cos)(
4
)0()(cos)(
)'('2
1
)0(4
12)(cos
4
12
)12(
)()'(''
2)(
lllll
ll
lllll
ll
lllll
ll
PPbr
arQ
PPrr
arbrdrr
b
Pl
Pl
rrl
arbrr
bdr
r
where
1)0( lP
br
rr
for r>b
rr
br
for r<b
For a<r<b,
For r>b,
1
4 0Q
a r
b ra5 r5 1 3 Cos2
4 b3 r3
3 a9 r9 3 30 Cos2 35 Cos464 b5 r5
5 a13 r13 5 105 Cos2 315 Cos4 231 Cos6256 b7 r7
35 a17 r17 35 1260 Cos2 6930 Cos4 12 012 Cos6 6435 Cos816 384 b9 r9
1
65 536 b11 r1163 a21 r21 63 3465 Cos2
30 030 Cos4 90 090 Cos6 109 395 Cos8 46 189 Cos10
Fig. Plot of the electric potential as a function of r (the first 20 terms, l≤20). red ( = 0) and
blue ( = /2). a = 1. b = 2. Q = 1. 0 = 1. ____________________________________________________________________________ REFERENCES G.B. Arfken and H.J. Weber, Mathematical Methods for Physicists, Sixth edition (Elsevier
Academic Press, New York, 2005). J.D. Jackson, Classical Electrodynamics third edition (John Wiley & Sons, Inc., New York,
1999). Susan M. Lea Mathematics for Physicists (Thomson Brooks/Cole, 2004).
1
4 0Q
a b
b ra5 b5 1 3 Cos2
4 b3 r3
3 a9 b9 3 30 Cos2 35 Cos464 b5 r5
5 a13 b13 5 105 Cos2 315 Cos4 231 Cos6256 b7 r7
35 a17 b17 35 1260 Cos2 6930 Cos4 12 012 Cos6 6435 Cos816 384 b9 r9
1
65 536 b11 r1163 a21 b21 63 3465 Cos2
30 030 Cos4 90 090 Cos6 109 395 Cos8 46 189 Cos10
1 2 3 4 5ra
0.01
0.02
0.03
0.04
0.05
F
_____________________________________________________________________________ APPENDIX From the Green's law, we have
SV
dad ')]|'|
1(')'()'('
|'|
1[
4
1'
)'(
|'|
1
4
1)( 3
00
nrr
rrrr
rr
rrr
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