講者： 許永昌 老師
1
Contents
 Orthogonality
 Expansion of functions, Legendre series
 Examples
 Alternate definitions of Legendre Polynomials
 Rodrigues’ Formula
 Schlaefli Integral (I’ll not describe it.)
2
Orthogonality (請預讀P756~P757)
 Since Pn(x) is the solution of
LPn  n  n  1 Pn ,
L
d
d
2
1

x
,


dx
dx
 and L is a Hermitian Operator, we can conclude that {Pn(x)}
will form an orthogonal complete set during
x[-1,1].
 Solve:
 Gotten from the integration of g2:
1
g
x
,
t
dx



 1
1 1  2tx  t 2 dx
1
2
1

g

 t Pn  x  & orthogonality ,
n

t 
2n
n 0
1
1
Pn2 dx
n 0
1 1t 2 1
1
1
1  2 n 1 2
  2 dy  ln 1  t   ln 1  t    t
,
2t 1t  y
t
t
t n 0
2n  1
1
2
P
P
dx

1 n m 2n  1  n,m .
3
Legendre Series (請預讀P757~P758)
 Any finite and continuous function in x[-1,1] can be
expanded by Legendre polynomials.
 The expansion is

f 
n 0
Pn


Pn | f
2n  1

Pn Pn | f   an Pn
Pn | Pn
2
n 0
n 0
 Usage:
 an can be determined by <Pn|f> directly when f is known.
 an sometimes are determined by matching boundary
conditions.
4
Example 12.3.1 (請預讀P758~P759)
 Earth’s Gravational Field:
 In fact, the Earth is pear-shaped instead of spherical.
 U  r,    G    r '  d '
r r'
 G 

 2
r 'l *

  4
ˆ
ˆ
ˆ



P
r
'
Y
r
'
Y
r






lm
 r ' dr ' d  '
 0  n n
  2l  1  r l 1 lm
n 2
l ,m



If  is  independent and

GM

R
 zd 0  at center of mass .
n 1
R 

R

a
P
cos

  .


n 
n 

 r n 2  r 

 The parameters are shown in the textbook.
5
Example 12.3.2 (請預讀P759~P761)
 Metal sphere in a uniform field:
1
1

 sin    V
 PDE: 2V  0    2  r r 2 r  2
r sin 
 r

 B.C.:


V(r0,,)=0,
V(r,,)=E0z=-E0rcos.
 Solve:


     1

V r  F n  r  Pn  cos  ,       2 
Assuming that( RzV=V):   
n 0
Substitue Eq.(2) into Eq.(1):
 1 d 2 d


r

n
n

1


 r 2 dr dr
 F n  0       3

Fn = Anrn+Bnrn1,
Code: grounded_ball.m
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Example 12.3.2 (請預讀P759~P761)
 From B.C.s, we can find
 <V>=0:
 V(r,,) =-E0rcos :
 V(r0,,)=0 :
 Therefore,
Fn(r0)=0 and A0= An>1=0
A0=0,
An>1=0 & A1=-E0,


r03 
V  r    E0  r  2  cos  .
r 

B0= Bn>1=0
 Induced Surface charge:
V
   0
r
 3 0 E0 cos  .
r0
 Induced dipole moment: p 
2
3
ˆ
r

r
d


z
4

r
0 0 E0 .
 0
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Example 12.3.3 (請預讀P761~P762)
 Electrostatic Potential of a Ring of Charge
 Total Charge: q.
 Radius: a .
 r>a.
 Solve:

an
 Assuming that V  r,     cn n1 Pn  cos  ,
r
n 0
 B.C.
n
2n
q
q   1 2 n  a 
 V  r,   0  

 2n Cn   ,
2
2
4 0 r  a
4 0 r
n 0
2
 Comparing Eqs. (1) & (2), we get
V  r,   
q
4 0 r


n 0
 1
n
2n
Cn2 n  a 
  P2 n  cos  .
22 n
r
r
     1
     2
*若直接積分還蠻麻煩的。
8
Rodrigues’ Formula (請預讀P767)
 Series form:
1
Pn  x   n
2
n
2
 
2 n 2 s n n 2 s

1
C
   n Cs x ,
s
s 0
 It can be written as
n
2
 
Pn  x   
s 0
 1
s
d n 2 n 2 s
C
x
,
n
n
2 n!
dx
n
s
 1 n 2 n2 s
d
 n  n Cs x
,
dx s 0 2 n !
n
n
s



d n  m 2 n 2 s
n
x

0
when
s

 2  
dx n m

n
1 dn 2
 n
x  1 .
n 
2 n ! dx
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Homework
 12.3.3 (11.3.3e)
 12.3.7 (11.3.7e)
 12.3.8 (11.3.8e)
 12.3.13 (11.3.13e)
 12.3.19 (11.3.19e)
 12.4.3 (11.4.3e)
 12.4.6 (11.4.6e)
 12.4.7 (11.4.7e)
10
Nouns
n
1 dn 2
Pn  x   n
x  1 .
n 
2 n ! dx
P767
P768
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